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Full text of "Mechanics"

ADDISON 
WESLEY 



■■■ 



MECHANICS 

Second Edition 
By Keith i;. Symon 




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THE AUTHOR 



Keith H Sya boS.B., A M ,etnd 

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Mechanics 



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MECHANICS 



This book is in the 
ADDISON-WESLEY SERIES IN PHYSICS 



MECHANICS 



by 

KEITH R. SYMON 

University of Wisconsin 



SECOND EDITION 



ADDISON-WESLEY PUBLISHING COMPANY, INC. 

BEADING, MASSACHUSETTS, U.S.A. 
LONDON, ENGLAND 



Copyright © 1953, 1960 
ADDISON-WESLEY PUBLISHING COMPANY, INC. 



Printed in the United States of America 

ALL EIGHTS EESEBVED. THIS BOOK, OR PARTS THERE- 
OF, MAY NOT BE REPRODUCED IN ANY FORM WITH- 
OUT WRITTEN PERMISSION OF THE PUBLISHER. 

Library of Congress Catalog Card No. 60-5164 
Second Edition 



To my Father 



PREFACE 

This text is intended as the basis for an intermediate course in mechanics 
at the undergraduate level. Such a course, as essential preparation for 
advanced work in physics, has several major objectives. It must develop 
in the student a thorough understanding of the fundamental principles of 
mechanics. It should treat in detail certain specific problems of primary 
importance in physics, for example, the harmonic oscillator, and the 
motion of a particle under a central force. The problems suggested and 
those worked out in the text have been chosen with regard to their in- 
terest and importance in physics, as well as to their instructive value. 
This book contains sufficient material for a two-semester course, and is 
arranged in such a way that, with appropriate omissions, it can be used 
for a single three- or four-hour course for one semester. The author has 
used the material in the first seven chapters in a three-hour course in 
mechanics. 

The choice of topics and their treatment throughout the book are in- 
tended to emphasize the modern point of view. Applications to atomic 
physics are made wherever possible, with an indication as to the extent of 
the validity of the results of classical mechanics. The inadequacies in 
classical mechanics are carefully pointed out, and the points of departure 
for quantum mechanics and for relativistic mechanics are indicated. The 
development, except for the last four chapters, proceeds directly from 
Newton's laws of motion, which form a suitable basis from which to attack 
most mechanical problems. More advanced methods, using Lagrange's 
equations and tensor algebra, are introduced in the last four chapters. 

An important objective of a first course in mechanics is to train the 
student to think about physical phenomena in mathematical terms. Most 
students have a fairly good intuitive feeling for mechanical phenomena in 
a qualitative way. The study of mechanics should aim at developing an 
almost equally intuitive feeling for the precise mathematical formulation 
of physical problems and for the physical interpretation of the mathe- 
matical solutions. The examples treated in the text have been worked 
out so as to integrate, as far as possible, the mathematical treatment with 
the physical interpretation. After working an assigned problem, the 
student should study it until he is sure he understands the physical inter- 
pretation of every feature of the mathematical treatment. He should de- 
cide whether the result agrees with his physical intuition about the prob- 
lem. If not, then either his solution or his intuition should be appropriately 
corrected. If the answer is fairly complicated, he should try to see whether 

vii 



Vlll PREFACE 

it can be simplified in certain special or limiting cases. He should try to 
formulate and solve similar problems on his own. 

Only a knowledge of differential and integral calculus has been presup- 
posed. Mathematical concepts beyond those treated in the first year of 
calculus are introduced and explained as needed. A previous course in 
elementary differential equations or vector analysis may be helpful, but it 
is the author's experience that students with an adequate preparation in 
algebra and calculus are able to handle the vector analysis and differential 
equations needed for this course with the explanations provided herein. 
A physics student is likely to get more out of his advanced courses in 
mathematics if he has previously encountered these concepts in physics. 

The text has been written so as to afford maximum flexibility in the 
selection and arrangement of topics to be covered. With certain obvious 
exceptions, many sections or groups of sections can be postponed or 
omitted without prejudice to the understanding of the remaining material. 
Where particular topics presented earlier are needed in later parts of the 
book, references to section and equation numbers make it easy to locate 
the earlier material needed. 

In the first chapter the basic concepts of mechanics are reviewed, and 
the laws of mechanics and of gravitation are formulated and applied to a 
few simple examples. The second chapter undertakes a fairly thorough 
study of the problem of one-dimensional motion. The chapter concludes 
with a study of the harmonic oscillator as probably the most important 
example of one-dimensional motion. Use is made of complex numbers to 
represent oscillating quantities. The last section, on the principle of super- 
position, makes some use of Fourier series, and provides a basis for certain 
parts of Chapters 8 and 12. If these chapters are not to be covered, Sec- 
tion 2-11 may be omitted Or, better, skimmed to get a brief indication of 
the significance of the principle of superposition and the way in which 
Fourier series are used to treat the problem of an arbitrary applied force 
function. 

Chapter 3 begins with a development of vector algebra and its use in 
describing motions in a plane or in space. Boldface letters are used for 
vectors. Section 3-6 is a brief introduction to vector analysis, which is 
used very little in this book except in Chapter 8, and it may be omitted 
or skimmed if Chapter 8 and a few proofs in some other chapters are 
omitted. The author feels there is some advantage in introducing the 
student to the concepts and notation of vector analysis at this stage, where 
the level of treatment is fairly easy; in later courses where the physical 
concepts and mathematical treatment become more difficult, it will be well 
if the notations are already familiar. The theorems stating the time rates 
of change of momentum, energy, and angular momentum are derived for 
a moving particle, and several problems are discussed, of which motion 



PREFACK ix 

under central forces receives major attention. Examples are taken from 
astronomical and from atomic problems. 

In Chapter 4 the conservation laws of energy, momentum, and angular 
momentum are derived, with emphasis on their position as cornerstones of 
present-day physics. They are then applied to typical problems, particu- 
larly collision problems. The two-body problem is solved, and the motion 
of two coupled harmonic oscillators is worked out. The general theory of 
coupled oscillations is best treated by means of linear transformations in 
vector spaces, as in Chapter 12, but the behavior of coupled oscillating 
systems is too important to be omitted altogether from even a one-semester 
course. The section on two coupled oscillators can be omitted or postponed 
until Chapter 12. The rigid body is discussed in Chapter 5 as a special 
kind of system of particles. Only rotation about a fixed axis is treated; 
the more general study of the motion of a rigid body is left to a later chap- 
ter, where more advanced methods are used. The section on statics treats 
the problem of the reduction of a system of forces to an equivalent simpler 
system. Elementary treatments of the equilibrium of beams, flexible 
strings, and of fluids are given in Sections 5-9, 5-10, and 5-11. 

The theory of gravitation is studied in some detail in Chapter 6. The 
last section, on the gravitational field equations, may be omitted without 
disturbing the continuity of the remaining material. The laws of motion 
in moving coordinate systems are worked out in Chapter 7, and applied 
to motion on the rotating earth and to the motion of a system of charged 
particles in a magnetic field. Particular attention is paid to the status 
in Newtonian mechanics of the "fictitious forces" which appear when 
moving coordinate systems are introduced, and to the role to be played 
by such forces in the general theory of relativity. 

The last five chapters cover more advanced material and are designed 
primarily to be used in the second semester of a two-semester course in 
intermediate mechanics. In a shorter course, any or all of the last five 
chapters may be omitted without destroying the unity of the course, 
although the author has found it possible to utilize parts of Chapter 8 or 
9 even in a one-semester course. In Chapter 8 an introductory treatment 
of vibrating strings and of the motion of fluids is presented, with emphasis 
on the fundamental concepts and mathematical methods used in treating 
the mechanics of continuous media. Chapter 9 on Lagrange's equations 
is intended as an introduction to the methods of advanced dynamics. 
Hamilton's equations and the concept of phase space are presented, since 
they are prerequisite to any later course in quantum mechanics or statis- 
tical mechanics, but the theory of canonical transformations and the use 
of variational principles are beyond the scope of this book. Chapter 10 
develops the algebra of tensors, including orthogonal coordinate trans- 
formations, which are required in the last two chapters. The inertia tensor 



X PREFACE 

and the stress tensor are described in some detail as examples. Section 
10-6 on the stress tensor will enable the reader to extend the discussion 
of ideal fluids in Chapter 8 to a solid or viscous medium. The methods 
developed in Chapters 9 and 10 are applied in Chapter 11 to the general 
rotation of a rigid body about a point, and in Chapter 12 to the study of 
small vibrations of a physical system about a state of equilibrium or of 
steady motion. 

The problems at the end of each chapter are arranged in the order 
in which the material is covered in the chapter, for convenience in as- 
signment. An attempt has been made to include a sufficient variety of 
problems to guarantee that anyone who can solve them has mastered the 
material in the text. The converse is not necessarily true, since most 
problems require more or less physical ingenuity in addition to an under- 
standing of the text. Many of the problems are fairly easy and should be 
tractable for anyone who has understood the material presented. A few 
are probably too difficult for most college juniors or seniors to solve with- 
out some assistance. Those problems which are particularly difficult or 
time-consuming are marked with an asterisk. 

The last three chapters and the last three sections of Chapter 9 have 
been added to the present edition in order to provide enough material for 
a full two-semester course in mechanics. Except for corrections and a 
few minor changes and additions, the first eight chapters and the first 
eight sections of Chapter 9 remain the same as in the first edition of this 
text. 

Grateful acknowledgment is made to Professor Francis W. Sears of 
Dartmouth College and to Professor George H. Vineyard of Brookhaven 
National Laboratory for their many helpful suggestions, and to Mr. Charles 
Vittitoe and Mr. Donald Roiseland for a critical reading of the last four 
chapters. The author is particularly grateful to the many teachers and 
students who have offered corrections and suggestions for improvement 
which have been incorporated in this revised edition. While space does not 
permit mentioning individuals here, I hope that each may find my thanks 
expressed in the changes that have been made in this edition. 

January, 1960 K. R. S. 



CONTENTS 

Chapter 1. Elements of Newtonian Mechanics 1 

1-1 Mechanics, an exact science 1 

1-2 Kinematics, the description of motion 4 

1-3 Dynamics. Mass and force 5 

1-4 Newton's laws of motion 7 

1-5 Gravitation 10 

1-6 Units and dimensions 11 

1-7 Some elementary problems in mechanics 13 

Chapter 2. Motion of a Particle in One Dimension .... 21 

2-1 Momentum and energy theorems 21 

2-2 Discussion of the general problem of one-dimensional motion . 22 

2-3 Applied force depending on the time 25 

2-4 Damping force depending on the velocity 28 

2-5 Conservative force depending on position. Potential energy . 30 

2-6 Falling bodies 35 

2-7 The simple harmonic oscillator 39 

2-8 Linear differential equations with constant coefficients ... 41 

2-9 The damped harmonic oscillator 47 

2-10 The forced harmonic oscillator 50 

2-11 The principle of superposition. Harmonic oscillator with arbitrary 

applied force 59 

Chapter 3. Motion of a Particle in Two ok Three Dimensions 68 

3-1 Vector algebra 68 

3-2 Applications to a set of forces acting on a particle .... 77 

3-3 Differentiation and integration of vectors 81 

3-4 Kinematics in a plane 87 

3-5 Kinematics in three dimensions 91 

3-6 Elements of vector analysis 95 

3-7 Momentum and energy theorems 100 

3-8 Plane and vector angular momentum theorems 101 

3-9 Discussion of the general problem of two- and three-dimensional 

motion 104 

3-10 The harmonic oscillator in two and three dimensions .... 106 

3-11 Projectiles 108 

3-12 Potential energy . 112 

3-13 Motion under a central force 120 

3-14 The central force inversely proportional to the square of the 

distance 125 

xi 



Xll CONTENTS 

3-15 Elliptic orbits. The Kepler problem 132 

3-16 Hyperbolic orbits. The Rutherford problem. Scattering cross 

section 135 

3-17 Motion of a particle in an electromagnetic field 139 

Chapter 4. The Motion of a System of Particles 155 

4-1 Conservation of linear momentum. Center of mass .... 155 

4-2 Conservation of angular momentum 158 

4-3 Conservation of energy 162 

4-4 Critique of the conservation laws 165 

4-5 Rockets, conveyor belts, and planets 168 

4-6 Collision problems 171 

4-7 The two-body problem 178 

4-8 Center-of-mass coordinates. Rutherford scattering by a charged 

particle of finite mass 181 

4-9 The IV-body problem 185 

4-10 Two coupled harmonic oscillators 188 

Chapter 5. Rigid Bodies. Rotation About an Axis. Statics . . 203 

5-1 The dynamical problem of the motion of a rigid body . . . 203 

5-2 Rotation about an axis 206 

5-3 The simple pendulum 208 

5-4 The compound pendulum 212 

5-5 Computation of centers of mass and moments of inertia . . . 215 

5-6 Statics of rigid bodies 225 

5-7 Statics of structures 231 

5-8 Stress and strain 232 

5-9 Equilibrium of flexible strings and cables 235 

5-10 Equilibrium of solid beams 239 

5-11 Equilibrium of fluids 245 

Chapter 6. Gravitation 257 

6-1 Centers of gravity for extended bodies 257 

6-2 Gravitational field and gravitational potential 259 

6-3 Gravitational field equations 262 

Chapter 7. Moving Coordinate Systems 269 

7-1 Moving origin of coordinates 269 

7-2 Rotating coordinate systems 271 

7-3 Laws of motion on the rotating earth 278 

7—4 The Foucault pendulum 280 

7-5 Larmor's theorem 283 

7-6 The restricted three-body problem 285 



CONTENTS Xiii 

Chapter 8. Introduction to the Mechanics of Continuous Media 294 
8-1 The equation of motion for the vibrating string . . . . . . 294 

8-2 Normal modes of vibration for the vibrating string .... 296 

8-3 Wave propagation along a string 300 

8-4 The string as a limiting case of a system of particles . . , . 305 
8-5 General remarks on the propagation of waves ...... 310 

8-6 Kinematics of moving fluids 313 

8-7 Equations of motion for an ideal fluid 321 

8-8 Conservation laws for fluid motion 323 

8-9 Steady flow 329 

8-10 Sound waves 332 

8-11 Normal vibrations of fluid in a rectangular box 337 

8-12 Sound waves in pipes 341 

8-13 The Mach number 343 

8-14 Viscosity 345 

Chapter 9. Lagrange's Equations 354 

9-1 Generalized coordinates 354 

9-2 Lagrange's equations . . 365 

9-3 Examples 3gg 

9-4 Systems subject to constraints 369 

9-5 Examples of systems subject to constraints 375 

9-6 Constants of the motion and ignorable coordinates .... 381 

9-7 Further examples 384 

9-8 Electromagnetic forces and velocity-dependent potentials . . 388 

9-9 Lagrange's equations for the vibrating string 391 

9-10 Hamilton's equations 396 

9-11 Liouville's theorem 399 

Chapter 10. Tensor Algebra. Inertia and Stress Tensors . . 406 

10-1 Angular momentum of a rigid body . 406 

10-2 Tensor algebra 407 

10-3 Coordinate transformations 414 

10-4 Diagonalization of a symmetric tensor 421 

10-5 The inertia tensor 430 

10-6 The stress tensor 438 

Chapter 11. The Rotation of a Rigid Body 450 

11-1 Motion of a rigid body in space 450 

11-2 Euler's equations of motion for a rigid body 451 

1 1-3 Poinsot's solution for a freely rotating body 455 

11-4 Euler's angles 458 

11-5 The symmetrical top 461 



Xiv CONTENTS 

Chapteb 12. Theokt of Small Vibrations 473 

12-1 Condition for stability near an equilibrium configuration . 473 

12-2 Linearized equations of motion near an equilibrium configuration 475 

12-3 Normal modes of vibration 477 

12-4 Forced vibrations 481 

12-5 Perturbation theory 484 

12-6 Small vibrations about steady motion 490 

12-7 Betatron oscillations in an accelerator 497 

12-8 Stability of Lagrange's three bodies 500 

Bibliography 515 

Answers to Odd-Numbered Problems 521 

List of Symbols 533 

Index 545 



CHAPTER 1 
ELEMENTS OF NEWTONIAN MECHANICS 

1-1 Mechanics, an exact science. When we say that physics is an 
exact science, we mean that its laws are expressed in the form of mathe- 
matical equations which describe and predict the results of precise quanti- 
tative measurements. The advantage in a quantitative physical theory is 
not alone the practical one that it gives us the power accurately to predict 
and to control natural phenomena. By a comparison of the results of 
accurate measurements with the numerical predictions of the theory, we 
can gain considerable confidence that the theory is correct, and we can 
determine in what respects it needs to be modified. It is often possible 
to explain a given phenomenon in several rough qualitative ways, and if 
we are content with that, it may be impossible to decide which theory is 
correct. But if a theory can be given which predicts correctly the results 
of measurements to four or five (or even two or three) significant figures, 
the theory can hardly be very far wrong. Rough agreement might be a 
coincidence, but close agreement is unlikely to be. Furthermore, there 
have been many cases in the history of science when small but significant 
discrepancies between theory and accurate measurements have led to the 
development of new and more far-reaching theories. Such slight discrep- 
ancies would not even have been detected if we had been content with a 
merely qualitative explanation of the phenomena. 

The symbols which are to appear in the equations that express the laws 
of a science must represent quantities which can be expressed in numerical 
terms. Hence the concepts in terms of which an exact science is to be 
developed must be given precise numerical meanings. If a definition of a 
quantity (mass, for example) is to be given, the definition must be such 
as to specify precisely how the value of the quantity is to be determined 
in any given case. A qualitative remark about its meaning may be helpful, 
but is not sufficient as a definition. As a matter of fact, it is probably not 
possible to give an ideally precise definition of every concept appearing in 
a physical theory. Nevertheless, when we write down a mathematical 
equation, the presumption is that the symbols appearing in it have precise 
meanings, and we should strive to make our ideas as clear and precise as 
possible, and to recognize at what points there is a lack of precision or 
clarity. Sometimes a new concept can be defined in terms of others whose 
meanings are known, in which case there is no problem. For example, 

momentum = mass X velocity 
1 



2 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1 

gives a perfectly precise definition of "momentum" provided "mass" and 
"velocity" are assumed to be precisely defined already. But this kind of 
definition will not do for all terms in a theory, since we must start some- 
where with a set of basic concepts or "primitive" terms whose meanings 
are assumed known. The first concepts to be introduced in a theory can- 
not be defined in the above way, since at first we have nothing to put on 
the right side of the equation. The meanings of these primitive terms 
must be made clear by some means that lies outside of the physical theories 
being set up. We might, for example, simply use the terms over and over 
until their meanings become clear. This is the way babies learn a language, 
and probably, to some extent, freshman physics students learn the same 
way. We might define all primitive terms by stating their meaning in 
terms of observation and experiment. In particular, nouns designating 
measurable quantities, like force, mass, etc., may be defined by specifying 
the operational process for measuring them. One school of thought holds 
that all physical terms should be defined in this way. Or we might simply 
state what the primitive terms are, with a rough indication of their physi- 
cal meaning, and then let the meaning be determined more precisely by 
the laws and postulates we lay down and the rules that we give for inter- 
preting theoretical results in terms of experimental situations. This is the 
most convenient and flexible way, and is the way physical theories are 
usually set up. It has the disadvantage that we are never sure that our 
concepts have been given a precise meaning. It is left to experience to 
decide not only whether our laws are correct, but even whether the con- 
cepts we use have a precise meaning. The modern theories of relativity 
and quanta arise as much from fuzziness in classical concepts as from in- 
accuracies in classical laws. 

Historically, mechanics was the earliest branch of physics to be developed 
as an exact science. The laws of levers and of fluids in static equilibrium 
were known to Greek scientists in the third century B.C. The tremendous 
development of physics in the last three centuries began with the discovery 
of the laws qf mechanics by Galileo and Newton. The laws of mechanics 
as formulated by Isaac Newton in the middle of the seventeenth century 
and the laws of electricity and magnetism as formulated by James Clerk 
Maxwell about two hundred years later are the two basic theories of classi- 
cal physics. Relativistic physics, which began with the work of Einstein 
in 1905, and quantum physics, as based upon the work of Heisenberg 
and Schroedinger in 1925-1926, require a modification and reformulation 
of mechanics and electrodynamics in terms of new physical concepts. 
Nevertheless, modern physics builds on the foundations laid by classical 
physics, and a clear understanding of the principles of classical mechanics 
and electrodynamics is still essential in the study of relativistic and quan- 
tum physics. Furthermore, in the vast majority of practical applications 
of mechanics to the various branches of engineering and to astronomy, the 



1-1] MECHANICS, AN EXACT SCIENCE 3 

laws of classical mechanics can still be applied. Except when bodies travel 
at speeds approaching the speed of light, or when enormous masses or 
enormous distances are involved, relativistic mechanics gives the same re- 
sults as classical mechanics; indeed, it must, since we know from experi- 
ence that classical mechanics gives correct results in ordinary applications. 
Similarly, quantum mechanics should and does agree with classical mechan- 
ics except when applied to physical systems of molecular size or smaller. 
Indeed, one of the chief guiding principles in formulating new physical 
theories is the requirement that they must agree with the older theories 
when applied to those phenomena where the older theories are known to 
be correct. 

Mechanics is the study of the motions of material bodies. Mechanics 
may be divided into three subdisciplines, kinematics, dynamics, and statics. 
Kinematics is the study and description of the possible motions of mate- 
rial bodies. Dynamics is the study of the laws which determine, among 
all possible motions, which motion will actually take place in any given 
case. In dynamics we introduce the concept of force. The central prob- 
lem of dynamics is to determine for any physical system the motions which 
will take place under the action of given forces. Statics is the study of 
forces and systems of forces, with particular reference to systems of forces 
which act on bodies at rest. 

We may also subdivide the study of mechanics according to the kind of 
physical system to be studied. This is, in general, the basis for the outline 
of the present book. The simplest physical system, and the one we shall 
study first, is a single particle. Next we shall study the motion of a sys- 
tem of particles. A rigid body may be treated as a special kind of system 
of particles. Finally, we shall study the motions of continuous media, 
elastic and plastic substances, solids, liquids, and gases. 

A great many of the applications of classical mechanics may be based 
directly on Newton's laws of motion. All of the problems studied in this 
book, except in Chapters 9-12, are treated in this way. There are, how- 
ever, a number of other ways of formulating the principles of classical 
mechanics. The equations of Lagrange and of Hamilton are examples. 
They are not new physical theories, for they may be derived from Newton's 
laws, but they are different ways of expressing the same physical theory. 
They use more advanced mathematical concepts, they are in some respects 
more elegant than Newton's formulation, and they are in some cases more 
powerful in that they allow the solutions of some problems whose solution 
based directly on Newton's laws would be very difficult. The more differ- 
ent ways we know to formulate a physical theory, the better chance we 
have of learning how to modify it to fit new kinds of phenomena as they 
are discovered. This is one of the main reasons for the importance of the 
more advanced formulations of mechanics. They are a starting point for 
the newer theories of relativity and quanta. 



ELEMENTS OP NEWTONIAN MECHANICS 



[CHAP. 1 



1-2 Kinematics, the description of motion. Mechanics is the science 
which studies the motions of physical bodies. We must first describe mo- 
tions. Easiest to describe are the motions of a particle, that is, an object 
whose size and internal structure are negligible for the problem with which 
we are concerned. The earth, for example, could be regarded as a particle 
for most problems in planetary motion, but certainly not for terrestrial 
problems. We can describe the position of a particle by specifying a point 
in space. This may be done by giving three coordinates. Usually, rec- 
tangular coordinates are used. For a particle moving along a straight line 
(Chapter 2) only one coordinate need be given. To describe the motion 
of a particle, we specify the coordinates as functions of time : 



one dimension : x(t), 
three dimensions: x(t), y(t), z(t). 



(1-D 



The basic problem of classical mechanics is to find ways to determine func- 
tions like these which specify the positions of objects as functions of time, 
for any mechanical situation. The physical meaning of the function x(t) 
is contained in the rules which tell us how to measure the coordinate re of a 
particle at a time t. Assuming we know the meaning of x(t), or at least 
that it has a meaning (this assumption, which we make in classical me- 
chanics, is not quite correct according to quantum mechanics), we can 
define the x-component of velocity v x at time t as* 

_ _ dx 
p Vx ~ x ~ ~dt 



». = * = ^> (1-2) 



~r 



A 



LA. 



and, similarly, 
dy 



v,= z = 



dz 
dt' 



three dimensions 



h 



one dimension 



Fig. 1-1. Rectangular coordinates 
specifying the position of a particle P 
relative to an origin 0. 



y-axis w e now d e fme the components of 
acceleration a x , a y , a,,, as the deriva- 
tives of the velocity components 
with respect to time (we list several 
equivalent notations which may be 
used) : 

dv x 
dt 

dVy 

dt 

dv z 

~dJ 



a x = v x = — 77 — x 



a v = v v = -j? = y = 



V, = -7T = Z 



dt* ' 

d*y 
dt* 
dS 

dt*' 



(1-3) 



* We shall denote a time derivative either by d/dt or by a dot. Both notations 
are given in Eq. (1-2). 



1-3] DYNAMICS. MASS AND FORCE 5 

For many purposes some other system of coordinates may be more con- 
venient for specifying the position of a particle. When other coordinate 
systems are used, appropriate formulas for components of velocity and 
acceleration must be worked out. Spherical, cylindrical, and plane polar 
coordinates will be discussed in Chapter 3. For problems in two and three 
dimensions, the concept of a vector is very useful as a means of represent- 
ing positions, velocities, and accelerations. A systematic development of 
vector algebra will be given in Section 3-1. 

To describe a system of particles, we may specify the coordinates of 
each particle in any convenient coordinate system. Or we may introduce 
other kinds of coordinates, for example, the coordinates of the center of 
mass, or the distance between two particles. If the particles form a rigid 
body, the three coordinates of its center of mass and three angular coordi- 
nates specifying its orientation in space are sufficient to specify its position. 
To describe the motion of continuous matter, for example a fluid, we would 
need to specify the density p(x, y, z, t) at any point (x, y, z) in space at each 
instant t in time, and the velocity vector v(x, y, z, t) with which the matter 
at the point (x, y, z) is moving at time t. Appropriate devices for describ- 
ing the motion of physical systems will be introduced as needed. 

1-3 Dynamics. Mass and force. Experience leads us to believe that 
the motions of physical bodies are controlled by interactions between them 
and their surroundings. Observations of the behavior of projectiles and 
of objects sliding across smooth, well-lubricated surfaces suggest the idea 
that changes in the velocity of a body are produced by interaction with its 
surroundings. A body isolated from all interactions would have a con- 
stant velocity. Hence, in formulating the laws of dynamics, we focus our 
attention on accelerations. 

Let us imagine two bodies interacting with each other and otherwise 
isolated from interaction with their surroundings. As a rough approxima- 
tion to this situation, imagine two boys, not necessarily of equal size, en- 
gaged in a tug of war over a rigid pole on smooth ice. Although no two 
actual bodies can ever be isolated completely from interactions with all 
other bodies, this is the simplest kind of situation to think about and one 
for which we expect the simplest mathematical laws. Careful experiments 
with actual bodies lead us to conclusions as to what we should observe 
if we could achieve ideal isolation of two bodies. We should observe that 
the two bodies are always accelerated in opposite directions, and that the 
ratio of their accelerations is constant for any particular pair of bodies no 
matter how strongly they may be pushing or pulling each other. If we 
measure the coordinates xi and x 2 of the two bodies along the line of their 
accelerations, then 

xi/x 2 = —ki2, (1^1) 



6 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1 

where k i2 is a positive constant characteristic of the two bodies concerned. 
The negative sign expresses the fact that the accelerations are in opposite 
directions. 

Furthermore, we find that in general the larger or heavier or more mas- 
sive body is accelerated the least. We find, in fact, that the ratio k i2 is 
proportional to the ratio of the weight of body 2 to that of body 1. The 
accelerations of two interacting bodies are inversely proportional to their 
weights. This suggests the possibility of a dynamical definition of what 
we shall call the masses of bodies in terms of their mutual accelerations. 
We choose a standard body as a unit mass. The mass of any other body 
is defined as the ratio of the acceleration of the unit mass to the accelera- 
tion of the other body when the two are in interaction : 

mi = k u = —xjxi, (1-5) 

where m,- is the mass of body i, and body 1 is the standard unit mass. 

In order that Eq. (1-5) may be a useful definition, the ratio fci 2 of the 
mutual accelerations of two bodies must satisfy certain requirements. If 
the mass defined by Eq. (1-5) is to be a measure of what we vaguely call 
the amount of matter in a body, then the mass of a body should be the sum 
of the masses of its parts, and this turns out to be the case to a very high 
degree of precision. It is not essential, in order to be useful in scientific 
theories, that physical concepts for which we give precise definitions should 
correspond closely to any previously held common-sense ideas. However, 
most precise physical concepts have originated from more or less vague 
common-sense ideas, and mass is a good example. Later, in the theory 
of relativity, the concept of mass is somewhat modified, and it is no longer 
exactly true that the mass of a body is the sum of the masses of its parts. 

One requirement which is certainly essential is that the concept of mass 
be independent of the particular body which happens to be chosen as 
having unit mass, in the sense that the ratio of two masses will be the 
same no matter what unit of mass may be chosen. This will be true be- 
cause of the following relation, which is found experimentally, between 
the mutual acceleration ratios denned by Eq. (1-4) of any three bodies: 

ki 2 k 23 k 31 = 1. (1-6) 

Suppose that body 1 is the unit mass. Then if bodies 2 and 3 interact 
with each other, we find, using Eqs. (1-4), (1-6), and (1-5), 

x 2 /x 3 = —k 23 

= - l/(*i2ft 81 ) (1-7) 

= —k 13 /ki2 

= — m 3 /m 2 . 



1-4] newton's laws of motion 7 

The final result contains no explicit reference to body 1, which was taken 
to be the standard unit mass. Thus the ratio of the masses of any two 
bodies is the negative inverse of the ratio of their mutual accelerations, in- 
dependently of the unit of mass chosen. 
By Eq. (1-7), we have, for two interacting bodies, 

m 2 x 2 = — m x z x . (1-8) 

This suggests that the quantity (mass X acceleration) will be important, 
and we call this quantity the force acting on a body. The acceleration of 
a body in space has three components, and the three components of force 
acting on the body are 

F x = mx, F y = my, F z = ml. (1-9) 

The forces which act on a body are of various kinds, electric, magnetic, 
gravitational, etc., and depend on the behavior of other bodies. In 
general, forces due to several sources may act on a given body, and it is 
found that the total force given by Eqs. (1-9) is the vector sum of the 
forces which would be present if each source were present alone. 

The theory of electromagnetism is concerned with the problem of de- 
termining the electric and magnetic forces exerted by electrical charges 
and currents upon one another. The theory of gravitation is concerned 
with the problem of determining the gravitational forces exerted by 
masses upon one another. The fundamental problem of mechanics is to 
determine the motions of any mechanical system, given the forces acting 
on the bodies which make up the system. 

1-4 Newton's laws of motion. Isaac Newton was the first to give a 
complete formulation of the laws of mechanics. Newton stated his famous 
three laws as follows:* 

(1) Every body continues in its state of rest or of uniform motion 
in a straight line unless.it is compelled to change that state by forces 
impressed upon it. 

(2) Rate of change of momentum is proportional to the impressed 
force, and is in the direction in which the force acts. 

(3) To every action there is always opposed an equal reaction. 

In the second law, momentum is to be defined as the product of the mass 
and the velocity of the particle. Momentum, for which we use the symbol 

* Isaac Newton, Mathematical Principles of Natural Philosophy and his System 
of the World, tr. by F. Cajori (p. 13). Berkeley: University of California Press, 
1934. 



8 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1 

p, has three components, denned along x-, y-, and z-axes by the equations 

p x = mv x , p y = mv y , p z = mv z . (1-10) 

The first two laws, together with the definition of momentum, Eqs. (1-10), 
and the fact that the mass is constant by Eq. (1-4),* are equivalent to 
Eqs. (1-9), which express them in mathematical form. The third law 
states that when two bodies interact, the force exerted on body 1 by body 2 
is equal and opposite in direction to that exerted on body 2 by body 1. 
This law expresses the experimental fact given by Eq. (1-4), and can 
easily be derived from Eq. (1-4) and from Eqs. (1-5) and (1-9). 

The status of Newton's first two laws, or of Eqs. (1-9), is often the sub- 
ject of dispute. We may regard Eqs. (1-9) as defining force in terms of 
mass and acceleration. In this case, Newton's first two laws are not laws 
at all but merely definitions of a new concept to be introduced in the theory. 
The physical laws are then the laws of gravitation, electromagnetism, etc., 
which tell us what the forces are in any particular situation. Newton's 
discovery was not that force equals mass times acceleration, for this is 
merely a definition of "force." What Newton discovered was that the 
laws of physics are most easily expressed in terms of the concept of force 
defined in this way. Newton's third law is still a legitimate physical law 
expressing the experimental result given by Eq. (1-4) in terms of the con- 
cept of force. This point of view toward Newton's first two laws is con- 
venient for many purposes and is often adopted. Its chief disadvantage is 
that Eqs. (1-9) define only the total force acting on a body, whereas we 
often wish to speak of the total force as a (vector) sum of component forces 
of various kinds due to various sources. The whole science of statics, 
which deals with the forces acting in structures at rest, would be unintelli- 
gible if we took Eqs. (1-9) as our definition of force, for all accelerations are 
zero in a structure at rest. 

We may also take the laws of electromagnetism, gravitation, etc., to- 
gether with the parallelogram law of addition, as defining "force. " Equa- 
tions (1-9) then become a law connecting previously defined quantities. 
This has the disadvantage that the definition of force changes whenever a 
new kind of force (e.g., nuclear force) is discovered, or whenever modifica- 
tions are made in electromagnetism or in gravitation. Probably the best 
plan, the most flexible at least, is to take force as a primitive concept of 



* In the theory of relativity, the mass of a body is not constant, but depends on 
its velocity. In this case, law (2) and Eqs. (1-9) are not equivalent, and it turns 
out that law (2) is the correct formulation. Force should then be equated to time 
rate of change of momentum. The simple definition (1-5) of mass is not correct 
according to the theory of relativity unless the particles being accelerated move 
at low velocities. 



1- 4] newton's laws of motion 9 

our theory, perhaps defined operationally in terms of measurements with 
a spring balance. Then Newton's laws are laws, and so are the laws of 
theories of special forces like gravitation and electromagnetism. 

Aside from the question of procedure in regard to the definition of force, 
there are other difficulties in Newton's mechanics. The third law is not 
always true. It fails to hold for electromagnetic forces, for example, when 
the interacting bodies are far apart or rapidly accelerated and, in fact, it 
fails for any forces which propagate from one body to another with finite 
velocities. Fortunately, most of our development is based on the first 
two laws. Whenever the third law is used, its use will be explicitly noted 
and the results obtained will be valid only to the extent that the third law 
holds. 

Another difficulty is that the concepts of Newtonian mechanics are not 
perfectly clear and precise, as indeed no concepts can probably ever be for 
any theory, although we must develop the theory as if they were. An 
outstanding example is the fact that no specification is made of the coordi- 
nate system with respect to which the accelerations mentioned in the first 
two laws are to be measured. Newton himself recognized this difficulty 
but found no very satisfactory way of specifying the correct coordinate 
system to use. Perhaps the best way to formulate these laws is to say 
that there is a coordinate system with respect to which they hold, leaving 
it to experiment to determine the correct coordinate system. It can be 
shown that if these laws hold in any coordinate system, they hold also in 
any coordinate system moving uniformly with respect to the first. This 
is called the principle of Newtonian relativity, and will be proved in 
Section 7-1, although the reader should find little difficulty in proving it 
for himself. 

Two assumptions which are made throughout classical physics are that 
the behavior of measuring instruments is unaffected by their state of 
motion so long as they are not rapidly accelerated, and that it is possible, 
in principle at least, to devise instruments to measure any quantity with 
as small an error as we please. These two assumptions fail in extreme 
cases, the first at very high velocities, the second when very small magni- 
tudes are to be measured. The failure of these assumptions forms the 
basis of the theory of relativity and the theory of quantum mechanics, 
respectively. However, for a very wide range of phenomena, Newton's 
mechanics is correct to a very high degree of accuracy, and forms the 
starting point at which the modern theories begin. Not only the laws but 
also the concepts of classical physics must be modified according to the 
modern theories. However, an understanding of the concepts of modern 
physics is made easier by a clear understanding of the concepts of classical 
physics. These difficulties are pointed out here in order that the reader 
may be prepared to accept later modifications in the theory. This is not to 



10 



ELEMENTS OF NEWTONIAN MECHANICS 



[CHAP. 1 



say that Newton himself (or the reader either at this stage) ought to have 
worried about these matters before setting up his laws of motion. Had 
he done so, he probably never would have developed his theory at all. It 
was necessary to make whatever assumptions seemed reasonable in order 
to get started. Which assumptions needed to be altered, and when, and 
in what way, could only be determined later by the successes and failures 
of the theory in predicting experimental results. 



1-5 Gravitation. Althoug 
the motions of the planets 
a property of physical bodies 
to formulate a mathematical 
Newton showed, by methods 
planets could be quantitatively 
every pair of bodies is 
masses and inversely 
them. In symbols, 



there had been previous suggestions that 

of falling bodies on earth might be due to 

by which they attract one another, the first 

theory of this phenomenon was Isaac Newton. 

be considered later, that the motions of the 

accounted for if he assumed that with 

associated a force of attraction proportional to their 

proportional to the square of the distance between 



and 



to] 



F = 



Gmim 2 



(1-H) 



where mi, m 2 are the masses of the attracting bodies, r is the distance be- 
tween them, and G is a universal constant whose value according to ex- 
periment is* 



G = (6.670 ± 0.005) X 10 



" 8 cm 3 -sec 2 -gm" 



(1-12) 



For a spherically symmetrical body, we shall show later (Section 6-2) that 
the force can be computed as if all the mass were at the center. For a 
small body of mass m at the surface of the earth, the force of gravitation 

is therefore 

F = mg, (1-13) 

where 

GM 



9 = 



fl2 



= 980.2 cm-sec 



—2 



(1-14) 



and M is the mass of the earth and R its radius. The quantity g has the 
dimensions of an acceleration, and we can readily show by Eqs. (1-9) and 
(1-13) that any freely falling body at the surface of the earth is accelerated 
downward with an acceleration g. 

The fact that the gravitational force on a body is proportional to its 
mass, rather than to some other constant characterizing the body (e.g., 
its electric charge), is more or less accidental from the point of view of 
Newton's theory. This fact is fundamental in the general theory of rela- 



* Smithsonian Physical Tables, 9th ed., 1954. 



1-6] UNITS AND DIMENSIONS 11 

tivity. The proportionality between gravitational force and mass is proba- 
bly the reason why the theory of gravitation is ordinarily considered a 
branch of mechanics, while theories of other kinds of force are not. 

Equation (1-13) gives us a more convenient practical way of measuring 
mass than that contemplated in the original definition (1-5). We may 
measure a mass by measuring the gravitational force on it, as in a spring 
balance, or by comparing the gravitational force on it with that on a stand- 
ard mass, as in the beam or platform balance; in other words, by weigh- 
ing it. 

1-6 Units and dimensions. In setting up a system of units in terms of 
which to express physical measurements, we first choose arbitrary standard 
units for a certain set of "fundamental" physical quantities (e.g., mass, 
length, and time) and then define further derived units in terms of the 
fundamental units (e.g., the unit of velocity is one unit length per unit of 
time). It is customary to choose mass, length, and time as the funda- 
mental quantities in mechanics, although there is nothing sacred in this 
choice. We could equally well choose some other three quantities, or even 
more or fewer than three quantities, as fundamental. 

There are three systems of units in common use, the centimeter-gram- 
second or cgs system, the meter-kilogram-second or mks system, and the 
foot-pound-second or English system, the names corresponding to' the 
names of the three fundamental units in each system.* Units for ether 
kinds of physical quantities are obtained from their defining equations by 
substituting the units for the fundamental quantities which occur. For 
example, velocity, by Eq. (1-2), 

dx 

Vx = Tt' 

is defined as a distance divided by a time. Hence the units of velocity are 
cm/sec, m/sec, and ft/sec in the three above-mentioned systems, respec- 
tively. 

Similarly, the reader can show that the units of force in the three sys- 
tems as given by Eqs. (1-9) are gm-cm-sec -2 , kgm-m-sec -2 , lb-ft-sec -2 . 
These units happen to have the special names dyne, newton, and poundal, 
respectively. Gravitational units of force are sometimes defined by re- 
placing Eqs. (1-9) by the equations 

F x = mx/g, F y = my/g, F z = mz/g, (1-15) 



* In the mks system, there is a fourth fundamental unit, the coulomb of electri- 
cal charge, which enters into the definitions of electrical units. Electrical units in 
the cgs system are all defined in terms of centimeters, grams, and seconds. Elec- 
trical units in the English system are practically never used. 



(1-11) 



12 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1 

where g = 980.2 cm-sec -2 = 9.802 m-sec -2 = 32.16 ft-sec -2 is the stand- 
ard acceleration of gravity at the earth's surface. Unit force is then that 
force exerted by the standard gravitational field on unit mass. The names 
gram-weight, kilogram-weight, pound-weight are given to the gravita- 
tional units of force in the three systems. In the present text, we shall 
write the fundamental law of mechanics in the form (1-9) rather than 
(1-15) ; hence we shall be using the absolute units for force and not the 
gravitational units. 

Henceforth the question of units will rarely arise, since nearly all our 
examples will be worked out in algebraic form. It is assumed that the 
reader is sufficiently familiar with the units of measurement and their 
manipulation to be able to work out numerical examples in any system of 
units should the need arise. 

In any physical equation, the dimensions or units of all additive terms 
on both sides of the equation must agree when reduced to fundamental 
units. As an example, we may check that the dimensions of the gravita- 
tional constant in Eq. (1-11) are correctly given in the value quoted in 
Eq. (1-12) : 

„ = Gm 1 m 2 

r 2 

We substitute for each quantity the units in which it is expressed : 

. _2 N (cm 3 -sec _2 -gm -1 )(gm)(gm) . _ 2x ,, 1C ^ 

(gm-cm-sec ) = - , „/ = (gm-cm-sec ). (1-16) 

The check does not depend on which system of units we use so long as we 
use absolute units of force, and we may check dimensions without any 
reference to units, using symbols I, m, t for length, mass, time : 

{mlr2) = (^- 2 m-*)(m)(m) = ( ^_ 2) (1 _ 1?) 

When constant factors like G are introduced, we can, of course, always 
make the dimensions agree in any particular equation by choosing appro- 
priate dimensions for the constant. If the units in the terms of an equa- 
tion do not agree, the equation is certainly wrong. If they do agree, this 
does not guarantee that the equation is right. However, a check on dimen- 
sions in a result will reveal most of the mistakes that result from algebraic 
errors. The reader should form the habit of mentally checking the dimen- 
sions of his formulas at every step in a derivation. When constants are 
introduced in a problem, their dimensions should be worked out from the 
first equation in which they appear, and used in checking subsequent steps. 



1-7] SOME ELEMENTARY PROBLEMS IN MECHANICS 13 

1-7 Some elementary problems in mechanics. Before beginning a sys- 
tematic development of mechanics based on the laws introduced in this 
chapter, we shall review a few problems from elementary mechanics in 
order to fix these laws clearly in mind. 

One of the simplest mechanical problems is that of finding the motion of 
a body moving in a straight line, and acted upon by a constant force. If 
the mass of the body is m and the force is F, we have, by Newton's second 
law, 

F = ma. (1-18) 



The acceleration is then constant: 

dv F 

a = "^7 = — ' 

at m 



(1-19) 



If we multiply Eq. (1-19) by dt, we obtain an expression for the change in 
velocity dv occurring during the short time dt: 

F 
dv = — dt. (1-20) 

Integrating, we find the total change in velocity during the time t: 

/ dv = / —dt, (1-21) 

Jvq Jo m 

v — v = — t, (1-22) 

where v is the velocity at t = 0. If a; is the distance of the body from a 
fixed origin, measured along its line of travel, then 

We again multiply by dt and integrate to find x : 

L dx =L( vo+ ^) dt > (i - 24) 



xo ■'0 

x = x + v t + \ — 1\ (1-25) 



F,2 



where x represents the position of the body at t = 0. We now have a 
complete description of the motion. We can calculate from Eqs. (1-25) 
and (1-22) the velocity of the body at any time t, and the distance it has 



14 



ELEMENTS OF NEWTONIAN MECHANICS 



[CHAP. 1 



traveled. A body falling freely near the surface of the earth is acted upon 
by a constant force given by Eq. (1-13), and by no other force if air re- 
sistance is negligible. In this case, if x is the height of the body above 
some reference point, we have 



F = —mg. 



(1-26) 



The negative sign appears because the force is downward and the positive 
direction of x is upward. Substituting in Eqs. (1-19), (1-22), and (1-25), 
we have the familiar equations 



a = 



v = v — gt, 

x = x + v t — %gt 2 . 



(1-27) 
(1-28) 
(1-29) 



In applying Newton's law of motion, Eq. (1-18), it is essential to 
decide first to what body the law is to be applied, then to insert the 
mass m of that body and the total force F acting on it. Failure to keep 
in mind this rather obvious point is the source of many difficulties, one of 
which is illustrated by the horse-and-wagon dilemma. A horse pulls upon 
a wagon, but according to Newton's third law the wagon pulls back with 
an equal and opposite force upon the horse. How then can either the 
wagon or the horse move? The reader who can solve Problem 4 at the 
end of this chapter will have no difficulty answering this question. 

Consider the motion of the system illustrated in Fig. 1-2. Two masses 
mi and m 2 hang from the ends of a rope over a pulley, and we will suppose 
that m 2 is greater than my. We take x as the distance from the pulley 




mi 



mig 



ro2 



mzg 



Fig. 1-2. Atwood's machine. 



1-7] SOME ELEMENTARY PROBLEMS IN MECHANICS 15 

to m 2 . Since the length of the rope is constant, the coordinate x fixes 
the positions of both m x and m 2 . Both move with the same velocity 

. ■ = !• d- 3 °) 

the velocity being positive when m^ is moving upward and m 2 is moving 
downward. If we neglect friction and air resistance, the forces on mj 
and m 2 are 

F x = -vug + t, (1-31) 

F 2 = m 2 g - t, (1-32) 

where t is the tension in the rope. The forces are taken as positive when 
they tend to produce a positive velocity dx/dt. Note that the terms involv- 
ing t in these equations satisfy Newton's third law. The equations of 
motion of the two masses are 

— m x q + r = m x a, (1-33) 

m 2 g — t = m 2 a, (1-34) 

where a is the acceleration dv/dt, and is the same for both masses. By 
adding Eqs. (1-33) and (1-34), we can eliminate r and solve for the accel- 
eration : 

d 2 x (m 2 — mi) 
a = di? = (m, + m 2 ) 9- d-35) 

The acceleration is constant and the velocity v and position x can be 
found at any time t as in the preceding example. We can substitute for 
a from Eq. (1-35) in either Eq. (1-33) or (1-34) and solve for the tension: 

2m,\m 2 
T = WH g - (1 ~ 36) 

As a check, we note that if mx = m 2 , then a = and 

t = mtf = m 2 g, (1-37) 

as it should if the masses are in static equilibrium. As a matter of interest, 
note that if m 2 » ntj, then 

a = g, (1-38) 

t = 2m ig . (1-39) 

The reader should convince himself that these two results are to be ex- 
pected in this case. 



16 



ELEMENTS OF NEWTONIAN MECHANICS 



[CHAP. 1 




\ (L 






V\ mjf sin 6 


mg cos 6/ 


^S> 






V mg / 



' Fig. 1-4. Resolution of forces into 

Fig. 1-3. Forces acting on a brick components parallel and perpendicular 
sliding down an incline. to the incline. 



When several forces act on a body, its acceleration is determined by the 
vector sum of the forces which act. Conversely, any force can be resolved 
in any convenient manner into vector components whose vector sum is the 
given force, and these components can be treated as separate forces acting 
on the body.* As an example, we consider a brick of mass m sliding down 
an incline, as shown in Fig. 1-3. The two forces which act on the brick are 
the weight mg and the force F with which the plane acts on the brick. 
These two forces are added according to the parallelogram law to give a 
resultant R which acts on the brick: 



R = ma. 



(1-40) 



Since the brick is accelerated in the direction of the resultant force, it is 
evident that if the brick slides down the incline without jumping off or 
penetrating into the inclined plane, the resultant force R must be directed 
along the incline. In order to find R, we resolve each force into com- 
ponents parallel and perpendicular to the incline, as in Fig. 1-4. The 
force F exerted on the brick by the plane is resolved in Fig. 1-4 into two 
components, a force N normal to the plane preventing the brick from 
penetrating the plane, and a force / parallel to the plane, and opposed to 



* A systematic development of vector algebra will be given in Chapter 3. Only 
an understanding of the parallelogram law for vector addition is needed for the 
present discussion. 



1-7] SOME ELEMENTARY PROBLEMS IN MECHANICS 17 

the motion of the brick, arising from the friction between the brick and the 
plane. Adding parallel components, we obtain 

R = mg sin 8 — f, (1^1) 

and 

= N — mg cos 0. (1-42) 

If the frictional force / is proportional to the normal force N, as is often 
approximately true for dry sliding surfaces, then 

/ = fiN = fjimg cos d, iX~4S) 

where p, is the coefficient of friction. Using Eqs. (1-43), (1-il), and (1-40), 
we can calculate the acceleration : 

a = g (sin — n cos 0). (1-44) 

The velocity and position can now be found as functions of the time t, 
as in the first example. Equation (1-44) holds only when the brick is 
sliding down the incline. If it is sliding up the incline, the force / will 
oppose the motion, and the second term in Eq. (1-44) will be positive. 
This could only happen if the brick were given an initial velocity up the 
incline. If the brick is at rest, the frictional force / may have any value 
up to a maximum fi,N: 

f < H*N, (1-45) 

where /*„, the coefficient of static friction, is usually greater than p. In 
this case R is zero, and 

/ = mg sin 6 < n s mg cos 0. (1-46) 

According to Eq. (1-46), the angle of the incline must not be greater than 
a limiting value 6 r , the angle of repose: 

tan < tan 6 r = ju s . (1-47) 

If is greater than 9 T , the brick cannot remain at rest. 

If a body moves with constant speed v around a circle of radius r, its 

acceleration is toward the center of the circle, as we shall prove in Chapter 

3, and is of magnitude 

v 2 
« = -• (1-48) 

Such a body must be acted on by a constant force toward the center. 
This centripetal force is given by 

F = ma = ?y- ■ (1-49) 



18 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1 

Note that mv 2 /r is not a "centrifugal force" directed away from the center, 
but is mass times acceleration and is directed toward the center, as is the 
centripetal force F. As an example, the moon's orbit around the earth is 
nearly circular, and if we assume that the earth is at rest at the center, then, 
by Eq. (1-11), the force on the moon is 

F = «M2 , (1-50) 

where M is the mass of the earth and m that of the moon. We can ex- 
press this force in terms of the radius R of the earth and the acceleration 
g of gravity at the earth's surface by substituting for GM from Eq. (1-14) : 

F= M^!. (1-51) 



The speed v of the moon is 



v = 2 ^, d-52) 



where T is the period of revolution. Substituting Eqs. (1-51) and (1-52) 
in Eq. (1-49), we can find r: 

<* = *%£■ («8> 

This equation was first worked out by Isaac Newton in order to check his 
inverse square law of gravitation.* It will not be quite accurate because 
the moon's orbit is not quite circular, and also because the earth does not 
remain at rest at the center of the moon's orbit, but instead wobbles 
slightly due to the attraction of the moon. By Newton's third law, this 
attractive force is also given by Eq. (1-51). Since the earth is much 
heavier than the moon, its acceleration is much smaller, and Eq. (1-53) 
will not be far wrong. The exact treatment of this problem is given in 
Section 4-7. Another small error is introduced by the fact that g, as de- 
termined experimentally, includes a small effect due to the earth's rota- 
tion. (See Section 7-3.) If we insert the measured values, 

g = 980.2 cm-sec -2 , 
R = 6,368 kilometers, 
T = 27J days, 

we obtain, from Eq. (1-53), 

r = 383,000 kilometers. 



! Isaac Newton, op. cit., p. 407. 



PROBLEMS 19 

The mean distance to the moon according to modern measurements is 

r = 385,000 kilometers. 

The values of r and R available to Newton would not have given such close 
agreement. 



Problems 

1. Compute the gravitational force of attraction between an electron and a 
proton at a separation of 0.5 A (1 A = 10 -8 cm). Compare with the electrostatic 
force of attraction at the same distance. 

2. The coefficient of viscosity ij is defined by the equation 

F dv 

where F is the frictional force acting across an area A in a moving fluid, and dv is 
the difference in velocity parallel to A between two layers of fluid a distance ds 
apart, ds being measured perpendicular to A. Find the units in which the vis- 
cosity ij would be expressed in the foot-pound-second, cgs, and mks systems. Find 
the three conversion factors for converting coefficients of viscosity from one of 
these systems to another. 

3. A motorist is approaching a green traffic light with speed »o, when the 
light turns to amber, (a) If his reaction time is r, during which he makes his 
decision to stop and applies his foot to the brake, and if his maximum braking 
deceleration is o, what is the minimum distance s mm from the intersection at the 
moment the light turns to amber in which he can bring his car to a stop? (b) If 
the amber light remains on for a time t before turning red, what is the maximum 
distance Smax from the intersection at the moment the light turns to amber such 
that he can continue into the intersection at speed vo without running the red 
light? (c) Show that if his initial speed vq is greater than 

V0 max = 2a(< — t), 

there will be a range of distances from the intersection such that he can neither 
stop in time nor continue through without running the red light, (d) Make some 
reasonable estimates of t, t, and o, and calculate vo max in miles per hour. If vo = 
§!>o max, calculate s mm and s ma x. 

4. A boy of mass m pulls (horizontally) a sled of mass M . The coefficient of 
friction between sled and snow is /x. (a) Draw a diagram showing all forces acting 
on the boy and on the sled, (b) Find the horizontal and vertical components oA 
each force at a moment when boy and sled each have an acceleration a. (c) If 
the coefficient of static friction between the boy's feet and the ground is ju„ what 
is the maximum acceleration he can give to himself and the sled, assuming trac- 
tion to be the limiting factor? 



20 ELEMENTS OF NEWTONIAN MECHANICS [CHAP. 1 

5. A floor mop of mass m is pushed with a force F directed along the handle, 
which makes an angle 6 with the vertical. The coefficient of friction with the floor 
is fi. (a) Draw a diagram showing all forces acting on the mop. (b) For given 
6, n, find the force F required to slide the mop with uniform velocity across the 
floor, (c) Show that if is less than the angle of repose, the mop cannot be started 
across the floor by pushing along the handle. Neglect the mass of the mop handle. 

6. A box of mass m slides across a horizontal table with coefficient of friction 
fi. The box is connected by a rope which passes over a pulley to a body of mass M 
hanging alongside the table. Find the acceleration of the system and the tension 
in the rope. 

7. The brick shown in Figs. 1-3 and 1-4 is given an initial velocity t>o up the 
incline. The angle 6 is greater than the angle of repose. Find the distance the 
brick moves up the incline, and the time required for it to slide up and back to its 
original position. 

8. A curve in a highway of radius of curvature r is banked at an angle 6 with 
the horizontal. If the coefficient of friction is /x 8 , what is the maximum speed with 
which a car can round the curve without skidding? 

9. Assuming the earth moves in a circle of radius 93,000,000 miles, with a 
period of revolution of one year, find the mass of the sun in tons. 

10. (a) Compute the mass of the earth from its radius and the values of g 
and G. (b) Look up the masses and distances of the sun and moon and compute 
the force of attraction between earth and sun and between earth and moon. 
Check your results by making a rough estimate of the ratio of these two forces 
from a consideration of the fact that the former causes the earth to revolve about 
the sun once a year, whereas the latter causes the earth to wobble in a small 
circle, approximately once a month, about the common center of gravity of the 
earth-moon system. 

11. The sun is about 25,000 light years from the center of the galaxy, and 
travels approximately in a circle at a speed of 175 mi/sec. Find the approximate 
mass of the galaxy by assuming that the gravitational force on the sun can be 
calculated as if all the mass of the galaxy were at its center. Express the result as 
a ratio of the galactic mass to the sun's mass. (You do not need to look up either 
G or the sun's mass to do this problem if you compare the revolution of the sun 
around the galactic center with the revolution of the earth about the sun.) 



CHAPTER 2 
MOTION OF A PARTICLE IN ONE DIMENSION 

2-1 Momentum and energy theorems. In this chapter, we study the 
motion of a particle of mass m along a straight line, which we will take to 
be the x-axis, under the action of a force F directed along the z-axis. The 
discussion will be applicable, as we shall see, to other cases where the 
motion of a mechanical system depends on only one coordinate, or where 
all but one coordinate can be eliminated from the problem. 

The motion of the particle is governed, according to Eqs. (1-9), by the 
equation 

m§=F. (2-1) 

Before considering the solution of Eq. (2-1), we shall define some concepts 
which are useful in discussing mechanical problems and prove some 
simple general theorems about one-dimensional motion. The linear mo- 
mentum p, according to Eq. (1-10), is denned as 

p = mo = m -37 • (2-2) 

From Eq. (2-1), using Eq. (2-2) and the fact that m is constant, we obtain 

dp w 



dt 



(2-3) 



This equation states that the time rate of change of momentum is equal 
to the applied force, and is, of course, just Newton's second law. We may 
call it the (differential) momentum theorem. If we multiply Eq. (2-3) 
by dt and integrate from ti to t 2 , we obtain an integrated form of the 
momentum theorem: 



P2 — Pi 



[ H Fdt. (2-4) 



Equation (2-4) gives the change in momentum due to the action of the 
force F between the times t t and t 2 - The integral on the right is called 
the impulse delivered by the force F during this time; F must be known 
as a function of t alone in order to evaluate the integral. If F is given 
as F(x, v, f), then the impulse can be computed for any particular given 
motion x(t), v(t). 

21 



22 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

A quantity which will turn out to be of considerable importance is the 
kinetic energy, denned (in classical mechanics) by the equation 

(2-5) 







T 


= imv 2 . 




If 


we multiply Eq. (2-1) by v, 


we 

mv 


obtain 

dv „ 
dt = FV > 




or 












s» 


mv 2 


. dT 
' dt 


Fv 



(2-6) 

Equation (2-6) gives the rate of change of kinetic energy, and may be 
called the (differential) energy theorem. If we multiply by dt and inte- 
grate from <i to t 2 , we obtain the integrated form of the energy theorem: 

rl 2 



T x = f 2 Fv dt. (2-7) 

Jti 



Equation (2-7) gives the change in energy due to the action of the force F 
between the times ti and t%. The integral on the right is called the work 
done by the force during this time. The integrand Fv on the right is the 
time rate of doing work, and is called the power supplied by the force F. 
In general, when F is given as F(x, v, t), the work can only be computed 
for a particular specified motion x(t), v(t). Since v = dx/dt, we can re- 
write the work integral in a form which is convenient when F is known 
as a function of x: 

T 2 - T x = f F dx. (2-8) 

Jxi 

2-2 Discussion of the general problem of one-dimensional motion. If 

the force F is known, the equation of motion (2-1) becomes a second-order 
ordinary differential equation for the unknown function x(t). The force F 
may be known as a function of any or all of the variables t, x, and v. For 
any given motion of a dynamical system, all dynamical variables (x, v, F, 
p, T, etc.) associated with the system are, of course, functions of the 
time t, that is, each has a definite value at any particular time t. However, 
in many cases a dynamical variable such as the force may be known to 
bear a certain functional relationship to x, or to v, or to any combination 
of x, v, and t. As an example, the gravitational force acting on a body 
falling from a great height above the earth is known as a function of the 
height above the earth. The frictional drag on such a body would depend 
on its speed and on the density of the air and hence on the height above 
the earth; if atmospheric conditions are changing, it would also depend 
on t. If F is given as F(x, v, t), then when x(t) and v(t) are known, these 



2-2] THE GENERAL PROBLEM 23 

functions can be substituted to give F as a function of the time t alone; 
however, in general, this cannot be done until after Eq. (2-1) has been 
solved, and even then the function F(t) may be different for different 
possible motions of the particle. In any case, if F is given as F(x, v, t) 
(where F may depend on any or all of these variables), then Eq. (2-1) 
becomes a definite differential equation to be solved: 

<^ = ±F(x,x,t). (2-9) 

This is the most general type of second-order ordinary differential equa- 
tion, and we shall be concerned in this- chapter with studying its solutions 
and their applications to mechanical problems. 

Equation (2-9) applies to all possible motions of the particle under the 
action of the specified force. In general, there will be many such motions, 
for Eq. (2-9) prescribes only the acceleration of the particle at every in- 
stant in terms of its position and velocity at that instant. If we know 
the position and velocity of a particle at a certain time, we can determine 
its position a short time later (or earlier). Knowing also its acceleration, 
we can find its velocity a short time later. Equation (2-9) then gives the 
acceleration a short time later. In this manner, we can trace out the past 
or subsequent positions and velocities of a particle if its position x and 
velocity v are known at any one time t . Any pair of values of x and v 
will lead to a possible motion of the particle. We call t Q the initial instant, 
although it may be any moment in the history of the particle, and the 
values of x and v at t we call the initial conditions. Instead of specifying 
initial values for x and v, we could specify initial values of any two quan- 
tities from which x and v can be determined; for example, we may specify 
x and the initial momentum p — mv . These initial conditions, together 
with Eq. (2-9), then represent a perfectly definite problem whose solu- 
tion should be a unique function x(t) representing the motion of the 
particle under the specified conditions. 

The mathematical theory of second-order ordinary differential equa- 
tions leads to results in agreement with what we expect from the nature 
of the physical problem in which the equation arises. The theory asserts 
that, ordinarily, an equation of the form (2-9) has a unique continuous 
solution x(t) which takes on given values x Q and v of x and x at any chosen 
initial value t of t. "Ordinarily" here means, as far as the beginning 
mechanics student is concerned, "in all cases of physical interest."* The 
properties of differential equations like Eq. (2-9) are derived in most 

* For a rigorous mathematical statement of the conditions for the existence of 
a solution of Eq. (2-9), see W. Leighton, An Introduction to the Theory of Differen- 
tial Equations. New York: McGraw-Hill, 1952. (Appendix 1.) 



24 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

treatises on differential equations. We know that any physical problem 
must always have a unique solution, and therefore any force function 
F(x, x, t) which can occur in a physical problem will necessarily satisfy the 
required conditions for those values of x, x, t of physical interest. Thus 
ordinarily we do not need to worry about whether a solution exists. How- 
ever, most mechanical problems involve some simplification of the actual 
physical situation, and it is possible to oversimplify or otherwise distort 
a physical problem in such a way that the resulting mathematical problem 
no longer possesses a unique solution. The general practice of physicists 
in mechanics and elsewhere is to proceed, ignoring questions of mathe- 
matical rigor. On those fortunately rare occasions when we run into diffi- 
culty, we then consult our physical intuition, or check, our lapses of rigor, 
until the source of the difficulty is discovered. Such a procedure may 
bring shudders to the mathematician, but it is the most convenient and 
rapid way to apply mathematics to the solution of physical problems. 
The physicist, while he may proceed in a nonrigorous fashion, should never- 
theless be acquainted with the rigorous treatment of the mathematical 
methods which he uses. 

The existence theorem for Eq. (2-9) guarantees that there is a unique 
mathematical solution to this equation for all cases which will arise in 
practice. In some cases the exact solution can be found by elementary 
methods. Most of the problems considered in this text will be of this 
nature. Fortunately, many of the most important mechanical problems 
in physics can be solved without too much difficulty. In fact, one of the 
reasons why certain problems are considered important is that they can 
be easily solved. The physicist is concerned with discovering and verify- 
ing the laws of physics. In checking these laws experimentally, he is free, 
to a large extent, to choose those cases where the mathematical analysis 
is not too difficult to carry out. The engineer is not so fortunate, since 
his problems are selected not because they are easy to solve, but because 
they are of practical importance. In engineering, and often also in physics, 
many cases arise where the exact solution of Eq. (2-9) is difficult or im- 
possible to obtain. In such cases various methods are available for obtain- 
ing at least approximate answers. The reader is referred to courses and 
texts on differential equations for a discussion of such methods.* From 
the point of view of theoretical mechanics, the important point is that 
a solution always does exist and can be found, as accurately as desired. 
We shall restrict our attention to examples which can be treated by 
simple methods. 



* W. E. Milne, Numerical Calculus. Princeton: Princeton University Press, 
1949. (Chapter 5.) 

H. Levy and E. A. Baggott, Numerical Solutions of Differential Equations. 
New York: Dover Publications, 1950. 



2-3] APPLIED FORCE DEPENDING ON THE TIME 25 

2-3 Applied force depending on the time. If the force F is given as a 
function of the time, then the equation of motion (2-9) can be solved in 
the following manner. Multiplying Eq. (2-9) by dt and integrating from 
an initial instant t to any later (or earlier) instant t, we obtain Eq. (2-4), 
which in this case we write in the form 



mv ^- mv 



/" F(t)dt. (2-10) 

J to 



Since F(t) is a known function of t, the integral on the right can, at least 
in principle, be evaluated and the right member is then a function of t 
(and t ). We solve for v : 

dr 1 /"' 

V = -dl = V o + m F ® dL ( 2 - U ) 

J to 

Now multiply by dt and integrate again from t to t: 

x — x = v (t — t ) +— I \ J F(t) dt 



m Lo L/io 



dt. (2-12) 



To avoid confusion, we may rewrite the variable of integration as t' in 
the first integral and t" in the second: 

x = xo + v (t - to) + — l dt" I F(t') dt'. (2-13) 



m I to J tB 



This gives the required solution x(t) in terms of two integrals which can 
be evaluated when F(t) is given. A definite integral can always be evalu- 
ated. If an explicit formula for the integral cannot be found, then at 
least it can always be computed as accurately as we please by numerical 
methods. For this reason, in the discussion of a general type of problem 
such as the one above, we ordinarily consider the problem solved when 
the solution has been expressed in terms of one or more definite integrals. 
In a practical problem, the integrals would have to be evaluated to obtain 
the final solution in usable form.* 



* The reader who has studied differential equations may be disturbed by the 
appearance of three constants, to, »o, and xo, in the solution (2-13), whereas the 
general solution of a second-order differential equation should contain only two 
arbitrary constants. Mathematically, there are only two independent constants 
in Eq. (2-13), an additive constant containing the terms xo — voto plus a term 
from the lower limit of the last integral, and a constant multiplying t containing 
the term wo plus a term from the lower limit of the first integral. Physically, we 
can take any initial instant to, and then just two parameters xo and t»o are re- 
quired to specify one out of all possible motions subject to the given force. 



26 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

Problems in which F is given as a function of t usually arise when we 
seek to find the behavior of a mechanical system under the action of some 
external influence. As an example, we consider the motion of a free elec- 
tron of charge — e when subject to an oscillating electric field along the 
ic-axis: 

E x = E cos (ut + 0). (2-14) 

The force on the electron is 

F = —eE x = —eE cos (ut + 0). (2-15) 

The equation of motion is 

m ^ = -eE cos (ut + 0). (2-16) 

We multiply by dt and integrate, taking t = : 

dx . eE sin eEo .,.,„■> ,„ ,« 

v = -jt = v H y sin (ut + 0). (2-17) 

dt mu mu 

Integrating again, we obtain 

eEn cos . ( . eE sin \ , . eE . . ,„ ,., 

x = x — u „ h V ^o H 1 1 ^ ^ cos (at + 6). (2-18) 

If the electron is initially at rest at x = 0, this becomes 

a; = ^— 5 1 -\ ^ cos (ut + 0). (2-19) 

mw 2 mu mu 2 

It is left to the reader to explain physically the origin of the constant term 
and the term linear in t in Eq. (2-19) in terms of the phase of the electric 
field at the initial instant. How do the terms in Eq. (2-19) depend on 
e, m, E , and w? Explain physically. Why does the oscillatory term turn 
out to be out of phase with the applied force? 

The problem considered here is of interest in connection with the propa- 
gation of radio waves through the ionosphere, which contains a high density 
of free electrons. Associated with a radio wave of angular frequency w is 
an electric field which may be given by Eq. (2-14). The oscillating term in 
Eq. (2-18) has the same frequency w and is independent of the initial 
conditions. This coherent oscillation of the free electrons modifies the 
propagation of the wave. The nonoscillating terms in Eq. (2-18) depend 
on the initial conditions, and hence on the detailed motion of each electron 
as the wave arrives. These terms cannot contribute to the propagation 
characteristics of the wave, since they do not oscillate with the frequency 
of the wave, although they may affect the leading edge of the wave which 



2-3] APPLIED FOBCE DEPENDING ON THE TIME 27 

arrives first. We see that the oscillatory part of the displacement x is 180° 
out of phase with the applied force due to the electric field. Since the elec- 
tron has a negative charge, the resulting electric polarization is 180° out 
of phase with the electric field. The result is that the dielectric constant 
of the ionosphere is less than one. (In an ordinary dielectric at low fre- 
quencies, the charges are displaced in the direction of the electric force 
on them, and the dielectric constant is greater than one.) Since the veloc- 
ity of light is 

v = c(ne)- 112 , (2-20) 

where c = 3 X 10 10 cm/sec and e and /x are the dielectric constant and 
magnetic permeability respectively, and since /x = 1 here, the (phase) 
velocity v of radio waves in the ionosphere is greater than the velocity c 
of electromagnetic waves in empty space. Thus waves entering the 
ionosphere at an angle are bent back toward the earth. The effect is seen 
to be inversely proportional to w 2 , so that for high enough frequencies, 
the waves do not return to the earth but pass out through the ionosphere. 



Only a slight knowledge of electromagnetic theory is required to carry this dis- 
cussion through mathematically.* The dipole moment of the electron displaced 
from its equilibrium position is 

e 2 2 

—ex = = E cos (tot + 0) = e —= E x (2-21) 

(wo 2 mw 2 

if we consider only the oscillating term. If there are N electrons per cm 3 , the 
total dipole moment per unit volume is 

Ne 2 
P x = 5 E x . (2-22) 

The electric displacement is 

D x = E x + 1tP x = (l - ^r) E *- (2-23) 

Since the dielectric constant is denned by 

D x = eE x , (2-24) 



we conclude that 



and since /i = 1, 



....(l-M^)" 1 ". (2-26) 



* See, e.g., G. P. Harnwell, Principles of Electricity and Electromagnetism, 2nd 
ed. New York: McGraw-Hill, 1949. (Section 2.4.) 



28 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

2-4 Damping force depending on the velocity. Another type of force 
which allows an easy solution of Eq. (2-9) is the case when F is a function 
of v alone : 

m ~ = F(v). (2-27) 

To solve, we multiply by [m^(y)] -1 dt and integrate from t to t: 

dv t — to 



f 



F(v) 



(2-28) 



The integral on the left can be evaluated, in principle at least, when F(v) 
is given, and an equation containing the unknown v results. If this equa- 
tion is solved for v (we assume in general discussions that this can always 
be done), we will have an equation of the form 



The solution for x is then 

x = x + 



C«( v -hr) dL (2 " 30) 



In the case of one-dimensional motion, the only important kinds of forces 
which depend on the velocity are frictional forces. The force of sliding or 
rolling friction between dry solid surfaces is nearly constant for a given 
pair of surfaces with a given normal force between them, and depends on 
the velocity only in that its direction is always opposed to the velocity. 
The force of friction between lubricated surfaces or between a solid body 
and a liquid or gaseous medium depends on the velocity in a complicated 
way, and the function F(v) can usually be given only in the form of a 
tabulated summary of experimental data. In certain cases and over 
certain ranges of velocity, the frictional force is proportional to some fixed 
power of the velocity: 

F = (=F)bv n . (2-31) 

If n is an odd integer, the negative sign should be chosen in the above 
equation. Otherwise the sign must be chosen so that the force has the 
opposite sign to the velocity v . The frictional force is always opposed to 
the velocity, and therefore does negative work, i.e., absorbs energy from 
the moving body. A velocity-dependent force in the same direction as 
the velocity would represent a source of energy; such cases do not often 
occur. 



2-4] DAMPING FORCE DEPENDING ON THE VELOCITY 29 

As an example, we consider the problem of a boat traveling with initial 
velocity v , which shuts off its engines at t = when it is at the position 
xo = 0. We assume the force of friction given by Eq. (2-31) with n = 1 : 

dv 
m-TT=—bv. (2-32) 

We solve Eq. (2-32), following the steps outlined above [Eqs. (2-27) 
through (2-30)]: 



rv 
Jvo V 



It, 

m 



= 


1. 


v e~ 


-btjm 


dt 


= 


mvQ 
b 


(1 


— e" 


—bt/m 



, v b 

In — = t, 

Vq m 

v = v e- btlm . (2-33) 

We see that as t — ► oo , v — * 0, as it should, but that the boat never comes 
completely to rest in any finite time. The solution for x is 



(2-34) 

As t — * oo , x approaches the limiting value 

*, = ^. (2-35) 

Thus we can specify a definite distance that the boat travels in stopping. 
Although according to the above result, Eq. (2-33), the velocity never 
becomes exactly zero, when t is sufficiently large the velocity becomes so 
small that the boat is practically stopped. Let us choose some small 
velocity v 3 such that when v < v„ we are willing to regard the boat as 
stopped (say, for example, the average random speed given to an anchored 
boat by the waves passing by it). Then we can define the time t a required 
for the boat to stop by 

v s = v e- bt '' m , < s = ^ln^- (2-36) 

Vg 

Since the logarithm is a slowly changing function, the stopping time t s will 
not depend to any great extent on precisely what value of v e we choose so 
long as it is much smaller than v . It is often instructive to expand solu- 
tions in a Taylor series in t. If we expand the right side of Eqs. (2-33) 



30 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

and (2-34) in power series in t, we obtain* 

v = v -^t + ---, (2-37) 

x = v t-± b -^t 2 + ---. (2-38) 

Note that the first two terms in the series for v and x are just the formulas 
for a particle acted on by a constant force —too, which is the initial value 
of the frictional force in Eq. (2-32). This is to be expected, and affords a 
fairly good check on the algebra which led to the solution (2-34). Series 
expansions are a very useful means of obtaining simple approximate for- 
mulas valid for a short range of time. 

The characteristics of the motion of a body under the action of a fric- 
tional force as given by Eq. (2-31) depend on the exponent n. In general, 
a large exponent n will result in rapid initial slowing but slow final stopping, 
and vice versa, as one can see by sketching graphs of F vs. v for various 
values of n. For small enough values of n, the velocity comes to zero in 
a finite time. For large values of n, the body not only requires an infinite 
time, but travels an infinite distance before stopping. This disagrees 
with ordinary experience, an indication that while the exponent n may be 
large at high velocities, it must become smaller at low velocities. The 
exponent n = 1 is often assumed in problems involving friction, particu- 
larly when friction is only a small effect to be taken into account approxi- 
mately. The reason for taking n = 1 is that this gives easy equations 
to solve, and is often a fairly good approximation when the frictional 
force is small, provided b is properly chosen. -— ' 

2-5 Conservative force depending on position. Potential energy. One 
of the most important types of motion occurs when the force F is a func- 



* The reader who has not already done so should memorize the Taylor series 
for a few simple functions like 



2 3 4 

l + x + X - + X ' X 



2 ' 2-3 ' 2-3-4 



ln(l + x) = *_|- + ?L_fL + 



(i + xr = i + ^ + ^^/ + w(w - 1)( r- 2) x 3 + ---. 

These three series are extremely useful in obtaining approximations to compli- 
cated formulas, valid when x is small. 



2-5] CONSERVATIVE FORCE DEPENDING ON POSITION 31 

tion of the coordinate x alone : 

dv 
m Tt = F(x). (2-39) 

We have then, by the energy theorem (2-8), 

imv 2 — \rrw\ = f" F(x) dx. (2-40) 

Jx 

The integral on the right is the work done by the force when the particle 
goes from x to x. We now define the potential energy V(x) as the work 
done by the force when the particle goes from x to some chosen standard 
point x s : 

V(x) = f ° F(x) dx= - f F(x) dx. (2-41) 

Jx Jx, 

The reason for calling this quantity potential energy will appear shortly. 
In terms of V(x), we can write the integral in Eq. (2-40) as follows: 

f F(x) dx = -V(x) + V(x ). (2-42) 

Jx 

With the help of Eq. (2-42), Eq. (2-40) can be written 

\mv 2 + V{x) = \mvl + V(x ). (2^3) 

The quantity on the right depends only on the initial conditions and is 
therefore constant during the motion. It is called the total energy E, and 
we have the law of conservation of kinetic plus potential energy, which 
holds, as we can see, only when the force is a function of position alone : 

\mo 2 + V(x) = T + V = E. (2-44) 

Solving for v, we obtain 

v =tt->M [E ~ nx)]U2 - <*-«) 

The function x(t) is to be found by solving for x the equation 



\?i.. 



[E - V(x)]- 112 dx = t- t . (2-46) 



In this case, the initial conditions are expressed in terms of the constants 
E and xq. 

In applying Eq. (2-46), and in taking the indicated square root in the 
integrand, care must be taken to use the proper sign, depending on whether 



32 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

the velocity v given by Eq. (2-45) is positive or negative. In cases where 
v is positive during some parts of the motion and negative during other 
parts, it may be necessary to carry out the integration in Eq. (2-46) 
separately for each part of the motion. 

From the definition (2-41) we can express the force in terms of the 
potential energy: 

This equation can be taken as expressing the physical meaning of the po- 
tential energy. The potential energy is a function whose negative 
derivative gives the force. The effect of changing the coordinate of the 
standard point x s is to add a constant to V(x). Since it is the derivative 
of V which enters into the dynamical equations as the force, the choice 
of standard point x, is immaterial. A constant can always be added to 
the potential V(x) without affecting the physical results. (The same 
constant must, of course, be added to E.) 

As an example, we consider the problem of a particle subject to a 
linear restoring force, for example, a mass fastened to a spring: 

F = -kx. (2-48) 

The potential energy, if we take x s — 0, is 

rx 

V{x) = — I —kx dx 
Jo 

= \kx 2 . (2-49) 

Equation (2-46) becomes, for this case, with t = 0, 

(E - %kx 2 )~ U2 dx = t. (2-50) 



\?ix„ 



'10 

Now make the substitutions 



sin d = x yjzjj ' (2-51) 

W = JA, (2-52) 



so that 



Jf f (E - %kx 2 )- 112 dx = i f de = Ue-e ), 

\ ^ Jxq w Je « 



>x 

and, by Eq. (2-50), 



6 = <at + O . 



2-5] CONSERVATIVE FORCE DEPENDING ON POSITION 33 

We can now solve for x in Eq. (2-51) : 

x = J-r- sin = A sin (wi + O ), (2-53) 



where 



F- ( 2 ~ 54 ) 



Thus the coordinate x oscillates harmonically in time, with amplitude A 
and frequency co/27r. The initial conditions are here determined by the 
constants A and O , which are related to E and x by 

E = %JcA 2 , (2-55) 

x Q = A sin O . (2-56) 

Notice that in this example we meet the sign difficulty in taking the 
square root in Eq. (2-50) by replacing (1 — sin 2 0) _1/2 by (cos0) _1 , a 
quantity which can be made either positive or negative as required by 
choosing in the proper quadrant. 

A function of the dependent variable and its first derivative which 
is constant for all solutions of a second-order differential equation, is called 
a first integral of the equation. The function %mx 2 + V(x) is called the 
energy integral of Eq. (2-39). An integral of the equations of motion of 
a mechanical system is also called a constant of the motion. In general, 
any mechanical problem can be solved if we can find enough first inte- 
grals, or constants of the motion. 

Even in cases where the integral in Eq. (2-16) cannot easily be evalu- 
ated or the resulting equation solved to give an explicit solution for x(t), 
the energy integral, Eq. (2-44), gives us useful information about the 
solution. For a given energy E, we see from Eq. (2-15) that the particle 
is confined to those regions on the x-axis where V(x) < E. Furthermore, 
the velocity is proportional to the square root of the difference between 
E and V(x). Hence, if we plot V(x) versus x, we can give a good qualita- 
tive description of the kinds of motion that are possible. For the potential- 
energy function shown in Fig. 2-1 we note that the least energy possible 
is E . At this energy, the particle can only be at rest at x Q . With a 
slightly higher energy E u the particle can move between xi and x 2 ; its 
velocity decreases as it approaches xi or x 2 , and it stops and reverses its 
direction when it reaches either Xi or x 2 , which are called turning points 
of the motion. With energy E 2 , the particle may oscillate between turn- 
ing points x 3 and x iy or remain at rest at x 5 . With energy E 3 , there are 
four turning points and the particle may oscillate in either of the two 



34 



MOTION OF A PARTICLE IN ONE DIMENSION 



[CHAP. 2 




xa xzx\ x« X2 Xi Xi 

Fig. 2-1. A potential-energy function for one-dimensional motion. 

potential valleys. With energy E t , there is only one turning point; if 
the particle is initially traveling to the left, it will turn at x 6 and return 
to the right, speeding up over the valleys at x and x 5 , and slowing down 
over the hill between. At energies above E 5 , there are no turning points 
and the particle will move in one direction only, varying its speed accord- 
ing to the depth of the potential at each point. 

A point where V(x) has a minimum is called a point of stable equilibrium. 
A particle at rest at such a point will remain at rest. If displaced a slight 
distance, it will experience a restoring force tending to return it, and it 
will oscillate about the equilibrium point. A point where V(x) has a maxi- 
mum is called a point of unstable equilibrium. In theory, a particle at 
rest there can remain at rest, since the force is zero, but if it is displaced 
the slightest distance, the force acting on it will push it farther away from 
the unstable equilibrium position. A region where V(x) is constant is 
called a region of neutral equilibrium, since a particle can be displaced 
slightly without suffering either a restoring or a repelling force. 

This kind of qualitative discussion, based on the energy integral, is 
simple and very useful. Study this example until you understand it well 
enough to be able to see at a glance, for any potential energy curve, the 
types of motion that are possible. 

It may be that only part of the force on a particle is derivable from a 
potential function V(x). Let F' be the remainder of the force: 



F = 



dV 
dx 



+ F'. 



(2-57) 



In this case the energy (T -\- V) is no longer constant. If we substitute F 



2-6] FALLING BODIES 35 

from Eq. (2-57) in Eq. (2-1), and multiply by dx/dt, we have, after 
rearranging terms, 

I (T + V) = F'v. (2-58) 

The time rate of change of kinetic plus potential energy is equal to the 
power delivered by the additional force F'. 

2-6 Falling bodies. One of the simplest and most commonly occurring 
types of one-dimensional motion is that of falling bodies. We take up 
this type of motion here as an illustration of the principles discussed in 
the preceding sections. 

A body falling near the surface of the earth, if we neglect air resistance, 
is subject to a constant force 

F = —mg, (2-59) 

where we have taken the positive direction as upward. The equation of 
motion is 

d 2 x 
™- d ~p = — mg. (2-60) 

The solution may be obtained by any of the three methods discussed 
in Sections 2-3, 2-4, and 2-5, since a constant force may be considered 
as a function of either t, v, or x. The reader will find it instructive to 
solve the problem by all three methods. We have already obtained the 
result in Chapter 1 [Eqs. (1-28) and (1-29)]. 

In order to include the effect of air resistance, we may assume a fric- 
tional force proportional to v, so that the total force is 

F = — mg — bv. (2-61) 

The constant & will depend on the size and shape of the falling body, 
as well as on the viscosity of the air. The problem must now be treated 
as a case of F(v): 

dv , 

m -^ = — mg — bv. (2-62) 

Taking v = at t = 0, we proceed as in Section 2-4 [Eq. (2-28)]: 

dv bt 



f 

Jo 



! v + (mg/b) m 

We integrate and solve for v : 



(2-63) 



v = - ?f (1 - e- bt/m ). (2-64) 



36 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

We may obtain a formula useful for short times of fall by expanding the 
exponential function in a power series : 

v = -gt + \ % t 2 + ■ ■ ■ . (2-65) 

Thus for a short time (t <£ m/b), v = —gt, approximately, and the effect 

of air resistance can be neglected. After a long time, we see from Eq. (2-64) 

that 

v = ~ > if t » -r- • 

b o 

The velocity mg/b is called the terminal velocity of the falling body in ques- 
tion. The body reaches within 1/e of its terminal velocity in a time 
t = m/b. We could use the experimentally determined terminal velocity 
to find the constant b. We now integrate Eq. (2-64), taking x — 0: 

b z \ m / 

By expanding the exponential function in a power series, we obtain 

If t « m/b, x = — \qt 2 , as in Eq. (1-29). When t » m/b, 

This result is easily interpreted in terms of terminal velocity. Why is 
the positive constant present? 

For small heavy bodies with large terminal velocities, a better approxi- 
mation may be 

F = bv 2 . (2-68) 

The reader should be able to show that with the frictional force given by 
Eq. (2-68), the result (taking x = v = at t = 0) is 

(2-69) 



x = -%gt 2 + i % t 3 + ■ ■ ■ . (2-67) 

771/ 



v = - 


-^«"(VS') 




\7Yl 

-gt, if *«^' 




.-# i! «»>!■ 



2-6] 



FALLING BODIES 



r In cosh 

b 



m / 



37 
(2-70) 



-&t 2 , 



if 



t« 



W 



m 



In 2 - 



t, 



if 



f » 



fm 

w 



Again there is a terminal velocity, given this time by (mg/b) 112 . The ter- 
minal velocity can always be found as the velocity at which the frictional 
force equals the gravitational force, and will exist whenever the frictional 
force becomes sufficiently large at high velocities. 

In the case of bodies falling from a great height, the variation of the 
gravitational force with height should be taken into account. In this case, 
we neglect air resistance, and measure x from the center of the earth. 
Then if M is the mass of the earth and m the mass of the falling body, 



F = 



and 



V(x) = 



mMG 



Fdx = - 



mMG 



(2-71) 
(2-72) 



where we have taken x s = oo in order to avoid a constant term in V(x). 
Equation (2-4:5) becomes 

12 

(2-73) 



-i^vio^r 



The plus sign refers to ascending motion, the minus sign to descending 
motion. 

The function V(x) is plotted in v{x) 

Fig. 2-2. We see that there are two 
types of motion, depending on 
whether E is positive or negative. 
When E is positive, there is no turn- 
ing point, and if the body is initially 
moving upward, it will continue to 
move upward forever, with decreas- 
ing velocity, approaching the limit- 
ing velocity 

(OP 

Vl = \ ~m ' ( 2_74) Fig. 2-2. Plot of V(x) = -(mMG/x). 



38 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

When E is negative, there is a turning point at a height 

mMG , n __. 

x T = —^ ■ (2-75) 

If the body is initially moving upward, it will come to a stop at xt, and 
fall back to the earth. The dividing case between these two types of 
motion occurs when the initial position and velocity are such that E = 0. 
The turning point is then at infinity, and the body moves upward forever, 
approaching the limiting velocity vi = 0. If E = 0, then at any height x, 
the velocity will be 

»• = >/—• (2_76) 

This is. called the escape velocity for a body at distance x from the center 
of the earth, because a body moving upward at height x with velocity v e 
will just have sufficient energy to travel upward indefinitely (if there is 
no air resistance). 

To find x(t), we must evaluate the integral 

dx /2 t, (2-77) 



(* + =F)" 



1/2 -\ m 



where x is the height at t = 0. To solve for the case when E is negative, 
we substitute 

cos *=VH- < 2 - 78) 

Equation (2-77) then becomes 



mMG 



(-E)- 



\z / 2 cos 2 Odd = J-t. (2-79) 



(We choose a positive sign for the integrand so that will increase when t 
increases.) We can, without loss of generality, take x to be at the turning 
point x T , since the body will at some time in its past or future career pass 
through x T if no force except gravity acts upon it, provided E < 0. Then 
d = 0, and 

mMG , ,. a , [2 . 

(0 + sin 6 cos 0) = ~ — t, 



(—E) 3 I 2 \m 

or 



+ isin20= x F^t, (2-80) 

V x T 



2-7] 



THE SIMPLE HARMONIC OSCILLATOR 



39 



and 



Xt COS 0. 



(2-81) 



This pair of equations cannot be solved explicitly for x(t). A numerical 
solution can be obtained by choosing a sequence of values of and finding 
the corresponding values of x and t from Eqs. (2-80) and (2-81). That 
part of the motion for which x is less than the radius of the earth will, of 
course, not be correctly given, since Eq. (2-71) assumes all the mass of the 
earth concentrated at x = (not to mention the fact that we have omitted 
from our equation of motion the forces which would act on the body when 
it collides with the earth). 

The solution can be obtained in a similar way for the cases when E is 
positive or zero. 

2-7 The simple harmonic oscillator. The most important problem in 
one-dimensional motion, and fortunately one of the easiest to solve, is the 
harmonic or linear oscillator. The simplest example is that of a mass m 
fastened to a spring whose constant is k. If we measure x from the re- 
laxed position of the spring, then the spring exerts a restoring force 



F = —kx. 



(2-82) 



The potential energy associated 
with this force is 

V(x) = ikx 2 . (2-83) 

The equation of motion, if we 
assume no other force acts, is 

m^~ + kx = 0. (2-84) 



-v.Q.QQCLCLP- 



m 



aa 






Fig. 2-3. Model of a simple har- 
monic oscillator. 



Equation (2-84) describes the free harmonic oscillator. Its solution was 
obtained in Section 2-5. The motion is a simple sinusoidal oscillation 
about the point of equilibrium. In all physical cases there will be some 
frictional force acting, though it may often be very small. As a good 
approximation in most cases, particularly when the friction is small, we 
can assume that the frictional force is proportional to the velocity. Since 
this is the only kind of frictional force for which the problem can easily 
be solved, we shall restrict our attention to this case. If we use Eq. (2-31) 
for the frictional force with n = 1, the equation of motion then becomes 



d 2 x , dx . , n 

m dP +b dt +kx = - 



(2-85) 



This equation describes the damped harmonic oscillator. Its motion, at 
least for small damping, consists of a sinusoidal oscillation of gradually 



40 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

decreasing amplitude, as we shall show later. If the oscillator is subject 
to an additional impressed force F(t), its motion will be given by 

m W + b Tt + kx = F(tl (2_86) 

If F(t) is a sinusoidally varying force, Eq. (2-86) leads to the phenomenon 
of resonance, where the amplitude of oscillation becomes very large when 
the frequency of the impressed force equals the natural frequency of the 
free oscillator. 

The importance of the harmonic oscillator problem lies in the fact that 
equations of the same form as Eqs. (2-84)-(2-86) turn up in a wide variety 
of physical problems. In almost every case of one-dimensional motion 
where the potential energy function V(x) has one or more minima, the 
motion of the particle for small oscillations about the minimum point fol- 
lows Eq. (2-84). To show this, let V(x) have a minimum at x = x , and 
expand the function V(x) in a Taylor series about this point: 

V(x) = V(x ) + (^\ { x - x ) + i(g\ o (* - *o) 2 

+ *(€?) (*-*o) 3 +---. (2-87) 



'\dx S J: 



*0 



The constant V(x ) can be dropped without affecting the physical results. 
Since x is a minimum point, 

Making the abbreviations 

* = (S). - <2 " 89) 

x' = x — x 0) (2-90) 

we can write the potential function in the form 

V(x') = %kx' 2 H . (2-91) 

For sufficiently small values of x', provided k ^ 0, we can neglect the terms 
represented by dots, and Eq. (2-91) becomes identical with Eq. (2-83). 
Hence, for small oscillations about any potential minimum, except in the 
exceptional case k = 0, the motion is that of a harmonic oscillator. 

When a solid is deformed, it resists the deformation with a force propor- 
tional to the amount of deformation, provided the deformation is not too 



2-8] LINEAR DIFFERENTIAL EQUATIONS 41 

great. This statement is called Hooke's law. It follows from the fact that 
the undeformed solid is at a potential-energy minimum and that the 
potential energy may be expanded in a Taylor series in the coordinate 
describing the deformation. If a solid is deformed beyond a certain point, 
called its elastic limit, it will remain permanently deformed; that is, its 
structure is altered so that its undeformed shape for minimum potential 
energy is changed. It turns out in most cases that the higher-order terms 
in the series (2-91) are negligible almost up to the elastic limit, so that 
Hooke's law holds almost up to the elastic limit. When the elastic limit 
is exceeded and plastic flow takes place, the forces depend in a complicated 
way not only on the shape of the material, but also on the velocity of de- 
formation and even on its previous history, so that the forces can no 
longer be specified in terms of a potential-energy function. 

Thus practically any problem involving mechanical vibrations reduces 
to that of the harmonic oscillator at small amplitudes of vibration, that is, 
so long as the elastic limits of the materials involved are not exceeded. 
The motions of stretched strings and membranes, and of sound vibrations 
in an enclosed gas or in a solid, result in a number of so-called normal 
modes of vibration, each mode behaving in many ways like an independent 
harmonic oscillator. An electric circuit containing inductance L, resist- 
ance R, and capacitance C in series, and subject to an applied electro- 
motive force E{t), satisfies the equation 

where q is the charge on the condenser and dq/dt is the current. This 
equation is identical in form with Eq. (2-86). Early work on electrical 
circuits was often carried out by analogy with the corresponding mechani- 
cal problem. Today the situation is often reversed, and the mechanical 
and acoustical engineers are able to make use of the simple and effective 
methods developed by electrical engineers for handling vibration prob- 
lems. The theory of electrical oscillations in a transmission line or in a 
cavity is similar mathematically to the problem of the vibrating string or 
resonating air cavity. The quantum-mechanical theory of an atom can 
be put in a form which is identical mathematically with the theory of a 
system of harmonic oscillators. 

2-8 Linear differential equations with constant coefficients. Equa- 
tions (2-84)-(2-86) are examples of second-order linear differential equa- 
tions. The order of a differential equation is the order of the highest 
derivative that occurs in it. Most equations of mechanics are of second 
order. (Why?) A linear differential equation is one in which there are 



42 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

no terms of higher than first degree in the dependent variable (in this 
case x) and its derivatives. Thus the most general type of linear differ- 
ential equation of order n would be 

a n (t) ^f + a„_i(«) gj^ + • • • + ai(t) g + a {t)x = b(t). (2-93) 

If b(t) = 0, the equation is said to be homogeneous; otherwise it is inhomo- 
geneous. Linear equations are important because there are simple general 
methods for solving them, particularly when the coefficients a , a\, . . . , a n 
are constants, as in Eqs. (2-84)-(2-86). In the present section, we shall 
solve the problem of the free harmonic oscillator [Eq. (2-84)], and at the 
same time develop a general method of solving any linear homogeneous 
differential equation with constant coefficients. This method is applied 
in Section 2-9 to the damped harmonic oscillator equation (2-85). In 
Section 2-10 we shall study the behavior of a harmonic oscillator under a 
sinusoidally oscillating impressed force. In Section 2-11 a theorem is 
developed which forms the basis for attacking Eq. (2-86) with any im- 
pressed force F(t), and the methods of attack are discussed briefly. 

The solution of Eq. (2-84), which we obtained in Section 2-5, we now 
write in the form 

x = A sin (o) t + 6), w = Vk/m . (2-94) 

This solution depends on two "arbitrary" constants A and 6. They are 
called arbitrary because no matter what values are given to them, the 
solution (2-94) will satisfy Eq. (2-84). They are not arbitrary in a phys- 
ical problem, but depend on the initial conditions. It can be shown that 
the general solution of any second-order differential equation depends on 
two arbitrary constants. By this we mean that we can write the solution 
in the form 

x = x(t; C u C 2 ), (2-95) 

such that for every value of C\ and C2, or every value within a certain 
range, x(t; C\, C%) satisfies the equation and, furthermore, practically 
every solution of the equation is included in the function x{t; C\, C2) for 
some value of Ci and C 2 .* If we can find a solution containing two arbi- 
trary constants which satisfies a second-order differential equation, then 
we can be sure that practically every solution will be included in it. The 
methods of solution of the differential equations studied in previous sec- 
tions have all been such as to lead directly to a solution corresponding to 



* The only exceptions are certain "singular" solutions which may occur in 
regions where the mathematical conditions for a unique solution (Section 2-2) 
are not satisfied. 



2-8] LINEAR DIFFERENTIAL EQUATIONS 43 

the initial conditions of the physical problem. In the present and subse- 
quent sections of this chapter, we shall consider methods which lead to 
a general solution containing two arbitrary constants. These constants 
must then be given the proper values to fit the initial conditions of the 
physical problem; the fact that a solution with two arbitrary constants 
is the general solution guarantees that we can always satisfy the initial 
conditions by proper choice of the constants. 

We now state two theorems regarding linear homogeneous differential 
equations: 

Theorem I. If x = x x (t) is any solution of a linear homogeneous differ- 
ential equation, and C is any constant, then x = Cx^t) is also a solution. 

Theorem II. // x = Xi(t) and x = x 2 (t) are solutions of a linear homo- 
geneous differential equation, then x = Xi(t) + x 2 (t) is also a solution. 

We prove these theorems only for the case of a second-order equation, 
since mechanical equations are generally of this type: 

a 2 (t) § + a 1 (t) g + a (t)x = 0. (2-96) 

Assume that x = x x (i) satisfies Eq. (2-96). Then 

a 2 (t) ^fel + ai (t) ^ + aoWCxt) = 



c [ a2(t) S 1 + fllW lit + a °® x i\ = °- 



Hence x = Cx^t) also satisfies Eq. (2-96). If x x {t) and x 2 (t) both satisfy 
Eq. (2-96), then 

,.v d 2 (Xt + X 2 ) d(X! + x 2 ) 

zW <P r ai{t) - h a {t)(xi + x 2 ) 

+ [ aa(i) %- + fll(<) W + a °® x >] = °- 

Hence x = x^t) + x 2 (t) also satisfies Eq. (2-96). The problem of finding 
the general solution of Eq. (2-96) thus reduces to that of finding any 
two independent "particular" solutions x^t) and x 2 {t), for then Theorems I 
and II guarantee that 

x = Ctx^t) + C 2 x 2 (t) (2-97) 



44 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

is also a solution. Since this solution contains two arbitrary constants, 
it must be the general solution. The requirement that xi(t) and x 2 (t) 
be independent means in this case that one is not a multiple of the other. 
If xi(<) were a constant multiple of x 2 (t), then Eq. (2-97) would really 
contain only one arbitrary constant. The right member of Eq. (2-97) 
is called a linear combination of xi and x 2 . 

In the case of equations like (2-84) and (2-85), where the coefficients 
are constant, a solution of the form x = e pt always exists. To show this, 
assume that a , a lt and a 2 are all constant in Eq. (2-96) and substitute 

x = e > H = pe ' W = v e ■ (2_98) 

We then have 

(a 2 p 2 + a lP + a )e pt = 0. (2-99) 

Canceling out e pt , we have an algebraic equation of second degree in p. 
Such an equation has, in general, two roots. If they are different, this 
gives two independent functions e pt satisfying Eq. (2-96) and our prob- 
lem is solved. If the two roots for p should be equal, we have found only 
one solution, but then, as we shall show in the next section, the function 

x = te pt (2-100) 

also satisfies the differential equation. The linear homogeneous equation 
of nth order with constant coefficients can also be solved by this method. 
Let us apply the method to Eq. (2-84). Making the substitution (2-98), 
we have 

mp 2 + k = 0, (2-101) 

whose solution is 

p = ± yP± = ±ia, , « = ^ • (2-102) 

This gives, as the general solution, 

x = de^ot _|_ Cze-^ot. (2-103) 

In order to interpret this result, we remember that 

e*» = cos 6 + * sin 0. (2-104) 

If we allow complex numbers x as solutions of the differential equation, 
then the arbitrary constants Ci and C 2 must also be complex in order for 
Eq. (2-103) to be the general solution. The solution of the physical 
problem must be real, hence we must choose Ci and C 2 so that x turns out 



2-8] LINEAR DIFFERENTIAL EQUATIONS 45 

to be real. The sum of two complex numbers is real if one is the complex 
conjugate of the other. If 

C = a + ib, (2-105) 

and 

C* = a- ib, (2-106) 

then 

C + C* = 2a, C -C* = 2ib. (2-107) 

Now e ia <>t i s the complex conjugate of e -iw °', so that if we set d = C, 
C 2 = C*, then x will be real: 

x = Ce iaot + C*e~ iu °K (2-108) 

We could evaluate x by using Eqs. (2-104), (2-105), and (2-106), but 
the algebra is simpler if we make use of the polar representation of a 
complex number: 

C = a + ib = re { °, (2-109) 

C* = a — ib = re~ ie , (2-110) 

where 

r= (a 2 + b 2 ) m , tan0 = ^, (2-111) 

a = r cos 0, b = r sin 8. (2-112) 

The reader should verify that these equations follow algebraically from 
Eq. (2-104). If we represent C as a point in the complex plane, then a 
and b are its rectangular coordinates, and r and are its polar coordinates. 
Using the polar representation of C, Eq. (2-108) becomes (we set r = %A) 

= A cos (u t + 0). (2-113) 

This is the general real solution of Eq. (2-84). It differs from the solu- 
tion (2-94) only by a shift of ir/2 in the phase constant 6. 

Setting B x = A cos 0, B 2 = —A sin 0, we can write our solution in 
another form: 

x = Bi cos u t + B 2 sin w <. (2-114) 

The constants A, 0, or B u B 2 , are to be obtained in terms of the initial 
values x Q , v at t = by setting 

x = A cos = B u (2-115) 

v = —o) A sin = u B 2 . (2-116) 



A = U + & 


r 


\ <»l) 


r 


tan 0= "° ■ , 
a; wo 




-Bi = x , 




B 2 = ^- 
w 





46 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

The solutions are easily obtained: 

/ „.2\l/2 

(2-117) 

(2-118) 

or 

(2-119) 

(2-120) 

Another way of handling Eq. (2-103) would be to notice that, since 
Eq. (2-84) contains only real coefficients, a complex function can satisfy 
it only if both real and imaginary parts satisfy it separately. (The proof 
of this statement is a matter of substituting x = u + iw and carrying 
out a little algebra.) Hence if a solution is (we set r = A) 

x = Ce iWot = Ae i{aat+6) 

= A cos (u < + 0) + iA sin (u t + 9), (2-121) 

then both the real and imaginary parts of this solution must separately be 
solutions, and we have either solution (2-113) or (2-94). We can carry 
through the solutions of linear equations like this, and perform any alge- 
braic operations we please on them in their complex form (so long as we 
do not multiply two complex numbers together), with the understanding 
that at each step what we are really concerned with is only the real part 
or only the imaginary part. This procedure is often useful in the treat- 
ment of problems involving harmonic oscillations, and we shall use it in 
Section 2-10. 

It is often very convenient to represent a sinusoidal function as a com- 
plex exponential : 

cos = real part of e 10 = ~ ' (2-122) 

it —it 

sin 6 = imaginary part of e te = ^ (2-123) 

Exponential functions are easier to handle algebraically than sines and 
cosines. The reader will find the relations (2-122), (2-123), and (2-104) 
useful in deriving trigonometric formulas. The power series for the sine 
and cosine functions are readily obtained by expanding e l ° in a power series 
and separating the real and imaginary parts. The trigonometric rule for 
sin (A + B) and cos (A + B) can be easily obtained from the algebraic 
rule for adding exponents. Many other examples could be cited. 



2-9] THE DAMPED HARMONIC OSCILLATOR 47 

2-9 The damped harmonic oscillator. The equation of motion for a 
particle subject to a linear restoring force and a frictional force proportional 
to its velocity is [Eq. (2-85)] 

mx + bx + kx = 0, (2-124) 

where the dots stand for time derivatives. Applying the method of Sec- 
tion 2-8, we make the substitution (2-98) and obtain 

mp 2 + bp + k = 0. (2-125) 

The solution is 

1/2 

(2-126) 



= _ _ + ±Y - a 

P 1m l\2mj m. 



[(, 



We distinguish three cases: (a) k/m > (6/2m) 2 , (b) k/m < (b/2m) 2 , and 
(c) k/m = (6/2m) 2 . 
In case (a), we make the substitutions 

"o ,= yj~ - (2-127) 

T = 2^' (2- 12 8) 

"i = («o - y 2 V 12 , (2-129) 

where 7 is called the damping coefficient and (w /27r) is the natural fre- 
quency of the undamped oscillator. There are now two solutions for p: 

p = — J ± iui. (2-130) 

The general solution of the differential equation is therefore 

x = C 1 e- /t+iait + C 2 e- yt - iWlt . (2-131) 

Setting 

d = Ue ie , C 2 = Ue~ U , (2-132) 

we have 

x = Ae~ yt cos (»i« + 0). (2-133) 

This corresponds to an oscillation of frequency (w!/27r) with an amplitude 
Ae- yt which decreases exponentially with time (Fig. 2-4). The constants 
A and depend upon the initial conditions. The frequency of oscillation 
is less than without damping. The solution (2-133) can also be written 

x = e~ yt (Bi cosgM + B 2 sinwiO- (2-134) 



48 



MOTION OF A PARTICLE IN ONE DIMENSION 



[CHAP. 2 




Fig. 2-4. Motion of damped harmonic oscillator. Heavy curve: x = 
Ae~ yt cos ut, 7 = w/8. Light curve: a; = ±Ae~ yt . 

In terms of the constants co and 7, Eq. (2-124) can be written 

x + 2lx + v>%x = 0. (2-135) 

This form of the equation is often used in discussing mechanical oscilla- 
tions. 

The total energy of the oscillator is 



E = hnx 2 + Jfcc 2 . 



(2-136) 



In the important case of small damping, 7 <C «o, we can set o>i = o>o and 
neglect 7 compared with w , and we have for the energy corresponding to 
the solution (2-133), approximately, 



E = ikA 2 e~ 2yt = E e- 2yt . 



(2-137) 



Thus the energy falls off exponentially at twice the rate at which the ampli- 
tude decays. The fractional rate of decline or logarithmic derivative of E is 



I §E 

E dt 



dlnE 
dt 



= —27. 



(2-138) 



We now consider case (b), (co < 7). In this case, the two solutions 
for p are 



2-9] THE DAMPED HARMONIC OSCILLATOR 49 

p=-y 1 = -y- (7 2 - cog) 1 ' 2 , 

v = -i% = -v + (t 2 - «S) 1/a . 
The general solution is 



(2-139) 



x = C ie ~ yit + C 2 e- y2t . (2-140) 

These two terms both decline exponentially with time, one at a faster rate 
than the other. The constants C\ and C 2 may be chosen to fit the initial 
conditions. The reader should determine them for two important cases: 
x ^ 0, v = and x = 0, v ?± 0, and draw curves x{f) for the two 
cases. 
In case (c), (co = T), we have only one solution for p: 

P = —y. (2-141) 

The corresponding solution for x is 

* = e~ yt . (2-142) 

We now show that, in this case, another solution is 

x = te~ yt . (2-143) 

To prove this, we compute 



x = e~ yt - yte~ y \ 



(2-144) 



* = -2Te- 7J + y 2 te~ yt . 

The left side of Eq. (2-135) is, for this x, 

x + 27^ + o%x = (cwg — y 2 )te~ yt . (2-145) 

This is zero if w = 7. Hence the general solution in case co = 7 is 

x = (C x + C 2 t)e~ n . (2-146) 

This function declines exponentially with time at a rate intermediate be- 
tween that of the two exponential terms in Eq. (2-140): 

?i > y > 7 2 . (2-147) 

Hence the solution (2-146) falls to zero faster after a sufficiently long time 
than the solution (2-140), except in the case C 2 = in Eq. (2-140). Cases 
(a), (b), and (c) are important in problems involving mechanisms which 
approach an equilibrium position under the action of a frictional damping 



50 



MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 




Fig. 2-5. Return of harmonic oscillator to equilibrium, (a) Underdamped. 
(b) Overdamped. (o) Critically damped. 

force, e.g., pointer reading meters, hydraulic and pneumatic spring returns 
for doors, etc. In most cases, it is desired that the mechanism move 
quickly and smoothly to its equilibrium position. For a given damping 
coefficient 7, or for a given u , this is accomplished in the shortest time 
without overshoot if co = 7 [case (c)]. This case is called critical damping. 
If «o < 7, the system is said to be overdamped; it behaves sluggishly and 
does not return as quickly to x = as for critical damping. If w > 7, 
the system is said to be underdamped; the coordinate x then overshoots the 
value x = and oscillates. Note that at critical damping, ui = 0, so 
that the period of oscillation becomes infinite. The behavior is shown in 
Fig. 2-5 for the case of a system displaced from equilibrium and released 
(#0 n* 0, v = 0). The reader should draw similar curves for the case 
where the system is given a sharp blow at t = (i.e., x = 0, v j£ 0). 



2-10 The forced harmonic oscillator. The harmonic oscillator subject 
to an external applied force is governed by Eq. (2-86). In order to sim- 
plify the problem of solving this equation, we state the following theorem : 

Theorem III. // x»(<) is a solution of an inhomogeneous linear equation 
[e.g., Eq. (2-86)], and Xh(t) is a solution of the corresponding homogeneous 
equation [e.g., Eq. (2-85)], then x(t) = Xi(t) + Xh{t) is also a solution of 
the inhomogeneous equation. 

This theorem applies whether the coefficients in the equation are constants 
or functions of t. The proof is a matter of straightforward substitution, 
and is left to the reader. In consequence of Theorem III, if we know the 
general solution Xh of the homogeneous equation (2-85) (we found this in 
Section 2-9), then we need find only one particular solution Xi of the in- 
homogeneous equation (2-86). For we can add Xi to Xh and obtain a solu- 
tion of Eq. (2-86) which contains two arbitrary constants and is therefore 
the general solution. 



2-10] THE FORCED HARMONIC OSCILLATOR 51 

The most important case is that of a sinusoidally oscillating applied 
force. If the applied force oscillates with angular frequency w and ampli- 
tude F , the equation of motion is 

d x dx 

m -£p + b -£ + kx = F cos (at + O ), (2-148) 

where 6 is a constant specifying the phase of the applied force. There are, 
of course, many solutions of Eq. (2-148), of which we need find only one. 
From physical considerations, we expect that one solution will be a steady 
oscillation of the coordinate x at the same frequency as the applied force: 

x = A a cos (at + 0.). (2-149) 

The amplitude A, and phase S of the oscillations in x will have to be de- 
termined by substituting Eq. (2-149) in Eq. (2-148). This procedure is 
straightforward and leads to the correct answer. The algebra is simpler, 
however, if we write the force as the real part of a complex function:* 

F(t) = Re(F e iat ), (2-150) 

F = Foe"". (2-151) 

Thus if we can find a solution x(t) of 

m ll& + b 7fi + kx= F ° e """' (2-152) 

then, by splitting the equation into real and imaginary parts, we can show 
that the real part of x(t) will satisfy Eq. (2-148). We assume a solution of 
the form 

ico t 

x = x e , 
so that 

x = twxoe*", x = -w 2 x e iw '. (2-153) 

Substituting in Eq. (2-152), we solve for x : 

Xo = _^ (2 _ 154) 

wo — w + 2iTco 

The solution of Eq. (2-152) is therefore 



x = x e-= 2 ^f^ • (2-155) 

Wo — w 2 + 2tTw 



* Note the use of roman type (F, x) to distinguish complex quantities from 
the corresponding real quantities {F, x). 



52 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

We are often more interested in the velocity 



t'coFo 



iut 



±=^ 1 1— T— • (2-156) 

TO w — w + 2i7u 

The simplest way to write Eq. (2-156) is to express all complex factors in 
polar form [Eq. (2-109)] : 

i = e" /2 , (2-157) 

co 2 , - co 2 + 2i7co = [(col - co 2 ) 2 + 4Y 2 co 2 ] 1/2 exp ' ' - 1 27w 



I i tan x 



2 2 

COq — w 



If we use these expressions, Eq. (2-156) becomes 



(2-158) 



where 



" m[(co 2 - co 2 ) 2 + 47V] 1 ' 2 ' ( 59) 

= I - tan" 1 -^- = tan" 1 »L=L«? , (2 -l 6 0) 
2 «o — w 2 27co 

^=[^yfky^ (2 " 161) 



By Eq. (2-159), 

x = Re(x) 

F co 



m [(cog - co 2 ) 2 + 47V] 
and 

a; = Re(x) = Re (k/ico) 

F 1 



— cos (cot + <?o + (3), (2-163) 



m [(a>i — a*)' + 47 2 co 2 ] 1/2 

This is a particular solution of Eq. (2-148) containing no arbitrary con- 
stants. By Theorem III and Eq. (2-133), the general solution (for the 
underdamped oscillator) is 

' = Ae ~" oos <-' + «> + m _ Jf + 47vi-» *" *-+*>+»• 

(2-165) 



2-10] THE FORCED HARMONIC OSCILLATOR 53 

This solution contains two arbitrary constants A, 6, whose values are de- 
termined by the initial values x , v at t — 0. The first term dies out ex- 
ponentially in time and is called the transient. The second term is called 
the steady state, and oscillates with constant amplitude. The transient de- 
pends on the initial conditions. The steady state which remains after the 
transient dies away is independent of the initial conditions. 

In the steady state, the rate at which work is done on the oscillator by 
the applied force is 

F 2 

xF(t) = -2 — " cos (cot + e ) cos (cot + B + 0) 

m [(co 2 — a>o) + 4tV] 1/2 

_ F% co cos /3 cos 2 (cot + fl ) F% w sin sin 2(a>t + O ) 



m [(a, 2 - w 2 ) 2 + 4Y 2 w 2 ] 1/2 2m [(co 2 - co 2 ) 2 + 4Y 2 co 2 ] 1/2 

(2-166) 

The last term on the right is zero on the average, while the average value 
of cos 2 (tot + O ) over a complete cycle is \. Hence the average power de- 
livered by the applied force is 

Pav = <**«»„ = ^^ " — - , (2-167) 

2m [(to 2 - col) 2 + 47V] 1/2 

or 

-Pav = %FoX m cos /3, (2-168) 

where x m is the maximum value of x. A similar relation holds for power 
delivered to an electrical circuit. The factor cos ft is called the power 
factor. In the electrical case, is the phase angle between the current and 
the applied emf. Using formula (2-162) for cos/3, we can rewrite Eq. 
(2-167) : 

F 2 yco 2 

m (co 2 — cog) 2 + 4T 2 co 2 

It is easy to show that in the steady state power is supplied to the oscillator 
at the same average rate that power is being dissipated by friction, as of 
course it must be. The power P av has a maximum for co ±= co . In Fig. 
2-6, the power P av (in arbitrary units) and the phase of ff of steady-state 
forced oscillations are plotted against co for two values of 7. The heavy 
curves are for small damping; the light curves are for greater damping. 
Formula (2-169) can be simplified somewhat in case y <JC co . In this case, 
P av is large only near the resonant frequency co , and we shall deduce a 
formula valid near co = co . Defining 

Aw = co — w , (2-170) 



54 



MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 



■P»v, (7 = J«o) 




-x/2 



Fig. 2-6. Power and phase of forced harmonic oscillations. 

and assuming Aw « w , we have 

(w 2 — wo) = (w + w ) Aw = 2w Aw, 

Hence 



(2-171) 
(2-172) 

(2-173) 



This simple formula gives a good approximation to P av near resonance. 
The corresponding formula for /3 is 

—Aw 



w 2 


= 


2 

w . 






-> 


= 


Fl 
4m 




7 


av 


(Aw) 


2 _j_ 7 2 



cos |3 = 



sin j3 = 



[(Aw) 2 + T2]i/2 ' "*" >- " [(Aw) 2 + Y 2 ] 1 / 2 
When w « w , = ir/2, and Eq. (2-164) becomes 

x +■■%>- cob (at +e ) = nr" 
worn K 



(2-174) 



(2-175) 



This result is easily interpreted physically; when the force varies slowly, 
the particle moves in such a way that the applied force is just balanced by 
the restoring force. When w » w , /3 = —tt/2, and Eq. (2-164) becomes 



x= - -4 s - cos (jut + 6 ) = ~^-- 
w z m w 2 m 



(2-176) 



2-10] THE FORCED HARMONIC OSCILLATOR 55 

The motion now depends only on the mass of the particle and on the fre- 
quency of the applied force, and is independent of the friction and the 
restoring force. This result is, in fact, identical with that obtained in Sec- 
tion 2-3 [see Eqs. (2-15) and (2-19)] for a free particle subject to an 
oscillating force. 

We can apply the result (2-165) to the case of an electron bound to an 
equilibrium position x = by an elastic restoring force, and subject to an 
oscillating electric field : 

E x = E cos tot, (2-177) 

F = — eE cos ut. (2-178) 



The motion will be given by 

x = Ac'" cos ( Wl < + 6) - <%> sin <■<* + A 

m [(co 2 - cog) 2 + 47 2 co 2 ] 1/2 



(2-179) 



The term of interest here is the second one, which is independent of the 
initial conditions and oscillates with the frequency of the electric field. 
Expanding the second term, we get 

„ _ eE sin ff cos wt eE cos /3 sin cot 

;£ — — - - - - — 



tn [(co 2 - cog) 2 + 47 2 co 2 ] 1/2 m [(co 2 - co 2 ) 2 + 4T 2 co 2 ] 1/2 



2 



-eE cos ut coq 



co 



tn [(co 2 - co 2 ) 2 + 47 V] 

eE sin ut 2Tco 

tn [(co 2 - co 2 ) 2 + 4T 2 co 2 ] 



(2-180) 



The first term represents an oscillation of x in phase with the applied force 
at low frequencies, 180° out of phase at high frequencies. The second term 
represents an oscillation of x that is 90° out of phase with the applied force, 
the velocity x for this term being in phase with the applied force. Hence 
the second term corresponds to an absorption of energy from the applied 
force. The second term contains a factor 7 and is therefore small, if 
7 <C co , except near resonance. If we imagine a dielectric medium con- 
sisting of electrons bound by elastic forces to positions of equilibrium, then 
the first term in Eq. (2-180) will represent an electric polarization propor- 
tional to the applied oscillating electric field, while the second term will 
represent an absorption of energy from the electric field. Near the resonant 
frequency, the dielectric medium will absorb energy, and will be opaque 
to electromagnetic radiation. Above the resonant frequency, the dis- 
placement of the electrons is out of phase with the applied force, and the 



56 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

resulting electric polarization will be out of phase with the applied electric 
field. The dielectric constant and index of refraction will be less than 
one. For very high frequencies, the first term of Eq. (2-180) approaches 
the last term of Eq. (2-18), and the electrons behave as if they were free. 
Below the resonant frequency, the electric polarization will be in phase 
with the applied electric field, and the dielectric constant and index of re- 
fraction will be greater than one. 



Computing the dielectric constant from the first term in Eq. (2-180), in the 
same manner as for a free electron [see Eqs. (2-20)-(2-26)], we find, for N elec- 
trons per unit volume: 

A-wNe cop — to 

€=1 + 71 2,2 ,^22 " < 2 - 181 > 

m (o>o — «■ ) + 47 to 

The index of refraction for electromagnetic waves (n = 1) is 

c , ,1/2 1/2 ,~ 100 , 

n = - = (tie) = e . (2-182) 

v 

For very high or very low frequencies, Eq. (2-181) becomes 

6=1 + ^f- , co « oo, (2-183) 



rruoQ 

2 
TOO) 2 



4irNe 
e = 1 - ^-|- , fa) » coo. (2-184) 



The mean rate of energy absorption per unit volume is given by Eq. (2-169): 

- = ^ 2 l\ 2 2 - (S-185) 

dt m (a) — coo) + 47 fa> 



The resulting dielectric constant and energy absorption versus frequency 
are plotted in Fig. 2-7. Thus the dielectric constant is constant and 
greater than one at low frequencies, increases as we approach the resonant 
frequency, falls to less than one in the region of "anomalous dispersion" 
where there is strong absorption of electromagnetic radiation, and then 
rises, approaching one at high frequencies. The index of refraction will 
follow a similar curve. This is precisely the sort of behavior which is 
exhibited by matter in all forms. Glass, for example, has a constant dielec- 
tric constant at low frequencies; in the region of visible light its index of 
refraction increases with frequency; and it becomes opaque in a certain 



2-10] THE FORCED HARMONIC OSCILLATOR 57 




Fig. 2-7. Dielectric constant and energy absorption for medium containing 
harmonic oscillators. 



band in the ultraviolet. X-rays are transmitted with an index of refrac- 
tion very slightly less than one. A more realistic model of a transmitting 
medium would result from assuming several different resonant frequencies 
corresponding to electrons bound with various values of the spring con- 
stant k. This picture is then capable of explaining most of the features in 
the experimental curves for e or n vs. frequency. Not only is there qualita- 
tive agreement, but the formulas (2-181 )-(2-185) agree quantitatively 
with experimental results, provided the constants JV, to , and 7 are properly 
chosen for each material. The success of this theory was one of the reasons 
for the adoption, until the year 1913, of the "jelly model" of the atom, in 
which electrons were imagined embedded in a positively charged jelly in 
which they oscillated as harmonic oscillators. The experiments of Ruther- 
ford in 1913 forced physicists to adopt the "planetary" model of the atom, 
but this model was unable to explain even qualitatively the optical and 
electromagnetic properties of matter until the advent of quantum mechan- 
ics. The result of the quantum-mechanical treatment is that, for the inter- 
action of matter and radiation, the simple oscillator picture gives essentially 
correct results when the constants are properly chosen.* 

We now consider an applied force F(t) which is large only during a 
short time interval U and is zero or negligible at all other times. Such a 
force is called an impulse, and corresponds to a sudden blow. We assume 
the oscillator initially at rest at x = 0, and we assume the time St so short 
that the mass moves only a negligibly small distance while the force is 
acting. According to Eq. (2-4), the momentum just after the force is 
applied will equal the impulse delivered by the force : 



mvo 



= Po = JF dt, (2-186) 

where v is the velocity just after the impulse, and the integral is taken 
over the time interval U during which the force acts. After the impulse, 

* See John C. Slater, Quantum Theory of Matter. New York: McGraw-Hill 
Book Co., 1951. (Page 378.) 



58 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

the applied force is zero, and the oscillator must move according to Eq. 
(2-133) if the damping is less than critical. We are assuming St so small 
that the oscillator does not move appreciably during this time, hence we 
choose = — (x/2) — ait , in order that x = at t = t , where t is 
the instant at which the impulse occurs : 

x = Ae~ yt sin [ai(t - t )]. (2-187) 

The velocity at t = t is 

v = aiAe"" . (2-188) 

Thus 

A = ^ e yt °. (2-189) 

The solution when an impulse po is delivered at t ?= t to an oscillator at 
rest is therefore 

(0, * < t 0) 
* = { _BL. e - T «-<o> sin [wi(< _ <o)] t > f (2-190) 

Here we have neglected the short time 8t during which the force acts. 

We see that the result of an impulse-type force depends only on the total 
impulse po delivered, and is independent of the particular form of the func- 
tion F(t), provided only that F(t) is negligible except during a very short 
time interval St. Several possible forms of F(t) which have this property 
are listed below: 

(0, t < t , 

Po/St, t < t < t + St, (2-191) 

0, t> t + St, 

m = sW^ exp i~ (1 J^\' ~™ <t<cc (2 " 193) 

The reader may verify that each of these functions is negligible except 
within an interval of the order of St around t , and that the total impulse 
delivered by each is p . The exact solution of Eq. (2-86) with F(f) given 
by any of the above expressions must reduce to Eq. (2-190) when St — * 
(see Problem 23). 



2-11] THE PRINCIPLE OF SUPERPOSITION 59 

2-11 The principle of superposition. Harmonic oscillator with arbi- 
trary applied force. An important property of the harmonic oscillator is 
that its motion x(t), when subject to an applied force F(t) which can be 
regarded as the sum of two or more other forces Fi(t), F 2 {t), . . . , is the 
sum of the motions x^t), x 2 (t), . . . , which it would have if each of the 
forces F n {t) were acting separately. This principle applies to small mechan- 
ical vibrations, electrical vibrations, sound waves, electromagnetic waves, 
and all physical phenomena governed by linear differential equations. The 
principle is expressed in the following theorem: 

Theorem IV. Let the {finite or infinite*) set of functions x n (t), n = 1,2, 3, 
. . . , be solutions of the equations 

mx n + bx n + kx n = F n (t), (2-194) 

and let 

F(t) = X) *"»(*)• (2-195) 

n 

*(0 = Z) x «® (2-196) 

n 

mx + bx + kx = F(t). (2-197) 



Then the function 
satisfies the equation 



To prove this theorem, we substitute Eq. (2-196) in the left side of Eq. 
(2-197): 

mx + bx + kx = m ^ x n + b ^ x n + k ^ x n 

n n n 

= ^ (mx n + bx n + kx n ) 

n 

= !>»(*> 

= F(t). 

This theorem enables us to find a solution of Eq. (2-197) whenever the 
force F(t) can be expressed as a sum of forces F n {t) for which the solutions 
of the corresponding equations (2-194) can be found. In particular, 
whenever F(t) can be written as a sum of sinusoidally oscillating terms : 

^(0 = £ C " cos ("»* + "»)> (2-198) 



* When the set of functions is infinite, there are certain mathematical restric- 
tions which need not concern us here. 



60 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

a particular solution of Eq. (2-197) will be, by Theorem IV and Eq. (2-164), 

* = D — 771 * , , a 2ll/2 sin {uj + 9 n + ft,), (2-199) 

V m [(u 2 - <4) + 47 2 co 2 ] 1/2 

2 2 

^ = tan ^^7"- 

The general solution is then 

a —~tt i . i a \ i v~* C« sin (co„£ + n + ft„) . . 

x = Ae cos (wi< + 6) + 2*,— TTl 2x2 , , 2 2,1/2 ' C 2-200 ) 

» m [(co 2 - co 2 ,) 2 + 47 2 co 2 ] 1/2 

where A and are, as usual, to be chosen to make the solution (2-200) fit 
the initial conditions. 

We can write Eqs. (2-198) and (2-199) in a different form by setting 

A n = C n cos 0„, B n = — C n sin 6 n . (2-201) 

Then 

F(t) = J2 (A n cos u n t + B n sin wj), (2-202) 

n 

and 

_ y-> A n sin (u n t + ff B ) — B n cos (co»< + ft>) f2-20*ri 

V m[(co 2 - cog) 2 + 47 2 co 2 ] 1 ' 2 ' ^ ^ 

An important case of this kind is that of a periodic force F(t), that is, a 
force such that 

F(t + T) = F(t), (2-204) 

where T is the period of the force. For any continuous function F(t) satis- 
fying Eq. (2-204) (and, in fact, even for only piecewise continuous func- 
tions), it can be shown that F(t) can always be written as a sum of sinus- 
oidal functions: 



where 



F(t) = ±A +J1 (i B cos ^ + B n sin ^) , (2-205) 

n = l 
rT 

An =Th 

rT 



F(t) cos ?^ dt, n = 0, 1, 2, 



F(t) sin ^ <ft, n = 1, 2, 3, 



(2-206) 



This result enables us, at least in principle, to solve the problem of the 
forced oscillator for any periodically varying force. The sum in Eq. (2-205) 



2-11] THE PRINCIPLE OF SUPERPOSITION 61 

is called a Fourier series* The actual computation of the solution by this 
method is in most cases rather laborious, particularly the fitting of the 
constants A, 6 in Eq. (2-200) to the initial conditions. However, the knowl- 
edge that such a solution exists is often useful in itself. If any of the fre- 
quencies 2irn/T coincides with the natural frequency w of the oscillator, 
then the corresponding terms in the series in Eqs. (2-199) or (2-203) will 
be relatively much larger than the rest. Thus a force which oscillates 
nonsinusoidally at half the frequency w may cause the oscillator to per- 
form a nearly sinusoidal oscillation at its natural frequency co . 

A generalization of the Fourier series theorem [Eqs. (2-205) and (2-206)] 
applicable to nonperiodic forces is the Fourier integral theorem, which 
allows us to represent any continuous (or piecewise continuous) function 
F(t), subject to certain limitations, as a superposition of harmonically 
oscillating forces. By means of Fourier series and integrals, we may solve 
Eq. (2-197) for almost any physically reasonable force F(t). We shall not 
pursue the subject further here. Suffice it to say that while the methods 
of Fourier series and Fourier integrals are of considerable practical value 
in solving vibration problems, their greatest importance in physics probably 
lies in the fact that in principle such a solution exists. Many important 
results can be deduced without ever actually evaluating the series or 
integrals at all. 

A method of solution known as Green's method is based on the solution 
(2-190) for an impulse-type force. We can think of any force F(t) as the 
sum of a series of impulses, each acting during a short time St and deliver- 
ing an impulse F(t) St: 

F(t) = £ F n (t), (2-207) 

n— — oo 

0, if t < t n , where t n = n St, 

Fn(t) = F(t n ), if t n < t < tn+u (2-208) 

0, if t > tn +1 . 

As St —> 0, the sum of all the impulse forces F n (t) will approach F(t). (See 
Fig. 2-8.) According to Theorem IV and Eq. (2-190), a solution of 
Eq. (2-197) for a force given by Eq. (2-207) is 



* For a proof of the above statements and a more complete discussion of 
Fourier series, see Dunham Jackson, Fourier Series and Orthogonal, Polynomials. 
Menasha, Wisconsin: George Banta Pub. Co., 1941. (Chapter 1.) 



62 



MOTION OF A PARTICLE IN ONE DIMENSION 



[CHAP. 2 



F(t) 




Fig. 2-8. Representation of a force as a sum of impulses. Heavy curve: 
F(f). Light curve: £„F»(fl. 



where t no < t < t no+1 . If we let St -» and write t„ = t', Eq. (2-209) 
becomes 

J — I 



x(ty 



ma)i 



e- 7( '- (,) sin [ Wl (« - t')] dt'. 



(2-210) 



The function 

G(t, t>) = \ 



0, if t' > t, 



(2-211) 



mu>\ 



sin[w!(< — t')], if t' < t, 



is called the Green's function for Eq. (2-197). In terms of Green's function, 
x (t) = f G(t, t')F(t') dt'. (2-212) 

J 00 

If the force F(t) is zero for t < t , then the solution (2-210) will give 
x(t) = f or t < t - This solution is therefore already adjusted to fit the 
initial condition that the oscillator be at rest before the application of the 
force. For any other initial condition, a transient given by Eq. (2-133), 
with appropriate values of A and 6, will have to be added. The solution 
(2-210) is useful in studying the transient behavior of a mechanical sys- 
tem or electrical circuit when subject to forces of various kinds. 



Problems 

1. A tug of war is held between two teams of five men each. Each man weighs 
160 lb and can initially pull on the rope with a force of 200 lb-wt. At first the 
teams are evenly matched, but as the men tire, the force with which each man 
pulls decreases according to the formula 



(200 lb-wt) e 



-tlT 



where the mean tiring time r is 10 sec for one team and 20 sec for the other. 
Find the motion, (g — 32 ft-sec -2 .) What is the final velocity of the two 
teams? Which of our assumptions is responsible for this unreasonable result? 



PROBLEMS 63 

2. A high-speed proton of electric charge e moves with constant speed v 
in a straight line past an electron of mass m, charge — e, initially at rest. The 
electron is at a distance a from the path of the proton, (a) Assume that the 
proton passes so quickly that the electron does not have time to move appre- 
ciably from its initial position until the proton is far away. Show that the 
component of force in a direction perpendicular to the line along which the 

proton moves is 

2 

F = 7~i~~i 2 2x3/2 ' (electrostatic or gaussian units) 
(o + vot ) 

where a is the distance of the electron from the path of the proton and t = 
when the proton passes closest to the electron, (b) Assume that the electron 
moves only along a line perpendicular to the path of the proton. Find the final 
kinetic energy of the electron, (c) Write the component of the force in a direc- 
tion parallel to the proton velocity, and calculate the net impulse in that direc- 
tion delivered to the electron. Does this justify the assumption in part (b)? 

3. A particle which had originally a velocity vo is subject to a force given 
by Eq. (2-192). (a) Find v(t) and x(t). (b) Show that as St -► 0, the motion 
approaches motion at constant velocity with an abrupt change in velocity at 
t = to of amount po/m. 

4. A particle initially at rest is subject, beginning at < = 0, to a force 

F = Foe~ 7( cos(cot+0). 

(a) Find its motion, (b) How does the final velocity depend on 6, and on w? 
[Hint: The algebra is simplified by writing cos (co«+ 0) in terms of complex expo- 
nential functions.] 

5. A boat with initial velocity vo is slowed by a frictional force 

F = -be a \ 

(a) Find its motion, (b) Find the time and the distance required to stop. 

6. A jet engine which develops a constant maximum thrust F is used to power 
a plane with a frictional drag proportional to the square of the velocity. If the 
plane starts at t = with a negligible velocity and accelerates with maximum 
thrust, find its velocity v(t). 

7. Find v(t) and x(t) for a particle of mass m which starts at xo = with 
velocity v , subject to a force given by Eq. (2-31) with n ^ 1. Find the time 
to stop, and the distance required to stop, and verify the remarks in the last 
paragraph of Section 2-4. 

8. (a) A body of mass m slides on a rough horizontal surface. The coefficient 
of static friction is y. a , and the coefficient of sliding friction is y.. Devise an 
analytic function F(v) to represent the frictional force which has the proper 
constant value at appreciable velocities and reduces to the static value at very 
low velocities, (b) Find the motion under the force you have devised if the 
body starts with an initial velocity vo. 

9. A particle of mass m is repelled from the origin by a force inversely pro- 
portional to the cube of its distance from the origin. Set up and solve the equa- 
tion of motion if the particle is initially at rest at a distance xq from the origin. 



64 



MOTION OF A PARTICLE IN ONE DIMENSION 



[CHAP. 2 



10. (a) A mass m is connected to the origin with a spring of constant k, 
whose length when relaxed is I. The restoring force is very nearly proportional 
to the amount the spring has been stretched or compressed so long as it is not 
stretched or compressed very far. However, when the spring is compressed too 
far, the force increases very rapidly, so that it is impossible to compress the 
spring to less than half its relaxed length. When the spring is stretched more 
than about twice its relaxed length, it begins to weaken, and the restoring 
force becomes zero when it is stretched to very great lengths, (a) Devise a force 
function F(x) which represents this behavior. (Of course a real spring is de- 
formed if stretched too far, so that F becomes a function of its previous history, 
but you are to assume here that F depends only on x.) (b) Find V(x) and 
describe the types of motion which may occur. 

11. A particle is subject to a force 

x 3 

(a) Find the potential V(x), describe the nature of the solutions, and find the 
solution x(t). (b) Can you give a simple interpretation of the motion when 
E 2 » hal 



v 


x) 




' \ 




i 


-Xi \ 




Xl 


-V<\ 







Figure 2-9 



12. An alpha particle in a nucleus is held by a potential having the shape 
shown in Fig. 2-9. (a) Describe the kinds of motion that are possible, (b) De- 
vise a function V(x) having this general form and having the values — Vo 
and Vi at x = and x = ±x\, and find the corresponding force. 

13. Derive the solutions (2-69) and (2-70) for a falling body subject to a 
frictional force proportional to the square of the velocity. 

14. A body of mass m falls from rest through a medium which exerts a fric- 
tional drag fee" 1 »'. (a) Find its velocity v(t). (b) What is the terminal velocity? 
(c) Expand your solution in a power series in t, keeping terms up to t 2 . (d) Why 
does the solution fail to agree with Eq. (1-28) even for short times <? 

15. A projectile is fired vertically upward with an initial velocity vo. Find its 
motion, assuming a frictional drag proportional to the square of the velocity. 
(Constant g.) 

16. Derive equations analogous to Eqs. (2-80) and (2-81) for the mo- 
tion of a body whose velocity is greater than the escape velocity. [Hint: Set 
sinh/3 = (Ex/mMG) 1 ' 2 .] 

17. Find the motion of a body projected upward from the earth with a velocity 
equal to the escape velocity. Neglect air resistance. 



PBOBLEMS 65 

18. Find the general solution for the motion of a body subject to a linear re- 
pelling force F = kx. Show that this is the type of motion to be expected in 
the neighborhood of a point of unstable equilibrium. 

19. The potential energy for the force between two atoms in a diatomic mole- 
cule has the approximate form: 

t// \ a . b 

where x is the distance between the atoms and a, b are positive constants, (a) 
Find the force, (b) Assuming one of the atoms is very heavy and remains at 
rest while the other moves along a straight line, describe the possible motions, 
(c) Find the equilibrium distance and the period of small oscillations about the 
equilibrium position if the mass of the lighter atom is m. 

20. A particle of mass m is subject to a force given by 



F = B (£. - ^ 4- 21a% \ 

\x 2 is 1 " iSf 



(a) Find and sketch the potential energy. (B and a are positive.) (b) Describe 
the types of motion which may occur. Locate all equilibrium points and de- 
termine the frequency of small oscillations about any which are stable, (c) A 
particle starts at x = 3a/2 with a velocity v = —vo, where Do is positive. What is 
the smallest value of wo for which the particle may eventually escape to a very 
large distance? Describe the motion in that case. What is the maximum velocity 
the particle will have? What velocity will it have when it is very far from its 
starting point? 

21. A particle of mass m moves in a potential well given by 

. , _ -V a 2 (a 2 + x 2 ) 
VW 8a* + x* ' 

(a) Sketch V(x) and F(x). (b) Discuss the motions which may occur. Locate 
all equilibrium points and determine the frequency of small oscillations about 
any that are stable, (c) A particle starts at a great distance from the potential 
well with velocity vo toward the well. As it passes the point x = a, it suffers a 
collision with another particle, during which it loses a fraction a of its kinetic 
energy. How large must a be in order that the particle thereafter remain trapped 
in the well? How large must a be in order that the particle be trapped in one 
side of the well? Find the turning points of the new motion if a = 1. 

22. Starting with e 2i> = (e ie ) 2 , obtain formulas for sin 20, cos 20 in terms of 
sin 0, cos 6. 

23. Find the general solutions of the equations: 

(a) mx + bx — kx = 0, 

(b) mx — bx + kx = 0. 

Discuss the physical interpretation of these equations and their solutions, assum- 
ing that they are the equations of motion of a particle. 



66 MOTION OF A PARTICLE IN ONE DIMENSION [CHAP. 2 

24. Show that when co 2 , — T 2 is very small, the underdamped solution (2-133) 
is approximately equal to the critically damped solution (2-146), for a short 
time interval. What is the relation between the constants C\, C2 and A, 61 This 
result suggests how one might discover the additional solution (2-143) in the 
critical case. 

25. A mass m subject to a linear restoring force —kx and damping — bx is dis- 
placed a distance xo from equilibrium and released with zero initial velocity. 
Find the motion in the underdamped, critically damped, and overdamped cases. 

26. Solve Problem 25 for the case when the mass starts from its equilibrium 
position with an initial velocity vo. Sketch the motion for the three cases. 

27. A mass of 1000 kgm drops from a height of 10 m on a platform of negligible 
mass. It is desired to design a spring and dashpot on which to mount the plat- 
form so that the platform will settle to a new equilibrium position 0.2 m below 
its original position as quickly as possible after the impact without overshooting. 

(a) Find the spring constant k and the damping constant 6 of the dashpot. 
Why does the result seem to contradict the remarks at the end of Section 2-9? 

(b) Find, to two significant figures, the time required for the platform to settle 
within 1 mm of its final position. 

28. A force Fo(l — e~ at ) acts on a harmonic oscillator which is at rest at 
t = 0. The mass is m, the spring constant k = ±ma 2 , and b = ma. Find the 
motion. Sketch x(t). 

*29. Solve Problem 28 for the case k = ma 2 , b = 2ma. 

30. A force Fo cos (cot + #o) acts on a damped harmonic oscillator beginning 
at t = 0. (a) What must be the initial values of x and v in order that there be 
no transient? (b) If xo = vq = 0, find the amplitude A and phase 9 of the 
transient in terms of Fo, 80. 

31. An undamped harmonic oscillator of mass m, natural frequency coo, is 
initially at rest and is subject at t = to a blow so that it starts from xo = 
with initial velocity vo and oscillates freely until t = 3ir/2coo. From this time 
on, a force F = B cos (cot + 0) is applied. Find the motion. 

32. An underdamped harmonic oscillator is subject to an applied force 

F = F e~ at cos (cot + 0). 

Find a particular solution by expressing F as the real part of a complex exponen- 
tial function and looking for a solution for x having the same exponential time 
dependence. 

33. (a) Find the motion of a damped harmonic oscillator subject to a constant 
applied force Fo, by guessing a "steady-state" solution of the inhomogeneous 
equation (2-86) and adding a solution of the homogeneous equation, (b) Solve 
the same problem by making the substitution x' = x — a, and choosing the 
constant a so as to reduce the equation in x' to the homogeneous equation (2-85). 
Hence show that the effect of the application of a constant force is merely to shift 
the equilibrium position without affecting the nature of the oscillations. 



* An asterisk is used, as explained in the Preface, to indicate problems which 
may be particularly difficult. 



PROBLEMS 67 

34. Find the motion of a mass m subject to a restoring force — kx, and to a 
damping force {±)nmg due to dry sliding friction. Show that the oscillations are 
isochronous (period independent of amplitude) with the amplitude of oscillation 
decreasing by 2v.qlu% during each half-cycle until the mass comes to a stop. 
[Hint: Use the result of Problem 33. When the force has a different algebraic 
form at different times during the motion, as here, where the sign of the damping 
force must be chosen so that the force is always opposed to the velocity, it is 
necessary to solve the equation of motion separately for each interval of time 
during which a particular expression for the force is to be used, and to choose 
as initial conditions for each time interval the final position and velocity of the 
preceding time interval.] 

35. An undamped harmonic oscillator (7 = 0), initially at rest, is subject to 
a force given by Eq. (2-191). (a) Find x(f). (b) For a fixed p Q , for what value 
of dt is the final amplitude of oscillation greatest? (c) Show that as St -» 0, your 
solution approaches that given by Eq, (2-190). 

36. Find the solution analogous to Eq. (2-190) for a critically damped har- 
monic oscillator subject to an impulse po delivered at t = to. 

37. (a) Find, using the principle of superposition, the motion of an under- 
damped oscillator [7 = (l/3)co ] initially at rest and subject, after t = 0, 
to a force 

F = A sin w t + B sin 3wot, 

where wo is the natural frequency of the oscillator, (b) What ratio of B to A is 
required in order for the forced oscillation at frequency 3wo to have the same 
amplitude as that at frequency too? 

38. Find, by the Fourier-series method, the steady-state solution for the 
damped harmonic oscillator subject to a force 



F(t) = I 0, if nT < '- ( n +*) r > 

\f , if (n + %)T < t < (n - 



(n + i)T < t < (n + 1)T, 

where n is any integer, and T - 6jr/«o, where w is the resonance frequency of the 
oscillator. Show that if 7 <JC coo, the motion is nearly sinusoidal with period T/3. 

39. An underdamped oscillator initially at rest is acted upon, beginning at 
t = 0, by a force 

F = Foe-"'. 

Find its motion by using Green's solution (2-210). 

40. Using the result of Problem 36, find by Green's method the motion of a 
critically damped oscillator initially at rest and subject to a force F(t). 



CHAPTER 3 

MOTION OF A PARTICLE IN TWO OR THREE DIMENSIONS 

3-1 Vector algebra. The discussion of motion in two or three dimensions 
is vastly simplified by the introduction of the concept of a vector. A 
vector is denned geometrically as a physical quantity characterized by a 
magnitude and a direction in space. Examples are velocity, force, and 
position with respect to a fixed origin. Schematically, we represent a 
vector by an arrow whose length and direction represent the magnitude 
and direction of the vector. We shall represent a vector by a letter in bold- 
face type. The same letter in ordinary italics will represent the magnitude 
of the vector. (See Fig. 3-1.) The magnitude of a vector may also be 
represented by vertical bars enclosing the vector symbol: 



IAI. 



(3-1) 



Two vectors are equal if they have the same magnitude and direction; 
the concept of vector itself makes no reference to any particular location.* 




Fig. 3-1. A vector A and its magnitude A. (c > 0) 





Fig. 3-2. Definition of multiplication of a vector by a scalar, (c > 0) 



* A distinction is sometimes made between "free" vectors, which have no par- 
ticular location in space; "sliding" vectors, which may be located anywhere along 
a line; and "fixed" vectors, which must be located at a definite point in space. We 
prefer here to regard the vector as distinguished by its magnitude and direction 
alone, so that two vectors may be regarded as equal if they have the same magni- 
tudes and directions, regardless of their positions in space. 

68 



3-1] VECTOR ALGEBRA 69 

A quantity represented by an ordinary (positive or negative) number is 
often called a scalar, to distinguish it from a vector. We define a product 
of a vector A and a positive scalar c as a vector cA in the same direction as 
A of magnitude cA. If c is negative, we define cA as having the magnitude 
\c\A and a direction opposite to A. (See Fig. 3-2.) It follows from this 
definition that 

|cA| = |c| |A|. (3-2) 

It is also readily shown, on the basis of this definition, that multiplication 
by a scalar is associative in the following sense: 

(cd)A = c(dA). (3-3) 

It is sometimes convenient to be able to write the scalar to the right of the 
vector, and we define Ac as meaning the same vector as cA: 

Ac = cA. (3-4) 

We define the sum (A + B) of two vectors A and B as the vector which 
extends from the tail of A to the tip of B when A is drawn with its tip 
at the tail of B, as in Fig. 3-3. This definition is equivalent to the usual 
parallelogram rule, and is more convenient to use. It is readily extended 
to the sum of any number of vectors, as in Fig. 3—4. 

On the basis of the definition given in Fig. 3-3, we can readily prove 
that vector addition is commutative and associative: 

A + B = B + A, (3-5) 

(A + B) + C=A + (B +.C). (3-6) 

According to Eq. (3-6), we may omit parentheses in writing a vector sum, 
since the order of adding does not matter. From the definitions given by 
Figs. 3-2 and 3-3, we can also prove the following distributive laws: 

c(A + B) = cA + cB, (3-7) 

(c + d)A = cA + dk. (3-8) 

These statements can be proved by drawing diagrams representing the 



A + B 





A+B+C+D 



Fig. 3-3. Definition of addition of Fig. 3-4. Addition of several vec- 
two vectors. tors. 



70 MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3 




(A + B) + C or A + (B + C) 

Fig. 3-5. Proof of Eq. (3-6). 

right and left members of each equation according to the definitions given. 
For example, the diagram in Fig. 3-5 makes it evident that the result of 
adding C to (A + B) is the same as the result of adding (B + C) to A. 

According to Eqs. (3-3) through (3-8), the sum and product we have 
defined have most of the algebraic properties of sums and products of 
ordinary numbers. This is the justification for calling them sums and 
products. Thus it is unnecessary to commit these results to memory. 
We need only remember that we can manipulate these sums and products 
just as we manipulate numbers in ordinary algebra with the one exception 
that the product defined by Fig. 3-2 can be formed only between a scalar 
and a vector, and the result is a vector. 

A vector may be represented algebraically in terms of its components or 
projections along a set of coordinate axes. Drop perpendiculars from the 
tail and tip of the vector onto the coordinate axes as in Fig. 3-6. Then 
the component of the vector along any axis is defined as the length of the 
segment cut off on the axis by these perpendiculars. The component is 
taken as positive or negative according to whether the projection of the 
tip of the vector lies in the positive or negative direction along the axis 




(a) 




Fig. 3-6. (a) Components of a vector in a plane, (b) Components of a vector 
in space. 



3-1] 



VECTOR ALGEBRA 



71 




Fig. 3-7. Diagrammatic proof of the formula A = Aj + A v j. 



from the projection of the tail. The components of a vector A along x-, y-, 
and z-axes will be written A x , A y , and A z . The notation (A x , A y , A z ) will 
sometimes be used to represent the vector A: 



A \A X , Ay, Az). 



(3-9) 



If we define vectors i, j, k of unit length along the x-, y-, z-axes respectively, 
then we can write any vector as a sum of products of its components 
with i, j, k: 

A = A x \ + A y ] + A z k. (3-10) 

The correctness of this formula can be made evident by drawing a dia- 
gram in which the three vectors on the right, which are parallel to the 
three axes, are added to give A. Figure 3-7 shows this construction for 
the two-dimensional case. 

We now have two equivalent ways of denning a vector: geometrically 
as quantity with a magnitude and direction in space, or algebraically as a 
set of three numbers [A x , A y , A z ), which we call its components.* The 
operations of addition and multiplication by a scalar, which are defined 
geometrically in Figs. 3-2 and 3-3 in terms of the lengths and directions of 
the vectors involved, can also be defined algebraically as operations on the 
components of the vectors. Thus cA is the vector whose components are 
the components of A, each multiplied by c: 



cA = (cA x , cA V) cA z ), 



(3-11) 



* These two ways of defining a vector are not quite equivalent as given here, 
for the algebraic definition requires that a coordinate system be set up, whereas 
the geometric definition does not refer to any particular set of axes. This flaw 
can be remedied by making the algebraic definition also independent of any 
particular set of axes. This is done by studying how the components change 
when the axes are changed, and defining a vector algebraically as a set of three 
quantities which transform in a certain way when the axes are changed. This 
refinement will not concern us in this chapter. 



72 



MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS 

y 



(A + B), 



[chap. 3 




Fig. 3-8. Proof of equivalence of algebraic and geometric definitions of 
vector addition. 



and A + B is the vector whose components are obtained by adding the 
components of A and B: 



A + B = (A x + B x , A y + B V) A, + B z ). 



(3-12) 



The equivalence of the definitions (3-11) and (3-12) to the corresponding 
geometrical definitions can be demonstrated by drawing suitable diagrams. 
Figure 3-8 constitutes a proof of Eq. (3-12) for the two-dimensional case. 
All vectors are drawn in Fig. 3-8 so that their components are positive; 
for a complete proof, similar diagrams should be drawn for the cases where 
one or both components of either vector are negative. The length of 
a vector can be defined algebraically as follows: 



|A| = (A* + AI+ At) 



2-.1/2 



(3-13) 



where the positive square root is to be taken. 

We can now give algebraic proofs of Eqs. (3-2), (3-3), (3-5), (3-6), 
(3-7), and (3-8), based on the definitions (3-11), (3-12), and (3-13). 
For example, to prove Eq. (3-7), we show that each component of the left 
side agrees with each component on the right. For the a:-component, the 
proof runs: 

[c(A + B)L = c(A + B), 



= c{A x + B x ) 
= cA x + cB x 
= (cA), + (cB), 
= (cA + cB) x . 



[by Eq. (3-11)] 
[by Eq. (3-12)] 

[by Eq. (3-11)] 
[by Eq. (3-12)] 



3-1] VECTOR ALGEBRA 73 





A-B> 

B " 5" 

Fig. 3-9. Two methods of subtraction of vectors. 

Since all components are treated alike in the definitions (3-11), (3-12), 
(3-13), the same proof holds for the y- and z-components, and hence the 
vectors on the left and right sides of Eq. (3-7) are equal. 

In view of the equivalence of the geometrical and algebraic definitions 
of the vector operations, it is unnecessary, for geometrical applications, to 
give both an algebraic and a geometric proof of each formula of vector 
algebra. Either a geometric or an algebraic proof, whichever is easiest, 
will suffice. However, there are important cases in physics where we have 
to consider sets of quantities which behave algebraically like the com- 
ponents of vectors although they cannot be interpreted geometrically as 
quantities with a magnitude and direction in ordinary space. In order 
that we may apply the rules of vector algebra in such applications, it is 
important to know that all of these rules can be proved purely algebraically 
from the algebraic definitions of the vector operations. The geometric 
approach has the advantage of enabling us to visualize the meanings of 
the various vector notations and formulas. The algebraic approach sim- 
plifies certain proofs, and has the further advantage that it makes possible 
wide applications of the mathematical concept of vector, including many 
cases where the ordinary geometric meaning is no longer retained. 

We may define subtraction of vectors in terms of addition and multi- 
plication by —1: 

A - B = A + (-B) = (A x - B x , A y - B y , A, - B z ). (3-14) 

The difference A — B may be found geometrically according to either of 
the two schemes shown in Fig. 3-9. Subtraction of vectors may be shown 
to have all the algebraic properties to be expected by analogy with sub- 
traction of numbers. 

It is useful to define a scalar product (A-B) of two vectors A and B as 
the product of their magnitudes times the cosine of the angle between them 
(Fig. 3-10): 

A-B = AB cos 6. (3-15) 

The scalar product is a scalar or number. It is also called the dot product 
or inner product, and can also be defined as the product of the magnitude of 
either vector times the projection of the other along it. An example of 



B. 




74 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

its use is the expression for the work done when a force F acts through a 
distance s not necessarily parallel to it: 

W = Fs cosd = F-s. 

We are entitled to call A-Ba product because it has the following alge- 
braic properties which are easily proved from the geometrical definition 
(3-15): 

(cA)-B = A-(cB) = c(A-B), (3-16) 
A-(B + C) = A-B + AC, (3-17) 

A-B = B-A, (3-18) 

Fig. 3-10. Angle between 

A-A = A 2 (3-19) two vectors. 

These equations mean that we can treat the dot product algebraically like 
a product in the algebra of ordinary numbers, provided we keep in mind 
that the two factors must be vectors and the resulting product is a scalar. 
The following statements are also consequences of the definition (3-15), 
where i, j, and k are the unit vectors along the three coordinate axes: 

i-i = j-j = k-k = 1, 
i-j = j-k = k-i = 0. 

A-B = AB, when A is parallel to B, (3-21) 

A-B = 0, when A is perpendicular to B. (3-22) 

Notice that, according to Eq. (3-22), the dot product of two vectors is 

zero if they are perpendicular, even though neither vector is of zero length. 

The dot product can also be defined algebraically in terms of components: 

A-B = A X B X + A y B y + A Z B Z . (3-23) 

To prove that Eq. (3-23) is equivalent to the geometric definition (3-15), 
we write A and B in the form given by Eq. (3-10), and make use of Eqs. 
(3-16), (3-17), (3-18), and (3-20), which follow from Eq. (3-15): 

A-B = (iA x + \A V + kA z )-(iB x + jB y + kB 2 ) 

= (i-i)A x B x + (i-j)A x B y + (i-k)4A + (j-i)A y B x + yjA y B y 
+ ybA y B, + k-iA l B x + k-jA z B v + k-kA 2 B z 

= A X B X + AyB y + A Z B Z . 
This proves Eq. (3-23). The properties (3-16) to (3-20) can all be proved 



(3-20) 



3-1] VECTOR ALGEBRA 75 

AxB 

shaded area = |A x B| 
B-— 




A 

Fig. 3-11. Definition of vector product. 

readily from the algebraic definition (3-23) as well as from the geometric 
definition (3-15). We can regard Eqs. (3-21) and (3-22) as algebraic 
definitions of parallel and perpendicular. 

Another product convenient to define is the vector product, also called 
the cross product or outer product. The cross product (A X B) of two 
vectors A and B is defined as a vector perpendicular to the plane of A and 
B whose magnitude is the area of the parallelogram having A and B as 
sides. The sense or direction of (A X B) is defined as the direction of 
advance of a right-hand screw rotated from A toward B. (See Fig. 3-11.) 
The length of (A X B), in terms of the angle between the two vectors, 
is given by 

| A X B| = AB sin 0. (3-24) 

Note that the scalar product of two vectors is a scalar or number, while 
the vector product is a new vector. The vector product has the following 
algebraic properties which can be proved from the definition given in 
Fig. 3-11:* 

A X B = — B x A, (3-25) 

(cA) X B = A X (cB) = c(A X B), (3-26) 

A x (B + C) = (A x B) + (A x C), (3-27) 

A x A = 0, (3-28) 

A X B = 0, when A is parallel to B, (3-29) 

|A X B| = AB, when A is perpendicular to B, (3-30) 

ixi = jxj = kxk = 0, 

i X j = k, j x k = i, k X i = j. (3-31) 



* Here stands for the vector of zero length, sometimes called the null vector. 
It has no particular direction in space. It has the properties: 

A + = A, A-0 = 0, A X = 0, A — A = 0, = (0, 0, 0). 



76 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

Hence the cross product can be treated algebraically like an ordinary- 
product with the exception that the order of multiplication must not be 
changed, and provided we keep in mind that the two factors must be vectors 
and the result is a vector. Switching the order of factors in a cross product 
changes the sign. This is the first unexpected deviation of the rules of 
vector algebra from those of ordinary algebra. The reader should there- 
fore memorize Eq. (3-25). Equations (3-29) and (3-30), as well as the 
analogous Eqs. (3-21) and (3-22), are also worth remembering. (It goes 
without saying that all geometrical and algebraic definitions should be 
memorized.) In a repeated vector product like (A X B) X (C X D), the 
parentheses cannot be omitted or rearranged, for the result of carrying out 
the multiplications in a different order is not, in general, the same. [See, 
for example, Eqs. (3-35) and (3-36).] Notice that according to Eq. (3-29) 
the cross product of two vectors may be null without either vector being 
the null vector. 

From Eqs. (3-25) to (3-31), using Eq. (3-10) to represent A and B, we 
can prove that the geometric definition (Fig. 3-11) is equivalent to the 
following algebraic definition of the cross product: 

A X B = (AyB, - A l B y , A,B X - A X B Z , A x B y - A y B x ). (3-32) 
We can also write A X B as a determinant: 



A x B = 



i J k 

A x Ay A z 
B x By B z 



(3-33) 



Expansion of the right side of Eq. (3-33) according to the ordinary rules 
for determinants yields Eq. (3-32). Again the properties (3-25) to (3-31) 
follow also from the algebraic definition (3-32). 
The following useful identities can be proved: 

A-(B X C) = (A X B)C, (3-34) 

A X (B X C) = B(A-C) - C(A-B), (3-35) 

(A x B) x C = B(A-C) - A(B-C), (3-36) 

i-(j X k) = 1. (3-37) 

The first three of these should be committed to memory. Equation (3-34) 
allows us to interchange dot and cross in the scalar triple product. The 
quantity A-(B x C) can be shown to be the volume of the parallelepiped 
whose edges are A, B, C, with positive or negative sign depending on 
whether A, B, C are in the same relative orientation as i, j, k, that is, 
depending on whether a right-hand screw rotated from A toward B would 



3-2] APPLICATIONS TO A SET OF FORCES ACTING ON A PARTICLE 77 

advance along C in the positive or negative direction. The triple vector 
product formulas (3-35) and (3-36) are easy to remember if we note that 
the positive term on the right in each case is the middle vector (B) times 
the scalar product (A-C) of the other two, while the negative term is the 
other vector within the parentheses times the scalar product of the other 
two. 

As an example of the use of the vector product, the rule for the force 
exerted by a magnetic field of induction B on a moving electric charge q 
(esu) can be expressed as 

F = 2 V x B, 

c 

where c is the speed of light and v is the velocity of the charge. This 
equation gives correctly both the magnitude and direction of the force. 
The reader will remember that the subject of electricity and magnetism is 
full of right- and left-hand rules. Vector quantities whose directions are 
determined by right- or left-hand rules generally turn out to be expressible 
as cross products. 

3-2 Applications to a set of forces acting on a particle. According to the 
principles set down in Section 1-3, if a set of forces Fi, F 2 , . . . , F„ act on a 
particle, the total force F, which determines its acceleration, is to be ob- 
tained by taking the vector sum of the forces F 1; F 2 , . . . , F„: 

F = F x + F 2 H + F„. (3-38) 

The forces F 1; F 2 , . . . , F„ are often referred to as component forces, and F 
is called their resultant. The term component is here used in a more gen- 
eral sense than in the preceding section, where the components of a vector 
were defined as the projections of the vector on a set of coordinate axes. 
When component is meant in this sense as one of a set of vectors whose sum 
is F, we shall use the term (vector) component. In general, unless other- 
wise indicated, the term component of a vector F in a certain direction will 
mean the perpendicular projection of the vector F on a line in that direc- 
tion. In symbols, the component of F in the direction of the unit vector n 
is 

F n = n-F. (3-39) 

In this sense, the component of F is not a vector, but a number. The com- 
ponents of F along the x-, y-, and 2-axes are the components in the sense of 
Eq. (3-39) in the directions i, j, and k. 

If the forces Fi, F 2 , . . . , F n are given, the sum may be determined 
graphically by drawing a careful scale diagram according to the definition 
of Fig. 3-3 or 3-4. The sum may also be determined analytically by 



78 MOTION OF PARTICLE IN TWO OK THREE DIMENSIONS [CHAP. 3 

F 2 




Fig. 3-12. Sum of two forces. 

drawing a rough sketch of the sum diagram and using trigonometry to 
calculate the magnitude and direction of the vector F. If, for example, 
two vectors are to be added, the sum can be found by using the cosine and 
sine laws. In Fig. 3-12, F u F 2 , and 6 are given, and the magnitude and 
direction of the sum F are calculated from 

F 2 = F\ + F% - 2F X F 2 cos 6, (3-40) 

Fi F* F 

1 - 2 - (3-41) 



sin $ sin a sin 9 

Note that the first of these equations can be obtained by squaring, in the 
sense of the dot product, the equation 

F = Fj + F 2 . (3-42) 

Taking the dot product of each member of this equation with itself, we 
obtain 

F-F = F 2 = Fi-F x + 2F!-F 2 + F 2 -F 2 
= F\ + Fl — 2F X F 2 cos 0. 

(Note that in Fig. 3-12 is the supplement of the angle between ¥ x and F 2 
as defined by Fig. 3-10.) This technique can be applied to obtain directly 
the magnitude of the sum of any number of vectors in terms of their 
lengths and the angles between them. Simply square Eq. (3-38), and 
split up the right side according to the laws of vector algebra into a sum of 
squares and dot products of the component forces. The angle between F 
and any of the component forces can be found by crossing or dotting the 
component vector into Eq. (3-38). For example, in the case of a sum of 
two forces, we cross ¥ x into Eq. (3^12) : 

FiXF = FiXF, + FiX F 2 . 

We take the magnitude of each side, using Eqs. (3-28) and (3-24): 

Frf sin a = F X F 2 sin 8, or -!—- = J^- ■ 

sin 6 sin a 



3-2] APPLICATIONS TO A SET OF FORCES ACTING ON A PARTICLE 79 

When a sum of more than two vectors is involved, it is usually simpler to 
take the dot product of the component vector with each side of Eq. (3-38). 
The vector sum in Eq. (3-38) can also be obtained by adding separately 
the components of Fi, . . . , F„ along any convenient set of axes: 



Fx = Fix + F 2x + • • • + F nx , 
F y = F ly + F 2y + ■ ■ . + F nv , 
F z = F u + F 2l + • • • + F nl . 



(3-43) 



When a sum of a large number of vectors is to be found, this is likely to be 
the quickest method. The reader should use his ingenuity in combining 
and modifying these methods to suit the problem at hand. Obviously, if 
a set of vectors is to be added which contains a group of parallel vectors, 
it will be simpler to add these parallel vectors first before trying to apply 
the methods of the preceding paragraph. 




Fig. 3-13. Force F acting at point P. 

Just as the various forces acting on a particle are to be added vectorially 
to give the total force, so, conversely, the total force, or any individual 
force, acting on a particle may be resolved in any convenient manner into 
a sum of (vector) component forces which may be considered as acting 
individually on the particle. Thus in the problem discussed in Section 1-7 
(Fig. 1-4), the reaction force F exerted by the plane on the brick is resolved 
into a normal component N and a frictional component /. The effect of 
the force F on the motion of the brick is the same as that of the forces N 
and / acting together. If it is desired to resolve a force F into a sum of 
(vector) component forces in two or three perpendicular directions, this 
can be done by taking the perpendicular projections of F in these directions, 
as in Fig. 3-6. The magnitudes of the vector components of F, along a set 
of perpendicular directions, are just the ordinary components of F in these 
directions in the sense of Eq. (3-39). 

If a force F in the a^-plane acts on a particle at the point P, we define 
the torque, or moment of the force F about the origin (Fig. 3-13) as the 
product of the distance OP and the component of F perpendicular to r: 



No = rF sin a. 



(3-44) 



80 



MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 



The moment No of the force F about the point is denned as positive when 
F acts in a counterclockwise direction about as in Fig. 3-13, and nega- 
tive when F acts in a clockwise direction. We can define in a similar way 
the moment about of any vector quantity located at the point P. The 
concept of moment will be found useful in our study of the mechanics of 
particles and rigid bodies. The geometrical and algebraic properties of 
torques will be studied in detail in Chapter 5. Notice that torque can be 
denned in terms of the vector product: 



No = ±|r X F|. 



(3-45) 



where the + or — sign is used according to whether the vector r X F points 
in the positive or negative direction along the z-axis. 




Fig. 3-14. Moment of a force about an axis in space. 

We can generalize the above definition of torque to the three-dimensional 
case by denning the torque or moment of a force F, acting at a point P, 
about an axis AB (Fig. 3-14). Let n be a unit vector in the direction of 
AB, a nd l et F be resolved into vector components parallel and perpendicu- 
lar to AB: 

F = F|| + F±, (»-46) 

where 

Fii = n(n-F), 

(3-47) 
Fj. = F - Fn. . 



We now define the moment of F about the axis AB as the moment, denned 
by Eq. (3-44) or (3-45), of the force F±, in a plane through the point P 



3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 81 

perpendicular to AB, about the point at which the axis ~AB passes through 
this plane: 

Nab = dbrFxsina = ±|r X F x |, (3-48) 

where the + or — sign is used, depending on whether r x Fx is in the same 
or opposite direction to n. According to this definition, a force like Fn 
parallel to AB has no torque or moment about A~B. Since r X Fn is per- 
pendicular to n, 

n-(r x F) = n-[r x (F M + Fx)] # 

= n-(r x Fn) +n-(r x Fx) 
= n-(r x F±) 

= ±|r x F x |. 

Hence we can define Nab in a neater way as follows: 

N AB = n-(r X F). (3-49) 

This definition automatically includes the proper sign, and does not require 
a resolution of F into Fn and Fx- Furthermore, r can now be drawn to P 
from any point on the axis AB, since a component of r parallel to A~B, like 
a component of F parallel to AB, gives a component in the cross product 
perpendicular to n which disappears from the dot product. 

Equation (3-49) suggests the definition of a vector torque or vector moment, 
about a point 0, of a force F acting at a point P, as follows: 

N = r X F, (3-50) 

where r is the vector from to P. The vector torque N has, according 
to Eq. (3-49), the property that its component in any direction is the 
torque, in the previous sense, of the force F about an axis through in that 
direction. Hereafter the term torque will usually mean the vector torque 
defined by Eq. (3-50). Torque about an axis AB in the previous sense 
will be called the component of torque along ~AB. We can define the 
vector moment of any vector located at a point P, about a point 0, by an 
equation analogous to Eq. (3-50). 

3-3 Differentiation and integration of vectors. A vector A may be a 
function of a scalar quantity, say t, in the sense that with each value of t a 
certain vector A(t) is associated, or algebraically in the sense that its com- 
ponents may be functions of t: 

A = A(<) = [A x (t), A y {t), AM- (3-51) 



82 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

The most common example is that of a vector function of the time; for 
example, the velocity of a moving particle is a function of the time: v(t). 
Other cases also occur, however; for example, in Eq. (3-76), the vector n 
is a function of the angle 6. We may define the derivative of the vector A 
with respect to t in analogy with the usual definition of the derivative of 
a scalar function (see Fig. 3-15) : 

§ = lim A « + *> ~ A ® • (3-52) 

dt a«-»o At 

(Division by At here means multiplication by 1/At.) We may also define 
the vector derivative algebraically in terms of its components: 



dA 

dt 



(dA x dA y dA z \ . dA x , ,dA y . ,dA z ., eo . 



As an example, if v(<) is the vector velocity of a particle, its vector accelera- 
tion a is 

a = dv/dl. 

Examples of the calculation of vector derivatives based on either definition 
(3-52) or (3-53) will be given in Sections 3-4 and 3-5. 

The following properties of vector differentiation can be proved by 
straightforward calculation from the algebraic definition (3-53), or they 
may be proved from the definition (3-52) in the same way the analogous 
properties are proved for differentiation of a scalar function: 

s< A + B > = f + f' ™ 

£(/A)_fA+/f, (3-55) 

|(A.B, = f.B + A.f, (3-56, 

|(AxB) = f XB + Axf- (3-57) 

These results imply that differentiation of vector sums and products obeys 
the same algebraic rules as differentiation of sums and products in ordinary 
calculus, except, however, that the order of factors in the cross product 
must not be changed [Eq. (3-57)]. To prove Eq. (3-55), for example, 
from the definition (3-53), we simply show by direct calculation that the 
corresponding components on both sides- of the equation are equal, making 
use of the definitions and properties of the vector operations introduced 



3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 83 

in the preceding section. For the z-component, the proof runs: 

d 



[>>!= 



dt 



(/A), [by Eq. (3-53)] 



= J t CM.) [by Eq. (3-11)] 

= it a 4. f dAx [standard rule of ordi- 
dt x dt nary calculus] 

dt A * + Kw) x ^y Eq. (3-53)] 



dt J 

Xdi A ) x + \ f -d7) x 

(df dA\ 
\dt A+J dt) x 



[by Eq. (3-11)] 
[by Eq. (3-12)] 



As another example, to prove Eq. (3-56) from the definition (3-52), we 
proceed as in the proof of the corresponding theorem for products of ordi- 
nary scalar functions. We shall use the symbol A to stand for the increment 
in the values of any function between t and t + At; the increment AA of a 
vector A is defined in Fig. 3-15. Using this definition of A, and the rules 
of vector algebra given in the preceding section, we have 

A(A-B) = (A + AA).(B + AB) - A-B 
At At 

= (AA)-B + A-(AB) + (AA)-(AB) 
At 



(AA)-B A-(AB) (AA)-(AB) 
At ^ At + At 

AA AB (AA)-(AB) 

At ^ At ^ At 



(3-58) 



When At — » 0, the left side of Eq. (3-58) approaches the left side of Eq. 
(3-56), and the first two terms on the right side of Eq. (3-58) approach the 



A(t + At) 




Fig. 3-15. Vector increment AA = A(t + At) — A(t). 



84 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 




Fig. 3-16. The position vector r of the point (x, y, z). 



two terms on the right of Eq. (3-56), while the last term on the right of 
Eq. (3-58) vanishes. The rigorous justification of this limit process is 
exactly similar to the justification required for the corresponding process 
in ordinary calculus. 

In treating motions in three-dimensional space, we often meet scalar 
and vector quantities which have a definite value at every point in space. 
Such quantities are functions of the space coordinates, commonly x, y, 
and z. They may also be thought of as functions of the position vector r 
from the origin to the point x, y, z (Fig. 3-16). We thus distinguish scalar 
point functions 

«(r) = u{x, y, z), 

and vector point functions 

A(r) = A(x, y, z) = [A x (x, y, z), A y (x, y, z), A z (x, y, z)]. 

An example of a scalar point function is the potential energy V(x, y, z) of a 
particle moving in three dimensions. An example of a vector point func- 
tion is the electric field intensity E(x, y, z). Scalar and vector point func- 
tions are often functions of the time t as well as of the point x, y, z in space. 
If we are given a curve C in space, and a vector function A defined at 
points along this curve, we may consider the line integral of A along C: 

f A-dr. 

Jc 

To define the line integral, imagine the curve C divided into small seg- 
ments, and let any segment be represented by a vector dr in the direction 
of the segment and of length equal to the length of the segment. Then 
the curve consists of the successive vectors dr laid end to end. Now for 
each segment, form the product A-dr, where A is the value of the vector 
function at the position of that segment. The line integral above is denned 
as the limit of the sums of the products A-dr as the number of segments 



3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 85 

increases without limit, while the length |dr| of every segment approaches 
zero. As an example, the work done by a force F, which may vary from 
point to point, on a particle which moves along a curve C is 



W 



= [ F-dr, 

Jc 



which is a generalization, to the case of a varying force and an arbitrary 
curve C, of the formula 

W = F-s, 

for a constant force acting on a body moving along a straight line seg- 
ment s. The reason for using the symbol dx to represent a segment of the 
curve is that if r is the position vector from the origin to a point on the 
curve, then dx is the increment in r (see Fig. 3-15) from one end to the 
other of the corresponding segment. If we write r in the form 

r = ix + \y + kz, (3-59) 

then 

dx = i dx + j dy + k dz, (3-60) 

where dx, dy, dz are the differences in the coordinates of the two ends of the 
segment. If s is the distance measured along the curve from some fixed 
point, we may express the line integral as an ordinary integral over the 
coordinate s: 

f A-dx = f A cos dds, (3-61) 

where 6 is the angle between A and the tangent to the curve at each point. 
(See Fig. 3-17.) This formula may be used to evaluate the integral if 
we know A and cos 6 as functions of s. We may also write the integral, 




Fig. 3-17. Elements involved in the line integral. 



86 



MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3 



using Eq. (3-60), as 

f A'dt = f (A x dx + Aydy + A. z dz). 
Jc Jc 



(3-62) 



One of the most convenient ways to represent a curve in space is to give 
the three coordinates (x, y, z) or, equivalently, the position vector r, as 
functions of a parameter s which has a definite value assigned to each 
point of the curve. The parameter s is often, though not necessarily, the 
distance measured along the curve from some reference point, as in Fig. 
3-17 and in Eq. (3-61). The parameter s may also be the time at which 
a moving particle arrives at any given point on the curve. If we know 
A(r) and r(s), then the line integral can be evaluated from the formula 



//■*- /(*•§)* 



/(^5 + ^ + ^S> 



(3-63) 



The right member of this equation is an ordinary integral over the vari- 
able s. 




Figure 3-18 



As an example of the calculation of a line integral, let us compute the 
work done on a particle moving in a semicircle of radius a about the origin 
in the a^-plane, by a force attracting the particle toward the point (a; = a, 
y = 0) and proportional to the distance of the particle from the point 
(a, 0). Using the notation indicated in Fig. 3-18, we can write down the 
following relations: 



= iOr - a), 



7T 
2 



e = ^ - P = ha, 



D 2 = 2a 2 (l - cos a), 



D — 2a sin • 



F = -fcD, 



JcD = 2ka sin -^ > 



s = o(ir — a). 



3-4] KINEMATICS IN A PLANE 87 

Using these relations, we can evaluate the work done, using Eq. (3-61): 

W = f F>dt 
Jc 

(■xa 

= F cosd ds 

Js=0 

= — J 2ka 2 sin ^ cos „ da 
= — 4fca 2 f sin cos dO 



2ka 



2 



In order to calculate the same integral from Eq. (3-63), we express r and 
F along the curve as functions of the parameter a: 



x = a cos a, y = a sin a, 
2 a 



F x = kD cos /S = 2ka sin 2 | = fca(l — cos a), 
f„ = — kD sin /3 = — 2ka sin - cos ^ = — &a sin a. 
The work is now, according to Eq. (3-63), 
W = f F-dr 

-/!('•£+'.£)*" 

/•o 
= / [— fca 2 (l — cos a) sin a — ka 2 sin a cos a] da 

J V 

= fca 2 / sin a da 
Jo 

= 2fta 2 . 

3-4 Kinematics in a plane. Kinematics is the science which describes 
the possible motions of mechanical systems without regard to the dynami- 
cal laws that determine which motions actually occur. In studying the 
kinematics of a particle in a plane, we shall be concerned with methods for 
describing the position of a particle, and the path followed by the particle, 
and with methods for rinding the various components of its velocity and 
acceleration. 



88 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 




Fig. 3-19. Position vector and rectangular coordinates of a point P in a plane. 



The simplest method of locating a particle in a plane is to set up two 
perpendicular axes and to specify any position by its rectangular coordi- 
nates x, y with respect to these axes (Fig. 3-19). Equivalently, we may 
specify the position vector r = (x, y) from the origin to the position of the 
particle. If we locate a position by specifying the vector r, then we need 
to specify in addition only the origin from which the vector is drawn. 
If we specify the coordinates x, y, then we must also specify the coordi- 
nate axes from which x, y are measured. 

Having set up a coordinate system, we next wish to describe the path 
of a particle in the plane. A curve in the xy-plane may be specified by 
giving y as a function of x along the curve, or vice versa: 



or 



y = v(x), 

x = x(y). 



(3-64) 
(3-65) 



Forms (3-64) and (3-65), however, are not convenient in many cases, for 
example when the curve doubles back on itself. We may also specify the 
curve by giving a relation between x and y, 



/(*, y) = 0, 



(3-66) 



such that the curve consists of those points whose coordinates satisfy this 
relation. An example is the equation of a circle: 



x 2 + y 2 



,2 _ 



0. 



One of the most convenient ways to represent a curve is in terms of a 
parameter s: 

x = x(s), y = y(s), (3-67) 

or 

r = r(s). 



The parameter s has a unique value at each point of the curve. As s 
varies, the point [x(s), y(s)] traces out the curve. The parameter s may, 



3-4] KINEMATICS IN A PLANE 89 

for example, be the distance measured along the curve from some fixed 
point. The equations of a circle can be expressed in terms of a parameter 6 
in the form 

x = a cos 6, 

y = a sin 6, 

where 6 is the angle between the z-axis and the radius a to the point (x, y) 
on the circle. In terms of the distance s measured around the circle, 

s 
x — a cos - > 
a 

. s 
y = a sin - • 
a 

In mechanical problems, the parameter is usually the time, in which 
case Eqs. (3-67) specify not only the path of the particle, but also the 
rate at which the particle traverses the path. If a particle travels with 
constant speed v around a circle, its position at any time t may be given by 

vt 
x = a cos — i 
a 

. vt 
y = a sin — • 
a 

If a particle moves along the path given by Eq. (3-67), we may specify 
its motion by giving s(i), or by specifying directly 

x = x(t), y = y(t), (3-68) 

or 

r = t{t). (3-69) 

The velocity and acceleration, and their components, are given by 



v = — = i — -I- i ^ 
dt dt^~ J dt 

dx dy 

v *=di' v « = Tt' 



dv _ d^r _ . d?x . . d?y 
dt ~ dl* ~ l dP + J dt 2 ' 

— d?£ _ d 2 y 

i* - dt 2 ' a y~dj2- 



(3-70) 



(3-71) 



90 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 



\* 



\V^ n 




T 

y 

_L 



n(9 + dB)f\<fa 



tL 




Fig. 3-20. Plane polar coordinates. 



Fig. 3-21. Increments in the vec- 
tors n and 1. 



Polar coordinates, shown in Fig. 3-20, are convenient in many prob- 
lems. The coordinates r, are related to x, y by the following equations: 



x = r cos i 



y = r sin 0, 



and 



= tan x - = sin 
x 



r= (x< + y*) 

-l y 

(x 2 + ?/2)l/2 



2x1/2 



(3-72) 
(3-73) 



= cos 



(* 2 + y 2 ) 1 ' 2 



We define unit vectors n, 1 in the directions of increasing r and 0, respec- 
tively, as shown. The vectors n, 1 are functions of the angle 6, and are 
related to i, j by the equations 



n = i cos 8 + j sin 0, 
1 = — i sin 9 + j cos 6. 



(3-74) 



Equations (3-74) follow by inspection of Fig. 3-20. Differentiating, we 
obtain the important formulas 



eta _ . rfl _ 

dd ~ ' dd~ 



-n. 



(3-75) 



Formulas (3-75) can also be obtained by studying Fig. 3-21 (remembering 
that |n| = |1| = 1). The position vector r is given very simply in terms 
of polar coordinates: 

r = rn(0). (3-76) 

We may describe the motion of a particle in polar coordinates by specifying 
r(t), 0(0, thus determining the position vector r(t). The velocity vector is 



dr dr . da dO . . M 
dt dt dd dt 



(3-77) 



3-5] KINEMATICS IN THREE DIMENSIONS 91 

Thus we obtain the components of velocity in the n, 1 directions: 

v r = f, v e = rd. (3-78) 

The acceleration vector is 

dt dO dt dO dt 

= (f - r^ 2 )n + (WJ + 2rt)L (3-79) 

The components of acceleration are 

a r = r — r0 2 , a„ = r0 + 2^. (3-80) 

The term rd 2 = v 2 /r is called the centripetal acceleration arising from motion 
in the direction. If r = f = 0, the path is a circle, and a r = —v 2 /r. 
This result is familiar from elementary physics. The term 2f6 is some- 
times called the coriolis acceleration. 

3-5 Kinematics in three dimensions. The development in the preceding 
section for kinematics in two dimensions utilizing rectangular coordinates 
can be extended immediately to the three-dimensional case. A point is 
specified by its coordinates x, y, z, with respect to chosen rectangular axes 
in space, or by its position vector r = (x, y, z) with respect to a chosen 
origin. A path in space may be represented in the form of two equations 
in x, y, and z: 

f(x, y, z) = 0, g(x, y, z) = 0. (3-81) 

Each equation represents a surface. The path is the intersection of the 
two surfaces. A path may also be represented parametrically: 

x = x(s), y = y(s), z = z(s). (3-82) 

Velocity and acceleration are again given by 

dr 



and 



v = ^i = w x + )v y + kv z , 


(3-83) 


dx dy dz 

v * = Tt' v « = Tt' v °=di' 


(3-84) 


dy . . . , , 
a = -^ = ia x + ja y + ka z , 


(3-85) 


d 2 x d 2 y d 2 z 

'■ ~ dp ' ay ~~dp' az ~ Ifi ' 


(3-86) 



Many coordinate systems other than cartesian are useful for special 
problems. Perhaps the most widely used are spherical polar coordinates 



92 MOTION OF PARTICLE IN TWO OK THREE DIMENSIONS [CHAP. 3 



< 


1 

j 


c 


A 


i/ 


z 


1 y^K 




x/ \ p 
/ V 





Fig. 3-22. Cylindrical polar coordinates. 

and cylindrical polar coordinates. Cylindrical polar coordinates (p, tp, z) 
are denned as in Fig. 3-22, or by the equations 



x — p cos tp, y = p sin tp, z = z, 
and, conversely, 

p= (x 2 + y 2 ) 112 , 



(3-87) 



^ -ill • -l V 

<P — tan - = sin , J 

a; (x 2 -f- 2/2)1/2 

2 = 2. 



= cos 



(x 2 + y 2 )i/ 2 



(3-88) 



A system of unit vectors h, m, k, in the directions of increasing p, tp, z, re- 
spectively, is shown in Fig. 3-22. k is constant, but m and h are func- 
tions of tp, just as in plane polar coordinates: 

h = i cos <p + j sin tp, m = — i sin <p + j cos <p, (3-89) 

and, likewise, 



dh 

dtp 



— m, 



dm 

dtp 



(3-90) 



The position vector r can be expressed in cylindrical coordinates in the 
form 

t = ph + zk. (3-91) 

Differentiating, we obtain for velocity and acceleration, using Eq. (3-90) : 
v = ^ = ph + p4>m + zk, (3-92) 

a = J = (p - P? 2 )h + (p£ + 2p£)m + zk. (3-93) 



3-5] 



KINEMATICS IN THREE DIMENSIONS 



93 



Since k, m, h form a set of mutually perpendicular unit vectors, any vector 
A can be expressed in terms of its components along k, m, h: 



A = Aph -f A^m + Ajs. 



(3-94) 



It must be noted that since h and m are functions of <p, the set of com- 
ponents {A p , A v , A z ) refers in general to a specific point in space at which 
the vector A is to be located, or at least to a specific value of the coordi- 
nate <p. Thus the components of a vector in cylindrical coordinates, and 
in fact in all systems of curvilinear coordinates, depend not only on the 
vector itself, but also on its location in space. If A is a function of a 
parameter, say t, then we may compute its derivative by differentiating 
Eq. (3-94), but we must be careful to take account of the variation of h 
and m if the location of the vector is also changing with t (e.g., if A is the 
force acting on a moving particle) : 



dk 
dt 



-te-^iM^+^s) 



dA z . 

m + ~dT k - 



(3-95) 



Formulas (3-92) and (3-93) are special cases of Eq. (3-95). A formula 
for dA/dt could have been worked out also for the case of polar coordinates 
in two dimensions considered in the preceding section, and would, in fact, 
have been exactly analogous to Eq. (3-95) except that the last term would 
be missing. 

Spherical polar coordinates (r, 6, tp) are defined as in Fig. 3-23 or by the 
equations 



r sin 6 cos <p, y = r sin 8 sin <p, z = r cos i 



(3-96) 



The expressions for x and y follow if we note that p = r sin 6, and 




Fig. 3-23. Spherical polar coordinates. 



94 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

use Eq. (3-87) ; the formula for z is evident from the diagram. Conversely, 

/ 2 1 2 r 2\l/2 

r = {x' + y' + z Y , 

! (x 2 4- v 2 ) 112 
= tan" 1 {X + g y ' , (3-97) 

<p = tan -1 -- 
x 

Unit vectors n, 1, m appropriate to spherical coordinates are indicated in 
Fig. 3-23, where m is the same vector as in cylindrical coordinates. The 
unit vector h is. also useful in obtaining relations involving n and 1. We 
note that k, h, n, 1, all lie in one vertical plane. From the figure, and 
Eq. (3-89), we have 

n = k cos + h sin 8 = k cos + i sin cos <p + j sin sin <p, 

1 = — k sin 8 + h cos 6 = — k sin 8 + i cos 8 cos <p + j cos 8 sin <p, (3-98) 

m = — i sin tp -\- j cos <p. 

By differentiating these formulas, or more easily by inspection of the dia- 
gram (as in Fig. 3-21), noting that variation of 8, with <p and r fixed, corre- 
sponds to rotation in the k, n, h, 1 plane, while variation of <p, with 8 and r 
fixed, corresponds to rotation around the z-axis, we find 



(3-99) 



88 


an . „ 

t- = m sm 8, 

dip 


dl 


91 

T- = m cos 8, 

0<p 


^=0 

ae ' 


dm . . n 
-r— = — h = — n sm 8 
o<p 



1 cos 8. 

In spherical coordinates the position vector is simply 

r = ra(0, <p). (3-100) 

Differentiating and using Eqs. (3-99), we obtain the velocity and accelera- 
tion: 

v = -j t = fn + rfa + (/y sin 0)m, (3-101) 

a = -^ = (r - r6 2 — rip 2 sin 2 0)n + (r9 + 2r6 - rip 2 sin 8 cos 0)1 

+ (rv sin 8 + 2f<p sin + 2rd<p cos 0)m. (3-102) 



3-6] ELEMENTS OF VECTOR ANALYSIS 95 

Again, n, m, 1 form a set of mutually perpendicular unit vectors, and any 
vector A may be represented in terms of its spherical components: 

A = Ara + A e l + Ajn. (3-103) 

Here again the components depend not only on A but also on its location. 
If A is a function of t, then 



dk _ f dA r 
dt ~\dt 



A e 


dd A ■ a ^¥> 1 










,(dA, 

^Xdt 


+ A 41- 
+ Ar dt 


-A, 


, cos t 


•*)' 




,(dA<p 
+ Vdt~ 


+ A r sin 6 


d<p 
dt 


+ A e 


-•S) 



m. (3-104) 

3-6 Elements of vector analysis. A scalar function u{x, y, z) has three 
derivatives, which may be thought of as the components of a vector point 
function called the gradient of u: 

We may also define grad u geometrically as a vector whose direction is the 
direction in which u increases most rapidly and whose magnitude is the 
directional derivative of u, i.e., the rate of increase of u per unit distance, 
in that direction. That this geometrical definition is equivalent to the 
algebraic definition (3-105) can be seen by taking the differential of u: 

du ^di dx + dy- dy + Tz dz - ( 3 - 106 ) 

Equation (3-106) has the form of a scalar product of grad u with the vector 
dr whose components are dx, dy, dz: 

du = dr-grad u. (3-107) 

Geometrically, du is the change in u when we move from the point r = 
(x, y, z) to a nearby point r + dt — {x + dx, y + dy, z + dz). By Eq. 
(3-15): 

du — \dt\ |grad u\ cos 6, (3-108) 

where is the angle between dt and grad u. Thus at a fixed small dis- 
tance \dx\ from the point r, the change in u is a maximum when dx is in the 
same direction as grad u, and then: 

|gradM| = -prT 



96 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

This confirms the geometrical description of grad u given above. An al- 
ternative geometrical definition of grad u is that it is a vector such that 
the change in u, for an arbitrary small change of position dx, is given by 
Eq. (3-107). 

In a purely symbolic way, the right member of Eq. (3-105) can be 
thought of as the "product" of a "vector": 

_ ( d d d\ . d . . d , , d , n ,._. 

V = U'^'^ = 1 ^ + , ^ + k ai' (3 - 109) 

with the scalar function u: 

gradw = Vm. (3-110) 

The symbol V is pronounced "del." V itself is not a vector in the geo- 
metrical sense, but an operation on a function u which gives a vector Vw. 
However, algebraically, V has properties nearly identical with those of a 
vector. The reason is that the differentiation symbols (d/dx, d/dy, d/dz) 
have algebraic properties like those of ordinary numbers except when they 
act on a product of functions: 



and 



d , , , du . dv d d d d . . 

_(„ + „) = _ + _, __ M = __ W , (3-111) 



f x (au) = a d £, (3-112) 



provided a is constant. However, 

d , n du dv . . 

_(„)«_, + „__. (3-H3) 

In this one respect differentiation operators differ algebraically from ordi- 
nary numbers. If d/dx were a number, d/dx(uv) would equal either 
u(d/dx)v or v{d/dx)u. Thus we may say that d/dx behaves algebraically 
as a number except that when it operates on a product, the result is a sum 
of terms in which each factor is differentiated separately, as in Eq. (3-113). 
A similar remark applies to the symbol V. It behaves algebraically as a 
vector, except that when it operates on a product it must be treated also 
as a differentiation operation. This rule enables us to write down a large 
number of identities involving the V symbol, based on vector identities. 
We shall require very few of these in this text, and shall not list them here. * 



* For a more complete treatment of vector analysis, see H. B. Phillips, Vector 
Analysis. New York: John Wiley & Sons, 1933. 



3-6] ELEMENTS OF VECTOR ANALYSIS 97 




Fig. 3-24. A volume V bounded by a surface 5. 

We can form the scalar product of V with a vector point function 
A(x, y, z). This is called the divergence of A: 

• A -* A -'£+%+'■&■ < 3 ->"> 

The geometrical meaning of div A is given by the following theorem, called 
the divergence theorem, or Gauss' theorem: 

jjj VA dV = Jfn>AdS, (3-115) 

v s 

where V is a given volume, S is the surface bounding the volume V, and n 
is a unit vector perpendicular to the surface S pointing out from the volume 
at each point of 5 (Fig. 3-24). Thus n-A is the component of A normal 
to S, and Eq. (3-115) says that the "total amount of V-A inside V" is 
equal to the "total flux of A outward through the surface 5." If v repre- 
sents the velocity of a moving fluid at any point in space, then 

//n.vdS 

a 

represents the volume of fluid flowing across S per second. If the fluid is 
incompressible, then according to Eq. (3-115), 



/// 



v-vdV 



would represent the total volume of fluid being produced within the vol- 
ume V per second. Hence v-v would be positive at sources from which 
the fluid is flowing, and negative at "sinks" into which it is flowing. We 
omit the proof of Gauss' theorem [Eq. (3-115)]; it may be found in any 
book on vector analysis.* 



Sec, e.g„ Phillips, op. cit. Chapter 3, Section 32. 



98 MOTION OF PARTICLE IN' TWO OB THREE DIMENSIONS [CHAP. 3 




Fig. 3-25. A surface S bounded by a curve C. 

We can also form a cross product of V with a vector point function 
A(x, y, z). This is called the curl of A: 

The geometrical meaning of the curl is given by Stokes' theorem: 

j jn-(v x A)dS = f A-dr, (3-117) 

where S is any surface in space, n is the unit vector normal to S, and C is 
the curve bounding S, dr being taken in that direction in which a man would 
walk around C if his left hand were on the inside and his head in the direc- 
tion of n. (See Fig. 3-25.) According to Eq. (3-117), curl A at any point 
is a measure of the extent to which the vector function A circles around that 
point. A good example is the magnetic field around a wire carrying an 
electric current, where the curl of the magnetic field intensity is propor- 
tional to the current density. We omit the proof of Stokes' theorem [Eq. 
(3-117)].* 

The reader should not be bothered by the difficulty of fixing these ideas 
in his mind. Understanding of new mathematical concepts like these 
comes to most people only slowly, as they are put to use. The definitions 
are recorded here for future use, One cannot be expected to be familiar 
with them until he has seen how they are used in physical problems. 



The symbolic vector V can also be expressed in cylindrical coordinates in terms 
of its components along h, m, k. (See Fig. 3-22.) We note that if u = u(p, tp, z), 

* ->+£*+£*■ (3 - 118) 



For the proof see Phillips, op. tit. Chapter 3, Section 29. 



3-6] ELEMENTS OF VECTOR ANALYSIS 99 

and, from Eqs. (3-91) and (3-90), 

dr = h dp + rap d<p + k dz, (3-119) 

a result whose geometric significance will be evident from Fig. 3-22. Hence, if we 
write 

_ . d . m d , , d 

we will have, since h, m, k are a set of mutually perpendicular unit vectors, 

du = dr-Vw, (3-121) 

as required by the geometrical definition of Vu = grad u. [See the remarks 
following Eq. (3-107).] A formula for V could have been worked out also for 
the case of polar coordinates in two dimensions and would have been exactly 
analogous to Eq. (3-120) except that the term in z would be missing. In apply- 
ing the symbol V to expressions involving vectors expressed in cylindrical co- 
ordinates [Eq. (3-94)], it must be remembered that the unit vectors h and m are 
functions of <p and subject to differentiation when they occur after d/d<p. 

We may also find the vector V in spherical coordinates (Pig. 3-23) by noting 
that 

, du , . du ,. , du , 

du = Tr dr + S6 M + dl P d *> ( 3 ~ 122 ) 

and 

dt = n dr + Xr dO + mr sin d<p. (3-123) 

Hence 

_ 3,13. m d 

V=n 3r + ^+rlu7fl^' ( 3 ~ 124 > 

in order that Eq. (3-121) may hold. Again we caution that in working with Eq. 
(3-124), th3 dependence of n, 1, m on 0, <p must be kept in mind. For example, 
the divergence of a vector function A expressed in spherical coordinates [Eq. 
(3-103)] is 

VA = n-— • + -••— + 



dr r dd r sin d<p 



dA 
dr 



'■ +i M+ A )+7h,(%+ *-•"+ *-•) 



= dAr 2Ar 1 dAt _JU_ , _1__ dA? 
dr r ~*~ r dd r tan r sin dip 

- 1 d /- 2 -^ i 1 d - • «^ , I dA„ 

-r-2fr (rA ' ) + 7^0r0 (sm6A ° ) + 7^re-dV- 

(In the above calculation, we use the fact that 1, m, n are a set of mutually per- 
pendicular unit vectors.) 



100 MOTION OP PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

3-7 Momentum and energy theorems. Newton's second law, as formu- 
lated in Chapter 1, leads, in two or three dimensions, to the vector equation 

m§ = F. (3-125) 

In two dimensions, this is equivalent to two component equations, in three 
dimensions, to three, which are, in cartesian coordinates, 



d 2 x „ d 2 y „ _ d 2 z 



m -- = F x , m^i = F m m — = F z . (3-126) 



In this section, we prove, using Eq. (3-125), some theorems for motion in 

two or three dimensions which are the vector analogs to those proved in 

Section 2-1 for one-dimensional motion. 

The linear momentum vector p of a particle is to be defined, according 

to Eq. (1-10), as follows: 

p = mv. (3-127) 

Equations (3-125) and (3-126) can then be written 

i <»*> = I = f > (3 - i28) 

or, in component form, 

d ^ = F x , *g = F n f = *V (3-129) 

If we multiply by dt, and integrate from t x to t 2 , we obtain the change in 
momentum between t\ and t^. 

p2 — Pi = «v 2 — ravi = J F dt. (3-130) 

The integral on the right is the impulse delivered by the force, and is a 
vector whose components are the corresponding integrals of the com- 
ponents of F. In component form: 

rh 
Px 2 — p Xl = / Fx dt, 
Jti 

Vv 2 - Pvi = [* Fydt, (3-131) 

Pz 2 — Pzi = f 2 p z dt. 

In order to obtain an equation for the rate of change of kinetic energy, 
we proceed as in Section 2-1, multiplying Eqs. (3-126) by v x , v y , v z , respec- 
tively, to obtain 

| (W) = F x v x , j t (inwl) = F y v y , j t (imv 2 ) = F z v z . (3-132) 



3-8] PLANE AND VECTOR ANGULAR MOMENTUM THEOREMS 101 

Adding these equations, we have 



d 

dt 
or 



- [*f»(»2 + V 2 y + Vf)\ = F X V X + FyVy + F z v z 



d ,, 2^ dT 

-(J TOy ) = W = F.v. (3-133) 

This equation can also be deduced from the vector equation (3-125) by 
taking the dot product with v on each side, and noting that 

d , 2i d . . dv , dv „ dv 
^(0=^(v.v) = -.v + v.^=2v.^. 

Thus, by Eq. (3-132), 

¥ ' v =™'Tt = * m dt=dt^ mv) - 

Multiplying Eq. (3-133) by dt, and integrating, we obtain the integrated 
form of the energy theorem: 

T 2 - T x = \mo\ - \mv\ = pF-vett. (3-134) 

Since v dt = dr, if F is given as a function of r, we can write the right 
member of Eq. (3-134) as a line integral: 

T a -T l = Pp-dr, (3-135) 

Jri 

where the integral is to be taken along the path followed by the particle 
between the points i x and r 2 . The integral on the right in Eqs. (3-134) and 
(3-135) is the work done on the particle by the force between the times 
ti and t 2 - Note how the vector notation brings out the analogy between 
the one- and the two- or three-dimensional cases of the momentum and 
energy theorems. 

3-8 Plane and vector angular momentum theorems. If a particle moves 
in a plane, we define its angular momentum Lo about a point as the 
moment of its momentum vector about the point 0, that is, as the product 
of its distance from times the component of momentum perpendicular 
to the line joining the particle to 0. The subscript will usually be omitted, 
except when moments about more than one origin enter into the discussion, 
but it must be remembered that angular momentum, like torque, refers 
to a particular origin about which moments are taken. The angular 
momentum L is taken as positive when the particle is moving in a counter- 
clockwise sense with respect to 0; L is expressed most simply in terms of 
polar coordinates with as origin. Let the particle have mass m. Then 



102 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 






Fig. 3-26. Components of velocity in a plane. 

its momentum is mv, and the component of momentum perpendicular to 
the radius vector from is mv t (Fig. 3-26), so that, if we use Eq. (3-78), 

L = rmv e = mrH. (3-136) 

If we write the force in terms of its polar components: 

F = nF r + IF,, (3-137) 

then in plane polar coordinates the equation of motion, Eq. (3-125), be- 
comes, by Eq. (3-80), 

ma r = mr — mrd 2 = F n (3-138) 

ma, = mrS + 2mr& = F e . (3-139) 

We now note that 

= 2mrf6 + mr 2 8. 



dt 
Thus, multiplying Eq. (3-139) by r, we have 

(mr 2 6) = rF e = N. (3-140) 



dL d , 2/n 



dt dt 

The quantity rF e is the torque exerted by the force F about the point 0. 
Integrating Eq. (3-140), we obtain the integrated form of the angular 
momentum theorem for motion in a plane: 

L 2 - L t = mr%6 2 - mr\»i = f" rF e dt. (3-141) 

We can generalize the definition of angular momentum to apply to three- 
dimensional motion by defining the angular momentum of a particle about 
an axis in space as the moment of its momentum vector about this axis, 
just as in Section 3-2 we denned the moment of a force about an axis. 



3-8] PLANE AND VECTOR ANGULAR MOMENTUM THEOREMS 



103 



The development is most easily carried out in cylindrical coordinates with 
the z-axis as the axis about which moments are to be taken. The generali- 
zation of theorems (3-140) and (3-141) to this case is then easily proved in 
analogy with the proof given above. This development is left as an exercise. 
As a final generalization of the concept of angular momentum, we define 
the vector angular momentum L about a point as the vector moment of 
the momentum vector about 0: 

L = r X p = m(r x v), (3-142) 

where the vector r is taken from the point as origin to the position of the 
particle of mass m. Again we shall omit the subscript when no confusion 
can arise. The component of the vector L in any direction is the moment 
of the momentum vector p about an axis in that direction through 0. 

By taking the cross product of r with both members of the vector equa- 
tion of motion [Eq. (3-125)], we obtain 

By the rules of vector algebra and vector calculus, 

dL d r , ,, 
It = dt [T * (mv)] 

= r X j t (mv) + v x (mv) 



-■"•(-S)- 



We substitute this result in Eq. (3-143) : 

dL/dt = r X F == N. (3-144) 

The time rate of change of the vector angular momentum of a particle is 
equal to the vector torque acting on it. The integral form of the angular 
momentum theorem is 






(3-145) 



The theorems for plane angular momentum and for angular momentum 
about an axis follow from the vector angular momentum theorems by tak- 
ing components in the appropriate direction. 



104 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

3-9 Discussion of the general problem of two- and three-dimensional 
motion. If the force F is given, in general as a function F(v, r, t) of position, 
velocity, and time, the equations of motion (3-126) become a set of three 
(or, in two dimensions, two) simultaneous second-order differential equa- 
tions: 

d 2 x 
m -Tp = F x (x, y, z, x, y, z, f), 

m § = F v (x, y,z,x,y,z,t), (3-146) 



dP 
dh 



= F z (x, y, z, x, y, z, t). 



If we are given the position r = (x , 2/o, Zo), and the the velocity v = 
(v X0 , v yo , v Z0 ) at any instant t , Eqs. (3-146) give us d 2 r/dt 2 , and from r, f , f , 
at time t, we can determine r, f a short time later or earlier at t + dt, thus 
extending the functions r, f , f, into the past and future with the help of 
Eqs. (3-146). This argument can be made mathematically rigorous, and 
leads to an existence theorem guaranteeing the existence of a unique solu- 
tion of these equations for any given position and velocity at an initial 
instant t . We note that the general solution of Eqs. (3-146) involves the 
six "arbitrary" constants x , yo, z , v xo , v yo , v 20 . Instead of these six con- 
stants, we might specify any other six quantities from which they can be 
determined. (In two dimensions, we will have two second-order differential 
equations and four initial constants.) 

In general, the solution of the three simultaneous equations (3-146) 
will be much more difficult than the solution of the single equation (2-9) 
for one-dimensional motion. The reason for the greater difficulty is that, 
in general, all the variables x, y, z and their derivatives are involved in all 
three equations, which makes the problem of the same order of difficulty 
as a single sixth-order differential equation. [In fact, the set of Eqs. (3-146) 
can be shown to be equivalent to a single sixth-order equation.] If each 
force component involved only the corresponding coordinate and its deriva- 
tives, 

F x = F x (x, x, t), 

F v = F v (y, y, t), (3-147) 

F z = F z {z, z, t), 

then the three equations (3-146) would be independent of one another. 
We could solve for x(t), y{t), z{t) separately as three independent problems 
in one-dimensional motion. The most important example of this case is 



3-9] DISCUSSION OF THE GENERAL PROBLEM 105 

probably when the force is given as a function of time only: 

F = F«) = [F x (t), F v (t), FM- (3-148) 

The x, y, and z equations of motion car} then each be solved separately by 
the method given in Section 2-3. The case of a frictional force propor- 
tional to the velocity will also be an example of the type (3-147). Other 
cases will sometimes occur, for example, the three-dimensional harmonic 
oscillator (e.g., a baseball in a tubful of gelatine, or an atom in a crystal 
lattice), for which the force is 

F y = —k y y, (3-149) 

F z = —k z z, 

when the axes are suitably chosen. The problem now splits into three 
separate linear harmonic oscillator problems in x, y, and z. In most cases, 
however, we are not so fortunate, and Eq. (3-147) does not hold. Special 
methods are available for solving certain classes of two- and three-dimen- 
sional problems. Some of these will be developed in this chapter. Prob- 
lems not solvable by such methods are always, in principle, solvable by 
various numerical methods of integrating sets of equations like Eqs. (3-146) 
to get approximate solutions to any required degree of accuracy. Such 
methods are even more tedious in the three-dimensional case than in the 
one-dimensional case, and are usually impractical unless one has the 
services of one of the large automatic computing machines. 

When we try to extend the idea of potential energy to two or three 
dimensions, we will find that having the force given as F(r), a function of 
r alone, is not sufficient to guarantee the existence of a potential-energy 
function V(t). In the one-dimensional case, we found that if the force is 
given as a function of position alone, a potential-energy function can al- 
ways be defined by Eq. (2-41). Essentially, the reason is that in one 
dimension, a particle which travels from xi to x 2 and returns to X\ must 
return by the same route, so that if the force is a function of position alone, 
the work done by the force on the particle during its return trip must nec- 
essarily be the same as that expended against the force in going from xi 
to x 2 . In three dimensions, a particle can travel from rj to r 2 and return 
by a different route, so that even if F is a function of r, the particle may be 
acted on by a different force on the return trip and the work done may 
not be the same. In Section 3-12 we shall formulate a criterion to determine 
when a potential energy V(t) exists. 

When V(r) exists, a conservation of energy theorem still holds, and the 
total energy (T + V) is a constant of the motion. However, whereas in 



106 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

one dimension the energy integral is always sufficient to enable us to solve 
the problem at least in principle (Section 2-5), in two and three dimensions 
this is no longer the case. If x is the only coordinate, then if we know a 
relation (T + V = E) between x and x, we can solve for x = f(x) and 
reduce the problem to one of carrying out a single integration. But with 
coordinates x, y, z, one relation between x, y, z, x, y, z is not enough. We 
would need to know five such relations, in general, in order to eliminate, for 
example, x, y, x, and y, and find z = f(z). In the two-dimensional case, we 
would need three relations between x, y, x, y to solve the problem by this 
method. To find four more relations like the energy integral from Eqs. 
(3-146) (or two more in two dimensions) is hopeless in most cases. In 
fact, such relations do not usually exist. Often, however, we can find other 
quantities (e.g., the angular momentum) which are constants of the motion, 
and thus obtain one or two more relations between x, y, z, x, y, z, which in 
many cases will be enough to allow a solution of the problem. Examples 
will be given later. 

3-10 The harmonic oscillator in two and three dimensions. In this 
section and the next, we consider a few simple problems in which the force 
has the form of Eqs. (3-147), so that the equations of motion separate into 
independent equations in x, y, and z. Mathematically, we then simply 
have three separate problems, each of the type considered in Chapter 2. 
The only new feature will be the interpretation of the three solutions 
x(t), y(t), z(t) as representing a motion in three-dimensional space. 

We first consider briefly the solution of the problem of the three-dimen- 
sional harmonic oscillator without damping, whose equations of motion 
are 

my = -k v y, (3-150) 

mz = —k z z. 

A model could be constructed by suspending a mass between three per- 
pendicular sets of springs (Fig. 3-27). The solutions of these equations, 
we know from Section 2-8: 

x = A x cos (a x t + 6 X ), o x = k x /m, 

y = A y cos (o)yt + By), w| = ky/m, (3-151) 

z = A z cos (u z t + e z ), oil = kz/m. 

The six constants (A x , A y , A z , d x , 6 y , d z ) depend on the initial values 
£o, 2/o, zo, %o, Vo, 2o- Each coordinate oscillates independently with simple 



3-10] 



THE HARMONIC OSCILLATOR 



107 




Fig. 3-27. Model of a three-dimensional harmonic oscillator. 

harmonic motion at a frequency depending on the corresponding restoring 
force coefficient, and on the mass. The resulting motion of the particle 
takes place within a rectangular box of dimensions 2A X X 2A y X 2A t 
about the origin. If the angular frequencies w xi u v , w z are commensurable, 
that is, if for some set of integers (n x , n^, n x ), 



n x 



U>y 



Ml 



(3-152) 



then the path of the mass m in space is closed, and the motion is periodic. 
If (n m n y , n a ) are chosen so that they have no common integral factor, 
then the period of the motion is 



t = 



27171, 



2irn y 



2-irrit 



W,T 



(3-153) 



During one period, the coordinate x makes n x oscillations, the coordinate y 
makes % oscillations, and the coordinate z makes n* oscillations, so that 
the particle returns at the end of the period to its initial position and 
velocity. In the two-dimensional case, if the path of the oscillating parti- 
cle is plotted for various combinations of frequencies^ and «„, and various 
phases B x and 9 V , many interesting and beautiful patterns are obtained. 
Such patterns are called Lissajous figures (Fig. 3-28), and may be pro- 
duced mechanically by a mechanism designed to move a pencil or other 
writing device according to Eqs. (3-151). Similar patterns may be ob- 
tained electrically on a cathode-ray oscilloscope by sweeping horizontally 



108 MOTION OF PABTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3 




Fig. 3-28. Lissajous figures. 



and vertically with suitable oscillating voltages. If the frequencies u x , 
u y , u z are incommensurable, so that Eq. (3-152) does not hold for any set 
of integers, the motion is not periodic, and the path fills the entire box 
2A X X 2A V X 2A Z , in the sense that the particle eventually comes ar- 
bitrarily close to every point in the box. The discussion can readily be 
extended to the cases of damped and forced oscillations in two and three 
dimensions. 

If the three constants k x , k v , k z are all equal, the oscillator is said to be 
isotropic, that is, the same in all directions. In this case, the three fre- 
quencies u x , u y , a z are all equal and the motion is periodic, with each coordi- 
nate executing one cycle of oscillation in a period. The path can be shown 
to be an ellipse, a straight line, or a circle, depending on the amplitudes 
and phases (A x , A y , A z , d x , 6 y , B z ). 

3-11 Projectiles. An important problem in the history of the science of 
mechanics is that of determining the motion of a projectile. A projectile 
moving under the action of gravity near the surface of the earth moves, if 



3 11] PROJECTILES 109 

air resistance is neglected, according to the equation 

d 2 x 
m dj2 = -™#K (3-154) 

where the 2-axis is taken in the vertical direction. In component form: 

d 2 x 
™^f = 0, (3-155) 

»^=0, (3-156) 

d 2 z 
m ^2 = —m- (3-157) 

The solutions of these equations are 

x — x + v xo t, (3-158) 

V = 2/o + » w «, (3-159) 

z = 2 + *>*„* — k< 2 , (3-160) 

or, in vector form, 

r = r + v t - igt 2 k. (3-161) 

We assume the projectile starts from the origin (0, 0, 0), with its initial 
velocity in the as-plane, so that v yo = 0. This is no limitation on the 
motion of the projectile, but merely corresponds to a convenient choice of 
coordinate system. Equations (3-158), (3-159), (3-160) then become 

* = Vx t, (3-162) 

V = 0, (3-163) 

z = Vzo t — \gt 2 . (3-164) 

These equations give a complete description of the motion of the projectile. 
Solving the first equation for t and substituting in the third, we have an 
equation for the path in the zz-plane: 

2 = — * -i-f-z 2 . (3-165) 

This can be rewritten in the form 



110 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

This is a parabola, concave downward, whose maximum altitude occurs at 

2 



v 
2g 



z m = 7~' (3-167) 



and which crosses the horizontal plane z = at the origin and at the point 

2 JW^o. (3 _ 168) 

9 

If the surface of the earth is horizontal, x m is the range of the projectile. 

Let us now take account of air resistance by assuming a frictional force 
proportional to the velocity: 

m g = _ m?k _ 6 |. ( 3 - 169 ) 

In component notation, if we assume that the motion takes place in the 
a;2-plane, 

m§=-hf t > 0-170) 

m w* = ~ m - 6 i' (3_171) 

It should be pointed out that the actual resistance of the air against a mov- 
ing projectile is a complicated function of velocity, so that the solutions we 
obtain will be only approximate, although they indicate the general nature 
of the motion. If the projectile starts from the origin at t = 0, the solu- 
tions of Eqs. (3-170) and (3-171) are (see Sections 2-4 and 2-6) 



,. _ ,. --Kim 




(3-172) 


x = m^o (1 _ e -M/») f 




(3-173) 


^ = (x + ^) e " 6i/m -?' 




(3-174) 


• - Op* + nr) c 1 - e ~ ulm) 


■mg , 

b *• 


(3-175) 



Solving Eq. (3-173) for t and substituting in Eq. (3-175), we obtain an 
equation for the trajectory: 

■ -(£ + ?*)' -# ta (=r^ta)- «» 76 > 



3-11] 



PROJECTILES HI 



^^rr 



Fig. 3-29. Trajectories for maximum range for projectiles with various muzzle 
velocities. 

For low air resistance, or short distances, when (bx)/(mv Xtt ) « 1, we may 
expand in powers of (bx)/(mv xo ) to obtain 

*-^*-*i* 2 -*-^* 3 ----. (3 _ 177) 

v x v XQ mvt 

Thus the trajectory starts out as a parabola, but for larger values of x 
(taking v Xo as positive), z falls more rapidly than for a parabola. Accord- 
ing to Eq. (3-176), as x approaches the value (mv xo )/b, z approaches minus 
infinity, i.e., the trajectory ends as a vertical drop at x = (mv xo )/b. From 
Eq. (3-174), we see that the vertical fall at the end of the trajectory takes 
place at the terminal velocity —mg/b. (The projectile may, of course, 
return to earth before reaching this part of its trajectory.) If we take the 
first three terms in Eq. (3-177) and solve for x when z = 0, we have ap- 
proximately, if aw « (mv X0 )/b, 

fc ***»-l*£ + .... (3-m) 

The second term gives the first-order correction to the range due to air re- 
sistance, and the first two terms will give a good approximation when the 
effect of air resistance is small. The extreme opposite case, when air 
resistance is predominant in determining range (Fig. 3-29), occurs when 
the vertical drop at x = (mv xo )/b begins above the horizontal plane z = 0. 
The range is then, approximately, 



b \mg / 



mg / < 3 - 179 ) 

We can treat (approximately) the problem of the effect of wind on the 
projectile by assuming the force of air resistance to be proportional to 
the relative velocity of the projectile with respect to the air: 

m d£= -Wl k - b (§ - v ») ' (3-180) 

where v w is the wind velocity. If v„ is constant, the term 6v„, in Eq. (3-180) 



112 MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3 

behaves as a constant force added to -mgk, and the problem is easily 
solved by the method above, the only difference being that there may be 
constant forces in addition to frictional forces in all three directions x, y, z. 
The air resistance to a projectile decreases with altitude, so that a better 
form for the equation of motion of a projectile which rises to appreciable 
altitudes would be 

where h is the height (say about five miles) at which the air resistance falls 
to 1/e of its value at the surface of the earth. In component form, 

mx = —b±e~" h , my = -bye-" h , (3-182) 

ml = —mg — bze~" . 

These equations are much harder to solve. Since z appears in the x and 
y equations, we must first solve the z equation for z(t) and substitute in 
the other two equations. The z equation is not of any of the simple types 
discussed in Chapter 2. The importance of this problem was brought out 
during the First World War, when it was discovered accidentally that 
aiming a cannon at a much higher elevation than that which had previ- 
ously been believed to give maximum range resulted in a great increase in 
the range of the shell. The reason is that the reduction in air resistance, 
at altitudes of several miles, more than makes up for the loss in horizontal 
component of muzzle velocity resulting from aiming the gun higher. 

3-12 Potential energy. If the force F acting on a particle is a function 
of its position r = (x, y, z), then the work done by the force when the 
particle moves from ri to r 2 is given by the line integral 



f 

Jii 



F-dr. 



It is suggested that we try to define a potential energy 7(r) = V(x, y, z) in 
analogy with Eq. (2^11) for one-dimensional motion, as the work done by 
the force on the particle when it moves from r to some chosen standard 
point r g : 

V(i) = - f F(r)-dr. (3-183) 

Such a definition implies, however, that the function V(i) shall be a func- 
tion only of the coordinates (x, y, z) of the point r (and of the standard 
point r s , which we regard as fixed), whereas in general the integral on the 



3 _ 12] POTENTIAL ENERGY 



113 



right depends upon the path of integration from r, to r. Only if the integral 
on the right is independent of the path of integration will the definition 
be legitimate. 

Let us assume that we have a force function F(x, y, z) such that the line 
integral in Eq. (3-183) is independent of the path of integration from r s to 
any point r. The value of the integral then depends only on r (and on r s ), 
and Eq. (3-183) defines a potential energy function V(r). The change in 
V when the particle moves from r to r + dr is the negative of the work 
done by the force F: 

dV = —F'dr. (3-184) 

Comparing Eq. (3-184) with the geometrical definition [Eq. (3-107)] of 
the gradient, we see that 

— F = grad V, 

F « -W. (3 " 185) 

Equation (3-185) may be regarded as the solution of Eq. (3-183) for F in 
terms of V. In component form, 

'--£■ '--%■ '--£■ <MW 

In seeking a condition to be satisfied by the function F(r) in order that 
the integral in Eq. (3-183) be independent of the path, we note that, since 
Eq. (3-28) can be proved from the algebraic definition of the cross product, 
it must hold also for the vector symbol V: 

V X V = 0. (3-187) 

Applying (V X V) to the function V, we have 

V X W = curl (grad 7) = 0. (3-188) 

Equation (3-188) can readily be verified by direct computation. From 
Eqs. (3-188) and (3-185), we have 

V X F = curl F = 0. (3-189) 

Since Eq. (3-189) has been deduced on the assumption that a potential 
function exists, it represents a necessary condition which must be satisfied 
by the force function F(x, y, z) before a potential function can be defined. 
We can show that Eq. (3-189) is also a sufficient condition for the existence 
of a potential by making use of Stokes' theorem [Eq. (3-117)]. By Stokes' 
theorem, if we consider any closed path C in space, the work done by the 



114 MOTION OF PARTICLE IN TWO OE THREE DIMENSIONS [CHAP. 3 




*2 



ri 
Fig. 3-30. Two paths between n and r 2 , forming a closed path. 

force F(r) when the particle travels around this path is 

f F-di = /7n-(V x F) dS, (3-190) 

s 

where S is a surface in space bounded by the closed curve C. If now 
Eq. (3-189) is assumed to hold, the integral on the right is zero, and we 
have, for any closed path. C, 



/« 



F-dt = 0. (3-191) 

c 



But if the work done by the force F around any closed path is zero, then 
the work done in going from r x to r 2 will be independent of the path fol- 
lowed. For consider any two paths between r x and r 2 , and let a closed 
path C be formed going from r x to r 2 by one path and returning to ri 
by the other (Fig. 3-30). Since the work done around C is zero, the work 
going from r x to r 2 must be equal and opposite to that on the return trip, 
hence the work in going from ri to r 2 by either path is the same. Applying 
this argument to the integral on the right in Eq. (3-183), we see that the 
result is independent of the path of integration from r s to r, and therefore 
the integral is a function V(r) of the upper limit alone, when the lower 
limit r, is fixed. Thus Eq. (3-189) is both necessary and sufficient for the 
existence of a potential function V(r) when the force is given as a function 
of position F(r). 

When curl F is zero, we can express the work done by the force when 
the particle moves from r x to r 2 as the difference between the values of the 
potential energy at these points: 

pF.dr= r-F-dt+TF-dr 
Jti Jti Ji, 

= V(ri) - F(r 9 ). (3-192) 

Combining Eq. (3-192) with the energy theorem (3-135), we have for any 
two times ti and Z 2 : 

T x + F(r0 = T 2 + V(r 2 ). (3-193) 



3-12] POTENTIAL ENERGY 115 

Hence the total energy (T + V) is again constant, and we have an energy 
integral for motion in three dimensions: 

T + V = %m(x 2 + y 2 + z 2 ) + V(x, y, z) = E. (3-194) 

A force which is a function of position alone, and whose curl vanishes, is 
said to be conservative, because it leads to the theorem of conservation of 
kinetic plus potential energy [Eq. (3-194)]. 

In some cases, a force may be a function of both position and time 
F(r, t). If at any time t the curl of F(r, t) vanishes, then a potential-energy 
function V(r, t) can be defined as 

V(i,t) = - f F(r,«)-dr, (3-195) 

and we will have, for any time t such that V X F(r, t) = 0, 

F(r, t) = -VF(r, t). (3-196) 

However, the conservation law of energy can no longer be proved, for 
Eq. (3-192) no longer holds. It is no longer true that the change in po- 
tential energy equals the negative of the work done on the particle, for 
the integral which defines the potential energy at time * is computed from 
the force function at that time, whereas the integral that defines the work 
is computed using at each point the force function at the time the particle 
passed through that point. Consequently, the energy T + V is not a 
constant when F and V are functions of time, and such a force is not to 
be called a conservative force. 

When the forces acting on a particle are conservative, Eq. (3-194) 
enables us to compute its speed as a function of its position. The energy E 
is fixed by the initial conditions of the motion. Equation (3-194), like 
Eq. (2-44), gives no information as to the direction of motion. This lack 
of knowledge of direction is much more serious in two and three dimensions, 
where there is an infinity of possible directions, than in one dimension, 
where there are only two opposite directions in which the particle may 
move. In one dimension, there is only one path along which the particle 
may move. In two or three dimensions, there are many paths, and unless 
we know the path of the particle, Eq. (3-194) alone allows us to say very 
little about the motion except that it can occur only in the region where 
V{x, y, z) < E. As an example, the potential energy of an electron in the 
attractive electric field of two protons (ionized hydrogen molecule H 2 + ) is 

2 2 

v = ~ k ~ k ' (esu) (3_197) 



116 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

-18 




Fig. 3-31. Potential energy of electron in electric field of two protons 2 A 
apart. (Potential energies in units of 10 -12 erg.) 

where r\, r 2 are the distances of the electron from the two protons. The 
function V(x, y) (for motion in the x2/-plane only) is plotted in Fig. 3-31 
as a contour map, where the two protons are 2 A apart at the points y = 0, 
x = ±1 A, and the figures on the contours of constant potential energy are 
the corresponding potential energies in units of 10 -12 erg. So long as 
E < —46 X 10~ 12 erg, the electron is confined to a region around one 
proton or the other, and we expect its motion will be either an oscillation 
through the attracting center or an orbit around it, depending on initial 
conditions. (These comments on the expected motion require some physical 
insight or experience in addition to what we can say from the energy integral 
alone.) For > E > —46 X 10 -12 erg, the electron is confined to a 
region which includes both protons, and a variety of motions are possible. 
For E > 0, the electron is not confined to any finite region in the plane. 
For E <<C —46 X 10~ 12 erg, the electron is confined to a region where the 
equipotentials are nearly circles about one proton, and its motion will be 
practically the same as if the other proton were not there. For E < 0, 
but \E\ <5C 46 X 10 -12 erg, the electron may circle in an orbit far from the 
attracting centers, and its motion then will be approximately that of an 
electron bound to a single attracting center of charge 2e, as the equi- 
potential lines far from the attracting centers are again very nearly circles. 



3-12] 



POTENTIAL ENERGY 



117 



Given a potential energy function V(x, y, z), Eq. (3-186) enables us to 
compute the components of the corresponding force at any point. Con- 
versely, given a force F(x, y, z), we may compute its curl to determine 
whether a potential energy function exists for it. If all components of 
curl F are zero within any region of space, then within that region, F may 
be represented in terms of a potential-energy function as —VV. The 
potential energy is to be computed from Eq. (3-183). Furthermore, since 
curl F = 0, the result is independent of the path of integration, and we 
may compute the integral along any convenient path. As an example, 
consider the following two force functions: 



(a) F x = axy, 



Fy = -az 2 , 



F,= - 



ax 



(b) F x = ay(y 2 - 3* 2 ), F y = 3ax(y 2 - z 2 ), F z = ~6axyz, 
where a is a constant. We compute the curl in each case: 



(a) 



1 \dy dz/^ J \dz dx/ + K \dx dyj 



= (2az)i + (2ax)j - (ax)k, 
(b) V X F = 0. 

In case (a) no potential energy exists. In case (b) there is a potential 
energy function, and we proceed to find it. Let us take r„ = 0, i.e., take 
the potential as zero at the origin. Since the components of force are given 
as functions of x, y, z, the simplest path of integration from (0, 0, 0) to 
(#0, Vo, zo) along which to compute the integral in Eq. (3-183) is one which 
follows lines parallel to the coordinate axes, for example as shown in 
Fig. 3-32: 

V(x ,y ,z ) = — f xo,Vo ' z ° p-dr = — f F-di — f F-dr — f F-dr. 

J(0,0,0) JCi Jc 2 Jc 3 





( 


•(xo,yo,zo) 




, 


c 3 




(0,0,0) 




y 






(nU)/ 


c 2 


Oo,W>,0) 



Fig. 3-32. A path of integration from (0, 0, 0) to (xo, Vo, «o). 



118 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

Now along Ci, we have 

y = z = 0, F x = F y = F z = 0, dr = i da;. 

Thus 

/" F-dr = f x °F x dx = 0. 

JCi JO 

Along C 2 , 











a; — X0) 


z = 


o, 




Thus 




F* 


= ay 


3 , Fy = 

dr = 


Zax Q y' 
\dy. 


, F,= 


= 0, 


J. 11U.O 

Along 


c 3) 




Jc 2 


F-dr = / 
Jo 

X = Xq, 


Fydy 
y = 


= ax y%- 

yo, 




F x 


= 02/0(2/0 


— 


3* 2 ), 


Fy — Sax (yl 


-z\ 


F z 


Thus 








dr = 


kdz. 




_2 



-6axo2/oz, 



f F«dr = f " F z dz = —Zax yoZ 2 . 
JCs Jo 

Thus the potential energy, if the subscript zero is dropped, is 

V(x, y, z) = —axy 3 + Saxyz 2 . 

It is readily verified that the gradient of this function is the force given by 
(b) above. In fact, one way to find the potential energy, which is often 
faster than the above procedure, is simply to try to guess a function whose 
gradient will give the required force. 

An important case of a conservative force is the central force, a force 
directed always toward or away from a fixed center O, and whose magni- 
tude is a function only of the distance from 0. In spherical coordinates, 
with as origin, 

F = nF(r). (3-198) 

The cartesian components of a central force are (since n = r/r) 

F x = ^F(r), 

F v = y - F(x), [r = (x 2 + y 2 + z 2 ) 112 ], (3-199) 

T 

F z = *-F(r). 



3-12] 



POTENTIAL ENERGY 



119 



(ro,8o,<po) 



(r,,6 s ,ip s ) 





Fig. 3-33. Path of integration for a central force. 

The curl of this force can be shown by direct computation to be zero, no 
matter what the function F(r) may be. For example, we find 

*** = x d ( F ( r A dr _ X V d (F(r)\ 
dy dr\ r ) by r dr\ r ) ' 

= d (F(r)\ dr ^ xy d (F(r)\ 
dr\ r ) dx r dr\ r ) 



111 
dx 



Therefore the g-component of curl F vanishes, and so, likewise, do the other 
two components. To compute the potential energy, we choose any stand- 
ard point r s , and integrate from r 8 to r along a path (Fig. 3-33) following 
a radius (Ci) from i„ whose coordinates are (r„ 0„ <p,), to the point 
0"o, 0«, <P»), then along a circle (C 2 ) of radius r about the origin to the 
point (r , e 0) <po). Along C u 

dr = n dr, 



Along C 2 , 



Thus 



f F-dr = f ° F(r) dr. 

JCi Jr, 

dt = ]r dd + tor sin d<p, 

f F-dr = 0. 
Jc 2 



V(to) = - r ?-dr = - f F-dr - f F- 

Jr, JCi JC 2 

= - f° F(r) dr. 

The potential energy is a function of r alone: 

V(t) = V(r) = - f F(r) dr. 

Jr, 



dr 



(3-200) 



120 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

3-13 Motion under a central force. A central force is a force of the form 
given by Eq. (3-198). Physically, such a force represents an attraction 
[if F(r) < 0] or repulsion [if F(r) > 0] from a fixed point located at the 
origin r = 0. In most cases where two particles interact with each other, 
the force between them is (at least primarily) a central force; that is, if 
either particle be located at the origin, the force on the other is given by 
Eq. (3-198). Examples of attractive central forces are the gravitational 
force acting on a planet due to the sun, or the electrical attraction acting 
on an electron due to the nucleus of an atom. The force between a proton 
or an alpha particle and another nucleus is a repulsive central force. In 
the most important cases, the force F(r) is inversely proportional to r 2 . 
This case will be treated in the next section. Other forms of the function 
F(r) occur occasionally; for example, in some problems involving the struc- 
ture and interactions of nuclei, complex atoms, and molecules. In this 
section, we present the general method of attack on the problem of a par- 
ticle moving under the action of a central force. 

Since in all these examples, neither of the two interacting particles 
is actually fastened to a fixed position, the problem we are solving, like 
most problems in physics, represents an idealization of the actual problem, 
valid when one of the particles can be regarded as practically at rest at 
the origin. This will be the case if one of the particles is much heavier 
than the other. Since the forces acting on the two particles have the same 
magnitude by Newton's third law, the acceleration of the heavy one will 
be much smaller than that of the lighter one, and the motion of the heavy 
particle can be neglected in comparison with the motion of the lighter one. 
We shall discover later, in Section 4-7, that, with a slight modification, our 
solution can be made to yield an exact solution to the problem of the 
motion of two interacting particles, even when their masses are equal. 

We may note that the vector angular momentum of a particle under 
the action of a central force is constant, since the torque is 

N = rxF=(rx n)F(r) = 0. (3-201) 

Therefore, by Eq. (3-144), 

f = 0. (3-202) 

As a consequence, the angular momentum about any axis through the 
center of force is constant. It is because many physical forces are central 
forces that the concept of angular momentum is of importance. 

In solving for the motion of a particle acted on by a central force, we 
first show that the path of the particle lies in a single plane containing the 
center of force. To show this, let the position r and velocity v<> be given 
at any initial time t , and choose the rc-axis through the initial position r 



3-13] MOTION UNDER A CENTRAL FORCE 121 

of the particle, and the z-axis perpendicular to the initial velocity v . Then 
we have initially: 

*o = jr |, 2/o = zo = 0, (3-203) 

Vx = v -i, v yo = v -j, v ls> = 0. (3-204) 

The equations of motion in rectangular coordinates are, by Eqs. (3-199), 

mx = j F(r), my = ^ F(r), mz = - F(r). (3-205) 

A solution of the z-equation which satisfies the initial conditions on z and 
Vz is 

z(t) = 0. (3-206) 

Hence the motion takes place entirely in the a;j/-plane. We can see phys- 
ically that if the force on a particle is always toward the origin, the particle 
can never acquire any component of velocity out of the plane in which it 
is initially moving. We can also regard this result as a consequence of 
the conservation of angular momentum. By Eq. (3-202), the vector 
L — m(i X v) is constant; therefore both r and v must always lie in a 
fixed plane perpendicular to L. 

We have now reduced the problem to one of motion in a plane with two 
differential equations and four initial conditions remaining to be satisfied. 
If we choose polar coordinates r, in the plane of the motion, the equations 
of motion in the r and directions are, by Eqs. (3-80) and (3-198), 

mr — mrd 2 = F(r), (3-207) 

mr8 + 2mr& = 0. (3-208) 

Multiplying Eq. (3-208) by r, as in the derivation of the (plane) angular 
momentum theorem, we have 

it ^» = d -tt = °- ( 3 - 209 > 

This equation expresses the conservation of angular momentum about the 
origin and is a consequence also of Eq. (3-202) above. It may be inte- 
grated to give the angular momentum integral of the equations of motion: 

mr 2 d = L = a constant. (3-210) 

The constant L is to be evaluated from the initial conditions. Another 
integral of Eqs. (3-207) and (3-208), since the force is conservative, is 

T + V = Imr* + imr 2 6 2 + V(r) = E, (3-211) 



122 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

where V(r) is given by Eq. (3-200) and E is the energy constant, to be 
evaluated from the initial conditions. If we substitute for 6 from Eq. 
(3-210), the energy becomes 

hmf 2 + £2 + V(r) = E. (3-212) 

We can solve for f: 



^I^-^-SfT- (3 - 2i3) 



Therefore 



/ 



dr = Jft- t. (3-214) 



/ / t2 \i/2 -\ m 

[ [ E - ™ - £^) 

The integral is to be evaluated and the resulting equation solved for r(t). 
We then obtain 0(0 from Eq. (3-210): 

' = '« + /„' Ji dt - < 3 - 215 > 

We thus obtain the solution of Eqs. (3-207) and (3-208) in terms of the 
four constants L, E, r , do, which can be evaluated when the initial position 
and velocity in the plane are given. 

It will be noted that our treatment based on Eq. (3-212) is analogous 
to our treatment of the one-dimensional problem based on the energy in- 
tegral [Eq. (2-44)]. The coordinate r here plays the role of x, and the 6 
term in the kinetic energy, when 6 is eliminated by Eq. (3-210), plays the 
role of an addition to the potential energy. We may bring out this analogy 
further by substituting from Eq. (3-210) into Eq. (3-207): 

mf - -^ = F(r). (3-216) 

If we transpose the term —L 2 /mr 3 to the right side, we obtain 

mr = F(r)+£- 3 - (3-217) 

This equation has exactly the form of an equation of motion in one dimen- 
sion for a particle subject to the actual force F(r) plus a "centrifugal force" 
L 2 /mr 3 . The centrifugal force is not really a force at all but a part of the 
mass times acceleration, transposed to the right side of the equation in 
order to reduce the equation for r to an equation of the same form as for 
one-dimensional motion. We may call it a "fictitious force." If we treat 



3-13] MOTION UNDER A CENTRAL FORCE 123 

Eq. (3-217) as a problem in one-dimensional motion, the effective "poten- 
tial energy" corresponding to the "force" on the right is 

>V>(r) = -f m dr-f£.dr 



The second term in 'V is the "potential energy" associated with the 
"centrifugal force." The resulting energy integral is just Eq. (3-212). 
The reason why we have been able to obtain a complete solution to our 
problem based on only two integrals, or constants of the motion (L and E), 
is that the equations of motion do not contain the coordinate 0, so that the 
constancy of L is sufficient to enable us to eliminate entirely from Eq. 
(3-207) and to reduce the problem to an equivalent problem in one-dimen- 
sional motion. 

The integral in Eq. (3-214) sometimes turns out rather difficult to eval- 
uate in practice, and the resulting equation difficult to solve for r(t). It is 
sometimes easier to find the path of the particle in space than to find its 
motion as a function of time. We can describe the path of the particle by 
giving r(0). The resulting equation is somewhat simpler if we make the 
substitution 

(3-219) 











1 

u = - 1 
r 


1 

r = -• 
u 




Then 


we 


have, 


using 
r 


Eq. (3-210), 

_ 1 du , 
«2 dd 

_ L du 
m de 


= -r 2 6^ 

de 










f 


L d 2 u * 
~ m~cW 


r2„,2 
L u 


d 2 u 

de 2 



(3-220) 



(3-221) 

Substituting for r and f in Eq. (3-217), and multiplying by —m/(L 2 u 2 ), we 
have a differential equation for the path or orbit in terms of u(fi) : 

In case L = 0, Eq. (3-222) blows up, but we see from Eq. (3-210) that in 
this case 6 is constant, and the path is a straight line through the origin. 



124 MOTION OF PARTICLE IN TWO OK THBEE DIMENSIONS [CHAP. 3 

Even in cases where the explicit solutions of Eqs. (3-214) and (3-215), 
or Eq. (3-222), are difficult to carry through, we can obtain qualitative 
information about the r motion from the effective potential 'V given by 
Eq. (3-218), just as in the one-dimensional case discussed in Section 2-5. 
By plotting 'V'(r), we can decide for any total energy E whether the 
motion in r is periodic or aperiodic, we can locate the turning points, and 
we can describe roughly how the velocity f varies during the motion. If 
'V'(r) has a minimum at a point r , then for energy E slightly greater than 
'V (r ), r may execute small, approximately harmonic oscillations about r 
with angular frequency given by 



CO 



_ 1 (d 2 'V'\ 
m \ dr 2 / 



2 = ^ ^ • (3-223) 



[See the discussion in Section 2-7 concerning Eq. (2-87).] We must re- 
member, of course, that at the same time the particle is revolving around 
the center of force with an angular velocity 

i = ~ (3-224) 

mr 2 

The rate of revolution decreases as r increases. When the r motion is 
periodic, the period of the r motion is not, in general, the same as the 
period of revolution, so that the orbit may not be closed, although it is 
confined to a finite region of space. (See Fig. 3-34.) In cases where the 
r motion is not periodic, then 6 — > as r — * oo , and the particle may or 
may not perform one or more complete revolutions as it moves toward 
r = oo , depending on how rapidly r increases. In the event the motion is 
periodic, that is, when the particle moves in a closed orbit, the period of 
orbital motion is related to the area of the orbit. This can be seen as 
follows. The area swept out by the radius from the origin to the particle 




Fig. 3-34. An aperiodic bounded orbit. 



3~14] INVERSE SQUARE LAW FORCE 125 




Fig. 3-35. Area swept out by radius vector. 

when the particle moves through a small angle dd is approximately 
(Fig. 3-35) 

dS = %r 2 dd. (3-225) 

Hence the rate at which area is swept out by the radius is, by Eq. (3-210), 

^12/ L 

Tt = ^ 6 = 2^- ( 3 - 22 6) 

This result is true for any particle moving under the action of a central 
force. If the motion is periodic, then, integrating over a complete period r 
of the motion, we have for the area of the orbit 

S = 2^ * < 3 - 227 ) 

If the orbit is known, the period of revolution can be calculated from this 
formula. 

3-14 The central force inversely proportional to the square of the 
distance. The most important problem in three-dimensional motion is 
that of a mass moving under the action of a central force inversely pro- 
portional to the square of the distance from the center: 

F = f 2 n, (3-228) 

for which the potential energy is 

V(r) = j , (3-229) 

where the standard radius r s is taken to be infinite in order to avoid an 
additional constant term in V(r). As an example, the gravitational force 
(Section 1-5) between two masses mi and m 2 a distance r apart is given by 
Eq. (3-228) with 

K = —Gm y m 2 , G = 6.67 X 10 -8 dyne-gm -2 -cm 2 , (3-230) 

where K is negative, since the gravitational force is attractive. Another 



126 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 



•V\r) 



-\{Khn/V) 




Fig. 3-36. Effective potential for central inverse square law of force. 



example is the electrostatic force between two electric charges gi and q 2 a 
distance r apart, given by Eq. (3-228) with 



K = qiq 2 , 



(3-231) 



where the charges are in electrostatic units, and the force is in dynes. The 
electrostatic force is repulsive when qi and q 2 have the same sign, other- 
wise attractive. Historically, the first problems to which Newton's me- 
chanics was applied were problems involving the motion of the planets 
under the gravitational attraction of the sun, and the motion of satellites 
around the planets. The success of the theory in accounting for such 
motions was responsible for its initial acceptance. 

We first determine the nature of the orbits given by the inverse square 
law of force. In Fig. 3-36 is plotted the effective potential 



'V'(r) = — + -=—■ 



(3-232) 



For a repulsive force (K > 0), there are no periodic motions in r; only 
positive total energies E are possible, and the particle comes in from r = oo 
to a turning point and travels out to infinity again. For a given energy 
and angular momentum, the turning point occurs at a larger value of r than 
for K = (no force"), for which the orbit would be a straight line. For an 
attractive force (K < 0) with L^O, the motion is also unbounded if 
E > 0, but in this case the turning point occurs at a smaller value of r than 
for K = 0. Hence the orbits are as indicated in Fig. 3-37. The light 



3-14] 



INVEBSE SQUABE LAW FOBCE 



127 




Fig. 3-37. Sketch of unbounded inverse square law orbits. 



lines in Fig. 3-37 represent the turning point radius or perihelion distance 
measured from the point of closest approach of the particle to the attracting 
or repelling center. For K < 0, and -\K 2 m/L 2 < E < 0, the coordinate 
r oscillates between two turning points. For.E= —\K 2 m/h 2 , the particle 
moves in a circle of radius r = L 2 /(—Km). Computation shows (see 
Problem 30 at the end of this chapter) that the period of small oscillations 
in r is the same as the period of revolution, so that for E near — \K 2 m/L 2 , 
the orbit is a closed curve with the origin slightly off center. We shall show 
later that the orbit is, in fact, an ellipse for all negative values of E if 
L ?* 0. If L = 0, the problem reduces to the one-dimensional motion of 
a falling body, discussed in Section 2-6. 

To evaluate the integrals in Eqs. (3-214) and (3-215) for the inverse 
square law of force is rather laborious. We shall find that we can obtain all 
the essential information about the motion more simply by starting from 
Eq. (3-222) for the orbit. Equation (3-222) for the orbit becomes, in this 
case, 

d 2 u . mK 

-j^ + u= - — ■ (3-233) 



d0 2 



L2 



This equation has the same form as that of a harmonic oscillator (of unit 
frequency) subject to a constant force, where B here plays the role of t. 
The homogeneous equation and its general solution are 



d 2 u . 



(3-234) 



u = A cos (0 — d ), 



(3-235) 



128 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

where A, O are arbitrary constants. An obvious particular solimon of the 
inhomogeneous equation (3-233) is the constant solution 

„=_*££ (3-236) 

Hence the general solution of Eq. (3-233) is 

u = - = - ^ + A cos (0 - O ). (3-237) 

r L 2 

This is the equation of a conic section (ellipse, parabola, or hyperbola) with 
focus at r = 0, as we shall presently show. The constant O determines 
the. orientation of the orbit in the plane. The constant A, which may be 
taken as positive (since O is arbitrary), determines the turning points of 
the r motion, which are given by 

1 = _^ + A) ! = _^_ A . (3-238) 

If A > —mK/L 2 (as it necessarily is for K > 0), then there is only one 
turning point, ri, since r cannot be negative. We cannot have A < mK/L 2 , 
since r could then not be positive for any value of 0. For a given E, the 
turning points are solutions of the equation 



<F W = 7 + & = ^ (3 " 239) 



The solutions are 



J^ mK \(?nK\ 2 , 2mffl 1/2 

f! _ L2 + L\ L2 / + L2 J 

1 mK_ \(mK\ 2 , 2mE] 112 

r 2 ~ V L\ L* ) + L 2 J 



(3-240) 



Comparing Eq. (3-238) with Eq. (3-240), we see that the value of A in 
terms of the energy and angular momentum is given by 

A2== ^ + 2gtf. (3 _ 241) 

The orbit is now determined in terms of the initial conditions. 

An ellipse is defined as the curve traced by a particle moving so that 
the sum of its distances from two fixed points F, F' is constant.* The 



* For a more detailed treatment of conic sections, see W. F. Osgood and 
W. C. Graustein, Plane and Solid Analytic Geometry. New York: Macmillan, 
1938. (Chapters 6, 7, 8, 10.) 



3-1 4] INVERSE SQUARE LAW FORCE 



129 




Fig. 3-38. Geometry of the ellipse. 



points F, F' are called the foci of the ellipse. Using the notation indicated 
in Fig. 3-38, we have 

r' + r = 2a, (3-242) 

where a is half the largest diameter (major axis) of the ellipse. In terms 
of polar coordinates with center at the focus F and with the negative z-axis 
through the focus F', the cosine law gives 

r' 2 = r 2 + 4a 2 e 2 + 4rae cos 0, (3-243) 

where ae is the distance from the center of the ellipse to the focus, e is 
called the eccentricity of the ellipse. If e = 0, the foci coincide and the 
ellipse is a circle. As € -* 1, the ellipse degenerates into a parabola or 
straight line segment, depending on whether the focus F' recedes to in- 
finity or remains a finite distance from F . Substituting r' from Eq. (3-242) 
in Eq. (3-243), we find 

r= l + ecosV < 3 " 244 ) 

This is the equation of an ellipse in polar coordinates with the origin at one 
focus. If b is half the smallest diameter (minor axis), we have, from 
Fig. 3-38, 

b = a(l- e 2 ) 1 ' 2 . (3-245) 

The area of the ellipse can be obtained in a straightforward way by in- 
tegration : 

S = wab. (3-246) 

A hyperbola is denned as the curve traced by a particle moving so that 
the difference of its distances from two fixed foci F, F' is constant (Fig. 
3-39). A hyperbola has two branches defined by 

r' — r = 2a (+ branch), 

(3-247) 
r — r = —2a (— branch). 



130 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

D 




+branch 



-branch 




Fig. 3-39. Geometry of the hyper- 
bola. 



Fig. 3-40. Geometry of the para- 
bola. 



We shall call the branch which encircles F the + branch (left branch in the 
figure), and the branch which avoids F, the — branch (right branch in the 
figure). Equation (3-243) holds also for the hyperbola, but the eccen- 
tricity e is now greater than one. The equation of the hyperbola becomes 
in polar coordinates: 

«(c 2 - 1) 



±1 + CCOS0 



(3-248) 



(The + sign refers to the + branch, the — sign to the — branch.) The 
asymptotes of the hyperbola (dotted lines in Fig. 3-39) make an angle a 
with the axis through the foci, where a is the value of for which r is infinite : 



cos a = ± 



1 



(3-249) 



A parabola is the curve traced by a particle moving so that its distance 
from a fixed line D (the directrix) equals its distance from a fixed focus F. 
From Fig. 3-40, we have 

a 
1 + cos ' (3-250) 



r = 



where a is the distance from the focus F to the directrix D. 

We can write the equations for all three conic sections in the standard 
form 



- = B + A cos 0, 
r 



(3-251) 



3_14 1 INVERSE SQUARE LAW FORCE 131 

where A is positive, and B and A are given as follows: 
B > A, ellipse, 

B = «(1 - «*) ' ^ = a( i - e2) J (3-252) 



B = A, parabola, 



* = £• ^ = £; (3-253) 



< B < A, hyperbola, + branch, 

—A<B<0, hyperbola, — branch, 

*= -«(</_ D' A = a(e2 *_ iy (3-255) 

The case B < —A cannot occur, since r would then not be positive for any 
value of 6. If we allow an arbitrary orientation of the curve with respect 
to the z-axis, then Eq. (3-251) becomes 

- = B + A cos (0 - 6 ), (3-256) 

where 6 is the angle between the z-axis and the line from the origin to the 
perihelion (point of closest approach of the curve to the origin). It will 
be noted that in all cases 

€ = -\Bj- (3-267) 



For an ellipse or hyperbola, 

B 



a = 



A* - £2 



(3-258) 



Equation (3-237) for the orbit of a particle under an inverse square law 
force has the form of Eq. (3-256) for a conic section, with [if we use 



Eq. (3-241)] 






-(■ 



(3-259) 

* + «!)"■. 



132 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

The eccentricity of the orbit, by Eq. (3-257), is 

--( i + 2 S) m - < 3 - 26o) 

For an attractive force (K < 0), the orbit is an ellipse, parabola, or hyper- 
bola, depending on whether E < 0, E = 0, or E > 0; if a hyperbola, it is 
the + branch. For a repulsive force (K > 0), we must have E > 0, and 
the orbit can only be the — branch of a hyperbola. These results agree 
with our preliminary qualitative discussion. For elliptic and hyperbolic 
orbits, the semimajor axis a is given by 



a = 



K 



2E 



(3-261) 



It is curious that this relation does not involve the eccentricity or the angu- 
lar momentum; the energy E depends only on the semimajor axis a, and 
vice versa. Equations (3-260) and (3-261) may be obtained directly from 
Eq. (3-239) for the turning points of the r motion. If we solve this equa- 
tion for r, we obtain the turning points 



ri - 2 = 2E ± 



[(0+£eT- ^ 



The maximum and minimum radii for an ellipse are 

r li2 = a(l ± e), (3-263) 

and the minimum radius for a hyperbola is 

ri = a(e =F 1), (3-264) 

where the upper sign is for the + branch and the lower sign for the — 
branch. Comparing Eqs. (3-263) and (3-264) with Eq. (3-262), we can 
read off the values of a and e. Thus if we know that the path is an ellipse 
or hyperbola, we can find the size and shape from Eq. (3-239), which fol- 
lows from the simple energy method of treatment, without going through 
the exact solution of the equation for the orbit. This is a useful point to 
remember. 

3-15 Elliptic orbits. The Kepler problem. Early in the seventeenth 
century, before Newton's discovery of the laws of motion, Kepler an- 
nounced the following three laws describing the motion of the planets, de- 
duced from the extensive and accurate observations of planetary motions 
by Tycho Brahe: 



3-1 5J ELLIPTIC ORBITS. THE KEPLER PROBLEM 133 

(1) The planets move in ellipses with the sun at one focus. 

(2) Areas swept out by the radius vector from the sun to a planet in 
equal times are equal. 

(3) The square of the period of revolution is proportional to the cube 
of the semimajor axis. 

The second law is expressed by our Eq. (3-226), and is a consequence of 
the conservation of angular momentum; it shows that the force acting on 
the planet is a central force. The first law follows, as we have shown, 
from the fact that the force is inversely proportional to the square of the 
distance. The third law follows from the fact that the gravitational force 
is proportional to the mass of the planet, as we now show. 

In the case of an elliptical orbit, we can find the period of the motion 
from Eqs. (3-227) and (3-246): 



2m . 1m 2/1 2sl/2 (ir 2 K 2 m\ 112 

r = ^ r 7rab = - r ^(l-er' 2 ={^^) , (3-265) 



or, using Eq. (3-261), 



T a = 4*V|£ 



(3-266) 



In the case of a small body of mass m moving under the gravitational at- 
traction [Eq. (3-230)] of a large body of mass M, this becomes 

2 47T 2 s 

T = MG a - (3 ~ 267 ) 

The coefficient of a 3 is now a constant for all planets, in agreement with 
Kepler's third law. Equation (3-267) allows us to "weigh" the sun, if we 
know the value of G, by measuring the period and major axis of any plane- 
tary orbit. This has already been worked out in Chapter 1, Problem 9, 
for a circular orbit. Equation (3-267) now shows that the result applies 
also to elliptical orbits if the semimajor axis is substituted for the radius. 
We have shown that Kepler's laws follow from Newton's laws of motion 
and the law of gravitation. The converse problem, to deduce the law of 
force from Kepler's laws and the law of motion, is an easier problem, and a 
very important one historically, for it was in this way that Newton de- 
duced the law of gravitation. We expect that the motions of the planets 
should show slight deviations from Kepler's laws, in view of the fact that 
the central force problem which was solved in the last section represents an 
idealization of the actual physical problem. In the first place, as pointed 
out in Section 3-13, we have assumed that the sun is stationary, whereas 
actually it must wobble slightly due to the attraction of the planets going 



134 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

around it. This effect is very small, even in the case of the largest planets, 
and can be corrected for by the methods explained later in Section 4-7. 
In the second place, a given planet, say the earth, is acted on by the gravi- 
tational pull of the other planets, as well as by the sun. Since the masses 
of even the heaviest planets are only a few percent of the mass of the sun, 
this will produce small but measurable deviations from Kepler's laws. The 
expected deviations can be calculated, and they agree with the very pre- 
cise astronomical observations. In fact, the planets Neptune and Pluto 
were discovered as a result of their effects on the orbits of the other planets. 
Observations of the planet Uranus for about sixty years after its discovery 
in 1781 showed unexplained deviations from the predicted orbit, even after 
corrections were made for the gravitational effects of the other known 
planets. By a careful and elaborate mathematical analysis of the data, 
Adams and Leverrier were able to show that the deviations could be ac- 
counted for by assuming an unknown planet beyond Uranus, and they cal- 
culated the position of the unknown planet. The planet Neptune was 
promptly discovered in the predicted place. 

The orbits of the comets, which are occasionally observed to move in 
around the sun and out again, are, at least in some cases, very elongated 
ellipses. It is not at present known whether any of the comets come from 
beyond the solar system, in which case they would, at least initially, have 
parabolic or hyperbolic orbits. Even those comets whose orbits are known 
to be elliptical have rather irregular periods due to the perturbing gravita- 
tional pull of the larger planets near which they occasionally pass. Be- 
tween close encounters with the larger planets, a comet will follow fairly 
closely a path given by Eq. (3-256), but during each such encounter, its 
motion will be disturbed, so that afterwards the constants A, B, and 
will have values different from those before the encounter. 

As noted in Section 3-13, we expect in general that the bounded orbits 
arising from an attractive central force F(r) will not be closed (Fig. 3-34). 
Closed orbits (except for circular orbits) arise only where the period of 
radial oscillations is equal to, or is an exact rational multiple of, the period 
of revolution. Only for certain special forms of the function F(r), of which 
the inverse square law is one, will the orbits be closed. Any change in the 
inverse square law, either a change in the exponent of r or an addition to 
F{r) of a term not inversely proportional to r 2 , will be expected to lead to 
orbits that are not closed. However, if the change is very small, then the 
orbits ought to be approximately elliptical. The period of revolution will 
then be only slightly greater or slightly less than the period of radial oscilla- 
tions, and the orbit will be approximately an ellipse whose major axis 
rotates slowly about the center of force. As a matter of fact, a slow pre- 
cession of the major axis of the orbit of the planet Mercury has been ob- 
served, with an angular velocity of 41 seconds of arc per century, over and 



3-16] HYPERBOLIC ORBITS. THE RUTHERFORD PROBLEM 



135 



above the perturbations accounted for by the gravitational effects of the 
other planets. It was once thought that this could be accounted for by 
the gravitational effect of dust in the solar system, but it can be shown 
that the amount of dust is far too small to account for the effect. It is 
now fairly certain that the effect is due to slight corrections to Newton's 
theory of planetary motion required by the theory of relativity.* 

The problem of the motion of electrons around the nucleus of an atom 
would be the same as that of the motion of planets around the sun, if 
Newtonian mechanics were applicable. Actually, the motion of electrons 
must be calculated from the laws of quantum mechanics. Before the dis- 
covery of quantum mechanics, Bohr was able to give a fair account of the 
behavior of atoms by assuming that the electrons revolve in orbits given 
by Newtonian mechanics. Bohr's theory is still useful as a rough picture 
of atomic structure, f 

3-16 Hyperbolic orbits. The Rutherford problem. Scattering cross 
section. The hyperbolic orbits are of interest in connection with the mo- 
tion of particles around the sun which may come from or escape to outer 
space, and also in connection with the collisions of two charged particles. 
If a light particle of charge q t encounters a heavy particle of charge q 2 at 
rest, the light particle will follow a hyperbolic trajectory past the heavy 
particle, according to the results obtained in Section 3-14. In the case 
of collisions of atomic particles, the region in which the trajectory bends 
from one asymptote to the other is very small (a few angstrom units or 
less), and what is observed is the deflection angle © = ir — 2a (Fig. 3-41) 
between the paths of the incident particle before and after the collision. 
Figure 3-41 is drawn for the case of a repelling center of force at F, but the 
figure may also be taken to represent the case of an attracting center at 
F'. By Eqs. (3-249) and (3-260), 

tan | = cot a = (e » - i)-i/» = Qgg) 1 ' 2 • (3-268) 

Let the particle have an initial speed v , and let it be traveling in such a 
direction that, if undeflected, it would pass a distance s from the center of 
force (F). The distance s is called the impact parameter for the collision. 
We can readily compute the energy and angular momentum in terms of 



* 



A. Einstein and L. Infeld, The Evolution of Physics. New York: Simon and 
Schuster, 1938. (Page 253.) For a mathematical discussion, see R. C. Tolman, 
Relativity, Thermodynamics, and Cosmology. Oxford: Oxford University Press, 
1934. (Section 83.) 

t M. Born, Atomic Physics, tr. by John Dougall. New York: Stechert, 1936. 
(Chapter 5.) 



136 



MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 




Fig. 3-41. A hyperbolic orbit. 

the speed and impact parameter: 

E = \mo%, (3-269) 

L = mv s. (3-270) 

Substituting in Eq. (3-268), we have for the scattering angle 0: 

tan-^ = J^- 2 - (3-271) 

2 msv 

If a light particle of charge q t collides with a heavy particle of charge q 2 , 
this is, by Eq. (3-231), 

tan ® = JM?L (3-272) 

2 msv 

In a typical scattering experiment, a stream of charged particles may be 
shot in a definite direction through a thin foil. Many of the particles 
emerge from the foil in a different direction, after being deflected or scat- 
tered through an angle 6 by a collision with a particle within the foil. 
To put Eq. (3-272) in a form in which it can be compared with experiment, 
we must eliminate the impact parameter s, which cannot be determined 
experimentally. In the experiment, the fraction of incident particles scat- 
tered through various angles is observed. It is customary to express 
the results in terms of a cross section denned as follows. If N incident 
particles strike a thin foil containing n scattering centers per unit area, the 
average number dN of particles scattered through an angle between and 



3-16] 



HYPERBOLIC ORBITS. THE RUTHERFORD PROBLEM 



137 




Fig. 3-42. Cross section for scattering. 

+ d® is given in terms of the cross section da by the formula 

dN 



N 



= nda. 



(3-273) 



da is called the cross section for scattering through an angle between and 
© + d®, and can be thought of as the effective area surrounding the scat- 
tering center which the incident particle must hit in order to be scattered 
through an angle between and © + d®. For if there is a "target area" 
da around each scattering center, then the total target area in a unit area 
is n da. If N particles strike one unit area, the average number striking 
the target area is Nn da, and this, according to Eq. (3-273), is just dN, the 
number of particles scattered through an angle between © and © + d®. 

Now consider an incident particle approaching a scattering center F 
as in Figs. 3-41 and 3-42. If the impact parameter is between s and 
s + ds, the particle will be scattered through an angle between and 
© + d®, where is given by Eq. (3-272), and d® is given by the differen- 
tial of Eq. (3-272) : 



2cos2(0/2) 



d® 



Igigzl 

ms 2 Vo 



(3-274) 



The area of the ring around F of inner radius s, outer radius s + ds, at 
which the incident particle must be aimed in order to be scattered through 
an angle between and + d®, is 

da = 27rs ds. (3-275) 

Substituting for s from Eq. (3-272), and for ds from Eq. (3-274) (omitting 



138 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

the negative sign), we obtain 

\2m»g/ sinM©/2) 

This formula can be compared with da determined experimentally as given 
by Eq. (3-273). Formula (3-276) was deduced by Rutherford and used in 
interpreting his experiments on the scattering of alpha particles by thin 
metal foils. He was able to show that the formula agrees with his experi- 
ments with gi = 2e (charge on alpha particle),* and q 2 = Ze (charge on 
atomic nucleus), so long as the perihelion distance (a + ae in Fig. 3-41) is 
larger than about 10 -12 cm, which shows that the positive charge on the 
atom must be concentrated within a region of radius less than 10~~ 12 cm. 
This was the origin of the nuclear theory of the atom. The perihelion 
distance can be computed from formula (3-262) or by using the conserva- 
tion laws for energy and angular momentum, and is given by 



3i«2 



2E 



\ mqilzj 



(3-277) 



The smallest perihelion distance for incident particles of a given energy 
occurs when L = (s = 0), and has the value 

r lmi „ = ^- (3-278) 

Hence if there is a deviation from Coulomb's law of force when the alpha 
particle grazes or penetrates the nucleus, it should show up first as a devia- 
tion from Rutherford's law [Eq. (3-276)] at large angles of deflection @, and 
should show up when the energy E is large enough so that 

E > ^ , (3-279) 

where r is the radius of the nucleus. The earliest measurements of nuclear 
radii were made in this way by Rutherford, and turn out to be of the order 
of 10 -12 cm. 

The above calculation of the cross section is strictly correct only when 
the alpha particle impinges on a nucleus much heavier than itself, since the 
scattering center is assumed to remain fixed. This restriction can be re- 
moved by methods to be discussed in Section 4-8. Alpha particles also 
collide with electrons, but the electron is so light that it cannot appreciably 
deflect the alpha particle. The collision of an alpha particle with a nucleus 



* Here e stands for the magnitude of the electronic charge. 



3-17] MOTION OF A PARTICLE IN AN ELECTROMAGNETIC FIELD 139 

should really be treated by the methods of quantum mechanics. The con- 
cept of a definite trajectory with a definite impact parameter s is no longer 
valid in quantum mechanics. The concept of cross section is still valid in 
quantum mechanics, however, as it should be, since it is denned in terms of 
experimentally determined quantities. The final result for the scattering 
cross section turns out the same as our formula (3-276).* It is a fortunate 
coincidence in the history of physics that classical mechanics gives the right 
answer to this problem. 

3-17 Motion of a particle in an electromagnetic field. The laws deter- 
mining the electric and magnetic fields due to various arrangements of 
electric charges and currents are the subject matter of electromagnetic 
theory. The determination of the motions of charged particles under given 
electric and magnetic forces is a problem in mechanics. The electric force 
on a particle of charge q located at a point r is 

F = qE(r), (3-280) 

where E(r) is the electric field intensity at the point r. The electric field 
intensity may be a function of time as well as of position in space. The 
force exerted by a magnetic field on a charged particle at a point r depends 
on the velocity v of the particle, and is given in terms of the magnetic 
induction B(r) by the equation.f 

F = 2 VX B(r), (3-281) 

where c = 3 X 10 10 cm/sec is the velocity of light, and all quantities are 
in gaussian units, i.e., q is in electrostatic units, B in electromagnetic units 
(gauss), and v and F are in cgs units. In mks units, the equation reads 

F = qv X B(r). (3-282) 

Equation (3-280) holds for either gaussian or mks units. We shall base 
our discussion on Eq. (3-281) (gaussian units), but the results are readily 
transcribed into mks units by omitting c wherever it occurs. The total 
electromagnetic force acting on a particle due to an electric field intensity E 
and a magnetic induction B is 

F = qE + I v X B. (3-283) 



* D. Bohm, Quantum Theory. New York: Prentice-Hall, 1951. (Page 537). 
t G. P. Harnwell, Principles of Electricity and Electromagnetism, 2nd ed. New 
York: McGraw-Hill, 1949. (Page 302.) 



140 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

If an electric charge moves near the north pole of a magnet, the magnet 
will exert a force on the charge given by Eq. (3-281) ; and by Newton's 
third law the charge should exert an equal and opposite force on the 
magnet. This is indeed found to be the case, at least when the velocity 
of the particle is small compared with the speed of light, if the magnetic 
field due to the moving charge is calculated and the force on the magnet 
computed. However, since the magnetic induction B is directed radially 
away from the pole, and the force F is perpendicular to B, the forces on 
the charge and on the pole are not directed along the line joining them, 
as in the case of a central force. Newton's third law is sometimes stated 
in the "strong" form in which action and reaction are not only equal and 
opposite, but are directed along the line joining the interacting particles. 
For magnetic forces, the law holds only in the "weak" form in which nothing 
is said about the directions of the two forces except that they are opposite. 
This is true not only of the forces between magnets and moving charges, 
but also of the magnetic forces exerted by moving charges on one another. 

If the magnetic field is constant in time, then the electric field intensity 
can be shown to satisfy the equation 

V X E = 0. (3-284) 

The proof of this statement belongs to electromagnetic theory and need 
not concern us here.* We note, however, that this implies that for static 
electric and magnetic fields, the electric force on a charged particle is con- 
servative. We can therefore define an electric potential 

*(r) = - pE-dr, (3-285) 

Jr , 
such that 

E = —V*. (3-286) 

Since E is the force per unit charge, <j> will be the potential energy per unit 
charge associated with the electric force: 

F(r) = g*(r). (3-287) 

Furthermore, since the magnetic force is perpendicular to the velocity, it 
can do no work on a charged particle. Consequently, the law of con- 
servation of energy holds for a particle in a static electromagnetic field: 

T + q<t> = E, (3-288) 

where E is a constant. 



* Harnwell, op. cit. (Page 340.) 



3-17] MOTION OF A PARTICLE IN AN ELECTROMAGNETIC FIELD 141 

A great variety of problems of practical and theoretical interest arise 
involving the motion of charged particles in electric and magnetic fields. 
In general, special methods of attack must be devised for each type of 
problem. We shall discuss two special problems which are of interest both 
for the results obtained and for the methods of obtaining those results. 

We first consider the motion of a particle of mass m, charge q, in a uni- 
form constant magnetic field. Let the z-axis be chosen in the direction of 
the field, so that 

B(r, t) = Bk, (3-289) 

where B is a constant. The equations of motion are then, by Eq. (3-281), 

mx = — y, my = — —x, mz = 0. (3-290) 

c c 

According to the last equation, the z-component of velocity is constant, 
and we shall consider the case when v z = 0, and the motion is entirely in 
the a;?/-plane. The first two equations are not hard to solve, but we can 
avoid solving them directly by making use of the energy integral, which in 
this case reads 

imv 2 = E. (3-291) 

The force is given by: 

F = ^vxk, (3-292) 

F = ^- (3-293) 

The force, and consequently the acceleration, is therefore of constant 
magnitude and perpendicular to the velocity. A particle moving with 
constant speed v and constant acceleration a perpendicular to its direction 
of motion moves in a circle of radius r given by Eq. (3-80) : 



2 

a = rd 2 = — = — • (3-294) 

r m 



,2 v< F 



We substitute for F from Eq. (3-293) and solve for r: 

The product Br is therefore proportional to the momentum and inversely 
proportional to the charge. 



142 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

This result has many practical applications. If a cloud chamber is placed in a 
uniform magnetic field, one can measure the momentum of a charged particle by 
measuring the radius of curvature of its track. The same principle is used in a 
beta-ray spectrometer to measure the momentum of a fast electron by the curva- 
ture of its path in a magnetic field. In a mass spectrometer, a particle is ac- 
celerated through a known difference of electric potential, so that, by Eq. (3-288), 
its kinetic energy is 

imv 2 = q(<j>o - «i). (3-296) 

It is then passed through a uniform magnetic field B. If q is known, and r, B, 
(<po — 4>i) are measured, we can eliminate v between Eqs. (3-295) and (3-296), 
and solve for the mass: 

- - qBV (3-297) 



2 C 2(4> - 0i) 



There are many variations of this basic idea. The historic experiments of 
J. J. Thomson which demonstrated the existence of the electron were essentially 
of this type, and by them Thomson succeeded in showing that the path traveled 
by a cathode ray is that which would be followed by a stream of charged particles, 
all with the same ratio q/m. In a cyclotron, charged particles travel in circles in 
a uniform magnetic field, and receive increments in energy twice per revolution 
by passing through an alternating electric field. The radius r of the circles there- 
fore increases, according to Eq. (3-295), until a maximum radius is reached, at 
which radius the particles emerge in a beam of definite energy determined by 
Eq. (3-295). The frequency v of the alternating electric field must be the same 
as the frequency v of revolution of the particles, which is given by 

v = 2jm>. (3-298) 

Combining this equation with Eq. (3-295), we have 

- - qB (3-299) 



2Tr?nc 



Thus if B is constant, v is independent of r, and this is the fundamental principle 
on which the operation of the cyclotron is based.* In the betatron, electrons 
travel in circles, and the magnetic field within the circle is made to increase. 
Since B is changing with time, V X E is no longer zero; the changing magnetic 
flux induces a voltage around the circle such that a net amount of work is done 
on the electrons by the electric field as they travel around the circle. The betatron 
is so designed that the increase of B at the electron orbit is proportional to the 
increase of mv, so that r remains constant. 



* According to the theory of relativity, the mass of a particle increases with 
velocity at velocities near the speed of light, and consequently the cyclotron 
cannot accelerate particles to such speeds unless v is reduced or B is increased as 
the particle velocity increases. [It turns out that Eq. (3-295) still holds in 
relativity theory.] 



3-17] MOTION OF A PARTICLE IN AN ELECTROMAGNETIC FIELD 143 

Finally, we consider a particle of mass m, charge q, moving in a uniform 
constant electric field intensity E and a uniform constant magnetic induc- 
tion B. Again let the z-axis be chosen in the direction of B, and let the 
y-axis be chosen so that E is parallel to the t/z-plane: 

B = Bk, E = E v j + EJk, (3-300) 

where B, E y , E z are constants. The equations of motion, by Eq. (3-283), 
are 

.. qB . 

mx= =c y ' (3-301) 

my = —2-& + q E y> (3-302) 

mz = qE z . (3-303) 

The z-component of the motion is uniformly accelerated: 

* = 20 + z t + | & t 2 . (3-304) 

To solve the x and y equations, we differentiate Eq. (3-301) and substitute 
in Eq. (3-302) in order to eliminate y. 

m 2 c ... qB . , _, 

Tr x = - , x + &»■ (3-305) 



qB'"~ c 

By making the substitutions 

qB 



U) = 

mc 



(3-306) 



qE v 



* = ^> (3-307) 



we can write Eq. (3-305) in the form 



-^p+u> £ x = aw. (3-308) 

This equation has the same form as the equation for a harmonic oscillator 
with angular frequency w subject to a constant applied "force" aw, except 
that x appears in place of the coordinate. The corresponding oscillator 
problem was considered in Chapter 2, Problem 33. The solution in this 
case will be 

* = £ + A x cos (uf + e x ), (3-309) 

where A x and 6 X are arbitrary constants to be determined. By eliminating x 



144 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

from Eqs. (3-301) and (3-302), in a similar way, we obtain a solution for y: 

y = A y cos (at + 6 V ). (3-310) 

We get x and y by integrating Eqs. (3-309) and (3-310): 

* = C x + - + ^ sin (co* + 8 X ), (3-311) 

CO CO 

y = C y + 4+ sin (co< + «,)• (3-312) 

CO 

Now a difficulty arises, for we have six constants A x , A v , 6 X , y , C x , and C y 
to be determined, and only four initial values Xq, yo, xq, yo to determine 
them. The trouble is that we obtained the solutions (3-311) and (3-312) by 
differentiating the original equations, and differentiating an equation may 
introduce new solutions that do not satisfy the original equation. Con- 
sider, for example, the very simple equation 

x = 3. 
Differentiating, we get 

x = 0, 
whose solution is 

x = C. 

Now only for one particular value of the constant C will this satisfy the 
original equation. Let us substitute Eqs. (3-311) and (3-312) or, equiva- 
lents, Eqs. (3-309) and (3-310) into the original Eqs. (3-301) and (3-302), 
using Eqs. (3-306) and (3-307): 

- 2? A x sin (cot + 6 X ) = ^- A y cos (co< + 0„), (3-313) 

c c 

- 2® Ay sin (at + e v ) = - 2? A x cos (at + 6 X ). (3-314) 

c c 

These two equations will hold only if A X9 A y> B x , and B y are chosen so that 

sin (cof + X ) = —cos (at + 8 V ), (3-316) 

cos (at + 6 X ) = sin (co< + 6 y ). (3-317) 
The latter two equations are satisfied if 

*»=«. + £• (3-318) 



3-17] MOTION OF A PARTICLE IN AN ELECTBOMAGNETIC FIELD 145 




Fig. 3-43. Orbits in the xy-plane of a charged particle subject to a magnetic 
field in the z-direction and an electric field in the ^-direction. 



Let us set 



A x 

e x 



Ay = coA, 



8,= e + %- 



Then Eqs. (3-311) and (3-312) become 

x = C x + A sin (cot +■ 6) + 

V = Cy + A cos (cot + 6). 



at 

CO 



(3-319) 
(3-320) 
(3-321) 

(3-322) 
(3-323) 



There are now only four constants, A, 6, C x , C y , to be determined by the 
initial values x , yo, x 0) y . The z-motion is, of course, given by Eq. (3-304). 
If E y = 0, the xy-motion is in a circle of radius A with angular velocity w 
about the point (C x , C y ) ; this is the motion considered in the previous 
example. The effect of E y is to add to this uniform circular motion a 
uniform translation in the x-direction! The resulting path in the :n/-plane 
will be a cycloid having loops, cusps, or ripples, depending on the initial 
conditions and on the magnitude of E v (Fig. 3-43). This problem is of 
interest in connection with the design of magnetrons. 



146 motion op particle in two or three dimensions [chap. 3 

Problems 

1. Prove, on the basis of the geometric definitions of the operations of vector 
algebra, the following equations. In many cases a diagram will suffice, (a) Eq. 
(3-7), (b) Eq. (3-17), (c) Eq. (3-26), (d) Eq. (3-27), *(e) Eq. (3-35). 

2. Prove, on the basis of the algebraic definitions of the operations of vector 
algebra in terms of components, the following equations: (a) Eq. (3^8), (b) Eq. 
(3-17), (c) Eq. (3-27), (d) Eq. (3-34), (e) Eq. (3-35). 

3. Derive Eq. (3-32) by direct calculation, using Eq. (3-10) to represent A 
and B, and making use of Eqs. (3-25) to (3-31). 

4. (a) Prove that A • (B X C) is the volume of the parallelepiped whose edges 
are A, B, C with positive or negative sign according to whether a right-hand screw 
rotated from A toward B would advance along C in the positive or negative direc- 
tion. A, B, C are any three vectors not lying in a single plane, (b) Use this 
result to prove Eq. (3-34) geometrically. Verify that the right and left members 
of Eq. (3-34) are equal in sign as well as in magnitude. 

5. Prove the following inequalities. Give a geometric and an algebraic proof 
(in terms of components) for each: 

(a) |A + B| < |A| + |B|. 

(b) |A • B| < |A| |B|. 
(0 |A X B| < |A| |B|. 

6. (a) Obtain a formula analogous to Eq. (3-40) for the magnitude of the 
sum of three forces Fi, F2, F3, in terms of Fi, F2, F3, and the angles 812, 023, 031 
between pairs of forces. [Use the suggestions following Eq. (3-40).] 

(b) Obtain a formula in the same terms for the angle ai, between the total 
force and the component force Fi. 

7. Prove Eqs. (3-54) and (3-55) from the definition (3-52) of vector differen- 
tiation. 

8. Prove Eqs. (3-56) and (3-57) from the algebraic definition (3-53) of vector 
differentiation. 

9. Give suitable definitions, analogous to Eqs. (3-52) and (3-53), for the 
integral of a vector function A(<) with respect to a scalar t: 



f 

J 4, 



A(«) dt 



Write a set of equations like Eqs. (3-54)-(3-57) expressing the algebraic proper- 
ties you would expect such an integral to have. Prove that on the basis of either 
definition 



d -f 

dtJo 



A(t) dt = A(t). 



10. A 45° isosceles right triangle ABC has a hypotenuse AB of length 4a. A 
particle is acted on by a force attracting it toward a point O on the hypotenuse a 
distance a from the point A. The force is equal in magnitude to k/r 2 , where r is 



PROBLEMS 147 

the distance of the particle from the point 0. Calculate the work done by this 
force when the particle moves from A to C to B along the two legs of the triangle. 
Make the calculation by both methods, that based on Eq. (3-61) and that based 
on Eq. (3-63). 

11. (a) A particle in the xy-pl&ne is attracted toward the origin by a force 
F = k/y, inversely proportional to its distance from the z-axis. Calculate the 
work done by the force when the particle moves from the point x = 0, y = a to 
the point x = 2a, y = along a path which follows the sides of a rectangle 
consisting of a segment parallel to the z-axis from x = 0, y = a to x = 2a, 
y = a, and a vertical segment from the latter point to the z-axis. (b) Calculate 
the work done by the same force when the particle moves along an ellipse of 
semiaxes a, 2a. [Hint: Set x = 2a sin 0, y = a cos 0.] 

12. (a) Find the components of d 3 i/dt 3 in spherical coordinates, (b) Find 
the components of d 2 A/dt 2 in cylindrical polar coordinates, where the vector A 
is a function of t and is located at a moving point. 

*13. (a) Plane parabolic coordinates /, h are defined in terms of cartesian 
coordinates x, y by the equations 

*-/-*» V = 2W 1 ' 2 , 

where / and h are never negative. Find / and h in terms of x and y. Let unit 
vectors f , h be defined in the directions of increasing / and h respectively. That 
is, f is a unit vector in the direction in which a point would move if its /-coordinate 
increases slightly while its ^-coordinate remains constant. Show that f and h are 
perpendicular at every point. [Hint: f = (i dx + j dy)[(dx) 2 + (dy) 2 ]~ 1/2 , when 
df > 0, dh = 0. Why?] 

(b) Show that f and h are functions of /, h, and find their derivatives with 
respect to / and h. Show that r = / 1/2 (/ + h) 1/2 f + h ll2 <J + h) V2 h. Find 
the components of velocity and acceleration in parabolic coordinates. 

14. A particle moves along the parabola 

y = 4/o — 4/oz, 

where /o is a constant. Its speed v is constant. Find its velocity and acceleration 
components in rectangular and in polar coordinates. Show that the equation of 
the parabola in polar coordinates is 

r cos - = /o. 

What is the equation of this parabola in parabolic coordinates (Problem 13)? 

15. A particle moves with varying speed along an arbitrary curve lying in the 
zy-plane. The position of the particle is to be specified by the distance s the par- 
ticle has traveled along the curve from some fixed point on the curve. Let r(s) 
be a unit vector tangent to the curve at the point s in the direction of increasing s. 
Show that 

dr v 



148 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

where v(s) is a unit vector normal to the curve at the point s, and r(s) is the radius 

of curvature at the point s, defined as the distance from the curve to the point of 

intersection of two nearby normals* Hence derive the following formulas for the 

velocity and acceleration of the particle: 

.2 

v = st, a = st -\ v . 

r 

16. Using the properties of the vector symbol V, derive the vector identities: 

curl (curl A) = grad (div A) — V 2 A, 
u grad v = grad (uv) — v grad u. 

Then write out the ^-components of each side of these equations and prove by 
direct calculation that they are equal in each case. (One must be very careful, 
in using the first identity in curvilinear coordinates, to take proper account of the 
dependence of the unit vectors on the coordinates.) 

17. Calculate curl A in cylindrical coordinates. 

18. Give a suitable definition of the angular momentum of a particle about an 
axis in space. Taking the specified axis as the z-axis, express the angular momen- 
tum in terms of cylindrical coordinates. If the force acting on the particle has 
cylindrical components F z , F„, F v , prove that the time rate of change of angular 
momentum about the z-axis is equal to the torque about that axis. 

19. A moving particle of mass m is located by spherical coordinates r(t), 0(t), 
<p(t). The force acting on it has spherical components F r , Ft, F v . Calculate the 
spherical components of the angular momentum vector and of the torque vector 
about the origin, and verify by direct calculation that the equation 

dL 

follows from Newton's equation of motion. 

20. Solve for the next term beyond those given in Eqs. (3-177) and (3-178). 

21. A projectile is to be fired from the origin in the zz-plane (z-axis vertical) 
with muzzle velocity wo to hit a target at the point x = xo, z = 0. (a) Neglect- 
ing air resistance, find the correct angle of elevation of the gun. Show that, in 
general, there are two such angles unless the target is at or beyond the maximum 
range, (b) Find the first-order correction to the angle of elevation due to air 
resistance. 

22. A projectile is fired from the origin with initial velocity vo = (v xo , v yo , v zo ). 
The wind velocity is v K = wj. Solve the equations of motion (3-180) for x, y, z 
as functions of t. Find the point x\, y\ at which the projectile will return to the 
horizontal plane, keeping only first-order terms in 6. Show that if air resistance 
and wind velocity are neglected in aiming the gun, air resistance alone will cause 
the projectile to fall short of its target a fraction 4bv 20 /3mg of the target distance, 



* W. F. Osgood, Introduction to the Calculus. New York: Macmillan, 1937, 
259. 



PROBLEMS 149 

and that the wind causes an additional miss in the ^-coordinate of amount 
2bwv 2 J(mg 2 ). 

23. Determine which of the following forces are conservative, and find the 
potential energy for those which are: 

(a) F x = 6abz 3 y — 20bx 3 y 2 , F y = 6abxz 3 — 10bx 4 y, 
F z = 18abxz 2 y. 

(b) F x = 18abyz 3 — 20bx 3 y 2 , F y = 18ate 3 — \Qbx*y, 
F z = 6abxyz 2 . 

(c) F = iF.(x) + jF y (y) + W.(«). 

24. Determine the potential energy for any of the following forces which are 
conservative: 

(a) F x = 2ax(z 3 + y 3 ), F y = 2ay(z 3 + y 3 ) + 3ay 2 (x 2 + y 2 ), 
F z = 3az 2 (x 2 + y 2 ). 



(b) 


F„ = ap 2 cos <p, F v = 


= ap 2 sin <p, F z = 2az 2 . 


(c) 


F r = — 2ar sin cos <p, 
F v = ar sin sin ip. 


F$ = — ar cos 6 cos <p, 



25. A particle is attracted toward the 2-axis by a force proportional to the 
square of its distance from the xy-plane and inversely proportional to its distance 
from the «-axis. Add an additional perpendicular force in such a way as to 
make the total force conservative, and find the potential energy. Be sure to 
write expressions for the forces and potential energy which are dimensionally 
consistent. 

26. Find the components of force for the following potential-energy functions: 

(a) V = axy 2 z 3 . 

(b) V = ikr 2 . 

(c) V = ik*x 2 + ik v y 2 + ik z z 2 . 

27. Find the force on the electron in the hydrogen molecule ion for which the 
potential is 

2 2 

e e 



where r\ is the distance from the electron to the point y = z = 0, x = — a, 
and t2 is the distance from the electron to the point y = z = 0, x = a. 

28. Show that F = nF(r) (where n is a unit vector directed away from the 
origin) is a conservative force by showing by direct calculation that the integral 



I 



r 2 

F-dr 



along any path between ri and T2 depends only on r\ and r-2. [Hint: Express F 
and dr in spherical coordinates.] 



150 MOTION OF PARTICLE IN TWO OB THREE DIMENSIONS [CHAP. 3 

29. The potential energy for an isotropic harmonic oscillator is 

V = ikr 2 . 

Plot the effective potential energy for the r-motion when a particle of mass m 
moves with this potential energy and with angular momentum L about the 
origin. Discuss the types of motion that are possible, giving as complete a descrip- 
tion as is possible without carrying out the solution. Find the frequency of 
revolution for circular motion and the frequency of small radial oscillations 
about this circular motion. Hence describe the nature of the orbits which differ 
slightly from circular orbits. 

30. Find the frequency of small radial oscillations about steady circular mo- 
tion for the effective potential given by Eq. (3-232) for an attractive inverse 
square law force, and show that it is equal to the frequency of revolution. 

31. Find r{t), 8(t) for the orbit of the particle in Problem 29. Compare with 
the orbits found in Section 3-10 for the three-dimensional harmonic oscillator. 

32. A particle of mass m moves under the action of a central force whose poten- 
tial is 

V(r) = Kr 4 , K > 0. 

For what energy and angular momentum will the orbit be a circle of radius a 
about the origin? What is the period of this circular motion? If the particle is 
slightly disturbed from this circular motion, what will be the period of small 
radial oscillations about r = o? 

33. According to Yukawa's theory of nuclear forces, the attractive force be- 
tween a neutron and a proton has the potential 

Ke~ ar 
V(r) = — — , K < 0. 
r 

(a) Find the force, and compare it with an inverse square law of force, (b) Dis- 
cuss the types of motion which can occur if a particle of mass m moves under 
such a force, (c) Discuss how the motions will be expected to differ from the 
corresponding types of motion for an inverse square law of force, (d) Find L 
and E for motion in a circle of radius a. (e) Find the period of circular motion 
and the period of small radial oscillations, (f) Show that the nearly circular 
orbits are almost closed when a is very small. 

34. (a) Discuss by the method of the effective potential the types of motion to 
be expected for an attractive central force inversely proportional to the cube of 
the radius: 

(b) Find the ranges of energy and angular momentum for each type of motion. 

(c) Solve the orbital equation (3-222), and show that the solution is one of the 
forms: 



PROBLEMS 151 



- = a cos me - e )), (i) 

r 

- = A cosh [0(0 - 6o)), (2) 

r 

1 

r 

1 

r 

1 1 ±09 



= A sinh LQ(<? - 0o)L (3) 

= A(d - 0o), (4) 



= - e ±p °. (5) 

- r tq 

(d) For what values of L and E does each of the above types of motion occur? 
Express the constants A and /? in terms of E and L for each case, (e) Sketch a 
typical orbit of each type. 

35. (a) Discuss the types of motion that can occur for a central force 

™ --*+%- 

Assume that K > 0, and consider both signs for K'. 

(b) Solve the orbital equation, and show that the bounded orbits have the form 
(if L 2 > —mK') 



1 + e cos ad 



(c) Show that this is a precessing ellipse, determine the angular velocity of pre- 
cession, and state whether the precession is in the same or in the opposite direc- 
tion to the orbital angular velocity. 

36. A comet is observed a distance of 1.00 X 10 8 km from the sun, travel- 
ing toward the sun with a velocity of 51.6 km per second at an angle of 45° 
with the radius from the sun. Work out an equation for the orbit of the comet 
in polar coordinates with origin at the sun and z-axis through the observed 
position of the comet. (The mass of the sun is 2.00 X 10 30 kgm.) 

37. It will be shown in Chapter 6 (Problem 5) that the effect of a uniform dis- 
tribution of dust of density p about the sun is to add to the gravitational attrac- 
tion of the sun on a planet, of mass m an additional attractive central force 

F' = —mhr, 
where 

(a) If the mass of the sun is M , find the angular velocity of revolution of the 
planet in a circular orbit of radius ro, and find the angular frequency of small 
radial oscillations. Hence show that if F' is much less than the attraction due 



152 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

to the sun, a nearly circular orbit will be approximately an ellipse whose major 
axis precesses slowly with angular velocity 

(b) Does the axis precess in the same or in the opposite direction to the orbital 
angular velocity? Look up M and the radius of the orbit of Mercury, and cal- 
culate the density of dust required to cause a precession of 41 seconds of arc 
per century. 

38. It can be shown (Chapter 6, Problems 15 and 19) that the correction to 
the potential energy of a mass m in the earth's gravitational field, due to the 
oblate shape of the earth, is approximately, in spherical coordinates, relative 
to the polar axis of the earth, 

, rimMGR 2 2 m 

V = =-= (1—3 cos 0), 

where M is the mass of the earth and 2R, 2R(l — ij) are the equatorial and 
polar diameters of the earth. Calculate the rate of precession of the perigee 
(point of closest approach) of an earth satellite moving in a nearly circular 
orbit in the equatorial plane. Look up the mass of the earth and the equatorial 
and polar diameters, and estimate the rate of precession in degrees per revolution 
for a satellite 400 miles above the earth. 

*39. Calculate the torque on an earth satellite due to the oblateness potential 
energy correction given in Problem 38. A satellite moves in a circular orbit of 
radius r whose plane is inclined so that its normal makes an angle a with the 
polar axis. Assume that the orbit is very little affected in one revolution, and 
calculate the average torque during a revolution. Show that the effect of such a 
torque is to make the normal to the orbit precess in a cone of half angle a about 
the polar axis, and find a formula for the rate of precession in degrees per revolu- 
tion. Calculate the rate for a satellite 400 miles above the earth, using suitable 
values for M , i\, and R. 

40. (a) A satellite is to be launched from the surface of the earth. Assume 
the earth is a sphere of radius R, and neglect friction with the atmosphere. The 
satellite is to be launched at an angle a with the vertical, with a velocity v o, so 
as to coast without power until its velocity is horizontal at an altitude hi above 
the earth's surface. A horizontal thrust is then applied by the last stage rocket 
so as to add an additional velocity Awi to the velocity of the satellite. The final 
orbit is to be an ellipse with perigee hi (point of closest approach) and apogee 
hi (point farthest away) measured from the earth's surface. Find the required 
initial velocity vq and additional velocity Ai>i, in terms of R, a, hi, h%, and g, 
the acceleration of gravity at the earth's surface. 

(b) Write a formula for the change Shi in perigee height due to a small error 
5/3 in the final thrust direction, to order (5/3) 2 . 

41. Two planets move in the same plane in circles of radii n, r^ about the 
sun. A space probe is to be launched from planet 1 with velocity v\ relative 
to the planet, so as to reach the orbit of planet 2. (The velocity vi is the relative 



PROBLEMS 153 

velocity after the probe has escaped from the gravitational field of the planet.) 
Show that f i is a minimum for an elliptical orbit whose perihelion and aphelion 
are r\ and r^. In that case, find v\, and the relative velocity V2 between the 
space probe and planet 2 if the probe arrives at radius T2 at the proper time to 
intercept planet 2. Express your results in terms of n, r%, and the length of 
the year Y\ of planet 1. Look up the appropriate values of r\ and r^, and esti- 
mate v\ for trips to Venus and Mars from the earth. 

42. A rocket is in an elliptical orbit around the earth, perigee r\, apogee r^, 
measured from the center of the earth. At a certain point in its orbit, its engine 
is fired for a short time so as to give a velocity increment Aw in order to put 
the rocket on an orbit which escapes from the earth with a final velocity wo 
relative to the earth. (Neglect any effects due to the sun and moon.) Show 
that Ay is a minimum if the thrust is applied at perigee, parallel to the orbital 
velocity. Find Aw in that case in terms of the elliptical orbit parameters e, a, 
the acceleration g at a distance R from the earth's center, and the final velocity 
wo- Can you explain physically why Aw is smaller for larger e? 

43. A satellite moves around the earth in an orbit which passes across the 
poles. The time at which it crosses each parallel of latitude is measured so that 
the function 0(t) is known. Show how to find the perigee, the semimajor axis, 
and the eccentricity of its orbit in terms of 0(0, and the value of g at the surface 
of the earth. Assume the earth is a sphere of radius R. 

44. It can be shown that the orbit given by the special theory of relativity for 
a particle of mass m moving under a potential energy V(r) is the same as the orbit 
which the particle would follow according to Newtonian mechanics if the poten- 
tial energy were 

where E is the energy (kinetic plus potential), and c is the speed of light. Discuss 
the nature of the orbits for an inverse square law of force according to the theory 
of relativity. Show by comparing the orbital angular velocity with the frequency 
of radial oscillations for nearly circular motion that the nearly circular orbits, 
when the relativistic correction is small, are precessing ellipses, and calculate 
the angular velocity of precession. 

45. A particle of mass m moves in an elliptical orbit of major axis 2a, eccentric- 
ity e, in such a way that the radius to the particle from the center of the ellipse 
sweeps out area at a constant rate 

dS 

dt ~ L ' 

and with period r independent of a and e. (a) Write out the equation of the ellipse 
in polar coordinates with origin at the center of the ellipse, (b) Show that the 
force on the particle is a central force, and find F(r) in terms of m, t. 

46. A rocket moves with initial velocity wo toward the moon of mass M , 
radius ro. Find the cross section a for striking the moon. Take the moon to be 
at rest, and neglect all other bodies. 



154 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 

47. Show that for a repulsive central force inversely proportional to the cube 
of the radius, 



K 

r z 



FW--J. K>0, 



the orbits are of the form (1) given in Problem 34, and express /8 in terms of K, E, 
L, and the mass m of the incident particle. Show that the cross section for 
scattering through an angle between and + d® for a particle subject to 
this force is 

&r-»LE *-e d®. 
mvl 2 (2tt - 0) 2 

48. A velocity selector for a beam of charged particles of mass m, charge e, 
is to be designed to select particles of a particular velocity vo- The velocity 
selector utilizes a uniform electric field E in the a;-direction and a uniform 
magnetic field B in the y-direction. The beam emerges from a narrow slit along 
the y-axis and travels in the 2-direction. After passing through the crossed 
fields for a distance I, the beam passes through a second slit parallel to the first 
and also in the yz-plane. 

(a) If a particle leaves the origin with a velocity »o at a small angle with the 
2-axis, find the point at which it arrives at the plane z = I. Assume that the 
initial angle is small enough so that second-order terms in the angle may be 
neglected. 

(b) What is the best choice of E, B in order that as large a fraction as possible 
of the particles with velocity vo arrive at the second slit, while particles of other 
velocities miss the slit as far as possible? 

(c) If the slit width is h, what is the maximum velocity deviation Sv from 
vo for which a particle moving initially along the 2-axis can pass through the 
second slit? Assume that E, B have the values chosen in part (b). 

49. A particle of charge q in a cylindrical magnetron moves in a uniform mag- 
netic field 

B = Bk, 

and an electric field, directed radially outward or inward from a central wire 

along the 2-axis, 

E = -h, 
P 

where p is the distance from the 2-axis, and h is a unit vector directed radially 
outward from the 2-axis. The constants a and B may be either positive or 
negative, (a) Set up the equations of motion in cylindrical coordinates, (b) 
Show that the quantity 

2. . qB 2 „ 
mp <p + ~ p = K 

is a constant of the motion, (c) Using this result, give a qualitative discussion, 
based on the energy integral, of the types of motion that can occur. Consider all 
cases, including all values of a, B, K, and E. (c) Under what conditions can 
circular motion about the axis occur? (d) What is the frequency of small radial 
oscillations about this circular motion? 



CHAPTER 4 
THE MOTION OF A SYSTEM OF PARTICLES 

4-1 Conservation of linear momentum. Center of mass. We consider 
in this chapter the behavior of mechanical systems containing two or more 
particles acted upon by internal forces exerted by the particles upon one 
another, and by external forces exerted upon particles of the system by 
agents not belonging to the system. We assume the particles to be point 
masses each specified by its position (a;, y, z) in space, like the single par- 
ticle whose motion was studied in the preceding chapter. 

Let the system we are studying contain N particles, and let them be 
numbered 1, 2, . . . , N. The masses of the particles we designate by 
mi, m 2 , . . . , mjv. The total force acting on the fcth particle will be the sum 
of the internal forces exerted on particle k by all the other (AT — 1) parti- 
cles in the system, plus any external force which may be applied to particle 
k. Let the sum of the internal forces on particle k be Ff., and let the total 
external force on particle k be F|. Then the equation of motion of the 
fcth particle will be 

m£ k = ¥i + F$, k = 1, 2, . . . , N. (4-1) 

The N equations obtained by letting fc in Eqs. (4-1) run over the num- 
bers 1, . . . , 2V are the equations of motion of our system. Since each of 
these 2V equations is itself a vector equation, we have in general a set of 32V 
simultaneous second-order differential equations to be solved. The solu- 
tion will be a set of functions rjt(<) specifying the motion of each particle in 
the system. The solution will depend on 62V "arbitrary" constants speci- 
fying the initial position and velocity of each particle. The problem of 
solving the set of equations (4-1) is very difficult, except in certain special 
cases, and no general methods are available for attacking the 2V-body 
problem, even in the case where the forces between the bodies are central 
forces. The two-body problem can often be solved, as we shall see, and 
some general theorems are available when the internal forces satisfy certain 
conditions. 

If P* = WftVfc is the linear momentum of the kth particle, we can write 
Eqs. (4-1) in the form 

^ = n + H, k=l,...,N. (4-2) 

155 



156 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

Summing the right and left sides of these equations over all the particles, 
we have 

Ef = |E^=EFl+EFi (4-3) 

Jfe=l Jfc=l Jb=l k=l 

We designate by P the total linear momentum of the particles, and by F 
the total external force: 

N N 

p = E p* = E m * v *> t 4-4 ) 

jb=i &=i 



N 
I 



F = £ **• (4-5) 



We now make the assumption, to be justified below, that the sum of the 
internal forces acting on all the particles is zero: 

E Pi = 0. (4-6) 

k=l 

When Eqs. (4-4), (4-5), and (4-6) are substituted in Eq. (4-3), it becomes 

dP 



dt 



= F. (4-7) 



This is the momentum theorem for a system of particles. It states that the 
time rate of change of the total linear momentum is equal to the total 
external force. An immediate corollary is the conservation theorem for 
linear momentum, which states that the total momentum P is constant 
when no external forces act. 

We now try to justify the assumption (4-6). Our first proof is based 
on Newton's third law. We assume that the force Fj. acting on particle k 
due to all the other particles can be represented as a sum of separate forces 
due to each of the other particles: 

Fi = X) FL*, . (4-8) 

where F\^ k is the force on particle k due to particle 1. According to New- 
ton's third law, the force exerted by particle I on particle k is equal and 
opposite to that exerted by k on I: 

FJLi = -Fi-*. (4-9) 



4-1] CONSERVATION OF LINEAR MOMENTUM. CENTER OP MASS 157 

Equation (4-9) expresses Newton's third law in what we may call the weak 
form; that is, it says that the forces are equal and opposite, but does not 
imply that the forces act along the line joining the two particles. If we 
now consider the sum in Eq. (4-6), we have 

E F ^=SE F ^ (4-10) 

fc=l k=l l^k 

The sum on the right is over all forces acting between all pairs of particles 
in the system. Since for each pair of particles k, I, two forces F i k _ >l and 
F!_»j; appear in the total sum, and by Eq. (4-9) the sum of each such pair 
is zero, the total sum on the right in Eq. (4-10) vanishes, and Eq. (4-6) 
is proved. 

Thus Newton's third law, in the form (4-9), is sufficient to guarantee the 
conservation of linear momentum for a system of particles, and it was for 
this purpose that the law was introduced. The law of conservation of 
momentum has, however, a more general validity than Newton's third 
law, as we shall see later. We can derive assumption (4-6) on the basis of 
a somewhat weaker assumption than Newton's third law. We do not need 
to assume that the particles interact in pairs. We assume only that the 
internal forces are such that they would do no net work if every particle in 
the system should be displaced the same small distance 5r from its position 
at any particular instant. An imagined motion of all the particles in the 
system is called a virtual displacement. The motion described, in which 
every particle moves the same small distance 5r, is called a small virtual 
translation of the system. We assume, then, that in any small virtual 
translation 5r of the entire system, the internal forces would do no net 
work. From the point of view of the general idea of conservation of energy, 
this assumption amounts to little more than assuming that space is homo- 
geneous. If we move the system to a slightly different position in space 
without otherwise disturbing it, the internal state of the system should be 
unaffected, hence in particular the distribution of various kinds of energy 
within it should remain the same and no net work can have been done by 
the internal forces. Let us use this idea to prove Eq. (4-6). The work 
done by the force FJ. in a small virtual translation St is 

8W k = Fl-Sr. (4-11) 

The total work done by all the internal forces is 

8W = X) SW " = « r '( E F *J ' ( 4 ~ 12 ) 

fc=i \fc=i / 



158 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

where we have factored out 5r from the sum, since it is the same for all 
particles. Assuming that dW = 0, we have 

Sr-[ V fA = 0. (4-13) 



<JH- ft 



Since Eq. (4-13) must hold for any 5r, Eq. (4-6) follows. 

We can put Eq. (4-7) in an illuminating form by introducing the con- 
cept of center of mass of the system of particles. The vector R which 
locates the center of mass is defined by the equation 



MR = X) m ^k, (4-14) 

where M is the total mass: 

N 



k=i 



The coordinates of the center of mass are given by the components of 
Eq. (4-14) : 

X = mYj mkXk > Y = m ^ mkn ' Z = M ^ m * *" ^ 4-16 ^ 

k=l fc=l fc=l 

The total momentum defined by Eq. (4—1) is, in terms of the center of mass, 

N 

so that Eq. (4-7) can be written 

MR = F. (4-18) 

This equation has the same form as the equation of motion of a particle of 
mass M acted on by a force F. We thus have the important theorem that 
[when Eq. (4-6) holds] the center of mass of a system of particles moves like 
a single particle, whose mass is the total mass of the system, acted on by a force 
equal to the total external force acting on the system. 

4-2 Conservation of angular momentum. Let us calculate the time rate 
of change of the total angular momentum of a system of N particles rela- 
tive to a point Q not necessarily fixed in space. The vector angular mo- 
mentum of particle k about a point Q, not necessarily the origin, is to be 
defined according to Eq. (3-142) : 

L*q = m k (r k — t q ) X (ft - i Q ), (4-19) 



4-2] CONSERVATION OF ANGULAR MOMENTUM 159 

where r e is the position vector of the point Q, and (r k — i Q ) is the vector 
from Q to particle k. Note that in place of the velocity ik we have 
written the velocity (ik — iq) relative to the point Q as origin, so that 
LkQ is the angular momentum of m^ calculated as if Q were a fixed 
origin. This is the most useful way to define the angular momentum about 
a moving point Q. Taking the cross product of (rj, — iq) with the equa- 
tion of motion (4-2) for particle k, as in the derivation of Eq. (3-144), we 
obtain 

(r* - r Q ) X ^ = (r» - i Q ) X F% + (r* - r Q ) X F|. (4-20) 

We now differentiate Eq. (4-19) : 

^p = (r* - r Q ) X ^ + m k (ik - f Q ) X (f* - f Q ) - m*(r ft - r ) X r Q . 

(4-21) 
The second term on the right vanishes. Therefore, by Eq. (4-20), 

^f = (r* - r Q ) x ¥% + (r k - t q ) x r k - m k (t k - r Q ) x t Q . 

(4-22) 

The total angular momentum and total external torque about the point Q 
are defined as follows: 



N e = £ (r* - r ) X FJ. (4-24) 



N. 

z 

Summed over all particles, Eq. (4-22) becomes, if we use Eq. (4-14), 



AT 

E 

k=i 



^f = N Q + E (** - '«) X F* - ^( R - r «) x f Q- (4-26) 



The last term will vanish if the acceleration of the point Q is zero or is 
along the line joining Q with the center of mass. We shall restrict the 
discussion to moments about a point Q satisfying this condition: 

(R - r Q ) X t Q = 0. (4-26) 

The most important applications will be to cases where Q is at rest, or 
where Q is the center of mass. If we also assume that the total internal 



160 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

torque vanishes: N 

£ (r* - t Q ) X Ft = 0, (4-27) 

then Eq. (4-25) becomes 

^ = Ng. (4-28) 

This is the angular momentum theorem for a system of particles. An 
immediate corollary is the conservation theorem for angular momentum, 
which states that the total angular momentum of a system of particles is 
constant if there is no external torque on the system. 

In order to prove Eq. (4-27) from Newton's third law, we need to assume 
a stronger version of the law than that needed in the preceding section, 
namely, that the force Ff.^ is not only equal and opposite to Fj'_, fc , but 
that these forces act along the line joining the two particles; that is, the 
two particles can only attract or repel each other. We shall assume, as 
in the previous section, that Fj. is the sum of forces due to each of the 
other particles : 

E (r fc - r e ) x Fit = £ £ < r * - r «) x F <-* 

k=l *=1 l*k 

= E £ [(r* - v Q ) x FL* + (r, - r ) x P^,]. 

fc=1 l=1 (4-29) 

In the second step, the sum of torques has been rearranged as a sum of 
pairs of torques due to pairs of forces which, according to Newton's third 
law, are equal and opposite [Eq. (4-9)], so that 

£ (r* - r Q ) X F| = f; £ [(r* - r ) - (r, - r«)] X FL* 

= E E (r* - r,) X FL*. (4-30) 

k=l 1=1 

The vector (r k — t{) has the direction of the line joining particle I with 
particle k. If F}_> k acts along this line, the cross product in Eq. (4-30) 
vanishes. Hence if we assume Newton's third law in the strong form, then 
assumption (4-27) can be proved. 

Alternatively, by assuming that no net work is done by the internal 
forces in a small virtual rotation about any axis through the point Q, we 
can show that the component of total internal torque in any direction is 
zero, and hence justify Eq. (4-27). 



4r-2] CONSERVATION OF ANGULAR MOMENTUM 161 




rxF 




L + dL 



Fig. 4-1. Motion of a simple gyroscope. 

As an application of Eq. (4-28), we consider the action of a gyroscope 
or top. A gyroscope is a rigid system of particles symmetrical about an 
axis and rotating about that axis. The reader can convince himself that 
when the gyroscope is rotating about a fixed axis, the angular momentum 
vector of the gyroscope about a point Q on the axis of rotation is directed 
along the axis of rotation, as in Fig. 4-1. The symmetry about the axis 
guarantees that any component of the angular momentum Lj of particle 
k that is perpendicular to the axis will be compensated by an equal and 
opposite component due to the diametrically opposite particle. Let us 
choose the point Q where the gyroscope axis rests on its support. If now 
a force F is applied downward on the gyroscope axis (e.g., the force of 
gravity), the torque (r x F) due to F will be directed perpendicular to r 
and to L, as shown in Fig. 4-1. By Eq. (4-28) the vector dL/dt is in the 
same direction, as shown in the figure, and the vector L tends to precess 
around the figure in a cone under the action of the force F. Now the 
statement that L is directed along the gyroscope axis is strictly true only 
if the gyroscope is simply rotating about its axis. If the gyroscope axis 
itself is changing its direction, then this latter motion will contribute an 
additional component of angular momentum. If, however, the gyroscope 
is spinning very rapidly, then the component of angular momentum along 
its axis will be much greater than the component due to the motion of the 
axis, and L will be very nearly parallel to the gyroscope axis. Therefore 
the gyroscope axis must also precess around the vertical, remaining essen- 
tially parallel to L. A careful analysis of the off-axis components of L 
shows that, if the gyroscope axis is initially stationary in a certain direction 
and is released, it will wobble slightly down and up as it precesses around 
the vertical. This will be shown in Chapter 11. The gyroscope does not 
"resist any change in its direction, " as is sometimes asserted, for the rate 



162 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

of change in its angular momentum is always equal to the applied torque, 
just as the rate of change of linear momentum is always equal to the ap- 
plied force. We can make the gyroscope turn in any direction we please 
by applying the appropriate torque. The importance of the gyroscope as 
a directional stabilizer arises from the fact that the angular momentum 
vector L remains constant when no torque is applied. The changes in 
direction of a well-made gyroscope are small because the applied torques 
are small and L is very large, so that a small dL gives no appreciable change 
in direction. Furthermore, a gyroscope only changes direction while a 
torque is applied; if it shifts slightly due to occasional small frictional 
torques in its mountings, it stops shifting when the torque stops. A large 
nonrotating mass, if mounted like a gyroscope, would acquire only small 
angular velocities due to frictional torques, but once set in motion by a 
small torque, it would continue to rotate, and the change in position might 
eventually become large. 

4-3 Conservation of energy. In many cases, the total force acting on 
any particle in a system of particles depends only on the positions of the 
particles in the system : 

F t = F*(ri, r 2 , . . . , ly), k = 1, 2, . . . , N. (4-31) 

The external force FJ, for example, might depend on the position r* of 
particle k, and the internal force Fj. might depend on the positions of the 
other particles relative to particle k. It may be that a potential function 

y( T it r 2> • • • > t n) exists such that 

F - dV F - dV F - dV h- 1 N 

Fkx --te k ' Fk «--w k ' Fk *-~df k ' k - 1> ---> N - 

(4-32) 

Conditions to be satisfied by the force functions Ffc(ri, . . . , r^) in order 
for a potential V to exist can be worked out, analogous to the condition 
(3-189) for a single particle. The result is rather unwieldy and of little 
practical importance, and we omit this development here. If a potential 
energy exists, we can derive a conservation of energy theorem as follows. 
By Eq. (4-32), the equations of motion of the fcth particle are 

(4-33) 

Multiplying Eqs. (4-33) by v kx , v ky , Vkz, respectively, and adding, we have 
for each k: 

d ft m „ 2 \ _1_ dV dXk _1_ dV d W° _j_ d V dz k n h , jj ,, oa\ 



dVicx 


_s>Z, 


dv ky 




dvkz 


_ dV 


m "-dT = 


dx k ' 


mk -w = 


— — , 

tyk' 


"*-* = 


dZk 



4-3] CONSERVATION OF ENERGY 163 

This is to be summed over all values of k : 

d V n 2 N i v^ IdV dzjc , dV dy k . dV dzA _ ,. __ N 

j t ^^) + Z\^- kW + ei kW + er kW ) = o. (4-35) 

The second term in Eq. (4-35) is dV/dt: 

$r = ^ (w dx± dV fa av dzj\ 

dt £l W dt d Vk dt dz k dt/' { axy) 

and the first term is the time derivative of the total kinetic energy 

AT 

T = X) i m ^- ( 4 - 37 ) 

Consequently, Eq. (4-35) can be written 

j t (T + V) = 0. (4-38) 

Hence we again have a conservation of energy theorem, 

T + V = E, (4-39) 

where E is constant. If the internal forces are derivable from a potential- 
energy function V, as in Eq. (4-32), but the external forces are not, the 
energy theorem will be 

j t (T + V) = JT ¥%-v k . (4-40) 

Suppose the internal force acting on any particle A; can be regarded as 
the sum of forces due to each of the other particles, where the force Ff^j. 
on k due to I depends only on the relative position (tk — rj) of particle k 
with respect to particle I: 

Ft = X) Fi-^bfob - r,). (4-41) 

It may be that the vector function F|_ >Jfc (rj ; — rj) is such that we can define 
a potential-energy function 

V k i(T k i) = - jf" FL*(r t ,)-<frw f (4-42) 

where 

m = n — Ti- (4-43) 

This will be true if F\^, k is a conservative force in the sense of Chapter 3, 



164 THE MOTION OF A SYSTEM OP PARTICLES [CHAP. 4 

that is, if 

curl FL* = 0, (4-44) 

where the derivatives are with respect to x k i, yu, zm- The gravitational and 
electrostatic forces between pairs of particles are examples of conservative 
forces. If Ff_j. is conservative, so that Vu can be denned, then* 

V i -dVki .dV„ dV k i 

*'-* * dx kl J dy kl dz k i 

~ l dx k 3 dy k * dz k ^*° ; 

If Newton's third law (weak form) holds, then 

oxjci dy k i dz k i 

= _i*Z*i_ j^lM -^-IM. (4-46) 

dxi dyi dzi 

Thus V k i will also serve as the potential-energy function for the force Fj^. 
We can now define the total internal potential energy V 1 for the system of 
particles as the sum of Vu over all the pairs of particles : 

F'(ri, . . . , r N ) = j^ *fj V k i(v k - r,). (4-47) 

k =i 1=1 

It follows from Eqs. (4-41), (4-45), and (4-46), that the internal forces are 
given by 

F£=-i^-J^-k^, k=l,...,N. (4-48) 
dx k dy k dz k 

In particular, if the forces between pairs of particles are central forces, the 
potential energy V k i(r k i) for each pair of particles depends only on the 
distance r k i between them, and is given by Eq. (3-200) ; the internal forces 
of the system are then conservative, and Eq. (4—48) holds. The energy 
theorem (4-40) will be valid for such a system of particles. If the external 
forces are also conservative, their potential energy can be added to V 1 , and 
the total energy is constant. 

If there is internal friction, as is often the case, the internal frictional 
forces depend on the relative velocities of the particles, and the conserva- 
tion law of potential plus kinetic energy no longer holds. 



* Note that V(x k i) = V(x kh y kh z u ) = V(x k — x h y k — yi, z k — z t ), so that 
dV/dx k = dV/dx H = — dV/dxi, etc. 



4 ~ 4 J CRITIQUE OP THE CONSERVATION LAWS 



165 



4-4 Critique of the conservation laws. We may divide the phenomena 
to which the laws of mechanics have been applied into three major classes. 
The motions of celestial bodies— stars, satellites, planets— are described 
with extremely great precision by the laws of classical mechanics. It was 
in this field that the theory had many of its important early successes. 
The motions of the bodies in the solar system can be predicted with great 
accuracy for periods of thousands of years. The theory of relativity pre- 
dicts a few slight deviations from the classically predicted motion, but these 
are too small to be observed except in the case of the orbit of Mercury, 
where relativity and observation agree in showing a slow precession of the 
axis of the elliptical orbit around the sun at an angular velocity of about 
0.01 degree per century. 

The motion of terrestrial bodies of macroscopic and microscopic size 
constitutes the second major division of phenomena. Motions in this class 
are properly described by Newtonian mechanics, without any significant 
corrections, but the laws of force are usually very complicated, and often 
not precisely known, so that the beautifully precise calculations of celestial 
mechanics cannot be duplicated here. 

The third class of phenomena is the motion of "atomic" particles: 
molecules, atoms, electrons, nuclei, protons, neutrons, etc. Early attempts 
to describe the motions of such particles were based on classical mechanics, 
and many phenomena in this class can be understood and predicted on this 
basis. However, the finer details of the behavior of atomic particles can 
only be. properly described in terms of quantum mechanics and, for high 
velocities, relativistic quantum mechanics must be introduced. We might 
add a fourth class of phenomena, having to do with the intrinsic structure 
of the elementary particles themselves (protons, neutrons, electrons, etc.). 
Even quantum mechanics fails to describe such phenomena correctly, and 
physics is now struggling to produce a new theory which will describe this 
class of phenomena. 

The conservation law for linear momentum holds for systems of celestial 
bodies as well as for bodies of macroscopic and microscopic size. The 
gravitational a'nd mechanical forces acting between such bodies satisfy 
Newton's third law, at least to a high degree of precision. Linear momen- 
tum is also conserved in most interactions of particles of atomic size, except 
when high velocities or rapid accelerations are involved. The electrostatic 
forces between electric charges at rest satisfy Newton's third law, but when 
the charges are in motion, their electric fields propagate with the velocity 
of light, so that if two charges are in rapid relative motion, the forces be- 
tween them may not at any instant be exactly equal and opposite. If a 
fast electron moves past a stationary proton, the proton "sees" the electron 
always a little behind its actual position at any instant, and the force on 
the proton is determined, not by where the electron is, but by where it was 



166 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

a moment earlier. When electric charges accelerate, they may emit electro- 
magnetic radiation and lose momentum in so doing. It turns out that the 
law of conservation of momentum can be preserved also in such cases, but 
only by associating momentum with the electromagnetic field as well as 
with moving particles. Such a redefinition of momentum goes beyond the 
original limits of Newtonian mechanics. 

Celestial bodies and bodies of macroscopic or microscopic size are ob- 
viously not really particles, since they have a structure which for many 
purposes is not adequately represented by merely giving to the body three 
position coordinates x, y, z. Nevertheless, the motion of such bodies, in 
problems where their structure can be neglected, is correctly represented 
by the law of motion of a single particle, 

mx = F. (4-49) 

This is often justified by regarding the macroscopic body as a system 
of smaller particles satisfying Newton's third law. For such a system, 
the linear momentum theorem holds, and can be written in the form of 
Eq. (4-18), which has the same form as Eq. (4-49). This is a very con- 
venient way of justifying the application of Eq. (4-49) to bodies of macro- 
scopic or astronomical size, provided our conscience is not troubled by the 
fact that according to modern ideas it does not make sense. If the particles 
of which the larger body is composed are taken as atoms and molecules, 
then in the first place Newton's third law does not invariably hold for such 
particles, and in the second place we should apply quantum mechanics, not 
classical mechanics, to their motion. The momentum theorem (4-18) can 
be derived for bodies made up of atoms by using the laws of electrodynam- 
ics and quantum mechanics, but this lies outside the scope of Newtonian 
mechanics. Hence, for the present, we must take the law of motion (4-49), 
as applied to macroscopic and astronomical bodies, as a fundamental 
postulate in itself, whose justification is based on experimental grounds 
or on the results of deeper theories. The theorems proved in Section 4-1 
show that this postulate gives a consistent theory of mechanics in the 
sense that if, from bodies satisfying this postulate, we construct a com- 
posite body, the latter body will also satisfy the postulate. 

The law of conservation of angular momentum, as formulated in Sec- 
tion 4-2 for a system of particles, holds for systems of celestial bodies 
(regarded as particles) and for systems of bodies of macroscopic size when- 
ever effects due to rotation of the individual bodies can be neglected. 
When rotations of the individual bodies enter into the motion, then a 
conservation law for angular momentum still holds, provided we include 
the angular momentum associated with such rotations; the bodies are then 
no longer regarded as particles of the simple type considered in the pre- 
ceding sections whose motions are completely described simply by specify- 



4-4] CRITIQUE OF THE CONSERVATION LAWS 167 

ing the function r(t) for each particle. The total angular momentum of 
the solar system is very nearly constant, even if the sun, planets, and 
satellites are regarded as simple particles whose rotations can be neglected. 
Tidal forces, however, convert some rotational angular momentum into 
orbital angular momentum of the planets and satellites, and so rotational 
angular momentum must be included if the law of conservation of angular 
momentum is to hold precisely. Some change in angular momentum occurs 
due to friction with interplanetary dust and rocks, but the effect is too 
small to be observed, and could in any case be included by adding the 
angular momentum of the interplanetary matter to the total. 

The law of conservation of total angular momentum, including rotation, 
of astronomical and terrestrial bodies can be justified by regarding each 
body as a system of smaller particles whose mutual forces satisfy Newton's 
third law (strong form). The argument of Section 4-2 then gives the law 
of conservation of total angular momentum, the rotational angular momen- 
tum of a body appearing as ordinary orbital angular momentum (r x p) of 
the particles of which it is composed. This argument is subject to the same 
criticism as applied above to the case of linear momentum. If the "par- 
ticles" of which a body is composed are atoms and molecules, then Newton's 
third law does not always hold, particularly in its strong form; moreover, 
the laws of quantum mechanics apply to such particles; and in addition 
atoms and molecules also possess rotational angular momentum which 
must be taken into account. Even the elementary particles — electrons, 
protons, neutrons, etc. — possess an intrinsic angular momentum which 
is not associated with their orbital motion. This angular momentum is 
called spin angular momentum from its analogy with the intrinsic angular 
momentum of rotation of a macroscopic body, and must be included if the 
total is to satisfy a conservation law. Thus we never arrive at the ideal 
simple particle of Newtonian mechanics, described by its position r(t) alone. 
We are left with the choice of accepting the conservation law of angular 
momentum as a basic postulate, or appealing for its justification to theories 
which go beyond classical mechanics. 

The gravitational forces acting between astronomical bodies are con- 
servative, so that the principle of conservation of mechanical energy holds 
very accurately in astronomy. In principle, there is a small loss of mechan- 
ical energy in the solar system due to friction with interplanetary dust and 
rocks, but the effect is too small to produce any observable effects on plane- 
tary motion, even with the high precision with which astronomical events 
are predicted and observed. There is also a very gradual but measurable 
loss of rotational energy of planets and satellites due to tidal friction. For 
terrestrial bodies of macroscopic or microscopic size, friction usually plays 
an important part, and only in certain special cases where friction may be 
neglected can the principle of conservation of energy in the form (4-39) 



168 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

or even (4-40) be applied. However, it was discovered by Joule that we 
can associate energy with heat in such a way that the law of conservation 
of energy of a system of bodies still applies to the total kinetic plus poten- 
tial plus heat energy. If we regard a body as composed of atoms and mole- 
cules, its heat energy turns out to be kinetic and potential energy of ran- 
dom motion of its atoms and molecules. The electromagnetic forces on 
moving charged particles are not conservative, and an electromagnetic 
energy must be associated with the electromagnetic field in order to pre- 
serve the conservation law of energy. Such extensions of the concept of 
energy to include heat and electromagnetic energy are, of course, outside 
the domain of mechanics. When the definition of energy is suitably ex- 
tended to include not only kinetic energy, but energy associated with the 
electromagnetic fields and any other force fields which may act, then a 
law of conservation of energy holds quite generally, in classical, relativistic, 
and quantum physics. 

The conservation laws of energy, momentum, and angular momentum 
are the cornerstones of present-day physics, being generally valid in all 
physical theories. It seems at present an idle exercise to attempt to prove 
them for material bodies within the framework of classical mechanics by 
appealing to an outmoded picture of matter as made up of simple New- 
tonian particles exerting central forces upon one another. The conserva- 
tion laws are in a sense not laws at all, but postulates which we insist must 
hold in any physical theory. If, for example, for moving charged particles, 
we find that the total energy, defined as (T + V), is not constant, we do 
not abandon the law, but change its meaning by redefining energy to in- 
clude electromagnetic energy in such a way as to preserve the law. We 
prefer always to look for quantities which are conserved, and agree to 
apply the names "total energy," "total momentum," "total angular mo- 
mentum" only to such quantities. The conservation of these quantities 
is then not a physical fact, but a consequence of our determination to de- 
fine them in this way. It is, of course, a statement of physical fact, which 
may or may not be true, to assert that such definitions of energy, momen- 
tum, and angular momentum can always be found. This assertion, has 
so far been true; a deeper justification will be suggested at the end of 
Section 9-6. 

4-5 Rockets, conveyor belts, and planets. There are many problems 
that can be solved by appropriate applications of the conservation laws 
of linear momentum, angular momentum, and energy. In solving such 
problems, it is necessary to decide which conservation laws are appropriate. 
The conservation laws of linear and angular momentum or, rather, the 
theorems (4-7) and (4-28) of which they are corollaries, are always appli- 
cable to any physical system provided all external forces and torques are 



4-5] ROCKETS, CONVEYOR BELTS, AND PLANETS 169 

taken into account, and application of one or the other is appropriate 
whenever the external forces or torques are known. The law of conserva- 
tion of kinetic plus potential energy is applicable only when there is no 
conversion of mechanical energy into other forms of energy. We cannot 
use the law of conservation of energy when there is friction, for example, 
unless there is a way to determine the amount of heat energy produced. 

The conservation laws of energy, momentum, and angular momentum 
refer always to a definite fixed system of particles. In applying the con- 
servation laws, care must be taken to decide just how much is included in 
the system to which they are to be applied, and to include all the energy 
and momentum of this system in writing down the equations. One may 
choose the system arbitrarily, including and excluding whatever particles 
may be convenient, but if any forces act from outside the system on parti- 
cles in the system, these must be taken into account. 

A typical problem in which the law of conservation of linear momentum 
is applicable is the conveyor belt problem. Material is dropped continu- 
ously from a hopper onto a moving belt, and it is required to find the force 
F required to keep the belt moving at constant velocity v (Fig. 4-2). Let 
the rate at which mass is dropped on the belt be dm/dt. If to is the mass 
of material on the belt, and M is the mass of the belt (which really does 
not figure in the problem), the total momentum of the system, belt plus 
material on the belt and in the hopper, is 

P = (to + M)v. (4-50) 

We assume that the hopper is at rest; otherwise the momentum of the 
hopper and its contents must be included in Eq. (4-50). The linear mo- 
mentum theorem requires that 

„ dP dm ,. _,. 

This gives the force applied to the belt. The power supplied by the force is 
This is twice the rate at which the kinetic energy is increasing, so that the 




_© ® © <s> 

Fig. 4-2. A conveyor belt. 



170 



THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 



conservation theorem of mechanical energy (4-40) does not apply here. 
Where is the excess half of the power going? 

The equation of motion of a rocket can be obtained from the law of 
conservation of momentum. Let the mass of the rocket at any given in- 
stant be M , and let its speed be v relative to some fixed coordinate system. 
If material is shot out of the rocket motor with an exhaust velocity u rela- 
tive to the rocket, the velocity of the exhaust relative to the fixed coordi- 
nate system is v + u. If an external force F also acts on the rocket, then 
the linear momentum theorem reads in this case: 

£ (Mv) - (v + u) ^ = F. (4-53) 

The first term is the time rate of change of momentum of the rocket. The 
second term represents the rate at which momentum is appearing in the 
rocket exhaust, where— (dM/dt) is the rate at which matter is being ex- 
hausted. The conservation law applies to a definite fixed system of par- 
ticles. If we fix our attention on the rocket at any moment, we must 
remember that at a time dt later this system will comprise the rocket plus 
the material exhausted from the rocket during that time, and both must 
be considered in computing the change in momentum. The equation can 
be rewritten: 

M| = u? + F. (4-54) 

dt dt 

The first term on the right is called the thrust of the rocket motor. Since 
dM/dt is negative, the thrust is opposite in direction to the exhaust veloc- 
ity. The force F may represent air resistance, or a gravitational force. 
Let us solve this equation for the special case where there is no external 
force: 

M$ = u^- (4-55) 

dt dt 

We multiply by dt/M and integrate, assuming that u is constant: 

v-v =-uln^- (4-56) 

The change of speed in any interval of time depends only on the exhaust 
velocity and on the fraction of mass exhausted during that time interval. 
This result is independent of any assumption as to the rate at which mass 
is exhausted. 

Problems in which the law of conservation of angular momentum is use- 
ful turn up frequently in astronomy. The angular momentum of the 
galaxy of stars, or of the solar system, remains constant during the course 



4-6] COLLISION PROBLEMS 171 

of its development provided no material is ejected from the system. The 
effect of lunar tides is gradually to slow down the rotation of the earth. 
As the angular momentum of the rotating earth decreases, the angular 
momentum of the moon must increase. The magnitude of the (orbital) 
angular momentum of the moon is 

L = mr 2 w, (4-57) 

where m is the mass, o> is the angular velocity, and r is the radius of the 
orbit of the moon. We can equate the mass times the centripetal accelera- 
tion to the gravitational force, to obtain the relation 

2 GMm . 

mrur = — — ' (4-58) 

where M is the mass of the earth. Solving this equation for w and substi- 
tuting in Eq. (4-57), we obtain 

L= (GMm 2 r) 1/2 . (4-59) 

Therefore, as the moon's angular momentum increases, it moves farther 
away from the earth. (In attempting to determine the rate of recession 
of the moon by equating the change of L to the change of the earth's rota- 
tional angular momentum, it would be necessary to determine how much 
of the slowing down of the earth's rotation by tidal friction is due to the 
moon and how much to the sun. The angular momentum of the moon 
plus the rotational angular momentum of the earth is not constant because 
of the tidal friction due to the sun. The total angular momentum of the 
earth-moon system about the sun is very nearly constant except for the 
very small effect of tides raised on the sun by the earth.) 

4-6 Collision problems. Many questions concerning collisions of 
particles can be answered by applying the conservation laws. Since the 
conservation laws are valid also in quantum mechanics,* results obtained 
with their use are valid for particles of atomic and subatomic size, as well 
as for macroscopic particles. In most collision problems, the colliding 
particles are moving at constant velocity, free of any force, for some time 
before and after the collision, while during the collision they are under the 
action of the forces which they exert on one another. If the mutual forces 
during the collision satisfy Newton's third law, then the total linear mo- 
mentum of the particles is the same before and after the collision. If 
Newton's third law holds in the strong form, the total angular momentum 



* P. A. M. Dirac, The Principles of Quantum Mechanics, 3rd ed. Oxford: 
Oxford University Press, 1947. (Page 115.) 



172 THE MOTION OF A SYSTEM OF PABTICLES [CHAP. 4 

is conserved also. If the forces are conservative, kinetic energy is con- 
served (since the potential energy before and after the collision is the same). 
In any case, the conservation laws are always valid if we take into account 
all the energy, momentum, and angular momentum, including that asso- 
ciated with any radiation which may be emitted and including any energy 
which is converted from kinetic energy into other forms, or vice versa. 

We consider first a collision between two particles, 1 and 2, in which 
the total kinetic energy and linear momentum are known to be conserved. 
Such a collision is said to be elastic. If we designate by subscripts 1 and 
2 the two particles, and by subscripts I and F the values of kinetic energy 
and momentum before and after the collision respectively, the conservation 
laws require 

PlJ + P2/ = PlF + P2F, (4-60) 

Tu + T 2I = T 1F + T 2F . (4-61) 

Equation (4-61) can be rewritten in terms of the momenta and masses of 
the particles: 

Vu + Pf/ = Pif_ + P2F_ . (4 _g 2) 



2mi 2wi2 2mi 2m 



2 



To specify any momentum vector p, we must specify three quantities, 
which may be either its three components along any set of axes, or its mag- 
nitude and direction (the latter specified perhaps by spherical angles 0, <p). 
Thus Eqs. (4-60) and (4-62) represent four equations involving the ratio 
of the two masses and twelve quantities required to specify the momenta 
involved. If nine of these quantities are given, the equations can be solved 
for the remaining four. In a typical case, we might be given the masses and 
initial momenta of the two particles, and the final direction of motion of 
one of the particles, say particle 1. We could then find the final momen- 
tum p 2 F of particle 2, and the magnitude of the final momentum pip (or 
equivalently, the energy) of particle 1. In many important cases, the mass 
of one of the particles is unknown, and can be computed from Eqs. (4-60) 
and (4-62) if enough is known about the momenta and energies before and 
after the collision. Note that the initial conditions alone are not enough 
to determine the outcome of the collision from Eqs. (4-60) and (4-62) ; we 
must know something about the motion after the collision. The initial 
conditions alone would determine the outcome if we could solve the equa- 
tions of motion of the system. 

Consider a collision of a particle of mass m 1; momentum pu, with a 
particle of mass ra 2 at rest. This is a common case. (There is actually no 
loss of generality in this problem, since, as we pointed out in Section 1-4 
and will show in Section 7-1, if m 2 is initially moving with a uniform veloc- 



4-6] COLLISION PROBLEMS 173 



Pu 




Fig. 4-3. Collision of particle mi with particle m 2 at rest. 

ity v 2 /, Newton's laws are equally applicable in a coordinate system mov- 
ing with uniform velocity v 2J , in which m 2 is initially at rest.) Let m x be 
"scattered" through an angle #1; that is, let &i be the angle between its 
final and its initial direction of motion (Fig. 4-3). The momentum p 2 i? 
must lie in the same plane as pu and pif since there is no component of 
momentum perpendicular to this plane before the collision, and there must 
be none after. Let p 2 F make an angle # 2 with the direction of pu. We 
write out Eq. (4-60) in components along and perpendicular to pu: 

Pu = Pif cos #1 + p 2P cos t? 2 , (4-63) 

= Pif sin &i — p 2F sin & 2 . (4-64) 

Equation (4-62) becomes, in the present case, 



2 2 2 

PlI — PlF __ P2F 

mi m 2 

If two of the quantities 

(Pif/ph, P2f/Pii, #i, #2, mi/m 2 ) 



(4-65) 



are known, the remaining three can be found. If the masses, the initial 
momentum pu, and the angle &i are known, for example, we can solve for 
Pif, P2F, &2 as follows. Transposing the first term on the right to the 
left side in Eqs. (4-63) and (4-64), squaring, and adding, we eliminate & 2 : 

Pit + Pif — ZpuPiF cos#i = pi F . (4-66) 

After substituting this in Eq. (4-65), we can solve for p 1F : 



***■ = —^— cos * ± [(— ^— Y cos 2 * + m * - H 1 ' 2 , 
Pu mi + m 2 l\mi + m 2 f m x + m 2 \ 

(4-67; 
and p 2 F can now be found from Eq. (4-66), and # 2 from Eq. (4-63) 



174 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

If mi > m 2 , the quantity under the radical is zero for #1 = d m , where 
# m is given by 2 

cos 2 # m = 1 - ^|, < d m < f • ' (4-68) 

mi 

If t?i > t? m (and #i < tt), then p\ F /pu is either imaginary or negative, 
neither of which is allowable physically, so that # m represents the maxi- 
mum angle through which m x can be scattered. If m t 3> m 2 , this angle 
is very small, as we know from experience. For t?j < & m , there are two 
values of Pif/Pii, the larger corresponding to a glancing collision, the 
smaller to a more nearly head-on collision; # 2 will be different for these 
two cases. The case t?i = may represent either no collision at all 
(p 1F = p u ) or a head-on collision. In the latter case, 

Pip = tm - m 2 f p 2J . = _2m 2 

Pu mi rf m 2 Pn mi + m 2 

If mi = m 2 , Eqs. (4-67), (4-66), and (4-64) reduce to 

^ = cos «?i, ^ = sin tfi, * a = (f - *i) • (4-70) 
Pu Pu \^ ' 

#i now varies from # x = for no collision to t>i = ir/2 for a head-on 
collision in which the entire momentum is transferred to particle 2. (Actu- 
ally, &i is undefined if p 1F = 0, but j?i — >7r/2 and p iF — * as the colli- 
sion approaches a head-on collision.) If m x < m 2 , all values of #i from 
to 7T are possible, and give a positive value for pir/pu if the plus sign is 
chosen in Eq. (4-67). The minus sign cannot be chosen, since it leads to a 
negative value for pwlpu- If #i = 0i then p XF = pu\ this is the case 
when there is no collision. The case &i = ir corresponds to a head-on 
collision, for which 

""' 0! = 7T, 

(4-71) 

#2 = 0. 



PlF 


m 2 — mi 


Pu 


mi + m 2 


P2F 


2m 2 


Pu 


mi + m 2 



If mi is unknown, but either pu or Tu can be measured or calculated, ob- 
servation of the final momentum of particle 2 (whose mass is assumed 
known) is sufficient to determine wii- As an example, if Tu = pf 7 /2mi 
is known, and T 2F is measured for a head-on collision, m x is given by 
Eq. (4-69) or (4-71) : 

m 2 J-2F L\l2F / -I 

We thus determine m t to within one of two possible values. If results for 



4-6] COLLISION PROBLEMS 175 

a collision with another particle of different mass m 2 , or for a different 
scattering angle, are known, mi is determined uniquely. Essentially this 
method was used by Chadwick to establish the existence of the neutron.* 
Unknown neutral particles created in a nuclear reaction were allowed to 
impinge on matter containing various nuclei of known masses. The ener- 
gies of two kinds of nuclei of different masses m 2 , ra 2 projected forward 
by head-on collisions were measured. By writing Eq. (4-72) for both 
cases, the unknown energy Tu could be eliminated, and the mass mi was 
found to be practically equal to that of the proton. 

We have seen that if we know the initial momenta of two colliding 
particles of known masses, and the angle of scattering #1 (or t? 2 ), all other 
quantities involved in the collision can be calculated from the conserva- 
tion laws. To predict the angles of scattering, we must know not only 
the initial momenta and the initial trajectories, but also the law of force 
between the particles. An example is the collision of two particles acted 
on by a central inverse square law of force, to be treated in Section 4-8. 
Such predictions can be made for collisions of macroscopic or astronomical 
bodies under suitable assumptions as to the law of force. For atomic par- 
ticles, which obey quantum mechanics, this cannot be done, although we 
can predict the probabilities of observing various angles #! (or # 2 ) for 
given initial conditions; that is, we can predict cross sections. In all cases 
where energy is conserved, the relationships between energies, momenta, 
and angles of scattering developed above are valid except at particle veloc- 
ities comparable with the velocity of light. In the latter case, Eqs. (4-60), 
(4-61), (4-63), and (4-64) are still valid, but the relativistic relationships 
between mass, momentum, and energy must be used, instead of Eq. (4-62). 
We quote without proof the relation between mass, momentum, and 
energy as given by the theory of relativity :f 

where c is the speed of light, and m is the rest mass of the particle, that is, 
the mass when the particle is at rest. The relativistic relations between 
kinetic energy, momentum, and velocity are 

T = mc 2 ( 1 - l\ > (4-74) 

Wi - 0> 2 A 2 ) / 

p mv = , (4-75) 

\/l — (y 2 A 2 ) 



* J. Chadwick, Nature, 129, 312 (1932). 

t P. G. Bergmann, Introduction to the Theory of Relativity. New York: Prentice- 
Hall, 1946. (Chapter 6.) 



176 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

which reduce to the classical relations (2-5) and (3-127), when !)«c. 
Unless v is nearly equal to c, the second term on the right in Eq. (4-73) 
is much smaller than the first, and this equation reduces to the classical 
one. With the help of Eq. (4-73), the conservation laws can be applied to 
collisions involving velocities near the speed Of light. 

Atoms, molecules, and nuclei possess internal potential and kinetic 
energy associated with the motion of their parts, and may absorb or re- 
lease energy on collision. Such inelastic collisions between atomic particles 
are said to be of the first kind, or endoergic, if kinetic energy of transla- 
tional motion is absorbed, and of the second kind, or exoergic, if kinetic 
energy is released in the process. It may also happen that in an atomic 
or nuclear collision, the final particles after the collision are not the same 
as the initial particles before collision. For example, a proton may collide 
with a nucleus and be absorbed while a neutron is released and flies away. 
There are a great many possible types of such processes. Two particles 
may collide and stick together to form a single particle or, conversely, a 
single particle may suddenly break up into two particles which fly apart. 
Two particles may collide and form two other particles which fly apart. 
Or three or more particles may be formed in the process and fly apart 
after the collision. In all these cases, the law of conservation of momen- 
tum holds, and the law of conservation of energy also if we take into 
account the internal energy of the atoms and molecules. We consider here 
a case in which a particle of mass mi collides with a particle of mass m 2 
at rest (Fig. 4-4). Particles of masses mz and m± leave the scene of the 
collision at angles # 3 and #4 with respect to the original direction of 
motion of mi. Let kinetic energy Q be absorbed in the process (Q > 
for an endoergic collision; Q — for an elastic collision; Q < for an 



pi 




Fig. 4-4. Collision of mi with TO2 at rest, resulting in the production of m% 
and m,4. 



4-6] COLLISION PROBLEMS 177 

exoergic collision). Then, applying the conservation laws of energy and 
momentum, we write 

Pi = Pz cos # 3 + Pi cos # 4 , (4-76) 

= p 3 sin # 3 — p 4 sin # 4 , (4-77) 

T t = T 3 + Ti + Q. (4-78) 

Since kinetic energy can be expressed in terms of momentum, if the masses 
are known, we may find any three of the quantities p it p 3 , p 4 , # 3 , # 4 , Q in 
terms of the other three. In many cases pi is known, p 3 and «? 3 are meas- 
ured, and it is desired to calculate Q. By eliminating # 4 from Eqs. (4-76) 
and (4-77), as in the previous example, we obtain 

pl = Pi + Pz — 2piPz cos t} 3 . (4-79) 

This may now be substituted in Eq. (4-78) to give Q in terms of known 
quantities: 

f) — T T T — Pl P3 Pl + P3 — 2pip 3 C0S ^3 

2mi 2m 3 2m 4 (4_80) 

or 

« = r. G - 2) - ^ (i + *) + 2 (a^) 1 '^ » 3 . 

(4-81) 

Every step up to the substitution for T\, T 3 , and r 4 is valid also for par- 
ticles moving at velocities of the order of the velocity of light. At high 
velocities, the relativistic relation (4-73) between T and p should be used 
in the last step. Equation (4-81) is useful in obtaining Q for a nuclear 
reaction in which an incident particle mi of known energy collides with a 
nucleus m 2 , with the result that a particle m 3 is emitted whose energy and 
direction of motion can be observed. Equation (4-81) allows us to deter- 
mine Q from these known quantities, taking into account the effect of the 
slight recoil of the residual nucleus m 4 , which is usually difficult to observe 
directly. 

Collisions of inert macroscopic bodies are always inelastic and endoergic, 
kinetic energy being converted to heat by frictional forces during the im- 
pact. Kinetic energy of translation may also be converted into kinetic 
energy of rotation, and conversely. (Exchanges of rotational energy are 
included in Q in the previous analysis.) Such collisions range from the 
nearly elastic collisions of hard steel balls, to which the above analysis of 
elastic collisions applies when rotation is not involved, to completely in- 
elastic collisions in which the two bodies stick together after the collision. 



178 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

Let us consider a completely inelastic collision in which a bullet of mass 
mi, velocity Vi strikes and sticks in an object of mass m 2 at rest. Let the 
velocity of the two after the collision be v 2 . Evidently the conservation 
of momentum implies that v 2 be in the same direction as Vi, and we have: 

miVi = (mi + m 2 )v 2 . (4-82) 

The velocity after the collision is 

v 2 = ^-Vl (4-83) 

mi + m 2 

Energy is not conserved in such a collision. The amount of energy con- 
verted into heat is 

Q = \ mx v\ - i(mx + m 2 )v\ = hm x v\ ( m ™+ m ^ ■ (4-84) 

In a head-on collision of two bodies in which rotation is not involved, it 
was found experimentally by Isaac Newton that the ratio of relative veloc- 
ity after impact to relative velocity before impact is roughly constant for 
any two given bodies. Let bodies mi,m 2 , traveling with initial velocities 
vn, v 2 i along the z-axis, collide and rebound along the same axis with 
velocities Vif, v 2 f. Then the experimental result is expressed by the 
equation* 

v 2F — vif = e(vir — v 2I ), (4-85) 

where the constant e is called the coefficient of restitution, and has a value 
between and 1. If e = 1, the collision is perfectly elastic; if e = 0, it is 
completely inelastic. Conservation of momentum yields, in any case, 

miVu + m 2 v 2 i = miViF + m 2 v 2F . (4-86) 

Equations (4-85) and (4-86) enable us to find the final velocities V\ F and 
v 2 p for a head-on collision when the initial velocities are known. 

4-7 The two-body problem. We consider in this section the motion of a 
system of two particles acted on by internal forces satisfying Newton's 
third law (weak form), and by no external forces, or by external forces 
satisfying a rather specialized condition to be introduced later. We shall 
find that this problem can be separated into two single-particle problems. 



* More recent experiments show that e is not really constant, but depends on 
the initial velocities, on the medium in which the collision takes place, and on 
the past history of the bodies. For a more complete discussion with references, 
see G. Barnes, "Study of Collisions," Am^ J. Phys. 26, 5 (January, 1958). 



4-7] THE TWO-BODY PROBLEM 179 

The motion of the center of mass is governed by an equation (4-18) of the 
same form as that for a single particle. In addition, we shall find that the 
motion of either particle, with respect to the other as origin, is the same 
as the motion with respect to a fixed origin, of a single particle of suitably 
chosen mass acted on by the same internal force. This result will allow 
application of the results of Section 3-14 to cases where the motion of the 
attracting center cannot be neglected. 

Let the two particles have masses mi and m 2 , and let them be acted on 
by external forces Ff , F|, and internal forces F\, F| exerted by each parti- 
cle on the other, and satisfying Newton's third law: 

Fj = -F 2 . (4-87) 

The equations of motion for the system are then 

mif j = FJ + F?, (4-88) 

m 2 f 2 = F 2 + FJ. (4-89) 
We now introduce a change of coordinates: 

„ _ rmtt + mar, , (4 _ 90) 

(4-91) 

r, (4-92) 

r, (4-93) 

where R is the coordinate of the center of mass, and r is the relative coor- 
dinate of m x with respect to m 2 . (See Fig. 4-5.) Adding Eqs. (4-88) and 
(4-89) and using Eq. (4-87), we obtain the equation of motion for R: 

(mi + m 2 )R = FI + F|. (4-94) 

Multiplying Eq. (4-89) by mi, and subtracting from Eq. (4-88) multiplied 







xs. - 


m 


i + m 2 






r = 


= r x - 


- r 2 . 


The 


inverse 


transformation is 










ri = 


R + 


m 2 




mi + m 2 






r 2 = 


R - 


mi 




mi + m 2 




0' 
Fig. 4-5. Coordinates for the two-body problem. 



180 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

by m 2 , using Eq. (4-87), we obtain the equation of motion for r: 

( F e ¥ e \ 

m^mg = (mi + m 2 )F[ + m x m 2 I — ) • (4-95) 

\Wll m 2 / 

We now assume that 

^- = ^-, (4-96) 

■mi m 2 

and introduce the abbreviations 

M = mi + m 2 , (4-97) 

<*= m T 2 ' (4-98) 

mi + m 2 

F = F\ + F|. (4-99) 

Equations (4-94) and (4-95) then take the form of single-particle equa- 
tions of motion : 

M& = F, (4-100) 

fix = F\. (4-101) 

Equation (4-100) is the familiar equation for the motion of the center of 
mass. Equation (4-101) is the equation of motion for a particle of mass n 
acted on by the internal force F\ that particle 2 exerts on particle 1. 
Thus the motion of particle 1 as viewed from particle 2 is the same as if 
particle 2 were fixed and particle 1 had a mass n (n is called the reduced 
mass). If one particle is much heavier than the other, /x is slightly less 
than the mass of the lighter particle. If the particles are of equal mass, /* 
is half the mass of either. We may now apply the results of Section 3-14 
to any two-body problem in which the two particles exert an inverse square 
law attraction or repulsion on each other, provided the external forces are 
either zero or are proportional to the masses, as required by Eq. (4-96). 

Equation (4-96) is satisfied if the external forces are gravitational forces 
exerted by masses whose distances from the two bodies mi and m 2 are 
much greater than the distance r from mi to m 2 . As an example, the motion 
of the earth-moon system can be treated, to a good approximation, by the 
method of this section, since the moon is much closer to the earth than 
either is to the sun (or to the other planets). Atomic particles are acted 
on by electrical forces proportional to their charges, and hence Eq. (4-96) 
holds ordinarily only if the external forces are zero. There is also the less 
important case where the two particles have the same ratio of charge to 
mass, and are acted on by external forces due to distant charges. We may 
remark here that although Eqs. (4-88) and (4-89) are not the correct equa- 
tions for describing the motions of atomic particles, the introduction of the 



4-8] CENTEH-OF-MASS COORDINATES 181 

coordinates R, r, and the reduction of the two-body problem to two one- 
body problems can be carried out in the quantum-mechanical treatment in 
a way exactly analogous to the above classical treatment, under the same 
assumptions about the forces. 

It is worth remarking that the kinetic energy of the two-body system 
can be separated into two parts, one associated with each of the two one- 
body problems into which we have separated the two-body problem. The 
center-of-mass velocity and the relative velocity are, according to Eqs. 
(4-90)-(4-93), related to the particle velocities by 



V — R — miVl + m2V2 

mi + m 2 


(4-102) 


v = f = vi — v 2 , 


(4-103) 


or 




V! = V + -£-V, 

mi 


(4-104) 


v 2 = V - -£- v. 

m 2 


(4-105) 


The total kinetic energy is 




T = \m\v\ + |m 2 f I 




= iMv 2 + iiw 2 - 


(4-106) 



The angular momentum can similarly be separated into two parts: 

L = mi(ri x v x ) + m 2 (r 2 x v 2 ) 
= M(R x V) + M(r x v). (4-107) 

The total linear momentum is, however, just 

P = mxVi + m 2 v 2 = MV. (4-108) 

There is no term /*v in the total linear momentum. 

4-8 Center-of-mass coordinates. Rutherford scattering by a charged 
particle of finite mass. By making use of the results of the preceding sec- 
tion, we can solve a two-body scattering problem completely, if we know 
the interaction force between the two particles, by solving the one-body 
equation of motion for the coordinate r. The result, however, is not in 
a very convenient form for application. The solution r(t) describes the 
motion of particle 1 with respect to particle 2 as origin. Since particle 2 
itself will be moving along some orbit, this is not usually a very convenient 
way of interpreting the motion. It would be better to describe the motion 
of both particles by means of coordinates r x (0, r 2 (i) referred to some fixed 



182 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

origin. Usually one of the particles is initially at rest; we shall take it to 
be particle 2, and call it the target particle. Particle 1, approaching the 
target with an initial velocity vu, we shall call the incident particle. The 
two particles are to be located by vectors r x and r 2 relative to an origin 
with respect to which the target particle is initially at rest. We shall call 
the coordinates r 1; r 2 the laboratory coordinate system. 

The translation from the coordinates R, r to laboratory coordinates is 
most conveniently carried out in two steps. We first introduce a center-of- 
mass coordinate system in which the particles are located by vectors x\, r 2 
with respect to the center of mass as origin: 

^ = fl ~ R ' (4-109) 

*2 = r 2 — K-i 

and, conversely, 

ri = r ' 1+R ' (4-110) 

r 2 = x\ + R. 

The relation between the center-of-mass coordinates and the relative co- 
ordinate r is obtained from Eqs. (4-92) and (4-93) : 

' n ' + Z ""' , <*-»» 

r * mi r = cL-r 

2 mi + m 2 m 2 ' 

The position vectors of the particles relative to the center of mass are con- 
stant multiples of the relative coordinate r. The center of mass has the 
advantage over particle 2, as an origin of coordinates, in that it moves 
with uniform velocity in collision problems where no external forces are 
assumed to act. 

In the center-of-mass coordinate system the total linear momentum is 
zero, and the momenta p\ and p| of the two particles are always equal and 
opposite. The scattering angles t?i and d{ between the two final directions 
of motion and the initial direction of motion of particle 1 are the supple- 
ments of each other, as shown in Fig. 4-6. 

We now determine the relation between the scattering angle © in the 
equivalent one-body problem and the scattering angle j?i in the laboratory 
coordinate system (Fig. 4-7). The velocity of the incident particle in the 
center-of-mass system is related to the relative velocity in the one-body 
problem, according to Eq. (4-111), by 

v{ = -£-v. (4-112) 

mi 



4-8] 



CENTER-OF-MASS COORDINATES 



183 





Fig. 4-6. Two-particle collision in Fig. 4-7. Orbits for two-body colli- 
center-of-mass coordinates. sion in the laboratory system. 



Since these two velocities are always parallel, the angle of scattering &[ 
of the incident particle in the center-of-mass system is equal to the angle 
of scattering © in the one-body problem. The incident particle velocities 
in the center-of-mass and laboratory systems are related by [Eq. (4-110)] 



Vl = vi + V, 



(4-113) 



where the constant velocity of the center of mass can be expressed in terms 
of the initial velocity in the laboratory system by Eq. (4-102) : 



V = 



mi a 

mi + m 2 m 2 



(4-114) 



The relation expressed by Eq. (4-113) is shown in Fig. 4-8, from which 
the relation between d\ = and #i can be determined: 



. . v\f sin @ 
tan#i = — — — , 

v\ F cos S + V 



(4-115) 




Fig. 4-8. Relation between velocities in laboratory and center-of-mass co- 
ordinate systems. 



184 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

or, with the help of Eqs. (4-112) and (4-114), 

sin /a iia\ 

tan th = ~ , , t r > (4-116) 

cos + (miVi/m 2 VF) 

where vz and vp are the initial and final relative speeds, and we have 
substituted i>j for vn, since initially the relative velocity is just the velocity 
of the incident particle. If the collision is elastic, the initial and final speeds 
are the same and Eq. (4-116) reduces to: 

sin © /, ,,« 

tani?! = „ , , -, — r- (4-117) 

1 cos + (wi/m 2 ) 

A similar relation for t? 2 can be worked out. 

If the incident particle is much heavier than the target particle, then 
t?i will be very small, no matter what value may have. This corre- 
sponds to the result obtained in Section 4-6, that t? x can never be larger 
than t> m given by Eq. (4-68), if m l > m 2 . If mi = m 2 , then Eq. (4-117) 
is easily solved for #i : 

x , sin 2 sin (0/2) cos (0/2) _ 

tan * 1 = cos + 1 = 2 cos* (0/2) - tan 2 ' 

0! = J0. (4-118) 

Since may always have any value between and w without violating the 
conservation laws in the center-of-mass system, the maximum value of «?i 
in this case is 7r/2, in agreement with the corresponding result of Section 
4-6. If the target mass m 2 is much larger than the incident mass mi, then 
tan«?i = tan©; this justifies rigorously our application to this case of 
Eq. (3-276) for the Rutherford cross section, deduced in Chapter 3 for 
the one-body scattering problem with an inverse square law force. 

According to the above developments, Eq. (3-276) applies also to the 
two-body problem for any ratio m,i/m 2 of incident mass to target mass, 
but must be interpreted as the angle of scattering in terms of relative 
coordinates, or else in terms of center-of-mass coordinates. That is, da 
in Eq. (3-276) is the cross section for a scattering process in which the 
relative velocity v after the collision makes an angle between and © + d© 
with the initial velocity. Since it is the laboratory scattering angle #i that 
is ordinarily measured, we must substitute for and d© in Eq. (3-276) 
their values in terms of #i and d&i as determined from Eq. (4-117). This 
is most easily done in case m t = m 2 , when, by Eq. (4-118), the Rutherford 
scattering cross section [Eq. (3-276)] becomes 

jr = (Ml) 2 15^*1 2t gin di ^ (4 _ n9) 

\2fwy sin 4 #i 



4-9] THE AT-BODY PROBLEM 185 

4-9 The TV-body problem. It would be very satisfactory if we could 
arrive at a general method of solving the problem of any number of parti- 
cles moving under the forces which they exert on one another, analogous to 
the method given in Section 4-7 by which the two-body problem was re- 
duced to two separate one-body problems. Unfortunately no such general 
method is available for systems of more than two particles. This does not 
mean that such problems cannot be solved. The extremely accurate cal- 
culations of the motions of the planets represent a solution of a problem 
involving the gravitational interactions of a considerable number of bodies. 
However, these solutions are not general solutions of the equations of 
motion, like the system of orbits we have obtained for the two-body case, 
but are numerical solutions obtained by elaborate calculations for specified 
initial conditions and holding over certain periods of time. Even the three- 
body problem admits of no general reduction, say, to three one-body prob- 
lems, or to any other manageable set of equations. 




0' 
Fig. 4-9. Center-of-mass and internal coordinates of a system of particles. 

However, we can partially separate the problem of the motion of a 
system of particles into two problems: first, to find the motion of the center 
of mass, and second, to find the internal motion of the system, that is, the 
motion of its particles relative to the center of mass. Let us define the in- 
ternal coordinate vector rjj. of the fcth particle as the vector from the center 
of mass to the fcth particle (Fig. 4-9) : 

4 = r* - R, k=l,...,N, (4-120) 

i k = R + 4, k=l,...,N. (4-121) 

In view of the definition (4-14) of the center of mass, the internal co- 
ordinates rjj. satisfy the equation 

£ m k ri = 0. (4-122) 



186 THE MOTION OF A SYSTEM OP PARTICLES [CHAP. 4 

We define the center-of-mass velocity and the internal velocities: 

V = % (4-123) 

vj = it = v fc - V. (4-124) 

The total internal momentum of a system of particles (i.e., the momentum 
* relative to the center of mass) vanishes by Eq. (4-122) : 

* 

H :JV 

£ m k vt = 0. (4-125) 

fc=i 

We first show that the total kinetic energy, momentum, and angular 
momentum can each be split up into a part depending on the total mass M 
and the motion of the center of mass, and an internal part depending only 
on the internal coordinates and velocities. The total kinetic energy of the 
system of particles is 

T=22 im*»f. (4-126) 



&=i 



By substituting for v& from Eq. (4-124), and making use of Eq. (4-125), 
we can split T into two parts: 



JV 



T =22 W^ 2 + 2V-v£ + vi a ) 



k=i 

= 22 2 m * v2 + E w* 8 + 12 w * v>v * 
&=i fc=i £=1 

= \MV 2 + J2 \m k vf + v-22 »*▼* 

k=l fc=l 

= \MV 2 + 22 i m ^l 2 - (4-127) 

The total linear momentum is, if we make use of Eqs. (4-124) and 

p = 22 mkVk 

iV N 



= 22 m * v + 2 mkV ' c 



k=l fc=l 

= MV. (4-128) 

The internal linear momentum is zero. 



4-9] THE 2V-BODY PROBLEM 187 

The total angular momentum about the origin is, if we use Eqs. (4-121), 
(4-122), (4-124), and (4-125), 



L = ^2 m k (r k X Vfc) 

N 

= J2 m *( R x V + 4 x V + R x v| + 4 x vl) 

= J2 m *( R xV) + (t ™* r * J x V + R x ( X) m * v * J 

fc=l \fc=l / \fc=l / 



I 

fc=l 



+ 53 m M x ▼*) 



N 
I 



= M(R X V) + £ »*('* x v *)- (4^129) 



Notice that the internal angular momentum depends only on the internal 
coordinates and velocities and is independent of the origin about which L 
is being computed (and from which the vector R is drawn). 

The position of particle k with respect to particle I is specified by the 
vector 

r* - r, = 4 - rf. (4-130) 

The relative positions of the particles with respect to each other depend 
only on the internal coordinates r^, and likewise the relative velocities, so 
that the internal forces Ff. will be expected to depend only on the internal 
coordinates x\, and possibly on the internal velocities. If there is a poten- 
tial energy associated with the internal forces, it likewise will depend only 
on the internal coordinates. 

Although the forces, energy, momentum, and angular momentum can 
each be split into two parts, a part associated with the motion of the center 
of mass and an internal part depending only on the internal coordinates and 
velocities, it must not be supposed that the internal motion and the center- 
of-mass motion are two completely separate problems. The motion of the 
center of mass, as governed by Eq. (4-18), is a separate one-body problem 
when the external force F is given. However, in most cases F will depend 
to some extent on the internal motion of the system. The internal equa- 
tions of motion contain the external forces except in special cases and, 
furthermore, they also depend on the motion of the center of mass. If we 
substitute Eqs. (4-121) in Eqs. (4-1), and rearrange, we have 

m k H = Pi + F£ - m fc R. (4-131) 



188 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

There are many cases, however, in which a group of particles forms a 
system which seems to have some identity of its own independent of other 
particles and systems of particles. An atomic nucleus, made up of neutrons 
and protons, is an example, as is an atom, made up of nucleus and electrons, 
or a molecule, composed of nuclei and electrons, or the collection of particles 
which make up a baseball. In all such cases, it turns out that the internal 
forces are much stronger than the external ones, and the acceleration R is 
small, so that the internal equations of motion (4-131) depend essentially 
only on the internal forces, and their solutions represent internal motions 
which are nearly independent of the external forces and of the motion of 
the system as a whole. The system viewed externally then behaves like 
a single particle with coordinate vector R, mass M, acted on by the 
(external) force F, but a particle which has, in addition to its "orbital" 
energy, momentum, and angular momentum associated with the motion 
of its center of mass, an intrinsic or internal energy and angular momentum 
associated with its internal motion. The orbital and intrinsic parts of the 
energy, momentum, and angular momentum can be identified in Eqs. 
(4-127), (4-128), and (4-129). The internal angular momentum is usually 
called spin and is independent of the position or velocity of the center of 
mass relative to the origin about which the total angular momentum is to 
be computed. So long as the external forces are small, this approximate 
representation of the system as a single particle is valid. Whenever the 
external forces are strong enough to affect appreciably the internal motion, 
the separation into problems of internal and of orbital motions breaks 
down and the system begins to lose its individuality. Some of the central 
problems at the frontiers of present-day physical theories are concerned 
with bridging the gap between a loose collection of particles and a system 
with sufficient individuality to be treated as a single particle. 

4-10 Two coupled harmonic oscillators. A very commonly occurring 
type of mechanical system is one in which several harmonic oscillators 
interact with one another. As a typical example of such a system, con- 
sider the mechanical system shown in Fig. 4-10, consisting of two masses 
mi, m 2 fastened to fixed supports by springs whose elastic constants are 
fci, k 2 , and connected by a third spring of elastic constant £3. We suppose 
the masses are free to move only along the rc-axis; they may, for example, 
slide along a rail. If spring A; 3 were not present, the two masses would 



1 



U-Zl-^l |~ X 2-] 



fa ] mi jfc 3 mi ; k 2 
Fig. 4-10. A simple model of two coupled harmonic oscillators. 



4-10] TWO COUPLED HARMONIC OSCILLATORS 189 

vibrate independently in simple harmonic motion with angular frequencies 
(neglecting damping) 

.0 [k~i /&2 



* = Vv w - = v^- (4 " 132) 

We wish to investigate the effect of coupling these two oscillators to- 
gether by means of the spring k 3 . We describe the positions of the two 
masses by specifying the distances xi and x 2 that the springs ki and k 2 
have been stretched from their equilibrium positions. We assume for 
simplicity that when springs fci and k 2 are relaxed (x\ = x 2 = 0), spring 
k 3 is also relaxed. The amount by which spring fc 3 is compressed is then 
(xi + x 2 )- The equations of motion for the masses mi, m 2 (neglecting 
friction) are 

miXi = — kiXi — k 3 (xi + x 2 ), (4-133) 



m 2 x 2 = —k 2 x 2 — k 3 (xi + x 2 ). 


(4-134) 


he form 




miXi + k'lXi + k a x 2 = 0, 


(4-135) 


m 2 x 2 + ¥ 2 x 2 + k 3 x x = 0, 


(4-136) 


k\ = *i + ^3, 


(4-137) 


k' 2 = k 2 + ^3- 


(4-138) 



where 



We have two second-order linear differential equations to solve simul- 
taneously. If the third terms were not present, the equations would be 
independent of one another, and we would have independent harmonic 
vibrations of X\ and x 2 at frequencies 

«io = >/J> (4-139) 

fcT 



W20 = v^ • ^ 14 °) 

These are the frequencies with which each mass would vibrate if the other 
were held fixed. Thus the first effect of the coupling spring is simply to 
change the frequency of independent vibration of each mass, due to the 
fact that each mass is now held in position by two springs instead of one. 
The third terms in Eqs. (4-135) and (4-136) give rise to a coupling between 
the motions of the two masses, so that they no longer move independently. 



190 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

We may solve Eqs. (4-135), (4-136) by an extension of the method of 
Section 2-8 applicable to any set of simultaneous linear differential equa- 
tions with constant coefficients. We assume that 

Xx = C l e pt , t (4-141) 

x 2 = C 2 e pt , (4-142) 

where Ci, C 2 are constants. Note that the same time dependence is as- 
sumed for both Xi and x 2 , in order that the factor e pt will cancel out when 
we substitute in Eqs. (4-135) and (4-136) : 

(m lP 2 + kUd + k 3 C 2 = 0, (4-143) 

(m 2 p 2 + k' 2 )C 2 + k 3 d = 0. (4-144) 

We now have two algebraic equations in the three unknown quantities 
C\, C 2 , p. We note that either Eq. (4-143) or (4-144) can be solved for 
the ratio C 2 /C\ : 

Cz = _ m x p 2 + fc'i = _ k 3 _ (4-145) 

Ci k 3 m 2 p 2 + k 2 

The two values of C 2 /Ci must be equal, and we have an equation for p: 

m *P 2 + *1 = *2 , (4-146) 

k 3 m 2 p 2 + fc'2 

which may be rearranged as a quadratic equation in p 2 , called the secular 
equation: 

m x m 2 p* + (m 2 k\ + mi k' 2 )p 2 + (k\k' 2 - fc|) = 0, (4-147) 
whose solutions are 

«* = _ if *L + *£} ± [if*L + *kY _ -MMl , fc| 1 1/2 

2 \mi m 2 J l4 \mi m 2 ) mim 2 m\m 2 \ 



I (,,%. _i_ ,..?^ j. Ti /,..?_ _ ,.<?.\ 2 _i_ fc 3 ] 1/2 

m\m 2 \ 



[i<«s. 



(«f + «ao) ± 7 («io - oioV + zPtr ■ (4-148) 



It is not hard to show that the quantity in brackets is less than the square 
of the first term, so that we have two negative solutions for p 2 . If we 
assume that «i > w 2 o, the solutions for p 2 are 

p 2 = _ w 2 = _ (w?o + i Aa)2)j 

2 2 • , 2 1 A Sv (4-149) 

P = — « 2 = — («20 — 2 A&> ), 



4-10] 

where 

Aw 2 = (w?o — w|o) 

with the abbreviation 



TWO COUPLED HARMONIC OSCILLATORS 



\ («fo - «Io)V 



fca 



191 



(4-150) 



(4-151) 



\/mim2 
where k is the coupling constant. If «i = «2o> Eq. (4-150) reduces to 

Aco 2 = 2k 2 . (4-152) 

The four solutions for p are 

p = ±r'coi, ±K02- (4-153) 

If p 2 = — cof, Eq. (4-145) can be written 



C 2 mi , 2 2 n Aw 

— (W], — WioJ = 



Ci 



&s 



Mi 

2/c 2 "\ m2 : 



and if p 2 = — cof, it can be written 



m 2 



Ci m 2 , 2 2 \ Aa> 



(4-154) 



(4-155) 



By substituting from Eq. (4-153) in Eqs. (4-141), (4-142), we get four 
solutions of Eqs. (4-135) and (4-136) provided the ratio C 2 /Ci is chosen 
according to Eq. (4-154) or (4-155). Each of these solutions involves one 
arbitrary constant (Ci or C 2 ). Since the equations (4-135), (4-136) are 
linear, the sum of these four solutions will also be a solution, and is in fact 
the general solution, for it will contain four arbitrary constants (say 

^D ^D ^2> ^2/ ' 

1 2k 2 \toi 2/c 2 \mi 

(4-156) 

Aco* /mi n j ai t , Aco l m Ar</„—i<'it_i_r<.^ i '"2tj_r<'.o— iu '2t 



(4-157) 

(4-158) 
(4-159) 



* 2= 2^>fe C ' ie 



+ ^ yj^- C"ie- tUlt + C 2 e lM2t + C' 2 e- 



In order to make xi and £2 real, we choose 

C 2 = iA 2 e J ' 92 , C" 2 = i^e - * 2 , 



192 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

so that 

A 2 I 

x x = Ai cos («i< + 6{) — 2^- -J^ 2 - A 2 cos (o>2< + B 2 ), (4-160) 
x 2 = ^- J2± Ai cos fat + 0i) + A 2 cos (w 2 < + 2 )- (4-161) 

This is the general solution, involving the four arbitrary constants 
Ai, A 2 , di, 6 2 . We see that the motion of each coordinate is a super- 
position of two harmonic vibrations at frequencies «i and o> 2 . The os- 
cillation frequencies are the same for both coordinates, but the relative 
amplitudes are different, and are given by Eqs. (4-154) and (4-155). 

If Ax or A 2 is zero, only one frequency of oscillation appears. The re- 
sulting motion is called a normal mode of vibration. The normal mode of 
highest frequency is given by 

x t = Ai cos (wi* + 0i), (4-162) 

* a = i£ VS A * cos (wi< + 6l) ' (4_163) 

d = wfo + iAw 2 . (4-164) 

The frequency of oscillation is higher than «i . By referring to Fig. 4-10, 
we see that in this mode of oscillation the two masses mi and m 2 are oscil- 
lating out of phase; that is, their displacements are in opposite directions. 
The mode of oscillation of lower frequency is given by 

Xi = ~ i£ -n/S a * cos (W2< + 02) ' (4_165) 

x 2 = A 2 cos (u 2 t + 9 2 ), (4-166) 

o>! = wlo — iAw 2 . (4-167) 

In this mode, the two masses oscillate in phase at a frequency lower 
than w 2 o- The most general motion of the system is given by Eqs. (4-160), 
(4-161), and is a superposition of the two normal modes of vibration. 

The effect of coupling is thus to cause both masses to participate in the 
oscillation at each frequency, and to raise the highest frequency and lower 
the lowest frequency of oscillation. Even when both frequencies are 
initially equal, the coupling results in two frequencies of vibration, one 
higher and one lower than the frequency without coupling. When the 
coupling is very weak, i.e., when 

k 2 « Kcofo - «io), (4-168) 



4-10] TWO COUPLED HARMONIC OSCILLATORS 193 

then Eq. (4-150) becomes 

Ac ° 2 - 2 2K 2 • (4- 169 ) 

w 10 — W20 

For the highest frequency mode of vibration, the ratio of the amplitude of 
vibration of mass m 2 to that of mass mi is then 

x 2 Aco 2 lm[ . k 2 



Thus, unless w 2 « Bi, the mass m 2 oscillates at much smaller amplitude 
than TO X . Similarly, it can be shown that for the low-frequency mode of 
vibration, m x oscillates at much smaller amplitude than m 2 . If two oscil- 
lators of different frequency are weakly coupled together, there are two 
normal modes of vibration of the system. In one mode, the oscillator of 
higher frequency oscillates at a frequency slightly higher than without 
coupling, and the other oscillates weakly out of phase at the same fre- 
quency. In the other mode, the oscillator of lowest frequency oscillates at 
a frequency slightly lower than without coupling, and the other oscillates 
weakly and in phase at the same frequency. At or near resonance, when 
the two natural frequencies coi and u 20 are equal, the condition for weak 
coupling [Eq. (4-168)] is not satisfied even when the coupling constant is 
very small. Aco 2 is then given by Eq. (4-152), and we find for the two nor- 
mal modes of vibration : 

(4-171) 
w 2 = w? ± k 2 . (4-172) 

The two oscillators oscillate in or out of phase with an amplitude ratio 
depending only on their mass ratio, and with a frequency higher or lower 
than the uncoupled frequency by an amount depending on the coupling 
constant. 

An interesting special case is the case of two identical oscillators 
(mi = m 2 , hi = k 2 ) coupled together. The general solution (4-160), 
(4-161) is, in this case, 

Xi — Ai cos (wi< + 81) — A 2 cos (co 2 < + 2 ), (4-173) 

x 2 = Ai cos (wi< + 0i) + A 2 cos (« 2 * + #2)) (4-174) 

where «i and « 2 are given by Eq. (4-172). If A 2 = 0, we have the high- 
frequency normal mode of vibration, and if Ai = 0, we have the low-fre- 
quency normal mode. Let us suppose that initially m 2 is at rest in its 
equilibrium position, while mi is displaced a distance A from equilibrium 



194 THE MOTION OF A SYSTEM OP PARTICLES [CHAP. 4 



and released at t = 0. The choice of constants which fits these initial 
conditions is 

61 = 2 = 0, 

Ai = —A 2 = %A, 



(4-175) 



so that Eqs. (4-173), (4-174) become 

xi = JA (coswi* + COSW2O, (4-176) 

x 2 = \A (coswi* — cosw 2 <), (4-177) 

which can be rewritten in the form 

Xl = A cos ("Up* t ) cos (^Jp> f ) , (4-178) 

x 2 = -A sin (e^p> ,) sin («L±«S ,) . (4-179) 

If the coupling is small, wi and w 2 are nearly equal, and xi and a; 2 oscillate 
rapidly at the angular frequency («i + w 2 )/2 == wi = w 2 , with an am- 
plitude which varies sinusoidally at angular frequency (w x — w 2 )/2. The 
motion of each oscillator is a superposition of its two normal-mode motions, 
which leads to beats, the beat frequency being the difference between the 
two normal-mode frequencies. This is illustrated in Fig. 4-11, where os- 
cillograms of the motion of x 2 are shown: (a) when the high-frequency 
normal mode alone is excited, (b) when the low-frequency normal mode 
is excited, and (c) when oscillator m x alone is initially displaced. In Fig. 
4-12, oscillograms of Xi and x 2 as given by Eqs. (4-178), (4-179) are 
shown. It can be seen that the oscillators periodically exchange their 
energy, due to the coupling between them. Figure 4-13 shows the same 
motion when the springs fci and k 2 are not exactly equal. In this case, 
oscillator mi does not give up all its energy to ra 2 during the beats. Figure 
4-14 shows that the effect of increasing the coupling is to increase the beat 
frequency «i — co 2 [Eq. (4-172)]. 

If a frictional force acts on each oscillator, the equations of motion 
(4-135) and (4-136) become 

miXi + bi±! + kiXt + k 3 x 2 = 0, (4-180) 

m 2 x 2 + b 2 x 2 + k' 2 x 2 + k 3 xi — 0, (4-181) 



4-10] 



TWO COUPLED HARMONIC OSCILLATORS 



195 




(a) lulfiikjtii 







Fig. 4-11. Motion of coupled har- Fig. 4-12. Motion of two identical 
monic oscillators, (a) High-frequency coupled oscillators, 
normal mode, (b) Low-frequency nor- 
mal mode, (c) mi initially displaced. 




Fig. 4-13. Motion of two nonidenti- Fig. 4-14. Motion of two coupled 
cal coupled oscillators. oscillators, (a) Weak coupling, (b) 

Strong coupling. 



196 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

where b% and b 2 are the respective friction coefficients. The substitution 
(4-141), (4-142) leads to a fourth-degree secular equation for p: 

m 1 m 2 p 4: + (m 2 bi + mib 2 )p 3 + (m 2 fci + mik' 2 + b x b 2 )p 2 

+ (&i*i + b 2 k\)p + (k\k' 2 - fc|) = 0. (4-182) 

This equation cannot be solved so easily as Eq. (4-147). The four roots 
for p are, in general, complex, and have the form (if &i and b 2 are not too 
large) 

p = — 7i ± t'wi, 

(4-183) 

p = —y 2 ± iu 2 . 

That the roots have this form with Ti and T 2 positive can be shown (though 
not easily) algebraically from a study of the coefficients in Eq. (4-182). 
Physically, it is evident that the roots have the form (4-183), since this will 
lead to damped vibrations, the expected result of friction. If 6i and b 2 are 
large enough, one or both of the pairs of complex roots may become a pair 
of real negative roots, the corresponding normal mode or modes being 
overdamped. A practical solution of Eq. (4-182) can, in general, be ob- 
tained only by numerical methods when numerical values for the constants 
are given, although an approximate algebraic solution can be found when 
the damping is very small. 

The problem of the motion of a system of two coupled harmonic oscil- 
lators subject to a harmonically oscillating force applied to either mass can 
be solved by methods similar to those which apply to a single harmonic 
oscillator. A steady-state solution can be found in which both oscillators 
oscillate at the frequency of the applied force with definite amplitudes and 
phases, depending on their masses, the spring constants, the damping, and 
the amplitude and phase of the applied force. The system is in resonance 
with the applied force when its frequency corresponds to either of the two 
normal modes of vibration, and the masses then vibrate at large amplitudes 
limited only by the damping. The general solution consists of the steady- 
state solution plus the general solution of the unforced problem. A super- 
position principle can be proved according to which, if a number of forces 
act on either or both masses, the solution is the sum of the solutions with 
each force acting separately. This theorem can be used to treat the prob- 
lem of arbitrary forces acting on the two masses. 

Other types of coupling between the oscillators are possible in addition 
to coupling by means of a spring as in the example above. The oscillators 
may be coupled by frictional forces. A simple example would be the case 
where one mass slides over the other, as in Fig. 4-15. We assume that the 
force of friction is proportional to the relative velocity of the two masses. 



4-10] 



TWO COUPLED HARMONIC OSCILLATORS 



197 



u_n— J 



KM 



mi 




mo i i ■" 

-^2 — ^ *'2 

Fig. 4-15. Frictional coupling. 







-h— "-> 



t3; 



-x 2 



-A 



Fig. 4-16. Coupling through a mass. 



The equations of motion of mi and m 2 are then 

m^i = — k\Xi — b{±i + x 2 ), 
m 2 x 2 = —k 2 x 2 — b(x 2 + ±i), 



or 



miXi + b±x + kiXi + bx 2 = 0, 
w 2 x 2 + bx 2 + k 2 x 2 + &±! = 0. 



(4-184) 
(4-185) 

(4-186) 
(4-187) 



The coupling is expressed in Eqs. (4-186), (4-187) by a term in the equation 
of motion of each oscillator depending on the velocity of the other. The 
oscillators may also be coupled by a mass, as in Fig. 4-16. It is left to the 
reader to set up the equations of motion. (See Problem 26 at the end of 
this chapter.) 

Two oscillators may be coupled in such a way that the force acting on 
one depends on the position, velocity, or acceleration of the other, or on 
any combination of these. In general, all three types of coupling occur to 
some extent; a spring, for example, has always some mass, and is subject 
to some internal friction. Thus the most general pair of equations for two 
coupled harmonic oscillators is of the form 

m x x\ + bi±i + k x xi + m c x 2 + b c x 2 + k c x 2 = 0, (4-188) 
m 2 x 2 + b 2 x 2 + k 2 x 2 + m c xi + b e x\ + k c x\ = 0. (4-189) 

These equations can be solved by the method described above, with similar 
results. Two normal modes of vibration appear, if the frictional forces 
are not too great. 

Equations of the form (4-188), (4-189), or the simpler special cases con- 
sidered in the preceding discussions, arise not only in the theory of coupled 
mechanical oscillators, but also in the theory of coupled electrical circuits. 
Applying Kirchhoff 's second law to the two meshes of the circuit shown in 
Fig. 4-17, with mesh currents i\, i 2 around the two meshes as shown, we 
obtain 



(L + Li)g! + (R + Rtfa + 



fe + £) qi 



1 



+ Lq 2 + Rq 2 + ^ q 2 = 0, 
(4-190) 



198 



THE MOTION OF A SYSTEM OF PARTICLES 
Ki R 2 



[chap. 4 




Ci C 2 

Fig. 4-17. Coupled oscillating circuits. 



and 



(L + L 2 )q 2 + (R+ R 2 )q 2 + 



(h+£ 



q 2 + Lqt + Rqi+jjqi = 0, 

(4-191) 

where qi and q 2 are the charges built up on Ci and C 2 by the mesh currents 
ii and i 2 . These equations have the same form as Eqs. (4-188), (4-189), 
and can be solved by similar methods. In electrical circuits, the damping 
is often fairly large, and finding the solution becomes a formidable task. 

The discussion of this section can be extended to the case of any number 
of coupled mechanical or electrical harmonic oscillators, with analogous 
results. The algebraic details become almost prohibitive, however, unless 
we make use of more advanced mathematical techniques. We therefore 
postpone further discussion of this problem to Chapter 12. 

All mechanical and electrical vibration problems reduce in the limiting 
case of small amplitudes of vibration to problems involving one or several 
coupled harmonic oscillators. Problems involving vibrations of strings, 
membranes, elastic solids, and electrical and acoustical vibrations in trans- 
mission lines, pipes, or cavities, can be reduced to problems of coupled 
oscillators, and exhibit similar normal modes of vibration. The treatment 
of the behavior of an atom or molecule according to quantum mechanics 
results in a mathematical problem identical with the problem of coupled 
harmonic oscillators, in which the energy levels play the role of oscillators, 
and external perturbing influences play the role of the coupling mechanism. 



199 

Problems 

1. Formulate and prove a conservation law for the angular momentum about 
the origin of a system of particles confined to a plane. 

2. Water is poured into a barrel at the rate of 120 lb per minute from a height 
of 16 ft. The barrel weighs 25 lb, and rests on a scale. Find the scale reading 
after the water has been pouring into the barrel for one minute. 

3. A scoop of mass mi is attached to an arm of length I and negligible weight. 
The arm is pivoted so that the scoop is free to swing in a vertical arc of radius I. 
At a distance I directly below the pivot is a pile of sand. The scoop is lifted until 
the arm is at a 45° angle with the vertical, and released. It swings down and 
scoops up a mass m,2 of sand. To what angle with the vertical does the arm of the 
scoop rise after picking up the sand? This problem is to be solved by considering 
carefully which conservation laws are applicable to each part of the swing of the 
scoop. Friction is to be neglected, except that required to keep the sand in the 
scoop. 

4. (a) A spherical satellite of mass m, radius a, moves with speed v through 
a tenuous atmosphere of density p. Find the frictional force on it, assuming 
that the speed of the air molecules can be neglected in comparison with v, 
and that each molecule which is struck becomes embedded in the skin of the 
satellite, (b) If the orbit is a circle 400 km above the earth (radius 6360 km), 
where p = 10 -11 kgm/m -3 , and if a = 1 m, m = 100 kgm, find the change 
in altitude and the change in period of revolution in one week. 

5. A two-stage rocket is to be built capable of accelerating a 100-kgm payload 
to a velocity of 6000 m/sec in free flight. (In a two-stage rocket, the first stage 
is detached after exhausting its fuel, before the second stage is fired.) Assume 
that the fuel used can reach an exhaust velocity of 1500 m/sec, and that struc- 
tural requirements imply that an empty rocket (without fuel or payload) will 
weigh 10 % as much as the fuel it can carry. Find the optimum choice of masses 
for the two stages so that the total take-off weight is a minimum. Show that it 
is impossible to build a single-stage rocket which will do the job. 

6. A rocket is to be fired vertically upward. The initial mass is Mo, the exhaust 
velocity — u is constant, and the rate of exhaust — (dM/dt) = A is constant. 
After a total mass A M is exhausted, the rocket engine runs out of fuel. Neglect- 
ing air resistance and assuming that the acceleration g of gravity is constant, 
set up and solve the equation of motion, and show that if Mo, u, and AM are 
fixed, then the larger the rate of exhaust A, that is, the faster it uses up its fuel, 
the greater the maximum altitude reached by the rocket. 

7. A uniform spherical planet of radius a revolves about the sun in a circular 
orbit of radius ro, and rotates about its axis with angular velocity coo, normal 
to the plane of the orbit. Due to tides raised on the planet, its angular velocity 
of rotation is decreasing. Find a formula expressing the orbit radius r as a func- 
tion of angular velocity to of rotation at any later or earlier time. [You will 
need formulas (5-9) and (5-91) from Chapter 5.] Apply your formula to the 
earth, neglecting the effect of the moon, and estimate how much farther the 
earth will be from the sun when the day has become equal to the present year. 
If the effect of the moon were taken into account, would the distance be greater 
or less? 



200 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

*8. A mass m of gas and debris surrounds a star of mass M . The radius of the 
star is negligible in comparison with the distances to the particles of gas and 
debris. The material surrounding the star has initially a total angular momentum 
L, and a total kinetic and potential energy E. Assume that m « Jlf , so that 
the gravitational fields due to the mass m are negligible in comparison with that 
of the star. Due to internal friction, the surrounding material continually loses 
mechanical energy. Show that there is a maximum energy AE which can be lost 
in this way, and that when this energy has been lost, the material must all lie 
on a circular ring around the star (but not necessarily uniformly distributed). 
Find AE and the radius of the ring. (You will need to use the method of La- 
grange multipliers.) 

9. A particle of mass mi, energy Tu collides elastically with a particle of mass 
m,2, at rest. If the mass «2 leaves the collision at an angle #2 with the original 
direction of motion of mi, find the energy T%f delivered to particle m2. Show 
that T2F is a maximum for a head-on collision, and that in this case the energy 
lost by the incident particle in the collision is 

T T 4miw 2 „ 

Tu ~ TlF = (mi + m 2 )2 Tu - 

10. A cloud-chamber picture shows the track of an incident particle which 
makes a collision and is scattered through an angle #1. The track of the target 
particle makes an angle #2 with the direction of the incident particle. Assuming 
that the collision was elastic and that the target particle was initially at rest, find 
the ratio mi/wi2 of the two masses. (Assume small velocities so that the classical 
expressions for energy and momentum may be used.) 

11. Show that an elastic collision corresponds to a coefficient of restitution 
e = 1, that is, show that for a head-on elastic collision between two particles, 
Eq. (4-85) holds with e = 1. 

12. Calculate the energy loss Q for a head-on collision between a particle of 
mass mi, velocity f 1 with a particle of mass wi2 at rest, if the coefficient of restitu- 
tion is e. 

13. A particle of mass mi, momentum pu collides elastically with a particle 
of mass m2, momentum P21 going in the opposite direction. If mi leaves the colli- 
sion at an angle <&i with its original course, find its final momentum. 

14. Find the relativistic corrections to Eq. (4-81) when the incident particle 
mi and the emitted particle W13 move with speeds near the speed of light. Assume 
that the recoil particle mi is moving slowly enough so that the classical relation 
between energy and momentum can be used for it. 

15. A particle of mass mi, momentum pi collides with a particle of mass m,2 at 
rest. A reaction occurs from which two particles of masses WI3 and mi result, 
which leave the collision at angles #3 and #4 with the original path of mi. Find 
the energy Q absorbed in the reaction in terms of the masses, the angles, and pi. 

16. A nuclear reaction whose Q is known occurs in a photographic plate in 
which the tracks of the incident particle mi and the two product particles m3 
and m4 can be seen. Find the energy of the incident particle in terms of mi, 
m3, m4, Q, and the measured angles #3 and #4 between the incident track and 
the two final tracks. What happens if Q = 0? 



PROBLEMS 201 

17. The Compton scattering of x-rays can be interpreted as the result of elastic 
collisions between x-ray photons and free electrons. According to quantum 
theory, a photon of wavelength X has a kinetic energy hc/\, and a linear momen- 
tum of magnitude h/\, where h is Planck's constant and c is the speed of light. 
In the Compton effect, an incident beam of x-rays of known wavelength Xj in a 
known direction is scattered in passing through matter, and the scattered radia- 
tion at an angle #1 to the incident beam is found to have a longer wavelength \f, 
which is a function of the angle t?i- Assuming an elastic collision between an inci- 
dent photon and an electron of mass m at rest, set up the equations expressing 
conservation of energy and momentum. Use the relativistic expressions for the 
energy and momentum of the electron. Show that the change in x-ray wave- 
length is 

Xf — X/ = — (1 — cos #i), 
mc 

and that the ejected electron appears at an angle given by 

sin j?i 



tan #2 = 



[1+ (h/\imc)](l — cos^x) 



18. Work out a correction to Eq. (3-267) which takes into account the motion 
of the central mass M under the influence of the revolving mass m. A pair of stars 
revolve about each other, so close together that they appear in the telescope as a 
single star. It is determined from spectroscopic observations that the two stars 
are of equal mass and that each revolves in a circle with speed v and period r 
under the gravitational attraction of the other. Find the mass m of each star by 
using your formula. 

19. Show that if the incident particle is much heavier than the target particle 
(wi 2> m,2), the Rutherford scattering cross section da [Eq. (3-276)] in laboratory 
coordinates is approximately 

dc ± I _£i^ J 2 i /22 2 o 1,2 2ir sin #1 d #i 

\2m 2 vV [1 - (1 - 7 2 # 2 ) 1/2 ] 2 (1 - 7 2 # 2 ) 1/2 

if 7#i < 1, where 7 = mi/m.2- Otherwise, da = 0. 

20. Find an expression analogous to Eq. (4-116) for the angle of recoil of the 
target particle (#2 in Fig. 4-7) in terms of the scattering angle in the equivalent 
one-body problem. Show that, for an elastic collision, 

$2 = 4(t - 0). 

21. Assume that m,2 2> mi, and that = #1 + 5, in Eq. (4-117). Find a 
formula for 5 in terms of t?i. Show that the first-order correction to the Ruther- 
ford scattering cross section [Eq. (3-276)], due to the finite mass of m,2, vanishes. 

22. Set up the equations of motion for Fig. 4-10, assuming that the relaxed 
length of each spring is I, and that the distance between the walls is 3(Z + a), so 
that the springs are stretched, even in the equilibrium position. Show that the 
equations can be put in the same form as Eqs. (4-135) and (4-136). 



202 THE MOTION OF A SYSTEM OF PARTICLES [CHAP. 4 

23. For the normal mode of vibration given by Eqs. (4-162) and (4-163), find 
the force exerted on mi through the coupling spring, and show that the motion 
of xi satisfies the equation for a simple harmonic oscillator subject to this driving 
force. 

24. The system of coupled oscillators shown in Fig. 4-10 is subject to an ap- 
plied force 

F = Fo cos cot, 

applied to mass mi. Set up the equations of motion and find the steady-state 
solution. Sketch the amplitude and phase of the oscillations of each oscillator as 
functions of w. 

25. Find the two normal modes of vibration for a pair of identical damped 
coupled harmonic oscillators [Eqs. (4-180), (4-181)]. That is, mi = m2, &i = 62, 
fci = fe. [Hint: If fe = 0, you can certainly find the solution. You will find this 
point helpful in factoring the secular equation.] 

26. Set up the equations of motion for the system shown in Fig. 4-16. The 
relaxed lengths of the two springs are h, h- Separate the problem into two 
problems, one involving the motion of the center of mass, and the other involving 
the "internal motion" described by the two coordinates x\, X2- Find the normal 
modes of vibration. 



CHAPTER 5 

RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 

5-1 The dynamical problem of the motion of a rigid body. In order to 
apply the theorems of the preceding chapter to the motion of a rigid body, 
we regard a rigid body as a system of many particles whose positions rela- 
tive to one another remain fixed. We may define a rigid body as a system 
of particles whose mutual distances are all constant. The forces which 
hold the particles at fixed distances from one another are internal forces, 
and may be imagined as exerted by rigid weightless rods connected be- 
tween all pairs of particles. Forces like this which maintain certain fixed 
relations between the particles of a system are called forces of constraint. 
Such forces of constraint can always be regarded as satisfying Newton's 
third law (strong form), since the constraints could be maintained by rigid 
rods fastened to the particles by frictionless universal joints. We may 
therefore apply the theorems of conservation of linear and angular momen- 
tum to the motion of a rigid body. For a perfectly rigid body, the theorem 
of conservation of mechanical energy holds also, since we can show by 
Newton's third law that the forces of constraint do no work in a rigid mo- 
tion of the system of particles. The work done by the force exerted by a 
moving rod on a particle at one end is equal and opposite to the work done 
by the force exerted by the rod on a particle at the other end, since both 
particles have the same component of velocity in the direction of the rod 
(Fig. 5-1): 

F 2 -»i-vi + Fi_ 2 -v 2 = F 2 _rvi — F 2 _»i-v 2 (5-1) 

= F 2 _ > i-(vi— v 2 ) 

= 0. 

We shall base our derivation of the equations of motion of a rigid body 
on these conservation laws. No actual solid body is ever perfectly rigid, 
so that our theory of the motion of rigid bodies will be an idealized ap- 
proximation to the motion of actual bodies. However, in most applica- 
tions the deviation of actual solid bodies from true rigidity is not sig- 
nificant. In a like spirit is our assumption that the ideal rigid body can 
be imagined as made up of ideal point particles held at fixed distances 
from one another. 

A solid body of ordinary size is composed of such a large number of 
atoms and molecules that for most purposes it is more convenient to repre- 

203 



204 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 




Fig. 5-1. Forces exerted by two particles connected by a rigid rod. 



sent its structure by specifying the average density p of mass per unit 
volume at each point in the body. The density is defined by 



P = 



dM 
dV' 1 



(5-2) 



where dM is the total mass in a volume dV which is to be chosen large 
enough to contain a large number of atoms, yet small enough so that the 
properties of the material are practically uniform within the volume dV. 
Only when a, dV satisfying these two requirements can be chosen in the 
neighborhood of a point in the body can the density p be properly defined 
at that point. Sums over all the particles, such as occur in the expressions 
for total mass, total momentum, etc., can be replaced by integrals over the 
volume of the body. For example, the total mass is 



M^J^rm^ fff p 



dV. 



(5-3) 



(body) 



Further examples will appear in the following sections. 

In order to describe the position of a rigid body in space, six coordinates 
are needed. We may, for example, specify the coordinates (x lt y lt z{) of 
some point Pi in the body. Any other point P 2 of the body a distance r 
from Pi will then lie somewhere on a sphere of radius r with center at 
(£11 2/1, zi). We can locate P 2 on this sphere with two coordinates, for 
example, the spherical coordinate angles 2 , <p 2 with respect to a set of 
axes through the point tei, j/i, z\). Any third point P 3 a distance a^O 
from the line through Pi and P 2 must now lie on a circle of radius a about 
this line. We can locate P3 on this circle with one coordinate. We thus 
require a total of six coordinates to locate the three points Pi, P 2 , P3 of 



5-1] THE DYNAMICAL PROBLEM OF THE MOTION OF A RIGID BODY 205 

the body, and when three noncollinear points are fixed, the locations of 
all points of a rigid body are fixed. There are many possible ways of choos- 
ing six coordinates by which the position of a body in space can be specified. 
Usually three of the six coordinates are used as above to locate some point 
in the body. The remaining three coordinates determine the orientation 
of the body about this point. 

If a body is not connected to any supports, so that it is free to move in 
any manner, it is convenient to choose the center of mass as the point to 
be located by three coordinates (X, Y, Z), or by the vector R. The motion 
of the center of mass R is then determined by the linear momentum theo- 
rem, which can be expressed in the form (4-18) : 

MR = F, (5-4) 

where M is the total mass and F is the total external force. The equation 
for the rotational motion about the center of mass is given by the angular 
momentum theorem (4-28) : 

Tt = N ' <™> 

where L is the angular momentum and N is the torque about the point R. 
If the force F is independent of the orientation of the body in space, as in 
the case of a body moving in a uniform gravitational field, the motion of 
the center of mass is independent of the rotational motion, and Eq. (5-4) 
is a separate equation which can be solved by the methods of Chapter 3. 
If the torque N is independent of the position R of the center of mass, or 
if R(t) is already known, so that N can be calculated as a function of time 
and of the orientation of the body, then the rotational motion about the 
center of mass may be determined from Eq. (5-5). In the more general 
case, when F and N each depend on both position and orientation, Eqs. 
(5-4) and (5-5) must be solved simultaneously as six coupled equations 
in some suitable set of coordinates; this case we shall not attempt to treat, 
although after the reader has studied Chapter 11, he will be able to set up 
for himself the six equations which must be solved. 

If the body is constrained by external supports to rotate about a fixed 
point 0, then moments and torques are to be computed about that point. 
We have to solve Eq. (5-5) for the rotation about the point 0. In this case 
Eq. (5-4) serves only to determine the constraining force required to 
maintain the point at rest. 

The difficulty in applying Eq. (5-5) lies in the choice of three coordinates 
to describe the orientation of the body in space. The first thought that 
comes to mind is to choose a zero position for the body, and to specify 
any other orientation by specifying the angles of rotation <p x , <p y , <p z , about 



206 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 



three perpendicular axes, required to bring the body to this orientation. 
However, a little experimenting with a solid body will convince anyone 
that no suitable coordinates of this sort exist. Consider, for example, 
the position specified by <p x = 90°, <p y = 90°, <p z = 0. If a body is first 
rotated 90° about the x-axis, and then 90° about the 2/-axis, the final posi- 
tion will be found to be different from that resulting from a 90° rotation 
about the y-axis followed by a 90° rotation about the x-axis. It turns out 
that no simple symmetric set of coordinates can be found to describe the 
orientation of a body, analogous to the coordinates x, y, z which locate the 
position of a point in space. We therefore postpone to Chapter 11 the 
treatment of the rather difficult problem of the rotation of a body around 
a point. We shall discuss here only the simple problem of rotation about 
a fixed axis. 

5-2 Rotation about an axis. It requires only one coordinate to specify 
the orientation of a body which is free to rotate only about a fixed axis. 
Let the fixed axis be taken as the z-axis, and let a line OA in the body, 
through the axis and lying in (or parallel to) the xy-plane, be chosen. We 
fix the position of the body by specifying the angle between the line OA 
fixed in the body and the x-axis. Choosing cylindrical coordinates to 
locate each particle in the body, we now compute the total angular momen- 
tum about the z-axis. (See Fig. 5-2.) We shall write r, instead of p» to 
represent the distance of particle ro 8 - from the z-axis, in order to avoid con- 
fusion with the density p : 

L = ]C m » r ^;- (5-6) 

i 

Let fa be the angle between the direction of the line OA in the body and 
the direction of the radius from the z-axis to the particle rm. Then, for a 




Fig. 5-2. Coordinates of a particle in a rigid body. 



5-2] ROTATION ABOUT AN AXIS 207 

rigid body, ft is constant, and 

<Pi = + Pi, (5-7) 

<Pi = e. (5-8) 

Substituting in Eq. (5-6), we have 

L = 2^ m i r i^ 



= (E rmr f\ 

= 1.6, (5-9) 



where 

The quantity 7« is a constant for a given body rotating about a given axis, 
and is called the moment of inertia about that axis. We may also express 
I t as an integral over the body: 



fffpr 2 dV. (5-11) 



(body) 

It is sometimes convenient to introduce the radius of gyration k z denned by 
the equation 

Mh\ = /,; (5-12) 

that is, k z is a radius such that if all the mass of the body were situated a 
distance k z from the axis, its moment of inertia would be I„. 

Using Eq. (5-9), we may write the component of Eq. (5-5) along the 
axis of rotation in the form 

^ = 1,9 = N z , (5-13) 

where N z is the total external torque about the axis. Equation (5-13) is 
the equation of motion for rotation of a rigid body about a fixed axis. It 
has the same form as Eq. (2-1) for the motion of a particle along a straight 
line. The problem of rotation of a body about a fixed axis is therefore 
equivalent to the problem treated in Chapter 2. All methods and results 
of Chapter 2 can be extended directly to the present problem according to 
the following scheme of analogy: 



208 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 



Rectilinear motion 



Rotation about a fixed axis 



position: 




X 




angular position : 


velocity : 




v = 


X 


angular velocity: w = 6 


acceleration : 




a = 


X 


angular acceleration : a = S 


force : 




F 




torque : N z 


mass: 




m 




moment of inertia: I z 


potential energy: 






potential energy: 


V(x) = - 


7* 


F(x) dx 




V(d) = - f" N z (e) de 


F(x) = - 


dV 
dx 






»m = - W 


kinetic energy: 


T = 


%mx 2 


kinetic energy: T = J/ 2 d 


linear momentum 


V = 


mx 


angular momentum : L = I z 6 



The only mathematical difference between the two problems is that the 
moment of inertia I z depends upon the location of the axis in the body, 
while the mass of a body does not depend on its position or on its motion. 
This does not affect the treatment of rotation about a single fixed axis. 
The rotational potential and kinetic energies defined by the equations, 



V{6) = - f N z {6) de, 



N z 
T 



_ dV 
de ' 

iij 2 , 



(5-14) 

(5-15) 
(5-16) 



are not merely analogous to the corresponding quantities defined by Eqs. 
(2-41), (2-47), and (2-5) for linear motion. They are, in fact, equal to the 
potential and kinetic energies, defined in Chapters 2 and 4, of the system 
of particles making up the rigid body. The potential energy defined by 
Eq. (5-14), for example, is the work done against the forces whose torque 
is N z , when the body is rotated through the angle — e . The kinetic 
energy defined by Eq. (5-16) is just the sum of the ordinary kinetic energies 
of motion of the particles making up the body. The proof of these state- 
ments is left as an exercise. 



5-3 The simple pendulum. As an example of the treatment of rota- 
tional motion, we consider the motion of a simple pendulum, consisting of 
a mass m suspended from a fixed point by a string or weightless rigid rod 



5-3] 



THE SIMPLE PENDULUM 



209 




Fig. 5-3. The simple pendulum. 



of length I. If a string supports the mass m, we must suppose that it 
remains taut, so that the distance I from m to remains constant; other- 
wise we cannot treat the system as a rigid one. We consider only motions 
of the pendulum in one vertical plane, in order to be able to apply the 
simple theory of motion about a single fixed axis through 0. We then 
have (Fig. 5-3) 

I z = ml 2 , (5-17) 



N z = — mgl sin 6, 



(5-18) 



where the z-axis is an axis through perpendicular to the plane in which 
the pendulum is swinging. The torque is taken as negative, since it acts 
in such a direction as to decrease the angle 6. Substituting in the equation 
of motion (5-13), we find 

(5-19) 



sin 6. 



This equation is not easy to solve. If, however, we consider only small 
oscillations of the pendulum (say <3C tt/2), then sin = 0, and we can 
write 

6 + | 6 = 0. (5-20) 



This is of the same form as Eq. (2-84) for the harmonic oscillator. Its 
solution is 

= k cos (cot + 0), (5-21) 

where 

(5-22) 



= (!)'". 



and k and are arbitrary constants which determine the amplitude and 
phase of the oscillation. Notice that the frequency of oscillation is 



210 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 



independent of the amplitude, provided the amplitude is small enough so 
that Eq. (5-20) is a good approximation. This is the basis for the use 
of a pendulum to regulate the speed of a clock. 

We can treat the problem of motion at large amplitudes by means of 
the energy integral. The potential energy associated with the torque 
given by Eq. (5-18) is 



V(0) = — / —mgl sin dd 
J'* 

= —mgl cos 0, 



(5-23) 



where we have taken a — ir/2 for convenience. We could have written 
down V(9) right away as the gravitational potential energy of a mass m, 
referred to the horizontal plane through as the level of zero potential 
energy. The energy integral is 



%ml 2 6 2 — mgl cos = E. 



(5-24) 



We could prove that E is constant from the equation of motion (5-13), 
but we need not, since the analogy described in the preceding section 
guarantees that all theorems for one-dimensional linear motion will hold in 
their analogous forms for rotational motion about an axis. The potential 
energy V(8) is plotted in Fig. 5-4. We see that for — mgl < E < mgl, the 
motion is an oscillating one, becoming simple harmonic motion for E 
slightly greater than —mgl. For E > mgl, the motion is nonoscillatory; 
steadily increases or steadily decreases, with oscillating between a maxi- 
mum and minimum value. Physically, when E > mgl, the pendulum has 
enough energy to swing around in a complete circle. (In this case, of 
course, the mass must be held by a rigid rod instead of a string, unless 6 is 
very large.) This motion is still a periodic one, the pendulum making 
one complete revolution each time increases or decreases by 2ir. In 




Fig. 5-4. Potential energy for simple pendulum. 



5-3] THE SIMPLE PENDULUM 211 

either case, the attempt to solve Eq. (5-24) for leads to the equation 

I 9 dO (2gY» 

J, (E/mgl + cos $)H» ~\l/ t - {d lb) 

The integral on the left must be evaluated in terms of elliptic functions. 
The period of the motion can be obtained by integrating between appro- 
priate limits. When the motion is oscillatory {E < mgl), the maximum 
value k of is given, according to Eq. (5-24), by 

E = — mgl cos k. (5-26) 

Equation (5-25) becomes, in this case, 

f de (2 9 y> 

J, (COS - COS K)l/2 _ \ I ) h & Zl) 



which can also be written 



/ 



*— -»(i) u \ M 



, [sin 2 (k/2) - sin 2 (tf/2)]i/» \l 

The angle oscillates between the limits ±k. We now introduce a new 
variable <p which runs from to 27T for one cycle of oscillation of 0: 



where 



sin 8/2 1 . ._ on . 

Sin * = ihW2 = a Sin 2' (5 - 29) 

a = sin | • (5-30) 



With these substitutions, Eq. (5-28) can be written 

(fa (g 



[ 

Jo 



* - / M/2 



o (1 - a 2 sin 2 v )i/2 \l 



t, (5-31) 



where we have taken O = 0, for convenience. The integral is now in a 
standard form for elliptic integrals. When a is small, the integrand can be 
expanded in a power series in a 2 : 



L 



[1 + £a 2 sin 2 <p + ■ ■ ■] d v = (jj t. (5-32) 



This can be integrated term by term: 



<P + ia 2 (2*> - sin2<p) -] = (j j t. (5-33) 



212 



BIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



The period of the motion is obtained by setting <p = 2ir : 



—©■V*+-) 



[chap. 5 



(5-34) 



Thus as the amplitude of oscillation becomes large, the period becomes 
slightly longer than for small oscillations, a prediction which is readily 
verified experimentally by setting up two pendulums of equal length and 
setting them to swinging at unequal amplitudes. Equation (5-33) can be 
solved approximately for <p by successive approximations, and the result 
substituted in Eq. (5-29), which can be solved for 8 by successive approxi- 
mations. The result, to a second approximation, is 



where 



= (k + j^) sin o>'t + ^ sin 3c/ 1, (5-35) 

"-*-(dVS+")" <-> 



If we neglect terms in /c 2 and k 3 , this solution agrees with Eq. (5-21). At 
larger amplitudes in second approximation, the frequency is slightly lower 
than at small amplitudes, and the motion of 8 contains a small third har- 
monic term. 

5-4 The compound pendulum. A rigid body suspended and free to 
swing about an axis is called a compound pendulum. We assume that the 
axis does not pass through the center of mass, and we specify the position 
of the body by the angle 8 between a vertical line and a perpendicular line 
drawn from a point on the axis, through the center of mass G (Fig. 5-5). 
In order to compute the total torque exerted by gravity, we anticipate a 




Fig. 5-5. The compound pendulum. 



5-4] THE COMPOUND PENDULUM 213 

theorem, to be proved later, that the total torque is the same as if the 
total gravitational force were applied at the center of mass G. We then 
have, using Eqs. (5-12) and (5-13), 

Mk%e = —Mgh sin 6, (5-37) 

where h is the distance OG. This equation is the same as Eq. (5-19) for a 
simple pendulum of length I, if we take 

* = f- (5-38) 

The point 0' a distance I from along the line through the center of 
mass G is called the center of oscillation. If all the mass M were at 0', 
the motion of the pendulum would be the same as its actual motion, for 
any given initial conditions. If the distance O'G is h', we have 

I = h + h', (5-39) 

hh' = k% — h 2 . (5-40) 

It will be shown in the next section that the moment of inertia about 
any axis equals the moment of inertia about a parallel axis through the 
center of mass G plus Mh 2 , where h is the distance from the axis to G. 
Let kg be the radius of gyration about G. We then have 

k 2 = k% + h 2 , (6-41) 

so that Eq. (5-40) becomes 

hh' = k%. (5-42) 

Since this equation is symmetrical in h and hi , we conclude that if the 
body were suspended about a parallel axis through 0', the center of oscilla- 
tion would be at 0. The acceleration g of gravity can be measured very 
accurately by measuring the period of small oscillations of a pendulum and 
using Eq. (5-22). If a compound pendulum is used, the radius of gyration 
must be known, or the period measured about two axes, preferably 0, 0', 
so that the radius of gyration can be eliminated from the equations. 

Consider a rigid body suspended from an axis about which it is free to 
move. Let it be struck a blow at a point 0' a distance I from the axis, 
the direction of the blow being perpendicular to the line 00' from the 
axis to 0'. Place 0' so that the line 00' passes through the center of mass 
G, and let h, h' be the distances OG, O'G (Fig. 5-6). The impulse delivered 
at the point 0' by the force F' during the blow is 



V = /> 



dt. (5-43) 



214 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 













) '" 


h 






Q» 


1 


h' 
0' 




I » 




/ r 



Fig. 5-6. Rigid body pivoted at and struck a blow at 0'. 

At the instant the blow is struck, a force F will, in general, have to be 
exerted on the body at the point on the axis in order to keep fixed. 
The impulse delivered to the body at is 



-/, 



dt. 



(5-44) 



An equal and opposite impulse — / is delivered by the body to the support 
at 0. The momentum theorem for the component P of linear momentum 
of the body in the direction of F is: 



f = J t Wh8) = F + F', 



(5-45) 



where 6 is the angular velocity of the body about 0. From this we have, 
for the momentum just after the blow, 



MU = J + J', 



(5-16) 



assuming that the body is initially at rest. The conservation theorem of 
angular momentum about is: 

§ =±(Mkle) = F>l. (H7) 

Integrating, we have, for the angular momentum just after the blow, 

Mkl6 = J'l. (5-48) 

We eliminate 6 between Eqs. (5-46) and (5-48) : 

= h% (l + j) • (5-49) 



hi 



5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 215 

We now ask for the condition that no impulsive force be exerted on the 
axis at at the instant of the blow, i.e., J = 0: 

hi = h%. (5-50) 

This equation is identical with Eq. (5-38) and may also be expressed in 
the symmetrical form [Eq. (5-42)] 

hh' = k%. (5-51) 

The point 0' at which a blow must be struck in order that no impulse be 
felt at the point is called the center of percussion relative to 0. We see 
that the center of percussion is the same as the center of oscillation relative 
to 0, and that is the center of percussion relative to 0'. If the body is 
unsupported, and is struck at 0', its initial motion will be a rotation about 
0. For example, a batter tries to hit a baseball at the center of percussion 
relative to his hands. If the ball hits very far from the center of percus- 
sion, the blow is transmitted to his hands by the bat. 

5-5 Computation of centers of mass and moments of inertia. We have 
given in Section 4-1 the following definition of center of mass for a system 
of particles : 

R = m E «**• ( 5 ~ 52 ) 

For a solid body, the sum may be expressed as an integral: 

R = jjfff P TdV, (5-53) 

or, in component form, 

X = ±fffpxdV, (5-54) 

Y = IfIfI pydV > (5_55) 

Z = ±fffpzdV. (5-56) 

The integrals can be extended either over the volume of the body, or over 
all space, since p = outside the body. These equations define a point G 
of the body whose coordinates are (X, Y, Z). We should first prove that 
the point G thus defined is independent of the choice of coordinate system. 
Since Eq. (5-52) or (5-53) is in vector form, and makes no reference to 
any particular set of axes, the definition of G certainly does not depend on 



216 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 




Fig. 5-7. Location of center of mass relative to two different origins. 

any particular choice of directions for the axes. We should prove, how- 
ever, that G is independent also of the choice of origin. Consider a system 
of particles, and let any particle ra; be located by vectors r t - and x\ with 
respect to any two origins and 0'. If a is the vector from to 0', the 
relation between r; and r£ is (Fig. 5-7) 



ft + a. 



(5-57) 



The centers of mass G, G' with respect to 0, 0' are located by the vectors 
R and R', where R' is defined by 



R ' = m £ m ^- 

i 

Using Eq. (5-57), we can rewrite Eq. (5-58) : 

i 

= MZ m ^-M a E m » 

i i 

= R - a. 



(5-58) 



(5-59) 



Thus R and R' are vectors locating the same point with respect to and 0' 
so that G and G' are the same point. 

General theorems like the one above can be proved either for a system of 
particles or for a body described by a density p. Whichever point of view 
is adopted in any proof, a parallel proof can always be given from the 
other point of view. 

Much of the labor involved in the calculation of the position of the 
center of mass from Eqs. (5-54), (5-55), (5-56) can often be avoided by 
the use of certain laborsaving theorems, including the theorem proved 



5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 217 

above which allows us a free choice of coordinate axes and origin. We 
have first the following theorem regarding symmetrical bodies: 

Theorem. If a body is symmetrical with respect to a plane, its 

center of mass lies in that plane. (5-60) 

When we say a body is symmetrical with respect to a plane, we mean that 
for every particle on one side of the plane there is a particle of equal mass 
located at its mirror image in the plane. For a continuously distributed 
mass, we mean that the density at any point equals the density at its 
mirror image in the plane. Choose the origin in the plane of symmetry, 
and let the plane of symmetry be the xy-plaue. Then in computing Z from 
Eq. (5-56) [or (5-52)], for each volume element (or particle) at a point 
(x, y, z) above the xy-plane, there is, by symmetry, a volume element of 
equal mass at the point (x, y, —z) below the zy-plane, and the contribu- 
tions of these two elements to the integral in Eq. (5-56) will cancel. Hence 
Z = 0, and the center of mass lies in the xy-p\a,ne. This proves Theorem 
(5-60). The theorem has a number of obvious corollaries: 

If a body is symmetrical in two planes, its center of mass lies on 

their line of intersection. (5-61) 

// a body is symmetrical about an axis, its center of mass lies on 

that axis. (5-62) 

If a body is symmetrical in three planes with one common point, 

that point is its center of mass. (5-63) 

7/ a body has spherical symmetry about a point (i.e., if the density 
depends only on the distance from that point), that point is its 
center of mass. (5-64) 

These theorems enable us to locate the center of mass immediately in some 
cases, and to reduce the problem to a computation of only one or two 
coordinates of the center of mass in other cases. One should be on the 
lookout for symmetries, and use them to simplify the problem. Other 
cases not included in these theorems will occur (e.g., the parallelepiped), 
where it will be evident that certain integrals will be equal or will cancel, 
and the center of mass can be located without computing them. 

Another theorem which often simplifies the location of the center of 
mass is that if a body is composed of two or more parts whose centers of 
mass are known, then the center of mass of the composite body can be 
computed by regarding its component parts as single particles located at 
their respective centers of mass. Let a body, or system of particles, be 
composed of n parts of masses M\, . . . , M n . Let any part M& be composed 
of Nk particles of masses muu ■ ■ ■ , mkN k , located at the points r^, . . . , tkN k - 



218 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 



Then the center of mass of the part Mu is located at the point 



and 



i _ k _ 



M k = X m k i. 
1=1 



(5-65) 



(5-66) 



The center of mass of the entire body is located at the point 
where 



k=i i=i 



h=l 1=1 



By Eq. (5-65), Eq. (5-67) becomes 

M 



k=i 



and by Eq. (5-66), Eq. (5-68) becomes 

M = J) M k . 



(5-67) 



(5-68) 



(5-69) 



(5-70) 



A-=l 



Equations (5-69) and (5-70) are the mathematical statement of the theo- 
rem to be proved. 




5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 219 



As an example, let us consider a uniform rectangular block with a cylin- 
drical hole drilled out, as shown in Fig. 5-8. By the symmetry about the 
two vertical planes bisecting the block parallel to its sides, we conclude 
that the center of mass lies along the vertical line AB through the centers 
of the top and bottom faces. Let the center of mass of the block lie a 
distance Z below A, and let the density of the block be p. If the hole were 
not cut out, the mass of the block would be 6 cm X 4 cm X 10 cm X p, 
and its center of mass would be at the midpoint of AB, 5 cm from A. The 
mass of the material drilled out is ir cm 2 X 6 cm X p, and its center of 
mass, before it was removed, was on AB, 2 cm below A. Hence the 
theorem (5-69) above allows us to write 

(6 cm X 4 cm X 10 cm X p) X 5 cm = (7r cm 2 X 6 cm X p) X 2 cm 

+ 6 cm X (4 cm X 10 cm — tt cm 2 ) X p X Z. 
The solution for Z is 

6X4X10X5 — 7TX6X2 
Z = 6X(4X10-tt) Cm - 

As a second example, we locate the center of mass of a hemisphere of 
radius a. By symmetry, if the density is uniform, the center of mass lies 
on the axis of symmetry, which we take as the z-axis. We have then to 





(a) 



(b) 





(c) (d) 

Fig. 5-9. Methods of integrating over a hemisphere. 



220 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

compute only the integral in Eq. (5-56). The integral can be set up in 
rectangular, cylindrical, or spherical coordinates (Fig. 5-9) : 

ra /-(a 2 -* 2 ) 1 ' 2 r^-z 2 -^) 1 ' 2 

Rectangular: Z = £ J ^ j^^^ J^^^u* >> z dx * *■ 

ra r2* r (a 2 -* 2 ) 1 ' 2 

Cylindrical : Z = -jj / / / pzr dr dip dz. 

Ml y 2== o J <p=o Jr=o 

ra r*l2 r2r 

Spherical : Z = -^ I J I (pr cos 0)r 2 sin dr dO dip. 
M. J r== o Je=,o J v=o 

Any one of these expressions can be used to evaluate Z for any density 
distribution. If p is uniform, we can also build up the hemisphere out of 
rings or disks and save one or two integrations. For example, building up 
the hemisphere out of disks perpendicular to the z-axis (this is equivalent 
to carrying out the integration over r and <p in cylindrical coordinates), we 
can write 

z ~h/,.o z '" ( ° 2 "* 2) * 
-GspsXt*)-* (5 " 71 > 

where the integrand is zp times the volume of a disk of thickness dz, radius 
(a 2 - z 2 ) 1 ' 2 . 

When the density p is uniform, the center of mass of a body depends 
only on its geometrical shape, and is given by 

K = ±fffrdV. (5-72) 

v 

The point G whose coordinate R is given by Eq. (5-72) is called the cen- 
troid of the volume V. If we replace the volume V by an area A or curve C 
in space, we obtain formulas for the centroid of an area or of a curve: 

R = jffrdA, (5-73) 

A 

R = i J r da, (5-74) 

where s is the length of the curve C. The following two theorems, due to 
Pappus, relate the centroid of an area or curve to the volume or area swept 
out by it when it is rotated about an axis: 



5-51 COMPUTATION OF CEN'TERS OF MASS AND MOMENTS OF INERTIA 221 





Fig. 5-10. Pappus' first theorem. 



Fig. 5-11. Sphere formed by rota- 
ting a semicircle. 



Theorem 1. If a plane curve rotates about an axis in its own 
plane which does not intersect it, the area of the surface of revolu- 
tion which it generates is equal to the length of the curve multiplied 
by the length of the path of its centroid, (5-75) 



Theorem 2. If a plane area rotates about an axis in its own plane 
which does not intersect it, the volume generated is equal to the area 
times the length of the path of its centroid. (Z 



-76) 



The proof of Theorem 1 is very simple, with the notation indicated in 
Fig. 5-10: 

A = f 2iryds= 2tt f y ds = 2ttYs, (5-77) 

where Y is the y-co ordinate of the centroid of the curve C, and s is its 
length. The proof of Theorem 2 is similar and is left to the reader. These 
theorems may be used to determine areas and volumes of figures sym- 
metrical about an axis when the centroids of the generating curves or 
areas are known, and conversely. We locate, for example, the position of 
the center of mass of a uniform semicircular disk of radius a, using Pappus' 
second theorem. If the disk is rotated about its diameter, the volume of 
the sphere generated, by Pappus' theorem (Fig. 5-11), is 



fra> = (^) (2*7), 



from which we obtain 



Y m 



ia 
3tt' 



(5-78) 



The moment of inertia / of a body about an axis is defined by Eq, (5-1 0) : 

/ = £ rmrl (5-79) 



222 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

P 




Fig. 5-12. Location of point P with respect to points and G. 



or 



1 = fffpr 2 dV, 



(5-80) 



where r is the distance from each point or particle of the body to the given 
axis. We first prove several laborsaving theorems regarding moments 
of inertia : 

Parallel Axis Theorem. The moment of inertia of a body 
about any given axis is the moment of inertia about a parallel axis 
through the center of mass, plus the moment of inertia about the 
given axis if all the mass of the body were located at the center of 
mass. (5-81) 

To prove this theorem, let Io be the moment of inertia about a z-axis 
through the point 0, and let Ig be the moment of inertia about a parallel 
axis through the center of mass G. Let r and r' be the vectors to any 
point P in the body, from and G, respectively, and let R be the vector 
from to G. The components of these vectors will be designated by 
(x, y, z), (x f , y', z'), and (X, Y, Z). Then, since (Fig. 5-12) 



r = r' + R, 



we see that 



x 2 + y 2 = (x' + X) 2 + (y' + Y) 2 

= x' 2 + y' 2 + X 2 + Y 2 + 2Xx' + 2Yy', 

so that the moment of inertia Io is 
Io = fff(x 2 + y 2 )pdV 

= fff(x' 2 + y' 2 )p dV + (X 2 + Y 2 ) Jffp dV + 2X fjfx'p dV 

+ 2Yfffy'pdV. (5-82) 



5-5] COMPUTATION OF CENTERS OF MASS AND MOMENTS OF INERTIA 223 

The first integral is la, and the integral in the second term is the total 
mass M of the body. The integrals in the last two terms are the same as 
the integrals occurring in Eqs. (5-54) and (5-55), and define the x- and 
^/-coordinates of the center of mass relative to G. Since G is the center of 
mass, these integrals are zero, and we have 

lo = Ig + M(X 2 + Y 2 ). (5-83) 

This is the mathematical statement of the Parallel Axis Theorem. If we 
know the moment of inertia about any axis, and can locate the center of 
mass, we can use this theorem to determine the moment of inertia about 
any other parallel axis. 

The moment of inertia of a composite body about any axis may be 
found by adding the moments of inertia of its parts about the same axis, a 
statement which is obvious from the definition of moment of inertia. This 
fact can be put to use in the same way as the analogous result for the 
center of mass of a composite body. 

A body whose mass is concentrated in a single plane is called a plane 
lamina. We have the following theorem for a plane lamina: 

Perpendicular Axis Theorem. The sum of the moments of 
inertia of a plane lamina about any two perpendicular axes in the 
plane of the lamina is equal to the moment of inertia about an axis 
through their point of intersection perpendicular to the lamina. (5-84) 

The proof of this theorem is very simple. Consider any particle of mass m 
in the xy-pl&ne. Its moments of inertia about the x- and y-axes are 

I x = my 2 , I y = mx 2 . (5-85) 

Adding these, we have the moment of inertia of m about the z-axis: 

h + I v = m(x 2 + y 2 ) = /*. (5-86) 

Since the moment of inertia of any lamina in the rcy-plane is the sum of 
the moments of inertia of the particles of which it is composed, we have 
theorem (5-84). 

We illustrate these theorems by finding the moments of inertia of a 
uniform circular ring of radius a, mass M, lying in the xy-p\&ne (Fig. 5-13). 
The moment of inertia about a z-axis perpendicular to the plane of the ring 
through its center is easily computed: 

7, = Ma 2 . (5-87) 

The moments I x and I y are evidently equal, and we have, therefore, by 

theorem (5-84), 

I x = \l t = %Ma 2 . (5-88) 



224 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 





Fig. 5-13. A ring of radius o. 



Fig. 5-14. Finding the moment of 
inertia of a disk. 




Fig. 5-15. Finding the moment of inertia of a solid sphere. 



The moment of inertia about an axis A tangent to the ring is, by the Parallel 

Axis Theorem, 

I A = l x + Ma 2 = §Ma 2 . (5-89) 

The moment of inertia of a solid body can be set up in whatever coordi- 
nate system may be convenient for the problem at hand. If the body is 
uniform and of simple shape, its moment of inertia can be computed by 
considering it as built up out of rods, rings, disks, etc. For example, the 
moment of inertia of a circular disk about an axis perpendicular to it 
through its center can be found by regarding the disk as made up of rings 
(Fig. 5-14) and using Eq. (5-87) : 



pa 

Jo 



r p2irr dr 



TOjO 



\Ma 2 . 



(5-90) 



The moment of inertia of a solid sphere can be calculated from Eq. (5-90) 
by regarding the sphere as made up of disks (Fig. 5-15) : 

I = f Sl^li {(ma 2 sin 2 0) d(a cos B) = ^f = \Ma 2 . (5-91) 



5-6] STATICS OF RIGID BODIES 225 

A body with a piece cut out can be treated by setting its moment of 
inertia equal to the moment of inertia of the original body minus the 
moment of inertia of the piece cut out, all moments being taken, of course, 
about the same axis. 

5-6 Statics of rigid bodies. The equations of motion of a rigid body 
are Eqs. (5-4) and (5-5) : 

MR = J2 F * (5-92) 

i 

^f=Z N t-o. (5-93) 

i 

Equation (5-92) determines the motion of the center of mass, located by 
the vector R, in terms of the sum of all external forces acting on the body. 
Equation (5-93) determines the rotational motion about a point 0, which 
may be the center of mass or a point fixed in space, in terms of the total 
external torque about the point 0. Thus if the total external force acting 
on a rigid body and the total external torque about a suitable point are 
given, its motion is determined. This would not be true if the body were 
not rigid, since then it would be deformed by the external forces in a man- 
ner depending on the particular points at which they are applied. Since 
we are concerned only with external forces throughout this section, we 
may omit the superscript e. It is only necessary to give the total torque 
about any one point 0, since the torque about any other point 0' can then 
be found from the following formula: 

J2 Ha,' = X) N i0 + (r G - to') X £ F f , (5-94) 

i i i 

where io, r < are vectors drawn to the points 0, 0' from any convenient 
origin. That is, the total torque about 0' is the total torque about plus 
the torque about 0' if the total force were acting at 0. The proof of 
Eq. (5-94) is very simple. Let r» be the vector from the origin to the point 
at which Fj acts. Then 

J2 N i0 ' = ]£ (ii - to-) X F s - 

i i 

= S (fi — to + TO — To') X F t - 
i 

= S ( r < - f o) X F»- + X) (To - T ') X Fi 

i i 

= £ N i0 + (r - to') X J2 *<• 



226 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

P i? 




0" 

Fig. 5-16. The torque is independent of where along its line of action a force acts. 

If, in particular, a rigid body is at rest, the left members of Eqs. (5-92) 
and (5-93) are zero, and we have 

i 

J2 N * = °- ( 5 ~ 96 ) 

i 

These are the conditions to be satisfied by the external forces and torques 
in order for a rigid body to be in equilibrium. They are not sufficient to 
guarantee that the body is at rest, for it might still be in uniform transla- 
tional and rotational motion, but if the body is initially at rest, it will 
remain at rest when these conditions are satisfied. It is sufficient for the 
total torque in Eq. (5-96) to be zero about any point, since then, by 
Eq. (5-94), it will be zero also about every other point if Eq. (5-95) holds. 
In computing the torque due to a force F, it is necessary to know not 
only the vector F (magnitude and direction), but also the point P of 
the body at which the force acts. But if we draw a line through P in 
the direction of F, then if F acts at any other point P' of this line, its 
torque will be the same, since, from the definition of the cross product, 
it can be seen (Fig. 5-16) that 

r P X F = r P > X F. (5-97) 

(The areas of the parallelograms involved are equal.) The line through P 
in the direction of F is called the line of action of the force. It is often 
convenient in computing torques to remember that the force may be con- 
sidered to act anywhere along its line of action. A distinction is some- 
times made in this connection between "free" and "sliding" vectors, the 
force being a "sliding" vector. The terminology is likely to prove confus- 
ing, however, since as far as the motion of the center of mass is concerned 
[Eq. (5-92)], the force is a "free" vector, i.e., may act anywhere, whereas 
in computing torques, the force is a "sliding" vector, and for a nonrigid 
body, each force must be localized at the point where it acts. It is better 
to define vector, as we defined it in Section 3-1, as a quantity having 
magnitude and direction, without reference to any particular location in 
space. Then, in the case of force, we need for some purposes to specify not 



5-6] STATICS OF RIGID BODIES 227 




Fig. 5-17. A single force C whose torque is the sum of the torques of A and B. 

only the force vector F itself, but in addition the point or line on which 
the force acts. 

A theorem due to Varignon states that if C = A + B, then the moment 
of C about any point equals the sum of the moments of A and B, provided 
A, B, and C act at the same point. The theorem is an immediate conse- 
quence of the vector identity given by Eq. (3-27) : 

rxC = rxA + rxB, if C = A + B. (5-98) 

This theorem allows us to compute the torque due to a force by adding 
the torques due to its components. Combining Varignon's theorem with 
the result of the preceding paragraph, we may reduce the torque due to 
two forces A, B acting in a plane, as shown in Fig. 5-17, to the torque 
due to the single force C, since both A and B may be considered to act at 
the intersection of their lines of action, and Eq. (5-98) then allows us to 
add them. We could now add C similarly to any third force acting in the 
plane. This process can be continued so long as the lines of action of the 
forces being added are not parallel, and is related to a more general theorem 
regarding forces in a plane to be proved below. 

Since, for a rigid body, the motion is determined by the total force and 
total torque, we shall call two systems of forces acting on a rigid body 
equivalent if they give the same total force, and the same total torque 
about every point. In view of Eq. (5-94), two systems of forces are then 
equivalent if they give the same total force, and the same total torque 
about any single point. It is of interest to know, for any system of forces, 
what is the simplest system of forces equivalent to it. 

If a system of forces F t - acting at points r t - is equivalent to a single force F 
acting at a point r, then the force F acting at r is said to be the resultant of 
the system of forces F,-. If F is the resultant of the system of forces F;, 
then we must have 

F = Z) Fi, (5-99) 

(r-ro)xF=^ (r« - r ) X F if (5-100) 



228 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

where r is any point about which moments are taken. By Eq. (5-94), if 
Eq. (5-99) holds, and Eq. (5-100) holds for any point r , it holds for every 
point r . The force — F acting at r is called the equilibrant of the system; 
if the equilibrant is added to the system of forces, the conditions for equi- 
librium are satisfied. 

An example of a system of forces having a resultant is the system of 
gravitational forces acting on a body near the surface of the earth. We 
shall show that the resultant in this case acts at the center of mass. Let 
the acceleration of gravity be g. Then the force acting on a particle m< is 

F< = m ig . (5-101) 

The total force is 

F = £ »« = M &> ( 5 - 102 ) 

i 

where M is the total mass. The total torque about any point is, with 
as origin, 

^2 n j0 = J2 ( r * x m *s) 

i i 

x g 



= te m * r ») 



= MR x g 

= R X Ig, (5-103) 

where R is the vector from to the center of mass. Thus the total torque 
is given by the force Mg acting at the center of mass. Because of this 
result, the center of mass is also called the center of gravity. We shall see 
in the next chapter that, in general, this result holds only in a uniform 
gravitational field, i.e., when g is the same at all points of the body. If 
the system of forces acting on a rigid body has a resultant, the forces may 
be replaced by this resultant in determining the motion of the body. 
A system of forces whose sum is zero is called a couple: 

2 F * = °- ( 5 " 104 ) 

i 

A couple evidently has no resultant, except in the trivial case where the 
total torque is zero also, in which case the resultant force is zero. By 
Eqs. (5-94) and (5-104), a couple exerts the same total torque about every 
point : 

X) N i0 < = X) N <°- ( 5 - 105 ) 



5-6] STATICS OF RIGID BODIES 229 

P 




p, 

Fig. 5-18. A simple couple. 

Thus a couple is characterized by a single vector, the total torque, and 
all couples with the same total torque are equivalent. The simplest sys- 
tem equivalent to any given couple, if we exclude the trivial case where 
the total torque is zero, is a pair of equal and opposite forces F, — F, acting 
at points P, P' separated by a vector r (Fig. 5-18) such that 

J2 Nio = r X F. (5-106) 

i 

Equation (5-106) states that the moment of the given couple about 
equals the moment of the couple (F, — F) about P'; the two systems 
are therefore equivalent, since the point about which the moment of a 
couple is computed is immaterial. The force F and the points P and P' 
are by no means uniquely determined. Since only the cross product r X F 
is determined by Eq. (5-106), we can choose P arbitrarily; we can choose 
the vector F arbitrarily except that it must lie in the plane perpendicular to 
the total torque; and we can then choose r as any vector lying in the same 
plane and determining with F a parallelogram whose area is the magnitude 
of the total torque. 

The problem of finding the simplest system equivalent to any given sys- 
tem of forces is solved by the following theorems : 

Theorem I. Every system of forces is equivalent to a single force 
through an arbitrary point, plus a couple (either or both of which 
may be zero). (5-107) 

To prove this, we show how to find the equivalent single force and couple. 
Let the arbitrary point P be chosen, let the sum of all the forces in the sys- 
tem be F, and let their total torque about the point P be N. Then, if we 
let the single force F act at P, and add a couple whose torque is N, we have 
a system equivalent to the original system. Since the couple can be com- 
posed of two forces, one of which may be allowed to act at an arbitrary 
point, we may let one force of the couple act at the point P, and add it to 
F to get a single force acting at P plus the other force of the couple. This 
proves 

Theorem II. Any system of forces can be reduced to an equiva- 
lent system which contains at most two forces. (5-108) 



230 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

The following theorem can be proved in two ways: 

Theorem III. A single nonzero force and a couple in the same 
plane (i.e., such that the torque vector of the couple is perpendicular 
to the single force) have a resultant and, conversely, a single force 
is equivalent to an equal force through any arbitrary point, plus a 
couple. (5-109) 

Since a couple with torque N is equivalent to a pair of equal and opposite 
forces, F, — F, where F may be chosen arbitrarily in the plane perpendicular 
to N, we may always choose F equal to the single force mentioned in the 
theorem. Furthermore, we may choose the point of action of F arbitrarily. 
Given a single nonzero force F acting at P, and a couple, we form a couple 
(F, — F) equivalent to the given couple, and let — F act at P; F and — F 
then cancel at P, and the remaining force F of the couple is the single re- 
sultant. The converse can be proved by a similar argument. 

The other method of proof is as follows. Let the given force F act at a 
point P, and let the total torque of the couple be N. Then the torque of 
the system about the point P is N. We take any vector r, in the plane 
perpendicular to N, which forms with F a parallelogram of area N, and 
let P' be the point displaced from P by the vector r. If the single force F 
acts at P', the torque about P will then be N, and hence this single force is 
equivalent to the original force F acting at P plus the couple. We can 
combine Theorems I and III to obtain 

Theorem IV. Every system of forces is equivalent to a single 
force plus a couple whose torque is parallel to the single force. 
(Or, alternatively, every system of forces is equivalent to a couple 
plus a single force perpendicular to the plane of the couple.) (5-110) 

To prove this, we use Theorem I to reduce any system to a single force 
plus a couple, and use Theorem III to eliminate any component of the 
couple torque perpendicular to the single force. The point of application 
of the single force mentioned in Theorem IV is no longer arbitrary, as its 
line of action will be fixed when we apply Theorem III. Either the single 
force or the couple may vanish in special cases. For a system of forces in 
a plane, all torques about any point in the plane are perpendicular to the 
plane. Hence Theorem IV reduces to 

Theorem V. Any system of forces in a plane has a resultant, 

unless it is a couple. (5-111) 

In practice, the reduction of a complicated system of forces to a simpler 
system is a problem whose simplest solution is usually obtained by an 
ingenious application of the various theorems and techniques mentioned 
in this section. One method which always works, and which is often the 



5-7] STATICS OF STRUCTURES 231 

simplest if the system of forces is very complicated, is to follow the pro- 
cedure suggested by the proofs of the above theorems. Find the total 
force F by vector addition, and the total torque N about some conven- 
iently chosen point P. Then F acting at P, plus a couple of torque N, 
together form a system equivalent to the original system. If F is zero, the 
original system reduces to a couple. If N is perpendicular to F, the system 
has a resultant, which can be found by either of the methods indicated in 
the proof of Theorem III. If N is not perpendicular to F, and neither is 
zero, then the system has no resultant, and can be reduced to a system of 
two forces, as in the derivation of Theorem II, or to a single force and a 
couple whose torque is parallel to it, as in Theorem IV. It is a matter of 
taste, or of convenience for the purpose at hand, which of these latter re- 
ductions is regarded as the simplest. In fact, for determining the motion 
of a body, the most convenient reduction is certainly just the reduction 
given by Theorem I, with the arbitrary point taken as the center of mass. 

5-7 Statics of structures. The determination of the forces acting at 
various points in a solid structure is a problem of utmost importance in 
all phases of mechanical engineering. There are two principal reasons for 
wanting to know these forces. First, the engineer must be sure that the 
materials and construction are such as will withstand the forces which will 
be acting, without breaking or crushing, and usually without suffering 
permanent deformation. Second, since no construction materials are 
really rigid, but deform elastically and sometimes plastically when subject 
to forces, it is necessary to calculate the amount of this deformation, and 
to take it into account, if it is significant, in designing the structure. When 
deformation or breaking of a structure is under consideration, the struc- 
ture obviously cannot be regarded as a rigid body, and we are interested 
in the actual system of forces acting on and in the structure. Theorems 
regarding equivalent systems of forces are not of direct interest in such 
problems, but are often useful as tools in analyzing parts of the structure 
which may, to a sufficient approximation, be regarded as rigid, or in sug- 
gesting possible equivalent redistributions of forces which would subject 
the structure to less objectionable stresses while maintaining it in equi- 
librium. 

If a structure is at rest, Eqs. (5-95) and (5-96) are applicable either to 
the structure as a whole, or to any part of it. It must be kept in mind that 
the forces and torques which are to be included in the sums are those 
which are external to and acting on whichever part of the structure is under 
consideration. If the structure is moving, the more general equations 
(5-92) and (5-93) are applicable. Either pair of vector equations repre- 
sents, in general, six component equations, or three if all forces lie in a 
single plane. (Why three?) It may be that the structure is so constructed 



232 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 




Fig. 5-19. The flagpole problem. 

that when certain of the external forces and their points of application are 
given, all the internal forces and torques acting on each part of the struc- 
ture can be determined by appropriate applications of Eqs. (5-95) and 
(5-96) (in the case of a structure at rest). Such a structure is said to be 
statically determinate. An elementary example is shown in Fig. 5-19, which 
shows a horizontal flagpole AB hinged at point A to a wall and supported 
by a cable BC. A force W acts on the pole as shown. When the force W 
and the dimensions of the structure are given, it is a simple matter to 
apply Eqs. (5-95) and (5-96) to the pole and to calculate the force F x ex- 
erted by the cable and the force F 2 acting through the hinge. Many ex- 
amples of statically determinate structures are given in any elementary 
physics textbook. 

Suppose now that the hinge at A in Fig. 5-19 were replaced by a welded 
joint, so that the flagpole would support the load even without the cable 
BC, provided the joint at A does not break. Then, given only the weight 
W, it is evidently impossible to determine the force F x exerted by the 
cable; Fi may have any value from zero to a rather large value, depending 
on how tightly the cable is drawn up and on how much stress is applied 
to the joint at A. Such a structure is said to be statically indeterminate. 
A statically indeterminate structure is one in which the forces acting on 
its parts are not completely determined by the external forces, but depend 
also on the distribution of stresses within the structure. To find the 
internal forces in an indeterminate structure, we would need to know the 
elastic characteristics of its parts and the precise way in which these parts 
are distorted. Such problems are usually far more difficult than problems 
involving determinate structures. Many methods of calculating internal 
forces in mechanical structures have been developed for application to en- 
gineering problems, and some of these are useful in a wide variety of 
physical problems. 



5-8 Stress and strain. If an imaginary surface cuts through any part 
of a solid structure (a rod, string, cable, or beam), then, in general, the 
material on one side of this surface will be exerting a force on the material 



5-8] 



STRESS AND STBAIN 



233 



F^ii F(- r 



(a) 



PWiFh 



(b) 



Fj~r 



(C) 



F,-w 



Fig. 5-20. Stresses in a beam, (a) Compression, (b) Tension, (c) Shear. 



on the other side, and conversely, according to Newton's third law. These 
internal forces which act across any surface within the solid are called 
stresses. The stress is defined as the force per unit area acting across any 
given surface in the material. If the material on each side of any surface 
pushes on the material on the other side with a force perpendicular to the 
surface, the stress is called a compression. If the stress is a pull perpen- 
dicular to the surface, it is called a tension. If the force exerted across the 
surface is parallel to the surface, it is called a shearing stress. Figure 5-20 
illustrates these stresses in the case of a beam. The vector labeled Fj_» r 
represents the force exerted by the left half of the beam on the right half, 
and the equal and opposite force F r _j is exerted on the material on the 
left by the material on the right. A stress at an angle to a surface can be 
resolved into a shear component and a tension or compression component. 
In the most general case, the stress may act in any direction relative to 
the surface, and may depend on the orientation of the surface. The de- 
scription of the state of stress of a solid material in the most general 
case is rather complicated, and is best accomplished by using the mathe- 
matical techniques of tensor algebra to be developed in Chapter 10. We 
shall consider here only cases in which either the stress is a pure com- 
pression, independent of the orientation of the surface, or in which only 
one surface is of interest at any point, so that only a single stress vec- 
tor is needed to specify the force per unit area across that surface. 

If we consider a small volume AV of any shape in a stressed material, 
the material within this volume will be acted on by stress forces exerted 
across the surface by the material surrounding it. If the material is not 
perfectly rigid, it will be deformed so that the material in the volume AV 
may have a different shape and size from that which it would have if there 
were no stress. This deformation of a stressed material is called strain. 



234 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

The nature and amount of strain depend on the nature and magnitude of 
the stresses and on the nature of the material. A suitable definition of 
strain, stating how it is to be measured, will have to be made for each kind 
of strain. A tension, for example, produces an extension of the material, 
and the strain would be defined as the fractional increase in length. 

If a wire of length I and cross-sectional area A is stretched to length 
I + Al by a force F, the definitions of stress and strain are 

stress = F/A, (5-112) 

strain = Al/l. (5-113) 

It is found experimentally that when the strain is not too large, the stress 
is proportional to the strain for solid materials. This is Hooke's law, and 
it is true for all kinds of stress and the corresponding strains. It is also 
plausible on theoretical grounds for the reasons suggested in the preliminary 
discussion in Section 2-7. The ratio of stress to strain is therefore constant 
for any given material if the strain is not too large. In the case of exten- 
sion of a material in one direction due to tension, this ratio is called Young's 
modulus, and is 

Y = stress = JL . (5_ n4 ) 

strain A Al 

If a substance is subjected to a pressure increment Ap, the resulting 
deformation will be a change in volume, and the strain will be denned by 

AV 
strain = -^=- ■ (5-115) 

The ratio of stress to strain in this case is called the bulk modulus B: 

„ stress ApV , . 

B = — — — = f==- > (5-116) 

strain AK 

where the negative sign is introduced in order to make B positive. 

In the case of a shearing stress, the stress is again denned by Eq. (5-112), 
where F is the force acting across and parallel to the area A. The result- 
ing shearing strain consists in a motion of A parallel to itself through a 
distance Al, relative to a plane parallel to A at a distance Ax from A (Fig. 
5-21). The shearing strain is then denned by 

strain = — = tan 0, (5-117) 

where 6 is the angle through which a line perpendicular to A is turned as 
a result of the shearing strain. The ratio of stress to strain in this case is 



5-9] 



EQUILIBRIUM OF FLEXIBLE STRINGS AND CABLES 



235 




Fig. 5-21. Shearing strain. 



called the shear modulus n: 



n = 



stress _ F 
strain — A tan 



(5-118) 



An extensive study of methods of solving problems in statics is outside 
the scope of this text. We shall restrict ourselves in the next three sec- 
tions to the study of three special types of problems which illustrate the 
analysis of a physical system, to determine the forces which act upon its 
parts and to determine the effect of these forces in deforming the system. 

5-9 Equilibrium of flexible strings and cables. An ideal flexible string 
is one which will support no compression or shearing stress, nor any bending 
moment, so that the force exerted across any point in the string can only 
be a tension directed along the tangent to the string at that point. Chains 
and cables used in many structures can be regarded for most purposes as 
ideal flexible strings. 

Let us first take a very simple problem in which a string of negligible 
weight is suspended between two points P and P 2 , and a force F x acts at a 
point Pi on the string (Fig. 5-22). Le t t b e the tension in the segment 
PoPi, and Ti the tension in the segment PiP 2 . Let lo and h be the lengths 
of these segments of the string, and let l 02 be the distance between P and 
P 2 . The angles a, /3 between the two segments of string and the line PqP 2 
are determined by the cosine law: 



cos a = 



02 



+ ll - l\ 
21q1 2 



cos /3 = 



02 



+ %- ll 

21iIq2 



(5-119) 



236 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 



Po 




To 

Fig. 5-22. A flexible string held at three points. 

so that the position of the point Pi is independent of the force F x , provided 
the string does not stretch. Since the bit of string at the point P x is in 
equilibrium, the vector sum of the three forces Fi, t , and Ti acting on 
the string at Pi must vanish, so that these forces form a closed triangle, 
as indicated in Fig. 5-22. The tensions are then determine d in t erms of 
the angle between the force F x and the direction of the line P0P2, by the 
sine law: 

sin (0 + 7) _ „ sin (7 - a) , .. 

T ° = Fl sin(a + /3) ' Tl - Fl sin(«+0) (5 " 120) 

Now suppose that the string stretches according to Hooke's law, so that 

l = l' (i + fc To ), h = HO. + kri), (5-121) 

where 1' , l[ are the unstretched lengths, and k is a constant [l//c would be 
Young's modulus, Eq. (5-114), multiplied by the cross-sectional area of 
the string]. The unknown quantities t , ti, l , and h can be eliminated 
from Eqs. (5-119) by substitution from Eqs. (5-120) and (5-121). We 
then have two rather complicated equations to be solved for the angles 
a and /3. The solution must be carried out by numerical methods when 
numerical values of 1' , l[, k, l 02 , F lt and 7 are given. When a and /3 are 
found, r , Ti, Z , and h can be found from Eqs. (5-120) and (5-121). One 
way of solving these equations by successive approximations is to assume 
first that the string does not stretch, so that l = 1' , l 1 = l' lt and to calcu- 
late a. and |3 from Eqs. (5-119), and t , Ti from Eqs. (5-120). Using these 
values of t , ti, we then calculate l , h from Eqs. (5-121). The new 
values of l , h can be used in Eqs. (5-119) to get better values for a, from 
which better values of t , t x can be calculated. These can be used to get 
still better values for l , h from Eqs. (5-121), and so on. As this process 
is repeated, the successive calculated values of a, /3, To, ti, Iq, h will con- 
verge toward the true values. If the string stretches only very little, the 



5-9] 



EQUILIBRIUM OF FLEXIBLE STRINGS AND CABLES 

y 



237 




Fig. 5-23. A flexible string hanging under its own weight. 

first few repetitions will be sufficient to give very close values. The method 
suggested here is an example of a very general class of methods of solution 
of physical problems by successive approximations. It is an example of 
what are called relaxation methods of solving statics problems. 

We next consider a string acted on by forces distributed continuously 
along the length of the string. A point on the string will be specified by 
its distance s from one end, measured along the string. Let f(s) be the 
force per unit length at the point s, that is, the force on a small segment of 
length ds is f ds. Then the total force acting on the length of string be- 
tween the end s = and the point s is zero if the string is in equilibrium: 



F + fids + T(s) = 0, 
Jo 



(5-122) 



where F is the supporting force at the end s = 0, and t(s) is a vector 
whose magnitude is the tension at the point s, oriented in the direction of 
increasing s. By differentiating Eq. (5-122) with respect to s, we obtain 
a differential equation for t(s) : 

dr 



ds 



(5-123) 



The simplest and most important application of Eq. (5-123) is to the 
case of a string having a weight w per unit length. If the string is acted 
on by no other forces except at the ends, it will hang in a vertical plane, 
which we take to be the xy-plane, with the z-axis horizontal and the y-axis 
vertical. Let d be the angle between the string and the x-axis (Fig. 5-23). 
Then the horizontal and vertical components of Eq. (5-123) become : 



^ (t sin 6) = w, 



^(rcos*) = 0. 



(5-124) 
(5-125) 



238 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

Equation (5-125) implies that 

t cos 6 = C. (5-126) 

The horizontal component of tension is constant, as it should be since the 
external forces on the string are all vertical, except at the ends. By divid- 
ing Eq. (5-124) by C, and using Eq. (5-126), we eliminate the tension: 

-*- = c' (5_127) 

If we represent the string by specifying the function y{x), we have the 
relations 

tan 6 = g = y', (5-128) 

ds = [(dx) 2 + (dy) 2 ] 112 = dx(l + y' 2 ) 1 ' 2 , (5-129) 

so that Eq. (5-127) becomes 

^ = g ( l + (5-130) 

This can be integrated, if w is constant : 

sinh- 1 y' = ^ + a, (5-132) 

where a is a constant. We solve for y' : 

^' = != sinh (ir + *)- (5 " 133) 

This can be integrated again, and we obtain 

y= /3 + ^cosh0£ + «)- (5-134) 

The curve represented by Eq. (5-134) is called a catenary, and is the form in 
which a uniform string will hang if acted on by no force other than its own 
weight, except at the ends. The constants C, /3, and a are to be chosen so 
that y has the proper value at the endpoints, and so that the total length 
of the string has the proper value. The total length is 

1 = fds = f x Xl (1 + y' 2 ) 112 dx = J' 1 cosh 0£ + a) dx 



JXq JXq 

C 

w 



[sinh (VZL + a)- sinh (^ + a)] • (5-135) 



5-10] EQUILIBRIUM OF SOLID BEAMS 239 

5-10 Equilibrium of solid beams. A horizontal beam subject to vertical 
forces is one of the simplest examples of a structure subject to shearing 
forces and bending moments. To simplify the problem, we shall consider 
only the case when the beam is under no compression or tension, and we 
shall assume that the beam is so constructed and the forces so applied that 
the beam bends in only one vertical plane, without any torsion (twisting) 
about the axis of the beam. We find first the stresses within the beam 
from a knowledge of the external forces, and then determine the distortion 
of the beam due to these stresses. 

Points along the beam will be located by a coordinate x measured hori- 
zontally from the left end of the beam (Fig. 5-24). Let vertical forces 
F\, . . . , F n act at the distances x\, . . . , x n from the left end. A force 
will be taken as positive if it is directed upward. Let A A' be a plane 
perpendicular to the beam at any distance x from the end. According 
to Theorem I (5-107), of Section 5-6, the system of forces exerted across 
the plane A A' by the material on the right against that on the left is 
equivalent to a single force S through any point in the plane, and a couple 
of torque N. (Note that in applying Theorem I, we are treating the 
plane A A' as a rigid body, that is, we are assuming that the cross-sectional 
plane AA' is not distorted by the forces acting on it.) In the case we are 
considering there is no compression or tension and all forces are vertical, so 
that S is directed vertically. We shall define the shearing force S as the 
vertical force acting across A A' from right to left ; S will be taken as positive 
when this force is directed upward, negative when it is downward.* By 
Newton's third law, the force acting across AA' from left to right is —S. 
Since we are assuming no torsion about the axis of the beam (z-axis), and 
since all the forces are vertical, the torque N will be directed horizontally 
and perpendicular to the beam. We shall define the bending moment N 



4 



^ 



-h' )n 



Sflfi A ' \ Fs 



Fig. 5-24. Forces acting on a beam. 



* This sign convention for S is in agreement with sign conventions throughout 
this book, where the upward direction is taken as positive. Sign conventions for 
shearing force and bending moment are not uniform in physics and engineering 
texts, and one must be careful in reading the literature to note what sign con- 
vention is adopted by each author. 



240 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

as the torque exerted from right to left across A A' about a horizontal axis 
in the plane AA'; N will be taken as positive when it tends to rotate the 
plane A A' in a counterclockwise direction. Since S is vertical, the torque 
will be the same about any horizontal axis in the plane AA'. 

The shearing force S and bending moment N can be determined by 
applying the conditions of equilibrium [Eqs. (5-95) and (5-96)] to the 
part of the beam to the left of the plane A A'. The total force and total 
torque about a horizontal axis in the plane A A' are, if we neglect the 
weight of the beam, 

X) F t + S = 0, (5-136) 

Xi<X 

-No - X) (* - x *> F i + N = °> ( 5 - 137 ) 

X{<X 

where the sums are taken over all forces acting to the left oi AA' and N 
is the bending moment, if any, exerted by the left end of the beam against 
its support. The torque iV will appear only if the beam is clamped or 
otherwise fastened at its left end. The force exerted by any clamp or 
other support at the end is to be included among the forces Fj. If the 
beam has a weight w per unit length, this should be included in the 
equilibrium equations : 



J2 Fi - f X w dx + S = 0, (5-138) 

x~ix J ° 

-N — J] (x — Xi)Fi + f X (x - x')w dx' + N = 0. (5-139) 

xllx J ° 

The shearing force and bending moment at a distance x from the end are 
therefore 

S= - J2 Fi+ [ X w dx, (5-140) 

x^tx J ° 

N = iV + Y* (* - x t )F t - f (x - x')w dx'. (5-141) 

If there is any additional force distributed continuously along the beam, 
this can be included in w as an additional weight per unit length. If the 
beam is free at its ends, the shearing force and bending moment must be 
zero at the ends. If we set aS = JV = at the right end of the beam, 
equations (5-140) and (5-141) may be solved for two of the forces acting 
on the beam when the others are known. If the beam is fastened or clamped 
at either end, S and N may have any values there. Equations (5-140) 
and (5-141) determine S and N everywhere along the beam when all 



5-10] 



EQUILIBRIUM OF SOLID BEAMS 



241 




Fig. 5-25. Distortion of a beam by shearing and bending, (a) Undistorted 
beam, (b) Beam in shear, (c) Beam bent and in shear. 



the forces are known, including the force and torque exerted through 
the clamp, if any, on the left end. The shearing force and bending mo- 
ment may be plotted as functions of a; whose slopes at any point are 
obtained by differentiating Eqs. (5-140) and (5-141) : 



dS 
dx 



= w, 



(5-142) 



dN - V F 
lx~ - ±> Fi 

Xi<X 



f 

JO 



w dx' = — S. 



(5-143) 



The shearing force increases by — Fi from left to right across a point no- 
where a force Fi acts. 

Let us now consider the distortion produced by the shearing forces and 
bending moments in a beam of uniform cross section throughout its 
length. In Fig. 5-25(a) is shown an undistorted horizontal beam through 
which are drawn a horizontal line 00' and a vertical plane AA'. In Fig. 
5-25 (b) the beam is under a shearing strain, the effect of which is to slide 
the various vertical planes relative to one another so that the line 00' 
makes an angle 6 with the normal to the plane A A'. According to Eq. 
(5-118), the angle is given in terms of the shearing force S and the shear 
modulus n by: 

S 



e 



nA 



(5-144) 



242 EIGID BODIES. ROTATION ABOUT AN AXIS. STATICS FcHAP. 5 

0' 





neutral 
layer 



Fig. 5-26. Strains in a bent beam. 



where A is the cross-sectional area, and we have made the approximation 
tan 6=6, since will be very small. In Fig. 5-25 (c), we show the further 
effect of bending the beam. The plane AA' now makes an angle <p with 
the vertical. It is assumed that the cross-sectional surface AA' remains 
plane and retains its shape when the beam is under stress, although this 
may not be strictly true near the points where forces are applied. In order 
to determine <p, we consider two planes A A' and BB' initially vertical and 
a small distance I apart. When the beam is bent, AA' and BB' will make 
angles <p and <p + A<p with the vertical (Fig. 5-26). Due to the bending, 
the fibers on the outside of the curved beam will be stretched and those on 
the inside will be compressed. Somewhere within the beam will be a neu- 
tral layer of unstretched fibers, and we shall agree to draw the line 00' so 
that it lies in this neutral layer. A line between A A' and BB' parallel to 
00' and a distance z above 00' will be compressed to a length I — Al, 
where (see Fig. 5-26) 

Al = z A<p. (5-145) 

The compressive force dF exerted across an element of area dA a distance z 
above the neutral layer 00' will be given by Eq. (5-114) in terms of Young's 
modulus: 



dF_ 
dA 



= rf=K* 



A<p 

T 



(5-146) 



or, if we let I = ds, an infinitesimal element of length along the line 00', 

(5-147) 



4E. = Yz ^ • 
dA ds 



This equation is important in the design of beams, as it determines the 
stress of compression or tension at any distance z from the neutral layer. 
The total compressive force through the cross-sectional area A of the beam 



5-10] EQUILIBRIUM OF SOLID BEAMS 243 

will be 

F = ffdF = Y^fJzdA. (5-148) 

A A 

Since we are assuming no net tension or compression of the beam, F = 0, 
and 

jjz dA = 0. (5-149) 



This implies that the neutral layer contains the centroid of the area A of 
the beam, and we may require that 00' be drawn through the centroid of 
the cross-sectional area of the beam. The bending moment exerted by 
the forces dF is 



N = If'dF=Y%ff*dA 



ds . 

A 



-.2 a d <P 



= Yk'A^, (5-150) 

where 

k 2 = i- jjz 2 dA, (5-151) 

A 

and k is the radius of gyration of the cross-sectional area of the beam about 
a horizontal axis through its centroid. The differential equation for <p is 
therefore 

d<p N 



ds Yk 2 A 



(5-152) 



Let the upward deflection of the beam from a horizontal z-axis be y(x), 
measured to the line 00' (Fig. 5-25). Then y(x) is to be determined by 
solving the equation 

^ = tan (6 + v ), (5-153) 

when and <p have been determined from Eqs. (5-144) and (5-152). If 
we assume that both and <p are very small angles, Eqs. (5-152) and 
(5-153) become 

S-7KT <-"*» 

|-» + ». (5-155) 



244 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

When there are no concentrated forces F, along the beam, we may differ- 
entiate Eq. (5-155) and make use of Eqs. (5-154), (5-144), (5-142), and 
(5-143) to obtain 

d 2 y w . N ( 1 ft . 

d^^^A + YWA' (5_156) 

d^y_J_dSv_ w _ « im 

«fa* _ nA dx 2 Yk 2 A K ' 

If bending can be neglected, as in a short, thick beam, Eq. (5-156) with 
N = becomes a second-order differential equation to be solved for y(x). 
For a longer beam, Eq. (5-157) must be used. These equations can also 
be used when concentrated loads F* are present, by solving them for each 
segment of the beam between the points where the forces Ft are applied, 
and fitting the solutions together properly at these points. The solutions 
on either side of a point Xi where a force F t - is applied must be chosen so 
that y, <p, N are continuous across Xi, while S, dN/dx, dy/dx, d 3 y/dx 3 
increase across the point X{ by an amount determined by Eqs. (5-140), 
(5-143), (5-155), and (5-156). The solution of Eq. (5-156) will contain 
two arbitrary constants, and that of Eq. (5-157), four, which are to be 
determined by the conditions at the ends of the beam or segment of beam. 
As an example, we consider a uniform beam of weight W, length L, 
clamped in a horizontal position (i.e., so that <p = 0)* at its left end 
(x = 0), and with a force Fi = —W exerted on its right end (x = L). 
In this case, Eq. (5-157) becomes 



t±-- W «-1«n 

dx* ~ YWAL {b M) 



The solution is 



V=- 2iYk*AL + iCsx3 + * C2x2 + ClX + C °- (5_159) 

To determine the constants Co, C\, C 2 , C3, we have at the left end of the 
beam: 

y = C = 0, (5-160) 

%-<>>-'-& — *££• <«•» 

where we have used Eqs. (5-155) and (5-144). We need two more con- 
ditions, which may be determined in a variety of ways. The easiest way 



* The condition <p = means that the plane A A' is vertical; that is, the beam 
would be horizontal if there were no shearing strain. 



5-11] EQUILIBRIUM OF FLUIDS 245 

in this case is to apply Eq. (5-156) and its derivative at the left end of the 
beam: 

d 2 y r _ W W'L + WL ( , 1ft9 . 

M> = ° 2 - ^AL TWA ' (5 ' 162) 

d*y _ 1 dN S_ _ W' + W 

dx* ~ U3 ~~ YWA dx ~ YWA ~ YWA ' v ° Q6> 

where we have used Eq. (5-143). The deflection of the beam at any point 
x is then 



_ L 3 

v YWA 



Wx 2 
AL 2 



V 3I t 6L«/ t 2L2 V 1 3 L/l 

-^i[-LV-2L) + -ry (5 " 164) 

The deflection at x = L is 

y = - Sx {W + * w ' ] -£a [w+ w ' ] - (5_i65) 

The first term in each equation is the deflection due to bending, and the 
second is that due to shear. The first term is proportional to L 3 , and in- 
versely proportional to k 2 . The second term is proportional to L and in- 
dependent of fc. Hence bending is more important for long, thin beams, 
and shear is more important for short, thick beams. Our analysis here is 
probably not very accurate for short, thick beams, since, as pointed out 
above, some of our assumptions are not valid near points of support or 
points where loads are applied (where "near" means relative to the cross- 
sectional dimensions of the beam). 

5-11 Equilibrium of fluids. A fluid is defined as a substance which will 
support no shearing stress when in equilibrium. Liquids and gases fit this 
definition, and even very viscous substances like pitch, or tar, or the mate- 
rial in the interior of the earth, will eventually come to an equilibrium in 
which shearing stresses are absent, if they are left undisturbed for a suffi- 
ciently long time. The stress F/A across any small area A in a fluid in 
equilibrium must be normal to A, and in practically all cases it will be a 
compression rather than a tension. 

We first prove that the stress F/A near any point in the fluid is inde- 
pendent of the orientation of the surface A. Let any two directions be 
given, and construct a small triangular prism with two equal faces A i = 
A 2 perpendicular to the two given directions. The third face A 3 is to form 
with A i and A 2 a cross section having the shape of an isosceles triangle 
(Fig. 5-27). Let Fj, F 2 , F3 be the stress forces perpendicular to the faces 



246 HIGID BODIES. BOTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 




Fig. 5-27. Forces on a triangular prism in a fluid. 



Ai, A 2 , A 3 . If the fluid in the prism is in equilibrium, 

F x + F 2 + F 3 = 0. 



(5-166) 



The forces on the end faces of the prism need not be included here, since 
they are perpendicular to F 1; F 2 , and F 3 , and must therefore separately 
add to zero. It follows from Eq. (5-166), and from the way the prism has 
been constructed, that F 1( F 2 , and F3 must form an isosceles triangle 
(Fig. 5-27), and therefore that 



Fi = F 2 



(5-167) 



Since the directions of Fx and F 2 are any two directions in the fluid, and 
since Ai = A 2 , the stress F/A is the same in all directions. The stress in 
a fluid is called the pressure p: 



V 



A x 



El 
A 2 



(5-168) 



Now suppose that in addition to the pressure the fluid is subject to an 
external force f per unit volume of fluid, that is, any small volume dV in 
the fluid is acted on by a force f dV. Such a force is called a body force; 
f is the body force density. The most common example is the gravita- 
tional force, for which 

f = Pg, (5-169) 

where g is the acceleration of gravity, and p is the density. In general, 
the body force density may differ in magnitude and direction at different 
points in the fluid. In the usual case, when the body force is given by 
Eq. (5-169), g will be constant and f will be constant in direction; if p is 
constant, f will also be constant in magnitude. Let us consider two nearby 
points P u P 2 in the fluid, separated by a vector dr. We construct a cylinder 
of length dt and cross-sectional area dA, whose end faces contain the points 
Pi and P 2 . Then the total component of force in the direction of dt acting 



5-11] EQUILIBRIUM OF FLUIDS 247 

on the fluid in the cylinder, since the fluid is in equilibrium, will be 

f 'dr dA + pi dA — p 2 dA = 0, 

where pi and p 2 are the pressures at Pi and P 2 . The difference in pressure 
between two points a distance dx apart is therefore 

dp = p 2 — pi = f-dr. (5-170) 

The total difference in pressure between two points in the fluid located by 
vectors rx and r 2 will be 



V2 



-pi = Pt-dr, (5-171) 

•'ri 



where the line integral on the right is to be taken along some path lying 
entirely within the fluid from rx to r 2 . Given the pressure pi at ri, Eq. 
(5-171) allows us to compute the pressure at any other point r 2 which can 
be joined to ri by a path lying within the fluid. The difference in pressure 
between any two points depends only on the body force. Hence any 
change in pressure at any point in a fluid in equilibrium must be accom- 
panied by an equal change at all other points if the body force does not 
change. This is Pascal's law. 

According to the geometrical definition (3-107) of the gradient, Eq. 
(5-170) implies that 

f = Vp. (5-172) 

The pressure gradient in a fluid in equilibrium must be equal to the body 
force density. This result shows that the net force per unit volume due to 
pressure is — Vp. The pressure p is a sort of potential energy per unit 
volume in the sense that its negative gradient represents a force per unit 
volume due to pressure. However, the integral of p dV over a volume 
does not represent a potential energy except in very special cases. Equa- 
tion (5-172) implies that the surfaces of constant pressure in the fluid are 
everywhere perpendicular to the body force. According to Eqs. (3-187) 
and (5-172), the force density f must satisfy the equation 

V X f = 0. (5-173) 

This is therefore a necessary condition on the body force in order for equi- 
librium to be possible. It is also a sufficient condition for the possibility 
of equilibrium. This follows from the discussion in Section 3-12, for if 
Eq. (5-173) holds, then it is permissible to define a function p(r) by the 
equation 

P(r) = Pi + f t-dt, (5-174) 



248 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

where Vi is the pressure at some fixed point rj, and the integral may be 
evaluated along any path from r x to r within the fluid. If the pressure in 
the fluid at every point r has the value p(r) given by (5-174), then Eq. 
(5-172) will hold, and the body force f per unit volume will everywhere 
be balanced by the pressure force — Vp per unit volume. Equation (5-174) 
therefore defines an equilibrium pressure distribution for any body force 
satisfying Eq. (5-173). 

The problem of finding the pressure within a fluid in equilibrium, if the 
body force density f (r) is given, is evidently mathematically identical with 
the problem discussed in Section 3-12 of finding the potential energy for a 
given force function F(r). We first check that V. X f is zero everywhere 
within the fluid, in order to be sure that an equilibrium is possible. We 
then take a point r x at which the pressure is known, and use Eq. (5-174) to 
find the pressure at any other point, taking the integral along any conven- 
ient path. 

The total body force acting on a volume V of the fluid is 



F 6 = 



[[ft dV. (5-175) 



The total force due to the pressure on the surface A of V is 

F p = - jfnpdA, (5-176) 

A 

where n is the outward normal unit vector at any point on the surface. 
These two must be equal and opposite, since the fluid is in equilibrium : 

F p = -F 6 . (5-177) 

Equation (5-176) gives the total force due to pressure on the surface of the 
volume V, whether or not V is occupied by fluid. Hence we conclude from 
Eq. (5-177) that a body immersed in a fluid in equilibrium is acted on by 
a force F p due to pressure, equal and opposite to the body force F& which 
would be exerted on the volume V if it were occupied by fluid in equi- 
librium. This is Archimedes' principle. Combining Eqs. (5-172), (5-175), 
(5-176), and (5-177), we have 

ffnp dA = fffvp dV. (5-178) 

A V 

This equation resembles Gauss' divergence theorem [Eq. (3-115)], except 
that the integrands are np and Vp instead of n-A and V-A. Gauss' the- 
orem can, in fact, be proved in a very useful general form which allows us to 
replace the factor n in a surface integral by V in the corresponding volume 



5-11] EQUILIBRIUM OF FLUIDS 249 

integral without any restrictions on the form of the integrand except that 
it must be so written that the differentiation symbol V operates on the 
entire integrand.* Given this result, we could start with Eqs. (5-175), 
(5-176), and (5-177), and deduce Eq. (5-172) : 

F 6 + F p = ffff dV - ffnp dA 

V A 

= f ff(i — Vp) dV = 0. (5-179) 

v 

Since this must hold for any volume V, Eq. (5-172) follows. 

So far we have been considering only the pressure, i.e., the stress, in a 
fluid. The strain produced by the pressure within a fluid is a change in 
volume per unit mass of the fluid or, equivalently, a change in density. 
If Hooke's law is satisfied, the change dV in a volume V produced by a 
small change dp in pressure can be calculated from Eq. (5-116), if the 
bulk modulus B is known : 

d -~=~ d i- (5-180) 

If the mass of fluid in the volume V is M , then the density is 

M 
P = y> (5-18D 

and the change dp in density corresponding to an infinitesimal change dV 
in volume is given by 

*=-££, (5-182) 

so that the change in density produced by a small pressure change dp is 

* = $• (5-183) 

After a finite change in pressure from p to p, the density will be 

p = p exp(/^)- (5-184) 

In any case, the density of a fluid is determined by its equation of state in 



* For the proof of this theorem, see Phillips, Vector Analysis. New York: 
John Wiley and Sons, 1933. (Chapter III, Section 34.) 



250 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

terms of the pressure and temperature. The equation of state for a per- 
fect gas is 

pV = RT, (5-185) 

where T is the absolute temperature, V is the volume per mole, and R is 
the universal gas constant: 

R = 8.314 X 10 7 erg-deg -1 C-mole -1 . (5-186) 

By substitution from Eq. (5-181), we obtain the density in terms of pres- 
sure and temperature : 

P = f| > (5-187) 

where M is the molecular weight. 

Let us apply these results to the most common case, in which the body 
force is the gravitational force on a fluid in a uniform vertical gravitational 
field [Eq. (5-169)]. If we apply Eq. (5-173) to this case, we have 

V X f = V X (/og) = 0. (5-188) 

Since g is constant, the differentiation implied by the V symbol operates 
only on p, and we can move the scalar p from one factor of the cross product 
to the other to obtain : 

(Vp) X g = 0, (5-189) 

that is, the density gradient must be parallel to the gravitational field. 
The density must be constant on any horizontal plane within the fluid. 
Equation (5-189) may also be derived from Eq. (5-188) by writing 
out explicitly the components of the vectors V X (pg) and (Vp) X g, and 
verifying that they are the same.* According to Eq. (5-172), the pres- 
sure is also constant in any horizontal plane within the fluid. Pressure 
and density are therefore functions only of the vertical height z within the 
fluid. From Eqs. (5-172) and (5-169) we obtain a differential equation 
for pressure as a function of z: 

f z = -pg. (5-190) 

If the fluid is incompressible, and p is uniform, the solution is 

V = Po — pgz, (5-191) 



* Equation (5-189) holds also in a nonuniform gravitational field, since 
T X g = 0, by Eq. (6-21). 



PROBLEMS 251 

where p is the pressure at z = 0. If the fluid is a perfect gas, either p or p 
may be eliminated from Eq. (5-190) by means of Eq. (5-187). If we 
eliminate the density, we have 

g = _f£,. (5 _ 192) 

As an example, if we assume that the atmosphere is uniform in temperature 
and composition, we can solve Eq. (5-192) for the atmospheric pressure 
as a function of altitude: 



V~ RT 7 



p = p exp i — -pf; z ) ■ (5-193) 



Problems 

1. (a) Prove that the total kinetic energy of the system of particles making up 
a rigid body, as defined by Eq. (4-37), is correctly given by Eq. (5-16) when the 
body rotates about a fixed axis, (b) Prove that the potential energy given by 
Eq. (5-14) is the total work done against the external forces when the body is 
rotated from d„ to 0, if N z is the sum of the torques about the axis of rotation due 
to the external forces. 

2. Prove, starting with the equation of motion (5-13) for rotation, that if N z 
is a function of alone, then T + V is constant. 

3. A wheel of mass M , radius of gyration k, spins smoothly on a fixed horizontal 
axle of radius a which passes through a hole of slightly larger radius at the hub of 
the wheel. The coefficient of friction between the bearing surfaces is /*. If the 
wheel is initially spinning with angular velocity wo, find the time and the number 
of turns that it takes to stop. 

4. The balance wheel of a watch consists of a ring of mass M, radius a, with 
spokes of negligible mass. The hairspring exerts a restoring torque N z = — kd. 
Find the motion if the balance wheel is rotated through an angle do and released. 

5. An airplane propeller of moment of inertia / is subject to a driving torque 

N = No(l + acoswoO, 
and to a frictional torque due to air resistance 

N f = — b&. 

Find its steady-state motion. 

6. A motor armature weighing 2 kgm has a radius of gyration of 5 cm. Its 
no-load speed is 1500 rpm. It is wound so that its torque is independent of its 
speed. At full load, it draws a current of 2 amperes at 110 volts. Assume that 
the electrical efficiency is 80%, and that the friction is proportional to the square 
of the angular velocity. Find the time required for it to come up to a speed of 
1200 rpm after being switched on without load. 

7. Derive Eqs. (5-35) and (5-36). 



252 



RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 



[CHAP. 5 



8. Assume that a simple pendulum suffers a frictional torque — mbid due to 
friction at the point of support, and a frictional force — 62V on the bob due to 
air resistance, where v is the velocity of the bob. The bob has a mass m, and is 
suspended by a string of length I. Find the time required for the amplitude to 
damp to 1/e of its initial (small) value. How should m, I be chosen if it is desired 
that the pendulum swing as long as possible? How should m, I be chosen if it 
is desired that the pendulum swing through as many cycles as possible? 

9. A compound pendulum is arranged to swing about either of two parallel 
axes through two points 0, 0' located on a line through the center of mass. The 
distances h, h! from 0, 0' to the center of mass, and the periods r, t' of small 
amplitude vibrations about the axes through and 0' are measured. and 0' 
are arranged so that each is approximately the center of oscillation relative to 
the other. If r = r', find a formula for g in terms of measured quantities. If 
t' = r(l + 8), where S <3C 1, find a correction to be added to your previous 
formula so that it will be correct to terms of order S. 

10. A baseball bat held horizontally at rest is struck at a point 0' by a ball 
which delivers a horizontal impulse /' perpendicular to the bat. Let the bat be 
initially parallel to the x-axis, and let the basbeall be traveling in the negative 
direction parallel to the «/-axis. The center of mass G of the bat is initially at the 
origin, and the point 0' is at a distance h' from G. Assuming that the bat is let 
go just as the ball strikes it, and neglecting the effect of gravity, calculate and 
sketch the motion x(t), y{t) of the center of mass, and also of the center of per- 
cussion, during the first few moments after the blow, say until the bat has 
rotated a quarter turn. Comment on the difference between the initial motion 
of the center of mass and that of the center of percussion. 

11. A circular disk of radius a lies in the zy-plane with its center at the origin. 
The half of the disk above the z-axis has a density a per unit area, and the half 
below the s-axis has a density 2<x. Find the center of mass G, and the moments of 
inertia about the x-, y-, and 2-axes, and about parallel axes through G. Make as 
much use of laborsaving theorems as possible. 

12. (a) Work out a formula for the moments of inertia of a cone of mass m, 
height h, and generating angle a, about its axis of symmetry, and about an axis 




Fig. 5-28. Frustum of a cone. 



PROBLEMS 



253 



45' 



Fig. 5-29. How much thread can be wound on this spool? 



through the apex perpendicular to the axis of symmetry. Find the center of mass 
of the cone, (b) Use these results to determine the center of mass of the frustum 
of a cone, shown in Fig. 5-28, and to calculate the moments of inertia about hori- 
zontal axes through each base and through the center of mass. The mass of the 
frustum is M . 

13. How many yards of thread 0.03 inch in diameter can be wound on the 
spool shown in Fig. 5-29? 

14. Given that the volume of a cone is one-third the area of the base times the 
height, locate by Pappus' theorem the centroid of a right triangle whose legs are 
of lengths a and 6. 

15. Prove that Pappus' second theorem holds even if the axis of revolution 
intersects the surface, provided that we take as volume the difference in the 
volumes generated by the two parts into which the surface is divided by the axis. 
What is the corresponding generalization of the first theorem? 

16. Find the center of mass of a wire bent into a semicircle of radius a. Find 
the three radii of gyration about x-, y-, and 2-axes through the center of mass, 
where z is perpendicular to the plane of the semicircle and x bisects the semicircle. 
Use your ingenuity to reduce the number of calculations required to a minimum. 

17. (a) Find a formula for the radius of gyration of a uniform rod of length I 
about an axis through one end making an angle a with the rod. (b) Using this 
result, find the moment of inertia of an equilateral triangular pyramid, con- 
structed out of six uniform rods, about an axis through its centroid and one of 
its vertices. 

18. Find the radii of gyration of a plane lamina in the shape of an ellipse of 
semimajor axis a, eccentricity e, about its major and minor axes, and about a 
third axis through one focus perpendicular to the plane. 

19. Forces 1 kgm-wt, 2 kgm-wt, 3 kgm-wt, and 4 kgm-wt act in sequence 
clockwise along the four sides of a square 0.5 X 0.5 m 2 . The forces are directed 
in a clockwise sense around the square. Find the equilibrant. 

20. An iceboat has a flat sail in the shape of a right triangle with a vertical 
leg of length o along the mast and a horizontal leg of length 6 along the boom. 
The force on the sail acts at its centroid and is given by F = fc[n • (w — v)]n, 



254 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 

1 4 lb 

3 1b 



[chap. 5 




Fig. 5-30. A system of forces acting on a cube. 

where n is a unit vector normal to the sail, w is the wind velocity, v is the velocity 
of the boat, and k is a constant. The sail makes an angle a with the center line 
of the boat. The angle a may have any value up to that for which F becomes 
zero. The center line of the boat makes an angle /3 with the direction ( — w) 
from which the wind is blowing. The runners are parallel to the center line. 
The coefficient of friction along the runners is n, and there is a force N per- 
pendicular to the runners sufficient to insure that v is parallel to the runners and 
constant in direction. Find v as a function of a, /?, w and the mass m of the 
boat. Find N and the point at which it acts. What value of a makes v a maxi- 
mum if ji is very small? 

21. (a) Reduce the system of forces acting on the cube shown in Fig. 5-30 to 
an equivalent single force acting at the center of the cube, plus a couple com- 
posed of two forces acting at two adjacent corners, (b) Reduce this system to 
a system of two forces, and state where these forces act. (c) Reduce this system 
to a single force plus a torque parallel to it. 

22. (a) A cable is connected in a straight line between two fixed points. By 
exerting a sidewise force W at the center of the cable, a considerably greater 
force r can be applied to the support points at each end bf the cable. Find a 
formula for t in terms of W, and the area A and Young's modulus Y of the 
cable, assuming that the angle through which the cable is pulled is small, (b) 
Show that this assumption is well satisfied if W = 100 lb, A = 3 in 2 , and 
Y = 60,000 lb-in- 2 . Find t. 

23. A cable is to be especially designed to hang vertically and to support a 
load Fata distance I below the point of support. The cable is to be made of a 
material having a Young's modulus F and a weight w per unit volume. Inas- 
much as the length I of the cable is to be fairly great, it is desired to keep the 
weight of the cable to a minimum by making the cross-sectional area A (z) of the 
cable, at a height z above the lower end, just great enough to support the load 
beneath it. The cable material can safely support a load just great enough to 
stretch it 1%. Determine the function A(z) when the cable is supporting the 
given load. 



PROBLEMS 



255 




Fig. 5-31. A suspension bridge. 



24. A cable 20 ft long is suspended between two points A and B, 15 ft apart. 
The line AB makes an angle of 30° with the horizontal (B higher). A weight of 
2000 lb is hung from a point C 8 ft from the end of the cable at A. (a) Find the 
position of point C, and the tensions in the cable, if the cable does not stretch, 
(b) If the cable is £ inch in diameter and has a Young's modulus of 5 X 10 5 lb- 
in -2 , find the position of point C and the tensions, taking cable stretch into 
account. Carry out two successive approximations, and estimate the accuracy 
of your result. 

25. (a) A cable of length I, weight w per unit length, is suspended from the 
points x = ±aon the x-axis. The y-axis is vertical. By requiring that y = at 
x = ±o, and that the total length of cable be I, show that a = in Eq. (5-134), 
and set up equations to be solved for /3 and C. (b) Show that the same results 
can be obtained for a and C by requiring that the cable be symmetrical about the 
2/-axis, and that the forces at its ends balance the weight of the cable. 

26. A bridge of weight w per unit length is to be hung from cables of negligible 
weight, as shown in Fig. 5-31. It is desired to determine the shape of the suspen- 
sion cables so that the vertical cables, which are equally spaced, will support 
equal weights. Assume that the vertical cables are so closely spaced that we can 
regard the weight w per unit length as continuously distributed along the 
suspension cable. The problem then differs from that treated in the text, where 
the string had a weight w per unit length s along the string, in that here there 
is a weight w per unit horizontal distance x. Set up a differential equation for 
the shape y (x) of the suspension cable, and solve for y(x) if the ends are at 
the points y = 0, x = ±§Z), and if the maximum tension in the cable is to 
be to- 

27. A cable of length I, weight w per unit length, is suspended from points 
x = =fca on the z-axis. The y-axts is vertical. A weight W is hung from the mid- 
point of the cable. Set up the equations from which /3, a, and C are to be de- 
termined. 

28. A seesaw is made of a plank of wood of rectangular cross section 2X12 in 2 
and 10 ft long, weighing 60 lb. Young's modulus is 1.5 X 10 6 lb-in -2 . The 
plank is balanced across a narrow support at its center. Two children weighing 
100 lb each sit one foot from the ends. Find the shape of the plank when it is 
balanced in a stationary horizontal position. Neglect shear. 

29. An empty pipe of inner radius a, outer radius b, is made of material with 
Young's modulus Y, shear modulus n, density p. A horizontal section of length 
L is clamped at both ends. Find the deflection at the center. Find the increase 
in deflection when the pipe is filled with a fluid of density po. 



256 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS [CHAP. 5 

30. An I-beam has upper and lower flanges of width a, connected by a center 
web of height 6. The web and flanges are of the same thickness c, assumed negligi- 
ble with respect to a and b, and are made of a material with Young's modulus Y, 
shear modulus n. The beam has a weight W , length L, and rests on supports at 
each end. A load W rests on the midpoint of the beam. Find the deflection of 
the beam at its midpoint. Separate the deflection into terms due to shear and 
to bending, and into terms due to the beam weight W and the load W. 

31. If the bulk modulus of water is B, and the atmospheric pressure at the sur- 
face of the ocean is po, find the pressure as a function of depth in the ocean, taking 
into account the compressibility of the water. Assume that B is constant. Look 
up B for water, and estimate the error that would be made at a depth of 5 miles 
if the compressibility were neglected. 

32. Find the atmospheric pressure as a function of altitude on the assumption 
that the temperature decreases with altitude, the decrease being proportional to 
the altitude. 



CHAPTER 6 

GRAVITATION 

6-1 Centers of gravity for extended bodies. You will recall that we 
formulated the law of gravitation in Section 1-5. Any two particles of 
masses wii and ra 2 , a distance r apart, attract each other with a force 
whose magnitude is given by Eq. (1-11) : 

F = ^ , (6-1) 

where 

G = 6.67 X 10 -8 dyne-cm 2 -gm -2 , (6-2) 

as determined by measurements of the forces between large lead spheres, 
carried out by means of a delicate torsion balance. Equation (6-1) can 
be written in a vector form which gives both the direction and magnitude 
of the attractive forces. Let ri and r 2 be the position vectors of the two 
particles. Then the gravitational force on m 2 due to m x is 

_ Gm 1 m 2 , r , ■ ( R o\ 

F i->2 = | fl _ r2 |3 ( f i - f 2)- (6-3) 

The vector (r x — r 2 ) gives the force the correct direction, and its magni- 
tude is divided out by the extra factor |r x — r 2 | in the denominator. 

The law of gravitation as formulated in Eq. (6-3) is applicable only to 
particles or to bodies whose dimensions are negligible compared with the 
distance between them; otherwise the distance |r x — r 3 | is not precisely 
denned, nor is it immediately clear at what points and in what directions 
the forces act. For extended bodies, we must imagine each body divided 
into pieces or elements, small compared with the distances between the 
bodies, and compute the forces on each of the elements of one body due 
to each of the elements of the other bodies. 

Consider now an extended body of mass M and a particle of mass m 
at a point P (Fig. 6-1). If the body of mass M is divided into small pieces 
of masses m,-, each piece is attracted toward m by a force which we shall 
call F,-. Now the system of forces F» can be resolved according to Theorem 
I of Section 5-6 (5-107) into a single force through an arbitrary point, 
plus a couple. Let this single force be F : 

F = J2 F,, (6-4) 

i 

257 



258 GRAVITATION [CHAP. 6 




Fig. 6-1. Gravitational attraction between a particle and an extended body. 

and let the arbitrary point be taken as the point P. Since none of the 
forces Fj exerts any torque about P, the total torque about P is zero, and 
the couple vanishes. The system of forces therefore has a resultant F 
acting along a line through the mass m. The force acting on m is — F, 
since Newton's third law applies to each of the forces F; in Eq. (6-4). 
We locate on this line of action of F a point G a distance r from P such 
that 

|F| = ^- (6-5) 

Then the system of gravitational forces between the body M and the par- 
ticle m is equivalent to the single resultant forces F on M and — F on m 
which would act if all the mass of the body M were concentrated at G. 
The point G is called the center of gravity of the body M relative to the 
point P; G is not, in general, at the center of mass of body M, nor even 
on the Tine joining P with the center of mass. The parts of the body close 
to P are attracted more strongly than those farther away, whereas in 
finding the center of mass, all parts of the body are treated alike. Further- 
more, the position of the point G will depend on the position of P. When 
P is far away compared with the dimensions of the body, the acceleration 
of gravity due to m will be nearly constant over the body and, in this 
case, we showed in Section 5-6 that G will coincide with the center of 
mass. Also, in the case of a uniform sphere or a spherically symmetrical 
distribution of mass, we shah show in the next section that the center of 
gravity always lies at the center of the sphere. The relative character of 
the concept of center of gravity makes it of little use except in the case of 
a sphere or of a body in a uniform gravitational held. 

For two extended bodies, no unique centers of gravity can in general 
be defined, even relative to each other, except in special cases, as when the 
bodies are far apart, or when one of them is a sphere. The system of 



6-2] GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL 259 

gravitational forces on either body due to the other may or may not have 
a resultant; if it does, the two resultants are equal and opposite and act 
along the same line. However, even in this case, we cannot define defi- 
nite centers of gravity (?i, G 2 for the two bodies relative to each other, 
since Eq. (6-5) specifies only the distance GiG 2 - 

The general problem of determining the gravitational forces between 
bodies is usually best treated by means of the concepts of the field theory 
of gravitation discussed in the next section. 

6-2 Gravitational field and gravitational potential. The gravitational 
force F m acting on a particle of mass m at a point r, due to other particles 
m» at points r*, is the vector sum of the forces due to each of the other 
particles acting separately: 

F - - Z, | r . _ r |3 (6-6) 

If, instead of point masses m*, we have mass continuously distributed in 
space with a density p(r), the force on a point mass m at r is 



-/// 



mG{ ?, ~ t (r0 dV. (6-7) 



The integral may be taken over the region containing the mass whose 
attraction we are computing, or over all space if we let p = outside 
this region. Now the force F m is proportional to the mass m, and we 
define the gravitational field intensity (or simply gravitational field) g(r), 
at any point r in space, due to any distribution of mass, as the force per 
unit mass which would be exerted on any small mass m at that point : 

g(r) = — . (6-8) 

m 

where F m is the force that would be exerted on a point mass m at the 
point r. We can write formulas for g(r) for point masses or continuously 
distributed mass: 

g(r) = E "f^ ^3 r) » (6-9) 

The field g(r) has the dimensions of acceleration, and is in fact the 
acceleration experienced by a particle at the point r, on which no forces 
act other than the gravitational force. 



260 GRAVITATION [CHAP. 6 

The calculation of the gravitational field g(r) from Eq. (6-9) or (6-10) 
is difficult except in a few simple cases, partly because the sum and in- 
tegral call for the addition of a number of vectors. Since the gravitational 
forces between pairs of particles are central forces, they are conservative, 
as we showed in Section 3-12, and a potential energy can be denned for 
a particle of mass m subject to gravitational forces. For two particles m 
and mi, the potential energy is given by Eqs. (3-229) and (3-230) : 

7 -< = ir?r7n (<W1) 

The potential energy of a particle of mass m at point r due to a system of 
particles m» is then 

We define the gravitational potential g(r) at point r as the negative of the 
potential energy per unit mass of a particle at point r: [This choice of 
sign in g(r) is conventional in gravitational theory.] 

8(r) = -^- (6-13) 

For a system of particles, 

* i *' 

If p(r) represents a continuous distribution of mass, its gravitational poten- 
tial is 

w-fff&k'*'- <6 - 15) 

Because it is a scalar point function, the potential g(r) is easier to work 
with for many purposes than is the field g(r). In view of the relation 
(3-185) between force and potential energy, g may easily be calculated, 
when g is known, from the relation 

g = VS. (6-16) 



The inverse relation is 



-f 



S(r) = g-dr. (6-17) 



The definition of g(r), like that of potential energy V(r), involves an arbi- 
trary additive constant or, equivalently, an arbitrary point r, at which 
g = 0. Usually r„ is taken at an infinite distance from all masses, as in 
Eqs. (6-14) and (6-15). 



6-2] GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL 261 




Fig. 6-2. Method of computing potential of a spherical shell. 

The concepts of gravitational field and gravitational potential are 
mathematically identical to those of electric field intensity and electro- 
static potential in electrostatics, except that the negative sign in Eq. (6-13) 
is conventional in gravitational theory, and except that all masses are 
positive and all gravitational forces are attractive, so that the force law 
has the opposite sign from that in electrostatics. The subject of potential 
theory is an extensive one, and we can give here only a very brief introduc- 
tory treatment. 

As an example of the use of the concept of potential, we calculate the 
potential due to a thin homogeneous spherical shell of matter of mass M, 
density a per unit area, and radius a: 

M = 4ira 2 a. (6-18) 

The potential at a point P is computed by integrating over a set of ring 
elements as in Fig. 6-2. The potential of a ring of radius a sin 0, width 
a dO, all of whose mass is at the same distance r from P, will be 

, _ GaQira sin 9)a dO 
and the total potential at P of the spherical shell is 
9(P) 



f G<r(2wa sin 0)a dd 
Jo r 

= MG J 

2 Jo 



sin dd 



(r% + a 2 — 2ar o cos0) 112 
= ^[(ro + o)- |ro-o|]- (6-19) 

We have two cases, according to whether P is outside or inside the shell: 

S(P) = ^ > r > a, g(P) = M£ , r < a. (6-20) 
"o a 



262 GRAVITATION [CHAP. 6 

Thus outside the shell the potential is the same as for a point mass M at 
the center of the shell. The gravitational field outside a spherical shell is 
then the same as if all the mass of the shell were at its center. The same 
statement then holds for the gravitational field outside any spherically 
symmetrical distribution of mass, since the total field is the sum of the 
fields due to the shells of which it is composed. This proves the statement 
made in the previous section; a spherically symmetrical distribution of 
mass attracts (and therefore is attracted by) any other mass outside it as 
if all its mass were at its center. Inside a spherical shell, the potential is 
constant, and it follows from Eq. (6-16) that the gravitational field is 
there zero. Hence a point inside a spherically symmetric distribution of 
mass at a distance r from the center is attracted as if the mass inside the 
sphere of radius r were at the center; the mass outside this sphere exerts 
no net force. These results would be somewhat more difficult to prove by 
computing the gravitational forces directly, as the reader can readily 
verify. Indeed, it took Newton twenty years! The calculation of the 
force of attraction on the moon by the earth described in the last section 
of Chapter 1 was made by Newton twenty years before he published his 
law of gravitation. It is likely that he waited until he could prove an 
assumption implicit in that calculation, namely, that the earth attracts any 
body outside it as if all the mass of the earth were concentrated at its center. 

6-3 Gravitational field equations. It is of interest to find differential 
equations satisfied by the functions g(r) and g(r). From Eq. (6-16) it 
follows that 

V X g = 0. (6-21) 

When written out in any coordinate system, this vector equation becomes 
a set of three partial differential equations connecting the components of 
the gravitational field. In rectangular coordinates, 

a^-^ =0 ' "to ~ to ~ °' to _ dy - °- (6 ~ 22) 

These equations alone do not determine the gravitational field, for they 
are satisfied by every gravitational field. To determine the gravitational 
field, we need an equation connecting g with the distribution of matter. 
Let us study the gravitational field g due to a point mass m. Consider 
any volume V containing the mass m, and let n be the unit vector normal 
at each point to the surface S that bounds V (Fig. 6-3). Let us compute 
the surface integral 

I = ffn-gdS. (6-23) 



6-3] GRAVITATIONAL FIELD EQUATIONS 263 




Fig. 6-3. A mass m enclosed in a volume V. 

The physical or geometric meaning of this integral can be seen if we in- 
troduce the concept of hues of force, drawn everywhere in the direction 
of g, and in such a manner that the number of lines per square centimeter 
at any point is equal to the gravitational field intensity. Then I is the 
number of lines passing out through the surface 5, and is called the flux 
of g through 8, The element of solid angle dti subtended at the position 
of m fay an clement of surface dS is defined as the area swept out on a 
sphere of unit radius by a, radius from m which sweeps over the surface 
element dS. This area is 

,„ dS cos B ,„ _,, 

dU = r2 ■ (6-24) 

Prom Fig. 6-3, we have the relation 

mG cos 9 
n-g= ^— ■ (6-25) 

When use is made of these two relations, the integral I [Eq. (0-23)] be- 
comes 

/ = f f—mGdtt = — 4nmG. (6-26) 

The integral 7 is independent of the position of m within the surface S. 
This result is analogous to the corresponding result in electrostatics that 
there are 4ir lines of force coming from every unit charge. Since the 
gravitational field of a number of masses is the sum of their individual 
fields, we have, for a surface 8 surrounding a set of masses rrii : 

I = f fn-gdS = — J^ 4irmiG. (6-27) 

8 i 

For a continuous distribution of mass within S, this equation becomes 

ffn-gdS^ - f f fiirGp dV. (i 5-28 ! 



264 GRAVITATION [CHAP. 6 

We now apply Gauss' divergence theorem [Eq. (3-115)] to the left side of 
this equation: 

Jfn-gdS = fffv-gdV. (6-29) 

s v 

Subtracting Eq. (6-28) from Eq. (6-29), we arrive at the result 

///(V-g + 4arGp) dV = 0. (6-30) 

v 

Now Eq. (6-30) must hold for any volume V, and this can only be true if 
the integrand vanishes: 

V • g = — 4irGp. (6-31) 

This equation in cartesian coordinates has the form 

?gi + d Jv + d Jl = -4arGp(x, y, z). (6-32) 

dx dy dz 

When p(x, y, z) is given, the set of equations (6-22) and (6-32) can be 
shown to determine the gravitational field (g x , g v , g z ) uniquely, if we add 
the boundary condition that g -» as |r| -> oo. Substituting from Eq. 
(6-16), we get an equation satisfied by the potential: 

V 2 g = -4tt(?p, (6-33) 



or 

i2, 



^8 + ^S + ^8 = _ 47r (?p. (6-34) 

dx 2 + dy2 ^ dz 2 p 

This single equation determines <3(x, y, z) uniquely if we add the condition 
that g -> as |r| -> oo. This result we quote from potential theory with- 
out proof. The solution of Eq. (6-33) is, in fact, Eq. (6-15). It is often 
easier to solve the partial differential equation (6-34) directly than to com- 
pute the integral in Eq. (6-15). Equations (6-33), (6-16), and (6-8) to- 
gether constitute a complete summary of Newton's theory of gravitation, 
as likewise do Eqs. (6-31), (6-21), and (6-8) ; that is, all the results of the 
theory can be derived from either of these sets of equations. 

Equation (6-33) is called Poisson's equation. Equations of this form 
turn up frequently in physical theories. For example, the electrostatic 
potential satisfies an equation of the same form, where p is the electric 
charge density. If p = 0, Eq. (6-33) takes the form 

V 2 9 = 0. (6-35) 

This is called Laplace's equation. An extensive mathematical theory of 



PROBLEMS 265 

Eqs. (6-33) and (6-35) has been developed.* A discussion of potential 
theory is, however, outside the scope of this text. 



Problems 

1. (a) Given Newton's laws of motion, and Kepler's first two laws of planetary 
motion (Section 3-15), show that the force acting on a planet is directed toward 
the sun and is inversely proportional to the square of the distance from the sun. 
(b) Use Kepler's third law to show that the forces on the planets are propor- 
tional to their masses, (c) If this suggests to you a universal law of attraction 
between any two masses, use Newton's third law to show that the force must be 
proportional to both masses. 

2. (a) Find the gravitational field and gravitational potential at any point z 
on the symmetry axis of a uniform solid hemisphere of radius a, mass M . The 
center of the hemisphere is at z = 0. (b) Locate the center of gravity of the 
hemisphere relative to a point outside it on the z-axis, and show that as 
z -> ±oo, the center of gravity approaches the center of mass. 

3. Assuming that the earth is a sphere of uniform density, with radius a, 
mass M , calculate the gravitational field intensity and the gravitational potential 
at all points inside and outside the earth, taking 9 = at an infinite distance. 

4. Assuming that the interior of the earth can be treated as an incompressible 
fluid in equilibrium, (a) calculate the pressure within the earth as a function of 
distance from the center, (b) Using appropriate values for the earth's mass and 
radius, calculate the pressure in tons per square inch at the center. 

5. Show that if the sun were surrounded by a spherical cloud of dust of uniform 
density p, the gravitational field within the dust cloud would be 



(MG . 4x _\r 
e= -\-W + Y pGr )r' 



where M is the mass of the sun, and r is a vector from the sun to any point in the 
dust cloud. 

6. Assume that the density of a star is a function only of the radius r measured 
from the center of the star, and is given by f 



Ma, 2 



2nr(r2+a 2 ) 2 



where M is the mass of the star, and a is a constant which determines the size of 
the star. Find the gravitational field intensity and the gravitational potential as 
functions of r. 



* O. D. Kellogg, Foundations of Potential Theory. Berlin: J. Springer, 1929. 

t The expression for p is chosen to make the problem easy to solve, not because 
it has more than a remote resemblance to the density variations within any actual 
star. 



266 GRAVITATION [CHAP. 6 

7. Set up the equations to be solved for the pressure as a function of radius 
in a spherically symmetric mass M of gas, assuming that the gas obeys the perfect 
gas laws and that the temperature is known as a function of radius. 

8. (a) Assume that ordinary cold matter collapses, under a pressure greater 
than a certain critical pressure po, to a state of very high density pi. A planet of 
mass M is constructed of matter of mean density po in its normal state. Assuming 
uniform density and conditions of fluid equilibrium, at what mass M o and radius 
ro will the pressure at the center reach the critical value po? (b) li M > Mo, the 
planet will have a very dense core of density pi surrounded by a crust of density 
po. Calculate the resulting pressure distribution within the planet in terms of the 
radius r\ of the core and the radius r 2 of the planet. Show that if M is somewhat 
larger than Mo, then the radius ri of the planet is less than ro. (The planet Jupiter 
is said to have a mass very nearly equal to the critical mass Mo, so that if it were 
heavier it might be smaller.) 

9. Find the pressure and temperature as functions of radius for the star of 
Problem 6 if the star is composed of a perfect gas of atomic weight A. 

10. Find the density and gravitational field intensity as a function of radius 
inside a small spherically symmetric planet, to order (1/B 2 ), assuming that the 
bulk modulus B is constant. The mass is M and the radius is a. [Hint: Calculate 
g(r) assuming uniform density; then find the resulting pressure p(r), and the 
density p(r) to order (1/B). Recalculate g(r) using the new p(r), and proceed 
by successive approximations to terms of order (1/B 2 ).] 

11. Consider a spherical mountain of radius a, mass M, floating in equilibrium 
in the earth, and whose density is half that of the earth. Assume that a is much 
less than the earth's radius, so that the earth's surface can be regarded as fiat in 
the neighborhood of the mountain. If the mountain were not present, the gravi- 
tational field intensity near the earth's surface would be go- (a) Find the differ- 
ence between go and the actual value of g at the top of the mountain, (b) If the 
top of the mountain is eroded flat, level with the surrounding surface of the earth, 
and if this occurs in a short time compared with the time required for the moun- 
tain to float in equilibrium again, find the difference between go and the actual 
value of g at the earth's surface at the center of the eroded mountain. 

12. (a) Find the gravitational potential and the field intensity due to a thin 
rod of length I and mass M at a point a distance r from the center of the rod in a 
direction making an angle 8 with the rod. Assume that ry> I, and carry the 
calculations only to second order in l/r. (b) Locate the center of gravity of the 
rod relative to the specified point. 

13. (a) Calculate the gravitational potential of a uniform circular ring of 
matter of radius a, mass M, at a distance r from the center of the ring in a direc- 
tion making an angle with the axis of the ring. Assume that r y> a, and cal- 
culate the potential only to second order in a/r. (b) Calculate to the same ap- 
proximation the components of the gravitational field of the ring at the specified 
point. 

14. A small body with cylindrical symmetry has a density p(r, 0) in spherical 
coordinates, which vanishes for r > a. The origin r = lies at the center of 
mass. Approximate the gravitational potential at a point r, far from the body 



PROBLEMS 267 

(r ^> a), by expanding in a power series in (a/r), and show that it has the form 

9(r,«) =^ + ^P2(cos0) + ^P 3 (cos0) + ---, 

where P2 (cos 6), P3 (cos 0) are quadratic and cubic polynomials in cos that 
do not depend on the body, and Q, E are constants which depend on the mass 
distribution. Find expressions for P2, P3, Q, and E, and show that Q is of the 
order of magnitude Ma 2 , and E of the order Ma 3 . It is conventional to normalize 
Pi so that the constant term is —\, and Pz so that the linear term is — § cos 0. 
The parameters Q, E are then called the quadrupole moment and the octopole 
moment of the body. The polynomials P2, P3, • ■ • are the Legendre polynomials. 

15. The earth has approximately the shape of an oblate ellipsoid of revolution 
whose polar diameter 2a(l — tj) is slightly shorter than its equatorial diameter 
2a. (ij = 0.0034.) To determine to first order in rj, the effect of the earth's 
oblateness on its gravitational field, we may replace the ellipsoidal earth by a 
sphere of radius R so chosen as to have the same volume. The gravitational 
field of the earth is then the field of a uniform sphere of radius R with the mass 
of the earth, plus the field of a surface distribution of mass (positive or negative), 
representing the mass per unit area which would be added or subtracted to form 
the actual ellipsoid. 

(a) Show that the required surface density is, to first order in rj, 

a = ^ijap(l — 3 cos 2 0), 

where is the colatitude, and p is the volume density of the earth (assumed 
uniform). Since the total mass thus added to the surface is zero, its gravitational 
field will represent the effect of the oblate shape of the earth. 

(b) Show that the resulting correction to the gravitational potential at a very 
great distance r^> a from the earth is, to order (a 3 //- 3 ), 

ss ~ \ v ^r~ (1 ~ 3 cos2 6) ' {r y> a) - 

16. (a) Use Gauss' theorem (6-26) to determine the gravitational field inside 
and outside a spherical shell of radius a, mass M, uniform density, (b) Cal- 
culate the resulting gravitational potential. 

17. (a) Find the gravitational field at a distance x from an infinite plane 
sheet of density a per unit area, (b) Compare this result with the field just 
outside a spherical shell of the same surface density. What part of the field 
comes from the immediately adjacent matter and what part from more distant 
matter? 

18. Show that the gravitational field equations (6-21), (6-31), and (6-33) 
are satisfied by the field intensity and potential which you calculated in Prob- 
lem 3. 

*19. (a) Show that SQ found in Problem 15(b) satisfies Laplace's equation 
(6-35). This, together with the fact that 59 has the same angular dependence 



268 GRAVITATION [CHAP. 6 

as the mass density which produces it, suggests that the formula given for 5Q 
may actually be valid everywhere outside the earth, (b) To show this, consider 
Poisson's equation (6-33) with p — /(r)(l — 3cos 2 0). Show that a solution 
9 = *(r)(l — 3 cos 2 0) will satisfy Eq. (6-33) with this form of p, provided 

d 2 h , 2 dh &h , _, 

(c) Show that h = r -3 satisfies this equation in the region where / = 0. Can 
you complete the proof that the formula for 5g found in Problem 15(b) is in 
fact valid everywhere outside the earth? 



CHAPTER 7 



MOVING COORDINATE SYSTEMS 



7-1 Moving origin of coordinates. Let a point in space be located by 
vectors r, r* with respect to two origins of coordinates 0, 0*, and let 0* be 
located by a vector h with respect to 
(Fig. 7-1). Then the relation be- 
tween the coordinates r and r* is 
given by 

r = r* + h, (7-1) 



r* = r 



(7-2) 



In terms of rectangular coordinates, 
with axes x*, y*, z* parallel to axes 
x, y, z, respectively, these equations 
can be written: 




Fig. 7-1. Change of origin of coordi- 
nates. 



x = x* + h x 



x* = x 



y = y* + h v , 
y* = y — h y , 



2 = z* + h z ; (7-3) 
z* = z- h z . (7-4) 



Now if the origin 0* is moving with respect to the origin 0, which we 
regard as fixed, the relation between the velocities relative to the two sys- 
tems is obtained by differentiating Eq. (7-1) : 



_ dx _ drj* cfli 
dt dt dt 
= v* + Vft, 



(7-5) 



where v and v* are the velocities of the moving point relative to and 0*, 
and Vh is the velocity of 0* relative to 0. We are supposing that the 
axes x*, y*, z* remain parallel to x, y, z. This is called a translation of the 
starred coordinate system with respect to the unstarred system. Written 
out in cartesian components, Eq. (7-5) becomes the time derivative of 
Eq. (7-3). The relation between relative accelerations is 



d 2 T 



2_* 



d't 



= a* + a A . 



d\ 
df2 



(7-6) 



Again these equations can easily be written out in terms of their rectangu- 
lar components. 

269 



270 MOVING COORDINATE SYSTEMS [CHAP. 7 

Newton's equations of motion hold in the fixed coordinate system, so 
that we have, for a particle of mass m subject to a force F: 

m § = F " ™ 

Using Eq. (7-6), we can write this equation in the starred coordinate sys- 
tem: 

m ^p- + ma h = F. (7-8) 

If 0* is moving at constant velocity relative to 0, then &h = 0, and we 
have 

m *±- = F. (7-9) 

Thus Newton's equations of motion, if they hold in any coordinate system, 
hold also in any other coordinate system moving with uniform velocity 
relative to the first. This is the Newtonian principle of relativity. It 
implies that, so far as mechanics is concerned, we cannot specify any 
unique fixed coordinate system or frame of reference to which Newton's 
laws are supposed to refer; if we specify one such system, any other system 
moving with constant velocity relative to it will do as well. This property 
of Eq. (7-7) is sometimes expressed by saying that Newton's equations of 
motion remain invariant in form, or that they are covariant, with respect to 
uniform translations of the coordinates. The concept of frame of reference 
is not quite the same as that of a coordinate system, in that if we make a 
change of coordinates that does not involve the time, we do not regard this 
as a change of frame of reference. A frame of reference includes all coordi- 
nate systems at rest with respect to any particular one. The principle 
of relativity proposed by Einstein asserts that the relativity principle is not 
restricted to mechanics, but holds for all physical phenomena. The special 
theory of relativity is the result of the application of this principle to all 
types of phenomena, particularly electromagnetic phenomena. It turns 
out that this can only be done by modifying Newton's equations of motion 
slightly and, in fact, even Eqs. (7-5) and (7-6) require modification, f 
For any motion of 0*, we can write Eq. (7-8) in the form 

= F — raa A . (7-10) 



dP 

This equation has the same form as the equation of motion (7-7) in a fixed 
coordinate system, except that in place of the force F, we have F — m&h- 



t P. G. Bergmann, Introduction to the Theory of Relativity. New York: Prentice- 
Hall, 1946. (Part 1.) 



7-2] ROTATING COORDINATE SYSTEMS 271 

The term —m&h we may call a fictitious force. We can treat the motion 
of a mass m relative to a moving coordinate system using Newton's equa- 
tions of motion if we add this fictitious force to the actual force which acts. 
From the point of view of classical mechanics, it is not a force at all, but 
part of the mass times acceleration transposed to the other side of the 
equation. The essential distinction is that the real forces F acting on m 
depend on the positions and motions of other bodies, whereas the fictitious 
force depends on the acceleration of the starred coordinate system with 
respect to the fixed coordinate system. In the general theory of relativity, 
terms like -maj are regarded as legitimate forces in the starred coordi- 
nate system, on the same footing with the force F, so that in all coordinate 
systems the same law of motion holds. This, of course, can only be done 
if it can be shown how to deduce the force —ma^ from the positions and 
motions of other bodies. The program is not so simple as it may seem 
from this brief outline, and modifications in the laws of motion are required 
to carry it through, f 

7-2 Rotating coordinate systems. We now consider coordinate systems 
x, y, z and x*, y*, z* whose axes are rotated relative to one another as in 
Fig. 7-2, where, for the present, the origins of the two sets of axes coin- 
cide. Introducing unit vectors i, j, k associated with axes x, y, z, and 
unit vectors i*, j*, k* associated with axes x*, y*, z*, we can express the 
position vector r in terms of its components along either set of axes: 

r = xx + y] + zk, (7-11) 

r = x *i* + y*j* + z*k*. (7-12) 

Note that since the origins now coincide, a point is represented by the 
same vector r in both systems; only the components of r are different 
along the different axes. The relations between the coordinate systems 




Fig. 7-2. Rotation of coordinate axes. 



t Bergmann, op. cit. (Part 2.) 



272 MOVING COORDINATE SYSTEMS [CHAP. 7 

can be obtained by taking the dot product of either the starred or the 

unstarred unit vectors with Eqs. (7-11) and (7-12). For example, if we 

compute i-r, j-r, k-r, from Eqs. (7-11) and (7-12) and equate the results, 

we obtain __ , _ „ 

x = s*(i*.i) + 2/*(j*-i) + 2*(k*-i), 

y = s*(i*-j) + y*(j*-J) + «*(k*'J), (7-13) 

z = x*(i*-k) + »*0*-k) + z*(k*-k). 

The dot products (i*-i), etc., are the cosines of the angles between the 
corresponding axes. Similar formulas for x*, y*, z* in terms of x, y, z can 
easily be obtained by the same process. These formulas are rather compli- 
cated and unwieldy, and we shall fortunately be able to avoid using them 
in most cases. Equations (7-11), (7-12), and (7-13) do not depend on the 
fact that the vector r is drawn from the origin. Analogous formulas apply 
in terms of the components of any vector A along the two sets of axes. 
The time derivative of any vector A was denned by Eq. (3-52) : 

dA =lim A(t + M)-A(t) . (y _ 14) 

at aj->o At 

In attempting to apply this definition in the present case, we encounter 
a difficulty if the coordinate systems are rotating with respect to each 
other. A vector which is constant in one coordinate system is not con- 
stant in the other, but rotates. The definition requires us to subtract 
A(0 from A(t + At). During the time At, coordinate system x*, y*, z* has 
rotated relative to x, y, z, so that at time t + At, the two systems will not 
agree as to which vector is (or was) A(0, i.e., which vector is in the same 
position that A was in at time t. The result is that the time derivative of 
a given vector will be different in the two coordinate systems. Let us use 
d/dt to denote the time derivative with respect to the unstarred coordinate 
system, which we regard as fixed, and d*/dt to denote the time derivative 
with respect to the rotating starred coordinate system. We make this 
distinction with regard to vectors only; there is no ambiguity with regard 
to numerical quantities, and we denote their time derivatives by d/dt, or 
by a dot, which will have the same meaning in all coordinate systems. 
Let the vector A be given by 

A = A x i + A y i + A*k, (7-15) 

A = A%i* + A*j* + A%*. (7-16) 

The unstarred time derivative of A may be obtained by differentiating 
Eq. (7-15), regarding i, j, k as constant vectors in the fixed system: 

^ = A x i + A v j + AX (7-17) 



7-2] 



ROTATING COORDINATE SYSTEMS 



273 



Similarly, the starred derivative of A is given in terms of its starred com- 
ponents by 



£p = i*i* + A%* -r Atk*. 



(7-18) 



We may regard Eqs. (7-17) and (7-18) as the definitions of unstarred and 
starred time derivatives of a vector. We can also obtain a formula for 
d/dt in starred components by taking the unstarred derivative of Eq. 
(7-16), remembering that the unit vectors i*, j*, k* are moving relative 
to the unstarred system, and have time derivatives: 



dA 
dt 



A%i* 



+ A*j* + Atk* + At^- + A 



*dj*_ 
y dt 



+ A ° dt 



(7-19) 



A similar formula could be obtained for d*A/dt in terms of its unstarred 
components. 

Let us now suppose that the starred coordinate system is rotating about 
some axis OQ through the origin, with an angular velocity w (Fig. 7-3). 
We define the vector angular velocity u as a vector of magnitude w directed 
along the axis OQ in the direction of advance of a right-hand screw rotat- 
ing with the starred system. Consider a vector B at rest in the starred 
system. Its starred derivative is zero, and we now show that its unstarred 
derivative is 

dB 



dt 



= w x B. 



(7-20) 



In order to subtract B(t) from B(t + At), we draw these vectors with 
their tails together, and it will be convenient to place them with their 
tails on the axis of rotation. (The time derivative depends only on the com- 
ponents of B along the axes, and not on the position of B in space.) We 




Fig. 7-3. Time derivative of a rotating vector. 



274 MOVING COORDINATE SYSTEMS [CHAP. 7 

first verify from Fig. 7-3 that the direction of dB/dt is given correctly by 
Eq. (7-20), recalling the definition [Eq. (3-24) and Fig. 3-11] of the cross 
product. The magnitude of dB/dt as given by Eq. (7-20) is 



dB 



dt 



\a X B| = uB sin 6. (7-21) 



This is the correct formula, since it can be seen from Fig. 7-3 that, when 
At is small, 

|AB| = (.Bsinfl)(wAO. 

When Eq. (7-20) is applied to the unit vectors i*, j*, k*, Eq. (7-19) be- 
comes, if we make use of Eqs. (7-18) and (7-16) : 

dA d*A 

^ = ~ + A* x (a x i*) + A* y (a x j*) + A* z (a x k*) 

d*A 
= ~ + a X A. (7-22) 

dt 

This is the fundamental relationship between time derivatives for rotating 
coordinate systems. It may be remembered by noting that the time de- 
rivative of any vector in the unstarred coordinate system is its derivative 
in the starred system plus the unstarred derivative it would have if it 
were at rest in the starred system. Equation (7-22) applies even when 
the angular velocity vector a is changing in magnitude and direction with 
time. Taking the derivative of right and left sides of Eq. (7-22), and 
applying Eq. (7-22) again to A and d*A/dt, we have for the second time 
derivative of any vector A : 

dA . da . 



d*A = d (d*A\ 
dt 2 ~ dt\dt ) 

d*A , (d*A . .\ , dw w . 

+ wX nr + » x \-dT + wxA ) + -di xA 



_rf* 2 A 
dt 2 

d* 2 A , „ d*A , , ., . dw . ,_ „„. 

= _ r + 2wX __ +wX(wxA )+^xA. (7-23) 

In view of Eq. (3-29), the starred and unstarred derivatives of any vector 
parallel to the axis of rotation are the same, according to Eq. (7-22). In 
particular, 

da _ d*a 

~dl ~ IT' 



7-2] ROTATING COORDINATE SYSTEMS 275 

It is to be noted that the vector w on both sides of this equation is the 
angular velocity of the starred system relative to the unstarred system, 
although its time derivative is calculated with respect to the unstarred 
system on the left side, and with respect to the starred system on the 
right. The angular velocity of the unstarred system relative to the starred 
system will be — to. 

We now show that the relations derived above for a rotating coordinate 
system are perfectly general, in that they apply to any motion of the 
starred axes relative to the unstarred axes. Let the unstarred rates of 
change of the starred unit vectors be given in terms of components along 
the starred axes by 

di* 

-£ = ani* + a 12 j* + ai 3 k*, 

d\* 

-jj- = a 21 i* + a 22 j* + a 23 k*, (7-24) 

dk* 

-jf = a 3 ii* + a 32 j* + a 33 k*. 

By differentiating the equation 

i*-i* = 1, (7-25) 

we obtain 

f -i* = 0. (7-26) 

From this and the corresponding equations for j* and k*, we have 

an = a 22 = a 33 = 0. (7-27) 

By differentiating the equation 

i*-k* = 0, (7-28) 

we obtain 



«K* = — 1 • ■ 

dt dt 



(7-29) 



From this and the other two analogous equations, we have 

«31 = — ai3, 0, 12 = — 021, 023 = —a32- (7-30) 

Let a vector to be defined in terms of its starred components by: 

w* = a 23 , 03% = o 31 , co^ = a 12 . (7-31) 



di* 
dt 


= 


CO 


X 


i*, 


dj* 
dt 


= 


CO 


X 


i*, 


dk* 
dt 


= 


w 


X 


k*. 



276 MOVING COORDINATE SYSTEMS [CHAP. 7 

Equations (7-24) can now be rewritten, with the help of Eqs. (7-27), 
(7-30), and (7-31), in the form 



(7-32) 



According to Eq. (7-20), these time derivatives of i*, j*, k* are just those 
to be expected if the starred unit vectors are rotating with an angular 
velocity a. Thus no matter how the starred coordinate axes may be 
moving, we can define at any instant an angular velocity vector «, given 
by Eq. (7-31), such that the time derivatives of any vector relative to the 
starred and unstarred coordinate systems are related by Eqs. (7-22) and 
(7-23). 

Let us now suppose that the starred coordinate system is moving so that 
its origin 0* remains fixed at the origin of the fixed coordinate system. 
Then any point in space is located by the same position vector r in both 
coordinate systems [Eqs. (7-11) and (7-12)]. By applying Eqs. (7-22) 
and (7-23) to the position vector r, we obtain formulas for the relation 
between velocities and accelerations in the two coordinate systems: 

dx d*T ,„ . 

dt == W + MXI > (7 " 33) 

d 2 i d* 2 t , , . , „ d*r , du ,„ „,. 

dii = -dW + » x (» x ^+ 2 « x W+dt XI - (7 " 34) 

Formula (7-34) is called Coriolis' theorem. The first term on the right is 
the acceleration relative to the starred system. The second term is called 
the centripetal acceleration of a point in rotation about an axis (centripetal 
means "toward the center"). Using the notation in Fig. 7-4, we readily 
verify that to X (« X r) points directly toward and perpendicular to the 
axis of rotation, and that its magnitude is 

\w X (« X r)| = u 2 r sine 

v 2 

(7-35) 



rsin 9 



where v = ur sin 6 is the speed of circular motion and (r sin 6) is the dis- 
tance from the axis. The third term is present only when the point r is 
moving in the starred system, and is called the coriolis acceleration. The 



7-2] ROTATING COORDINATE SYSTEMS 277 




Fig. 7-4. Centripetal acceleration. 

last term vanishes for a constant angular velocity of rotation about a 
fixed axis. 

If we suppose that Newton's law of motion (7-7) holds in the unstarred 
coordinate system, we shall have in the starred system: 

m^~ + ma X (w X r) + 2wuo X ^~ + m^ Xi = F. (7-36) 

Transposing the second, third, and fourth terms to the right side, we ob- 
tain an equation of motion similar in form to Newton's equation of motion : 

to -—- = F — mwx(wxr)— 2mo» X -3- m -^ X r. (7-37) 

The second term on the right is called the centrifugal force (centrifugal 
means "away from the center"); the third term is called the coriolis force. 
The last term has no special name, and appears only for the case of non- 
uniform rotation. If we introduce the fictitious centrifugal and coriolis 
forces, the laws of motion relative to a rotating coordinate system are the 
same as for fixed coordinates. A great deal of confusion has arisen regard- 
ing the term "centrifugal force." This force is not a real force, at least 
in classical mechanics, and is not present if we refer to a fixed coordinate 
system in space. We can, however, treat a rotating coordinate system as 
if it were fixed by introducing the centrifugal and coriolis forces. Thus a 
particle moving in a circle has no centrifugal force acting on it, but only a 
force toward the center which produces its centripetal acceleration. How- 
ever, if we consider a coordinate system rotating with the particle, in this 
system the particle is at rest, and the force toward the center is balanced 
by the centrifugal force. It is very often useful to adopt a rotating coordi- 
nate system. In studying the action of a cream separator, for example, it 
is far more convenient to choose a coordinate system in which the liquid 



278 MOVING COORDINATE SYSTEMS [CHAP. 7 

is at rest, and use the laws of diffusion to study the diffusion of cream 
toward the axis under the action of the centrifugal force field, than to try 
to study the motion from the point of view of a fixed observer watching 
the whirling liquid. 

We can treat coordinate systems in simultaneous translation and rota- 
tion relative to each other by using Eq. (7-1) to represent the relation be- 
tween the coordinate vectors r and r* relative to origins 0, 0* not neces- 
sarily coincident. In the derivation of Eqs. (7-32), no assumption was 
made about the origin of the starred coordinates, and therefore Eqs. (7-22) 
and (7-23) may still be used to express the time derivatives of any vector 
with respect to the unstarred coordinate system in terms of its time de- 
rivatives with respect to the starred system. Replacing dt*/dt, d 2 t*/dt in 
Eqs. (7-5) and (7-6) by their expressions in terms of the starred deriva- 
tives relative to the starred system as given by Eqs. (7-33) and (7-34), 
we obtain for the position, velocity, and acceleration of a point with re- 
spect to coordinate systems in relative translation and rotation : 

r = r* + h, (7-38) 

7-3 Laws of motion on the rotating earth. We write the equation of 
motion, relative to a coordinate system fixed in space, for a particle of 
mass ra subject to a gravitational force rag and any other nongravitational 
forces F: 

m^p=F + rag. (7-41) 

Now if we refer the motion of the particle to a coordinate system at rest 
relative to the earth, which rotates with constant angular velocity u, and 
if we measure the position vector r from the center of the earth, we have, 
by Eq. (7-34) : 

d* 2 r d*i 

= ra -jTg- + raw X (« X r) + 2mu X -3- » (7-42) 

which can be rearranged in the form 

d* 2 r d*r 

m — ■ = F + ra[g - » X (w X r)] - 2ra« X ~- • (7-43) 



7-3] 



LAWS OF MOTION ON THE ROTATING EARTH 



279 



This equation has the same form as Newton's equation of motion. We 
have combined the gravitational and centrifugal force terms because both 
are proportional to the mass of the particle and both depend only on the 
position of the particle; in their mechanical effects these two forces are 
indistinguishable. We may define the effective gravitational acceleration 
g e at any point on the earth's surface by: 



g.(r) = g(r) - co X (co X r). 



(7-44) 



The gravitational force which we measure experimentally on a body of 
mass m at restf on the earth's surface is mg e . Since — w X (co X r) points 
radially outward from the earth's axis, g e at every point north of the 
equator will point slightly to the south of the earth's center, as can be 
seen from Fig. 7-5. A body released near the earth's surface will begin 
to fall in the direction of g e , the direction determined by a plumb line is 
that of g e , and a liquid will come to equilibrium with its surface perpen- 
dicular to g e - This is why the earth has settled into equilibrium in the 
form of an oblate ellipsoid, flattened at the poles. The degree of flattening 
is just such as to make the earth's surface at every point perpendicular to 
g e (ignoring local irregularities). 

Equation (7-43) can now be written 



d* 2 r d*T 



(7-45) 



The velocity and acceleration which appear in this equation are unaffected 
if we relocate our origin of coordinates at any convenient point at the sur- 
face of the earth; hence this equation applies to the motion of a particle of 
mass m at the surface of the earth relative to a local coordinate system at 
rest on the earth's surface. The only unfamiliar term is the coriolis force 



« x (w x r) 




Fig. 7-5. Effective acceleration of gravity on the rotating earth. 



t A body in motion is subject also to the coriolis force. 



280 MOVING COORDINATE SYSTEMS [CHAP. 7 

which acts on a moving particle. The reader can convince himself by a 
few calculations that this force is comparatively small at ordinary veloci- 
ties d*i/dt. It will be instructive to try working out the direction of the 
coriolis force for various directions of motion at various places on the 
earth's surface. The coriolis force is of major importance in the motion 
of large air masses, and is responsible for the fact that in the northern 
hemisphere tornados and cyclones circle in the direction south to east to 
north to west. In the northern hemisphere, the coriolis force acts to de- 
flect a moving object toward the right. As the winds blow toward a low 
pressure area, they are deflected to the right, so that they circle the low 
pressure area in a counterclockwise direction. An air mass circling in this 
way will have a low pressure on its left, and a higher pressure on its right. 
This is just what is needed to balance the coriolis force urging it to the 
right. An air mass can move steadily in one direction only if there is a 
high pressure to the right of it to balance the coriolis force. Conversely, 
a pressure gradient over the surface of the earth tends to develop winds 
moving at right angles to it. The prevailing westerly winds in the northern 
temperate zone indicate that the atmospheric pressure toward the equator 
is greater than toward the poles, at least near the earth's surface. The 
easterly trade winds in the equatorial zone are due to the fact that any air 
mass moving toward the equator will acquire a velocity toward the west 
due to the coriolis force acting on it. The trade winds are maintained by 
high pressure areas on either side of the equatorial zone. 

7-4 The Foucault pendulum. An interesting application of the theory 
of rotating coordinate systems is the problem of the Foucault pendulum. 
The Foucault pendulum has a bob hanging from a string arranged to swing 
freely in any vertical plane. The pendulum is started swinging in a defi- 
nite vertical plane and it is observed that the plane of swinging gradually 
precesses about the vertical axis during a period of several hours. The 
bob must be made heavy, the string very long, and the support nearly 
frictionless, in order that the pendulum can continue to swing freely for 
long periods of time. If we choose the origin of coordinates directly below 
the point of support, at the point of equilibrium of the pendulum bob of 
mass m, then the vector r will be nearly horizontal, for small amplitudes of 
oscillation of the pendulum. In the northern hemisphere, a points in the 
general direction indicated in Fig. 7-6, relative to the vertical. Writing t 
for the tension in the string, we have as the equation of motion of the bob, 
according to Eq. (7—15) : 

d* 2 i d*i 

m -^p- = r + mg e — 2mu X -^- • (7-46) 

If the coriolis force were not present, this would be the equation for a 



7-4] 



THE FOUCAULT PENDULUM 



281 




mge 



simple pendulum on a nonrotating 
earth. The coriolis force is very 
small, less than 0.1% of the gravita- 
tional force if the velocity is 5 mi/hr 
or less, and its vertical component is 
therefore negligible in comparison 
with the gravitational force. (It is 
the vertical force which determines 
the magnitude of the tension in the 
string.) However, the horizontal 
component of the coriolis force is 
perpendicular to the velocity d*r/dt, 
and as there are no other forces in 
this direction when the pendulum 
swings to and fro, it can change the 
nature of the motion. Any force with a horizontal component perpendic- 
ular to d*r/dt will make it impossible for the pendulum to continue to 
swing in a fixed vertical plane. In order to solve the problem including 
the coriolis term, we use the experimental result as a clue, and try to find 
a new coordinate system rotating about the vertical axis through the 
point of support at such an angular velocity that in this system the 
coriolis terms, or at least their horizontal components, are missing. Let 
us introduce a new coordinate system rotating about the vertical axis 
with constant angular velocity kQ, where k is a vertical unit vector. We 
shall call this precessing coordinate system the primed coordinate system, 
and denote the time derivative with respect to this system by d'/dt. 
Then we shall have, by Eqs. (7-33) and (7-34) : 



Fig. 7-6. The Foucault pendulum. 



d< 2 dl 2 
Equation (7-46) becomes 

d' 2 T 
m ~M2 = T + m &° 



d*I d'T , n . ^ 

J*2_ j/2_ 

= ^ + a 2 kx(kxr)+2!lkx 



dt ' 



(7-47) 
(7-48) 



2m« X t ^ + Qk X i) 

m£2 2 k X (k X r) — 2mGk X ^f 



d't 



dt 

r + mg e — 2mO&> x (k x r) — mfi 2 k X (k X r) 

d't 



2m(w + kfi) X 



dt 



(7-49) 



282 MOVING COORDINATE SYSTEMS [CHAP. 7 

We expand the triple products by means of Eq. (3-35) : 



d' 2 r 



2i, 



m-375- = t + mg e — m(2fiwr + fi k-r)k 



<8 a 



+ m(2fik-w + fi 2 )r - 2m(« + kO) X ^ • (7-50) 



Every vector on the right side of Eq. (7-50) lies in the vertical plane con- 
taining the pendulum, except the last term. Since, for small oscillations, 
d'r/dt is practically horizontal, we can make the last term lie in this vertical 
plane also by making (« + kfi) horizontal. We therefore require that 

k-( w + kSi) = 0. (7-51) 

This determines : 

U = — w cos 0, (7-52) 

where w is the angular velocity of the rotating earth, fi is the angular 
velocity of the precessing coordinate system relative to the earth, and 6 is 
the angle between the vertical and the earth's axis, as indicated in Fig. 7-6. 
The vertical is along the direction of — g e , and since this is very nearly the 
same as the direction of — g (see Fig. 7-5), 6 will be practically equal to the 
colatitude, that is, the angle between r and w in Fig. 7-5. For small oscilla- 
tions, if fi is determined by Eq. (7-52), the cross product in the last term 
of Eq. (7-50) is vertical. Since all terms on the right of Eq. (7-50) now 
lie in a vertical plane containing the pendulum, the acceleration d' 2 r/dt 2 
of the bob in the precessing system is always toward the vertical axis, and 
if the pendulum is initially swinging to and fro, it will continue to swing 
to and fro in the same vertical plane in the precessing coordinate system. 
Relative to the earth, the plane of the motion precesses with angular 
velocity Q of magnitude and sense given by Eq. (7-52). In the northern 
hemisphere, the precession is clockwise looking down. 

Since the last three terms on the right in Eq. (7-50) are much smaller 
than the first two, the actual motion in the precessing coordinate system 
is practically the same as for a pendulum on a nonrotating earth. Even at 
large amplitudes, where the velocity d'r/dt has a vertical component, care- 
ful study will show that the last term in Eq. (7-50), when Q. is chosen 
according to Eq. (7-52), does not cause any additional precession relative 
to the precessing coordinate system, but merely causes the bob to swing in 
an arc which passes slightly east of the vertical through the point of sup- 
port. At the equator, Q. is zero, and the Foucault pendulum does not pre- 
cess; by thinking about it a moment, perhaps you can see physically why 
this is so. At the north or south pole, £2 = ±w, and the pendulum merely 
swings in a fixed vertical plane in space while the earth turns beneath it. 



7-5] lamor's theorem 283 

Note that we have been able to give a fairly complete discussion of the 
Foucault pendulum, by using Coriolis' theorem twice, without actually 
solving the equations of motion at all. 

7-5 Laimor's theorem. The coriolis force in Eq. (7-37) is of the same 
form as the magnetic force acting on a charged particle (Eq. 3-281), in 
that both are given by the cross product of the velocity of the particle 
with a vector representing a force field. Indeed, in the general theory of 
relativity, the coriolis forces on a particle in a rotating system can be re- 
garded as due to the relative motion of other masses in the universe in a 
way somewhat analogous to the magnetic force acting on a charged par- 
ticle which is due to the relative motion of other charges. The similarity 
in form of the two forces suggests that the effect of a magnetic field on a 
system of charged particles may be canceled by introducing a suitable 
rotating coordinate system. This idea leads to Larmor's theorem, which 
we state first, and then prove: 

Larmor's theorem. If a system of charged particles, all having the same 
ratio q/m of charge to mass, acted on by their mutual (central) forces, and 
by a central force toward a common center, is subject in addition to a weak 
uniform magnetic field B, its possible motions will be the same as the motions 
it could perform without the magnetic field, superposed upon a slow pre- 
cession of the entire system about the center of force with angular velocity 

- = ~ db B (7 " 53) 

The definition of a weak magnetic field will appear as the proof is de- 
veloped. We shall assume that all the particles have the same charge q 
and the same mass m, although it will be apparent that the only thing that 
needs to be assumed is that the ratio q/m is constant. Practically the 
only important applications of Larmor's theorem are to the behavior of 
an atom in a magnetic field. The particles here are electrons of mass m, 
charge q = — e, acted upon by their mutual electrostatic repulsions and 
by the electrostatic attraction of the nucleus. 

Let the central force acting on the fcth particle be ¥%, and let the sum of 
the forces due to the other particles be F\. Then the equations of motion 
of the system of particles, in the absence of a magnetic field, are 

m^i = F c k + Ft, k = 1, . . . , N, (7-54) 

where N is the total number of particles. The force F% depends only on 
the distance of particle fc from the center of force, which we shall take as 
origin, and the forces F^ depend only on the distances of the particles from 



284 MOVING COORDINATE SYSTEMS [CHAP. 7 

one another. When the magnetic field is applied, the equations of motion 
become, by Eq. (3-281) : 



TO 



% = n + n + l^xB, k=l,...,N. (7-55) 



In order to eliminate the last term, we introduce a starred coordinate sys- 
tem with the same origin, rotating about this origin with angular velocity 
«. Making use of Eqs. (7-33) and (7-34), we can write the equations of 
motion in the starred coordinate system: 

m -^ = F| + Pi - ma X (a X i k ) + 2 (« x t t ) X B 

We can make the last term vanish by setting 

Equation (7-56) then becomes 

m IP* = F * + F * + li B x < B x r *)> k=l,...,N. (7-58) 

The forces F£ and F^ depend only on the distances of the particles from the 
origin and on their distances from one another, and these distances will be 
the same in the starred and unstarred coordinate systems. Therefore, if 
we neglect the last term, Eqs. (7-58) have exactly the same form in 
terms of starred coordinates as Eqs. (7-54) have in unstarred coordinates. 
Consequently, their solutions will then be the same, and the motions of 
the system expressed in starred coordinates will be the same as the motions 
of the system expressed in unstarred coordinates in the absence of a mag- 
netic field. This is Larmor's theorem. 

The condition that the magnetic field be weak means that the last term 
in Eq. (7-58) must be negligible in comparison with the first two terms. 
Notice that the term we are neglecting is proportional to B 2 , whereas the 
term in Eq. (7-55) which we have eliminated is proportional to B. Hence, 
for sufficiently weak fields, the former may be negligible even though the 
latter is not. The last term in Eq. (7-58) may be written in the form 

2 

-^ B x (B X r k ) = mu X (« X r»). (7-59) 

Another way of formulating the condition for a weak magnetic field is to 



7-6] THE RESTRICTED THREE-BODY PROBLEM 285 

say that the Larmor frequency o>, given by Eq. (7-57), must be small com- 
pared with the frequencies of the motion in the absence of a magnetic 
field. 

The reader who has understood clearly the above derivation should be 
able to answer the following two questions. The cyclotron frequency, 
given by Eq. (3-299), for the motion of a charged particle in a magnetic 
field is twice the Larmor frequency, given by Eq. (7-57). Why does not 
Larmor's theorem apply to the charged particles in a cyclotron? Equa- 
tion (7-58) can be derived without any assumption as to the origin of co- 
ordinates in the starred system. Why is it necessary that the axis of rota- 
tion of the starred coordinate system pass through the center of force of 
the system of particles? 

7-6 The restricted three-body problem. We pointed out in Section 4-9 
that the three-body problem, in which three masses move under their 
mutual gravitational forces, cannot be solved in any general way. In this 
section we will consider a simplified problem, the restricted problem of 
three bodies, which retains many features of the more general problem, 
among them the fact that there is no general method of solving it. In the 
restricted problem, we are given two bodies of masses M\ and M 2 that 
revolve in circles under their mutual gravitational attraction and around 
their common center of mass. The third body of very small mass m moves 
in the gravitational field of Mi and M 2 . We are to assume that m is so 
small that the resulting disturbance of the motions of M i and M 2 can be 
neglected. We will further simplify the problem by assuming that m 
remains in the plane in which Mi and M 2 revolve. The problem thus 
reduces to a one-body problem in which we must find the motion of m in 
the given (moving) gravitational field of the other two. An obvious ex- 
ample would be a rocket moving in the gravitational fields of the earth and 
the moon, which revolve very nearly in circles about their common center 
of mass. 

If M\ and M 2 are separated by a distance a, then according to the 
results of Section 4-7, their angular velocity is determined by equating the 
gravitational force to mass times acceleration in the reduced problem, in 
which Mi is at rest and M 2 has mass n as given by Eq. (4-98) : 



so that 



2 MiM 2 G (n Rm 

u2= (Mi + M 2 )G 
a 6 



The center of mass divides the distance a into segments that are propor- 
tional to the masses. 



286 MOVING COORDINATE SYSTEMS [CHAP. 7 

We now introduce a coordinate system rotating with angular velocity 
w about the center of mass of M x and M 2 . In this system, M t and M 2 are 
at rest, and we will take them to be on the ar-axis at the points 

M 2 M\ 

Xl = M 1 + M 2 a > X2== ~M 1+ M 2 a - ff" 62 ) 

The angular velocity to is taken to be along the 2-axis. Then m moves in 
the xy-pl&ne, and its equation of motion is 

J*2 ,* 

m -^ = Fx + F 2 - mu X (» X r) - 2ww X ~ > (7-63) 

where F x and F2 are the gravitational attraction of Mi and M 2 on m. 
Written in terms of components, the two equations become 

M,G{x - Xl ) M 2 G(x - x 2 ) (Mj + M 2 )Gx . 

[(x-a;i)a+y2]8/a [(3 _ X2 )2 + ^3/2 f a a T ^^> 

.. = _ M x Gy M 2 Gy (M 1 + M 2 )Gy _ . 

[(X - Xtf + 2/2J3/2 [( x _ X2 )2 +2/ 2]3/2^ a 3 ^"^ 

(7-64) 
Note that the mass m cancels in these equations. 

Since the coriolis force is perpendicular to the velocity, it does no 
'work' in this moving coordinate system. Moreover, the centrifugal force 
has zero curl and can be derived from the 'potential energy' 

V c = -imcoV + y 2 ). (7-65) 

Therefore the total 'energy' in the moving coordinate system is a constant 
of the motion : 

'E' = \m(x 2 + y 2 ) + 'V, (7-66) 

where 

, v , __ _ mM-iG mM 2 G 



[{X — Zi)2 + 2/2]l/2 [( x — X2 )2 _|_ ^1/2 

m(M 1 + M 2 )G(x 2 + y 2 ) 
2a,3 



(7-67) 



The energy equation (7-66) enables us to make certain statements about 
the kinds of orbits that may be possible. In order to simplify the algebra, 
let us set 

£ = x/a, v = y/a, (7-68) 

^ ~ MT+m' h= ~ Mi+M 2 = *i - !■ C 7 " 69 ) 



7-6] THE RESTRICTED THREE-BODY PROBLEM 

Then Eq. (7-67) can be written as 

m{M l + M 2 )G I £ 2 



287 



'V 



Utt 



fl )2 + ,2]l/2 
Si 



1 



[(* - £a) 2 + >7 2 ] 1/2 2 ( * + 



» 2 )} 



(7-70) 



In order to see the nature of this function, let us first look for its singular 
points, where d'V'/d£ and d'V'/dy both vanish: 



&(« - €i) 



[(« - Si) 2 + » 2 ] 3/2 
€2*1 



+ 



Si CI - £2) 



[(« 



£ 2 )2 + ,2]3/2 

Siq 

K* - Si) 2 + v 2 ) 3 ' 2 ' [(S - S2) 2 + v 2 ] 312 



+ 



— t\ 



s = o, 

0. 



(7-71) 



A point Or, j/) for which these equations are satisfied is an equilibrium 
point for the mass m (in the rotating coordinate system), since Eqs. (7-64) 
are evidently satisfied if m is at rest at this point. We first consider points 
on the rj = axis. The second equation is then satisfied, and the first 
becomes 



|S~S 



Si) ■ Si(S - S2) 

I q "T" 1 > > I a 



S = 0. 



(7-72) 



In Fig. 7-7, we plot the function 'V, as given by Eq. (7-70), along the 
ij = axis. The roots of Eq. (7-72) are the maxima of ' V'(f, 0) in Fig. 7-7, 
where it can be seen that there are three such roots. Let us call them 
£a, Sb> £c as in the figure. Each is the root of a quintic equation which 
may be derived from Eq. (7-72). It is not difficult to show that 
d 2 'V'/dt dr,= 0, d 2 'V'/dt 2 < 0, and d 2 'V'/dr, 2 > at these points A, B, 




Fig. 7-7. A plot of 'F'(S, 0). 



288 



MOVING COORDINATE SYSTEMS 

y 



[chap. 7 




Fig. 7-8. Equipotential contours for 'V'(x, y). 



and C. If we expand 'V in a Taylor series about any one of these points, 
and consider only the quadratic terms, we see that the curves of constant 
'V are hyperbolas in the £ij-plane in the neighborhood of points A, B, C, 
as shown in Fig. 7-8, where we plot the contours of constant 'V. These 
points are saddlepoints of 'V; that is, 'V has a local maximum along the 
{-axis and a minimum along a line perpendicular to the {-axis at each of 
these points A, B, C. If i\ ?* 0, it can be factored from the second of 
Eqs. (7-71). We then multiply the second of Eqs. (7-71) by (£ — |j) 
and subtract from the first of these equations. After some manipulation 
and using Eq. (7-69), we obtain 



and, similarly, 



(I - £ 2 ) 2 + v 2 = 1, 
U - li) 2 + v 2 = l. 



(7-73) 
(7-74) 



These equations show that there are two singular points D, E, off the 
ij = axis, which lie at unit distance from (f i, 0) and (£ 2 , 0) which are 



7-6] THE RESTRICTED THREE-BODY PROBLEM 289 

themselves separated by a unit distance. By expanding 'V in a Taylor 
series about point D or E, we can show that curves of constant 'V are 
ellipses in the neighborhood of D or E, and that 'V has a maximum at D 
and E. Knowing the behavior near the singular points, we can easily 
sketch the general appearance of the contours of constant 'V, as shown 
in Fig. 7-8. The curves are numbered in order of increasing 'V. 

If this were a fixed coordinate system, we could immediately conclude 
that equilibrium points A, B, C, D, E are all unstable, since the force 
—V'V is directed away from each equilibrium point when m is at some 
nearby points. However, this argument does not hold here because it 
neglects the coriolis force in Eqs. (7-64). If we expand the right members 
of Eqs. (7-64) in powers of the displacements (say x — xo,y — yo) 
from one of the equilibrium points (say D), and retain only linear terms, 
we may determine approximately the motion near the equilibrium point. 
If this is done near point D (or E), for example, we find that in linear 
approximation, the motion near D is stable if one of the masses M i or 
M 2 contains more than about 96% of the total mass (Mi + M 2 ). (See 
Problem 16.) For motions very near to point D, we may expect the linear 
approximation to yield a solution which is valid for very long times. 
Whether those motions which are stable in linear approximation are truly 
stable, in the sense that they remain near point D for all time, is one of 
the unsolved problems of classical mechanics. This matter is discussed 
further at the end of Section 12-6.* 

It is not difficult to show that, even in linear approximation, the equi- 
librium points A, B,C are unstable. If the motion in linear approximation 
is unstable, then the exact solution is certainly unstable. That is, regard- 
less of how close m is initially to the equilibrium point (but not at it), it 
will not, in general, remain as close but will move exponentially away, at 
least at first. The neglected nonlinear terms may, of course, eventually 
prevent the solution from going more than some finite distance from the 
equilibrium point. 

The only rigorous statements we can make about the motion of m, for 
very long periods of time, are those which can be derived from the energy 
equation (7-66). Given an initial position and velocity of m, we can 
calculate 'E'. The orbit then must remain in the region where 'V < 'E'. 
For example, motions which start near either mass Mi or M 2 , with 
'E' < V 3 , must remain confined to a region near that mass. Motions 
with 'E' > V5 may go to arbitrarily large distances; whether they actually 
do, we cannot say from energy arguments. However, the studies which 



* A more complete discussion of the problem of three bodies, on a more ad- 
vanced level than the present text, will be found in Aurel Wintner, The Analytical 
Foundations of Celestial Mechanics. Princeton: Princeton University Press, 1947. 



290 MOVING COORDINATE SYSTEMS [CHAP. 7 

have so far been made of the three-body problem make it very plausible, 
though it has not been proved, that except for special cases (e.g., M 2 = 0) 
or for special initial conditions, most orbits eventually wander throughout 
the region that is 'energetically' allowed. 

If we could find another constant of the motion, say F(x, y, x, y), we 
could solve the problem by methods like those used in Chapter 3 for the 
central force problem, where the angular momentum is also constant. 
Unfortunately, no other such constant is known, and in view of the last 
sentence of the preceding paragraph, it seems likely that none exists. 
This problem has been studied very extensively.* 

Faced with this situation, we may turn to the possibility of computing 
particular orbits from given initial conditions. This can be done either 
analytically, by approximation methods, or numerically, and in principle 
can be done to any desired accuracy and for any desired finite period of 
time. 

In Chapter 12 we shall discuss a closely related special case of the three- 
body problem. 

Problems 

1 . (a) Solve the problem of the freely falling body by introducing a translating 
coordinate system with an acceleration g. Set up and solve the equations of 
motion in this accelerated coordinate system and transform the result back to 
a coordinate system fixed relative to the earth. (Neglect the earth's rotation.) 
(b) In the same accelerated coordinate system, set up the equations of motion 
for a falling body subject to an air resistance proportional to its velocity (rela- 
tive to the fixed air). 

2. A mass m is fastened by a spring (spring constant k) to a point of support 
which moves back and forth along the z-axis in simple harmonic motion at 
frequency oj, amplitude a. Assuming the mass moves only along the a>axis, set 
up and solve the equation of motion in a coordinate system whose origin is at 
the point of support. 

3. Generalize Eq. (5-5) to the case when the origin of the coordinate system 
is moving, by adding fictitious torques due to the fictitious force on each particle. 
Express the fictitious torques in terms of the total mass M, the coordinate R* 
of the center of mass, and ah. Compare your result with Eq. (4-25). 

4. Derive a formula for d 3 A./dt 3 in terms of starred derivatives relative to a 
rotating coordinate system. 

5. Westerly winds blow from west to east in the northern hemisphere with an 
average speed v. If the density of the air is p, what pressure gradient is required 
to maintain a steady flow of air from west to east with this speed? Make reason- 
able estimates of v and p, and estimate the pressure gradient in lb-in _2 -mile _1 . 



: See A. Wintner, op. cit. 



PROBLEMS 291 

6. (a) It has been suggested that birds may determine their latitude by 
sensing the coriolis force. Calculate the force a bird must exert in level flight 
at 30 mi/hr against the sidewise component of coriolis force in order to fly in a 
straight line. Express your result in g's, that is, as a ratio of coriolis force to 
gravitational force, as a function of latitude and direction of flight. 

(b) If the bird's flight path is slightly circular, a centrifugal force will be 
present, which will add to the coriolis force and produce an error in estimated 
latitude. At 45° N latitude, how much may the flight path bend, in degrees per 
mile flown, if the latitude is to be determined within ±100 miles? (Assume the 
sidewise force is measured as precisely as necessary!) 

7. A body is dropped from rest at a height h above the surface of the earth, 
(a) Calculate the coriolis force as a function of time, assuming it has a negligible 
effect on the motion. Neglect air resistance, and assume h is small so that g e 
can be taken as constant, (b) Calculate the net displacement of the point of 
impact due to the calculated coriolis force. 

*8. Find the answer to Problem 7(b) by solving for the motion in a non- 
rotating coordinate system. What approximations are needed to arrive at the 
same result? 

9. A gyroscope consists of a wheel of radius r, all of whose mass is located on 
the rim. The gyroscope is rotating with angular velocity 6 about its axis, which 
is fixed relative to the earth's surface. We choose a coordinate system at rest 
relative to the earth whose z-axis coincides with the gyroscope axis and whose 
origin lies at the center of the wheel. The angular velocity u of the earth lies 
in the zz-plane, making an angle a with the gyroscope axis. 

Find the x-, y-, and z-components of the torque N about the origin, due to the 
coriolis force in the xyz-coordinate system, acting on a mass m on the rim of the 
gyroscope wheel whose polar coordinates in the zt/-plane are r, 6. Use this 
result to show that the total coriolis torque on the gyroscope, if the wheel has 
a mass M, is 

N = jMr 2 u8 sin a. 

This equation is the basis for the operation of the gyrocompass. 

*10. A mass m of a perfect gas of molecular weight M, at temperature T, is 
placed in a cylinder of radius a, height h, and whirled rapidly with an angular 
velocity a about the axis of the cylinder. By introducing a coordinate system 
rotating with the gas, and applying the laws of static equilibrium, assuming 
that all other body forces are negligible compared with the centrifugal force, 
show that 



RT 

V = -jf Po exp 



/ Mo?r 2 \ 
\2RT )' 



where p is the pressure, r is the distance from the axis, and 

mMia 
P0 = 2-rhRT[exp (Ma 2 a 2 /2RT) — 1] ' 



292 MOVING COORDINATE SYSTEMS [CHAP. 7 

*11. A particle moves in the xy-pl&ne under the action of a force 

F = — At, 

directed toward the origin. Find its possible motions by introducing a coordinate 
system rotating about the z-axis with angular velocity w chosen so that the 
centrifugal force just cancels the force F, and solving the equations of motion 
in this coordinate system. Describe the resulting motions, and show that your 
result agrees with that of Problem 31, Chapter 3. 

12. A ball of mass m slides without friction on a horizontal plane at the sur- 
face of the earth. Show that it moves like the bob of a Foucault pendulum 
of length equal to the earth's radius, provided it remains near the point of 
tangency. 

13. The bob of a pendulum is started so as to swing in a circle. By substitut- 
ing in Eq. (7-46), find the angular velocity and show that the contribution due 
to the coriolis force is given very nearly by Eq. (7-52). Neglect the vertical 
component of the coriolis force, after showing that it is zero on the average for 
the assumed motion. 

14. An electron revolves about a fixed proton in an ellipse of semimajor axis 
10 ~ 8 cm. If the corresponding motion occurs in a magnetic field of 10,000 
gausses, show that Larmor's theorem is applicable, and calculate the angular 
velocity of precession of the ellipse. 

15. Write down a potential energy for the last term in Eq. (7-58). If the 
plane of the orbit in Problem 14 is perpendicular to B, and if the orbit is very 
nearly circular, calculate (by the methods of Chapter 3) the rate of precession 
of the ellipse due to the last term in Eq. (7-58) in the rotating coordinate system. 
Is this precession to be added to or subtracted from that calculated in Problem 14? 

16. Find the three second derivatives of 'V with respect to £, ij for the point 
D in Fig. 7-7. Expand the equations of motion (7-64), keeping terms linear in 
£' = £ — £ and rj = i\ — ?;o. Using the method of Section 4-10, find the 
condition on Mi, M% in order that the normal modes of oscillation be stable. 
If M\ > M.2, what is the minimum value of Mi/ (Mi + M2)? 

17. Prove the statements made in Section 7-6 regarding the second deriva- 
tives of 'V at points A, B, and C in Fig. 7-7. Expand the equations of motion 
about points A and B, keeping terms linear in jj and £' = £ — £a,b- Show by 
the method of Section 4-10 that some of the solutions are unstable for any 
values of the masses. (You cannot find the second derivatives explicitly, but the 
proof depends only on their signs.) 

*18. (a) Write out the quintic equation which must be solved for iU in Fig. 7-7. 
Show that if M 2 = 0, the solution is % A = —1- (b) Solve numerically for £ A 
to two decimal places for the earth-moon system, (c). Find the minimum 
launching velocity from the surface of the earth for which it is 'energetically' 
possible for a rocket to leave the earth-moon system. Compare with the escape 
velocity from the earth. 

19. Two planets, each of mass M and radius R, revolve in circles about each 
other at a distance a apart. Find the minimum velocity with which a rocket 
might leave one planet to arrive at the other. Show that the rocket must have a 



PROBLEMS 293 

larger velocity than would be calculated if the motion of the planets were neg- 
lected. 

20. (a) Locate all fixed points in the limiting case M 2 — » 0, and sketch Fig. 7-7 
for this case. Show that the results in Section 7-6 applied to this case are con- 
sistent with the complete solution given in Section 3-14. 

(b) Show from this example for which the complete solution is known, that 
the minimum 'energetically' possible launching velocity for escape calculated 
as in Problem 18(c) is not necessarily the true minimum escape velocity. 



CHAPTER 8 

INTRODUCTION TO THE MECHANICS 
OF CONTINUOUS MEDIA 

In this chapter we begin the study of the mechanics of continuous 
media, solids, fluids, strings, etc. In such problems, the number of parti- 
cles is so large that it is not practical to study the motion of individual 
particles, and we instead regard matter as continuously distributed in 
space and characterized by its density. We are interested primarily in 
gaining an understanding of the concepts and methods of treatment which 
are useful, rather than in developing in detail methods of solving practical 
problems. In the first four sections, we shall treat the vibrating string, 
using concepts which are a direct generalization of particle mechanics. In 
the remainder of the chapter, the mechanics of fluids will be developed in 
a way less directly related to particle mechanics. 

8-1 The equation of motion for the vibrating string. In this section we 
shall study the motion of a string of length I, stretched horizontally and 
fastened at each end, and set into vibration. In order to simplify the 
problem, we assume the string vibrates only in a vertical plane, and that 
the amplitude of vibration is small enough so that each point on the string 
moves only vertically, and so that the tension in the string does not change 
appreciably during the vibration. 

We shall designate a point on the string by giving its horizontal distance 
x from the left-hand end (Fig. 8-1). The distance the point x has moved 
from the horizontal straight line representing the equilibrium position of 
the string will be designated by u(x). Thus any position of the entire 
string is to be specified by specifying the function u(x) for < x < I. 
This is precisely analogous, in the case of a system of N particles, to speci- 
fying the coordinates x t , ?/,-, z», for i = 1, . . . , N. In the case of the 
string, x is not a coordinate, but plays the same role as the subscript i; it 
designates a point on the string. Our idealized continuous string has 




Fig. 8-1. The vibrating string. 
294 



8-1] THE EQUATION OF MOTION FOR THE VIBRATING STRING 295 

infinitely many points, corresponding to the infinitely many values of x 
between and I. For a given point x, it is u{x) that plays the role of a 
coordinate locating that point, in analogy with the coordinates x iy yi, z t of 
particle i. Just as a motion of the system of particles is to be described 
by functions Xi(t), Vi(t), Zi(t), locating each particle at every instant of 
time, so a motion of the string is to be described by a function u{x, t), 
locating each point x on the string at every instant of time. 

In order to obtain an equation of motion for the string, we consider a 
segment of string of length dx between x and x + dx. If the density of the 
string per unit length is <r, then the mass of this segment is <r dx. The 
velocity of the string at any point is du/dt, and its slope is du/dx. The 
vertical component of tension exerted from right to left across any point in 
the string is 

r„ = t sin 6, (8-1) 

where 6 is the angle between the string and the horizontal (Fig. 8-1). We 
are assuming that 6 is very small and, in this case, 

t sin 6 = r tan = t — • (8-2) 

The net upward force dF due to the tension, on the segment dx of string, 
is the difference in the vertical component r„ between the two ends of the 
segment : 

dF = [r M ] x+ dx — [tu]x 

(8-3) 

dx \ dx/ 

If we do not limit ourselves to very small slopes du/dx, then a segment of 
string may also have a net horizontal component of force due to tension, 
and the segment will move horizontally as well as vertically, a possibility 
we wish to exclude. If there is, in addition, a vertical force / per unit 
length, acting along the string, the equation of motion of the segment dx 
will be 



d 2 u d 



VlTx) 



<rdx W = dx'\ T ^l dx+fdx - (8_4) 

For a horizontal string acted on by no horizontal forces except at its ends, 
and for small amplitudes of vibration, the tension is constant, and 
Eq. (8-4) can be rewritten: 

d 2 u d 2 u , . , a _. 



296 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

The force / may be the gravitational force acting on the string, which is 
usually negligible unless the tension is very small. The force / may also 
represent an external force applied to the string to set it into vibration. 
We shall consider only the case/ = 0, and we rewrite Eq. (8-5) in the form 

d 2 u 1 d 2 u , . 

ex*-T>W = > (8_6) 



where 



=(r- 



(8-7) 



The constant c has the dimensions of a velocity, and we shall see in Sec- 
tion 8-3 that it is the velocity with which a wave travels along the string. 
Equation (8-6) is a partial differential equation for the function u(x, t) ; 
it is the mathematical expression of Newton's law of motion applied to 
the vibrating string. We shall want to find solutions u(x, t) to Eq. (8-6), 
for any given initial position u (x) of the string, and any given initial 
velocity v {x) of each point along the string. If we take the initial instant 
at t = 0, this means that we want a solution u(x, t) which satisfies the 
initial conditions: 

u(x, 0) = u (x), 

(8-8) 



f-1 = 



v (x). 



The solution must also satisfy the boundary conditions: 

«(0, t) = u(l, t) = 0, (8-9) 

which express the fact that the string is tied at its ends. From the nature 
of the physical problem, we expect that there should be just one solution 
u(x, t) of Eq. (8-6) which satisfies Eqs. (8-8) and (8-9), and this solution 
will represent the motion of the string with the given initial conditions. 
It is therefore reasonable to expect that the mathematical theory of partial 
differential equations will lead to the same conclusion regarding the num- 
ber of solutions of Eq. (8-6), and indeed it does. 

8-2 Normal modes of vibration for the vibrating string. We shall first 
try to find some solutions of Eq. (8-6) which satisfy the boundary condi- 
tions (8-9), without regard to the initial conditions (8-8). This is analo- 
gous to our treatment of the harmonic oscillator, in which we first looked 
for solutions of a certain type and later adjusted these solutions to fit the 
initial conditions of the problem. The method of finding solutions which 



8-2] NORMAL MODES OF VIBRATION FOR THE VIBRATING STRING 297 

we shall use is called the method of separation of variables. It is one of 
the few general methods so far devised for solving partial differential equa- 
tions, and many important equations can be solved by this method. Un- 
fortunately, it does not always work. In principle, any partial differential 
equation can be solved by numerical methods, but the labor involved in 
doing so is often prohibitive, even for the modern large-scale automatic 
computing machines. 

The method of separation of variables consists in looking for solutions 
of the form 

u(x, t) = X(z)0(<), (8-10) 

that is, u is to be a product of a function X of x and a function © of t. The 
derivatives of u will then be 

d 2 u „d 2 X d 2 u v d 2 @ 

w = @ !xT' W = X HP- < 8 - n) 

If these expressions are substituted in Eq. (8-6), and if we divide through 
by ©X, then Eq. (8-6) can be rewritten : 

ld 2 X_l_d 2 ®. ~ 

X dx* _ dP (8_12) 

The left member of this equation is a function only of x, and the right 
member is a function only of t. If we hold t fixed and vary x, the right 
member remains constant, and the left member must therefore be inde- 
pendent of x. Similarly, the right member must actually be independent 
of t. We may set both members equal to a constant. It is clear on physical 
grounds that this constant must be negative, for the right member of 
Eq. (8-12) is the acceleration of the string divided by the displacement, and 
the acceleration must be opposite to the displacement or the string will not 
return to its equilibrium position. We shall call the constant -co 2 : 

1 d 2 ® 2 c 2 d 2 X a 

*r = -" ' l¥=- w ' < 8 " 13 > 

The first of these equations can be rewritten as 

d 2 % 



dP 



+ o J e = 0, (8-14) 



which we recognize as the equation for the harmonic oscillator, whose gen- 
eral solution, in the form most suitable for our present purpose, is 

© = A cos wt + B sin wt, (8-15) 



298 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

where A and B are arbitrary constants. The second of Eqs. (8-13) has a 
similar form : 

d 2 X -- 2 

dx 2 
and has a similar solution : 



+ ^- X = 0, (8-16) 



X-Ccos— + Z>sin — ■ (8-17) 

c c 

The boundary condition (8-9) can hold for all times t only if X satisfies 
the conditions 

X(0) = C = 0, 

X(l) = C cos - + D sin - = 0. (8-18) 

c c 

The first of these equations determines C, and the second then requires 
that 

sin - = 0. (8-19) 

c 

This will hold only if w has one of the values 

Wn = 2p , n= 1, 2, 3, ... . (8-20) 

Had we taken the separation constant in Eqs. (8-13) as positive, we would 
have obtained exponential solutions in place of Eq. (8-17), and it would 
have been impossible to satisfy the boundary conditions (8-18). 

The frequencies v n = o) n /2ir given by Eq. (8-20) are called the normal 
frequencies of vibration of the string. For a given n, we obtain a solution by 
substituting Eqs. (8-15) and (8-17) in Eq. (8-10), and making use of Eqs. 
(8-18), (8-20): 

... . . nirx nirct , „ . nwx . rnrct , a 01 , 

u{x, t) = A sin — j- cos — j \- B sin — r— sin — j- > (8-21) 

where we have set D = 1. This is called a normal mode of vibration of 
the string, and is entirely analogous to the normal modes of vibration 
which we found in Section 4-10 for coupled harmonic oscillators. Each 
point on the string vibrates at the same frequency u n with an amplitude 
which varies sinusoidally along the string. Instead of two coupled oscil- 
lators, we have an infinite number of oscillating points, and instead of two 
normal modes of vibration, we have an infinite number. 

The initial position and velocity at t = of the nth normal mode of 
vibration as given by Eq. (8-21) are 



8-2] NORMAL MODES OF VIBRATION FOR THE VIBRATING STRING 299 

Uo\%) = A SID. -y- 

, s nircB . nirx 
v (x) = — z — sin —j- ■ 



T' 

(8-22) 



Only for these very special types of initial conditions will the string vibrate 
in one of its normal modes. However, we can build up more general solu- 
tions by adding solutions; for the vibrating string, like the harmonic oscil- 
lator, satisfies a principle of superposition. Let Ui(x, t) and u 2 (x, t) be any 
two solutions of Eq. (8-6) which satisfy the boundary conditions (8-9). 
Then the function 

u(x, t) = «i (#, + U2(x, t) 

also satisfies the equation of motion and the boundary conditions. This 
is readily verified simply by substituting u{x, t) in Eqs. (8-6) and (8-9), 
and making use of the fact that u x {x, t) and u 2 (x, t) satisfy these equations. 
A more general solution of Eqs. (8-6) and (8-9) is therefore to be obtained 
by adding solutions of the type (8-21), using different constants A and B 
for each normal frequency : 

u(x, t) = 2~i\ An sin ~T~ cos — z 1" B n sm ~T Sln ~T~ ) ' (8-23) 

The initial position and velocity for this solution are 
«oW = Zj An Sln ~T~ ' 

n=l * 

,, , v-\ rnrcB n . nirx 
Vo(x) = 2^ — i — Sln "J" - 



(8-24) 



n=l 



Whether or not Eq. (8-23) gives a general solution to our problem depends 
on whether, with suitable choices of the infinite set of constants A n , B n , 
we can make the functions u (x) and v (x) correspond to any possible 
initial position and velocity for the string. Our intuition is not very clear 
on this point, although it is clear that we now have a great variety of possi- 
ble functions WoO^) and v (x). The answer is provided by the Fourier series 
theorem, which states that any continuous function u (x) for (0 < x < I), 
which satisfies the boundary conditions (8-9), can be represented by the 
sum on the right in Eq. (8-24), if the constants A n are properly chosen.* 

* R. V. Churchill, Fourier Series and Boundary Value Problems. New York: 
McGraw-Hill, 1941. (Pages 57-70.) Even functions with a finite number of dis- 
continuities can be represented by Fourier series, but this point is not of great 
interest in the present application. 



300 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

Similarly, with the proper choice of the constants B n , any continuous func- 
tion v (x) for (0 < x < I) can be represented.* The expressions for A n 
and B n are, in this case, 



r i 

2 - 
IJo 

r i 
nirc Jo 



A„ = j J u {x) sin —j- dx, 

1 (8-25) 



"» = ZZZ I "oW Sln —y~ ® x - 

The most general motion of the vibrating string is therefore a superposi- 
tion of normal modes of vibration at the fundamental frequency V\ = c/2l 
and its harmonics v n = nc/2l. 

8-3 Wave propagation along a string. Equations (8-14) and (8-16) 
have also the complex solutions 

= Ae ±iat , (8-26) 

x = gi'O/o)*. (8-27) 
Hence Eq. (8-6) has complex solutions of the form 

u(x, t) = Ae ±iialcnx±ct \ (8-28) 

By taking the real part, or by adding complex conjugates and dividing 
by 2, we obtain the real solutions 

u(x, t) = A cos - (x - ct), (8-29) 

c 

u(x, t) = A cos - (x + ct). (8-30) 

By taking imaginary parts, or by subtracting complex conjugates and 
dividing by 2i, we could obtain similar solutions with cosines replaced by 
sines. These solutions do not satisfy the boundary conditions (8-9), but 
they are of considerable interest in that they represent waves traveling 
down the string, as we now show. 

A fixed point x on the string will oscillate harmonically in time, accord- 
ing to the solution (8-29) or (8-30), with amplitude A and angular fre- 
quency a. At any given instant t, the string will be in the form of a sinus- 



* The Fourier series theorem was quoted in Section 2-11 in a slightly different 
form. The connection between Eqs. (8-24) and (2-205) is to be made by replacing 
t by x and T by 21 in Eq. (2-205). Both sine and cosine terms are then needed to 
represent an arbitrary function uo(x) in the interval (0 < x < 21), but only sine 
terms are needed if we want to represent uo(x) only in the interval (0 < x < I). 
[Cosine terms alone would also do for this interval, but sine terms are appropriate 
if uo(x) vanishes at x = and x = I.] 



8-3] WAVE PROPAGATION ALONG A STRING 301 

oidal curve with amplitude A and wavelength X (distance between successive 
maxima) : 

X = ^- (8-31) 

We now show that this pattern moves along the string with velocity c, to 
the right in solution (8-29), and to the left in solution (8-30). Let 

£ = x — ct, (8-32) 

so that Eq. (8-29) becomes 

u = A cos — » (8-33) 

where { is called the phase of the wave represented by the function u. For 
a fixed value of £ , u has a fixed value. Let us consider a short time interval 
dt and find the increment dx required to maintain a constant value of |: 

d£ = dx — c dt = 0. (8-34) 

Now if dx and dt have the ratio given by Eq. (8-34), 

dx 



dt 



= c, (8-35) 



then the value of u at the point x + dx at time t + dt will be the same as 
its value at the point x at time t. Consequently, the pattern moves along 
the string with velocity c given by Eq. (8-7). The constant c is the phase 
velocity of the wave. Similarly, the velocity dx/dt for solution (8-30) is — c. 
It is often convenient to introduce the angular wave number k defined by 
the equation 

where k is taken as positive for a wave traveling to the right, and negative 
for a wave traveling to the left. Then both solutions (8-29) and (8-30) 
can be written in the symmetrical form 

u = A cos (kx — «<)• (8-37) 

The angular wave number k is measured in radians per centimeter, just as 
the angular frequency « is measured in radians per second. The expres- 
sion for u in Eq. (8-37) is the real part of the complex function 

u = Ae i(kx - at \ (8-38) 

This form is often used in the study of wave motion. 



302 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

The possibility of superposing solutions of the form (8-29) and (8-30) 
with various amplitudes and frequencies, together with the Fourier series 
theorem, suggests a more general solution of the form 

u(x, t) = f(x - ct) + g(x + ct), (8-39) 

where /(£) and g(y) are arbitrary functions of the variables £ = x — ct, 
and q = x + ct. Equation (8-39) represents a wave of arbitrary shape 
traveling to the right with velocity c, and another traveling to the left. 
We can readily verify that Eq. (8-39) gives a solution of Eq. (8-6) by 
calculating the derivatives of u: 

dx _ dt, dx dr, dx di dr, ' 

d 2 u __ d 2 f d 2 g 
dx 2 ~~ d? + dr, 2 ' 

du = dfdl.dgih = _ d[ dg } 
dt d£ dt dv dt d£ dri ' 

d 2 u _ ,dV 2 <Pg 
dt 2 C dp dr, 2 ' 

When these expressions are substituted in Eq. (8-6), it is satisfied iden- 
tically, no matter what the functions /(f) and g(rj) may be, provided, of 
course, that they have second derivatives. Equation (8-39) is, in fact, the 
most general solution of the equation (8-6) ; this follows from the theory 
of partial differential equations, according to which the general solution 
of a second-order partial differential equation contains two arbitrary func- 
tions. We can prove this without resorting to the theory of partial differ- 
ential equations by assuming the string to be of infinite length, so that 
there are no boundary conditions to concern us, and by supposing that the 
initial position and velocity of all points on the string are given by the 
functions u (x), v {x). If the solution (8-39) is to meet these initial con- 
ditions, we must have, at t — : 

«(*, 0) = Six) + g(x) = u (x), (8-40) 

At t — 0, £ = i\ = x, so that Eq. (8-41) can be rewritten : 

£[-f(x)+g(.x)) = "-*&, (8-42) 



8-3] WAVE PROPAGATION ALONG A STRING 303 

which can be integrated to give 

rx 

-/(*) + g(x) = - / v (x) dx + C. (8-43) 

C J 

By adding and subtracting Eqs. (8-40) and (8-43), we obtain the functions 
/andg: 



f(x) = i\u (x) / v (x) dx — C), 

9(x) = i( u (x) + - J v (x) dx + Cj- 



(8-44) 



The constant C can be omitted, since it will cancel out in u = f + g, and 
we can replace x by £ and ij respectively in these equations : 



/(*) = i(«ott) -\j o »o«)dt) 



(8-45) 



This gives a solution to Eq. (8-6) for any initial position and velocity of 
the string. 

Associated with a wave 

u = f{x — ct), (8-46) 

there is a flow of energy down the string, as we show by computing the 
power delivered from left to right across any point x on the string. The 
power P is the product of the upward velocity of the point x and the up- 
ward force [Eq. (8-2)] exerted by the left half of the string on the right 
half across the point x: 

„ du du ,„ ,_. 

P =- T 6x-di- <W 

If u is given by Eq. (8-46), this is 

P = cr(|)\ (8-48) 

which is always positive, indicating that the power flow is always from 
left to right for the wave (8-46). For a wave traveling to the left, P will 
be negative, indicating a flow of power from right to left. For a sinusoidal 
wave given by Eq. (8-37), the power is 

P = IccotA 2 sin 2 (fee — at), (8-49) 

or, averaged over a cycle, 

(P)„ = ±kwrA 2 . (8-50) 



304 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

We now consider a string tied at x = and extending to the left from 
a; = to a; = — oo. The solution (8-39) must now satisfy the boundary 
condition 

«(0,0 = f(-ct) + y(ct) = 0, (8-51) 

or 

/(-*) = -9(S), (8-52) 

for all values of £. The initial values uo(x) and v (x) will now be given 
only for negative values of x, and Eqs. (8-45) will define /(£) and g(ri) only 
for negative values of | and »j. The values of /(£) and g(rj) for positive 
values of £ and jj can then be found from Eq. (8-52) : 

/(I) = -*(-*), gin) = -/(-i). (8-53) 

Let us consider a wave represented by f(x — ct) traveling toward the end 
x = 0. A particular phase £o> for which the wave amplitude is/(£o)> will 
at time t be at the point 

x = So + ct . (8-54) 

Let us suppose that £ and t are so chosen that x is negative. At a later 
time <i, the phase £ will be at the point 

xi = So + c<i = cc + c(«i — t ). (8-55) 

At <i = t Q — (x /c), Xi = and the phase f reaches the end of the 
string. At later times x x will be positive, and f{x x — ct{) will have no 
physical meaning, since the string does not extend to positive values of x. 
Now consider the phase i) » of the leftward traveling wave g{x + ct), 
denned by 

jjo = x + ct = — £ o- (8-56) 

The amplitude of the leftward wave g(iio) f° r the phase ij is related to the 
amplitude of the rightward wave/(£ ) for the corresponding phase £ by 
Eq. (8-53): 

fOlo) = -/(«o). (8-57) 

At time ti, the phase ij will be at the point 

X2 = Vo — c&i = —x — c(h — t ). (8-58) 

If (<x — t ) > —x /c, x 2 is negative and 0(170) represents a wave of equal 
and opposite amplitude to /(£o), traveling to the left. Thus the wave 
fix — ct) is reflected out of phase at x = and becomes an equal and 
opposite wave traveling to the left. (See Fig. 8-2.) The total distance 



8-4] STRING AS LIMITING CASE OF SYSTEM OF PARTICLES 305 





t = h 

Fig. 8-2. A wave reflected at x = 0. 

traveled by the wave during the time (ii — t ), from x = x to x = 
and back to x = x 2 is, by Eq. (8-58), 

— x — x 2 = c(«i — t ), (8-59) 

as it should be. 

The solution (8-39) can also be fitted to a string of finite length fastened 
at x = and x = I. In this case, the initial position and velocity u (x) 
and v (x) are given only for (0 < x < I). The functions /(£) and g(t)) are 
then defined by Eq. (8-45) only for (0 < £ < I, < i? < I). If we define 
/(|) and g(v) for negative values of £ and ij by Eq. (8-53), in terms of their 
values for positive { and t\, then the boundary condition (8-51) will be satis- 
fied at x = 0. By an argument similar to that which led to Eq. (8-53), 
we can show ;hat the boundary condition (8-9) for x — I will be satisfied if, 
for all value* of £ and t\, 

/(£ + = -gQ-i), 

(8-60) 
9(V + I) = -fd -v). 

By means of Eqs. (8-53) and (8-60), we can find/(£) and g{i\) for all values 
of £ and v, once their values are given [by Eqs. (8-45)] for < £ < I, 
< i\ < I. Thus we find a solution for the vibrating string of length I hi 
terms of waves traveling in opposite directions and continuously being 
reflected at x = and x — I. The solution is equivalent to the solution 
given by Eqs. (8-23) and (8-25) in terms of standing sinusoidal waves. 

8-4 The string as a limiting case of a system of particles. In the first 
three sections of this chapter, we have considered an idealized string char- 
acterized by a continuously distributed mass with density a and tension t. 
An actual string is made up of particles (atoms and molecules) ; our treat- 
ment of it as continuous is valid because of the enormously large number of 
particles in the string. A treatment of an actual string which takes into 
account the individual atoms would be hopelessly difficult, but we shall 
consider in this section an idealized model of a string made up of a finite 
number of particles, each of mass m. Figure 8-3 shows this idealized 
string, in which an attractive force r acts between adjacent particles 



306 



THE MECHANICS OP CONTINUOUS MEDIA 



[CHAP. 8 




Fig. 8-3. A string made up of particles. 

along the line joining them. The interparticle forces are such that in 
equilibrium the string is horizontal, with the particles equally spaced a 
distance h apart. The string is of length (N + l)h, with N + 2 particles, 
the two end particles being fastened at the x-axis. The N particles which 
are free to move are numbered 1,2, ... ,N, and the upward displacement 
of particle j from the horizontal axis will be called Uj. It will be assumed 
that the particles move only vertically and that only small vibrations are 
considered, so that the slope of the string is always small. Then the 
equations of motion of this system of particles are 



m 



d uj 
dt 2 



. u )+i 



h 



Uj Uj My — 1 



j= 1,...,N, (8-61) 



where the expression on the right represents the vertical components of the 
forces r between particle j and the two adjacent particles, and we are sup- 
posing that the forces r are equal between all pairs of particles. Now let 
us assume that the number N of particles is very large, and that the dis- 
placement of the string is such that at any time t, a smooth curve u(x, t) 
can be drawn through the particles, so that 



u(jh, t) = Uj{t). 



(8-62) 



We can then represent the system of particles approximately as a contin- 
uous string of tension t, and of linear density 



<T = -r 



m 
h ' 



The equations of motion (8-61) can be written in the form 

d 2 Uj _ T 1 ( Uj + i — Uj _ Uj — Uj-j \ 

dt 2 ~ <r h\ h h ) ' 



(8-63) 



(8-64) 



Now if the particles are sufficiently close together, we shall have, approxi- 
mately, 

Uj+i — uj _^_ \du\ 

h Ldx] x =(j+l/2)h 



Uj — Uj—i j_ dw 

h L.dxXc=(j-ll2)h 



(8-65) 



8-4] STRING AS LIMITING CASE OF SYSTEM OF PARTICLES 307 

and hence / \ r, 2 i 

1 ( ttj+i — «; u i — u i-i ) ^. \ d u \ . C8-66') 

h\ h h ) ldx 2 ] x=jh v ; 

The function u(x, I) therefore, when h is very small, satisfies the equation 

d u T a u /0 „-n 

dt 2 a dx 2 

which is the same as Eq. (8-6) for the continuous string. 

The solutions of Eqs. (8-61) when JV is large will be expected to approxi- 
mate the solutions of Eq. (8-6). If we were unable to solve Eq. (8-6) 
otherwise, one method of solving it numerically would be to carry out the 
above process in reverse, so as to reduce the partial differential equation 
(8-67) to the set of ordinary differential equations (8-61), which could then 
be solved by numerical methods. The solutions of Eqs. (8-61) are of 
some interest in their own right. Let us rewrite these equations in the 
form 

m 1& + T U * ~ I {Uj+1 + Uj ~ l] = °' j = *' • • • ' N - (8_68) 

These are the equations for a set of harmonic oscillators, each coupled to 
the two adjacent oscillators. We are led, either by our method of treat- 
ment of the coupled oscillator problem or by considering our results for the 
continuous string, to try a solution of the form 

Uj = aje ±iat . (8-69) 

If we substitute this trial solution in Eqs. (8-68), the factor e ±iat cancels 
out, and we get a set of algebraic equations: 

(2t ,\ t t 

-j — mw J a, — ^ a i+ i — ^ a y _i = 0, j = 1, . . . , N. (8-70) 

This is a set of linear difference equations which could be solved for o J+ i 
in terms of a, and o,_i. Since a = 0, if a± is given we can find the values 
of the remaining constants a,- by successive applications of these equations. 
A neater method of solution is to notice the analogy between the linear 
difference equations (8-70) and the linear differential equation (8-16), 
and to try the solution 

aj = Ae ipj , j=l,...,N. (8-71) 

When this is substituted in Eqs. (8-70), we get, after canceling the factor 
Ae ip >: 

(?1 - mJ) - \ ie iv + e~ ip ) = 0, (8-72) 



308 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

OT J, 2 

cos p = 1 — • (8-73) 



If a is less than 



\mh/ 



(8-74) 



there will be real solutions for p. Let a solution be given by 

p = kh, < kh < ir. (8-75) 

Then another solution is 

V = —kh. (8-76) 

All other solutions for p differ from these by multiples of 27r, and in view 
of the form of Eq. (8-71), they lead to the same values of ay, so we can re- 
strict our attention to values of p given by Eqs. (8-75) and (8-76). 

If we substitute Eq. (8-71) in Eq. (8-69), making use of Eq. (8-75), we 
have a solution of Eqs. (8-68) in the form 

Uj = Ae± iikik -"\ (8-77) 

Since the horizontal distance of particle j from the left end of the string is 

Xj = jh, (8-78) 

we see that the solution (8-77) corresponds to our previous solution (8-38) 
for the continuous string, and represents traveling sinusoidal waves. By 
combining the two complex conjugate solutions (8-77) and using Eq. 
(8-78), we obtain the real solution 

My = A cos (kx, — (tit), (8-79) 

which corresponds to Eq. (8-37). We thus have sinusoidal waves which 
may travel in either direction with the velocity [Eq. (8-36)] 

u ha ._ „„. 

where p is given by Eq. (8-73). If u <3C u c [Eq. (8-74)], then p will be 
nearly zero, and we can expand cos p in Eq. (8-73) in a power series: 

t p 2 . mhu> 2 



-(?)'"• 



|p|=col — ) , (8-81) 



8-4] STRING AS LIMITING CASE OF SYSTEM OF PARTICLES 309 

and 



*(£)"■• 



(8-82) 



which agrees with Eq. (8-7) for the continuous string, in view of Eq. 
(8-63). However, for larger values of u, the velocity c is smaller than for 
the continuous string, and approaches 

c^jfcf (8-83) 

7T IT \m/ 

as w — * w c . (w c = oo for the continuous string for which mh = ah 2 = 0.) 
Since the phase velocity given by Eq. (8-80) depends upon the frequency, 
we cannot superpose sinusoidal solutions to obtain a general solution of the 
form (8-39). If a wave of other than sinusoidal shape travels along the 
string, the sinuosidal components into which it may be resolved travel with 
different velocities, and consequently the shape of the wave changes as it 
moves along. This phenomenon is called dispersion. 
When (>) > w e , Eq. (8-73) has only complex solutions for p, of the form 

p = ir ± i7. (8-84) 

These lead to solutions Uj of the form 

Uj = (- lYAe ±yj cos ut. (8-85) 

There is then no wave propagation, but only an exponential decline in 
amplitude of oscillation to the right or to the left from any point which 
may be set in oscillation. The minimum wavelength [Eq. (8-36)] which 
is allowed by Eq. (8-75) is 

X c = f£ = 2h. (8-86) 

It is evident that a wave of shorter wavelength than this would have no 
meaning, since there would not be enough particles in a distance less than 
X c to define the wavelength. The wavelength X c corresponds to the fre- 
quency w c , for which 

Uj = Ae iJT e ±ia ° l = (-lj^e**"'. (8-87) 

Adjacent particles simply oscillate out of phase with amplitude A. 
We can build up solutions which satisfy the boundary conditions 

« = u N+ i = (8-88) 

by adding and subtracting solutions of the form (8-77). We can, by suit- 
ably combining solutions of the form (8-77), obtain the solutions 



310 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

Uj = A sin pj cos ut + B sin pj sin cat + C cos pj cos cot 

+ D cos pj sin wf . (8-89) 
In order to satisfy the conditions (8-88), we must set 
C = D = 0, 

TITT 

P = j\m ' n=l,2,...,N, (8-90) 

where the limitation n < N arises from the limitation on p in Eq. (8-75). 
The normal frequencies of vibration are now given by Eq. (8-73) : 



Cn = [^V~ C0S WTv\ ' 



n = 1, 2, . . . , N. (8-91) 



If n <<C N, we can expand the cosine in a power series, to obtain 
Wn ~ lmh(N - 



h{N + 1) 2 J 

which agrees with Eq. (8-20) for the continuous string. 

A physical model which approximates fairly closely the string of particles 
treated in this section can be constructed by hanging weights m at intervals 
h along a stretched string. The mass m of each weight must be large in 
comparison with that of a length h of the string. 

8-5 General remarks on the propagation of waves. If we designate 
by F the upward component of force due to tension, exerted from left to 
right across any point in a stretched string, and by v the upward velocity 
of any point on the string, then, by Eq. (8-2), we have 

By Eq. (8-4), if there is no other force on the string, we have 

Tt = -*te' (8 ' 95) 

and by differentiating Eq. (8-93) with respect to t, assuming t is constant 



8-5] GENERAL REMARKS ON THE PROPAGATION OF WAVES 311 

in time, we obtain „ 

-^-r-. (8-96) 

Equations (8-95) and (8-96) are easily understood physically. The ac- 
celeration of the string will be proportional to the difference in the upward 
force F at the ends of a small segment of string. Likewise, since F is pro- 
portional to the slope, the time rate of change of F will be proportional to 
the difference in upward velocities of the ends of a small segment of the 
string. The power delivered from left to right across any point in the 
string is 

P = Fv. (8-97) 

Equations (8-95) and (8-96) are typical of many types of small ampli- 
tude wave propagation which occur in physics. There are two quantities, 
in this case F and v, such that the time rate of change of either is propor- 
tional to the space derivative of the other. For large amplitudes, the equa- 
tions for wave propagation may become nonlinear, and new effects like the 
development of shock fronts may occur which are not described by the 
equations we have studied here. When there is dispersion, linear terms in 
v and F, or terms involving higher derivatives than the first, may appear. 
Whenever equations of the form (8-95) and (8-96) hold, a wave equation 
of the form (8-6) can be derived for either of the two quantities. For ex- 
ample, if we differentiate Eq. (8-95) with respect to t, and Eq. (8-96) 
with respect to x, assuming a, t to be constant, we obtain 



or 

dx 2 c 2 dt 2 



d 2 F 

dxdt 


d 2 v 
T dx 2 


d 2 v 


,2 

d V 


1 d 2 v 





where 



<r- 



0, (8-98) 

(8-99) 



Similarly, we can show that 

Usually one of these two quantities can be chosen so as to be analogous to 
a force (F), and the other to the corresponding velocity (v), and then the 
power transmitted will be given by an equation like Eq. (8-97). Likewise, 
all other quantities associated with the wave motion satisfy a wave equa- 
tion, as, for example, u, which satisfies Eq. (8-6). 



312 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

As a further example, the equations for a plane sound wave traveling in 
the z-direction, which will be derived in Section 8-10, can be written in 
the form 

dv _ I dp' dp' _ dv 

di--p~-dx-' ~dt-- B dlc' (8_101) 

where p' is the excess pressure (above atmospheric), v is the velocity, in 
the ^-direction, of the air at any point, and where p is the density and B 
the bulk modulus. The physical meaning of these equations is clear almost 
without further discussion of the motion of gases. Both p' and v satisfy 
wave equations, easily derived from Eqs. (8-101) : 



aV l ay d\ l a 2 y 
dx 2 c 2 at 2 ' ex 2 c 2 at 2 


(8-102) 


where 




(b\ 112 




-(!) ' 


(8-103) 


and the power transmitted in the codirection per unit area is 




P = p'v. 


(8-104) 



In the case of a plane electromagnetic wave traveling in the x-direction 
and linearly polarized in the ^-direction, the analogous equations can be 
shown to be (gaussian units) 

dB z 6E V dE y dB z fo_ii\K\ 

-T7- = —C — ) -77" = — c — — > (8-105) 

dt dx dt dx 

where E u and B z are the y- and z-components of electric and magnetic field 
intensities, and c is the speed of light. The components E v and B z satisfy 
wave equations with wave velocity c, and the power transmitted in the 
a;-direction per unit area is 

As a final example, on a two-wire electrical transmission line, the volt- 
age E across the line and the current i through the line satisfy the equa- 
tions 

dE _ 1 di di _ 1 bE ,„„„ 

'di-~cYx , di--Ldx-' (8_107) 

where C is the shunt capacitance per unit length, and L is the series in- 
ductance per unit length. Again we can derive wave equations for i and E 



8-6] KINEMATICS OF MOVING FLUIDS 313 

with the wave velocity 



- (*r- 



(8-108) 



and again the power transmitted in the z-direction is 

P = Ei. (8-109) 

Thus the study of wave propagation in a string is applicable to a wide 
variety of physical problems, many of them of greater practical and theo- 
retical importance than the string itself. In many cases, our discussion 
of the string as made up of a number of discrete particles is also of interest. 
The electrical transmission line, for example, can be considered a limiting 
case of a series of low-pass filters. An electrical network made up of series 
inductances and shunt capacitances can be described by a set of equations 
of the same form as our Eqs. (8-61), with analogous results. In the case 
of sound waves, we are led by analogy to expect that at very high fre- 
quencies, when the wavelength becomes comparable to the distance be- 
tween molecules, the wave velocity will begin to depend on the frequency, 
and that therefwill be a limiting frequency above which no wave propaga- 
tion is possible. 

8-6 Kinematics of moving fluids. In this section we shall develop the 
kinematic concepts useful in studying the motion of continuously distrib- 
uted matter, with particular reference to moving fluids. One way of 
describing the motion of a fluid would be to attempt to follow the motion 
of each individual point in the fluid, by assigning coordinates x, y, z to each 
fluid particle and specifying these as functions of the time. We may, for ex- 
ample, specify a given fluid particle by its coordinates, x , y , z , at an ini- 
tial instant t = t . We can then describe the motion of the fluid by means 
of functions x(x , y , Zo, 0. y( x o, Vo, *o, t), z(x , y 0> z , t) which determine 
the coordinates x, y, z at time t of the fluid particle which was at x , y , z Q 
at time t - This would be an immediate generalization of the concepts of 
particle mechanics, and of the preceding treatment of the vibrating string. 
This program originally due to Euler leads to the so-called "Lagrangian 
equations" of fluid mechanics. A more convenient treatment for many 
purposes, due also to Euler, is to abandon the attempt to specify the his- 
tory of each fluid particle, and to specify instead the density and velocity 
of the fluid at each point in space at each instant of time. This is the 
method which we shall follow here. It leads to the "Eulerian equations" 
of fluid mechanics. We describe the motion of the fluid by specifying the 
density p(x, y, z, t) and the vector velocity v(x, y, z, t), at the point x, y, z 
at the time t. We thus focus our attention on what is happening at a par- 
ticular point in space at a particular time, rather than on what is happening 
to a particular fluid particle. 



314 THE MECHANICS OP CONTINUOUS MEDIA [CHAP. 8 

Any quantity which is used in describing the state of the fluid, for ex- 
ample the pressure p, will be a function [p{x, y, z, t)] of the space coordi- 
nates x, y, z and of the time t; that is, it will have a definite value at each 
point in space and at each instant of time. Although the mode of descrip- 
tion we have adopted focuses attention on a point in space rather than on 
a fluid particle, we shall not be able to avoid following the fluid particles 
themselves, at least for short time intervals dt. For it is to the particles, 
and not to the space points, that the laws of mechanics apply. We shall be 
interested, therefore, in two time rates of change for any quantity, say p. 
The rate at which the pressure is changing with time at a fixed point in 
space will be the partial derivative with respect to time (dp/dt) ; it is itself 
a function of x, y, z, and t. The rate at which the pressure is changing with 
respect to a point moving along with the fluid will be the total derivative 

dv^djp^dpdxdpdydpdz 

dt dt ~ r dx dt^~ dy dt ^ dz dt' {0 V) 

where dx/dt, dy/dt, dz/dt are the components of the fluid velocity v. The 
change in pressure, dp, occurring during a time dt, at the position of a 
moving fluid particle which moves from x, y, z to x -\- dx, y + dy, z -\- dz 
during this time, will be 

dp — p(x + dx,y + dy, z + dz, t + dt) — p(x, y, z, t) 

and if dt — * 0, this leads to Eq. (8-110). We can also write Eq. (8-110) in 
the forms: 

dt ~ dt +Vx dx + Vy dy + V *dz 



= £ + *.£ + *£ + '.£ (8-Hl) 



and 



1 = 1 + '-". <nw> 



where the second expression is a shorthand for the first, in accordance with 
the conventions for using the symbol V. The total derivative dp/dt is 
also a function of x, y, z, and t. A similar relation holds between partial 
and total derivatives of any quantity, and we may write, symbolically, 

l=4+ v - v ' < 8 - ii3 > 

where total and partial derivatives have the meaning defined above. 

Let us consider now a small volume dV of fluid, and we shall agree that 
dV always designates a volume element which moves with the fluid, so 



8-6] 



KINEMATICS OF MOVING FLUIDS 



315 




Fig. 8-4. A moving, expanding element of fluid. 

that it always contains the same fluid particles. In general, the volume 
SV will then change with time, and we wish to calculate this rate of change. 
Let us assume that SV is in the form of a rectangular box of dimensions 
Sx, Sy, Sz (Fig. 8-4) : 

SV = Sx Sy Sz. (8-114) 

The a;-component of fluid velocity v x may be different at the left and right 
faces of the box. If so, Sx will change with time at a rate equal to the 
difference between these two velocities : 



and, similarly, 



dt 



Sx 



dv x 
dx 



Sx, 



d „ dVy „ 

d . dv z 
-77 Sz = — Sz. 
dt dz 



(8-115) 



The time rate of change of SV is then 



— SV = Sy Sz 37 Sx + Sx Sz^-Sy+ Sx Sy -j: Sz 
dt dt dt dt 



= (t + ^ + ?^*^ 



dy 



dz 



) 



and finally, 



dt 



SV = V-v 57. 



(8-116) 



This derivation is not very rigorous, but it gives an insight into the 
meaning of the divergence V-v. The derivation can be made rigorous by 
keeping careful track of quantities that were neglected here, like the de- 



316 THE MECHANICS OF CONTIXUOCS MEDIA [CHAP. 8 

pendence of v x upon y and 2, and showing that we arrive at Eq. (8-1 1G) 
in the limit as SV — » 0. However, there is an easier way to give a more 
rigorous proof of Eq. (8-116). Let us consider a volume V of fluid which 
is composed of a number of elements 8V: 

V = E SV. (8-117) 

If we sum the left side of Eq. (8-116), we have 

Ea**- »£"-£• (s-118) 

The summation sign lie re really represents an integration, since we mean 
to pass to the limit SV — ► 0, but the algebraic steps in Eq. (8-118) would 
look rather unfamiliar if the integral sign were used. Now let us sum the 
right side of Eq. (8-116), this time passing to the limit and using the in- 
tegral sign, in order that we may apply Gauss' divergence theorem [Eq. 
(3-115)]: 

£ Vt«F= ff fv-vdV 



= I Jn*v dS, (8-119) 



where & is the surface bounding the volume V, and n is the outward 
normal unit vector. Since n-v is the outward component of velocity 
of the surface element dS, the volume added to V by the motion of dS in a 
time dt will be n-v dt dS (Fig. 8-5), and hence the last line in Eq. (8-119) 
is the proper expression for the rate of increase in volume : 



dV r r 

—■ = j Jn-vdS. (8-120) 



s 
Therefore Eq. (8-116) must be the correct expression for the rate of 




Fig. 8-5. Increase of volume due to motion of surface. 



8-6] KINEMATICS OP MOVING FLUIDS 317 

increase of a volume element, since it gives the correct expression for the 
rate of increase of any volume V when summed over V. Note that the 
proof is independent of the shape of SV. We have incidentally derived an 
expression for the time rate of change of a volume V of moving fluid: 



= fffV'VdV. (8-121) 



dV 

dt 

v 



If the fluid is incompressible, then the volume of every element of fluid 
must remain constant : 

j t &V = 0, (8-122) 

and consequently, by Eq. (8-116), 

Vv = 0. (8-123) 

No fluid is absolutely incompressible, but for many purposes liquids may 
be regarded as practically so and, as we shall see, even the compressibility 
of gases may often be neglected. 

Now the mass of an element of fluid is 

$m = p8V, (8-124) 

and this will remain constant even though the volume and density may 
not: 

|a» = |<pl7) = 0. (8-125) 

Let us carry out the differentiation, making use of Eq. (8-116) : 

or, when SV is divided out, 

^ + p vv = 0. (8-126) 

By utilizing Eq. (8-113), we can rewrite this in terms of the partial deriva- 
tives referred to a fixed point in space: 

-£ + v- Vp + pV-v = 0. 



318 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

The last two terms can be combined, using the properties of V as a symbol 
of differentiation: 

g+V(pv) = 0. (8-127) 

This is the equation of continuity for the motion of continuous matter. It 
states essentially that matter is nowhere created or destroyed; the mass 
8m in any volume SV moving with the fluid remains constant. 



We shall make frequent use in the remainder of this chapter of the properties 
of the symbol V, which were described briefly in Section 3-6. The operator V 
has the algebraic properties of a vector and, in addition, when a product is in- 
volved, it behaves like a differentiation symbol. The simplest way to perform 
this sort of manipulation, when V operates on a product, is first to write a sum 
of products in each of which only one factor is to be differentiated. The factor 
to be differentiated may be indicated by underlining it. Then each term may be 
manipulated according to the rules of vector algebra, except that the underlined 
factor must be kept behind the V symbol. When the underlined factor is the only 
one behind the V symbol, or when all other factors are separated out by paren- 
theses, the underline may be omitted, as there is no ambiguity as to what factor 
is to be differentiated by the components of V. As an example, the relation 
between Eqs. (8-126) and (8-127) is made clear by the following computation: 

V-(pv) = V-(pv) + V-(pv) 
= (Vp)«v+ pV-v 
= (Vp)-v+ pV-v 
= vVp+pV-v. (8-128) 

Any formulas arrived at in this way can always be verified by writing out both 
sides in terms of components, and the reader should do this a few times to con- 
vince himself. However, it is usually far less work to make use of the properties 
of the V symbol. 

We now wish to calculate the rate of flow of mass through a surface S 
fixed in space. Let dS be an element of surface, and let n be a unit vector 
normal to dS. If we construct a cylinder by moving dS through a distance 
v dt in the direction of — v, then in a time dt all the matter in this cylinder 
will pass through the surface dS (Fig. 8-6). The amount of mass in this 
cylinder is 

pn-v dt dS, 

where n-v dt is the altitude perpendicular to the face dS. The rate of 
flow of mass through a surface S is therefore 



dm 
dt 



ffpn-vdS = JJn- (pv) dS. (8-129) 



8-6] 



KINEMATICS OF MOVING FLUIDS 



319 




Fig. 8-6. Flow of fluid through a surface element. 



If n-v is positive, the mass flow across S is in the direction of n; if n-v 
is negative, the mass flow is in the reverse direction. We see that pv, the 
momentum density, is also the mass current, in the sense that its com- 
ponent in any direction gives the rate of mass flow per unit area in that 
direction. We can now give a further interpretation of Eq. (8-127) by 
integrating it over a fixed volume V bounded by a surface S with outward 
normal n: 



V V 



(8-130) 



Since the volume V here is a fixed volume, we can take the time differenti- 
ation outside the integral in the first term. If we apply Gauss' divergence 
theorem to the second integral, we can rewrite this equation: 



dt 



JffpdV= -Jfn-(pv)dS. 



(8-131) 



This equation states that the rate of increase of mass inside the fixed vol- 
ume V is equal to the negative of the rate of flow of mass outward across 
the surface. This result emphasizes the physical interpretation of each 
term in Eq. (8-127). In particular, the second term evidently represents 
the rate of flow of mass away from any point. Conversely, by starting 
with the self-evident equation (8-131) and working backwards, we have 
an independent derivation of Eq. (8-127). 

Equations analogous to Eqs. (8-126), (8-127), (8-129), and (8-131) 
apply to the density, velocity, and rate of flow of any physical quantity. 
An equation of the form (8-127) applies, for example, to the flow of electric 
charge, if p is the charge density and pv the electric current density. 



320 



THE MECHANICS OF CONTINUOUS MEDIA 



[CHAP. 8 




/ 'i 



(a) 




(b) 



Fig. 8-7. Meaning of nonzero curl v. (a) A vortex, (b) A transverse velocity 
gradient. 

The curl of the velocity V X v is a concept which is useful in describing 
fluid flow. To understand its meaning, we compute the integral of the 
normal component of curl v across a surface S bounded by a curve C. By 
Stokes' theorem (3-117), this is 



J fn'(V X v)dS = f v-dr, 



(8-132) 



where the line integral is taken around C in the positive sense relative to 
the normal n, as previously defined. If the curve C surrounds a vortex in 
the fluid, so that v is parallel to dr around C (Fig. 8-7), then the line in- 
tegral on the right is positive and measures, in a sense, the rate at which 
the fluid is whirling around the vortex. Thus V X v is a sort of measure 
of the rate of rotation of the fluid per unit area; hence the name curl v. 
Curl v has a nonzero value in the neighborhood of a vortex in the fluid. 
Curl v may also be nonzero, however, in regions where there is no vortex, 
that is, where the fluid does not actually circle a point, provided there is 
a transverse velocity gradient. Figure 8-7 illustrates the two cases. In 
each case, the line integral of v counterclockwise around the circle C will 
have a positive value. If the curl of v is zero everywhere in a moving fluid, 
the flow is said to be irrotational. Irrotational flow is important chiefly 
because it presents fairly simple mathematical problems. If at any point 
V X v = 0, then an element of fluid at that point will have no net 
angular velocity about that point, although its shape and size may be 
changing. 

We arrive at a more precise meaning of curl v by introducing a co- 
ordinate system rotating with angular velocity u. If v' designates the 
velocity of the fluid relative to the rotating system, then by Eq. (7-33), 

v = v' + to X r, 



where r is a vector from the axis of rotation (whose location does not mat- 
ter in this discussion) to a point in the fluid. Curl v is now 



8-7] EQUATIONS OF MOTION FOR AN IDEAL FLUID 321 

Vxv=vxv' + Tx(«xr) 
= V X v' 4- wV-r — wT r 
= V X v' + 3« — <o 

= V X v' + 2«, 

where the second line follows from Eq. (3-35) for the triple cross product, 
and the third line by direct calculation of the components in the second 
and third terms. If we set 

w = a V X v, (8-133) 

then 

V X v' = 0. (8-134) 

Thus if V X v j£ at a point P, then in a coordinate system rotating with 
angular velocity « = |V X y, the fluid flow is irrotational at the point P. 
We may therefore interpret |V X v as the angular velocity of the fluid 
near any point. If V X v is constant, then it is possible to introduce a 
rotating coordinate system in which the flow is irrotational everywhere. 

8-7 Equations of motion for an ideal fluid. For the remainder of this 
chapter, except in the last section, we shall consider the motion of an ideal 
fluid, that is, one in which there are no shearing stresses, even when the 
fluid is in motion. The stress within an ideal fluid consists in a pressure p 
alone. This is a much greater restriction in the case of moving fluids than 
in the case of fluids in equilibrium (Section 5-11). A fluid, by definition, 
supports no shearing stress when in equilibrium, but all fluids have some 
viscosity and therefore there are always some shearing stresses between 
layers of fluid in relative motion. An ideal fluid would have no viscosity, 
and our results for ideal fluids will therefore apply only when the viscosity 
is negligible. 

Let us suppose that, in addition to the pressure, the fluid is acted on 
by a body force of density f per unit volume, so that the body force acting 
on a volume element SV of fluid is f 8V. We need, then, to calculate the 
force density due to pressure. Let us consider a volume element SV = 
Sx 8y 8z in the form of a rectangular box (Fig. 8-8). The force due to 
pressure on the left face of the box is p 8y Sz, and acts in the ^-direction. 
The force due to pressure on the right face of the box is also p by 8z, and 
acts in the opposite direction. Hence the net a>component of force 8F X 
on the box depends upon the difference in pressure between the left and 
right faces of the box: 



8F X = ( - |£ Sx) Sy 8z. (8-135) 



322 



THE MECHANICS OF CONTINUOUS MEDIA 



[CHAP. 8 



Sx 



Fig. 8-8. Force on a volume element due to pressure. 



A similar expression may be derived for the components of force in the 
y- and z-directions. The total force on the fluid in the box due to pressure 
is then 



8F 



-(- 



.dp _ .dp 

dx dy 



SV 



= — Vp SV. 



(8-136) 



The force density per unit volume due to pressure is therefore — Vp. 
This result was also obtained in Section 5-11 [Eq. (5-172)]. 

We can now write the equation of motion for a volume element SV 
of fluid: 



p5y^= f 57 - VpSV. 



(8-137) 



This equation is usually written in the form 

p-+Vp = f. 



(8-138) 



By making use of the relation (8-113), we may rewrite this in terms of 
derivatives at a fixed point : 



+ V- W -\ Vp = - : 

P P 



at 



(8-139) 



where f/p is the body force per unit mass. This is Euler's equation of 
motion for a moving fluid. 

If the density p depends only on the pressure p, we shall call the fluid 
homogeneous. This definition does not imply that the density is uniform. 
An incompressible fluid is homogeneous if its density is uniform. A com- 



8-8] CONSERVATION LAWS FOR FLUID MOTION 



323 



pressible fluid of uniform chemical composition and uniform temperature 
throughout is homogeneous. When a fluid expands or contracts under the 
influence of pressure changes, work is done by or on the fluid, and part of 
this work may appear in the form of heat. If the changes in density occur 
sufficiently slowly so that there is adequate time for heat flow to maintain 
the temperature uniform throughout the fluid, the fluid may be considered 
homogeneous within the meaning of our definition. The relation between 
density and pressure is then determined by the equation of state of the 
fluid or by its isothermal bulk modulus (Section 5-11). In some cases, 
changes in density occur so rapidly that there is no time for any appreciable 
flow of heat. In such cases the fluid may also be considered homogeneous, 
and the adiabatic relation between density and pressure or the adiabatic 
bulk modulus should be used. In cases between these two extremes, the 
density will depend not only on pressure, but also on temperature, which, 
in turn, depends upon the rate of heat flow between parts of the fluid at 
different temperatures. 

In a homogeneous fluid, there are four unknown functions to be de- 
termined at each point in space and time, the three components of velocity 
v, and the pressure p. We have, correspondingly, four differential equations 
to solve, the three components of the vector equation of motion (8-139), 
and the equation of continuity (8-127). The only other quantities ap- 
pearing in these equations are the body force, which is assumed to be 
given, and the density p, which can be expressed as a function of the 
pressure. Of course, Eqs. (8-139) and (8-127) have a tremendous variety 
of solutions. In a specific problem we would need to know the conditions 
at the boundary of the region in which the fluid is moving and the values 
of the functions v and p at some initial instant. In the following sections, 
we shall confine our attention to homogeneous fluids. In the intermediate 
case mentioned at the end of the last paragraph, where the fluid is in- 
homogeneous and the density depends on both pressure and temperature, 
we have an additional unknown function, the temperature, and we will 
need an additional equation determined by the law of heat flow. We shall 
not consider this case, although it is a very important one in many 
problems. 

8-8 Conservation laws for fluid motion. Inasmuch as the laws of fluid 
motion are derived from Newton's laws of motion, we may expect that 
appropriate generalizations of the conservation laws of momentum, energy, 
and angular momentum also hold for fluid motion. We have already had 
an example of a conservation law for fluid motion, namely, the equation 
of continuity [Eq. (8-127) or (8-131)], which expresses the law of con- 
servation of mass. Mass is conserved also in particle mechanics, but we 
did not find it necessary to write an equation expressing this fact. 



324 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

A conservation law in fluid mechanics may be written in many equiva- 
lent forms. It will be instructive to study some of these in order to get a 
clearer idea of the physical meaning of the various mathematical expres- 
sions involved. Let p be the density of any physical quantity: mass, 
momentum, energy, or angular momentum. Then the simplest form of 
the conservation law for this quantity will be equation (8-125), which 
states that the amount of this quantity in an element SV of fluid remains 
constant. If the quantity in question is being produced at a rate Q per 
unit volume, then Eq. (8-125) should be generalized: 

1 (p SV) = Q SV. (8-140) 

This is often called a conservation law for the quantity p. It states that 
this quantity is appearing in the fluid at a rate Q per unit volume, or disap- 
pearing if Q is negative. In the sense in which we have used the term in 
Chapter 4, this should not be called a conservation law except when 
Q = 0. By a derivation exactly like that which led to Eq. (8-127), we 
can rewrite Eq. (8-140) as a partial differential equation: 

f+V-(pv) = Q. (8-141) 

at 

This is probably the most useful form of conservation law. The meaning 
of the terms in Eq. (8-141) is brought out by integrating each term over a 
fixed volume V and using Gauss' theorem,* as in the derivation of Eq. 
(8-131) : 

^ t fffpdV + ffn-vpdS = IJJQdV. (8-142) 



According to the discussion preceding Eq. (8-129), this equation states 
that the rate of increase of the quantity within V, plus the rate of flow 
outward across the boundary S, equals the rate of appearance due to 
sources within V. Another form of the conservation law which is some- 
times useful is obtained by summing equation (8-140) over a volume V 
moving with the fluid : 



* If p is a vector, as in the case of linear or angular momentum density, then a 
generalized form of Gauss' theorem [mentioned in Section 5-11 in connection 
with Eq. (5-178)] must be used. 



8 ~ 8 J CONSERVATION LAWS FOR FLUID MOTION 325 

If we pass to the limit SV -> 0, the summations become integrations: 

ifIfpdV = JffQdV. (8-144) 

v v 

The surface integral which appears in the left member of Eq. (8-142) 
does not appear in Eq. (8-144) ; since the volume V moves with the fluid, 
there is no flow across its boundary. Since Eqs. (8-140), (8-141), (8-142), 
and (8-144) are all equivalent, it is sufficient to derive a conservation law 
in any one of these forms. The others then follow. Usually it is easiest to 
derive an equation of the form (8-140), starting with the equation of 
motion in the form (8-138). We can also start with Eq. (8-139) and de- 
rive a conservation equation in the form (8-141), but a bit more manipula- 
tion is usually required. 

In order to derive a conservation law for linear momentum, we first 
note that the momentum in a volume element SV is pv SV. The mo- 
mentum density per unit volume is therefore pv, and this quantity will 
play the role played by p in the discussion of the preceding paragraph. In 
order to obtain an equation analogous to Eq. (8-140), we start with the 
equation of motion in the form (8-138), which refers to a point moving 
with the fluid, and multiply through by the volume 57 of a small fluid 
element : 

dv 
p SV It + Vp SV = f SV - (8-145) 

Since p SV = Sm is constant, we may include it in the time derivative: 

- (pv SV) = (f - Vp) SV. (8-146) 

The momentum of a fluid element, unlike its mass, is not, in general, con- 
stant. This equation states that the time rate of change of momentum 
of a moving fluid element is equal to the body force plus the force due to 
pressure acting upon it. The quantity f — Vp here plays the role of Q 
in the preceding general discussion. Equation (8-146) can be rewritten 
in any of the forms (8-141), (8-142), and (8-144). For example, we may 
write it in the form (8-144) : 



d 
dt 



] I ' JpvdV = ffffdV - fffvpdV. (8-147) 



We can now apply the generalized form of Gauss' theorem [Eq. (5-178)] 
to the second term on the right, to obtain 



326 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

| fffpv dV = ffft dV + ff-np dS, (8-148) 

where S is the surface bounding V. 

This equation states that the time rate of change of the total linear 
momentum in a volume V of moving fluid is equal to the total external 
force acting on it. This result is an immediate generalization of the linear 
momentum theorem (4-7) for a system of particles. The internal forces, 
in the case of a fluid, are represented by the pressure within the fluid. By 
the application of Gauss' theorem, we have eliminated the pressure within 
the volume V, leaving only the external pressure across the surface of V. 
It may be asked how we have managed to eliminate the internal forces 
without making explicit use of Newton's third law, since Eq. (8-138), 
from which we started, is an expression only of Newton's first two laws. 
The answer is that the concept of pressure itself contains Newton's third 
law implicitly, since the force due to pressure exerted from left to right 
across any surface element is equal and opposite to the force exerted from 
right to left across the same surface element. Furthermore, the points of 
application of these two forces are the same, namely, at the surface ele- 
ment. Both forces necessarily have the same line of action, and there is no 
distinction between the weak and strong forms of Newton's third law. 
The internal pressures will therefore also be expected to cancel out in the 
equation for the time rate of change of angular momentum. A similar 
remark applies to the forces due to any kind of stresses in a fluid or a solid; 
Newton's third law in strong form is implicitly contained in the concept of 
stress. 

Equations representing the conservation of angular momentum, analo- 
gous term by term with Eqs. (8-140) through (8-144), can be derived by 
taking the cross product of the vector r with either Eq. (8-138) or (8-139), 
and suitably manipulating the terms. The vector r is here the vector from 
the origin about which moments are to be computed to any point in the 
moving fluid or in space. This development is left as an exercise. The 
law of conservation of angular momentum is responsible for the vortices 
formed when a liquid flows out through a small hole in the bottom of a 
tank. The only body force here is gravity, which exerts no torque about 
the hole, and it can be shown that if the pressure is constant, or depends 
only on vertical depth, there is no net vertical component of torque across 
any closed surface due to pressure. Therefore the angular momentum of 
any part of the fluid remains constant. If a fluid element has any angular 
momentum at all initially, when it is some distance from the hole, its angu- 
lar velocity will have to increase in inverse proportion to the square of its 
distance from the hole in order for its angular momentum to remain con- 
stant as it approaches the hole. 



8-8] CONSERVATION LAWS FOR FLUID MOTION 327 

In order to derive a conservation equation for the energy, we take the 
dot product of v with Eq. (8-146), to obtain 

j t (ipv 2 SV) = v(f - Vp) SV. (8-149) 

This is the energy theorem in the form (8-140). In place of the density p, 
we have here the kinetic energy density \pv 2 . The rate of production of 
kinetic energy per unit volume is 

Q = v (f - Vp). (8-150) 

In analogy with our procedure in particle mechanics, we shall now try 
to define additional forms of energy so as to include as much as possible 
of the right member of Eq. (8-149) under the time derivative on the left. 
We can see how to rewrite the second term on the right by making use of 
Eqs. (8-113) and (8-116): 

l(p d V) = ^SV + p^^ 



so that 



dv 
= m 6V + v-Vp87 + pV-vSV, (8-151) 



-v Vp SV = - ~ (p 57) + & SV + pV-v SV. (8-152) 

Let us now assume that the body force f is a gravitational force: 

f = Pg = pvg, (8-153) 

where g is the gravitational potential [Eq. (6-16)], i.e., the negative poten- 
tial energy per unit mass due to gravitation. The first term on the right 
in Eq. (8-149) is then 

vf SV = (v vg)p SV = (^ - g )p SV 

= !<P9«F)-pf£-a7, (8-154) 

since p SV = Sm is constant. With the help of Eqs. (8-152) and (8-154), 
Eq. (8-149) can be rewritten : 



dt 



; [(*Pf 2 + V ~ P9) SV) = (^ - pg) SV + pv-v SV. (8-155) 
The pressure p here plays the role of a potential energy density whose 



328 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

negative gradient gives the force density due to pressure [Eq. (8-136)]. 
The time rate of change of kinetic energy plus gravitational potential 
energy plus potential energy due to pressure is equal to the expression on 
the right. 

Ordinarily, the gravitational field at a fixed point in space will not 
change with time (except perhaps in applications to motions of gas clouds 
in astronomical problems). If the pressure at a given point in space is 
constant also, then the first term on the right vanishes. What is the sig- 
nificance of the second term? For an incompressible fluid, V-v = 0, 
and the second term would vanish also. We therefore suspect that it 
represents energy associated with compression and expansion of the fluid 
element SV. Let us check this hypothesis by calculating the work done in 
changing the volume of the element SV. The work dW done by the fluid 
element SV, through the pressure which it exerts on the surrounding fluid 
when it expands by an amount d SV, is 

dW = pd SV. (8-156) 

The rate at which energy is supplied by the expansion of the fluid element 
is, by Eq. (8-116), 

dW d SV _ _ T7 . / 1C » N 

-dT = p -dT = pV ' v8V ' (8 ~ 157) 

which is just the last term in Eq. (8-155). So far, all our conservation 
equations are valid for any problem involving ideal fluids. If we restrict 
ourselves to homogeneous fluids, that is, fluids whose density depends only 
on the pressure, we can define a potential energy associated with the ex- 
pansion and contraction of the fluid element SV. We shall define the 
potential energy u 8m on the fluid element SV as the negative work done 
through its pressure on the surrounding fluid when the pressure changes 
from a standard pressure p to any pressure p. The potential energy per 
unit mass u will then be a function of p : 



-f 

JPo 



uSm= - / pd SV. (8-158) 

J Pa 

The volume SV = Sm/p is a function of pressure, and we may rewrite this 
in various forms: 



J Pa / 



<Pa P* 



dp 



I 
I 



^f-dp (8-159) 

po P 2 d V 

V 
PaP B 



8-9] STEADY FLOW 329 

where the last step makes use of the definition of the bulk modulus [Eq. 
(5-116)]. The time rate of change of u is, by Eqs. (8-158) or (8-159) 
and (8-116), 

d{u 8m) d 8V 

~dT~ = ~ p -dT= -p v ' v sv - ( 8 -!60) 

We can now include the last term on the right in Eq. (8-155) under the 
time derivative on the left : 

j t l(W + p-ps + pu) SV] = (& _ p g?) hV . (8-161) 

The interpretation of this equation is clear from the preceding discussion. 
It can be rewritten in any of the forms (8-141), (8-142), and (8-144). 

If p and § are constant at any fixed point in space, then the total kinetic 
plus potential energy of a fluid element remains constant as it moves 
along. It is convenient to divide by 5m = p 57 in order to eliminate refer- 
ence to the volume element : 

This is Bernoulli's theorem. The term dQ/dt is practically always zero; 
we have kept it merely to make clear the meaning of the term (l/p)(dp/dt), 
which plays a similar role and is not always zero. When both terms on the 
right are zero, as in the case of steady flow, we have, for a point moving 
along with the fluid, 

y2 , V 

2" + - — 9 + w=a constant. (8-163) 

Other things being equal, that is if u, g, and p are constant, the pressure of 
a moving fluid decreases as the velocity increases. For an incompressible 
fluid, p and u are necessarily constant. 

The conservation laws of linear and angular momentum apply not only 
to ideal fluids, but also, when suitably formulated, to viscous fluids and 
even to solids, in view of the remarks made above regarding Newton's 
third law and the concept of stress. The law of conservation of energy 
(8-162) will not apply, however, to viscous fluids, since the viscosity is 
due to an internal friction which results in a loss of kinetic and potential 
energies, unless conversion of mechanical to heat energy by viscous 
friction is included in the law. [Equation (8-155) applies in any case.] 

8-9 Steady flow. By steady flow of a fluid we mean a motion of the 
fluid in which all quantities associated with the fluid, velocity, density, 
pressure, force density, etc., are constant in time at any given point in 



330 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

space. For steady flow, all partial derivatives with respect to time can be 
set equal to zero. The total time derivative, which designates the time 
rate of change of a quantity relative to a point moving with the fluid, will 
not in general be zero, but, by Eq. (8-113), will be 

^ = v-V. (8-164) 

at 

The path traced out by any fluid element as it moves along is called a 
streamline. A streamline is a line which is parallel at each point (x, y, z) 
to the velocity v(x, y, z) at that point. The entire space within which the 
fluid is flowing can be filled with streamlines such that through each point 
there passes one and only one streamline. If we introduce along any 
streamline a coordinate s which represents the distance measured along 
the streamline from any fixed point, we can regard any quantity associated 
with the fluid as a function of s along the streamline. The component of 
the symbol V along the streamline at any point is d/ds, as we see if we 
choose a coordinate system whose z-axis is directed along the streamline 
at that point. Equation (8-164) can therefore be rewritten: 

it = 4s- (8 - l65) 

This equation is also evident from the fact that v = ds/dt. For example, 
Eq. (8-162), in the case of steady flow, can be written: 

i(£+- P -s +u )=°- (8 - i66) 

The quantity in parentheses is therefore constant along a streamline. 
The equation of continuity (8-127) in the case of steady flow becomes 

V-(pv) = 0. (8-167) 

If we integrate this equation over a fixed volume V, and apply Gauss' 
theorem, we have 

dS = 0, (8-168) 



J Jn-(pv) 



where S is the closed surface bounding V. This equation simply states 
that the total mass flowing out of any closed surface is zero. 

If we consider all the streamlines which pass through any (open) sur- 
face S, these streamlines form a tube, called a tube of flow (Fig. 8-9). The 
walls of a tube of flow are everywhere parallel to the streamlines, so that 
no fluid enters or leaves it. A surface S which is drawn everywhere per- 
pendicular to the streamlines and through which passes each streamline in 



8-9] STEADY FLOW 331 




Fio. 8-9. A tube of flow. 

a tube of flow, will be called a cross section of the tube. If we apply Eq. 
(8-168) to the closed surface bounded by the walls of a tube of flow and 
two cross sections St and S 2 , then since n is perpendicular to v over the 
walls of the tube, and n is parallel or antiparallel to v over the cross sec- 
tions, we have 

Jfpv dS - J jpv dS = 0, (8-169) 

or 

J J pv dS = I = a constant, (8-170) 

8 

where S is any cross section along a given tube of flow. The constant I 
ia called the fluid current through the tube. 

The energy conservation equation (8-161), when rewritten in the form 
(8-141), becomes, in the case of steady flow, 

V-[(W + p - pg + pu)v] = 0. (8-171) 

This equation lias the same form as Eq. (8-167), and we can conclude in 
the same way that the energy current is the same through any cross sec- 
tion S of a tube of flow: 

/ ' Jdpv 2 + V ~ P£ + pv)» dS = a constant. (8-172) 

5 

This result is closely related to Eq. (8-166). 



If the flow ia not only steady, but also irrotational, then 

V X v = (8-173) 

everywhere. This equation ia analogous in form to Eq. (3-189) for a conservative 
force, and we can proceed as in Section 3-12 to show that if Eq. (8-173) holds, it 



332 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

is possible to define a velocity potential function <j>(x, y, z) by the equation 

0(r) = fv-dr, (8-174) 

where r„ is any fixed point. The velocity at any point will then be 

v = V<f>. (8-175) 

Substituting this in Eq. (8-167), we have an equation to be solved for <t>: 

V-(pV<rt = 0. (8-176) 

In the cases usually studied, the fluid can be considered incompressible, and this 
becomes 

V 2 4> = 0. (8-177) 

This equation is identical in form with Laplace's equation (6-35) for the gravita- 
tional potential in empty space. Hence the techniques of potential theory may 
be used to solve problems involving irrotational flow of an incompressible fluid. 

8-10 Sound waves. Let us assume a fluid at rest with pressure p , 
density po, in equilibrium under the action of a body force f , constant in 
time. Equation (8-139) then becomes 

fv Po = ^- (8-178) 

Po Po 

We may note that this equation agrees with Eq. (5-172) deduced in Sec- 
tion 5-11 for a fluid in equilibrium. Let us now suppose that the fluid is 
subject to a small disturbance, so that the pressure and density at any 
point become 

V = Po + V', (8-179) 

P = Po + p', (8-180) 

where p'«p and p' <C p. We assume that the resulting velocity v and 
its space and time derivatives are everywhere very small. If we substi- 
tute Eqs. (8-179) and (8-180) in the equation of motion (8-139), and 
neglect higher powers than the first of p', p', v and their derivatives, mak- 
ing use of Eq. (8-178), we obtain 

%=-&*• (8 " 181) 

Making a similar substitution in Eq. (8-127), we obtain 

22-= -poV-v - Wpo. (8-182) 



8-10] SOUND WAVES 333 

Let us assume that the equilibrium density po is uniform, or nearly so, so 
that Vp is zero or very small, and the second term can be neglected. 

The pressure increment p' and density increment p' are related by the 
bulk modulus according to Eq. (5-183) : 

p' p' 

This equation may be used to eliminate either p' or p' from Eqs. (8-181) 
and (8-182). Let us eliminate p' from Eq. (8-182) : 

dp' 

-~= -BV-v. (8-184) 

Equations (8-181) and (8-184) are the fundamental differential equations 
for sound waves. The analogy with the form (8-101) for one-dimensional 
waves is apparent. Here again we have two quantities, p' and v, such that 
the time derivative of either is proportional to the space derivatives of the 
other. In fact, if v = iv x and if v x and p' are functions of x alone, then 
Eqs. (8-181) and (8-184) reduce to Eqs. (8-101). 

We may proceed, in analogy with the discussion in Section 8-5, to 
eliminate either v or p' from these equations. In order to eliminate v, 
we take the divergence of Eq. (8-181) and interchange the order of differ- 
entiation, again assuming p nearly uniform : 

JL(v.v) = -lvV. (8-185) 

We now differentiate Eq. (8-184) with respect to t, and substitute from 
Eq. (8-185): 

V V - i d -X = 0, (8-186) 



c 2 dt 2 
where 



Vpo/ 



(8-187) 



This is the three-dimensional wave equation, as we shall show presently. 
Formula (8-187) for the speed of sound waves was first derived by Isaac 
Newton, and applies either to liquids or gases. For gases, Newton as- 
sumed that the isothermal bulk modulus B = p should be used, but Eq. 
(8-187) does not then agree with the experimental values for the speed of 
sound. The sound vibrations are so rapid that they should be treated as 
adiabatic, and the adiabatic bulk modulus B = Jp should be used, where 
7 is the ratio of specific heat at constant pressure to that at constant 



334 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

volume.* Formula (8-187) then agrees with the experimental values of c. 
If we eliminate p' by a similar process, we obtain a wave equation for v: 

^ - ±§ = 0. (8-188) 

In deriving Eq. (8-188), it is necessary to use the fact that V X v = 0. It 
follows from Eq. (8-181) that V X v is in any case independent of time, so 
that the time-dependent part of v which is present in a sound wave is ir- 
rotational. [We could add to the sound wave a small steady flow with 
V X v ^ 0, without violating Eqs. (8-181) and (8-182).] 

In order to show that Eq. (8-186) leads to sound waves traveling with 
speed c, we note first that if p' is a function of x and t alone, Eq. (8-186) 
becomes 

This is of the same form as the one-dimensional wave equation (8-6), and 
therefore has solutions of the form 

p' = f(x — ct). (8-190) 

This is called a plane wave, for at any time t the phase x — ct and the pres- 
sure p' are constant along any plane (x = a constant) parallel to the yz- 
plane. A plane wave traveling in the direction of the unit vector n will 
be given by 

p' = /(n-r - ct), (8-191) 

where r is the position vector from the origin to any point in space. To 
see that this is a wave in the direction n, we rotate the coordinate system 
until the z-axis lies in this direction, in which case Eq. (8-191) reduces 
to Eq. (8-190). The planes / = a constant, at any time t, are now per- 
pendicular to n, and travel in the direction of n with velocity c. We can 
see from the argument just given that the solution (8-191) must satisfy 
Eq. (8-186), or we may verify this by direct computation, for any coor- 
dinate system: 

v r' = % v * = %*' < 8 " 192 > 

where 

$ = n-r - ct, (8-193) 



* Millikan, Roller, and Watson, Mechanics, Molecular Physics, Heat, and 
Sound. Boston: Ginn and Co., 1937. (Pages 157, 276.) 



8-10] SOUND WAVES 335 

and, similarly, 

W-g».V« = g..n = g, (8-194) 

dp dp W c dp (8 195) 

so that Eq. (8-186) is satisfied, no matter what the function /(£) may be. 
Equation (8-188) will also have plane wave solutions: 

v = h(n'.r - ct), (8-196) 

corresponding to waves traveling in the direction n' with velocity c, where 
h is a vector function of £' = n'-r — ct. To any given pressure wave of 
the form (8-191) will correspond a velocity wave of the form (8-196), re- 
lated to it by Eqs. (8-181) and (8-182). If we calculate dv/dt from Eq. 
(8-196), and Vp' from Eq. (8-191), and substitute in Eq. (8-181), we will 
have 

— — n it 

de ~ (Bpo) 1/2 d*' (8-197) 

Equation (8-197) must hold at all points r at all times t. The right mem- 
ber of this equation is a function of £ and is constant for a constant £. 
Consequently, the left member must be constant when £ is constant, and 
must be a function only of £, which implies that f ' = { (or at least that 
£' is a function of £), and hence n' = n. This is obvious physically, that 
the velocity wave must travel in the same direction as the pressure wave. 
We can now set £' = £, and solve Eq. (8-197) for h: 

h = (Sk 1 **' < 8 ~ 198 > 

where the additive constant is zero, since both p' and v are zero in a region 
where there is no disturbance. Equations (8-198), (8-196), and (8-190) 
imply that for a plane sound wave traveling in the direction n, the pressure 
increment and velocity are related by the equation 

V= (5^ n ' < 8 - 199 > 

where v, of course, is here the velocity of a fluid particle, not that of the 
wave, which is en. The velocity of the fluid particles is along the direction 
of propagation of the sound wave, so that sound waves in a fluid are longi- 
tudinal. This is a consequence of the fact that the fluid will not support 
a shearing stress, and is not true of sound waves in a solid, which may be 
either longitudinal or transverse. 



336 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

A plane wave oscillating harmonically in time with angular frequency w 
may be written in the form 

p' = A cos (k-r - cot) = Re Ae*^ 1 -"* , (8-200) 

where k, the wave vector, is given by 

k = -n. (8-201) 

If we consider a surface perpendicular to n which moves back and 
forth with the fluid as the wave goes by, the work done by the pressure 
across this surface in the direction of the pressure is, per unit area per unit 
time, 

P = pv. (8-202) 

If v oscillates with average value zero, then since p = p + p', where p is 
constant, the average power is 

Pav = {p'vU = ~^^2 ' (8-203) 

where we have made use of Eq. (8-199). This gives the amount of energy 
per unit area per second traveling in the direction n. 

The three-dimensional wave equation (8-186) has many other solutions 
corresponding to waves of various forms whose wave fronts (surfaces of 
constant phase) are of various shapes, and traveling in various directions. 
As an example, we consider a spherical wave traveling out from the origin. 
The rate of energy flow is proportional to p' 2 (a small portion of a spherical 
wave may be considered plane), and we expect that the energy flow per 
unit area must fall off inversely as the square of the distance, by the 
energy conservation law. Therefore p' should be inversely proportional to 
the distance r from the origin. We are hence led to try a wave of the form 

P' = ^f(r-ct). (8-204) 

This will represent a wave of arbitrary time-dependence, whose wave 
fronts, £ = r — ct = a constant, are spheres expanding with the velocity 
c. It can readily be verified by direct computation, using either rectangular 
coordinates, or using spherical coordinates with the help of Eq. (3-124), 
that the solution (8-204) satisfies the wave equation (8-186). 



A slight difficulty is encountered with the above development if we attempt to 
apply to a sound wave the expressions for energy flow and mass flow developed 
in the two preceding sections. The rate of flow of mass per unit area per second, 



8-11] NORMAL VIBRATIONS OF FLUID IN A RECTANGULAR BOX 337 

by Eqs. (8-199), (8-180), and (8-183), is 

pv = po(l + ^)^ T7 ,n. 

We should expect that pv would be an oscillating quantity whose average value 
is zero for a sound wave, since there should be no net flow of fluid. If we average 
the above expression, we have 

1/2 
<pv)av = g^ «p' 2 )av + B(p% v )ll, 

so that there is a small net flow of fluid in the direction of the wave, unless 

(P')av=-^^- (8-205) 

If Eq. (8-205) holds, so that there is no net flow of fluid, then it can be shown 
that, to second-order terms in p' and v, the energy current density given by Eq. 
(8-161) is, on the average, for a sound wave, 

(dpv 2 + V - P9 + P«)v)av = { p QB l7/2 n > (8-206) 

in agreement with Eq. (8-203). When approximations are made in the equa- 
tions of motion, we may expect that the solutions will satisfy the conservation 
laws only to the same degree of approximation. By adding second-order (or 
higher) terms like (8-205) to a first-order solution, we can of course satisfy the 
conservation laws to second-order terms (or higher). 

8-11 Normal vibrations of fluid in a rectangular box. The problem of 
the vibrations of a fluid confined within a rigid box is of interest not only 
because of its applications to acoustical problems, but also because the 
methods used can be applied to problems in electromagnetic vibrations, 
vibrations of elastic solids, wave mechanics, and all phenomena in physics 
which are described by wave equations. In this section, we consider a 
fluid confined to a rectangular box of dimensions L x L y L z . 

We proceed as in the solution of the one-dimensional wave equation in 
Section 8-2. We first assume a solution of Eq. (8-186) of the form 

V' = U(x,y,z)@(t). (8-207) 

Substitution in Eq. (8-186) leads to the equation 

Again we argue that since the left side depends only on x, y, and z, and the 



338 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

right side only on t, both must be equal to a constant, which we shall call 
-u 2 /c 2 : 



d*e 



+ « J © = 0, (8-209) 



dP 

2 

V 2 C/+\[/ = 0. (8-210) 

The solution of Eq. (8-209) can be written : 

= A cos ut + B sin ut, (8-211) 

or 

© = Ae-*", (8-212) 



where A and B are constant. The form (8-212) leads to traveling waves 
of the form (8-200). We are concerned here with standing waves, and we 
therefore choose the form (8-211). In order to solve Eq. (8-210), we again 
use the method of separation of variables, and assume that 

U(x, y, z) = X(x)Y(y)Z(z). (8-213) 

Substitution in Eq. (8-210) leads to the equation 

1 £X 1 d*Y 1 d*Z J 

X dx 2 ** Y dyz ~*~ Z dz* c 2 " ^ 8 ; 

This can hold for all x, y, z only if each term on the left is constant. We 
shall call these constants — fc 2 ., — fc 2 , — fc 2 , so that 

g + er=0, + fc 2 7 = O, g + fc 2 Z = 0, (8-215) 

where 2 

kl + fc 2 + fc 2 = ^ • (8-216) 

The solutions of Eqs. (8-215) in which we are interested are 

X = C x cos k x x + D x sin k x x, 

Y = C y cos k y y + D y sin k v y, (8-217) 

Z = C 2 cos k z z + D z sin k z z. 

If we choose complex exponential solutions for X, Y, Z, and 0, we arrive at 
the traveling wave solution (8-200), where k X) k v , k z are the components 
of the wave vector k. 



8-11] NORMAL VIBRATIONS OF FLUID IN A RECTANGULAR BOX 339 

We must now determine the appropriate boundary conditions to be 
applied at the walls of the box, which we shall take to be the six planes 
x = 0, x = L x , y = 0, y = L v , z = 0, z = L z . The condition is evidently 
that the component of velocity perpendicular to the wall must vanish at 
the wall. At the wall x = 0, for example, v x must vanish. According to 
Eq. (8-181), 

dv x 1 dp' 

af=-^ar <Mi8) 

We substitute for p' from Eqs. (8-207), (8-211), (8-213), and (8-217): 

dv k YZ 

-JX = ~ — (A cos cot + B sin cot)(—C x sin k x x + D x cos k x x). 

(8-219) 
Integrating, we have 

k YZ 

v x = ~— (A sin tot — B cos ut)(—C x sin k x x + D x cos k x x) 

(8-220) 

plus a function of x, y, z, which vanishes, since we are looking for oscillating 
solutions. In order to ensure that v x vanishes at x = 0, we must set 
D x = 0, i.e., choose the cosine solution for X in Eq. (8-217). This means 
that the pressure p' must oscillate at maximum amplitude at the wall. 
This is perhaps obvious physically, and could have been used instead of the 
condition v x = 0, which, however, seems more self-evident. The velocity 
component perpendicular to a wall must have a node at the wall, and the 
pressure must have an antinode. Similarly, the pressure must have an 
antinode (maximum amplitude of oscillation) at the wall x — L x : 

cos k x L x = ±1, (8-221) 

so that 

k x = 1 ^-, I = 0, 1, 2, . . . . (8-222) 

By applying similar considerations to the four remaining walls, we con- 
clude that D y — D z — 0, and 

7 m7r r> t o 

k y =-rj~, m = 0, 1, 2, . . . , 

Lly 

(8-223) 
fc z = j- , n = 0, 1, 2, 

Li z 

For each choice of three integers I, m, n, there is a normal mode of vibra- 



340 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

tion of the fluid in the box. The frequencies of the normal modes of vibra- 
tion are given by Eqs. (8-216), (8-222), and (8-223): 



Wj mn 



\Li x ,Li v Liz J 



The three integers I, m, n cannot all be zero, for this gives w = and does 
not correspond to a vibration of the fluid. If we combine these results 
with Eqs. (8-217), (8-213), 8-211), and (8-207), we have for the normal 
mode of vibration characterized by the numbers I, m, n: 

i i a ^in- .a frra miry nirz , 00 _, 

p = (A cos wi mn t + B sin Ul mn t) cos -f— cos -= — cos -f— > (8-225) 

Lt x Liy '-'z 

where we have suppressed the superfluous constant C x C y C z . The corre- 
sponding velocities are 

Vk , . . . t> ,s ■ tox miry nirz 

V x = j (A Sin bilmnt — B COS U>l mn t) sin -j— COS -j — COS -j— > 

J^xPo^lmn ^x Liy i-'z 

m-K , . . *n ,\ It™ ■ fniry nirz 

V y = j (A Sin Ulmnt — B cos o>i mn t) cos -j— sin —f — cos -=— > 

LiyPcfUlmn Li x Lly Li z 

nir , . . „ jX Itx rrvwy . nirz 

Vz = J (A Sin Uimnt — B COS iOlmnt) COS -j- COS -= Sin -j— ■ 

L/zPo^lmn J-i x L/y Li z 

(8-226) 

These four equations give a complete description of the motion of the fluid 
for a normal mode of vibration. The walls x = 0, x = L x , and the (I — 1) 
equally spaced parallel planes between them are nodes for v x and anti- 
nodes for p', v y , and v z . A similar remark applies to nodal planes parallel 
to the other walls. 

It will be observed that the normal frequencies are not, in general, har- 
monically related to one another, as they were in the case of the vibrating 
string. If, however, one of the dimensions, say L x , is much larger than the 
other two, so that the box becomes a long square pipe, then the lowest 
frequencies will correspond to the case where m = n = and I is a small 
integer, and these frequencies are harmonically related. Thus, in a pipe, 
the first few normal frequencies above the lowest will be multiples of the 
lowest frequency. This explains why it is possible to get musical tones 
from an organ pipe, as well as from a vibrating string. Our treatment here 
applies only to a closed organ pipe, and a square one at that. The treat- 
ment of a closed circular pipe is not much more difficult than the above 
treatment and the general nature of the results is similar. The open ended 



8-12] SOUND WAVES IN PIPES 341 

pipe is, however, much more difficult to treat exactly. The difficulty lies 
in the determination of the boundary condition at the open end; indeed, 
not the least of the difficulties is in deciding just where the boundary is. 
As a rough approximation, one may assume that the boundary is a plane 
surface across the end of the pipe, and that this surface is a pressure node. 
The results are then similar to those for the closed pipe, except that if one 
end of a long pipe is closed and one open, the first few frequencies above 
the lowest are all odd multiples of the lowest. 

The general solution of the equations for sound vibrations in a rectangu- 
lar cavity can be built up, as in the case of the vibrating string, by adding 
normal mode solutions of the form (8-225) for all normal modes of vibra- 
tion. The constants A and B for each mode of vibration can again be 
chosen to fit the initial conditions, which in this case will be a specification 
of p' and dp' /dt (or p' and v) at all points in the cavity at some initial 
instant. We shall not carry out this development here. [In the above 
discussion, we have omitted the case I = m = n = 0, which corresponds 
to a constant pressure increment p' . Likewise, we omitted steady velocity 
solutions v(z, y, z) which do not oscillate in time. These solutions would 
have to be included in order to be able to fit all initial conditions.] 

For cavities of other simple shapes, for example spheres and cylinders, 
the method of separation of variables used in the above example works, 
but in these cases instead of the variables x, y, z, coordinates appropriate 
to the shape of the boundary surface must be used, for example spherical 
or cylindrical coordinates. In most cases, except for a few simple shapes, 
the method of separation of variables cannot be made to work. Approxi- 
mate methods can be used when the shape is very close to one of the simple 
shapes whose solution is known. Otherwise the only general methods of 
solution are numerical methods which usually involve a prohibitive amount 
of labor. It can be shown, however, that the general features of our re- 
sults for rectangular cavities hold for all shapes; that is, there are normal 
modes of vibration with characteristic frequencies, and the most general 
motion is a superposition of these. 

8-12 Sound waves in pipes. A problem of considerable interest is the 
problem of the propagation of sound waves in pipes. We shall consider 
a pipe whose axis is in the z-direction, and whose cross section is rectangu- 
lar, of dimensions L x L y . This problem is the same as that of the preceding 
section except that there are no walls perpendicular to the z-axis. 

We shall apply the same method of solution, the only difference being 
that the boundary conditions now apply only at the four walls x = 0, 
x — L x , y = 0, y — L y . Consequently, we are restricted in our choice of 
the functions X(x) and Y(y), just as in the preceding section, by Eqs. 
(8-217), (8-222), and (8-223). There are no restrictions on our choice of 



342 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

solution of the Z-equation (8-215). Since we are interested in solutions 
representing the propagation of waves down the pipe, we choose the ex- 
ponential form of solution for Z: 

Z = e ik °*, (8-227) 

and we choose the complex exponential solution (8-212) for @. Our solu- 
tion for p', then, for a given choice of the integers I, m, is 

JU X Ly 

= A cos -j— cos -=r-^ cos (k z z — at). (8-228) 

L x Ly 

This represents a harmonic wave, traveling in the z-direction down the 
pipe, whose amplitude varies over the cross section of the pipe according 
to the first two cosine factors. Each choice of integers I, m corresponds 
to what is called a mode of propagation for the pipe. (The choice I = 0, 
m = is an allowed choice here.) For a given I, m and a given frequency w, 
the wave number k z is determined by Eqs. (8-216), (8-222), and (8-223): 

The plus sign corresponds to a wave traveling in the +z-direction, and 
conversely. For I = m = 0, this is the same as the relation (8-201) for a 
wave traveling with velocity c in the z-direction in a fluid filling three- 
dimensional space. Otherwise, the wave travels with the velocity 



Clm — 



|fc. 



-[-(^-(srr —> 



which is greater than c and depends on w. There is evidently a minimum 
frequency 



Wim 



[(#" + (")?' 



below which no propagation is possible in the I, m mode; for k z would be 
imaginary, and the exponent in Eq. (8-227) would be real, so that instead 
of a wave propagation we would have an exponential decline in amplitude 
of the wave in the z-direction. Note the similarity of these results to those 
obtained in Section 8-4 for the discrete string, where, however, there was 
an upper rather than a lower limit to the frequency. Since ci m depends on 
<a, we again have the phenomenon of dispersion. A wave of arbitrary 



8-13] THE MACH NUMBER 343 

shape, which can be resolved into sinusoidally oscillating components of 
various frequencies a, will be distorted as it travels along the pipe because 
each component will have a different velocity. We leave as an exercise the 
problem of calculating the fluid velocity v, and the power flow, associated 
with the wave (8-228). 

Similar results are obtained for pipes of other than rectangular cross 
section. Analogous methods and results apply to the problem of the 
propagation of electromagnetic waves down a wave guide. This is one 
reason for our interest in the present problem. 

8-13 The Mach number. Suppose we wish to consider two problems 
in fluid flow having geometrically similar boundaries, but in which the 
dimensions of the boundaries, or the fluid velocity, density, or compressi- 
bility are different. For example, we may wish to investigate the flow of a 
fluid in two pipes having the same shape but different sizes, or we may be 
concerned with the flow of a fluid at different velocities through pipes of 
the same shape, or with the flow of fluids of different densities. We might 
be concerned with the relation between the behavior of an airplane and the 
behavior of a scale model, or with the behavior of an airplane at different 
altitudes, where the density of the air is different. Two such problems in- 
volving boundaries of the same shape we shall call similar problems. Under 
what conditions will two similar problems have similar solutions? 

In order to make this question more precise, let us assume that for each 
problem a characteristic distance s is defined which determines the geo- 
metrical scale of the problem. In the case of similar pipes, s might be a 
diameter of the pipe. In the case of an airplane, s might be the wing span. 
We then define dimensionless coordinates x', y', z' by the equations 

x' = x/s , y' = y/s 0) z' = z/s . (8-232) 

The boundaries for two similar problems will have identical descriptions in 
terms of the dimensionless coordinates x', y', z'\ only the characteristic 
distance s will be different. In a similar way, let us choose a characteristic 
speed v associated with the problem. The speed v Q might be the average 
speed of flow of fluid in a pipe, or the speed of the airplane relative to the 
stationary air at a distance from it, or v might be the maximum speed of 
any part of the fluid relative to the pipe or the airplane. In any case, we 
suppose that v is so chosen that the maximum speed of any part of the 
fluid is not very much larger than v . We now define a dimensionless veloc- 
ity v', and a dimensionless time coordinate t' : 

v' = v/v , (8-233) 

t' = v t/s . (8-234) 



344 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

We now say that two similar problems have similar solutions if the solu- 
tions are identical when expressed in terms of the dimensionless velocity v' 
as a function of x', y', z', and t'. The fluid flow pattern will then be the 
same in both problems, differing only in the distance and time scales de- 
termined by s and v . We need also to assume a characteristic density 
p and pressure p . In the case of the airplane, these would be the density 
and pressure of the undisturbed atmosphere; in the case of the pipe, they 
might be the average density and pressure, or the density and pressure at 
one end of the pipe. We shall define a dimensionless pressure increment 
p" as follows: 

p" = P ~ Po • (8-235) 

PoVo 

We shall now assume that the changes in density of the fluid are small 
enough so that we can write 

9 = Po + % (P ~ Po) ' (8_236) 

where higher order terms in the Taylor series for p have been neglected. 
By making use of the definition (8-235) for p", and of the bulk modulus B 
as given by Eq. (5-183), this can be written 

p = p (l + M 2 p"), (8-237) 

where 

M =*°(p!r /2 =7- < 8 - 238) 

Here M is the ratio of the characteristic velocity v to the velocity of sound 
c and is called the Mach number for the problem. In a similar way, we can 
expand 1/p, assuming that \p — p \ « p : 

- = — (1 - M 2 p"). (8-239) 

P Po 

With the help of Eqs. (8-237) and (8-239), we can rewrite the equation 
of continuity and the equation of motion in terms of the dimensionless 
variables introduced by Eqs. (8-232) to (8-235). The equation of con- 
tinuity (8-127), when we divide through by the constant po*>oAo and 
collect separately the terms involving M , becomes 

V'-v' + M 2 [^ + V- (p'V)J = 0, (8-240) 



where 



V ' = i l + >w + k is- < 8 - 241 > 



8-14] viscosity 345 

The equation of motion (8-139), when we divide through by t^/s , be- 
comes, in the same way, 

%■ + v'- VV + (1 - M 2 p") V'p" = ^ - • (8-242) 

dt »o P 

Equations (8-240) and (8-242) represent four differential equations to be 
solved for the four quantities p', v', subject to given initial and boundary 
conditions. If the body forces are zero, or if the body forces per unit mass 
f/p are made proportional to vl/s , then the equations for two similar prob- 
lems become identical if the Mach number M is the same for both. Hence, 
similar problems will have similar solutions if they have the same Mach 
number. Results of experiments on scale models in wind tunnels can be 
extrapolated to full-sized airplanes flying at speeds with corresponding 
Mach numbers. If the Mach number is much less than one, the terms in 
M 2 in Eqs. (8-240) and (8-242) can be neglected, and these equations then 
reduce to the equations for an incompressible fluid, as is obvious either 
from Eq. (8-240) or (8-237). Therefore at fluid velocities much less than 
the speed of sound, even air may be treated as an incompressible fluid. 
On the other hand, at Mach numbers near or greater than one, the com- 
pressibility becomes important, even in problems of liquid flow. Note that 
the Mach number involves only the characteristic velocity v , and the 
velocity of sound, which in turn depends on the characteristic density p 
and the compressibility B. Changes in the distance scale factor s have no 
effect on the nature of the solution, nor do changes in the characteristic 
pressure p except insofar as they affect p and B. 

It must be emphasized that these results are applicable only to ideal 
fluids, i.e., when viscosity is unimportant, and to problems where the den- 
sity of the fluid does not differ greatly at any point from the characteristic 
density p . The latter condition holds fairly well for liquids, except when 
there is cavitation (formation of vapor bubbles), and for gases except at 
very large Mach numbers. 

8-14 Viscosity. In many practical applications of the theory of fluid 
flow, it is not permissible to neglect viscous friction, as has been done in 
the preceding sections. When adjacent layers of fluid are moving past one 
another, this motion is resisted by a shearing force which tends to reduce 
their relative velocity. Let us assume that in a given region the velocity 
of the fluid is in the z-direction, and that the fluid is flowing in layers 
parallel to the zz-plane, so that v x is a function of y only (Fig. 8-10). 
Let the positive y-axis be directed toward the right. Then if dv x /dy is 
positive, the viscous friction will result in a positive shearing force F x 
acting from right to left across an area A parallel to the xz-plane. The 



346 



THE MECHANICS OF CONTINUOUS MEDIA 



[CHAP. 8 




Fig. 8-10. Velocity distribution in the definition of viscosity. 



coefficient of viscosity r\ is defined as the ratio of the shearing stress to the 
velocity gradient : 

F x /A 



dv x /dy 



(8-243) 



When the velocity distribution is not of this simple type, the stresses due 
to viscosity are more complicated. (See Section 10-6.) 

We shall apply this definition to the important special case of steady 
flow of a fluid through a pipe of circular cross section, with radius a. We 
shall assume laminar flow; that is, we shall assume that the fluid flows in 
layers, as contemplated in the definition above. In this case, the layers 
are cylinders. The velocity is everywhere parallel to the axis of the pipe, 
which we take to be the z-axis, and the velocity v z is a function only of r, 
the distance from the axis of the pipe. (See Fig. 8-11.) If we consider a 
cylinder of radius r and of length I, its area will be A = 2-irrl, and accord' 
ing to the definition (8-243), the force exerted across this cylinder by the 
fluid outside on the fluid inside the cylinder is 



F.= 



dv z 



*CM>¥" 



(8-244) 



Since the fluid within this cylinder is not accelerated, if there is no body 
force the viscous force must be balanced by a difference in pressure 



8-14] 



VISCOSITY 



347 




Fig. 8-1 1 . Laminar flow in a pipe. 

between the two ends of the cylinder: 

Ap{irr 2 ) + F, = 0, 



(8-245) 



where Ap is the difference in pressure between the two ends of the cylinder 
a distance I apart, and we assume that the pressure is uniform over the 
cross section of the pipe. Equations (8-244) and (8-245) can be combined 
to give a differential equation for v z : 



dv z 
dr 



r Ap 

2 v l ' 



(8-246) 



We integrate outward from the cylinder axis : 



L dv >=-wL rdr > 



Vz = V 



r 2 Ap 



(8-247) 



where v is the velocity at the axis of the pipe. We shall assume that the 
fluid velocity is zero at the walls of the pipe : 



r i ° 2 A P n 

[Vz] r =a = Vq -^- = 0, 



(8-248) 



although this assumption is open to question. 



348 THE MECHANICS OF CONTINUOUS MEDIA [CHAP. 8 

Then 

(8-249) 

tT/fr 

and 





Vo = 


a 2 Ap 
4ul 




z 


Ap 


(a 2 - 


r 2 ) 



(8-250) 

The total fluid current through the pipe is 

I = Jjpv z dS = 2wp J" v z r dr. (8-251; 

We substitute from Eq. (8-250) and carry out the integration : 

This formula is called Poiseuille's law. It affords a convenient and simple 
way of measuring i\. 

Although we will not develop now the general equations of motion for 
viscous flow, we can arrive at a result analogous to that in Section 8-13, 
taking viscosity into account, without actually setting up the equations 
for viscous flow. Suppose that we are concerned, as in Section 8-13, with 
two similar problems in fluid flow, and let s , v , p , p be a characteristic 
distance, velocity, pressure, and density, which again define the scale in 
any problem. However, let us suppose that in this case viscosity is to be 
taken into account, so that the equation of motion (8-139) is augmented 
by a term corresponding to the force of viscous friction. We do not at 
present know the precise form of this term, but at any rate it will consist 
of rj multiplied by various derivatives of various velocity components, 
and divided by p [since Eq. (8-139) has already been divided through 
by p]. When we introduce the velocity v', and the dimensionless coordi- 
nates x', y', z', t', as in Section 8-13, and divide the equation of motion 
by v\/s , we will obtain just Eq. (8-242), augmented by a term involving 
the coefficient of viscosity. Since all the terms in Eq. (8-242) are dimen- 
sionless, the viscosity term will be also, and will consist of derivatives of 
components of v' with respect to x', y', z', multiplied by numerical factors 
and by a dimensionless coefficient consisting of rj times some combination 
of vo and s , and divided by p = p (l + M 2 p") [Eq. (8-237)]. Now the 
dimensions of r\, as determined by Eq. (8-243), are 

^ = length a xtime' ( 8 " 253 > 

and the only combination of p , v , and s having these dimensions 



8-14] viscosity 349 

is PoVqS - Therefore the viscosity term will be multiplied by the coeffi- 
cient 

1 



R(l + M 2 p") 
where R is the Reynolds number, defined by 



(8-254) 



^ = M o£o_ (8255) 

v 

We can now conclude that when viscosity is important, two similar prob- 
lems will have the same equation of motion in dimensionless variables, and 
hence similar solutions, only if the Reynolds number R, as well as the 
Mach number M, is the same for both. If the Mach number is very small, 
then compressibility is unimportant. If the Reynolds number is very large, 
then viscosity may be neglected. It turns out that there is a critical value 
of Reynolds number for any given problem, such that the nature of the 
flow is very different for R larger than this critical value than for smaller 
values of R. For small Reynolds numbers, the flow is laminar, as the 
viscosity tends to damp out any vortices which might form. For large 
Reynolds numbers, the flow tends to be turbulent. This will be the case 
when the viscosity is small, or the density, velocity, or linear dimensions 
are large. Note that the Reynolds number depends on s , whereas the 
Mach number does not, so that the distance scale of a problem is im- 
portant when the effects of viscosity are considered. Viscous effects are 
more important on a small scale than on a large scale. 

It may be noted that the expression (8-255) for the Reynolds number, 
together with the fact that Eq. (8-139) is divided by vo/s to obtain the 
dimensionless equation of motion, implies that the viscosity term to be 
added to Eq. (8-139) has the dimensions of (nD )/(p sl). This, in turn, 
implies that the viscous force density must be equal to 17 times a sum of 
second derivatives of velocity components with respect to x, y, and z. 
This is perhaps also evident from Eq. (8-243), since in calculating the 
total force on a fluid element, the differences in stresses on opposite faces 
of the element will be involved, and hence a second differentiation of 
velocities relative to x, y, and z will appear in the expression for the force. 
An expression for the viscous force density will be developed in Chapter 10. 



350 the mechanics of continuous media [chap. 8 

Problems 

1. A stretched string of length I is terminated at the end z = I by a ring of 
negligible mass which slides without friction on a vertical rod. (a) Show that the 
boundary condition at this end of the string is 



[-1 = 



0. 



(b) If the end x = is tied, find the normal modes of vibration. 

2. Find the boundary condition and the normal modes of vibration in Prob- 
lem 1 if the ring at one end has a finite mass m. What is the significance of the 
limiting cases m = and m = «> ? 

3. The midpoint of a stretched string of length I is pulled a distance u — Z/10 
from its equilibrium position, so that the string forms two legs of an isosceles 
triangle. The string is then released. Find an expression for its motion by the 
Fourier series method. 

4. A piano string of length I, tension t, and density a, tied at both ends, and 
initially at rest, is struck a blow at a distance o from one end by a hammer of 
mass m and velocity Do. Assume that the hammer rebounds elastically with 
velocity — vo, and that its momentum loss is transferred to a short length Al 
of string centered around x = a. Find the motion of the string by the Fourier 
series method, assuming that Al is negligibly small. If the finite length of Al 
were taken into account, what sort of effect would this have on your result? If 
it is desired that no seventh harmonic of the fundamental frequency be present 
(it is said to be particularly unpleasant), at what points a may the string be 
struck? 

5. A string of length I is tied at x = I. The end at x = is forced to move 
sinusoidally so that 

m(0, t) = A sin <iit. 

(a) Find the steady-state motion of the string; that is, find a solution in which 
all points on the string vibrate with the same angular frequency u. (b) How 
would you find the actual motion if the string were initially at rest? 

6. A force of linear density 

,, .-. , . nwx 
f(x, t) = /o sin — j— cos cot, 

where n is an integer, is applied along a stretched string of length I. (a) Find 
the steady-state motion of the string. [Hint: Assume a similar time and space 
dependence for u(x, t), and substitute in the equation of motion.] (b) Indicate 
how one might solve the more general problem of a harmonic applied force 

f&y t) = fo(x) cos ut, 

where fo(x) is any function. 

7. Assume that the friction of the air around a vibrating string can be repre- 
sented as a force per unit length proportional to the velocity of the string. Set 



PROBLEMS 351 

up the equation of motion for the string, and find the normal modes of vibration 
if the string is tied at both ends. 

8. Find the motion of a horizontal stretched string of tension t, density cr, 
and length I, tied at both ends, taking into account the weight of the string. The 
string is initially held straight and horizontal, and dropped. [Hint: Find the 
steady-state "motion" and add a suitable transient.] 

9. A long string is terminated at its right end by a massless ring which slides 
on a vertical rod and is impeded by a frictional force proportional to its velocity. 
Set up a suitable boundary condition and discuss the reflection of a wave at the 
end. How does the reflected wave behave in the limiting cases of very large and 
very small friction? For what value of the friction constant is there no reflected 
wave? 

10. Discuss the reflection of a wave traveling down a long string terminated 
by a massless ring, as in Problem 1. 

11. Find a solution to Problem 3 by superposing waves f(x — ct) and 
g(x + ct) in such a way as to satisfy the initial and boundary conditions. Sketch 
the appearance of the string at times t = 0, %1/c, \l/c, and l/c. 

12. (a) A long stretched string of tension t and density <n is tied at x = 
to a string of density 02- If the mass of the knot is negligible, show that u 
and du/dx must be the same on both sides of the knot. 

(b) A wave A cos {k\x — ojt) traveling toward the right on the first string 
is incident on the junction. Show that in order to satisfy the boundary conditions 
at the knot, there must be a reflected wave traveling to the left in the first string 
and a transmitted wave traveling to the right in the second string, both of the 
same frequency as the incident wave. Find the amplitudes and phases of the 
incident and reflected waves. 

(c) Check your result in part (b) by calculating the power in the transmitted 
and reflected waves, and showing that the total is equal to the power in the 
incident wave. 

13. Derive directly from Eq. (8-139) an equation expressing the conservation 
of angular momentum in a form analogous to Eq. (8-141). 

14. Derive an equation expressing the law of conservation of angular mo- 
mentum for a fluid in a form analogous to Eq. (8-140). From this, derive equa- 
tions analogous to Eqs. (8-141), (8-142), and (8-144). Explain the physical 
meaning of each term in each equation. Show that the internal torques due to 
pressure can be eliminated from the integrated forms, and derive an equation 
analogous to Eq. (8-148). 

15. Derive and interpret the following equation: 



it jjj {ipt)2 ~ p9 ~ pu) dV + IS n ' v( * p " 2 ~~ p9 ~ pu) dS 



dt 
v 



j P ^ dV > 

S V 



where 7 is a fixed volume bounded by a surface S with normal n. 



352 THE MECHANICS OP CONTINUOUS MEDIA [CHAP. 8 

16. (a) A mass of initially stationary air at 45° N latitude flows inward toward 
a low-pressure spot at its center. Show that the coriolis torque about the low- 
pressure center depends only on the radial component of velocity. Hence, show 
that if frictional torques are neglected, the angular momentum per unit mass at 
radius r from the center depends only on r and on the initial radius ro at which 
the air is stationary, but does not depend on the details of the motion. 

(b) Calculate the azimuthal component of velocity around the low as a func- 
tion of initial and final radius. If this were a reasonable model of a tornado, 
what would be the initial radius ro if the air at 264 ft from the center has a 
velocity of 300 mi/hr? 

17. Evaluate the potential energy u per unit mass as a function of p for a 
perfect gas of molecular weight M at temperature T. For the steady isothermal 
flow of this gas through a pipe of varying cross section and varying height above 
the earth, find expressions for the pressure, density, and velocity of the gas as 
functions of the cross section S of the pipe, the height h, and the pressure po 
and velocity vq at a point in the pipe at height h = where the cross section 
is So- Assume p, v, and p uniform over the cross section. 

18. Work Problem 17 for an incompressible fluid of density po- 

19. The function <t> = a/r, where a is a constant and r is the distance from a 
fixed point, satisfies Laplace's equation (8-177), except at r = 0, because it has 
the same form as the gravitational potential of a point mass. If this is a velocity 
potential, what is the nature of the fluid flow to which it leads? 

20. (a) Verify by direct computation that the spherical wave (8-204) satisfies 
the wave equation (8-186). (b) Write an analogous expression for a cylindrical 
wave of arbitrary time dependence, traveling out from the z-axis, independent 
of z and with cylindrical symmetry. Make the amplitude depend on the dis- 
tance from the axis in such a way as to satisfy the requirement of conservation 
of energy. Show that such a wave cannot satisfy the wave equation. (It is a 

• general property of cylindrical waves that they do not preserve their shape.) 
*21. Show that the normal mode of vibration given by Eqs. (8-225) and (8-226) 
can be represented as a superposition of harmonically oscillating plane waves 
traveling in appropriately chosen directions with appropriate phase relation- 
ships. Show that in the normal vibrations of a fluid in a box, the velocity oscil- 
lates 90° out of phase with the pressure at any point. How can this be reconciled 
with the fact that in a plane wave the velocity and pressure are in phase? 

22. Find the normal modes of vibration of a square organ pipe with one end 
open and the other closed, on the assumption that the open end is a pressure node. 

23. (a) Calculate the fluid velocity v for the wave given by Eq. (8-228). (b) 
Calculate the mean rate of power flow through the pipe. 

*24. Show that the expression (8-228) for a sound wave in a pipe can be repre- 
sented as a superposition of plane waves traveling with speed c in appropriate 
directions, and being reflected at the walls. Explain, in terms of this representa- 
tion, why there is a minimum frequency for any given mode below which a wave 
cannot propagate through the pipe in this mode. 

25. If the sound wave given by Eq. (8-228) is incident on a closed end of the 
pipe at z = 0, find the reflected wave. 



PROBLEMS 353 

26. Develop the theory of the propagation of sound waves in a circular pipe, 
using cylindrical coordinates and applying the method of separation of variables. 
Carry the solution as far as you can. You are not required to solve the equation 
for the radial part of the wave, but you should indicate the sort of solutions you 
would expect to find. 

27. A fluid of viscosity i\ flows steadily between two infinite parallel plane 
walls a distance I apart. The velocity of the fluid is everywhere in the same direc- 
tion, and depends only on the distance from the walls. The total fluid current 
between the walls in any unit length measured along the walls perpendicular to 
the direction of flow is I. Find the velocity distribution and the pressure gradient 
parallel to the walls, assuming that the pressure varies only in the direction of 
flow. 

28. Prove that the only combination of po, fo, so having the dimensions of 
viscosity is po^oso- 



CHAPTER 9 
LAGRANGE'S EQUATIONS 

9-1 Generalized coordinates. Direct application of Newton's laws to 
a mechanical system results in a set of equations of motion in terms of the 
cartesian coordinates of each of the particles of which the system is com- 
posed. In many cases, these are not the most convenient coordinates in 
terms of which to solve the problem or to describe the motion of the system. 
For example, in the problem of the motion of a single particle acted on by 
a central force, which we treated in Section 3-13, we found it convenient to 
introduce polar coordinates in the plane of motion of the particle. The 
reason was that the force in this case can be expressed more simply in terms 
of polar coordinates. Again in the two-body problem, treated in Section 
4-7, we found it convenient to replace the coordinates ri, r 2 of the two 
particles by the coordinate vector R of the center of mass, and the relative 
coordinate vector r which locates particle 1 with respect to particle 2. We 
had two reasons for this choice of coordinates. First, the mutual forces 
which the particles exert on each other ordinarily depend on the relative 
coordinate. Second, in many cases we are interested in a description of 
the motion of one particle relative to the other, as in the case of planetary 
motion. In problems involving many particles, it is usually convenient 
to choose a set of coordinates which includes the coordinates of the center 
of mass, since the motion of the center of mass is determined by a relatively 
simple equation (4-18). In Chapter 7, we found the equations of motion 
of a particle in terms of moving coordinate systems, which are sometimes 
more convenient to use than the fixed coordinate systems contemplated 
in Newton's original equations of motion. 

We shall include coordinate systems of the sort described above, to- 
gether with cartesian coordinate systems, under the name generalized co- 
ordinates. A set of generalized coordinates is any set of coordinates by 
means of which the positions of the particles in a system may be specified. 
In a problem requiring generalized coordinates, we may set up Newton's 
equations of motion in terms of cartesian coordinates, and then change to 
the generalized coordinates, as in the problems studied in previous chap- 
ters. It would be very desirable and convenient, however, to have a gen- 
eral method for setting up equations of motion directly in terms of any 
convenient set of generalized coordinates. Furthermore it is desirable to 
have uniform methods of writing down, and perhaps of solving, the 
equations of motion in terms of any coordinate system. Such a method 
was invented by Lagrange and is the subject of this chapter. 

354 



9-1] GENERALIZED COORDINATES 355 

In each of the cases mentioned in the first paragraph, the number of 
coordinates in the new system of coordinates introduced to simplify the 
problem was the same as the number of cartesian coordinates of all the 
particles involved. We may, for example, replace the two cartesian co- 
ordinates x, y of a particle moving in a plane by the two polar coordinates 
r, 0, or the three space coordinates x, y, z by three spherical or cylindrical 
coordinates. Or we may replace the six coordinates x lt y x , z lt x 2 , y 2 , z 2 
of a pair of particles by the three coordinates X, Y, Z of the center of 
mass plus the three coordinates x, y, z of one particle relative to the other. 
Or we may replace the three coordinates of a particle relative to a fixed 
system of axes by three coordinates relative to moving axes. (A vector 
counts as three coordinates.) 

In our treatment of the rotation of a rigid body about an axis (Section 
5-2), we described the position of the body in terms of the single angular 
coordinate 6. Here we have a case where we can replace a great many 
cartesian coordinates, three for each particle in the body, by a single 
coordinate 0. This is possible because the body is rigid and is allowed to 
rotate only about a fixed axis. As a result of these two facts, the position 
of the body is completely determined when we specify the angular position 
of some reference line in the body. The position of a free rigid body can 
be specified by six coordinates, three to locate its center of mass, and three 
to determine its orientation in space. This is a vast simplification com- 
pared with the 3N cartesian coordinates required to locate its N particles. 
A rigid body is an example of a system of particles subject to constraints, 
that is, conditions which restrict the possible sets of values of the coordi- 
nates. In the case of a rigid body, the constraint is that the distance 
between any two particles must remain fixed. If the body can rotate 
only about a fixed axis, then in addition the distance of each particle from 
the axis is fixed. This is the reason why specifying the value of the single 
coordinate is sufficient to determine the position of each particle in the 
body. We shall postpone the discussion of systems like this which involve 
constraints until Section 9-4. In this section, and the next, we shall set 
up the theory of generalized coordinates, assuming that there are as many 
generalized coordinates as cartesian coordinates. We shall then find, in 
Section 9-4, that this theory applies also to the motion of constrained 
systems. 

When we want to speak about a physical system described by a set of 
generalized coordinates, without specifying for the moment just what the 
coordinates are, it is customary to designate each coordinate by the letter q 
with a numerical subscript. A set of n generalized coordinates would be 
written as q lt q 2 , . . . , q n - Thus a particle moving in a plane may be 
described by two coordinates q u q 2 , which may in special cases be the 
cartesian coordinates x, y, or the polar coordinates r, 0, or any other suit- 



356 



lagrange's equations 



[chap. 9 



able pair of coordinates. A particle moving in space is located by three 
coordinates, which may be cartesian coordinates x, y, z, or spherical 
coordinates r, 0, <p, or cylindrical coordinates p, z, <p, or, in general 

5l, «2, ?3- 

The configuration of a system of TV particles may be specified by the 
37V cartesian coordinates xi, y\, Z\, x 2 , y 2 , z 2 , ■ ■ ■ , xx, y^, z^ of its par- 
ticles, or by any set of 37V generalized coordinates q u q 2 , . . . , q3N- Since 
for each configuration of the system, the generalized coordinates must 
have some definite set of values, the coordinates g 1( . . . , q3w will be func- 
tions of the cartesian coordinates, and possibly also of the time in the 
case of moving coordinate systems: 



Qi = 2i(zi, 2/i, zi, x 2 , 2/2, • • ■ , Vn, z n ; t), 
?2 = 92(^1, 2/i, , ZN)t), 

13N = q3N{Xl, 2/1, , z N ;t). 



(9-1) 



Since the coordinates q\, . . . , qz n specify the configuration of the system, 
it must be possible also to express the cartesian coordinates in terms of 
the generalized coordinates: 



Xi = xi(qi, q 2 , ■ ■ ■ ,Q3n; t), 
2/1 = 2/1(91, , Q3n; t), 

zn = zjv(<7i, , Q3N', t). 



(9-2) 



If Eqs. (9-1) are given, they may be solved for x\, 2/1, . . . , zjv to obtain 
Eqs. (9-2), and vice versa. 



The mathematical condition that this solution be (theoretically) possible is 
that the Jacobian determinant of Eqs. (9-1) be different from zero at all points, 
or nearly all points: 

dgi bq 2 dq3N 

dxi dxi dxi 



d(qi, . . . , g3iv) 



d(xi, 3/i, 



, zjv) 



dgi 


dq 2 
dyi 


dq3N 
dyi 


dgi 
dztf 


dq 2 


dq3N 
dzif 



j* 0. 



(9-3) 



If this inequality does not hold, then Eqs. (9-1) do not define a legitimate set 
of generalized coordinates. In practically all cases of physical interest, it will 



9-1] GENERALIZED COORDINATES 357 

be evident from the geometrical definitions of the generalized coordinates 
whether or not they are a legitimate set of coordinates. Thus we shall not 
have any occasion to apply the above test to our coordinate systems. [For a 
derivation of the condition (9-3), see W. F. Osgood, Advanced Calculus, New 
York: Macmillan, 1937, p. 129.] 



As an example, we have the equations (3-72) and (3-73) connecting 
the polar coordinates r, 6 of a single particle in a plane with its cartesian 
coordinates x, y. As an example of a moving coordinate system, we 
consider polar coordinates in which the reference axis from which 6 is 
measured rotates counterclockwise with constant angular velocity w 
(Fig. 9-1): 

r=(z 2 + t/ 2 ) 1/2 , 

6 = tan -1 ^ — at, (9-4) 

and conversely, 

x — r cos (0 + ut), 

y = rsin (0 + ut). (9-5) 

As an example of generalized coordinates for a system of particles, we 
have the center of mass coordinates X, Y, Z and relative coordinates 
x, y, z of two particles of masses m l and m 2 , as denned by Eqs. (4-90) 
and (4-91), where X, Y, Z are the components of R, and x, y, z are the 
components of r. Because the transformation equations (4-90) and 
(4-91) do not contain the time explicitly, we regard this as a fixed co- 
ordinate system, even though x, y, z are the coordinates of mi referred 
to a moving origin located on m 2 . The rule which defines the coordinates 
X, Y, Z, x, y, z is the same at all times. 

If a system of particles is described by a set of generalized coordinates 
flii • • • > Q3N, we shall call the time derivative q k , of any coordinate <?&, 
the generalized velocity associated with this coordinate. The generalized 
velocity associated with a cartesian coordinate a;,- is just the correspond- 




Fig. 9-1. A rotating polar coordinate system. 



358 lagrange's equations [chap. 9 

ing component Xi of the velocity of the particle located by that coordinate. 
The generalized velocity associated with an angular coordinate is the 
corresponding angular velocity 0. The velocity associated with the co- 
ordinate X in the preceding example is X, the x-component of velocity 
of the center of mass. The generalized velocities can be computed in terms 
of cartesian coordinates and velocities, and conversely, by differentiating 
Eqs. (9-1) or (9-2) with respect to t according to the rules for differentiat- 
ing implicit functions. For example, the cartesian velocity components 
can be expressed in terms of the generalized coordinates and velocities 
by differentiating Eqs. (9-2): 

¥k dxi . . dxi 
i 0-6) 






dZN . i d*N 



As an example, we have, from Eqs. (9-5): 

x = f cos (0 + at) — r0 sin (0 + at) — m sin (0 + ait), 

(9—7) 
y = r sin (0 + at) + rO cos (0 + at) + ru cos (0 + at). 

The kinetic energy of a system of N particles, in terms of cartesian 
coordinates, is 

T = j^ *»»<(*? + y' + *?)■ (»-8) 

By substituting from Eqs. (9-6), we obtain the kinetic energy in terms 
of generalized coordinates. If we rearrange the order of summation, the 
result is 

fc=l 1=1 k=l 

where 



9-1] GENEEALIZED COORDINATES 359 

The coefficients Am, B^, and T§ are functions of the coordinates 
<h> • • • > Izn, and also of t for a moving coordinate system. If A k i is 
zero except when k = I, the coordinates are said to be orthogonal. The coef- 
ficients B k and T Q are zero when t does not occur explicitly in Eqs. (9-1), 
i.e., when the generalized coordinate system does not change with time. 
We see that the kinetic energy, in general, contains three sets of terms: 

T = T 2 + T x + T , (9-13) 

where T 2 contains terms quadratic in the generalized velocities, Ti con- 
tains linear terms, and T is independent of the velocities. The terms 
T x and T appear only in moving coordinate systems; for fixed coordinate 
systems, the kinetic energy is quadratic in the generalized velocities. 

As an example, in plane polar coordinates [Eqs. (3-72)], the kinetic 
energy is 

T = %m(x 2 + y 2 ) 

= \{mf 2 + mr 2 6 2 ), (9-14) 

as may be obtained by direct substitution from Eqs. (3-72), or as a 
special case of Eq. (9-9), where 



dx doc 

-=cos0, -= -rsinfl, 

dy ■ » dy 

f r = sme, f e = rco S e. 



(9-15) 



If we take the moving coordinate system defined by Eqs. (9-5), we find, 
by substituting from Eqs. (9-7), or by using Eq. (9-9), 

T = £m(x 2 + V 2 ) 

= i(mf 2 + mr 2 6 2 ) + mr 2 o)6 + %mr 2 w 2 . (9-16) 

In this case, a term linear in 6 and a term independent of f and 6 appear. 
The kinetic energy for the two-particle system can also easily be written 
down in terms of X, Y, Z, x, y, z, defined by Eqsi (4-90) and (4-91). 

Instead of finding the kinetic energy first in cartesian coordinates and 
then translating into generalized coordinates, as in the examples above, 
it is often quicker to work out the kinetic energy directly in terms of 
generalized coordinates from a knowledge of their geometrical meaning. 
It may then be possible to start a problem from the beginning with a 
suitable set of generalized coordinates without writing out explicitly the 
transformation equations (9-1) and (9-2) at all. For example, we may 
obtain Eq. (9-14) immediately from the geometrical meaning of the 



360 lagrange's equations [chap. 9 

coordinates r, (see Fig. 3-20) by noticing that the linear velocity associ- 
ated with a change in r is f and that associated with a change in is r&. 
Since the directions of the velocities associated with r and are perpen- 
dicular, the square of the total velocity is 

v 2 = r 2 + r 2 8 2 , (9-17) 

from which Eq. (9-14) follows immediately. 

Care must be taken in applying this method if the velocities associated 
with changes of the various coordinates are not perpendicular. For exam- 
ple, let us consider a pair of coordinate axes u, w making an angle a less 
than 90° with each other, as in Fig. 9-2. Let u and w be the sides of a par- 
allelogram formed by these axes and by lines parallel to the axes through 
the mass m as shown. Let a and b be unit vectors in the directions of 
increasing u and w. Using u and w as coordinates, the velocity of the 
mass m is 

v = wa + wb. (9-18) 

The kinetic energy is 

T = jtov-v = \mu 2 + \mw 2 + muw cos a. (9-19) 

This is an example of a set of nonorthogonal coordinates in which a cross 
product term in the velocities appears in the kinetic energy. The reason 
for using the term orthogonal, which means perpendicular, is clear from 
this example. 

When systems of more than one particle are described in terms of 
generalized coordinates, it is usually safest to write out the kinetic energy 
first in cartesian coordinates and transform to generalized coordinates. 
However, in some cases, it is possible to write the kinetic energy directly 

in general coordinates. For exam- 
ple, if a rigid body rotates about an 
axis, we know that the kinetic 
energy is J/w 2 , where w is the angu- 
lar velocity about that axis and / is 
the moment of inertia. Also, we can 
use the theorem proved in Section 
4-9 that the total kinetic energy of 
a system of particles is the kinetic 
energy associated with the center of 
mass plus that associated with the 
internal coordinates. [See Eq. 
Fig. 9-2. A nonorthogonal coordi- (4-127).] As an example, the kinetic 
nate system. energy of the two-particle system in 




9-1] GENERALIZED COORDINATES 361 

terms of the coordinates X, Y, Z, x, y, z, defined by Eqs. (4-90) and (4-91) 
is 

T = W(X 2 + f 2 + Z 2 ) + i M (x 2 + y 2 + z 2 ), (9-20) 

where M and n are given by Eqs. (4-97) and (4-98). The result shows 
that this is an orthogonal coordinate system. If the linear velocity of 
each particle in a system can be written down directly in terms of the 
generalized coordinates and velocities, then the kinetic energy can im- 
mediately be written down. 

We now note that the components of the linear momentum of particle i, 
according to Eq. (9-8), are 

dT dT . dT , n .,, 

p ix = mxi = — , p iy = myi = — > Piz = mz t = — • (9-21) 
aXi dyi dZi 

In the case of a particle moving in a plane, the derivatives of T with 
respect to f and 6, as given by Eq. (9-14), are 

p r = m r= —, Ve = mr 2 e = -^ , (9-22) 

where p r is the component of linear momentum in the direction of in- 
creasing r, and p e is the angular momentum about the origin. Similar 
results will be found for spherical and cylindrical coordinates in three 
dimensions. In fact, it is not hard to show that for any coordinate q^ 
which measures the linear displacement of any particle or group of par- 
ticles in a given direction, the linear momentum of that particle or group 
in the given direction is dT/dqk) and that for any coordinate q k which 
measures the angular displacement of a particle or group of particles 
about an axis, their angular momentum about that axis is dT/dq k . This 
suggests that we define the generalized momentum pk associated with the 
coordinate q^ by* 

a/77 

p k = Wk - (9-23) 

If qk is a distance, pk is the corresponding linear momentum. If q k is an 



* The kinetic energy T is denned by Eq. (9-9) as a function of qi, . . . , J3at; 
gi, . . . , q3N, and perhaps of t. The derivatives of this function T with respect 
to these variables will be denoted by the symbols for partial differentiation. 
Since qi, . . . , qw\ ?i, . . . , qsN are all functions of the time t for any given 
motion of the system, T is also a function of t alone for any given motion. The 
derivative of T with respect to time in this sense will be denoted by d/dt. The 
same remarks apply to any other quantity which may be written as a function 
of the coordinates and velocities and perhaps of t, and which is also a function 
of t alone for any given motion. 



362 lagrange's equations [chap. 9 

angle, pk is the corresponding angular momentum. In other cases, pk 
will have some other corresponding physical significance. According to 
Eq. (9-9), the generalized momentum pk is 

Pk = X) A "ih + B k- (9-24) 

i=i 

In the case of the coordinates X, Y, Z, x, y, z for the two-particle sys- 
tem, this definition gives 

p x = MX, p Y = MY, p z = MZ, 

p x = fix, p v = ny, p z = nz, (9-25) 

where px , Py, Pz are the components of the total linear momentum of 
the two particles, and p x , p y , p z are the linear momentum components 
in the equivalent one-dimensional problem in x, y, z to which the two- 
body problem was reduced in Section 4-7. We shall see in the next sec- 
tion that the analogy between the generalized momenta pk and the 
cartesian components of linear momentum can be extended to the equa- 
tions of motion in generalized coordinates. 

If forces Fix, F Xy , F u , . . . , Fnz act on the particles, the work done 
by these forces if the particles move from the positions x\, y x , z x , . . . , zm 
to nearby points X\ + hx x , y x + 6y lt Z\ + hz\, . . . , z N + 8zn is 

iv 
SW = X) (*"<* Sxi + F iv 8y { + F iz to,). (9-26) 

i=i 

The small displacements Sx^ 8y{, 8zt may be expressed in terms of gen- 
eralized coordinates: 

^ dxi $ 

where 8qi, . . . , 5g 3 jv are the differences in the generalized coordinates 
associated with the two sets of positions of the particles. We call this 
a virtual displacement of the system because it is not necessary that it 
represent any actual motion of the system. It may be any possible mo- 
tion of the system. In the case of a moving coordinate system, we regard 
the time as fixed; that is, we specify the changes in position in terms of 



9-1] GENERALIZED COORDINATES 363 

the coordinate system at a particular time t. If we substitute Eqs. (9-27) 
in Eq. (9-26), we have, after rearranging terms: 

32V 



SW = 2 Qk dq k , (9-28) 



fc=i 



where 



The coefficients Q k depend on the forces acting on the particles, on the 
coordinates q\, . . . , q SN , and possibly also on the time t. In view of the 
similarity in form between Eqs. (9-26) and (9-28), it is natural to call 
the quantity Q k the generalized force associated with the coordinate q k . 
We can define the generalized force Q k directly, without reference to the 
cartesian coordinate system, as the coefficient which determines the 
work done in a virtual displacement in which q k alone changes: 

SW = Q k Sq k , (9-30) 

where SW is the work done when the system moves in such a way that 
q k increases by Sq k , all other coordinates remaining constant. Notice 
that the work in Eq. (9-26), and therefore also in Eq. (9-30), is to be 
computed from the values of the forces for the positions x x , . . . , z N , or 
Qi> • • • , qzN) that is, we do not take account of any change in the forces 
during the virtual displacement. 

If the forces F lx , . . . , F Nz are derivable from a potential energy 
V(x u ..., z N ) [Eqs. (4-32)], then 

SW = -SV 

If V is expressed in terms of generalized coordinates, then 

SW = —SV 

3N r)V 

By comparing this with Eq. (9-28), we see that 

Q*=-g, (9-33) 

which shows that in this sense also the definition of Q k as a generalized 



364 lagrange's equations [chap. 9 

force is a natural one. Equation (9-33) may also be verified by direct 
calculation of dV/dq k : 

dV = y. /W dxi dVdyt dV dzA 
dq k ~~ £-J \dxi dq k dyt dq k dz» dqj 

= -Q k . 

As an example, let us calculate the generalized forces associated with 
the polar coordinates r, 0, for a particle acted on by a force 

F = iF x + W v = nF r + lF e . (9-34) 

If we use the definition (9-29), we have, using Eqs. (9-15) : 

= F x cos + Fy sin 

= F r , (9-35) 

_ dx dy 

Qe ~ * x dd + tv de 

= —rF x sin 6 + rF v cos 
= rF,. 

We see that Q r is the component of force in the r-direction, and Q e is 
the torque acting to increase 0. It is usually quicker to use the defini- 
tion (9-30), which enables us to bypass the cartesian coordinates altogether. 
If we consider a small displacement in which r changes to r + 5r, with 6 
remaining constant, the work is 

SW = F r 5r, (9-36) 

from which the first of Eqs. (9-35) follows. If we consider a displace- 
ment in which r is fixed and increases by 50, the work is 

SW = F e r SO, (9-37) 

from which the second of Eqs. (9-35) follows. In general, if q k is a co- 
ordinate which measures the distance moved by some part of the mechani- 
cal system in a certain direction, and if F k is the component in this direc- 
tion of the total force acting on this part of the system, then the work done 



9-2] lagbange's equations 365 

when q k increases by dq k , all other coordinates remaining constant, is 

SW = F k Sq k . (9-38) 

Comparing this with Eq. (9-30), we have 

Qk = Ft. (9-39) 

In this case, the generalized force Q k is just the ordinary force F k . If q k 
measures the angular rotation of a certain part of the system about a 
certain axis, and if N k is the total torque about that axis exerted on this 
part of the system, then the work done when q k increases by Sq k is 

SW = N k 8q k . (9-40) 

Comparing this with Eq. (9-30), we have 

Qk = N k . (9-4:1) 

The generalized force Q k associated with an angular coordinate q k is the 
corresponding torque. 

9-2 Lagrange's equations. The analogy which led to the definitions of 
generalized momenta and generalized forces tempts us to suspect that 
the generalized equations of motion will equate the time rate of change 
of each momentum p k to the corresponding force Q k . To check this 
suspicion, let us calculate the time rate of change of p k : 

We will need to start with Newton's equations of motion in cartesian form: 

rriiVi = Fiy, [i = 1, . . . , N] (9^3) 

rtii'Zi = F iz . 



Therefore we express T in cartesian coordinates [Eq. (9-8)]. We then 
have 

dT 

dq k 



— = >,m I la;,TT i + Vi -^ *■ + & t — * I , (9-44) 



where x u y u . . . , z N are given as functions of q u . . . , q^N', ii, ■ • ■ , q>3N; t 
by Eqs. (9-6). Since dXi/dq k and dXi/dt are functions only of q u . . . , q 3 ^; 
t, we have, by differentiating Eqs. (9-6) : 



366 lagrange's equations [chap. 9 

dii dXi 

dqk dq k ' 

W = d £'~ H=l,--.,N;k=l,...,M] (9-45) 

oq k °qk 

dZj _ dzj 
dqk dq k 

By substituting from Eqs. (9-45) in Eq. (9~44), and differentiating 
again with respect to t, we obtain 

dt f{ \ dq k a dq k dq k / 

. v-> ( . d dx{ . . d dyi . . d dzi\ /n .„. 

According to Newton's equations of motion (9-43), and the definition 
(9-29), the first term in Eq. (9-46) is 

V^ ™ (* dXi j- r, d y* _i_ -4 Bz <\ — V 1 (p dXi MF d ^ i a- w bz *\ 
f{ \ dq k dq k dq k / ££ \ dq k " dq k dq k / 

= Qu. (9-47) 

The derivatives appearing in the last term in Eq. (9-46) are calculated 
as follows: 

d dXi _ -1^ d 2 Xi . d 2 Xj _ _d_ /^ dx^ . . dxA _ dii 

dtdq k ~ £{ dqkdqi Ql + dq k dt ~ dq k [^ dq t Ql + ~>t J ~ dq k ' 

(9-48) 

where we have made use of Eq. (9-6). Similar expressions hold for y 
and z. Thus the last sum in Eq. (9-46) is 

V* i • — ^i j- • — ^Mi a- • — d g< i 
£, m t \Xi j ( ^ + y t dt d ^ + Zi dt dq J 

El . dXi . . dyi . . dZ{\ d -r^ 1 /-2 , .2 i .2\ 

.. . mi \ Xi eq- k + yi i- k + Zi WJ = W^ l hmi{Xi + yi+ Zi) 

(9-4) 
We have finally: 



dT_ 
dqk 



f = ^ + S' *=1.-".^- (9-50) 



9-2] lagrange's equations 367 

Our original expectation was not quite correct, in that we must add to 
the generalized force Q k another term dT/dq k in order to get the rate of 
change of momentum j> k - To see its meaning, consider the kinetic energy 
of a particle in terms of plane polar coordinates, as given by Eq. (9-14). 
In this case, 

-gj = mre 2 , (9-51) 

and if we make use of Eqs. (9-22) and (9-35), the equation of motion 
(9-50) for qk = r is 

mf = F r + mr0 2 . (9-52) 

If we compare this with Eq. (3-207), which results from a direct applica- 
tion of Newton's law of motion, we see that the term dT/dr is part of 
the mass times acceleration which appears here transposed to the right 
side of the equation. In fact, dT/dr is the "centrifugal force" which 
must be added in order to write the equation of motion for r in the form 
of Newton's equation for motion in a straight line. Had we been a bit more 
clever originally, we should have expected that some such term might 
have to be included. We may call dT/dqk a "fictitious force" which ap- 
pears if the kinetic energy depends on the coordinate q k . This will be 
the case when the coordinate system involves "curved" coordinates, that 
is, if constant generalized velocities fa, ... , fan result in curved motions 
of some parts of the mechanical system. Equations (9-50) are usually 
written in the form 

i(H)-ir«* * = i."-.3*-: (9-53) 

If a potential energy exists, so that the forces Qk are derivable from 
a potential energy function [Eq. (9-33)], we may introduce the Lagrangian 
function 

L(Qi, ..., Qsn; fa, ■■■ , fan) t) = T — V, (9-54) 

where T depends on both q u . . . , q 3N and fa, ... , q 3N , but V depends 
only on q i} . . . , q 3 jf (and possibly t), so that 



A <H± = A <>T. 
dt dfa dt dfa 



(9-55) 



dL dT dV dT ..... 

^ — = 5 n — = ^ — + Qk- (9-56) 

dq k dq k dq k dq k 

Hence Eqs. (9-53) can be written in this case in the form 

a(i)-t = °- *-'.-.»• mi 



368 lagrange's equations [chap. 9 

In nearly all cases of interest in physics (although not in engineering), 
the equations of motion can be written in the form (9-57). The most 
important exception is the case where frictional forces are involved, but 
such forces do not usually appear in atomic or astronomical problems. 
Since Lagrange's equations have been derived from Newton's equations 
of motion, they do not represent a new physical theory, but merely a differ- 
ent but equivalent way of expressing the same laws of motion. As the 
example of Eqs. (9-52) and (3-207) illustrates, the equations we get by 
Lagrange's method can also be obtained by a direct application of Newton's 
law of motion. However, in complicated cases it is usually easier to work 
out the kinetic energy and the forces or potential energy in generalized 
coordinates, and write the equations in Lagrangian form. Particularly in 
problems involving constraints, as we shall see in Section 9-4, the La- 
grangian method is much easier to apply. The chief value of Lagrange's 
equations is, however, probably a theoretical one. From the manner in 
which they were derived, it is evident that Lagrange's equations (9-57) or 
(9-53) hold in the same form in any system of generalized coordinates. It 
can also be verified by direct computation (see Problem 24) that if Eqs. 
(9-57) hold in any coordinate system for any function L(g 1 , . . . , qzN',Qi, 
■ ■ • , izN) t), then equations of the same form hold in any other coordinate 
system. The Lagrangian function L has the same value, for any given set 
of positions and velocities of the particles, no matter in what coordinate 
system it may be expressed, but the form of the function L may be different 
in different coordinate systems. The fact that Lagrange's equations have 
the same form in all coordinate systems is largely responsible for their 
theoretical importance. Lagrange's equations represent a uniform way of 
writing the equations of motion of a system, which is independent of the 
kind of coordinate system used. They form a starting point for more ad- 
vanced formulations of mechanics. In developing the general theory of 
relativity, in which cartesian coordinates may not even exist, Lagrange's 
equations are particularly important. 

9-3 Examples. We first consider a system of particles m 1; . . . , mx, 
located by cartesian coordinates, and show that in this case Lagrange's 
equations become the Newtonian equations of motion. The kinetic en- 
ergy is 

N 

T = Y, i m ^ + y 2 i + if), (9-58) 



and 



*L = *L = *1 = o (9-59) 

dxt dy { dZi ' 

Wi = m iXi , — = mm , — = miZi . (9-60) 



9-4] SYSTEMS SUBJECT TO CONSTRAINTS 369 

The generalized force associated with each cartesian coordinate is just 
the ordinary force, as we see either from Eq. (9-29), or by comparing 
Eq. (9-28) with Eq. (9-26). Hence the equations of motion (9-53) are 



d(ar\ 

dt \dXi/ 

dt KdyJ dyi 

a (mi\ _ *i_ 

dt \dzi) dzi 



0L— •• - p 

n — TfliXi — r ix } 



dT = m iyi = F iv , [i = 1, . . . , N] (9-61) 



rriiZi = F iz . 



For a particle moving in a plane, the kinetic energy in polar coordi- 
nates is given by Eq. (9-14), and the forces Q r and Q e by Eqs. (9-35). 
The Lagrange equations are 

mr — mrti 2 = F r> (9-62) 

± ( mr 2 6) = rF e . (9-63) 

These equations were obtained in Section 3-13 by elementary methods. 
We now consider the rotating coordinate system denned by Eqs. (9-4) 
or (9-5). The kinetic energy is given by Eq. (9-16), and the generalized 
forces Q r and Q t will be the same as in the previous example. Lagrange's 
equations in this case are 

mr — mrt) 2 — 2rruor6 — mo) 2 r = F r , (9-64) 

j (mr 2 d) + 2mwrf = rF e . (9-65) 

The reader should verify that the third term on the left in Eq. (9-64) 
is the negative of the coriolis force in the r-direction due to the rotation 
of the coordinate system, and that the fourth term is the negative of 
the centrifugal force. The second term in Eq. (9-65) is the negative of 
the coriolis torque in the 0-direction. Thus the necessary fictitious forces 
are automatically included when we write Lagrange's equations in a 
moving coordinate system. It must be noticed, however, that we use the 
actual kinetic energy [Eq. (9-16)] with respect to a coordinate system 
at rest, expressed in terms of the rotating coordinates, and not the kinetic 
energy as it would appear in the rotating system if we ignored the motion 
of the coordinate system. 

9-4 Systems subject to constraints. One important class of mechanical 
problems in which Lagrange's equations are particularly useful comprises 
systems which are subject to constraints. 



370 laghange's equations [chap. 9 

A rigid body is a good example of a system of particles subject to 
constraints. A constraint is a restriction on the freedom of motion of a 
system of particles in the form of a condition which must be satisfied by 
their coordinates, or by the allowed changes in their coordinates. For 
example, a very simple hypothetical rigid body would be a pair of par- 
ticles connected by a rigid weightless rod of length I. These particles are 
subject to a constraint which requires that they remain a distance I apart. 
In terms of their cartesian coordinates, the constraint is 

[(* 2 - *i) 2 + (2/2 - 2/i) 2 + (*a - zi) 2 ] 1/2 = 1 (9-66) 

If we use the coordinates X, Y, Z of the center of mass and spherical 
coordinates r, 6, <p to locate particle 2 with respect to particle 1 as origin, 
the constraint takes the simple form: 

r=l. (9-67) 

There are thus only five coordinates X, Y, Z, 6, <p left to determine. Each 
constraint which can be expressed in the form of an equation like (9-66) 
enables us to eliminate one of the coordinates by choosing coordinates in 
such a manner that one of them is held constant by the constraint. For a 
rigid body, the constraints require that the mutual distances of all pairs 
of particles remain constant. For a body containing N particles, there 
are %N(N — 1) pairs of particles. However, it is not hard to show that 
it is sufficient to specify the mutual distances of 3N — 6 pairs, UN > 3. 
Hence we can replace the 3N cartesian coordinates of the N particles by 
3iV — 6 mutual distances, 3 coordinates of the center of mass, and 3 coor- 
dinates describing the orientation of the body. Since the 3N — 6 mutual 
distances are all constant, the problem is reduced to one of finding the 
motion in terms of six coordinates. Another example of a system subject 
to a constraint is that of a bead sliding on a wire. The wire is situated 
along a certain curve in space, and the constraints require that the posi- 
tion of the bead lie on this curve. Since the coordinates of the points 
along a space curve satisfy two equations (e.g., the equations of two 
surfaces which intersect along the curve), there are two constraints, and 
we can locate the position of the bead by a single coordinate. (Can you 
suggest a suitable coordinate?) If the wire is moving, we have a moving 
constraint, and our single coordinate is relative to a moving system of 
reference. Constraints which can be expressed in the form of an equa- 
tion relating the coordinates are called holonomic. All the above examples 
involve holonomic constraints. 

Constraints may also be specified by a restriction on the velocities, 
rather than on the coordinates. For example, a cylinder of radius a, 
rolling and sliding down an inclined plane, with its axis always horizontal, 



9-4] 



SYSTEMS SUBJECT TO CONSTRAINTS 



371 





Fig. 9-3. A cylinder rolling down Fig. 9-4. A disk rolling on a hori- 
an incline. zontal plane. 

can be located by two coordinates s and 0, as in Fig. 9-3. The coordi- 
nate s measures the distance the cylinder has moved down the plane, 
and the coordinate 9 is the angle that a fixed radius in the cylinder has 
rotated from the radius to the point of contact with the plane. Now 
suppose that the cylinder is rolling without slipping. Then the velocities 
s and 6 must be related by the equation 



s = a&, 



which may also be written 



ds = a d$. 



This equation can be integrated : 



— ad 



C, 



(9-68) 
(9-69) 

(9-70) 



where C is a constant. This equation is of the same type as Eq. (9-66), 
and shows that the constraint is holonomic, although it was initially ex- 
pressed in terms of velocities. If a constraint on the velocities, like Eq. 
(9-68), can be integrated to give a relation between the coordinates, like 
Eq. (9-70), then the constraint is holonomic. There are systems, however, 
in which such equations of constraint cannot be integrated. An example 
is a disk of radius a rolling on a horizontal table, as in Fig. 9-4. For sim- 
plicity, we assume that the disk cannot tip over, and that the diameter 
which touches the table is always vertical. Four coordinates are required 
to specify the position of the disk. The coordinates x and y locate the 
point of contact on the plane; the angle <p determines the orientation of 
the plane of the disk relative to the z-axis; and the angle 6 is the angle 
between a radius fixed in the disk and the vertical. If we now require 
that the disk roll without slipping (it can also rotate about the vertical 
axis), this implies two equations of constraint. The velocity of the point 
of contact perpendicular to the plane of the disk must be zero: 



x sin <p + y cos tp = 0, 



(9-71) 



372 lagrange's equations [chap. 9 

and the velocity parallel to the plane of the disk must be 

x cos <p — y sin <p = ad. (9-72) 

It is not possible to integrate these equations to get two relations between 
the coordinates x, y, 8, <p. To see this, we note that by rolling the disk 
without slipping, and by rotating it about a vertical axis, we can bring 
the disk to any point x, y, with any angle <p between the plane of the disk 
and the x-axis, and with any point on the circumference of the disk in 
contact with the table, i.e., any angle 0. For if the disk is at any point 
x, y, and the desired point on the circumference is not in contact with the 
table, we may roll the disk around a circle whose circumference is of 
proper length, so that when it returns to x, y, the desired point will be in 
contact with the table. It may then be rotated to the desired angle <p. 
This shows that the four coordinates x, y, 6, <p are independent of one 
another, and there cannot be any relation between them. It must there- 
fore be impossible to integrate Eqs. (9-71) and (9-72), and consequently 
this is an example of a nonholonomic constraint. 

The number of independent ways in which a mechanical system can 
move without violating any constraints which may be imposed is called 
the number of degrees of freedom of the system. To be more precise, the 
number of degrees of freedom is the number of quantities which must be 
specified in order to determine the velocities of all particles in the system 
for any motion which does not violate the constraints. For example, a 
single particle moving in space has three degrees of freedom, but if it is 
constrained to move along a certain curve, it has only one. A system 
of N free particles has 3N degrees of freedom, a rigid body has 6 degrees 
of freedom (three translational and three rotational), and a rigid body 
constrained to rotate about an axis has one degree of freedom. The 
disk shown in Fig. 9-4 has four degrees of freedom if it is allowed to 
slip on the table, because we need then to specify x, y, 0, <p. But if the 
disk is required to roll without slipping, there are only two degrees of 
freedom, because if <i> and any one of the velocities x, y, 6 are given, the 
remaining two can be found from Eqs. (9-71) and (9-72). The disk is 
only free to roll, and to rotate about a vertical axis. For holonomic sys- 
tems, the number of degrees of freedom is equal to the minimum number 
of coordinates required to specify the configuration of the system when 
coordinates held constant by the constraints are eliminated. Nonholo- 
nomic constraints occur in some problems in which bodies roll without 
slipping, but they are not of very great importance in physics. We shall 
therefore restrict our attention to holonomic systems. 

For a holonomic system of N particles subject to c independent con- 
straints, we can express the constraints as c relations which must hold 



9-4] SYSTEMS SUBJECT TO CONSTRAINTS 373 

between the 3iV cartesian coordinates (including possibly the time if the 
constraints are changing with time) : 

hi(xi, 2/i, ... , zn) t) = a x , 

hzfri, 2/i, ■ ■ • , zn) t) = a 2 , (9 _ ?3 . 

h c {xi, 2/i, . . . ,z N ;t) = a c , 

where hi, . . . , h c are c specified f mictions. The number of degrees of 
freedom will be 

/ = 3JV - c. (9-74) 

As Eqs. (9-73) are independent, we may solve them for c of the SN car- 
tesian coordinates in terms of the other 3N — c coordinates and the 
constants oi, . . . , a c . Thus only ZN — c coordinates need be specified, 
and the remainder can be found from Eqs. (9-73) if the constants 
ai, . . . , a c are known. We may take as generalized coordinates these 
3iV" — c cartesian coordinates and the c quantities oi, . . . , a c defined by 
Eqs. (9-73), and held constant by the constraints. Or we may define 
3N — c generalized coordinates qi, . . . , q/, in any convenient way: 

?i = ?i(*i, 2/i, • • • , zn] t), 

q 2 = q 2 (xi, J/i, . . . ,z N ;t), , g 

Qf = Qf(xi, 2/i, ... , zn; t). 

Equations (9-73) and (9-75) define a set of 3iV coordinates qi, . . . , q/; 
ai, . . . , a c , and are analogous to Eqs. (9-1). They may be solved for the 
cartesian coordinates: 



(9-76) 



Now let Qi, . . . , Qf, Qf + i, . . . , Qf+ C be the generalized forces corre- 
sponding to the coordinates qi, . . . , gy; Oi, . . . , a c . We have then a set 
of Lagrange equations for the constrained coordinates and another for 
the unconstrained coordinates: 

,/, (9-77) 

..,c;c+f=3N. (9-78) 



*\ = xi(qi, ■ • 


• . Qf, Oi, • • 


. , a c ; t), 


2/i = 2/i(«?i, • ■ 


• , Q/; 0,1, . . 


. , a c ; t), 


zn — ZN(qi, ■ • 


. , q f ; a lt . . 


• , a c ; t). 



d dT 
dt dqk 


dT -O 

-dq~ k - Qk ' 


k = 1, . 


d dT 
dt ddj 


dT a 


3= h 



374 lagrange's equations [chap. 9 

The importance of this separation of the problem into two groups of 
equations is that the forces of constraint can be so chosen that they do 
no work unless the constraints are violated, as we shall show in the next 
paragraph. If this is true, then according to the definition (9-30) of the 
generalized force, the forces of constraint do not contribute to the gen- 
eralized force Qk associated with an unconstrained coordinate q^. Since 
the values of the constrained coordinates a\, . . . , a c are held constant, 
we can solve Eqs. (9-77) for the motion of the system in terms of the 
coordinates g 1( . . . , q/, treating oi, . . . , a c as given constants, without 
knowing the forces of constraint. This is a great advantage, for the 
forces of constraint depend upon how the system is moving, and cannot, 
in general, be determined until after the motion has been found. All 
we usually know about the constraining forces is that they have whatever 
values are required to maintain the constraints. Having solved Eqs. (9-77) 
for q x (<),..., q/(t), we may then, if we wish, substitute these functions 
in Eqs. (9-78) and calculate the forces of constraint. This may be a 
matter of considerable interest to the engineer who needs to verify that 
the constraining members are strong enough to withstand the constrain- 
ing forces. Lagrange's equations thus reduce the problem of finding the 
motion of any holonomic system with / degrees of freedom to the problem 
of solving / second-order differential equations (9-77). When we speak 
of the generalized coordinates, the constrained coordinates ai, . . . , a c 
may or may not be included, as convenient. 

If a bead slides on a frictionless wire, the wire can only exert constrain- 
ing forces perpendicular to itself, so that no work is done on the bead 
so long as it stays on the wire.* If there is friction, we can separate the 
force on the bead into a component perpendicular to the wire which 
holds the bead on the wire without doing any work, and a frictional 
component along the wire which does work and will therefore have to be 
included in the generalized force associated with motion along the wire. 
If the. frictional component depends on the perpendicular component, as 
it does for dry sliding friction, then we cannot solve Eqs. (9-77) first, 
independently of Eqs. (9-78), and one great advantage of the Lagrangian 
method is lost. If two particles are held a fixed distance apart by a rigid 
rod, then by Newton's third law, the force exerted by the rod on one 
particle is equal and opposite to that on the other. It was shown in 



* If the wire is moving, the force exerted by the wire may do work on the 
bead, but the virtual displacements in terms of which the generalized forces 
have been defined are to be imagined as taking place at a fixed instant of time, 
and for such a displacement which does not violate the constraints, no work 
is done. Hence even in the case of moving constraints, the constraining forces 
do not appear in the generalized forces associated with the unconstrained 
coordinates. 



9-5] 



EXAMPLES OF SYSTEMS SUBJECT TO CONSTRAINTS 



375 



Section 5-1 that no net work is done on the system by the rod so long as 
the constraint is not violated, that is, so long as the rod is not stretched 
or compressed. A similar situation will be found in all other cases; the 
constraints could always be maintained by forces which do no work. 

If the forces Qi, . . . , Qf are derivable from a potential energy func- 
tion, then we can define a Lagrangian function L(q l; . . . , q f ; q u . . . , q/) 
which may in some cases depend on t, and which may also depend on the 
constants a\, . . . , a c . The first / Lagrange equations (9-77) can then 
be written in the form 



A ik _ Ok = o 

dt dq k dq k ~ ' 



h= 1, ...,/. 



(9-79) 



9-5 Examples of systems subject to constraints. A simple mechanical 
system involving constraints is the Atwood's machine shown in Fig. 9-5. 
Weights mj, jre 2 are connected by a rope of length I over a fixed pulley. 
We assume the weights move only vertically, so that we have only one 
degree of freedom. We take as coordinates the distance x of m\ below 
the pulley axle, and I, the length of the rope. The coordinate I is con- 
strained to have a constant value, and could be left out of consideration 
from the start if we wish only to find the motion. If we also want to 
find the tension in the string, we must include I as a coordinate. The 
kinetic energy is 

T = %m x x 2 + %m 2 (i - x) 2 . (9-80) 

The only forces acting on mi and m 2 are the tension t in the rope and the 
force of gravity. The work done when x increases by Sx, I remaining 




\m,2 

Fig. 9-5. Atwood's machine. 



376 lagrange's equations [chap. 9 

constant, is 

SW = (mig — t) 8x — (m 2 g — r) 8x 

= (mi — m 2 )g Sx = Q x Sx, (9-81) 

so that 

Q x = (mi - m a )g. (9-82) 

Note that Q x is independent of t. The work done when I increases by 
81, x remaining constant, is 

SW = {m 2 g - t) 81 = Qi 81, (9-83) 

so that 

Qi = m 2 g - t. (9-84) 

Notice that in order to obtain an equation involving the force of con- 
straint r, we must consider a motion which violates the constraint. This 
is also true if we wish to measure a force physically; we must allow at 
least a small motion in the direction of the force. The Lagrange equations 
of motion are (since I = l = 0) 

i\w~^ = (mi + m2)£ = (mi _ mz)9 > (9_85) 

d (dT\ dT .. , n ... 

It \a) ~ aT = ~ m2X = m * g ~ T - (9 " 86) 

The first equation is to be solved to find the motion: 

x = Xo + Vot + $ Wl ~^ 2 gt\ (9-87) 

mi -f- m 2 

The second equation can then be used to find the tension t necessary to 

maintain the constraint : 

, . .., 2mim 2 , , 

r = m 2 (g + x) = ^^^ g. (9-88) 

In this case the tension is independent of time and can be found from 
Eqs. (9-85) and (9-86) immediately, although in most cases the con- 
straining forces depend on the motion and can be determined only after 
the motion is found. Equations (9-85) and (9-86) have an obvious phys- 
ical interpretation and could be written down immediately from elemen- 
tary considerations, as was done in Section 1-7. 

A problem of little practical importance, but which is quite instructive, 
is that in which one cylinder rolls upon another, as shown in Fig. 9-6. 
The cylinder of radius a is fixed, and the cylinder of radius aa rolls around 
it under the action of gravity. Suppose we are given that the coefficient 



9-5] 



EXAMPLES OF SYSTEMS SUBJECT TO CONSTRAINTS 



377 




Fig. 9-6. One cylinder rolling on another. 



of static friction between the cylinders is n, the coefficient of sliding 
friction is zero,* and that the moving cylinder starts from rest with its 
center vertically above the center of the fixed cylinder. We shall assume 
that the axis of the moving cylinder remains horizontal during the motion. 
It is advisable in all problems, and essential in this one, to think carefully 
about the motion before attempting to find the mathematical solution. 
It is clear that the moving cylinder cannot roll all the way around the 
fixed cylinder, for the normal force F which is exerted by the fixed cyl- 
inder on the moving one can only be directed outward, never inward. 
Therefore at some point, the moving cylinder will fly off the fixed one. 
The point at which it flies off is the point at which 



F = 0. 



(9-89) 



Furthermore, the cylinder cannot continue to roll without slipping right 
up to the point at which it flies off, for the frictional force f which pre- 
vents slipping is limited by the condition 



f<tf, 



(9-90) 



and will certainly become too small to prevent slipping before the point 
at which Eq. (9-89) holds. The motion therefore is divided into three 
parts. At first the cylinder rolls without slipping through an angle t 
determined by the condition 

/ = nF. (9-91) 



* This implies that the moving cylinder either rolls without slipping, if the 
static friction is great enough, or slips without any friction at all. The latter 
assumption is made to simplify the problem. 



378 lagrange's equations [chap. 9 

Beyond the angle 0i, the cylinder slides without friction until it reaches 
the angle 2 determined .Jby Eq. (9-89), after which it leaves the fixed 
cylinder and falls freely. We may anticipate some mathematical difficulties 
with the initial part of the motion due to the fact that the initial posi- 
tion of the moving cylinder is one of unstable equilibrium. Physically 
there is no difficulty, since the slightest disturbance will cause the cylinder 
to roll down, but mathematically there may be a difficulty which we must 
watch out for, inasmuch as the needed slight disturbance will not appear 
in the equations. 

Let us find that part of the motion when the moving cylinder rolls with- 
out slipping. There is then only one degree of freedom, and we shall 
specify the position of the cylinder by the angle between the vertical and 
the line connecting the centers of the two cylinders. In order to compute 
the kinetic energy, we introduce the auxiliary angle <p through which the 
moving cylinder has rotated about its axis. The condition that the cylin- 
der roll without slipping leads to the equation of constraint: 

ad = aa(<p — 0), (9-92) 

which can be integrated in the form 

(1 + a)6 = a<p. (9-93) 

If we were concerned only with the rolling motion, we could now proceed 
to set up the Lagrange equation for 6, but inasmuch as we need to know 
the forces of constraint F and f, it is necessary to introduce additional co- 
ordinates which are maintained constant by these constraining forces. 
The frictional force f maintains the constraint (9-93), and an appropriate 
coordinate is 

7=0- -25L. . (9-94) 

So long as the cylinder rolls without slipping, 7 = 0; 7 measures the angle 
of slip around the fixed cylinder. The normal force F maintains the dis- 
tance r between the centers of the cylinders: 

r = a + aa= (1 + a)a. (9-95) 

The kinetic energy of the rolling cylinder is the energy associated with the 
motion of its center of mass plus the rotational energy about the center of 
mass: 

T = %m(r 2 + rH 2 ) + %I<p 2 . (9-96) 

After substituting <p from Eq. (9-94), and since I = %ma 2 a 2 , for a solid 
cylinder of radius aa, we have 

T = imr 2 + imr 2 6 2 + |m(l + a) 2 a 2 (6 2 - 270 + 7 2 ). (9-97) 



9-5] EXAMPLES OF SYSTEMS SUBJECT TO CONSTRAINTS 379 

The equations of constraint [Eq. (9-95) and 7=0] must not be used until 
after the equations of motion are written down. The generalized forces 
are most easily determined with the help of Eq. (9-30); they are* 

Q, = mgr sin 6, (9-98) 

Qy = -fad + ex), (9-99) 

Q r = F — mg cos 0. (9-100) 

The Lagrange equations for 6, 7, and r are now 

m[r 2 + \a?(l + a) 2 ]S + 2mrf6 — - Jma 2 (l + a) 2 7 = mgr sin 6, (9-101) 

-Jma 2 (l + a) 2 S + ima 2 (l + a) 2 y = ~fa(l + a), (9-102) 

mr — mrd 2 = F — mg cos 0. (9-103) 

We can now insert the constraints 7=0 and r = (1 + a)a, so that these 
equations become 

f (1 + a) 2 ma 2 = (1 + a)mga sin d, (9-104) 

/ = i(l + a)ma§, (9-105) 

F = mg cosd — (1 + a)ma6 2 . (9-106) 

Had we ignored the terms involving y in the kinetic energy, the 6 equation, 
which determines the motion, would have come out correctly, but the 
equation for the constraining force f would have been missing a term. 
This happens when the constrained coordinates are not orthogonal to the 
unconstrained coordinates, since a cross term (70) then appears in the 
kinetic energy. 

The equation of motion (9-104) can be solved by the energy method. 
The total energy, so long as the cylinder rolls without slipping, is 

4(1 + a) 2 maH 2 + (1 + a)mga cos B = E, (9-107) 

and is constant, as can easily be shown from Eq. (9-104), and as we 
know anyway since the gravitational force is conservative and the forces 
of constraint do no work. Since the moving cylinder starts from rest 
at 6 = 0, 

E = (1 + a)mga. (9-108) 



! The reader will find it an instructive exercise to verify these formulas. 



380 lagrange's equations [chap. 9 

We substitute this in Eq. (9-107) and solve for 0: 

6 = 2 (^ J'' 2 sin | , (9-109) 



where 



^mr^y (9 - 110) 



We can now integrate to find 6(t): 

f A* = (*L) 1,a f &, ( 9-iii) 

Jo sin 6/2 \a/ Jo ' v 

[ toto ll -(?)""'■ (9 - 112) 

When we substitute the lower limit = 0, we run into a difficulty, for 
In = — oo ! This is the expected difficulty due to the fact that = 
is a point of equilibrium, albeit unstable. If there is no disturbance what- 
ever, it will take an infinite time for the cylinder to roll off the equilibrium 
point. Let us suppose, however, that it does roll off due to some slight 
disturbance, and let us take the time t = as the time when the angle 
has some small value O . There is now no difficulty, and we have 

tan | = (tan ^) exp [(f) 1 '' *] ■ (9-113) 

As t — > oo, — > 2ir, and the moving cylinder rolls all the way around 
the fixed one, if the constraints continue to hold. The rolling constraint 
holds, however, only so long as Eq. (9-90) holds. When we substitute 
from Eqs. (9-105), (9-106), and (9-109), Eq. (9-90) becomes 

%mg sin < ifimg(7 cos — 4). (9-114) 

At 6 = 0, this certainly holds, so that the cylinder does initially roll, 
as we have supposed. At 8 = 7r/2, however, it certainly does not hold, 
since the left member is then positive and the right, negative. The angle 0i 
at which slipping begins is determined by the equation 

sin 0i = n(7 cos 0i — 4), (9-115) 

whose solution is 

™=a 28m 2 + [1 + 33 M 2 ] 1/2 • , Q 11ft , 

C0S 01 = 1 + 49m 2 ( } 



9-6] CONSTANTS OF THE MOTION AND IGNOBABLE COOBDINATES 381 

The second part of the motion, during which the moving cylinder slides 
without friction around the fixed one, can be found by solving Eqs. (9-101) 
and (9-102) for 6(t), 7(0, with / = and with only the single constraint 
r = (1 + <x)a, and with initial values 6 = 6 U = 1; determined from 
Eqs. (9-116) and (9-109). The solution can be found without essential 
difficulty, and the angle 2 at which the moving cylinder leaves the fixed 
one can then be determined from Eqs. (9-106) and (9-89). These calcu- 
lations are left to the reader. 

9-6 Constants of the motion and ignorable coordinates. We remarked 
in Chapter 3 that one general method for solving dynamical problems is 
to look for constants of the motion, that is, functions of the coordinates 
and velocities which are constant in time. One common case in which 
such constants can be found arises when the dynamical system is charac- 
terized by a Lagrangian function in which some coordinate ft does not 
occur explicitly. The corresponding Lagrange equation (9-57) then re- 
duces to 

d 

dt \dft> 



(i)=°- <•"»") 



This equation can be integrated immediately: 

T~r- = Pk = a constant. (9-118) 

"ft 

Thus, whenever a coordinate ft does not occur explicitly in the Lagrangian 
function, the corresponding momentum pu is a constant of the motion. 
Such a coordinate ft is said to be ignorable. If ft is ignorable, we can 
solve Eq. (9-118) for ft in terms of the other coordinates and velocities, 
and of the constant momentum p k , and substitute in the remaining 
Lagrange equations to eliminate ft and reduce by one the number of vari- 
ables in the problem; (ft was already missing from the equations, since 
it was assumed ignorable.) When the remaining variables have been 
found, they can be substituted in Eq. (9-118), to give ft as a function 
of t; ft is then obtained by integration. If all but one of the coordinates 
are ignorable, the problem can thus be reduced to a one-dimensional 
problem and solved by the energy integral method, if L does not depend 
on the time t explicitly. 

For example, in the case of central forces, the potential energy depends 
only on the distance r from the origin, so that if we use polar coordinates 
r, in a plane, V is independent of 0. Since T is also independent of 
according to Eq. (9-14) (T depends of course on 6), we will have 

fe = h < r - y ) = °« < 9 - 119 > 



382 lagrange's equations [chap. 9 

and hence 

= mr 6 = p 9 = a constant, (9-120) 



86 

a result which we obtained in Section 3-13 by a different argument. We 
see that the constancy of p e is a result of the fact that the system is sym- 
metrical about the origin, so that L cannot depend on 8. If a system of 
particles is acted on by no external forces, then if we displace the whole 
system in any direction, without changing the velocities and relative 
positions of the particles, there will be no change in T or V, or in L. If 
X, Y, and Z are rectangular coordinates of the center of mass, and if 
the remaining coordinates are relative to the center of mass, so that 
changing X corresponds to displacing the whole system, then 

H = 0, (9-121) 

and therefore px, the total linear momentum in the x-direction, will be 
constant, a result we proved in Section 4-1 by a different method. 

It is of interest to see how to show from Lagrange's equations that the 
total energy is a constant of the motion. In order to find an energy inte- 
gral of the equations of motion in Lagrangian form, it is necessary to 
know how to express the total energy in terms of the Lagrangian func- 
tion L. To this end, let us consider a system described in terms of a fixed 
system of coordinates, so that the kinetic energy T is a homogeneous 
quadratic function of the generalized velocities <ji, . . . , <j/ [i.e., T x = 
T = in Eq. (9-13)]. By Euler's theorem,* we have 

t & g = 2f. (9-122) 

k=l * K 

Thus if 

L = T 2 - V, (9-123) 

where V is a function of the coordinates q lt . . . , q/ alone, then, by Eq. 
(9-122), 

1 £i h £-L=T+V = E. (9-124) 

We now consider the time derivative of the left member of Eq. (9-124). 
For greater generality, we shall at first allow L to depend explicitly on t. 



* W. F. Osgood, Advanced Calculus. New York: Macmillan, 1937. (Page 121.) 
The reader unfamiliar with Euler's theorem can readily verify Eq. (9-122) for 
himself by substituting for T = T% from Eq. (9-9). 



9-6] CONSTANTS OF THE MOTION AND IGNOKABLE COORDINATES 383 

In the case we have considered, L does not depend explicitly on t. There 
are cases, however, when a system is subject to external forces that change 
with time and that can be derived from a potential V that varies with 
time. An example would be an atom subject to a varying external electric 
field. In such cases, the equations of motion can be written in the La- 
grangian form (9-57) with the Lagrangian depending explicitly on the 
time t. In the case of moving coordinate systems also, the Lagrangian 
may depend on the time even though the forces are conservative. The 
time derivative of the left member of Eq. (9-124) is 

d ( <A . dL T \ <AL dL . . d/dL\ dL . dL .. 1 dL 
^A . \d (dL\ dL\ dL dL , n 10rN 

If L does not depend explicitly on t, the right side of Eq. (9-125) is zero, 
and 

^2 § k ~p L = & constant. (9-126) 

When L has the form (T 2 — V), as in a stationary coordinate system, 
this is the conservation of energy theorem. Regardless of the form of L, 
Eq. (9-126) represents an integral of Lagrange's equations (9-57), when- 
ever L does not contain t explicitly, but the constant quantity on the 
left is not always the total energy. Note the analogy between the con- 
servation of generalized momentum p k when L is independent of qt, and 
the conservation of energy when L is independent of t. There are many 
ways in which the relation between time and energy is analogous to the 
relation between a coordinate and the corresponding momentum. 

We have seen that the familiar conservation laws of energy, momentum, 
and angular momentum can be regarded as consequences of symmetries 
exhibited by the mechanical systems to which they apply; that is, they are 
consequences of the fact that the Lagrangian function L, which determines 
the equations of motion, is independent of time and of the position and 
orientation of the entire system in space. This result, derived here for 
classical mechanics, holds generally throughout physics. In quantum 
mechanics and in relativity theory, even when we include electromagnetic 
and other kinds of force fields, conservation laws are associated with sym- 
metries in the fundamental equations. We might, for example, define 
energy as that quantity which is constant because the laws of physics are 
always the same (if indeed they are!). 



384 lagrange's equations [chap. 9 

9-7 Further examples. The spherical pendulum is a simple pendulum 
free to swing through the entire solid angle about a point. The pendulum 
bob is constrained to move on a spherical surface of radius R. We locate 
the bob by the spherical coordinates 8, <p (Fig. 9-7). We may include 
the length R of the pendulum as a coordinate if we wish to find the ten- 
sion in the string, but we omit it here, as we are concerned only with 
finding the motion. If the bob swings above the horizontal, we will sup- 
pose that it still remains on the sphere, which would be true if the string 
were replaced by a rigid rod. Otherwise the constraint disappears when- 
ever a compressional stress is required to maintain it, since a string will 
support only a tension and not a compression. The velocity of the bob is 

v = Rd\ + R sin 8 <pm. (9-127) 

Hence the kinetic energy is 

T = \mv 2 = %mR 2 8 2 + %mR 2 sin 2 <p 2 . (9-128) 

The potential energy due to gravity, relative to the horizontal plane, is 

V = mgR cos 0. (9-129) 

Hence the Lagrangian function is 

L = T — V = %mR 2 8 2 + %mR 2 sin 2 <p 2 — mgR cos 0. (9-130) 

The Lagrange equations are 

j 

(mR 2 6) — mR 2 <p 2 sin 6 cos 6 — mgR sin 8=0, (9-131) 



dt 

dt 



^ (mR 2 sm 2 d<p) = 0. (9-132) 



The coordinate <p is ignorable, and the second equation can be integrated 
immediately: 

mR 2 sin 2 <p = p v = a constant. (9-133) 

Also, since 

^ = 0, (9-134) 

the quantity 

*?* + *?£•-■&= %mR 2 6 2 + %mR 2 sin 2 <p 2 + mgR cos 8 (9-135) 
60 o<p 

is constant, by Eq. (9-126). We recognize the quantity on the right as 
the total energy, as it should be, since we are using a fixed coordinate 



9-7] 



FURTHER EXAMPLES 



385 




MgR 



-MgR 




Fig. 9-7. A spherical pendulum. 



Fig. 9-8. Effective potential 'V'(fi) 
for spherical pendulum. 



system. Calling this constant E, and substituting for <p from Eq. (9-133), 
we have 



vl 



hnR 2 r + J*. 2Q +. mgR cos = E. 
2m R 2 sin 2 6 

We may introduce an effective potential 'V'(6) for the motion: 

„2 



'V'(d) = mg R cos + 



so that 



2mR 2 sin 2 
%mR 2 2 = E — l V'(d). 



(9-136) 

(9-137) 
(9-138) 



Since the left member cannot be negative, the motion is confined to those 
values of for which 'V'(6) < E. The effective potential 'V'(O) is plotted 
in Fig. 9-8. We see that for p v = 0, 'V'(0) is the potential curve for a 
simple pendulum, with a minimum at = tt and a maximum at = 0. 
For E = —mgR, the pendulum is at rest at 6 = ir. For mgR > E > 
—mgR, the pendulum oscillates about = ir. For E > mgR, the pendulum 
swings in a circular motion through the top and bottom points 6 = 
and ir. When p v 5* 0, the motion is no longer that of a simple pendulum, 
and 'V'(6) now has a minimum at a point 6 between tc/2 and ir, and 
rises to infinity at = and = ir. The larger p v , the larger the mini- 
mum value of 'F(0), and the closer O is to w/2. If E = 'V'(6 ), then 
is constant and equal to O , and the pendulum swings in a circle about 
the vertical axis. As p r — > oo, the pendulum swings more and more 
nearly in a horizontal plane. For E > 'V'(0 O ), oscillates between a 
maximum and minimum value while the pendulum swings about the 
vertical axis. The reader should compare these results with his mechan- 
ical intuitions or his experience regarding the motion of a spherical 



386 lagranue's equations [chap. 9 

pendulum. The solution of Eq. (9-138) for 0(0 cannot be carried out 
in terms of elementary functions, but we can treat circular and nearly 
circular motions very easily. The relation between p f and O for uniform 
circular motion of the pendulum about the z-axis is 

[d'V'~\ D ■ a vl cos O , „ Q v 

— — = —mgR sin O 1 . , „ = 0. (9-139) 

L dd ie v mR 2 sin 8 O 

It is evident from this equation that O > ir/2, and that O — > tt/2 as 
p v — > oo. By substituting from Eq. (9-133), we obtain a relation be- 
tween tp and O for uniform circular motion: 

Z? = .£ ?_^ . (9-140) 

^ R (— cos0 o ) 

The energy for uniform circular motion at an angle O , if we use Eqs. 
(9-136) and (9-139), and the fact that = 0, is 

mgR / 2-3sin 2 0<A . } 

2 \ cos O / 

For an energy slightly larger than E , and an angular momentum p v 
given by Eq. (9-139), the angle will perform simple harmonic oscilla- 
tions about the value O . For if we set 



k 



\d 2 'V> 
I d02 . 



mgi? (1 + 3 cos 2 O ), (9-142) 



Je —cos O 

then, for small values of — O , we can expand 'V'{B) in a Taylor series: 

'F(0) = E + Jfc(0 - O ) 2 . (9-143) 

The energy equation (9-138) now becomes 

%mRH 2 + 4fc(tf - O ) 2 = E - E Q . (9-144) 

This is the energy for a harmonic oscillator with energy E — E , coordi- 
nate — O) mass mR 2 , spring constant k. The frequency of oscillation 
in is therefore given by 

2 k g 1 + 3 cos 2 0p , s 

W = m ~R? = R -cos0 o * (9 ~ M5) 

This oscillation in is superposed upon a circular motion around the 
z-axis with an angular velocity <p given by Eq. (9-133); <p will vary 
slightly as 6 oscillates, but will remain very nearly equal to the constant 
value given by Eq. (9-140). It is of interest to compare <j> and w: 



9-7] 



FURTHER EXAMPLES 



387 



■ 2 



1 



1 + 3 C0S2 6 



(9-146) 



Since 6 > t/2, this ratio is less than 1, so that co > <j>, and the pendulum 
wobbles up and down as it goes around the circle. At O = tt/2, <p = w, 
and the pendulum moves in a circle whose plane is tilted slightly from the 
horizontal; this case occurs only in the limit of very large values of p v . It 
is clear physically that when p v is so large that gravity may be neglected, 
the motion can be a circle in any plane through the origin. Can you show 
this mathematically? Near o = 0, w = 2<p, so that 6 oscillates twice per 
revolution and the pendulum bob moves in an ellipse whose center is on 
the z-axis. This corresponds to the motion of the two-dimensional har- 
monic oscillator discussed in Section 3-10, with equal frequencies in the 
two perpendicular directions. 

As a last example, we consider a system in which there are moving 
constraints. A bead of mass m slides without friction on a circular hoop 
of radius a. The hoop lies in a vertical plane which is constrained to 
rotate about a vertical diameter with constant angular velocity w. There 
is just one degree of freedom, and inasmuch as we are not interested in 
the forces of constraint, we choose a single coordinate 8 which measures 
the angle around the circle from the bottom of the vertical diameter to 
the bead (Fig. 9-9). The kinetic energy is then 



T = ±ma 2 d 2 + frnaW sin 2 0, 



and the potential energy is 



V = —mga cos 6. 



The Lagrangian function is 



%ma 2 6~ 2 + Jmo 2 w 2 sin 2 6 + mga cos 0. 



(9-147) 



(9-148) 



(9-149) 




'1 


/> 








+mga- 

















t/2 y^ / 






a) < a*,* 




7T 


— mga- 






V*GJ > (tic 





Fig. 9-9. A bead sliding on a rota- Fig. 9-10. Effective potential en- 
ting hoop. ergy for system shown in Fig. 9-9. 



388 lagrange's equations [chap. 9 

The Lagrange equation of motion can easily be written out, but this is 
unnecessary, for we notice that 

at u ' 

and therefore, by Eq. (9-126), the quantity 

0~- L= fynaH 2 - 4mo V sin 2 6 - mga cos = 'E' (9-150) 
do 

is constant. The constant 'E' is not the total energy T + V, for the 
middle term has the wrong sign. The total energy is evidently not con- 
stant in this case. (What force does the work which produces changes 
in T + V?) We may note, however, that we can interpret Eq. (9-149) 
as a Lagrangian function in terms of a fixed coordinate system with the 
middle term regarded as part of an effective potential energy: 

<y(0) = —^ma 2 u 2 s in 2 6 — mga cos 0. (9-151) 

The energy according to this interpretation is 'E'. The first term in 'V'(d) 
is the potential energy associated with the centrifugal force which must 
be added if we regard the rotating system as fixed. The effective poten- 
tial is plotted in Fig. 9-10. The shape of the potential curve depends on 
whether w is greater or less than a critical angular velocity 

w c = (g/a) 1 ' 2 . (9-152) 

It is left to the reader to show this, and to discuss the nature of the motion 
of the bead in the two cases. 

9-8 Electromagnetic forces and velocity-dependent potentials. If the 

forces acting on a dynamical system depend upon the velocities, it may 
be possible to find a function U(qi, . . . , q/; qi, . . . , q/; t) such that 

*-!*-«' s=1 '■ (M53) 

If such a function U can be found, then we can define a Lagrangian 
function 

L= T - U, (9-154) 

so that the equations of motion (9-53) can be written in the form (9-57) : 



9-8] ELECTROMAGNETIC FORCES, VELOCITY- DEPENDENT POTENTIALS 389 

The function U may be called a velocity-dependent potential. If there are 
also forces derivable from an ordinary potential energy V(q u . . . , q/), V 
may be included in U, since Eq. (9-153) reduces to Eq. (9-33) for those 
terms which do not contain the velocities. The function U may depend 
explicitly on the time t. If it does not, and if the coordinate system is a 
fixed one, then L will be independent of t, and the quantity 

E = Z & f| - L > (9-156) 

will be a constant of the motion, according to Eq. (9-126). In this case, 
we may say that the forces are conservative even though they depend on 
the velocities. It is clear from this result that it cannot be possible to 
express frictional forces in the form (9-153), for the total energy is not 
constant when there is friction unless we include heat energy, and heat 
energy cannot be defined in terms of the coordinates and velocities 
Qu ■ ■ ■ > Qf', Qi, ■ ■ ■ , if, and hence cannot be included in Eq. (9-156). 
It is not hard to show that if the velocity-dependent parts of U are linear 
in the velocities, as they are in all important examples, the energy E 
defined by Eq. (9-156) is just T + V, where V is the ordinary potential 
energy and contains the terms in U that are independent of the velocities. 
As an example, a particle of charge q subject to a constant magnetic 
field B is acted on by a force (guassian units) 

F = I v X B, (9-157) 



c 



or 



F* = ? (SB. ~ ZB V ), 



c 



F v = l (zB x - *B.), (9-158) 



c 



F z = I (xB v - yB x ). 



Equations (9-158) have the form (9-153) if 

U = | (zyB x + xzBy + yxB,). (9-159) 

It is, in fact, possible to express the electromagnetic force in the form 
(9-153) for any electric and magnetic field. The electromagnetic force 
on a particle of charge q is given by Eq. (3-283): 

F = gE + ^ v X B. (9-160) 

c 



390 lagrange's equations [chap. 9 

It is shown in electromagnetic theory* that for any electromagnetic field, 
it is possible to define a scalar function 4>{x, y, z, t) and a vector function 
A(x, y, z, t) such that 

E= _ V 4,_i^, (9-161) 

c at 

B = V X A. (9-162) 

The function <j> is called the scalar potential, and A is called the vector 
potential. If these expressions are substituted in Eq. (9-160), we obtain 

F = _ gT0 _ 1 d A + 2 v x ( V x A). (9-163) 

c at c 

The last term can be rewritten using formula (3-35) for the triple cross 
product : 

F=-gV*-^-^vVA + ? V(v-A). (9-164) 

c di c c 

[The components of v are (x, y, z) and are independent of x, y, z, so that v 
is not differentiated by the operator V.] The two middle terms can be 
combined according to Eq. (8-113): 

F= -qV4>- g ^ + g V(vA), (9-165) 

c at c 

where dk/dt is the time derivative of A evaluated at the position of the 
moving particle. It may now be verified by direct computation that the 
potential function 

U = q4> - ^ v-A, (9-166) 

when substituted in Eqs. (9-153), with q u q 2 , qz — x, y, z, yields the 
components of the force F given by Eq. (9-165). It is also easy to show 
that the energy E denned by Eq. (9-156) with L = T — U is 

E = T + q<j>. (9-167) 

If A and <£ are independent of t, then L is independent of t in a fixed 
coordinate system and the energy E is constant, a result derived by more 
elementary methods in Section 3-17 [Eq. (3-288)]. 

When there is a velocity dependent potential, it is customary to define 
the momentum in terms of the Lagrangian function, rather than in terms 



* See, e.g., Slater and Frank, Electromagnetism. New York: McGraw-Hill 
Book Co., 1947. (Page 87.) 



Px = 


mx -f- - A x , 
c 


Py = 


m v + f A v> 


Pz = 


mz + - A z . 
c 



9-9] lagrange's equations for the vibrating string 391 

of the kinetic energy: 

If the potential is not velocity dependent, then this definition is equivalent 
to Eq. (9-23). In any case, it is dL/dq^ whose time derivative occurs in 
the Lagrange equation for q k , and which is constant if q k is ignorable. 
In the case of a particle subject to electromagnetic forces, the momentum 
components p x , p v , p z will be, by Eqs. (9-168) and (9-166), 



(9-169) 



The second terms play the role of a potential momentum. 

It appears that gravitational forces, electromagnetic forces, and indeed 
all the fundamental forces in physics can be expressed in the form (9-153), 
for a suitably chosen potential function U. (Frictional forces we do not 
regard as fundamental in this sense, because they are ultimately reducible 
to electromagnetic forces between atoms, and hence are in principle also 
expressible in the form (9-153) if we include all the coordinates of the 
atoms and molecules of which a physical system is composed.) Therefore 
the equations of motion of any system of particles can always be expressed 
in the Lagrangian form (9-155), even when velocity dependent forces are 
present. It appears that there is something fundamental about the form 
of Eqs. (9-155). One important property of these equations, as we have 
already noted, is that they retain the same form if we substitute any new 
set of coordinates for q lt . . . , q f . This can be verified by a straightforward, 
if somewhat tedious, calculation. Further insight into the fundamental 
character of the Lagrange equations must await the study of a more ad- 
vanced formulation of mechanics utilizing the calculus of variations, which 
is beyond the scope of this book.* 

9-9 Lagrange's equations for the vibrating string. The Lagrange 
method can be extended also to the motion of continuous media. We shall 
consider only the simplest example, the vibrating string. Using the nota- 
tion of Section 8-1, we could take u{x) as a set of generalized coordinates 
analogous to q k . In place of the subscript k denoting the various degrees 
of freedom, we have the position coordinate x denoting the various points 



* See, e.g., H. Goldstein, Classical Mechanics. Reading, Mass.: Addison- 
Wesley, 1950. (Chapter 2.) 



392 lagbange's equations [chap. 9 

on the string. The number of degrees of freedom is infinite for an ideal 
continuous string. The generalization of the Lagrange method to deal 
with a continuous index x denoting the various degrees of freedom intro- 
duces mathematical complications which we wish to avoid here.* There- 
fore we make use of the possibility of representing the function u(x) as a 
Fourier series. 

According to the Fourier series theorem quoted in Section 8-2, if the 
string is tied at the ends x = 0, I, we can represent its position u(x) by 
the series (8-24) : 

«(*) = f>sin*P- (9-170) 



The coefficients qu are given by Eq. (8-25) 

.1 
2 / 



= 7 / u(x) sin ^ dx, k = 1, 2, 3, . . . . (9-171) 

l Jo i 



Since the coefficients g* give a complete description of the position of the 
string, they represent a suitable set of generalized coordinates. When the 
string vibrates, the coordinates gj. become functions of t: 

u(x,t) = f>*(«)sin^- (9-172) 

We have still an infinite number of coordinates q^, but they depend on the 
discrete subscript k and can be treated exactly like the generalized coordi- 
nates considered earlier in this chapter. Since the string could in principle 
be treated as a system with a very large number of particles, and since we 
are allowed to describe the system by any suitable set of generalized co- 
ordinates, we need only express the Lagrangian function in terms of the 
coordinates qk in order to write down the equations of motion. 
We first need to calculate the kinetic energy, which is evidently 

r = /o**(l) V (9 " 173) 

If we differentiate Eq. (9-172) with respect to t and square, we obtain 

©^ggM^in^sin^. (9-174) 

* For a treatment of this problem, see H. Goldstein, op. cit. (Chapter 11.) 



9-9] lagrange's equations for the vibrating string 393 

We now multiply by %<r dx and integrate from to I term by term.* Since 



/. 



'0 

the result obtained is 



farx . jirx , \U, j = k, 
i —j— sin J -r r - dx = \ 

0, j * k, 



sin — j— sin —f- ax = {- - ' / Q 17 r-\ 

I I In i ■ ^ h (9-175) 



T = J2 && (9-176) 



4=1 



We next calculate the generalized force Q k . If coordinate q k increases by 
Sq k , while the rest are held fixed, a point x on the string moves up a dis- 
tance given by Eq. (9-170) : 

8u = Sq k sin ^ • (9-177) 

The upward force on an element dx of string is given by Eq. (8-3). The 
work done is therefore 



Jo dx\ dx/ 



w = QkS «» = J irxVix)* udx - c 9 - 178 ) 

We substitute for du/dx from Eq. (9-170), and for Su from Eq. (9-177), 
and integrate term by term, to obtain (assuming t constant) : 

Qk = -ih (jrf qk. (9-179) 

The forces Q k are obviously derivable from the potential energy function, 

V=±ilr(^f) 2 ql (9-180) 

It will be instructive to calculate V directly by calculating the work done 
against the tension t in moving the string from its equilibrium position to 
the position u(x). At the same time we shall verify that this work is in- 
dependent of how we move the string to the position u(x). Let u(x, t) be 
the position of the string at any time t while the string is being moved to 



* In order to differentiate and integrate infinite series term by term, and to 
rearrange orders of summation, as we shall do freely in this section, we must 
require that the series all converge uniformly. This will be the case if u{x, t) 
and its derivatives are continuous functions. (For a precise statement and 
derivation of the conditions for manipulation of infinite series, see a text on 
advanced calculus, e.g., W. Kaplan, Advanced Calculus. Reading, Mass.: 
Addison- Wesley, 1952. Chapter 6.) 



394 lagkange's equations [chap. 9 

u(x). [The function u(x, t) is not necessarily a solution of the equation of 
motion, since we wish to consider an arbitrary manner of moving the string 
from u = to u = u(x).] At t = 0, the string is in its equilibrium 
position: 

u(x, 0) = 0. (9-181) 

Let t = ti be the time the string arrives at its final position: 

u(x, tj) = u(x). (9-182) 

The work done against the vertical components of tension [Eq. (8-3)] 
during the interval dt is 

f d_ ( du\ (du \ dx 

~~ Jx=o dx \ dx/ \dt ) 

We integrate by parts, remembering that u and du/dt are at x — 0, I: 

fl 2 

,„ / du du , ,. 

dV = I t— -rr^- dx dt 
J x =o dxotax 



The total work done is then 



dV 



where in the last expression, u = u{x) corresponds to the final position of 
the string. The result depends only on the final position of the string — an 
independent proof that the tension forces are conservative. 

The work done against the tension is stored as potential energy in the 
stretched string. By substituting in Eq. (9-184) from Eq. (9-170), we 
again can obtain Eq. (9-180). In Eq. (8-61) for a string of particles, the 
right member contains two terms that represent the vertical components 
of force between adjacent pairs of particles. A third way of deriving the 
potential energy is to find the potential energy function between a pair of 



9-9] lageange's equations for the vibrating string 395 

particles which yields this force. It must then be shown that, when this 
is summed over all pairs of adjacent particles, the result approaches Eq. 
(9-184) in the limit h -> 0. 
The Lagrangian function for the vibrating string can now be written as 

L = T - V = £ [fat* ~ & (t) 2 «*] • ( 9 - 185 ) 

The resulting Lagrange equation for q k is 

ilaqk + ¥t (jr) qk = 0, (9-186) 

whose general solution is 

qk = A k cos u k t + B k sin u k t, (9-187) 

-tG)""-* 

This result can be substituted in Eq. (9-172) to obtain the solution 

u(x, t) — 2_j \A k sin —j- cos u k t + B k sin -y- sin u k t) > (9-189) 

which is in agreement with Eq. (8-23). If u = u {x) and du/dt = v (x) 
are given at t = 0, we can use Eqs. (9-171) and (9-187) to find the 
constants A k , B k : 



where 



2 / kirx 

A k = q k (0) = j u o(x) sin— —dx, 



I 



D q k (0) 2 / . . . kwx , 
B k = 2-^— = — : / v {x) sm—r-dx, 

Wj; U) k l JO I 



(9-190) 



in agreement with Eqs. (8-25). 

The coordinates q k defined by Eqs. (9-170) and (9-171) are called the 
normal coordinates for the vibrating string. Each coordinate evidently 
represents one normal mode of vibration. The normal coordinates are also 
very useful in treating the case where a force f(x, I) is applied along the 
string (see Problem 26 at the end of this chapter). Mathematically, the 
normal coordinates have the property that the Lagrangian L becomes a 
sum of terms, each term involving only one degree of freedom. Thus 
in normal coordinates the problem is subdivided into separate problems, 
one for each degree of freedom. 



396 lagrange's equations [chap. 9 

It was, of course, rather fortunate that the coordinates g* which were 
chosen at the beginning of the problem turned out to be the normal co- 
ordinates. In general, this does not happen. For example, consider a 
string whose density varies along its length according to 

o - = o"o + asin-p- (9-191) 

This string is heaviest near its center. We will use the same coordinates 
q k as defined by Eqs. (9-170) and (9-171). We substitute Eqs. (9-191) 
and (9-172) in Eq. (9-173) and, instead of Eq. (9-176), we obtain (after 
some calculation), 

T = E E %T kj q k q h (9-192) 

&=i j— i 

where 

Ala k 2 
T ki = &(To + — 4fc2 _ x ' if * = h 

«V» - - £ [ < M . j) ,- 1 &- a ,-n ' i!k " * »""» 

and k, j are both even or both odd; otherwise 

T ki = 0. 

For this string, the q k 's are evidently not normal coordinates. In the 
Lagrange equations the q k 's, with k even, are all coupled together, as are 
those with k odd. The problem is then much more difficult, and a solution 
will not be attempted here. 

9-10 Hamilton's equations. The discussion in this section will be 
restricted to mechanical systems obeying Lagrange's equations in the form 
(9-57). The Lagrangian L is a function of the coordinates qk, of the veloci- 
ties q k , and perhaps of t. The state of the mechanical system at any time, 
that is, the positions and velocities of all its parts, is specified by giving the 
generalized coordinates and velocities qk, qk- Lagrange's equations are 
second-order equations which relate the accelerations q k to the coordinates 
and velocities. The state of the system could equally well be specified by 
giving the coordinates q k and the momenta p k defined by Eq. (9-168) : 

Pk = Wk' fc= 1,2> ""' / ' (9_194) 

These equations specify pk in terms of qi, . . . , g/; fa, . . . , fa. They can, 
in principle, be solved for q k in terms of qi, . . . , q/; p u . . . , p/. 



9-10] Hamilton's equations 397 

It is an interesting exercise to try to write equations of motion in terms 
of the coordinates qk and momenta pk- Note first that, by use of the 
definition (9-194) and the equations of motion (9-57), we have 

= 22 (Pk dq k + Pk dq k ) + -rr dt. (9-195) 

We next define a function H(q lt . . . , gy; p lf . . . , p f ; t) by 

/ 
H = J2 Vkik - L, (9-196) 

where for the velocities gj we substitute their expressions in terms of co- 
ordinates and momenta. Then we have 

dH = J2 (& dp k - p k dq k ) - ~ dt. (9-197) 

The definition (9-196) is chosen so that dH depends explicitly upon dp k , 
dqk, and dt. By inspection of Eq. (9-197), we see that 

tiff riff 

and 

dH = _dL 

dt dt ' 



(9-199) 



Equations (9-198) are the desired equations of motion that express q k and 
pk in terms of the coordinates and momenta. 

Equations (9-198) are Hamilton's equations of motion for a mechanical 
system. The function H, defined by Eq. (9-196), is called the Hamiltonicm 
function. We see from Eq. (9-124) that when V is a function only of the 
coordinates, for a stationary coordinate system, H is just the total energy 
expressed in terms of coordinates and momenta. For a moving coordinate 
system, where T is given by Eq. (9-13), the Hamiltonian is 

H = T 2 + V - T , (9-200) 

with T 2 expressed in terms of coordinates and momenta. According to 
Section 9-8, H will also be the total energy in a stationary coordinate 
system when electromagnetic forces are present. 



398 lagrange's equations [chap. 9 

When L does not contain the time explicitly, neither does H according 
to Eq. (9-199), as is also obvious from the way in which H was denned. 
According to Eq. (9-125), H is a constant of the motion in this case. This 
can also be proved directly from Eqs. (9-198), since it is easy to show that 

^ = ^ , (9-201) 

dt dt 

as the reader may verify. 

If any coordinate qk does not appear explicitly in H, then Eqs. (9-198) 

give 

p k = a constant, (9-202) 

in agreement with Eq. (9-118). Since H does not contain q k , we may take 
p k as a given constant, and the 2(/ — 1) equations (9-198) for the other 
coordinates and momenta are then the Hamiltonian equations for a sys- 
tem of / — 1 degrees of freedom. Thus degrees of freedom corresponding 
to coordinates that do not appear in H simply drop out of the problem. 
This is the origin of the term "ignorable coordinate. " After the remaining 
equations of motion have been solved for the nonignorable coordinates and 
momenta, any ignorable coordinate is given by Eqs. (9-198) as an integral 
over t: 

q k (t) = «*«>) + / ff dt. (9-203) 

JO °Pk 

Hamilton's equations are simply a new formulation of Newton's laws 
of motion. In simple cases, they reduce to equations which could have been 
written immediately from Newton's laws. In the harmonic oscillator, for 
example, with coordinate x, the momentum is 

p = mi. (9-204) 

The Hamiltonian function is therefore 

H = T + V = ^ + ikx 2 . (9-205) 

Equations (9-198) become 

x = 2-, p = —fee. (9-206) 

m 

The first of these is the definition of p, and the second is Newton's equation 
of motion. 

Although they are of comparatively little value as a means of writing 
the equations of motion of a system, Hamilton's equations are important 
for two general reasons. First, they provide a useful starting point in 
setting up the laws of statistical mechanics and of quantum mechanics. 



9-11] liouville's theorem 399 

Hamilton orginally developed his equations by analogy with a similar 
mathematical formulation which he had found useful in optics. It is not 
surprising that Hamilton's equations should form the starting point for 
wave mechanics! Second, there are a number of methods of solution of 
mechanical problems based on Hamilton's formulation of the equations of 
motion. It is clear from the way in which they were derived that Hamil- 
ton's equations (9-198), like Lagrange's equations, are valid for any set 
of generalized coordinates qi, . . . , q/ together with the corresponding 
momenta Pi, . . . , p/, defined by Eq. (9-194). In fact Hamilton's equations 
are valid for a much wider class of coordinate systems obtained by de- 
fining new coordinates and momenta as certain functions of the original 
coordinates and momenta. This is the basis for the utility of Hamilton's 
equations in the solution of mechanical problems. A further discussion of 
these topics is beyond the scope of this book.* We will, however, prove 
one general theorem in the next section which gives some insight into the 
importance of the variables pk and qu. 

9-11 Liouville's theorem. We may regard the coordinates q u . . . , qy 
as the coordinates of a point in an /-dimensional space, the configuration 
space of the mechanical system. To each point in the configuration space 
there corresponds a configuration of the parts of the mechanical system. 
As the system moves, the point Qi, . . . , Q/ traces a path in the configura- 
tion space. This path represents the history of the system. 

If we wish to specify both the configuration and the motion of a system 
at any given instant, we must specify the coordinates and velocities, or 
equivalently, the coordinates and momenta. The 2/-dimensional space 
whose points are specified by the coordinates and momenta Qi, . . • , Q/', 
Pi, . . . , pf is called the phase space of the mechanical system. As the sys- . 
tem moves, the phase point gi, ...,<?/; Pi, •••, p/ traces out a path in the 
phase space. The velocity of the phase point is given by Hamilton's 
equations (9-198). 

Each phase point represents a possible state of the mechanical system. 
Let us imagine that each phase point is occupied by a "particle" which 
moves according to the equations of motion (9-198). These particles 
trace out paths that represent all possible histories of the mechanical 
system. The theorem of Liouville states that the phase "particles" move 
as an incompressible fluid. More precisely, the phase volume occupied by 
a set of "particles" is constant. 

To prove Liouville's theorem, we make use of theorem (8-121) gener- 
alized to a space of 2/ dimensions. We may either generalize the argument 
which led to Eq. (8-116), or we may use the generalization of Gauss' 



* See H. Goldstein, op. cit. (Chapters 7, 8, 9.) 



4Q0 lagrange's equations [chap. 9 

divergence theorem which is valid in any number of dimensions. In either 
case, we have for a volume V in phase space, moving with the "particles": 



dV 
dt 



- /•; •/ S (t +*£)*■- * *• • ■ • *" fr*" 



which is Eq. (8-121) written for the 2/-dimensional phase space. We now 
substitute the velocities from Hamilton's equations (9-198) : 

(9-208) 

This is Liouville's theorem, and it should be noted that this theorem holds 
even when H depends explicitly on t. 

In the case of a harmonic oscillator, the phase space is a plane with co- 
ordinate axes x and p. The phase points move around ellipses H = con- 
stant, given by Eq. (9-205), and with velocities as given by Eq. (9-206). 
According to Liouville's theorem, the motion is that of a two-dimensional 
incompressible fluid. In particular, a set of points that lie in a region of 
area A will at any later time lie in another region of area A. 

Liouville's theorem makes the coordinates and momenta more useful 
for many purposes than coordinates and velocities. Because of this 
theorem, the concept of phase space is an important tool in statistical 
mechanics. Imagine a large number of mechanical systems identical to a 
given one, but with different initial conditions. Let each system be repre- 
sented by a point in their common phase space, and let these points move 
according to Hamilton's equations. The statistical properties of this col- 
lection of systems may be specified at any time t by giving the density 
P(<Zi. • • • > If) Pi, ■ • • > Pf', m the phase space of system points per unit 
volume. Liouville's theorem implies that the density p in the immediate 
neighborhood of any system point must remain constant as that point 
moves through the phase space. (Why?) If we define statistical equi- 
librium as a distribution in which p is constant in time at each fixed point 
in the phase space, then clearly the necessary and sufficient condition for 
equilibrium is that p be uniform along the flow lines of the system points. 
(Why?) 

We have been able in this section to give only a narrow glimpse of the 
power of the Hamiltonian methods. 



401 



Problems 



1. Coordinates u, w are defined in terms of plane polar coordinates r, by the 
equations 

u = In (r/a) — cot f , 

w = In (r/a) + 6 tan J", 

where a and f are constants. Sketch the curves of constant u and of constant w. 
Find the kinetic energy for a particle of mass m in terms of u, w, u, ib. Find 
expressions for Q u , Q m in terms of the polar force components F r , Ft. Find p u , p w . 
Find the forces Q a , Q w required to make the particle move with constant speed s 
along a spiral of constant u = uo. 

2. Two masses' mi and m.2 move under their mutual gravitational attraction 
in a uniform external gravitational field whose acceleration is g. Choose as 
coordinates the cartesian coordinates X, Y, Z of the center of mass (taking Z in 
the direction of g), the distance r between mi and wi2, and the polar angles 
and <p which specify the direction of the line from mi to wi2- Write expres- 
sions for the kinetic energy, the six forces Qx, . . . ,Q V , and the six momenta. 
Write out the six Lagrange equations of motion. 

3. (a) Set up the expression for the kinetic energy of a particle of mass m in 
terms of plane parabolic coordinates /, h, as defined in Problem 13 of Chapter 3. 
Find the momenta p/ and p*. (b) Write out the Lagrange equations in these 
coordinates if the particle is not acted. on by any force. . 

4. (a) Find the forces Q; and Qh required to make the particle in Problem 3 
move along a parabola / = /o = a constant, with constant generalized velocity 
ft, = ho, starting from h = at t = 0. (b) Find the corresponding forces F x 
and F v relative to a cartesian coordinate system. 

5. (a) Set up the Lagrange equations of motion in spherical coordinates r, 6, <p, 
for a particle of mass m subject to a force whose spherical components are 
F r , Fe, F v . 

(b) Set up Lagrange equations of motion for the same particle in a system of 
spherical coordinates rotating with angular velocity co about the 3-axis. 

(c) Identify the generalized centrifugal and coriolis forces 'Q r ', 'Qe', and 'Q v ' 
by means of which the equations in the rotating system can be made to take the 
same form as in the fixed system. Calculate the spherical components 'F r ', 'Ff', 
'F v ' of these centrifugal and coriolis forces, and show that your results agree 
with the expressions derived in Chapter 7. 

6. Set up the Lagrangian function for the mechanical system shown in 
Fig. 4-16, using the coordinates x, x\, X2 as shown. Derive the equations of 
motion, and show that they are equivalent to the equations that would be 
written down directly from Newton's law of motion. 

7. Choose suitable coordinates and write down the Lagrangian function for 
the restricted three-body problem. Show that it leads to the equations of mo- 
tion obtained in Section 7-6. 

8. Masses m and 2m are suspended from a string of length h which passes 
over a pulley. Masses 3m and 4m are similarly suspended by a string of length 



402 lagrange's equations [chap. 9 

h over another pulley. These two pulleys hang from the ends of a string of length 
h over a third fixed pulley. Set up Lagrange's equations, and find the accelera- 
tions and the tensions in the strings. 

9. A massless tube is hinged at one end. A uniform rod of mass m, length I, 
slides freely in it. The axis about which the tube rotates is horizontal, so that the 
motion is confined to a plane. Choose a suitable set of generalized coordinates, 
one for each degree of freedom, and set up Lagrange's equations. 

10. Set up Lagrange's equations for a uniform door whose axis is slightly out 
of plumb. What is the period of small vibrations? 

11. A double pendulum is formed by suspending a mass mi by a string of 
length h from a mass m\ which in turn is suspended from a fixed support by a 
string of length l\. (a) Choose a suitable set of coordinates, and write the 
Lagrangian function, assuming the double pendulum swings in a single vertical 
plane. 

(b) Write out Lagrange's equations, and show that they reduce to the equa- 
tions for a pair of coupled oscillators if the strings remain nearly vertical. 

(c) Find the normal frequencies for small vibrations of the double pendulum. 
Describe the nature of the corresponding vibrations. Find the limiting values 
of these frequencies when wu >J> W2, and when W2 » mi. Show that these 
limiting values are to be expected on physical grounds by considering the nature 
of the normal modes of vibration when either mass becomes vahishingly small. 

12. A ladder rests against a smooth wall and slides without friction on wall 
and floor. Set up the equation of motion, assuming that the ladder maintains 
contact with the wall. If initially the ladder is at rest at an angle a with the 
floor, at what angle, if any, will it leave the wall? 

13. One end of a uniform rod of mass M makes contact with a smooth vertical 
wall, the other with a smooth horizontal floor. A bead of mass m and negligible 
dimensions slides on the rod. Choose a suitable set of coordinates, set up the 
Lagrangian function, and write out the Lagrange equations. The rod moves in a 
single vertical plane perpendicular to the wall. 

14. A ring of mass M rests on a smooth horizontal surface and is pinned at a 
point on its circumference so that it is free to swing about a vertical axis. A 
bug of mass m crawls around the ring with constant speed, (a) Set up the equa- 
tions of motion, taking this as a system with two degrees of freedom, with the 
force exerted by the bug against the ring to be determined from the condition 
that he moves with constant speed. 

(b) Now set up the equation of motion, taking this as a system with one de- 
gree of freedom, the bug being constrained to be at a certain point on the ring 
at each instant of time. Show that the two formulations of the problem are 
equivalent. 

15. A pendulum bob of mass m is suspended by a string of length I from a 
point of support. The point of support moves to and fro along a horizontal z-axis 
according to the equation 

x = a cos ciit. 

Assume that the pendulum swings only in a vertical plane containing the z-axis. 
Let the position of the pendulum be described by the angle which the string 



PROBLEMS 



403 



makes with a line vertically downward, (a) Set up the Lagrangian function and 
write out the Lagrange equation. 

(b) Show that for small values of 6, the equation reduces to that of a forced 
harmonic oscillator, and find the corresponding steady-state motion. How does 
the amplitude of the steady-state oscillation depend on m, I, a, and co? 

16. A pendulum bob of mass m is suspended by a string of length I from a car 
of mass M which moves without friction along a horizontal overhead rail. The 
pendulum swings in a vertical plane containing the rail, (a) Set up the Lagrange 
equations, (b) Show that there is an ignorable coordinate, eliminate it, and dis- 
cuss the nature of the motion by the energy method. 

17. Find the tension in the string for the spherical pendulum discussed in 
Section 9-7, as a function of E, p ? , and 8. Determine, for a given E and p v , 
the angle Si at which the string will collapse. 

18. A particle of mass m slides over the inner surface of an inverted cone of 
half-angle a. The apex of the cone is at the origin, and the axis of the cone ex- 
tends vertically upward. The only force acting on the particle, other than the 
force of constraint, is the force pf gravity, (a) Set up the equations of motion, 
using as coordinates the horizontal distance p of the particle from the axis, and 
the angle <p measured in a horizontal circle around the cone. Show that <p is 
ignorable, and discuss the motion by the method of the effective potential. 

(b) For a given radius po, find the angular velocity <f>o of revolution in a hori- 
zontal circle, and the angular frequency o> of small oscillations about this cir- 
cular motion. Show that the small oscillations are a wobbling or an up-and- 
down spiraling motion, depending on whether the angle a is greater than or less 
than the angle 

-i J_ 



sin 



19. A flyball governor for a steam engine is shown in Fig. 9-11. Two balls, 
each of mass m, are attached by means of four hinged arms, each of length I, 
to sleeves which slide on a vertical rod. The upper sleeve is fastened to the rod; 
the lower sleeve has mass M and is free to slide up and down the rod as the balls 




Fig. 9-11. A flyball governor. 



404 laghange's equations [chap. 9 

move out from or toward the rod. The rod-and-ball system rotates with con- 
stant angular velocity w. (a) Set up the equation of motion, neglecting the weight 
of the arms and rod. Discuss the motion by the energy method. 

(b) Determine the value of the height z of the lower sleeve above its lowest 
point as a function of o> for steady rotation of the balls, and find the frequency 
of small oscillations of z about this steady value. 

20. Discuss the motion of the governor described in Problem 19 if the shaft 
is not constrained to rotate at angular velocity «, but is free to rotate, without 
any externally applied torque, (a) Find the angular velocity of steady rotation 
for a given height z of the sleeve, (b) Find the frequency of small vibrations 
about this steady motion, (c) How does this motion differ from that of Prob- 
lem 19? 

21. A rectangular coordinate system with axes x, y, z is rotating with uniform 
angular velocity w about the z-axis. A particle of mass m moves under the action 
of a potential energy V(x, y, z). (a) Set up the Lagrange equations of motion. 

(b) Show that these equations can be regarded as the equations of motion of a 
particle in a fixed coordinate system acted on by the force — VF, and by a force 
derivable from a velocity dependent potential U. Hence find a velocity de- 
pendent potential for the centrifugal and coriolis forces. Express U in spherical 
coordinates r, 6, <p, f, 6, <p, and verify that it gives rise to the forces 'Q r ', 'Qt, 'Q v ' 
found in Problem 5. 

22. Show that a uniform magnetic field B in the z-direction can be represented 
in cylindrical coordinates (Fig. 3-22) by the vector potential 

A = \Bp m. 

Write out the Lagrangian function for a particle in such a field. Write down 
the equations of motion, and show that there are three constants of the motion. 
Compare with Problem 49 of Chapter 3. 

23. The kinetic part of the Lagrangian function for a particle of mass m in 
relativistic mechanics is 

L k mc 2 [l - (y/c) 2 ] 1/2 . 

Show that this gives the proper formula (4-75) for the components of momentum. 
Show that if the potential function for electromagnetic forces Eq. (9-166) is 
subtracted, and if A and <f> do not depend explicitly on t, then T -j- q<f> is con- 
stant, with T given by formula (4-74). 

*24. Show by direct calculation that if Eqs. (9-155) hold for some function 
L(qi, . . . , <?/; 51, ... , fa; t), and we introduce new coordinates q*, . . . , q*, 
where 

qk = Mqi, ■ ■ ■ ,qf,t), k = 1, . . . , /, 
then 

lit w* a * — ' ' — *> • • ■ >/> 

at dq t dq t 

where L*(q%, . . . , q*; q*, . . . , q*; t) = L{qi, ...,qf, qi, ... , q f ; t) is obtained 
by substitution of /*(?*, . . . , q*; t) for q k . 



PROBLEMS 405 

25. Derive formula (9-184) by writing down a potential energy which gives 
the interparticle forces for the string of particles studied in Section 8-4, and 
passing to the limit h —* 0. 

26. A stretched string is subject to an externally applied force of linear density 
f(x, t). Introduce normal coordinates qk, and find an expression for the gen- 
eralized applied force Qk(t). Use the Lagrangian method to solve Problem 6(a), 
Chapter 8. 

27. Solve Problem 7, Chapter 8, by using the coordinates g* defined by 
Eqs. (9-170) and (9-171). 

28. Write down the Hamiltonian function for the spherical pendulum. Write 
the Hamiltonian equations of motion, and derive from them Eq. (9-136). 

*29. Work out the relativistic Hamiltonian function for a particle subject to 
electromagnetic forces, using the Lagrangian function given in Problem 23. 
Write out the Hamiltonian equations of motion and show that they are equiva- 
lent to the Lagrange equations. 

30. Work out the Hamiltonian function H(qk, pic) for the vibrating string, 
starting from Eq. (9-185). Write down the equations which relate the momenta 
Pk to the function u(x, t) which describes the motion of the string. Hence show 
that H = T + V, with T and V given by Eqs. (9-173) and (9-184). 

31. Write down the Hamiltonian function for Problem 2. Write out Hamil- 
ton's equations. Identify the ignorable coordinates and show that there remain 
two separate one degree of freedom problems, each of which can be solved (in 
principle) by the energy method. What are the corresponding two potential- 
energy functions? 

32. A beam of electrons is directed along the z-axis. The electrons are uni- 
formly distributed over the beam cross section, which is a circle of radius oo, 
and their transverse momentum components (p x , Pv) are distributed uniformly 
in a circle (in momentum space) of radius po- If the electrons are focused by 
some lens system so as to form a spot of radius oi, find the momentum distribu- 
tion of electrons arriving at the spot. 

33. A group of particles all of the same mass m, having initial heights and 
vertical momenta lying in the square — a < z < a, — b < p < b, fall freely 
in the earth's gravitational field for a time t. Find the region in the phase space 
within which they lie at time t, and show by direct calculation that its area is still 
4o6. 

34. In an electron microscope, electrons scattered from ah object of height 
zo are focused by a lens at distance Do from the object and form an image of 
height zi at a distance Dj behind the lens. The aperture of the lens is A. Show 
by direct calculation that the phase area in the (z, p z ) phase plane occupied by 
electrons leaving the object (and destined to pass through the lens) is the same 
as the phase area occupied by electrons arriving at the image. Assume that 
zo <3C Do and zi <C Dr. 



CHAPTER 10 

TENSOR ALGEBRA. INERTIA AND STRESS TENSORS 

In this chapter we shall develop the algebra of linear vector functions, 
or tensors, as a mathematical tool which is useful in treating many prob- 
lems. In particular, we shall need tensors in the study of the general 
motion of a rigid body and in the formulation of the concept of stress in 
a solid, or in a viscous fluid. 

10-1 Angular momentum of a rigid body. The equation of motion for 
the rotation of a rigid body is given by Eq. (5-5) and restated here: 

ft = N ' < 1(H > 

where L is the angular momentum and N is the torque about a point P 
which may be either fixed or the center of mass of the body. In Section 
5-2 we studied the rotation of a rigid body about a fixed axis. In order 
to treat the general problem of the rotation of a body about a point P, 
we must find the relation between the angular momentum vector L and 
the angular velocity vector u. 

Consider a body made up of point masses m k situated at points r k 
relative to an origin of coordinates at P. We have shown in Section 7-2 
that the most general motion of the body about the point P is a rotation 
with angular velocity u, and that the velocity Vfc of each particle in the 
body is given by 

v fc = to X r k . (10-2) 

We sum the angular momentum given by Eq. (3-142) over all particles: 



L = "22 in k r k X v& 

k=l 

N 

= X) m * r * X (« X r *)- (10-3) 

Equation (10-3) expresses L as a function of a>, L(«). By substitution in 
Eq. (10-3) it is readily verified that the function L(«), for any two vectors 
w, w', and any scalar c, satisfies the following relations: 

L(cw) = cL(«), (10-4) 

L(» + «') = L(») + L(<o'). (10-5) 

406 



10-2] TENSOR ALGEBRA 407 

A vector function L(w) with the properties (10-4), (10-5) is called a 
linear vector function. Linear vector functions are important because they 
occur frequently in physics, and because they have simple mathematical 
properties. 

In order to develop an analogy between Eq. (10-3) and Eq. (5-9) for 
the case of rotation about an axis, we make use of Eq. (3-35) : 



N 

I 

4=1 



L = ^2 [m k rlw — m k i k {r k ■ «)]. (10-6) 



The factor u is independent of k and can be factored from the sum over 
the first term. In a purely formal way, we may also factor « from the sum 
over the second term: 

L = (X) m * r *) w — (S m * r fc r fc) • «• (10-7) 

The second term has no meaning, of course, since the juxtaposition Tkik 
of two vectors has not yet been denned. We shall try to supply a mean- 
ing in the next section. 

10-2 Tensor algebra. The dyad product AB of two vectors is denned 
by the following equation, where C is any vector: 

(AB) • C = A(B • C). (10-8) 

The right member of this equation is expressed in terms of products de- 
fined in Section 3-1. The left member is, by definition, the vector given by 
the right member. Note that the dyad AB is defined only in terms of its dot 
product with an arbitrary vector C. We can readily show, from definition 
(10-8), that multiplication of a vector by a dyad is a linear operation in 
the sense that 

(AB) • (cC) = c[(AB) • C], (10-9) 

(AB) • (C + D) = (AB) • C + (AB) • D. (10-10) 

For fixed vectors A, B, the dyad AB therefore defines a linear vector func- 
tion F(C) : 

F(C) = (AB) • C. (10-11) 

The dyad AB is an example of a linear vector operator, that is, it represents 
an operation which may be performed on any vector C to yield a new 
vector (AB) • C, which is a linear function of C. 



408 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

A linear vector operator is also called a tensor * Tensors will be repre- 
sented by sans-serif boldface capitals, A, B, C, etc. We may for example 
let T be the tensor represented by the dyad AB: 

T = AB. (10-12) 

The meaning of the tensor T is specified by the definition,! 

T • C = A(B • C), (10-13) 

which gives the result of applying T to any vector C. We can form more 
general linear vector operations by taking sums of dyads. The sum of 
two dyads, or tensors S, T, is denned as follows: 

(S + T) • C = S • C + T • C. (10-14) 

Note that all definitions of algebraic operations on tensors, like the above 
definition of (S + T), are formulated in terms of the application of the 
tensors to an arbitrary vector C. The sum of one or more dyads is called 
a dyadic. According to the definition (10-14), the dyadic (AB + DE) 
operating on C yields the vector 

(AB + DE) • C = A(B • C) + D(E • C). (10-15) 

We can readily show that the sum of two linear operators is a linear 
operator; therefore dyadics are also linear vector operators and we have for 
any dyadic or tensor T, 

T • (cC) = c(T • C), (10-16) 

T • (C + D) = T • C + T • D. (10-17) 

The linearity relations (10-16), (10-17), together with the definition 
(10-14), guarantee that dyad products, sums of tensors, and dot products 
of tensors with vectors satisfy all the usual algebraic rules for sums and 
products. We can also define a dot product of a dyad with a vector on the 
left in the obvious way, 

C • (AB) = (C • A)B, (10-18) 



* More precisely, a linear vector operator may be called a second-rank tensor, 
to distinguish it from third- and higher-rank tensors obtained as linear combina- 
tions of triads ABC, etc. We shall be concerned in this book only with second- 
rank tensors, which we shall refer to simply as tensors. 

t The result of applying a tensor T to a vector C is often denoted by TC, 
without the dot. We shall use the dot throughout this book. 



10-2] TENSOR ALGEBRA 409 

and correspondingly for sums of dyads. Note that the dot product of a 
dyadic with a vector is not commutative; 

T • C = C • T (10-19) 

does not hold in general. We can define, in an obvious way, a product cT 
of a tensor by a scalar, with the expected algebraic properties (see Problem 

1). 
A very simple tensor is given by the dyadic 

1 = ii + j j + (10-20) 

where i, j, k are a set of perpendicular unit vectors along x-, y-, and z- 
axes. We calculate, using the definitions (10-14) and (10-8), 

1 • A = \A X + ]A V + kA z = A. (10-21) 

The tensor 1 is called the unit tensor; it may be defined as the operator 
which, acting on any vector, yields that vector itself. Evidently 1 is one 
of the special cases for which 

1A = A1. (10-22) 

If c is any scalar, the product cl is called a constant tensor, and has the 
property 

(cl) • A = A • (cl) = cA. (10-23) 

Using the definitions above, we can now write Eq. (10-7) in the form 

L = 1 • to, (10-24) 

where I is the inertia tensor of the rigid body, defined by 



N 

I 

4=1 



I = g {m k rl\ - m k r k T k ). (10-25) 



The inertia tensor I is the analog, for general rotations, of the moment of 
inertia for rotations about an axis. Note that L and a are not in general 
parallel. We will study the inertia tensor in more detail after we have 
developed the necessary properties of tensors. 

If we write all vectors in terms of their components, 

C = C x i + C v j + CM, (10-26) 

then it is clear that by multiplying out dyad products and collecting terms, 
any dyadic can be written in the form: 



410 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

T = T xx ii + T xy ij + T xz ik 
+ T vx ji + Tyyjj + T ye jb 
+ TJU + T zy kj + T zz kk. (10-27) 

Just as any vector A can be represented by its three components (A x , A y , 
A z ), so any dyadic can be specified by giving its nine components T xx , 
. . . , T zz . These may conveniently be written in the form of a square array 
or matrix: 

/ -* XX ■* X 

(10-28) 




As an example, the reader may verify that the components of the inertia 
tensor (10-25) are 

N N 

Ixx = X) m k (yl + zl), I xv = — ^ m k x k y k , etc. (10-29) 
fc=i fc=i 

In order to simplify writing the tensor components, it often will be con- 
venient to number the coordinate axes xi, x 2 , £3 instead of using x, y, z: 

x = x h y — x 2 , z = x 3 . (10-30) 

We shall write the corresponding unit vectors as e*: 

i = e 1( j = e 8 , k = e 3 . (10-31) 

Equations (10-26) and (10-27) can now be written as 

C = X) Ctfii, (10-32) 

i—X 

and 

3 
T = J^ Tifi^i- (10-33) 

Another advantage of this notation is that it allows the discussion to be 
generalized to vectors and tensors in a space of any number of dimensions 
simply by changing the summation limit. 

By using the definitions of dyad products and sums, we can express the 
components of the vector T • C in terms of the components of T and C: 

(T-C),-= YiTifij, (10-34) 

3=1 



10-2] TENSOR ALGEBRA 

as the reader should verify. Similarly, 

(C • Vi = E OF* 

3=1 

We note that, by Eq. (10-33), 

T ti = e< • (T • e,-) = (e,- • T) • e y . 



411 



(10-35) 



(10-36) 



We may omit the parentheses, since the order in which the multiplications 
are carried out does not matter. 

We can now show that any linear vector function can be represented by 
a dyadic. Let F(C) be any linear function of C. Consider first the case 
when C is a unit vector ey, and let T t j be the components of F in that case: 



(10-37) 





F(e,) = 


3 


Now 


any vector C can be written 


as 




C = 


= E c &- 

3=1 


By use of the linear property of F(C), we have therefore 




F(C) = 


E F ^' e ^ 

3-1 




= 


E c >x&) 

3=1 




= 


3 



(10-38) 



(10-39) 



i, j— i 



Thus the components of F(C) can be expressed in terms of the numbers 



If we define the dyadic 



[F(C)] t - = E Ttfii 

3=1 






we see from Eqs. (10-40) and (10-34) that 

F(C) = T • C. 



(10-40) 
(10-41) 

(10-42) 



412 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

Thus the concepts of dyadic and linear vector operator or tensor are 
identical, and are equivalent to the concept of linear vector function 
in the sense that every linear vector function defines a certain tensor or 
dyadic, and conversely. 
We can define a dot product of two tensors as follows: 

(T • S) • C = T • (S • C). (10^3) 

Application of the operator T • S to any vector means first applying S, 
and then T. We now calculate in terms of components, using the definition 
(10-43), 

(T • S) • C = T ■ X) SrftiBj 

3 3 

i=i j,k=i 

= E IE (E T i&>) c*l •*• ( 10 ^ 4 ) 

i=l L ft=1 \ ]=1 / J 

Comparing this result with Eq. (10-34), we see that 

3 

<J • S)* = IE TijSju. (HM5) 



Equation (10-45) also results if we simply evaluate T • S in a formal way 
by writing the dot and dyad products and collecting terms: 



3 



T • S = ^ TijSkie&j • e k ei 

ijhl=l 
3 

= X) TijSjiPfii, (10-46) 

ijl—l 

and this shows that our definition (10-43) is consistent with the ordinary 
rules of algebra. If T, S are written as matrices according to Eq. (10-28), 
then Eq. (10-45) is the usual mathematical rule for multiplying matrices. 
We can similarly show that the definition (10-14) implies that tensors 
are added by adding their component matrices according to the rule: 

(T + S) i3 - = Tn + Sij. (10-47) 

Sums and products of tensors obey all the usual rules of algebra except 
that dot multiplication, in general, is not commutative: 



10-2] TENSOR ALGEBRA 413 

T + S = S + T, (10-48) 

T • (S + P) = T • S + T • P, (10-49) 

T • (S • P) = (T • S) • P, (10-50) 

1 . T = T • 1 = T, (10-51) 

and so on, but 

T • S 5* S • T, in general. (10-52) 

It is useful to define the transpose V of a tensor T as follows: 

T' • C = C ■ T. (10-53) 

In terms of components, 

Tlj = T it . (10-54) 

The transpose is often written T, but the notation T' is preferable for 
typographical reasons. The following properties are easily proved: 

(T + S)« = T* + $', (10-55) 

(T • S) f = S* • T', (10-56) 

(¥*)' = T. (10-57) 
A tensor is said to be symmetric if 

T 1 = T. (10-58) 

For example, the inertia tensor, given by Eq. (10-25) is symmetric. For 
a symmetric tensor, 

Ta = T tJ . (10-59) 

A symmetric tensor may be specified by six components; the remaining 
three are then determined by Eq. (10-59). 
A tensor is said to be antisymmetric if 

V = -T. (10-60) 

The components of an antisymmetric tensor satisfy the equation 

Tji = -Tij. (10-61) 

Evidently the three diagonal components Tu are all zero, and if three off- 
diagonal components are given, the three remaining components are given 



414 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

by Eq. (10-61). An antisymmetric tensor has only three independent 
components (in three-dimensional space). An example is the linear operator 
defined by 

T • C = to X C, (10-62) 

where to is a fixed vector. Comparing Eq. (10-62) with Eq. (7-20), we 
see that the operator T can be interpreted as giving the velocity of any 
vector C rotating with an angular velocity a. Comparing Eq. (10-62) 
with Eq. (10-34), we see that the components of T are: 

Tn = T22 = T33 = 0, 

T21 = — T12 = W3, 

(10-63) 
T S2 = — T23 = «i, 

^13 = —T31 = CO2. t 

Since an antisymmetric tensor, like a vector, has three independent com- 
ponents, we may associate with every antisymmetric tensor T a vector 
to (in three-dimensional space only!) whose components are related to 
those of T by Eq. (10-63). The operation T • will then be equivalent to 
u X , according to Eq. (10-62). 

Given any tensor T, we can define a symmetric and an antisymmetric 
tensor by 

T. = i(T + TO, (10-64) 

T« = i(T - T'), (10-65) 

such that 

T = T 8 + T„. (10-66) 

We saw, in the preceding paragraph, that an antisymmetric tensor could 
be represented geometrically by a certain vector w. We will see in Sec- 
tion 10-4 how to represent a symmetric tensor. Since antisymmetric and 
symmetric tensors have rather different geometric properties, tensors 
which occur in physics are usually either symmetric or antisymmetric 
rather than a combination of the two. In three-dimensional space, the 
introduction of an antisymmetric tensor can always be avoided by the 
use of the associated vector. It is therefore not a coincidence that the two 
principal examples of tensors in this chapter, the inertia tensor and the 
stress tensor, are both symmetric. 

10-3 Coordinate transformations. We saw in the previous section that 
a tensor T may be defined geometrically as a linear vector operator by 
specifying the result of applying T to any vector C. Alternatively, the 



10-3] COORDINATE TRANSFORMATIONS 415 

tensor may be specified algebraically by giving its components Tij. A 
discrepancy exists between the two definitions of a tensor, in that the 
algebraic definition appears to depend upon the choice of a particular co- 
ordinate system. A similar discrepancy in the case of a vector was noted 
in Section 3-1. We will now remove the discrepancy by learning how to 
transform the components of vectors and tensors when the coordinate 
system is changed. We will restrict the discussion to rectangular co- 
ordinates. 

Let us consider two coordinate systems, x\, x 2l x 3 , and x[, x' 2 , x 3 , hav- 
ing the same origin. The coordinates of a point in the two systems are 
related by Eqs. (7-13) : 

3 

x'i = 23 GijXj, (10-67) 

3=1 

where 

aij = e^- • e,- (10-68) 

is the cosine of the angle between the x<- and a^-axes. Likewise, 

3 

Xj = ^2 a H x 'i- (10-69) 

The relations between the primed and unprimed components of any vector 

c = X) c'm = E c>i ( 10 - 7 °) 

i=l i=l 

may be obtained in a similar manner by dotting ej or e^- into Eq. (10-70) : 

C't = X) a <A. ( 10 " 71 ) 



3 

i=l 



Cj = 2 and. (10-72) 

i=l 

We can now define a vector algebraically as a set of three components 
(Ci, C 2 , C 3 ) which transform like the coordinates {x u x 2 , x 3 ) when the co- 
ordinate system is changed. By referring to all coordinate systems, this 
definition avoids giving preferential treatment to any particular coordinate 
system. In the same way, the primed and unprimed components of a 
tensor 

T = J2 T '«*& = £ T i*>+ 1 (10 " 73) 

»,*=! 3.1=1 



416 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

are related by [see Eq. (10-36)] 

3 

T'i k = el • T • e' k = £ a^cuiT,,, (10-74) 

3,1=1 
3 

T jt = e r T-e,= £ a i} a k iT' ik . (10-75) 

i,fc=l 

A tensor may be defined algebraically as a set of nine components (Tji) 
that transform according to the rule given in Eqs. (10-74) and (10-75). 
Note the distinction between a tensor and a matrix. The concept of a 
matrix is purely mathematical; matrices are arrays of numbers which may 
be added and multiplied according to the rules (10-45) and (10-47). The 
concept of a tensor is geometrical; a tensor may be represented in any 
particular coordinate system by a matrix, but the matrix must be trans- 
formed according to a definite rule if the coordinate system is changed. 

The coefficients a,-y defined by Eq. (10-68) are the components of the 
unit vectors e* in the unprimed system, and conversely: 

3 

e'i = X) a t fii, (10-76) 



and 



-i 



£ a v*'i- ( 1( >-77) 



Since e[, e' 2t e£ are a set of perpendicular unit vectors, we see that the 
numbers 0,7 must satisfy the equations: 

3 

e » • e * = 23 a « afc > = Sik > (10-78) 

3=1 

where 5,* is a shorthand notation for 

l« = (° *'**' (10-79) 

U if i = k. 

There are six relations (10-78) among the nine coefficients a ik . Hence, 
if three of the constants 0,7 are specified, the rest may be determined from 
Eqs. (10-78). It is clear that three independent constants must be specified 
to locate the primed axes relative to the unprimed (or vice versa). For 
the x[-bx\s may point in any direction and two coordinates are there- 
fore required to locate it. Once the position of the a^-axis is determined, 



10-3] COORDINATE TRANSFORMATIONS 417 

the position of the Xg-axis, which may be anywhere in a plane perpendicu- 
lar to x[, may be specified by one coordinate. The position of the Zg-axis 
is then determined (except for sign). We can write additional relations 
between the o,-/s by the use of such relations as 

ej • e* = 5ji, e\ x e' 2 = ±e' 3) ei • (e 2 X e 3 ) = ±1, etc. 

(10-80) 

Since at least three of the a,/s must be independent, it is clear that the 
relations obtained from Eqs. (10-80) are not independent but could be 
obtained algebraically from Eqs. (10-78). An interesting relation is ob- 
tained from 

an <*2i <*3i 



e x • (e 2 X e 3 ) = 



a 12 a 22 a 32 
«13 *23 «33 



= ±1, (10-81) 



where the result is +1 if the primed and unprimed systems are both 
right- or both left-handed and is —1 if one is right-handed and the other 
left-handed. Hence the determinant |a,y[ is +1 or —1 according to whether 
the handedness of the coordinate system is or is not changed. 



In a left-handed system, the cross product is to be defined using the left in 
place of the right hand. [In Eq. (10-81), the triple product on the left is to be 
evaluated in the primed system.] The algebraic definition is then the same 
in either case: 

(A X B) = (A 2 B 3 - A3B2, A3B1 - A1B3, A1B2 - A2B1). (10-82) 

This definition implies that the cross product A X B of two ordinary vectors is 
not itself an ordinary vector, since its direction reverses when we change the 
handedness of the coordinate system. An ordinary vector that has a direction 
independent of the coordinate system is called a polar vector. A vector whose 
sense depends upon the handedness of the coordinate system is called an axial 
vector or pseudovector. The angular velocity vector to is an axial vector, and so 
is any other vector whose sense is defined by a "right-hand rule." The vector 
associated with an (ordinary) antisymmetric tensor is an axial vector. The cross 
product w X C of an axial with a polar vector is itself a polar vector. The dis- 
tinction between axial and polar vectors arises only if we wish to consider both 
right- and left-handed coordinate systems. In the applications in this book, we 
need only consider rotations of the coordinate system. Since rotations do not 
change the handedness of the system, we shall not be concerned with this dis- 
tinction. 



418 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

The transformation denned by Eqs. (10-67), (10-71), and (10-74), 
where the coefficients satisfy Eq. (10-78), is called orthogonal. As the name 
implies, an orthogonal transformation enables us to change from one set 
of perpendicular unit vectors to another. 

The right member of Eq. (10-71) is formally similar to the right mem- 
ber of Eq. (10-34). This suggests an alternative interpretation of Eqs. 
(10-71). Let us define a tensor A with components 

Atj = ay, (10-83) 

and consider the vector 

C = A • C. (10-84) 

The components C- of C are given by Eq. (10-71). Similarly, by Eq. 

(10-72), 

C = A' • C. (10-85) 

Thus Eqs. (10-71) and (10-72) may be interpreted alternatively as repre- 
senting the result of operating with the tensors A, A' upon the vectors C, 
C, respectively. In the original interpretation, C ., C'i are components of 
the same vector C in two different coordinate systems. In the alternative 
interpretation, Cj, C\ are the components of two different vectors C, C 
in the same coordinate system. We are primarily interested in the first 
interpretation, in which these equations represent a coordinate trans- 
formation. However, the latter interpretation will often be useful in 
deriving certain algebraic properties of Eqs. (10-71), (10-72), which, of 
course, are independent of how we choose to interpret them. In case the 
primed axes are fixed in a rotating rigid body, either interpretation is use- 
ful. If the primed axes initially coincide with the unprimed axes, then we 
may interpret Eqs. (10-71), (10-72) as expressing the transformation from 
one coordinate system to the other. Alternatively, we may interpret A 
as the tensor which represents the operation of rotating the body from 
its initial to its present position, i.e., a vector fixed in the body and ini- 
tially coinciding with C will be rotated so as to coincide with C = A • C. 
Making use of Eqs. (10-84), (10-85), and (10-43), we deduce that, 

A* • (A • C) = (A* • A) ■ C = C. (10-86) 

Hence, by Eq. (10-22), 



and similarly 



A'-A = 1, (10-87) 

A A ( = I. (10-88) 



A tensor having this property is said to be orthogonal. Equation (10-87) 
is evidently equivalent to Eq. (10-78). In the second interpretation, Eqs. 
(10-74) and (10-75) can be written as 



10-3] COORDINATE TRANSFORMATIONS 419 

T' = A • T • A', (10-89) 

T = A' • T' • A. (10-90) 

The orthogonal tensor is the only example we shall have of a tensor with 
a definite geometrical significance, which is neither symmetric nor anti- 
symmetric ; it has, instead, the orthogonality property given by Eq. (10-87) . 
In view of the fact that the various vector operations were defined with- 
out reference to a coordinate system, it is clear that all algebraic rules for 
computing sums, products, transposes, etc., of vectors and tensors will be 
unaffected by an orthogonal transformation of coordinates. Thus, for 
example, 

(B + C)' y = B'i + d, (10-91) 

(T-C)}= t y 5A ( 10 - 92 ) 

1=1 

<J% = T' H . (10-93) 

We can also verify directly the above equations, and others like them, 
by using the transformation equations and the rules of vector and tensor 
algebra. This is most easily done by taking advantage of the second inter- 
pretation of the transformation equations. For example, we can prove 
Eq. (10-93) by noting that 



(T*)' = A • T' • A' 


[by Eq. (10-89)] 


= A.(A-T)< 


[by Eqs. (10-56) and (10-57)] 


= [(A • T) ■ A*]' 


[by Eqs. (10-56) and (10-57)] 


= (T')',Q.E.D. 


[by Eq. (10-89)]. 



Any property or relation between vectors and tensors which is expressed 
in the same algebraic form in all coordinate systems has a geometrical 
meaning independent of the coordinate system and is called an invariant 
property or relation. 

Given a tensor T, we may define a scalar quantity called the trace of T 
as follows: 

tr(T) = X) T »- ( 10 - 94 ) 

Since this definition is in terms of components, we must show that the 
trace of T is the same in all coordinate systems. In the primed system, 
we have 



420 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

tr(T) = J2 T '« 

1=1 

3 3 

= Z £ *U*nTji [by Eq. (10-74)] 

= Z 2 aijO-u \Tji [rearranging sums] 

3 

= J2 T H h i l [as in Eq. (10-78)] 

}\i=i 

= J2 T »> Q- ED - [ b y Ec l- (10-79)]. 

Another invariant scalar quantity associated with a tensor is the deter- 
minant 

Tn T12 T13 

det(T) = T ai T 22 T 23 , (10-95) 

Tzi ^32 ^33 

as may also be verified by direct computation. 

Let us now study the result of carrying out two coordinate transforma- 
tions in succession. The primed coordinates are defined by Eq. (10-67), 
in terms of the unprimed coordinates. Let double-primed coordinates be 
defined by 

x'i = Z a *i x 'i- ( 10 " 96 ) 

We substitute for x' t from Eq. (10-67) to obtain the double-primed co- 
ordinates in terms of the unprimed coordinates: 

3 3 
X]c = / , / , djciflijXi 

= E E a '^ a a \ x i 

i=l L i=l J 

= 5>jfcry, (10-97) 

y-i 

where the coefficients of the transformation x — » x" are given by 

3 

a'kj = X) °W ( 1Q - 98 ) 




10-4] DIAGONALIZATION OF A SYMMETRIC TENSOR 421 

Thus the matrix of coefficients a'^- is obtained by multiplying the matrices 
a 'ki> a a according to the rule for matrix multiplication. If we interpret the 
transformation coefficients as the components of tensors A, A', A", we 
then see from Eqs. (10-45) and (10-98) that 

A" = A' • A. (10-99) 

This result also follows immediately from Eq. (10-84), applied twice, and 
we therefore have an alternative way to derive Eq. (10-98). 

10-4 Diagonalization of a symmetric tensor. The constant tensor, 
defined by Eq. (10-23), has in every coordinate system* the matrix: 



(10-100) 



A nonconstant tensor may, in a particular coordinate system, have the 
matrix: 

/J 1 . \ 
T = 1 T 2 I • (10-101) 

\0 tJ 

The tensor T is then said to be in diagonal form. We do not call T a diagonal 
tensor, because the property (10-101) applies only to a particular co- 
ordinate system; after a change of coordinates [Eq. (10-74)], T will usually 
no longer be in diagonal form. If T is in diagonal form, then its effect on 
a vector is given simply by 

(T • C)< = T&, i = 1, 2, 3. (10-102) 

The importance of the diagonal form lies in the following fundamental 
theorem: 

Any symmetric tensor can be brought into diagonal form by 
an orthogonal transformation. The diagonal elements are then 
unique except for their order, and the corresponding axes are 
unique except for degeneracy. (10-103) 

Before proving this important theorem, let us try to understand its 
significance. The theorem states that, given any symmetric tensor T, 
we can always choose the coordinate axes so that T is represented by a 
diagonal matrix. Furthermore, this can be done in essentially only one 
way; there is only one diagonal form (10-101) for a given tensor T, except 



: See Problem 10 at the end of this chapter. 



422 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

for the order in which the diagonal elements T\, T 2 , T 3 appear, and each 
element is associated with a unique axis in space, except for degeneracy, 
that is, except when two or three of the diagonal elements are equal. The 
axes ei, e 2 , e3 in the coordinate system in which the tensor has a diagonal 
form are called its principal axes. The diagonal elements T\, T 2 , T% are 
called the eigenvalues or characteristic values of T. Whenever we write a 
tensor element with a single subscript, we shall mean it to be an eigenvalue. 
Any vector C parallel to a principal axis is called an eigenvector of T. 
An eigenvector, according to Eq. (10-102), has the property that opera- 
tion by T reduces to multiplication by the corresponding eigenvalue: 

T • C = TiC, (10-104) 

where Ti is the eigenvalue associated with the principal axis e» parallel 
toC. 

The theorem (10-103) allows us to picture a symmetric tensor T as a 
set of three numbers attached to three definite directions in space. If 
we think of T as applied to each vector C in the vector space, then formula 
(10-102) shows that the effect is a stretching or a compression along each 
principal axis, together with a reflection if T,- is negative. In Section 10-2 
we saw that a symmetric tensor may be specified by giving six components 
Tij in any arbitrarily chosen coordinate system. We now see that we can 
alternatively specify T by specifying the principal axes (this requires three 
numbers, as we have seen), and the three associated eigenvalues. 

If two or three of the eigenvalues are equal, we say the eigenvalue is 
doubly or triply degenerate. If the eigenvalue is triply degenerate, the 
tensor clearly is a constant tensor [Eq. (10-100)] and diagonal in every 
coordinate system. The principal axes are no longer unique; any axis 
is a principal axis. Every vector is an eigenvector of a constant tensor. 
If two eigenvalues are equal, say T x — T 2> then if we consider rotating 
the coordinate axes in the eie 2 -plane, we can see that the tensor will re- 
main in diagonal form; the four elements referring to this plane behave 
like a constant tensor in that plane. Again the principal axes are not 
unique, since two of them may lie anywhere in the eie2-plane. The third 
axis e3 associated with the nondegenerate eigenvalue T3, however, is 
unique. We can prove that every axis in the e^-plane is a principal 
axis by considering the effect of T on any vector 

C = dd + C 2 e 2 (10-105) 

in this plane. In view of Eq. (10-102), if T x = T 2 , we have 

T • C = Tide! + T 2 C 2 e 2 

= TiC, (10-106) 



10-4] 



DIAGONALIZATION OF A SYMMETRIC TENSOR 



423 



so that C is an eigenvector of T. Every vector in the eie 2 -plane is an 
eigenvector of T with eigenvalue T t . If we like, we may say that there 
is a principal plane associated with a doubly degenerate eigenvalue. 

We will now prove the theorem (10-103) by showing how the principal 
axes can be found. Let a symmetric tensor T be given in terms of its com- 
ponents Tn in some coordinate system, which we will call the initial co- 
ordinate system. To find a principal axis, we must look for an eigenvector 
of T. Let C be such an eigenvector, and T' the corresponding eigenvalue. 
We can rewrite Eq. (10-104) in the form 

(T — T'l)-C = 0. (10-107) 

If we write this equation in terms of components, we obtain 

(fii - T')C l + T 12 C 2 + T 13 C 3 = 0, 

TaiCi + (T 22 - T')C 2 + T 23 C 3 = 0, (10-108) 

TnCi + T 32 C 2 + (T 33 - T')C 3 = 0. 

These equations for the unknown vector C have, of course, the trivial 
solution C = 0. If we write the solution for C,- in terms of determinants, 
we see that C = is the only solution unless the determinant 



Tu - T' T 12 T l3 

rrp rpt rp 

21 * 22 — J- l 23 

rp rp rp rpt 

1 31 *■ 32 i 33 — 1 



o, 



(10-109) 



in which case the solution for C,- is indeterminate. In this case it is shown 
in the theory of linear equations* that Eqs. (10-108) have also nontrivial 
solutions d. It is clear that Eqs. (10-108) cannot determine the numbers 
Ci, C 2 , C 3 uniquely, but only their ratios to one another, Ci:C 2 :C 3 . This 
is also clear from Eq. (10-107), from which we began. Geometrically, 
only the direction of C is determined, not its magnitude (nor its sense). 
Equation (10-109), called the secular equation, represents a cubic equation 
to be solved for the eigenvalue T'. In general there will be three roots 
T[, T 2 , T' 3 . Given any root T', we can then substitute it in Eqs. (10-108) 
and solve for the ratios Ci:C 2 '-C 3 . Any vector whose components are in 
the ratio C\:C 2 :C 3 is an eigenvector of T corresponding to the eigenvalue 
T'. For each eigenvalue T' it we can then take a unit vector e'j along the 
direction of the corresponding eigenvectors. The axes e[, e' 2> e 3 are then 



* See, for example, Knebelman and Thomas, Principles of College Algebra. 
New York: Prentice-Hall, Inc., 1942. (Chapter IX, Theorem 10.) 



424 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

the principal axes of T. When we solve Eqs. (10-108) for the components 
of e'j for a given T' jt we get three numbers ay,- (=C» for T' = T' } ) which 
are the components of e^ along the axes e,- of the initial coordinate system: 

e'y = E *&*■ (10-110) 

»=i 

This is just Eq. (10-76), hence the numbers a# are the coefficients of the 
orthogonal transformation from the initial coordinate system to the prin- 
cipal axes. We say that the transformation with coefficients a,i diago- 
nalizes T. 

In order to be sure we can carry out the above program, we must prove 
three lemmas, as the reader may have noted. First, we must prove that 
the roots T' of the secular equation (10-109) are real; otherwise we cannot 
find real solutions of Eqs. (10-108) for C u C 2 , C 3 . Second, we must prove 
that the vectors ej, obtained from Eqs. (10-108) for the different eigen- 
values T'j, are perpendicular; otherwise we do not obtain a set of perpen- 
dicular unit vectors. Third, we must show that in the degenerate case, 
two (or three) perpendicular unit vectors ej can be found that correspond 
to a doubly (or triply) degenerate eigenvalue. 

Lemma 1. The roots of the secular equation (10-109) for a 
symmetric tensor are real. (10-111) 

Equation (10-109) is obtained from the eigenvalue equation (10-104) : 

T • C = T'C. (10-112) 

To prove the lemma, let us first allow T' to be complex. We will need 
also to allow the components C» of the vector C to be complex. A vector 
C with complex components, has no geometric meaning in the usual sense, 
of course, but we can regard all the algebraic definitions of the various 
vector operations as applying also to vectors with complex components. 
The various theorems of vector algebra will hold also for vectors with 
complex components. [There is one exception to these statements. The 
length of a complex vector cannot be defined by Eq. (3-13), but instead 
must be defined by 

|A| = (A* • A)*. (10-113) 

This definition will not be required here.] We will denote by C* the vector 
whose components are the complex conjugates of those of C. Let us multi- 
ply C* into Eq. (10-112): 

C* • T • C = 2T(C* • C). (10-114) 



10-4] DIAGONALIZATION OF A SYMMETRIC TENSOR 425 

If we take the complex conjugate of this equation, we have 

C • T • C* = r*(C* • C), (10-115) 

since T is real, and in view of Eq. (3-18). Now by definition (10-53), 

C* • T = T' • C*. (10-116) 

Hence 

C* • T • C = (C* • T) • C 

= (¥' • C*) • C [by Eq. (10-53)] 

= C • T' • C* [by Eq. (3-18)]. (10-117) 

For a symmetric tensor, T = T', so that the left members of Eqs. (10-114) 
and (10-115) are equal, and 

T' = T'*, (10-118) 

so that T' is real. 

Lemma 2. The eigenvectors of a symmetric tensor correspond- 
ing to different eigenvalues are perpendicular. (10-119) 

To prove this lemma, let us assume that T[, T 2 are two eigenvalues of T 
corresponding to the eigenvectors C 1( C 2 : 

T Ci = TiC 1( (10-120) 

T C 2 = r 2 C 2 . (10-121) 
We multiply C 2 into Eq. (10-120), and Ci into Eq. (10-121): 

C 2 • T • Ci = 2"i(C a • Ci), (10-122) 

d • T • C 2 = T' 2 (C 2 • Ci). (10-123) 
Since T is symmetric, the left members are equal, and we have 

(Ti - r'a)(C a • d) = 0. (10-124) 

If the eigenvalues T[, T' 2 are unequal, the eigenvectors C 2 , C x are per- 
pendicular. 

Lemma 3. In the case of double or triple degeneracy, Eqs. 
(10-108) have two or three mutually perpendicular solutions 
for the vector C. (10-125) 

For the proof of (10-125), suppose that T[ = T 2 . If we substitute 
T' = T[ in Eqs. (10-108), then by the theorem referred to in the 



426 



TENSOR ALGEBRA. INERTIA AND STRESS TENSORS 



[CHAP. 10 



footnote on page 423, there is at least one nontrivial solution C\, C2, 
C 3 . Let e( be a unit vector parallel to the vector (C\, C 2 , C 3 ). Then 



T • e' x = TWi. 



(10-126) 



Now, choose any pair of perpendicular unit vectors e 2 ' , e 3 ' perpendicular 
to ej, and use Eqs. (10-68) and (10-74) to transform the components of 
T into the double-primed coordinate system e^ , e 2 ', e 3 '. By a comparison 
of Eqs. (10-126) and (10-36), we see that we must get 



T'1'1 = T' u Till = 0, r'3'1 



0. 



(10-127) 



Since T is symmetric, its double-primed components must therefore be 
given by 

(T' x \ 
T = I T' 2 ' 2 r'2'3 • (10-128) 

\o r 2 ' 3 T'J 


Furthermore, the secular equation 




T\ - T' 






T'2'2 - T' r'2'3 


= (10-129) 




T' 2 ' 3 n's - T' 





must have the same roots as Eq. (10-109). This is true since the left mem- 
bers of both equations are the determinants of the same tensor (T — T'l), 
expressed in the unprimed and double-primed coordinate systems, and we 
noted at the end of Section 10-3 that the determinant of a tensor has the 
same value in all coordinate systems. If we expand the determinant 
(10-129) by minors of the first row, we obtain 



(Ti - T') 



22 



T 



T'l 



23 



23 



n'3 - r 



= 0. 



(10-130) 



Since T[ is a double or triple root of this equation, it must be a root of the 
equation 

rpil TV mil 

(10-131) 





T'2'2 - T' T' 2 ' 3 


= 0. 




rpn rptt rpt 

J- 23 * 33 — ■*■ 




Therefore the equations 


(T'2'2 - Ti)C' 2 ' + TWi = 0, 


T 


V'hC'i + (T'3'3 - T'x)C 


{ = 0, 



(10-132) 



10-4] 



DIAGONALIZATION OF A SYMMETRIC TENSOR 



427 



have a nontrivial solution which defines an eigenvector (0, C 2 , C 3 ) in the 
e^'eg -plane w i tn tne eigenvalue T[. We have therefore a second unit 
eigenvector e 2 parallel to (0, C 2 , C 3 ) and perpendicular to e[. If we take 
a third unit vector e^ perpendicular to e[, e 2 , then in this primed coor- 
dinate system, we must have 



T'u = T lt T' 21 = 0, 

T' 12 = 0, T' 22 = n, 
Thus T must have the components 



Tii = 0, 



(10-133) 



32 



T = 



fTi > 
Ti 

\0 n/ 



(10-134) 



and e{, e 2 , e^ are principal axes. If T[ were a triple root of Eq. (10-109), 
it would also be a triple root of the secular equation 



Ti 











T'i 











V 



(Ti - T'XTi - T')(T 3 - T') = 0. 

(10-135) 



Therefore T' 3 = T[, and we have three perpendicular eigenvectors cor- 
responding to the triple root T[ = T 2 = T' 3 . 

The above three lemmas complete the proof of the fundamental theorem 
(10-103). 

The algebra in this section may be generalized to vector spaces of any 
number of dimensions, with analogous results regarding the existence of 
principal axes of a symmetric tensor. 



At the end of Section 10-3 we noted that the trace and the determinant of 
a tensor T have the same value in all coordinate systems. We see from Eq. 
(10-101) that the trace is the sum of the eigenvalues of T: 



tr(T) = T x + T 2 + T 3> 
and the determinant is the product of the eigenvalues: 

det(T) = T1T2T3. 



(10-136) 



(10-137) 



We can form a third invariant scalar quantity associated with a symmetric 
tensor by summing the products of pairs of eigenvalues : 



M(X) = TiT 2 + T 2 T 3 + T 3 Tl 



(10-138) 



428 



TENSOR ALGEBRA. INERTIA AND STRESS TENSORS 



[CHAP. 10 



We can evaluate M (T) in any coordinate system by solving the secular equation 
(10-109) for the three roots T x , T 2 , T 3 and using Eq. (10-138). The solution of 
Eq. (10-109) can be avoided by noting that the sum (10-138) must be the 
coefficient of T' in Eq. (10-109), which is the sum of the diagonal minors of the 
determinant of T : 

T33T31 



M(T) = 



TuT 12 
T21T22 



+ 



T22T23 
T32T33 



+ 



T13T11 



(10-139) 



We could also show by direct calculation that M(T) as given by Eq. (10-139) 
has the same value after a coordinate transformation given by Eq. (10-74). 

For any tensor T, the determinant of (T — T'\) must have the same value 
in all coordinate systems. Therefore, in particular, the roots T' of Eq. (10-109) 
will be the same in all coordinate systems, even for a tensor T that is not sym- 
metric. We still call the roots T' the eigenvalues of T. If T is not symmetric, 
one eigenvalue will be real and the other two will be a conjugate complex pair. 
For the real eigenvalue we can find an eigenvector. For the complex eigen- 
values we cannot, in general, find eigenvectors. (That is, not unless we admit 
vectors with complex components, which have only algebraic significance. Even 
then, we cannot prove in general that the eigenvectors are orthogonal.) In any 
case, the expressions given by Eqs. (10-136), (10-137), (10-138), and (10-139) 
are still real and independent of the coordinate system. 



As an example of the diagonalization procedure, let us diagonalize the 
tensor 

T = AA + BD + DB, 

which obviously is symmetric. We will take 

A = 4aei, 

B = 7ae 2 + ae 3 , 

D = ae2 — 0,63. 

The tensor T is then represented in this coordinate system by the matrix: 



T = 






14a 2 



-6a' —2a' 
In this case, the secular equation (10-109) is 



16a 2 - T 







14a 2 - T 

-6a 2 





-6a 2 

-2a 2 - T 



= (16a 2 



T') 



X (T' 2 + 12a 2 T' - 64a 4 ) = 0. 



10-4] DIAGONALIZATION OF A SYMMETRIC TENSOR 429 

The roots (necessarily real) are 

T\ = 16a 2 , T' 2 = 16a 2 , T' 3 = -4a 2 . 
Equations (10-108), for the doubly degenerate root T' = 16a 2 , are 

0=0, 

-2a 2 C 2 - 6a 2 C 3 = 0, 

-6a 2 C 2 - 18a 2 C 3 = 0. 

Clearly, Ci is arbitrary, and the last two equations are both satisfied if 

C/2 == — 3t7 3 . 

Therefore any vector of the form 

C = Ciei - 3C 3 e 2 + C 3 e 3 

is an eigenvector for arbitrary C\, C 3 . Thus we have a two-parameter 
family of possible eigenvectors, from which we may select for e[, e 2 
any two perpendicular unit vectors. We will take 

ei = ei, 

,31 

©2 — — ;= 62 7= e 3 . 

Vio vlo 

We could have guessed e[ from the form of T. The reader should verify 
that ej and e 2 and, in fact, any vector in the eie2-plane satisfy Eq. 
(10-104) with T{ = 16a 2 . For T' = —4a 2 , Eqs. (10-108) become 

20a 2 C! = 0, 
18a 2 C 2 - 6a 2 C 3 = 0, 
-6a 2 C 2 + 2a 2 C 3 = 0. 
Now there is just a one-parameter family of solutions 

C\ = 0, C 3 = 3C 2 , 
of which there is one unit eigenvector (except for sign) : 

1 , 3 

e 3 = . — e 2 -\ — e 3 , 

Vio Vio 

where positive signs were chosen so that ej, e 2 , e^ would form a right- 
handed system. The vector e^ is perpendicular to the eje 2 -plane as it 



430 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

must be according to lemma 1. It may be verified that e 3 is an eigen- 
vector of T with the eigenvalue —4a 2 . 

By reference to Eq. (10-76), we may write the coefficients of the trans- 
formation to the principal axes of T: 

(an ai2 ai3\ A ° ° \ 

a 2 i a 22 a 23 )=lo 3/VW -l/VIo). 
a 3 i o 32 a 33 / \0 1/VlO 3/VlO / 

The reader should verify that these coefficients satisfy Eqs. (10-78) ; that 
the vectors ej are properly transformed according to Eqs. (10-71) and 
(10-72), which in this case are 

3 3 

Sjfc = 2~i °*»*jF»> ^a = 2^i aki ^'*> 

where e£ is the rth component of e'j in the unprimed coordinate system 
and 8jk is the fcth component of ej in the primed coordinate system; and 
also that T is properly transformed according to Eq. (10-74) from its 
original form to its diagonal form. 

10-5 The inertia tensor. The inertia tensor of a rigid body is given by 
Eq. (10-25). For a body of density p(x, y, z), we may rewrite the inertia 
tensor as ,,, 

| = jjfp{r 2 \ - it) dV, (10-140) 

where we have used the subscript "o" to remind us that the inertia tensor 
is calculated with respect to a set of axes with origin at O. We will omit the 
subscript except when the discussion concerns more than one origin. The 
diagonal components of I are just the moments of inertia [Eq. (5-80)] 
about the three axes: 

l xx = fff P (y 2 + z 2 ) dV, 

Iyy = ///P(* 2 + x 2 ) dV, (10-141) 

hz = fffp(x 2 + y 2 ) dV. 

The off-diagonal components, often called products of inertia, are 

Ixy = Iyx= — fffpxy dV > 

I V z ==/.„=- fffpyz dV, (10-142) 

Izx = Ixz = — fffP ZX dV - 



10-5] 



THE INERTIA TENSOR 



431 



Since we may use Eq. (10-74) to calculate the components of the 
inertia tensor relative to any other set of axes through the same origin 0, 
we see from Eqs. (10-74) and (10-141) that the moment of inertia about 
any axis through 0, in a direction designated by the unit vector n, is 



J„ = n 



n. 



(10-143) 



It often is easier to calculate the components of the inertia tensor with 
respect to a conveniently chosen set of axes and then use Eq. (10-143), 
than to calculate 7 n directly, if the axis n is not an axis of symmetry of the 
body. 

We can obtain a useful analog to the Parallel Axis Theorem (5-81) for 
the moment of inertia by calculating the inertia tensor \ relative to an 
arbitrary origin of coordinates in terms of the inertia tensor Iff rela- 
tive to the center of mass G. Let r and r' be position vectors of any point 
P in the body relative to O and G respectively, and let R be the coordinate 
of G relative to (Fig. 5-12), 



r = r' + R. 



(10-144) 



Then we have, from Eq. (10-140), 

\ = ///p[(r' + R) ■ (f + R)l - Of + R)(r> + R)] dV 
= fffplif ■ r')l - r'r'] dV + [(R • R)l - RR]///p dV 
+ 21 [R • ///pr' dV] - [fff P r> dV] R - R///pr' dV. 



(10-145) 



In view of the definition (5-53) of the center of mass, we have 

jjfpr' dV = 0. (10-146) 



Equation (10-145) therefore reduces to 

l„ = l G + M(fl 2 l — RR). 



(10-147) 



Note that both the statement and proof of this theorem are in precise 
analogy with the Parallel Axis Theorem (5-83) for the moment of inertia. 

It is evident from the definition (10-140) that the inertia tensor of a 
composite body may be obtained by summing the inertia tensors of its 
parts, all relative to the same origin. 

If a body rotates, the components of its inertia tensor, relative to sta- 
tionary axes, will change with time. The components relative to axes 
fixed in the body, of course, will not change if the body is rigid. We may 
think of the inertia tensor I as rotating with the body. If the (constant) 



432 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

components along axes fixed in the body are given, the (changing) com- 
ponents along stationary axes are then given by Eq. (10-89), where A 
represents the transformation from, body axes to space axes. The most 
convenient set of axes in the body for most purposes are the principal 
axes of the inertia tensor, also called the principal axes of the body. The 
eigenvalues of the inertia tensor are called the principal moments of in- 
ertia. We will learn more in the next chapter of the dynamical significance 
of the principal axes, but we may note here that, according to Eq. (10-24), 
if the body rotates about a principal axis, the angular momentum is paral- 
lel to the angular velocity. We may always choose arbitrary axes, compute 
I, and then use the method of Section 1-4 to find the principal axes. It is 
often possible, however, to simplify the problem by choosing to begin with 
a coordinate system in which one or all of the axes are principal axes. 

In many cases, a body will have some symmetry, so that we can see 
that certain of the products of inertia (10-142) will vanish if the axes are 
chosen in a certain way. For example, we can prove the following theorem: 

Any plane of symmetry of a body is perpendicular to a prin- 
cipal axis. (10-148) 

If we choose the yz-plane as the plane of symmetry, then 

p(-x, y, z) = p(x, y, z). (10-149) 

It is easy to show that because of Eq. (10-149) the integrals (10-142) for 
I X y and I iX will vanish. Therefore the a;-axis is a principal axis in this case. 
In a similar way, we can prove the theorem: 

Any axis of symmetry of a body is a principal axis. The plane 
perpendicular to this axis is a principal plane corresponding 
to a degenerate principal moment of inertia. (10-150) 

A sphere, or a body with spherical symmetry has, evidently, a constant 
inertia tensor. 

As an example, consider the right triangular pyramid shown in Fig. 
10-1. The components of the inertia tensor relative to the axes (x, y, z) 
are to be calculated from the formula: 

/•Jo ra-%t j.a-y-%z /y 2 -\- z 2 — X y —ZX \ 

I = / / / pi — xy z 2 + x 2 —yz I dx dy dz, 

J z=0 J y=0 Jx=0 \ 2 I 2/ 

\ — zx —yz x + y I 

where each component of I is to be obtained by evaluating the indicated 
integral over the corresponding component of the matrix. The density 
p is given in terms of the mass M by 

M = i a 3 p. 



10-5] 



THE INERTIA TENSOR 



433 




Fig. 10-1. A right triangular pyramid. 



Because of the symmetry between x and y, it is necessary to evalute only 
the four integrals 

J i = fllPZ 2 dx dy dz = [[[py 2 dx dy dz = -fa Ma 2 , 

J 2 = [I I pz 2 dx dydz = ^g Ma 2 , 

Jz = lijpxy dx dydz = -g$ Ma 2 , 

J 4 = III pxz dx dy dz = Iffpyz dx dy dz = ^ Ma 2 . 

The inertia tensor is then given by 

(Ji + J 2 -Js -J*\ I 13 -2 -3) 




:;)■(: 




Ji + J 2 -J 4 ) = \ -2 
-J 4 2JJ \-3 

where the notation means that each element of the matrix is to be mul- 
tiplied by Ma 2 /40. Let us find the principal axes. By symmetry [theorem 
(10-148)] the axis x" shown in Fig. 10-1 is a principal axis. Let us there- 
fore first transform to the axes x", y", z. The coefficients of the trans- 
formation are, by Eq. (10-68), 

(ax», a x - v a x »\ jl/y/2 -1/V2 0\ 
o»»« ay. y <V'J=(l/\/2 1/V2 0l- 
a-zx o. zy a zl / \0 1/ 



434 



TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 



Using Eq. (10-74), we now calculate the inertia tensor components along 
the x"-, y"-, and z-axes: 



I 



r 15 
11 



> 

-3V2 



Ma 2 
40 



\ -3V2 8 

We see that the x"-axis is indeed a principal axis. The secular equation is 
15 - X 



11 - X -3V2 =0, T' = 

-3V2 8 - X 

and the roots are 

\ x > = 15, \ v ' = 5, X z < = 14, 



Ma' 
40 



X, 



or 



2V = £ Ma 2 , T y . = i Ma 2 , TV = £> Ma 2 



Equations (10-108) can be solved for the components of the unit vectors 
i', j', k' in terms of i", j", k: 

i' = i", 

j' = 4V3J" + *V6k, 
k' = -fVoj" + iVSk. 

As a second example, let us find the inertia tensor about the point of 
the object shown in Fig. 10-2. The object is composed of three flat disks 





Fig. 10-2. Three disks. 



Fig. 10-3. A circular disk with its 
principal axes. 




10-5] THE INERTIA TENSOR 435 

of mass M and radius a. By symmetry, the principal axes are the indi- 
cated axes x, y, z. We first calculate the inertia tensor of a single disk 
about its center, relative to its principal axes x', y', z', as shown in Fig. 
10-3. The moment of inertia I z ' is given by Eq. (5-90), and the moments 
of inertia I x >, I y - are half I z >, by the Perpendicular Axis Theorem (5-84). 
We can therefore write the inertia tensor of a disk, relative to its principal 
axes x', y', z', as 

/l 0\ „, 2 

(10-151) 



For the bottom disk, the principal axes are parallel to x, y, z, and we need 
only apply theorem (10-147) to obtain its inertia tensor relative to the 
x-, y-, z-axes with origin at O: 

l = l ff + M(3a 2 l - 3a 2 kk) 

/l3 0\ ,, 2 
= o 13 )*%-. 
\0 2/ 4 

For the right-hand disk, with the axes x', y', z' oriented as shown, we first 
apply theorem (10-147) to obtain the inertia tensor about 0, relative to 
axes parallel to x', y', z'\ 

(b 0\ M 2 
l.cv.')=(0 10^- 
\0 6/ 

The transformation from x'-, y'-, z'-axes to x-, y-, z-axes is given by 

(a xx > a xy ' a xz \ /l \ 

a yx ' a vv > a yz > I = I J \y/Z I • 
a zx ' a zy > a zz >l \0 —\\/Z J / 

We now use Eq. (10-74). It is perhaps easier to carry out the process in 
two steps, according to Eq. (10-89):* 

(5 \ „ 2 

i 3V3)^- 
-i\/3 3 / 



* Matrices may be multiplied conveniently according to the rule (10-45) 
by noting that the element (T • S)i* is obtained by summing the products of 
pairs of elements across row i in T and down column k in S. 



436 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

Now 



\/l 
v/3 



(A • l.(,v.',) • A' = I i 3V3 1 1 J 
\0 -iV3 3 /\0 iV3 

= ■.<*,.) = 10 ^ |V3) Ma 





2 



We may interpret this algebra as a computation of l relative to a new 
set of axes. Alternatively, we may interpret A as a tensor which rotates 
the disk through an angle of 60° about the z-axis; l (xv*') * s * nen tne 
moment of inertia of a disk whose principal axes are parallel to x, y, z, and 
the algebra is a computation of the effect of rotating the disk to its final 
position. The left-hand disk, correspondingly, has the inertia tensor 

\ 7M- 2 

V -*V5)^- 

-iV3 




l 



The inertia tensors of the three disks may now be added and we obtain 

Ma 2 



f23 0\ „ 2 



l = I 22J 

\0 6i/ ' 

Let us calculate the moment of inertia of the object shown in Fig. 10-2 
about the y'-axis through 0. By Eq. (10-143), we have 

h = J' • U • j' = lOpfa 2 . 

We could use theorem (10-147) to obtain I about the center of gravity G, 
which is at the intersection of the z- and z'-axes. It is clear from symmetry 
that any axis perpendicular to one of the three disks through its center is 
a principal axis relative to G. This can only be true if the inertia tensor rel- 
ative to G has a double degeneracy in the yzy'z' '-plane. The reader should 
check this by carrying out the translation of l„ to the center of mass G. 
It should be noted that the principal axes of the inertia tensors of a body 
relative to two different points and 0', in general, will not be parallel, 
as experimentation with Eq. (10-147) will show. 

The kinetic energy T of a rotating rigid body can also be expressed con- 
veniently in terms of the inertia tensor. From Eqs. (10-2) and (10-3) 
and using the rules of vector algebra, we have 



10-5] 



THE INERTIA TENSOR 



437 



TV 



T = J2 W* 

N 



N 
7c=l 

= i« • l. 

Therefore T can be expressed in the form 

T = \u> ■ I • w. 



(10-152) 



(10-153) 



Equation (10-153) expressed in terms of components along any set of 
axes is then 

(10-154) 

This is the equation of a family of quadric surfaces in w-space, each sur- 
face the locus of angular velocities for which the kinetic energy has a con- 
stant value T. If Eq. (10-153) is written in terms of components along 
principal axes x', y', z', 



w; + \iw + wt = t, 



i r',,' 2 — 



(10-155) 



then we see that these surfaces are ellipsoids, since the moments of inertia 
are necessarily positive. If we define a vector 



_ a 



where a is a constant, then Eq. (10-153) can be written as 



r • I • r = o 



(10-156) 



(10-157) 



This is the equation of the inertia ellipsoid. The constant a determines the 
size of the ellipsoid. It is customary to set a = 1 in whatever units are 
being used, for example, a = 1 cm-erg-sec. In this case, we note that the 
size of the ellipsoid (but not its shape) depends on the units being used. 

The inertia ellipsoid of a body, like its inertia tensor, is relative to a 
particular origin about which moments are computed. The six coefficients 
of. the quadratic form on the left of Eq. (10-157) are the components of 
the inertia tensor: 



I xx x 2 + I yv y 2 + I zl z 2 + 2I xy xy + 2I yz yz + 2I tx zx = a 2 , (10-158) 



438 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

so that the inertia tensor is uniquely characterized by the corresponding 
inertia ellipsoid. This gives us another convenient geometrical way of 
picturing the inertia tensor. 

By comparing Eq. (10-157) with Eq. (10-143), we see that the radius 
to any point on the inertia ellipsoid is 

r = air*, (10-159) 

where I t is the moment of inertia about an axis parallel to r. In particular, 
the principal moments of inertia are related by Eq. (10-159) to the semi- 
principal axes of the inertia ellipsoid. We see that if there is double de- 
generacy, the inertia ellipsoid is an ellipsoid of revolution. If the principal 
moments of inertia are all equal, the ellipsoid of inertia is a sphere. 

For any symmetric tensor T, we can form a quadratic equation of the 
form (10-157) which defines a quadric surface that uniquely charac- 
terizes T. The principal axes of T are the principal axes of its associated 
quadric surface. If the eigenvalues of T are all positive, the surface is an 
ellipsoid. Otherwise, it will be a hyperboloid or a cylinder. If all the 
eigenvalues are negative, we would need to write —a 2 for the right member 
of the quadratic equation in order to define a real surface. 

10-6 The stress tensor. Let us represent any small surface element in 
a continuous medium by a vector dS whose magnitude dS is equal to the 
area of the surface element and whose direction is perpendicular to the 
surface element. To specify the sense of dS, we will distinguish between 
the two sides of the surface element, calling one the back and the other 
the front. The sense of dS is then from the back to the front. We may 
then describe the state of stress of the medium at any point Q by specify- 
ing the force P(dS) exerted across any surface element dS at Q by the 
matter at the back on the matter at the front of dS. We understand, of 
course, that the surface element dS is infinitesimal. That is, all statements 
we make are intended to be correct in the limit when all elements dS — > 0. 
For a sufficiently small surface element, the force P may depend on the 
area and orientation of the surface element, but not on its shape. Thus P 
is indeed a function only of the vector rfS at any particular point Q in the 
medium. We will show that P(dS) is a linear function of dS. We may 
therefore represent the function P(dS) by a tensor P, the stress tensor:* 

P(eZS) = P • dS. (10-160) 



* The reader is cautioned that many authors define the stress tensor with the 
opposite sign from the definition adopted here, so that a tension is a positive 
stress and a pressure, a negative stress. The latter convention is almost universal 
in engineering practice, whereas the definition adopted here is more common in 
works on theoretical physics. 



10-6] THE STRESS TENSOR 439 

P(dS 2 ) 




> P(-dS! — dS 2 ) 
-(dSj + dS 2 ) 
Fig. 10-4. A triangular prism in a continuous medium. 

To show that P(dS) is a linear vector function, we note first that if dS 
is small enough so that the state of stress of the medium does not change 
over the surface element, then the force P will be proportional to the 
area dS so long as the orientation of the surface is kept fixed. Thus for 
a positive constant c, 

P(cdS) = cP(dS). (10-161) 

If the direction of dS is reversed, the back and front of the surface ele- 
ment are interchanged, and therefore by Newton's third law, P(— dS) = 
— P(dS), so that Eq. (10-161) holds also if c is negative. Now, given any 
two vectors dS x , dS 2 , let us imagine a triangular prism in the medium with 
two sides dS u dS 2 , as in Fig. 10-4. If the end faces are perpendicular to 
the sides, then the third side is — dSi — dS 2 , as shown. If the length of 
the prism is made much greater than the cross-sectional dimensions, we 
may neglect the forces on the end faces, and the total force on the prism is 

dF = P(dSx) + P(dS 2 ) + P(-dSt - dS 2 ). (10-162) 

If the density is p, the acceleration of the prism is given by Newton's 
law of motion: 

p dVa = dF. (10-163) 

Now if we reduce all linear dimensions of the prism by a factor a, the areas 
dSi are multiplied by a 2 ; hence by Eq. (10-161), dF is multiplied by a 2 , 



440 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

and dV is multiplied by a 3 , so that 

ap dV&(a) = dF, (10-164) 

where a(a) is the acceleration of a prism a times smaller. Now as a — * 0, 
the acceleration should not become infinite; hence we conclude that 

dF = 0, (10-165) 

from which, by Eqs. (10-162) and (10-161), 

P(dS!) + P(eZS 2 ) = P(dS x + dS a ). (10-166) 

Equations (10-161) and (10-166) show that the function P(dS) is linear. 
Note that Eqs. (10-166) and (10-162) imply that there is no net force on 
the prism if the stress function P(rfS) is the same at all faces. Any net 
force can only result from differences in the stress at different points of 
the medium; such differences reduce to zero as a — ► 0. 

By considering small square prisms, and recognizing that the angular 
acceleration must not become infinite as the size shrinks to zero, we can 
show by a very similar argument (see Problem 32) that P must be a 
symmetric tensor. The stresses at each point Q in a medium are therefore 
given by specifying six components of the symmetric stress tensor P. 

If the medium is an ideal fluid whose only stress is a pressure p in all 
directions, the stress tensor is evidently just 

P = pi. (10-167) 

Note that we did not prove in Chapter 8 that in an ideal fluid, that is, 
one which can support no shearing stress, the pressure is the same in all 
directions. This was proved only in Chapter 5 for a fluid in equilibrium. 
This logical defect can now be remedied. (See Problem 33.) 

According to the definition of P, the total force due to the stress across 
any surface S is the vector sum of the forces on its elements: 



//' 



'P-dS. (10-168) 

s 

If S is the closed surface surrounding a volume V of the medium, and if 
we take n to be the conventional outward normal unit vector, then the 
total force exerted on the volume V by the matter outside it is 

F = - //n • P dS, (10-169) 

s 
and by the generalized Gauss' theorem, [see discussion below Eq. (5-178)], 

F = - fffv ■ P dV. (10-170) 

v 



10-6] THE STRESS TENSOR 441 

Since V is any volume in the medium, the force density due to stress is 

f s = -V P. (10-171) 

In agreement with an earlier discussion, we see that this force density 
arises only from differences in stress at different points in the medium. 
Equation (10-171) may also be derived by summing the forces on a small 
rectangular volume element. 

The equation of motion (8-138) may now be generalized to apply to any 
continuous medium: 

P ^ + V ■ P = f. (10-172) 

This equation may also be rewritten in the form (8-139) : 

J + T .Vv + Jv.P = f (10-173) 

at p p 

This equation, together with the equation of continuity (8-127), de- 
termines the motion of the medium when the body force density f and the 
stress tensor P are given. The stress P at any point Q may be a function 
of the density and temperature, of the relative positions and velocities 
of the elements near Q, and perhaps also of the previous history of the 
medium, which may be a solid (elastic or plastic) or a fluid (ideal or 
viscous). 

From Eqs. (10-172) and (10-173) we can derive conservation equa- 
tions analogous to those derived in Section 8-8. The conservation equation 
for energy analogous to Eq. (8-149) is, for example, 

j t {W 5F) = v • (f - V • P) bV: (10-174) 

The further manipulations of the energy equation carried out in Sec- 
tion 8-8 cannot all be carried through in the same way for Eq. (10-174) 
because of the difference in form between the stress term here and the 
pressure term in Eq. (8-149), as the reader may verify. The energy 
changes associated with changes in volume and shape of an element in a 
continuous medium are in general more complicated than those associated 
with expansion and contraction of an ideal fluid. 

In a viscous fluid, the stress tensor P will be expected to depend on the 
velocity gradients in the fluid. This is consistent with the dimensional 
arguments in Section 8-14, where we saw that the term V • P in 
Eq. (10-172) must consist of the coefficient of viscosity i\ multiplied by 
some combination of second derivatives of the velocity components with 
respect to x, y, and z. If the fluid is isotropic, as we shall assume, then 
the relation between P and the velocity gradients must not depend on the 
orientation of the coordinate system. We can guarantee that this will 



442 



TENSOR ALGEBRA. INERTIA AND STRESS TENSORS 



[CHAP. 10 



be so by expressing the relation in a vector form that does not refer ex- 
plicitly to components. The dyad 



j dv x 
dx 



Vv = 



dVy 

dx 



dx 



f\ 



dv x dv y dv z 
dy dy by 



\ dv x 
\dz 



dVy 
dz 



dv z 
dz 



(10-175) 



has as its components the nine possible derivatives of the components of 
v with respect to x, y, and z. Hence we must try to relate P to Vv. 
The dyad (10-175) is not symmetric, but we can separate it into a sym- 
metric and an antisymmetric part, as in Eqs. (10-64) through (10-66): 



Vv = (W). + (Vv)„, 

(Vv). = ivv + i(w)' f 

(Vv)„ = JT7- i(Vv)'. 



(10-176) 
(10-177) 
(10-178) 



The antisymmetric part is related, as in Eqs. (10-62) and (10-63), to a 
vector 

«=|VXv, (10-179) 

such that for any vector dx, 



(Vv)„ • dx = u X dx. 



(10-180) 



If dx is the vector from a given point Q to any nearby point Q', we see that 
the tensor (Vv) a selects out those parts of the velocity differences between 
Q and Q' which correspond to a (rigid) rotation of the fluid around Q with 
angular velocity a. This is in agreement with the discussion of Eq. (8-133), 
which is identical with Eq. (8-179). Since no viscous forces will be asso- 
ciated with a pure rotation of the fluid, the viscous forces must be expres- 
sible in terms of the tensor (Vv).. 
Since P is also symmetric, we are tempted to write simply 



P = C(Vv)., 



(10-181) 



where C is a constant. In the simple case depicted in Fig. 8-10, the only 
nonzero component of Vv is dv x /dy, and Eqs. (10-181) and (10-177) then 
give 

iC% o\ 

It » o 

dy 



\%c 












(10-182) 



. 



10-6] THE STRESS TENSOR 443 

and the viscous force across dS = j dS will be 

dF = P • dS = i C —^ dSi, (10-183) 

in agreement with Eq. (8-243) if C = —2i\. A negative sign is clearly 
needed, since the viscous force opposes the velocity gradient. However, 
Eq. (10-181) is not the most general linear relation between P and Vv 
that is independent of the coordinate system. For we can further de- 
compose (Vv), into a constant tensor and a traceless symmetric tensor 
in the following way: 

(Vv), = (Vv) c + (Vv)„, (10-184) 

(Vv) c = i7V(W).l = iV • vl, (10-185) 

(Vv)«. = (Vv), - iV • vl. (10-186) 

This decomposition is independent of the coordinate system, since we 
have shown that the trace is an invariant scalar quantity. We see by 
Eq. (8-116) that the tensor (Vv)„ measures the rate of expansion or con- 
traction of the fluid. The tensor (Vv) <8 , with five independent com- 
ponents, specifies the way in which the fluid is being sheared. We are 
therefore free to set 

P = -2 v (Vv)» - WV • vl, (10-187) 

with a coefficient r; which characterizes the viscous resistance to shear, 
and a coefficient 17' which characterizes a viscous resistance, if any, to ex- 
pansion and contraction. The last term corresponds to a uniform pressure 
(or tension) in all directions at the given point. The coefficient i\' is small 
and not very well determined experimentally for actual fluids. According 
to the kinetic theory of gases, ij' is zero for an ideal gas. To the viscous 
stress due to velocity gradients, given by formula (10-187), must be added 
a hydrostatic pressure which may also be present and which depends on 
the density, temperature, and composition of the fluid. If we lump the 
last term in Eq. (10-187) together with the hydrostatic pressure into a 
total pressure p, then the complete stress tensor is 

P = pi - v [Vv + (Vv)' - §V • vl]. (10-188) 

The reader may readily write this out in terms of components. 

Formula (10-188) is the most general expression for the stress in an 
isotropic fluid in which there is a hydrostatic pressure, plus viscous forces 
proportional to the velocity gradient. It is possible to imagine that the 
stress might also contain nonlinear terms in the velocity gradients, or even 
high-order derivatives of the velocity, but such terms could be expected 



444 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

to be small in comparison with the linear terms. Experimentally, the 
viscous stresses in fluids are given very accurately in most cases by 
formula (10-188). 

We have decomposed the tensor Vv, with nine independent com- 
ponents, into a sum of three tensors with one, three, and five independent 
components, each a linear combination of the components of Vv. A 
similar decomposition is clearly possible for any tensor. The reader may 
well ask whether any further decomposition is possible. This is a problem 
in group theory. We state without proof the result. Neither an anti- 
symmetric tensor nor a symmetric traceless tensor can be further de- 
composed in a manner independent of the coordinate system. The reader 
can convince himself that this is plausible by a little experimentation. 

Let us now consider an elastic solid. Let the solid be initially in an 
unstrained position, and let each point in the solid be designated by its 
position vector r relative to any convenient origin. Now let the solid be 
strained by moving each point r to a new position given by the vector 
r + p(r) relative to the same origin. We will designate the components 
of r by (x, y, z) and of p by (£, j/, f). If p were independent of r, the mo- 
tion would be a uniform displacement without deformation. Hence the 
strain at any point may be specified by the gradient dyad 



Vp 



dx dx dx \ 

a| dv at 

dy dy dy 

\dz dz dz j 



(10-189) 



Now Vp can again be decomposed into an antisymmetric part which 
corresponds to a rigid rotation about the point r + P and into a sym- 
metric part which describes the deformation of the solid in the neighbor- 
hood of each point : 

S = iVp + i(Vp)'. (10-190) 

The symmetric part can be further decomposed into a constant tensor 
describing a volume compression or expansion and a symmetric traceless 
tensor which describes the shear: 

S C = JV- P 1 =i^Pl, (10-191) 

S, t = £V P + i(Vp)' - Jv • pi. (10-192) 

If the solid is isotropic, then this is the most general possible decomposi- 



PROBLEMS 445 

tion. Furthermore, if Hooke's law holds, the stress should be proportional 
to the strain: 

P = -|oV -pi —bS, t . (10-193) 

The constants a and b are evidently related to the bulk modulus and the 
shear modulus. If the solid is not isotropic, as for example, a crystal, 
then the relation between P and S may depend on the choice of axes, and 
must therefore be written: 

Pa = Yj c iwSu. (10-194) 

k,l=l 

Since P and S have six independent components each, there are thirty-six 
constants dju- By using the fact that there is an elastic potential energy 
which is a function of the strain, it can be shown that in the most general 
case there are twenty-one independent constants Cijki- 



Problems 

1. The product ct of a tensor by a scalar has been used in the text without 
formal definition. Remedy this defect by supplying a suitable definition and 
proving that this product has the expected algebraic properties. 

2. Show that the centrifugal force in Eq. (7-37) is a linear function of the 
position vector r of the particle, and find an expression for the corresponding 
tensor in dyadic form. Write out the matrix of its coefficients. 

3. Define time derivatives dt/dt and dTt/dt relative to fixed and rotating 
coordinate systems, as was done in Chapter 7 for derivatives of vectors. Prove 
that 

where the cross product of a vector with a tensor is defined in the obvious way. 

4. Write out the relations between the coefficients 0,7 corresponding to the 
relations (10-80). Write down another relation between the unit vectors, in- 
volving a triple cross product, and write out the corresponding relations be- 
tween the coefficients. 

5. Transform the tensor 

T = AB -f- BA, 

where 

A = 5i - 3j + 2k, B = 5j + 10k, 

into a coordinate system rotated 45° about the «-axis, using Eq. (10-74). Trans- 
form the vectors A and B, using Eq. (10-71), and show that the results agree. 



446 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

6. Write down and prove two additional relations like those in Eqs. (10-91) 
through (10-93), involving algebraic properties which are preserved by trans- 
formations of coordinates. 

7. Prove Eqs. (10-91) and (10-92). 

8. Write down the matrix for the orthogonal tensor A which produces a 
rotation by an angle a about the 2-axis. Decompose A into a symmetric and 
an antisymmetric tensor as in Eq. (10-66). What is the geometrical interpre- 
tation of this decomposition? 

9. Prove that Det (T) [Eq. (10-95)] is the same in all coordinate systems. 

10. (a) Prove that the tensor given by formula (10-100) has the property 
given by Eq. (10-23). (b) Prove by direct calculation that this tensor is repre- 
sented by the same matrix in all coordinate systems. 

11. Prove by direct calculation that the quantity M(T) denned by Eq. (10-139) 
has the same value after the coordinate transformation (10-74). 

12. Diagonalize the tensor in Problem 5. (That is, find its eigenvalues and 
the corresponding principal axes.) 

13. Diagonalize the tensor 

(7 V6 -V3\ 

VQ 2 -5V2 J • 

-\/3 — 5\/2 —3 / 

(Hint: The secular equation can be factored; the roots are all integers.) 

14. What are the principal axes and corresponding eigenvalues of the tensor 
in Problem 2? Interpret physically. 

15. Verify the statements made in the last paragraph of Section 10-4 regard- 
ing the principal axis transformation found in the worked-out example. 

16. Prove that if two tensors S and T have a set of principal axes in common, 
then S • T = T • S. (The converse is also true.) 

17. Prove that if a tensor T satisfies an algebraic equation 

aj' n -\ h a 2 T 2 + aiT + a \ = 0, 

where T"' means T • T • • • T (n factors), then its eigenvalues must satisfy the 
same equation. The null tensor is defined in the obvious way. 

18. Use the result of Problem 17 to show that the eigenvalues of the tensor 
A representing a 180° rotation about some axis can only be ±1. [Hint: Consider 
the result of applying A twice.] Show that the roots cannot all be +1. Then 
show that —1 must be a double root. [Hint: Use Eqs. (10-137) and (10-81).] 
Can you guess the corresponding eigenvectors? This entire problem is to be 
answered by using general arguments, without writing down the matrix for A. 

19. Show that the eigenvalues of an orthogonal tensor [Eq. (10-87)] are com- 
plex (or real) numbers of unit magnitude. [Hint: Let C be an eigenvector (pos- 
sibly complex) of T, and consider the quantity (T • C) • (T • C*).] Hence show 
that one eigenvalue must be ±1, and the other two are of the form exp (±i a), 
for some angle a. 



PROBLEMS 



447 



20. Write out the components of the orthogonal tensor A corresponding to a 
rotation by an angle 6 about the z-axis. Find its eigenvalues. Find and interpret 
the eigenvectors corresponding to the real eigenvalue. 

21. Find the components of the tensor corresponding to a rotation by an 
angle 6 about the z-axis, followed by a rotation by an angle ip about the y-axis. 
Find its eigenvalues. (Hint: According to Problem 19, one eigenvalue is ±1; 
hence you can factor the secular equation.) Show that the result implies that 
this transformation is equivalent to a simple rotation about some axis. (You 
are not asked to find the axis.) Find the angle of rotation by comparing your 
result with the eigenvalues found in Problem 20. 

22. Show that the eigenvalues of an antisymmetric tensor are pure imaginary 
(or zero) . Hence show that an antisymmetric tensor must have one zero eigen- 
value and two conjugate imaginary eigenvalues. Find the eigenvectors corre- 
sponding to the zero eigenvalue for the tensor (10-62). 

23. Find the inertia tensor of a straight rod of length I, mass m, about its 
center. Use this result to find the inertia tensor about the centroid of an equi- 
lateral pyramid constructed out of six uniform rods. Show that this tensor can 
be written down immediately from symmetry considerations, given the result 
of Problem 17, Chapter 5. 

24. Translate to the center of mass G the inertia tensor calculated about the 
origin for the three disks in Fig. 10-2. Verify the statement made in the text 
regarding the double degeneracy of lg. 

25. Calculate the moment of inertia of a circular cone about a slant height. 
[Hint: Calculate the inertia tensor about the apex relative to principal axes, 
and use Eq. (10-143).] 

26. Formulate and prove the most comprehensive theorem you can with 
regard to the inertia tensor of a plane lamina. What can you say about the 
principal axes and principal moments of inertia? 

27. Find, by whatever method requires the least algebraic labor, the inertia 
tensor of a uniform rectangular block of mass M , dimensions a X b X c, about 
a set of axes through its center, of which the z-axis is parallel to side c, and the 
2/-axis is a diagonal of the rectangle o X b. 

28. (a) A uniform sphere of mass M , radius 
a, has two point masses \M, ^M, located on 
its surface and separated by an angular dis- 
tance of 45°. Find the principal axes and 
principal moments of inertia about the center 
of the sphere, (b) Find the inertia tensor 
about parallel axes through the center of 
mass. Are they still principal axes? 

29. (a) Find the inertia tensor of a plane 
rectangle of mass M, dimensions a X 6. (b) 
Use this result to find the inertia tensor about 
the center of mass of the house of cards shown 
in Fig. 10-5. Each card has mass M, dimen- 
sions a X b(a < b). Use principal axes. Fig. 10-5. A house of cards. 




448 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [CHAP. 10 

30. Find the equation for the ellipsoid of inertia of a uniform rectangular 
block of dimensions I X w X h. 

31. Find the equation for the ellipsoid of inertia of an object in the shape 
of an ellipsoid whose equation is 

2 2 2 

I 2 ^ w 2 ^ h 2 

32. Prove that the stress tensor P is symmetric. 

33. Prove that if there is no shear on any surface element at some point, then 
the stress tensor P at that point is a constant tensor [Eq. (10-23)]. 

34. Derive Eq. (10-171) by calculating the net force on a rectangular volume 
element. 

35. Derive from Eq. (10-173) the equation 

^+V.(pw+P)=f, 

which expresses the conservation of linear momentum. Show from this equation 
that the momentum current tensor (pw + P) represents the flow of momentum, 
and interpret physically the two terms in this tensor. 

36. Derive from Eq. (10-172) a law of conservation of angular momentum 
in a form analogous to Eq. (8-148). 

37. Write the equations of motion (10-173) in cylindrical components for a 
moving viscous fluid. Use these equations, together with suitable assumptions, 
to derive Poiseuille's law (8-252) for steady viscous flow in a pipe. Write the 
stress tensor for this case, in cylindrical coordinates, as a function of r, z, <p. 

38. Write out the components of the stress tensor P in a viscous fluid. 

39. (a) Show that the rate of production of kinetic energy per unit volume 
due to stresses in a moving medium is 

= -v(V-P). 

(b) Show that the rate at which work is done by the stresses on the medium, 
per unit volume, is 

— =-V.(P-v). 

{Hint: Calculate the work done across the surface of any volume V and use 
Gauss' theorem.) 

(c) Using these results, calculate the rate at which energy is dissipated per 
unit volume by viscous stresses in a moving fluid. Write it out in terms of 
components. 

40. Find the relation between the constants a and b in Eq. (10-193) and the 
bulk modulus B and shear modulus n defined by Eqs. (5-116) and (5-118). 
[Set up S and P for the situations used in defining B and n.] 

41. A solid is subject to a stress consisting of a pure tension t per unit area 
in one direction. Find the strain S in terms of r, a, and 6. Using this, and the 



PROBLEMS 449 

result of Problem 40, express Young's modulus Y [Eq. (5-114)] in terms of B 
and n. [Hint: Use symmetry to determine the form of S.] 

*42. Find the most general linear relation between S and P for a nonisotropic 
elastic substance which possesses cylindrical symmetry relative to a specified 
direction. 

*43. (a) Assume that in a nonisotropic elastic solid, there is an elastic potential 
energy V per unit volume, which is a quadratic function of the strain com- 
ponents. Show that there are 21 constants required to specify V. 

(b) Show that if the strain in a solid in equilibrium is increased by SS, the work 
done per unit volume against the stresses (exclusive of any work done against 
body forces) is 



$W J Pa dS fj : 



[Hint: Calculate the work done on a volume element by the stresses on its 
surface and use Gauss' theorem.] 

(c) Combine results (a) and (b) to show that P is a linear function of S in- 
volving 21 independent constants, in general. 



cflP 
dt : 


= F, 


dL 
dt 


= N, 


P = 


MV, 


L = 


l«, 



CHAPTER 11 

THE ROTATION OF A RIGID BODY 

11-1 Motion of a rigid body in space. The motion of a rigid body in 
space is determined by Eqs. (5-4) and (5-5) : 

(11-1) 

(H-2) 

where 

(11-3) 

(11-4) 

F and N are the total force on the body and the total torque about a 
suitable point 0, V is the velocity of the center of mass, and I and w are 
the inertia tensor and the angular velocity about the point 0. For an un- 
constrained body moving in space, the point is to be taken as the center 
of mass. If the body is constrained by external supports to rotate about 
a fixed point, that point is to be taken as the point 0. If the point is 
constrained to move in some fashion, the reader may supply the appro- 
priate equation of motion. (See Chapter 7, Problem 3.) 

Equations (11-2) and (11-4) for the rotation of a rigid body bear a 
formal analogy to Eqs. (11-1) and (11-3) for the motion of a point mass 
M. There are, however, three differences which spoil the analogy. In the 
first place, Eq. (11-4) involves a tensor I, whereas Eq. (11-3) involves a 
scalar M; thus P is always parallel to V, while L is not in general parallel 
to u. A more serious difference is the fact that the inertia tensor I is not 
constant with reference to axes fixed in space, but changes as the body 
rotates, whereas M is constant (in Newtonian mechanics). Finally, and 
perhaps most serious, is the fact that no symmetrical set of three coordi- 
nates analogous to X, Y, Z exist with which to describe the orientation of 
a body in space. This point was made in Section 5-1, and it is suggested 
that the reader review the last paragraph in that section. For these rea- 
sons, we cannot proceed to solve the problem of rotation of a rigid body 
by analogy with the methods of Chapter 3. 

There are two general approaches to the problem. We shall first, in 
Sections 11-2 and 11-3, try to obtain as much information as possible 

450 



11-2] euler's equations of motion for a rigid body 451 

from the vector equations (11-2), (11-4) without introducing a set of 
coordinates to describe the orientation of the body. We shall then, in 
Sections 11-4 and 11-5, use Lagrange's equations to determine the motion 
in terms of a set of angular coordinates suggested by Euler. 

11-2 Euler's equations of motion for a rigid body. The difficulty that 
I changes as the body rotates may be avoided by referring Eq. (11-2) to 
a set of axes fixed in the body. If we let "d'/dt" denote the time derivative 
with reference to axes fixed in the body, then by Eq. (7-22), Eq. (11-2) 
becomes 

^ + «XL = N. (11-5) 

Since I is constant relative to body axes, we may substitute from Eq. (11-4) 
to obtain 

I • — + « X (I • «) = N. (11-6) 

(Recall that d'u/dt = du/dt.) It is most convenient to choose as body 
axes the principal axes, ei, e 2 , e 3 of the body. Then Eq. (11-6) becomes 

JlWl + (h — -J2)W 3 W2 = N U 

1 2«2 + Vi — 7 3 )co 1 w 3 = N 2 , (11-7) 

J3W3 + (I2 — Jl)<«>2«l = N 3 . 

These are Euler's equations for the motion of a rigid body. If one point 
in the body is held fixed, that point is to be taken as the origin for the body 
axes, and the moments of inertia and torques are relative to that point. 
If the body is unconstrained, the center of mass is to be taken as origin 
for the body axes. 

In order to derive the energy theorem from Euler's equations, we multi- 
ply Eq. (11-6) by«: 

u . 1 . 4» = „ . N. (11-8) 

Since I is symmetric, the left member is 

dw da . Id',.. dT . _, 

B ' l 'I = I- | ' B= 2rf( (tt,|,B) = Hi' (U " 9) 

where T is given by Eq. (10-153). Here we have used the fact that 
d/dt, d'/dt have the same meaning when applied to a scalar quantity. 



452 THE ROTATION OF A RIGID BODY [CHAP. 11 

Comparing Eqs. (11-8) and (11-9), we obtain the energy theorem: 

5 = « • N, (H-10) 

at 

in analogy with theorem (3-133) for the motion of a particle. 

From Eqs. (11-7) we note immediately that a body cannot spin with 
constant angular velocity w, except about a principal axis, unless external 
torques are applied. If du/dt = 0, Eq. (11-6) becomes 

to X (I • a) = N. (11-H) 

The left member is zero only if I • <o is parallel to w, that is, if <o is along a 
principal axis of the body. If a wheel is to spin freely without exerting 
forces and torques on its bearings, then it must be not only statically 
balanced, i.e., with its center of mass on the axis of rotation, but also 
dynamically balanced, i.e., the axis of rotation must be a principal axis 
of the inertia tensor, as any automobile mechanic knows. 

In order to solve Eqs. (11-7) for u(t), we would need to know the com- 
ponents of torque along the (rotating) principal axes, an uncommon situ- 
ation, except for the case N = 0. We now consider a freely rotating sym- 
metrical body, with no applied torque. Let the symmetry axis of the 
body be e 3 , so that I\ = 7 2 - Then the third of Eqs. (11-7) is 

/ 3 W3 = 0, (11-12) 

and « 3 is constant. The first two equations may be written 

"i + /3o>3« 2 = 0, <o 2 — |8«3«i = 0, (H-13) 



where 



18 = h 7 Zl • (H-14) 



Equations (11-13) are a pair of coupled linear first-order equations in 
o>i, <o 2 . Let us look for a solution by setting 

»! = Axe 3 ", « a = A 2 e pt . (11-15) 

We readily verify that formulas (11-15) satisfy Eqs. (11-13) provided that 

p = ±tp» 8 , (11-16) 

and 

A 2 = Tiitj. (11-17) 



11-2] euler's equations of motion for a rigid body 453 

We have found a complex conjugate pair of solutions, 

Wl = e ±tfi»,t f W2 = =F fe ±W-.« f (ii_i 8 ) 

and these may be superposed with arbitrary constant multipliers to form 
the real solution : 

ax = A cos (/3w 3 « + 0), co 2 = A sin (fio) 3 t + 8). (11-19) 

The angular velocity vector a therefore precesses in a circle of radius A 
about the e 3 -axis, with angular velocity /8w 3 . The precession is in the same 
sense as o> 3 if I 3 > Ii, and in the opposite sense otherwise. The magni- 
tude of <o is 

co= [o>§ + A 2 ] 1 ' 2 , (H-20) 

and is constant, a result which can also be proved by direct calculation of 
d(co 2 )/dt from Eq. (11-7). The constants co 3 , A, 6 are determined by the 
initial conditions. There are three arbitrary constants, since Euler's 
equations are three first-order differential equations. Since an uncon- 
strained rotating rigid body has three rotational degrees of freedom, we 
should expect a total of six arbitrary constants to be determined by the 
initial conditions. What are the other three? 

The instantaneous axis of rotation, determined by the vector a, traces 
out a cone in the body (the body cone) as it precesses around the axis of 
symmetry. The half -angle a& of the body cone is given by 

tana 6 =— • (11-21) 

w 3 

Alternatively, if the body is initially rotating with angular velocity « 
about an axis making an angle aj, with the symmetry axis, then the con- 
stants « 3 and A are given by 

A = u sin ab, w 3 = w cos aj,. (11-22) 

In order to find the motion in space, we need to locate the <o-axis with 
respect to a direction fixed in space. We could do this by tracing out step 
by step the motion relative to space axes, allowing the body to rotate 
with constant angular velocity about an axis in the body cone which 
precesses with angular velocity /3to 3 . It is easier to locate w relative to 
L, since by Eq. (11-2) L is constant if N = 0. The angle a„ between <o 
and L is given by 

«-L <ol« 2T . . 

cos a > = -^l = —^r = ^l • (11_23) 

Since by Eq. (11-10) T is constant, the angle a s is constant. The axis of 



454 THE ROTATION OF A RIGID BODY [CHAP. 11 

rotation therefore traces out a cone in space, the space cone. The space 
cone has a half-angle a a given by Eq. (11-23) and its axis is the direction 
of the angular momentum vector L. The line of contact between the 
space cone and the body cone at any instant is the instantaneous axis of 
rotation. Since this axis in the body is instantaneously at rest, the body 
cone rolls without slipping around the space cone. This gives a complete 
description of the motion (see Fig. 11-1). 
We can express a a in terms of the constants co, a&. We have 

I = (eiei + e 2 e 2 )/i + e 3 e 3 / 3 
= hi + e 3 e 3 (/ 3 - h). (H-24) 

By substituting from Eqs. (11-14), (11-22), and (11-24) into Eqs. (11-4), 
(10-153), and (11-23), and with u = wn, we obtain: 

2T = w 2 /i[l + p cos 2 a b ], (11-25) 

L = ali[n + P cos a 6 e 3 ], (11-26) 

1 + ft cos 2 a b n , _, 

C0S "> = [1 + (20 + ]8»)cos» « 6 ]i/» ' (11_27) 

Note that a„ depends on only a& and not on w. It is clear from Eq. (11-26) 
that the space cone lies inside the body cone if > and outside if /3 < 
(see Fig. 11-1). This is clear also if, when the body cone rolls on the space 
cone, the precession of the axis of rotation is to have the sense given by 
Eq. (11-19). [The reader should check this, remembering that Eq. (11-19) 
describes the motion of the axis relative to the body.] 

Next, consider the case when the inertia tensor is nondegenerate. We 
shall number the principal axes so that 7 3 > 7 2 > Ii- It was shown 
above that a body may rotate freely about a principal axis. Let us study 
small deviations from this steady rotation. If <o is not along a principal 
axis, then it cannot remain constant. Let us assume that u lies very close 
to a principal axis, say to e 3 , so that o> 3 ~2> o>i and w 3 » o> 2 . Then if 
N = 0, we see from the third of Eqs. (11-7) that w 3 is constant to first 
order in «i and w 2 . The first two equations then become a pair of coupled 
linear equations in u\, co 2 , which we solve as in the preceding example 
to obtain 

«i = A[I 2 (I 3 - / 2 )] 1/2 cos (/3w 3 < + 0), 

o> 2 = Alhih - h)] 1 ' 2 sin ((3a> 3 t + B), 



(11-28) 



where A and 6 are arbitrary constants and 

p = [ (J3 ~ Z %\ ~ /2) ] 1/2 - dl-29) 



11-3] 



poinsot's solution for a freely rotating body 



455 




Fig. 11-1. Free rotation of a symmetrical body. 

The vector w therefore moves counterclockwise (looking down from the 
positive e 3 -axis) in a small ellipse about the e3-axis. In a similar manner, 
we can show that if u is nearly parallel to the ei-axis, it moves clockwise 
in a small ellipse about that axis, and that if a is nearly parallel to the 
e2-axis, the solution is of an exponential character. In the latter case, of 
course, the components «i and a>3 will not remain small, and the approxi- 
mation that «2 is constant will hold only during the initial part of the 
motion. We conclude that rotation about the axes of maximum and 
minimum moments of inertia is stable, while rotation about the inter- 
mediate axis is unstable. This result is readily demonstrated by tossing 
a tennis racket in the air and attempting to make it spin about any prin- 
cipal axis. The general solution of Eqs. (11-7) for u, when N = 0, can 
also in principle be obtained. We shall solve the problem in the next 
section by a different method. 

11-3 Poinsot's solution for a freely rotating body. If there are no 
torques, N = 0, then Eqs. (11-2) and (11-10) yield four integrals of the 
equations of motion: 

L = I • w = a constant, (11-30) 



T = \ <o • I • « 



a constant. 



(11-31) 



456 THE ROTATION OF A RIGID BODY [CHAP. 11 

Poinsot* has obtained a geometrical representation of the motion based 
on these constants, and utilizing the inertia ellipsoid. Let us imagine the 
inertia ellipsoid (10-157) rigidly fastened to the body and rotating with 
it. If we let r be the vector from the origin to the point where the axis of 
rotation intersects the inertia ellipsoid at any instant, 



r 
- «, 

CO 



then comparison of Eqs. (11-31) and (10-157) shows that 



2 2 

a co 



(11-32) 



(11-33) 



2r 2 
The normal to the ellipsoid at the point r is parallel to the vector 

V(r • I • r) = 2I i x 1 e 1 + 2I 2 x 2 e 2 + 2I 3 x 3 e 3 = 2 £ L, (11-34) 

where x x , x 2 , x 3 are the components of r along the principal axes. The 
tangent plane to the ellipsoid at the point r is therefore perpendicular to 
the constant vector L (see Fig. 11-2). Let I be the perpendicular distance 
from the origin to this tangent plane: 



r-L r co-l-w a(2T) 112 . , , 1t „- 

I = -—=- = = = v T ' — = a constant. (11-35) 



The tangent plane is therefore fixed in space (relative to the origin 0) and 
is called the invariable plane. Its position is determined by the initial con- 
ditions. Moreover, since the point of contact between the ellipsoid and 
the plane lies on the instantaneous axis of rotation, the ellipsoid rolls on 
the plane without slipping. The angular velocity at any instant has the 
magnitude 

(2r)1/2 r. (11-36) 



co = 



This gives a complete description of the motion. ^ 

As the inertia ellipsoid rolls on the invariable plane, with its center 
fixed at the origin, the point of contact traces out a curve called the 
polhode on the inertia ellipsoid, and a curve called the herpolhode on the 
invariable plane. This is illustrated in Fig. 11-2. The polhode is a closed 
curve on the inertia ellipsoid, denned as the locus of points r where the 
tangent planes lie a fixed distance I from the center of the ellipsoid. In 
Fig. 11-3 are shown various polhodes on a nondegenerate inertia ellipsoid. 
Note that the topological features of the diagram are in agreement with 



* Poinsot, Theorie Nouvelle de la Rotation des Corps, 1834. 



11-3] poinsot's solution for a freely rotating body 457 



Inertia ellipsoid 




Herpolhode /Invariable plane 



Fig. 11-2. The inertia ellipsoid rolls on the invariable plane. 




Fig. 11-3. Polhodes on a nondegenerate inertia ellipsoid. 



the conclusions at the end of the preceding section. In general, the herpol- 
hode is not closed but fills an annular ring in the invariable plane. 

In the case of a symmetrical body it can be shown (Problem 7) that the 
polhodes are circles about the symmetry axis and the herpolhodes are 
circles in the invariable plane. In that case, r and therefore, by Eq. (11-36), 
co (but not to!) are constant during the motion. Poinsot's description of 
the motion in this case agrees with that in the preceding section. The 



458 



THE ROTATION OF A RIGID BODY 



[CHAP. 11 



polhode and herpolhode are the intersections of the body and the space 
cones with the inertia ellipsoid and the invariable plane, respectively. 

11-4 Euler's angles. The results in Sections 11-2 and 11-3 regarding 
the motion of a rigid body were obtained without the use of any coordi- 
nates to describe the orientation of the body. In order to proceed further 
with the discussion, it is necessary to introduce a suitable set of coordi- 
nates. We choose a set of axes fixed in the body, which are most con- 
veniently taken as the principal axes, with origin at the center of mass, 
or at the fixed point if one exists. These axes will be labeled with sub- 
scripts 1, 2, 3, as before. If there is an axis of symmetry, it will be num- 
bered 3; otherwise, the axes may be numbered in any order. We need 
three coordinates to specify the orientation of the body axes with respect 
to a fixed set of space axes x, y, z. The relation between the two sets of 
axes could be specified by giving the coefficients of the transformation 
from coordinates x, y, z to xi, x 2 , x 3 . There are nine coefficients but only 
three of them are independent, as we have seen, and to try to use three of 
the coefficients as coordinates is not convenient. As was pointed out in 
Section 5-1, there is no symmetric set of coordinates analogous to x, y, z 
with which to describe the orientation of a body. Among the various 
coordinate systems that have been introduced for this purpose, one of 
the most useful is due to Euler. 

In Fig. 11-4, the Euler angles 6, <j>, $ are shown. These are used to spec- 
ify the position of the body axes 1, 2, 3 relative to the space axes x, y, z. 
The body axes 1, 2, 3 are shown as heavy lines; the space axes x, y, z are 
lighter. The angle is the angle between the 3-axis and the z-axis. 
Since the 3-axis is thus singled out for special treatment, if the body has 
an axis of symmetry, it should be taken as the 3-axis. Likewise, if the 
external torques possess an axis of symmetry in space, that axis should 
be taken as the z-axis. The intersection of the 1, 2-plane with the x?/-plane 




Fig. 11-4. Euler's angles. 



11-4] euler's angles 459 

is called the line of nodes, labeled £ in the diagram. The angle <f> is meas- 
ured in the xy-pla,ne from the z-axis to the line of nodes, as shown. The 
angle ^ is measured in the 1, 2-plane from the line of nodes to the 1-axis. 
We are assuming that both sets of axes x, y, z and 1, 2, 3 are right-handed. 
It will be convenient also to introduce a third (right-handed) set of axes, 
£, ri, f, of which £ is the line of nodes, f coincides with the body axis 3, 
and ij is in the 1, 2-plane. | 

In order to express the angular velocity vector o> in terms of Euler's 
angles, we first prove that angular velocities may be added like vec- 
tors, in the sense of the following theorem: 

Given a primed coordinate system rotating with angular velocity 
«i with, respect to an unprimed system, and a starred coordinate 
system rotating with angular velocity u 2 relative to the primed 
system, the angular velocity of the starred system relative to the 
unprimed system is «i + «2. (11-37) 

To prove this theorem, let A be any vector at rest in the starred system: 

Then by theorem (7-22), its velocity relative to the primed system is 

"=« 2 XA. (11-39) 

Now applying theorem (7-22) again, we find the velocity of A relative to 
the unprimed system: 

^=^+ Wl XA=(<o 1 + » 2 )XA. (llHtO) 

A final comparison with theorem (7-22) shows that («i + «2) is the 
angular velocity of the starred system relative to the unprimed one. 

Now consider Figure 11-4 and suppose that the body is moving so that 
6, <j>, ^ are changing with time. If alone changes, while <£, \p are fixed, 
the body rotates around the line of nodes with angular velocity tej. If <j> 
alone changes, the body rotates around the z-axis with angular velocity 
<j>k. If ^ alone changes, the body rotates around its 3-axis with angular 
velocity ^e 3 . Now if we consider a primed coordinate system rotating with 



f The reader is cautioned that the notation for Euler's angles, as well as the 
convention as to axes from which they are measured, and even the use of right- 
handed coordinate axes, are not standardized in the literature. It is therefore 
necessary to note carefully how each author defines the angles. The conventions 
adopted here are very common, but not universal. 



460 THE ROTATION OF A RIGID BODY [CHAP. 11 

angular velocity <£k about the 2-axis, and let the i-,»j,f-system rotate with 
angular velocity 0ej relative to this primed system, then by theorem 
(11-37), the angular velocity of the £,ij,f-system is 0e { + <£k. The axes 
1, 2, 3 rotate with angular velocity ^e 3 relative to £,ij,f, hence the angular 
velocity of the body is 

« = fe{ + <£k + ^e 3 . (11-41) 

We have, from Fig. 11-4, the relations 

ej = ei cos ^ — e 2 sin ^, 

e, = ei sin ^ + e 2 cos if/, (11-42) 

e r = e 3 , 



and 



k = e ? cos 6 + e, sin 6 
= ei sin 6 sin \p + e 2 sin B cos 4> + e 3 cos 6. (11-43) 



We may therefore express a in terms of its components along the principal 
axes: 

Wi = 6 cos ^ + <j> sin 6 sin \[/, 

w 2 = —6 sin $ + <j> sin cos ^, (11-44) 

w 3 = ^ + 4> cos 0. 

The kinetic energy is now given by Eq. (10-153) : 

T = i/i«? + \U<4 + */*»§. (11-45) 

The kinetic energy is a rather complicated expression involving 6, <j>, 4>, 0, 
and 4/. Note that 0, <£, ^ are not orthogonal coordinates, i.e., cross terms 
involving 6<j> and ^ appear in T. In the case of a symmetrical body 
(Ii = 7 2 ), the expression for T simplifies to the form: 

T = %I 1 6 2 + i/^ 2 sin 2 + J/gGJ- + <£ cos 0) 2 . (11-46) 

The generalized forces Qe, Q«, Q* are easily shown to be the torques 
about the £-, z-, and 3-axes. 

We are now in a position to write down Lagrange's equations for the 
rotation of a rigid body subject to given torques. If the torques are de- 
rivable from a potential energy V(6,<j>,ip), then there will be an energy 
integral. If V is independent of <f>, then inspection of Eqs. (11-44) shows 
that <j> will be an ignorable coordinate. Unfortunately, this is not enough 
to enable us to give a general solution of the problem. However, for a 
symmetrical body, if V is independent of yp also, we see from Eq. (11-46) 



11-5} 



THE SYMMETRICAL TOP 



461 



that both <f> and $ are ignorable. We have then three constants of the 
motion, enough to solve the problem. This case will be solved in the 
next section. A few other special cases are known for which the problem 
can be solved,* but for the general problem of the motion of an unsym- 
metrical body under the action of external torques, as for the many-body 
problem, there are no generally applicable methods of solution, except by 
numerical integration of the equations of motion. 

11-5 The symmetrical top. The symmetrical top, represented in 
Fig. 11-5, is a body for which Ii = 7 2 . It pivots around a fixed point 
that lies on the axis of symmetry a distance I from the center of mass G 
which also lies on the axis of symmetry. The only external forces are the 
forces of constraint at and the force of gravity. Therefore, by Eq. (11-46), 
the Lagrangian function is 

L = \I X P + J/^ 2 sin 2 6 + J/ 8 (tf + cos 0) 2 - mgl cos 8. (11-47) 

The coordinates ^ and <j> are ignorable, and we have therefore three 
integrals of the motion: 

dpj, dL 

~dt ~ w 



0, 



where 



d£* = — = n 
dt d<t> 

dE__dL =a 
dt ~~ dt ~ ' 



V* = ^(^ + 4> cos 0), 

p* = Ii4> sin 2 + 7 3 cos 6(4/ + 4 cos 9), 

E = ilj 2 + J/^ 2 sin 2 6 

+ fclaOA + <£ cos 0) 2 + mgl cos 0. 



(11-48) 

(11^9) 

(11-50) 

(11-51) 
(11-52) 

(11-53) 



We use Eqs. (11-51) and (11-52) to eliminate i, <£ from Eq. (11-53): 

E = w2 + (?*-^ n y) 2 + A + ^ cos ,. (11 -54) 



* See, for example, E. J. Routh, The Advanced Part of a Treatise on the Dy- 
namics of a System of Rigid Bodies, 6th ed. London: Macmillan, 1905. (Also 
New York: Dover, 1955.) 



462 



THE ROTATION OF A RIGID BODY 



[CHAP. 11 




Fig. 11-5. Coordinates for the symmetrical top. 



We can now solve the problem by the energy method. If we set 



W = E - 



~2 

V* 
2J» 



(P* ~ V* cos 0) '' 



+ mgl cos 6, 



then 



2/i sin* 6 

= \j-[E' -<F(f»]} 1/2 , 
and 6 is given, in principle, by computing the integral 

f * =(L\» t 

Je [E' - T(8)]W \2y/ 



(11-55) 
(11-56) 

(11-57) 
(11-58) 



and solving for 6(t). The constant O is the initial value of 0. Once 6{t) 
is known, Eqs. (11-51) and (11-52) can be solved for ^ and <f> and inte- 
grated to give ^(t), <t>(t). 

Comparison of Eqs. (11-44) and (11-51) shows that 



p+ = I3013, 



(11-59) 



so that w 3 is a constant of the motion. If W3 = 0, then Eq. (11-56) re- 
duces essentially to the formula (9-137) for a spherical pendulum, as it 



11-5] 



THE SYMMETRICAL TOP 



463 




0o */2 



Fig. 11-6. Effective potential energy for the symmetrical top. 

should. In Fig. 11-6, 'F'(0) is plotted versus for o> 3 ^ 0. The 'torque' 
associated with the 'potential energy' 'V'(d) is 



•N' = - 



d'V 



mgl sin — 



(p» — V* cos 0)(p» — p» cos 0) 
1 1 sin3 



(11-60) 



Inspection of Eq. (11-60) shows that, in general (if p* j£ p+), the 'torque' 
W is positive f or 8 = and negative for = 7r, and has one zero between 
and ir. Hence 'V has one minimum, as shown in Fig. 11-6, at a point 6 
satisfying the equation 

mgl 1 1 sin 4 O — (p# — V* cos o )(p* — P* cos O ) = 0. (11-61) 

If E' = 'V'(8 ), the axis of the top precesses uniformly at an angle O 
with the vertical, and with angular velocity 

P* ~ V* cos flp 
1 1 sin 2 O 



<£o 



(11-62) 



Solving Eq. (11-61) for (p* — p* cos O ), and using Eq. (11-59), we 
obtain 



(p — p* COS O ) = i/ 3 W3 



sin 2 0p 
cos 0Q 



(11-63) 



We see that if 0o < n"/2, there is a minimum spin angular velocity below 
which the top cannot precess uniformly at the angle O : 



Wmin = J ^ COS O J 



(11-64) 



464 THE ROTATION OF A RIGID BODY [CHAP. 11 

For « 3 > w min , there are two roots (11-63) and hence two possible values 
of <>o, a slow and a fast precession, both in the same direction as the spin 
angular velocity « 3 . For w 3 5>> w m i n , the fast and slow precessions occur 
at angular velocities 

4, ^h J**- (11-65) 

vo 7i cos O 

4,0 ~j£l' (ll_66) 



and 



It is the slow precession which is ordinarily observed with a rapidly spin- 
ning top. For O > t/2 (top hanging with its axis below the horizontal), 
there is one positive and one negative value for 4> . (To what do these 
motions of uniform precession reduce when oo s — » 0?) 

Study of Fig. 11-6 shows us that the more general motion involves a 
nidation or oscillation of the axis of the top in the 0-direction as it pre- 
cesses. The axis oscillates between angles 0i and 2 which satisfy the 
equation 

E> = ^-jX^ + m91 C0S *> {U ^ 7) 

where p^,, p+, and E' are determined from the initial conditions. If we 
multiply Eq. (11-67) by sin 2 0, it becomes a cubic equation in cos 0. We 
see from Fig. 11-6 that there must be two real roots cos 1; cos 2 between 
—1 and +1. The third root for cos must lie outside the physical range 
— 1 to +1. In fact, inspection of Eq. (11-67) will show that the third 
root is greater than +1. (In the case of uniform precession discussed 
in the preceding paragraph, the two physical roots coincide, cos 0i = 
cos 2 = cos O .) If initially 0=0, then the initial value cos 0i of cos 
satisfies Eq. (11-67) ; knowing one root of a cubic equation, we may factor 
the equation and find all three roots. During nutation, the precession 
velocity varies according to Eq. (1 1-52) : 

1 _ P* — P+ COS fii_RO\ 

+ - hsm'0 ' (11_68) 

If \p+\ < \p+\, we can define an angle 3 as follows: 

cos 3 = £* • (11-69) 

For > 3 , 4> has the same sign as w 3 , and for < 3 , it has an opposite 
sign. The derivative with respect to of the right member of Eq. (11-67) 
is negative at = 3 ; hence we see from Fig. 11-6 that 3 < 02, where 
2 is the largest angle satisfying Eq. (11-67). In fact, 3 < O . If 3 < 0i 



11-5] 



THE SYMMETRICAL TOP 



465 




(a) (b) (c) 

Fig. 11-7. Locus of top axis (3) on unit sphere. 

(or if |p^| > |p*| and p^, p* have the same sign), then <j> has the same sign 
as « 3 throughout the nutation, and the top axis traces out a curve like 
that shown in Fig. 11-7 (a). If 6 3 > U <£ changes sign during the nutation 
and the top axis moves as in Fig. ll-7(b). It is clear that if the top is set 
in motion initially above the horizontal plane with <f> opposite in sign to 
« 3 , the motion necessarily will be like that shown in Fig. ll-7(b). 

An important special case occurs when the top, spinning about its axis 
with angular velocity w 3 , is held with its axis initially at rest at an angle 
0i and then released. Initially, we have 

e = e u 6 = o, 4> = o, ^ = w 3 . (n-70) 

We substitute in Eqs. (11-51), (11-52), and (11-53) to find 

pj, = 7 3 W3, p* =. I 3 u 3 cos 0i, E' = mgl cos Oi. (11-71) 

In this case, we see that 8 3 = B\, and the motion is as shown in Fig. 11-7 (c). 
An elementary discussion of this case, based on the conservation of angu- 
lar momentum, was given in Section 4-2. Now Eq. (11-56) becomes 



'V 



where 



2/i L 



(cos 6 1 — cos fl) 2 



sin 2 6 
2Iimgl 

i%<4 ' 



+ a cos 6 



]■ 



(11-72) 



(11-73) 



The turning points for the nutation are the roots of Eq. (11-67), which 
becomes in this case, if we multiply by sin 2 6, 



(cos 8 1 — cos 6) 2 — a(cos 8 y — cos 0)(1 — cos 2 6) = 0. (11-74) 



466 THE ROTATION OF A RIGID BODY [CHAP. 11 

The roots are 

cos 6 = cos $i, 

(11-75) 

cos = ± [1 ± (1 - 4a cos 0i + 4a 2 ) 1 ' 2 ]. 
2a 

The angle 2 is given by the second formula, using the minus sign in the 
bracketed expression. The plus sign gives a root for cos greater than 
+1. Let us consider the case of a rapidly spinning top, that is, when 
a « 1. We then have 

cos 02 = cos 0i — a sin 2 X . (11-76) 

The angle 2 is only slightly greater than 0i, and the amplitude of nuta- 
tion is proportional to a. If we set 

2 = 0o + a, 0i = O — a, (11-77) 

and substitute in Eq. (11-76), we find that, to first order in a and a, 

a = \a sin 0i. (11-78) 

We now set 

6 = O + S = 0i + a + 8, (11-79) 

and substitute in Eq. (11-72), which becomes, to second order in a and 5, 



t2 



'V = F(0 O ) + i ^ «! S 2 . (11-80) 

^i 

The first term is constant, and the second leads to harmonic oscillations 
in 3 with a frequency 

<o 9 = ^ « 3 . (H-81) 

The nutation is given by 

= 0i + a — acosaet. (11-82) 

We substitute in Eq. (11-68) to obtain <j> to first order in a: 

1 1 sin Pi 
The average angular velocity of precession is 

W „ * ^tfa« "KL (11 _ 84 ) 

XYVa Ii sin 0i /3W3 



11-5] 



THE SYMMETRICAL TOP 



467 




Fig. 11-8. Effective potential energy when p^ = p+. 



The top axis therefore precesses very slowly and nutates very rapidly with 
very small amplitude. In practice, the frictional torques which we have 
neglected usually damp out the nutation fairly quickly, leaving only the 
uniform precession. 

As a final example, consider the case when the top is initially spinning 
with its symmetry axis vertical. In this case, so long as the 3- and z-axes 
coincide, the line of nodes is indeterminate. We see from Fig. 11-4 that 
the angle ^ + <f> is determined as the angle between the x- and 1-axes, 
although \f/, <t> separately are indeterminate. Hence we have initially 



V* = liU + 4) = -*3W3, 

P+ = l3(t + 4>) = V+- 



(11-85) 
(11-86) 



Equation (11-56) in this case becomes 



, v , _ jj»ira 

¥ ~~ 27i L tin* 9 



cos 0) 



+ a cos 6 



(11-87) 



where a is given by Eq. (11-73). This, of course, is just a special case of 
Eq. (11-72). In Fig. 11-8 we plot 'V for the case when p* = p*. The 
form of the curve depends upon the value of a. We see that a rapidly 
spinning top (a < £) can spin stably about the vertical axis; if disturbed, 
it will exhibit a small nutation about the vertical axis. A slowly spinning 
top (a > J) cannot spin stably about a vertical axis, but will execute a 
large nutation between 0i = and 8 2 given by Eq. (11-75). In this case, 



cos 6 2 = 



(11-88) 



468 THE ROTATION OF A RIGID BODY [CHAP. 11 

The minimum spin angular velocity below which the top cannot spin 
stably about a vertical axis occurs when a = \, or, by Eq. (11-73), 



Wmin - hr. 



(11-89) 



Note that this formula agrees with Eq. (11-64). If initially w 3 > w m i n , 
a top will spin with its axis vertical, but when friction reduces w 3 below 
w m in, it will begin to wobble. 

All of the above conclusions about the behavior of a symmetrical top 
under various initial conditions can easily be verified experimentally with 
a top or with a gyroscope. 



Problems 

1. Use the result of Problem 3, Chapter 10, to derive Eq. (11-6) directly from 
the equation 

2. (a) Assume that the earth is a uniform rigid ellipsoid of revolution, look 
up its equatorial and polar diameters, and calculate the angular velocity of 
precession of the North Pole on the earth's surface assuming that the polar axis 
(i.e., the axis of rotation) deviates slightly from the axis of symmetry. (An 
irregular precession of roughly this sort is observed with an amplitude of a few 
feet, and a period of 427 days.) 

(b) Assume that the earth is a rigid sphere and that a mountain of mass 10~ 9 
times the mass of the earth is added at a point 45° from the polar axis. Describe 
the resulting motion of the pole. How long does the pole take to move 1000 
miles? 

(c) For a rigid ellipsoidal earth, as in part (a), how massive a "mountain" 
must be placed on the equator in order to make the polar precession unstable? 

The earth is, of course, not of uniform density, but is more dense near its 
center. Even more important, the earth is not rigid, but behaves as an elastic 
spheroid for short times, and can deform plastically over long times. The results 
in this problem are therefore only suggestive and do not correspond to the 
actual motion of the earth. For example, the observed precession period of 427 
days is longer than would be calculated for a rigid earth. When plastic deforma- 
tion is taken into account, an appreciable wandering of the pole can result 
even for an ellipsoidal earth with a much smaller "mountain" than that cal- 
culated in part (c).* 



* An excellent short discussion of the rotation of the earth, treated as an 
elastic and plastic ellipsoid, will be found in an article by D. R. Inglis, Review 
of Modern Physics, vol. 29, p. 9 (1957). 



PROBLEMS 469 

3. Show that the axis of rotation of a freely rotating symmetrical rigid body 
precesses in space with an angular velocity 

«n = (|3 + sec a&)co3, 

where the notation is that used in Section 11-2. 

4. Show that if the only torque on a symmetrical rigid body is about the axis 
of symmetry, then (<u? + o>!) is constant, where «i and a>2 are angular velocity 
components along axes perpendicular to the symmetry axis. If #3(0 is given, 
show how to solve for coi, 012, and C03. 

5. A symmetrical rigid body moving freely in space is powered with jet engines 
symmetrically placed with respect to the 3-axis of the body, which supply a 
constant torque JV3 about the symmetry axis. Find the general solution for the 
angular velocity vector as a function of time, relative to body axes, and describe 
how the angular velocity vector moves relative to the body. 

6. (a) Consider a charged sphere whose mass m and charge e are both dis- 
tributed in a spherically symmetrical way. Show that if this body rotates in a 
uniform magnetic field B, the torque on it is 

N = rr^- L X B (gaussian units), 
2mc 

where g is a numerical constant, which is one if the mass density is everywhere 
proportional to the charge density. 

(b) Write an equation of motion for the body, and show that by introducing 
a suitably rotating coordinate system, you can eliminate the magnetic torque. 

(c) Compare this result with Larmor's theorem (Chapter 7). Why is no 
assumption needed here regarding the strength of the magnetic field? 

(d) Describe the motion. What points in the body are at rest in the rotating 
coordinate system? 

7. Prove (without using the results of Section 1-2) that if two principal 
moments of inertia are equal, the polhode and the herpolhode are both circles. 

8. (a) Obtain equations, in terms of principal coordinates xi, X2, xz, for two 
quadric surfaces whose intersection is the polhode. Your equations should 
contain the parameters 7i, I2, I3, I- 

(b) Find the equation for the projection of the polhode on any coordinate 
plane and show that the polhodes are closed curves around the major and minor 
poles of the ellipsoid, but that they are of hyperbolic type near the intermediate 
axis, as shown in Fig. 11-3. 

(c) Find the radii of the circles on the invariable plane which bound the 
herpolhode. 

9. Find the matrix (ay) which transforms the components of a vector from 
space axes to body axes. Express an in terms of Euler's angles. [Hint: The 
transformation can be made up of three consecutive rotations by angles 9, 4>> lA> 
about suitable axes, and taken in proper order.] 

10. Write out the Hamiltonian function in terms of 6, \p, <£, p«, p*, p* for a 
freely rotating unsymmetrical rigid body. Express the coefficients in terms 
of the parameters Ii, I3, (I2 — li)- 



470 THE BOTATION OF A RIGID BODY [CHAP. 11 

11. Use Lagrange's equations to treat the free rotation of an unsymmetrical 
rigid body near one of its principal axes, and show that your results agree with 
the last paragraph of Section 11-2. 

12. Set up Lagrange's equations for a symmetrical top, the end of whose 
axis slides without friction on a smooth table. Discuss carefully the differences 
in the motions between this case and the case when the end of the top axis pivots 
about a fixed point. 

13. A gyroscope is constructed of a disk of radius a, mass M, fastened rigidly 
at the center of an axle of length (3a/2), mass (2M/7), negligible cross section, 
and mounted inside two perpendicular rings, each of radius (3o/2), mass (M/3). 
The axle rotates in frictionless bearings at the intersection points of the rings. 
One of these intersection points pivots without friction about a fixed point 0. 
Set up the Lagrangian function and discuss the kinds of motion which may 
occur (under the action of gravity) . 

14. Discuss the free rotation of a symmetrical rigid body, using the La- 
grangian method. Find the angular velocity for uniform precession and the 
frequency of small nutations about this uniform precession. Describe the mo- 
tion and show that your results agree with the solutions found in Section 11-2 
and in Problem 3. 

15. A top consists of a disk of mass M, radius r, mounted at the center of a 
cylindrical axle of length I, radius a, where a <3C I, and negligible mass. The end 
of the axle rests on a table, as shown in Fig. 11-9. The coefficient of friction 
is fi. The top is set spinning about its symmetry axis with a very great angular 
velocity «3o, and released with its axis at an angle 0i with the vertical. Assume 
that C03 is great enough compared with all other motions of the top so that the 
edge of the axle in contact with the table slides on the table in a direction per- 
pendicular to the top axis, with the sense determined by «3. Write the equa- 
tions of motion for the top. Assume that the nutation is small enough to be 
neglected, and that the friction is not too great, so that the top precesses slowly 
at an angle 0o which changes slowly due to the friction with the table. Show 




Fig. 11-9. A simple top. 



PROBLEMS 471 

that the top axis will at first rise to a vertical position, and find approximately 
the time required and the number of complete revolutions of precession during 
this time. Describe the entire motion of the top relative to the table during 
this process. How long will it remain vertical before beginning to wobble? 

16. Obtain a toy gyroscope, and make the necessary measurements in order to 
predict the rate at which it will precess, when spinning at its top speed, if its 
axis pivots about a fixed point at an angle of 45° with the vertical. Calculate 
the amplitude of nutation if the axis is held at an angle of 45° and released. Per- 
form the experiment, and compare the measured rate of precession with the 
predicted rate. 

17. A planet consists of a uniform sphere of radius a, mass M, girdled at its 
equator by a ring of mass m. The planet moves (in a plane) about a star of 
mass M' . Set up the Lagrangian function, using as coordinates the polar co- 
ordinates r, a in the plane of the orbit, and Euler's angles 6, <t>, yf/, relative to 
space axes of which the z-axis is perpendicular to the plane of the orbit, and the 
x-axis is parallel to the axis from which a is measured. You may assume that 
r ^> a, and use the result of Problem 13, Chapter 6. Find the ignorable co- 
ordinates, and show that the period of rotation of the planet is constant. 

18. Assume that the planet of Problem 17 revolves in a circle of radius r about 
the star, although this does not quite satisfy the equations of motion. Assume 
that the period of revolution is short in comparison with any precession of the 
axis of rotation, so that in studying the rotation it is permissible to average over 
the angle a. Show that uniform (slow) precession of the polar axis may occur if 
the axis is tilted at an angle 0o from the normal to the orbital plane, and find 
the angular velocity of precession in terms of the masses M, m, M', the radii 
a, r, the angle 0o, and the angular velocity of rotation. Show that if the day is 
much shorter than the year, the above assumption regarding the period of revo- 
lution and the rate of precession is valid. Find the frequency of small nutations 
about this uniform precession and show that when the day is much shorter 
than the year, it corresponds to the free precession whose angular velocity is 
given in Problem 3. 

19. Find the masses M, m required to give the planet in Problem 17 the same 
principal moments of inertia as a uniform ellipsoid of the same mass and shape 
as the earth. Show that, with the approximations made in Problem 18, if the 
sun and moon lie in the earth's orbital plane (they do very nearly), the effect 
of both sun and moon on the earth's rotation can be taken into account simply 
by adding the precession angular velocities that would be caused by each 
separately. The equator makes an angle of 23.5° with the orbital plane. Find 
the resulting total period of precession. (The measured value is 26,000 years.) 

*20. Write Lagrangian equations of motion for the rigid body in Problem 5. 
Carry the solution as far as you can. (Make use of the results of Problem 5 if 
you wish.) Show that you can obtain a second order differential equation in- 
volving d alone. Can you find any particular solutions, or approximate solutions, 
of this equation for special cases? Describe the corresponding motions. (Note 
that this problem, to the extent that it can be solved, gives the motion of the 
body in space, in contrast to Problem 5, where we found the angular velocity 
relative to the body.) 



472 THE ROTATION OF A RIGID BODY [CHAP. 11 

21. An electron may for some purposes be regarded as a spinning charged 
sphere like that considered in Problem 6, with g very nearly equal to 2. Show 
that if g were exactly 2, and the electron spin angular momentum is initially 
parallel to its linear velocity, then as the electron moves through any magnetic 
field, its spin angular momentum would always remain parallel to its velocity. 

22. An earth satellite consists of a spherical shell of mass 20 kgm, diameter 
1 m. It is directionally stabilized by a gyro consisting of a 4-kgm disk, 20 cm 
in diameter, mounted on an axle of negligible mass whose frictionless bearings 
are fastened at the opposite ends of a diameter of the shell. The shell is initially 
not rotating, while the gyro rotates at angular velocity wo- A one-milligram 
dust grain traveling perpendicular to the gyro axis with a velocity of 
3 X 10 4 m/sec buries itself in the shell at one end of the axis. What must be 
the rotation frequency of the gyro in order that the gyro axis shall thereafter 
remain within 0.1 degree of its initial position? An accuracy of two significant 
figures in the result will be satisfactory. 

23. A gyrocompass is a symmetrical rigid body mounted so that its axis is 
constrained to move in a horizontal plane at the earth's surface. Choose a 
suitable pair of coordinate angles and set up the Lagrangian function if the 
gyrocompass is at a fixed point on the earth's surface of colatitude 0o- Neglect 
friction. Show that the angular velocity component &>3 along the symmetry 
axis remains constant, and that if «3 > (Ii/hfao sin do, where wo is the angular 
velocity of the earth, then the symmetry axis oscillates in the horizontal plane 
about a north-south axis. Find the frequency of small oscillations. In an actual 
gyrocompass, the rotor must be driven to make up for frictional torques about 
the symmetry axis, while frictional torques in the horizontal plane damp the 
oscillations of the symmetry axis, which comes to rest in a north-south line. 



CHAPTER 12 

THEORY OF SMALL VIBRATIONS 

An important and frequently recurring problem is to determine whether 
a given motion of a dynamical system is stable, and if it is, to determine 
the character of small vibrations about the given motion. The simplest 
problem of this kind is that of the stability of a point of equilibrium, 
which we shall discuss first. In this case, we can use the machinery of 
tensor algebra developed in Chapter 10 to give an elegant method of solu- 
tion for the small oscillations. A more general problem occurs when we 
are given any particular solution to the equations of motion. We may 
then ask whether that solution is stable, in the sense that every solution 
which starts from initial conditions near enough to those of the given 
solution will remain near that solution. This problem will be discussed in 
Section 12-6. Methods of solution will be given for the special case of 
steady motion. 

12-1 Condition for stability near an equilibrium configuration. Let us 

consider a mechanical system described by generalized coordinates 
xi, . . . , Xf, and subject to forces derivable from a potential energy 
V(xi, . . . , Xf) independent of time. If the system is subject to con- 
straints, we will suppose the coordinates chosen such that xi, . . . , x/ 
are unconstrained. The coordinate system is to be fixed in time, therefore 
the kinetic energy has the form 

k.i=i * 
Lagrange's equations then become 

V-* d /-Mr ■ \ V" 1 dMlm * * j_ dV — n h — i f 

1=1 l ' m=1 (12-2) 

These equations have a solution corresponding to an equilibrium con- 
figuration for which the coordinates all remain constant if they can be 
solved when all velocity-dependent terms are set equal to zero. The 
system can therefore be in equilibrium in any configuration for which 
the generalized forces vanish: 

|^=0, fc=l,...,/. (12-3) 

dXk 

473 



474 THEORY OF SMALL VIBRATIONS [CHAP. 12 

These / equations are to be solved for the equilibrium points, if any, of 
the system. 

The question of stability is easily answered in this case. If V(xi, . . . , x/) 
is a minimum for an equilibrium configuration x\, . . . , Xy relative to all 
nearby configurations x\ + Sxi, . . . , Xy + Sxy, then this is a stable 
configuration. The total energy 

E = T+V (12H1) 

is constant. Let 

E = V(x\, ...,xf) + iE (12-5) 

be the energy corresponding to any initial conditions x\ + 5x\, . . . , 
Xf + 5x°; xi, . . . ,x/ near equilibrium. Then if 5x°, . . . , Sx 1 }; x°, . . . , x° 
are small enough, we can make BE as small as we please. Since T is never 
negative, the motion is restricted by Eq. (12^) to a region in the configura- 
tion space for which 

V(x u ...,*/)< V(xl ■ ■ ■ , xf) + SE. (12-6) 

Since V is a minimum at (#?, . . . , x°), if 8E is sufficiently small, the 
motion is restricted to a small region near x\, . . . , x/. Furthermore, 
since 

T < SE, (12-7) 

the velocities x 1( . . . , xj are limited to small values. Therefore the equi- 
librium is stable in the sense that motions at small velocities near the 
equilibrium configuration remain near the equilibrium configuration. 

Conversely, if V is not a minimum near xi, . . . , x°, then it is plausible 
that the equilibrium is unstable, because in some direction away from 
x° u . . . , x°, V will decrease. If we can choose the coordinates so that x\, 
say, corresponds to that direction, and so that X\ is orthogonal to the 
other coordinates, then Eq. (12-2) for Xi is 

5 <*«*i> - JC 2 -axT XlXm = - SI • (12 " 8) 

For small enough velocities such that quadratic terms in the velocities are 
negligible, this becomes 

M lA = -g- (12-9) 

But as we move away from equilibrium in the x i -direction, dV/dxi be- 
comes negative, and x x has a positive acceleration away from the equi- 
librium point. In Section 12-3 we shall present a more rigorous proof 
that the equilibrium is unstable if V is not a minimum there. 



12-2] 



LINEARIZED EQUATIONS OP MOTION 



475 



*? 



Here the test for a minimum point should be recalled. If x\, . 
is an equilibrium configuration for which Eq. (12-3) holds, then it is a 
minimum of V(xi, . . . , x/) relative to nearby configurations, provided 
that all the determinants in the following sequence are positive: 



d V n 
dxl 



d 2 V 
dxl 



d 2 V 



dxidXf 



d 2 V d 2 V 



dXzdXi 



dx% 



> 0, 



where the derivatives are evaluated at xl, 



d 2 V 
dx\ 


d 2 V 
dx^Xf 


d 2 v 


d 2 V 


dx/dxi 


dxj 


.,*?.* 





> o, 



(12-10) 



12-2 Linearized equations of motion near an equilibrium configuration. 

We wish now to study the motion of a system in the neighborhood of an 
equilibrium configuration. The coordinates will be chosen so that the 
equilibrium configuration lies at the origin x x = • • • = x/ = 0. The 
potential energy V is to be expanded in a Taylor series in x x , . . . , x/. 
The constant term 7(0, . . . , 0) may be omitted as it does not enter 
into the equations of motion. The linear terms are absent, in view of 
Eqs. (12-3). If our study is restricted to small values of x u . . . , x/, we 
may neglect cubic and higher-order terms in x u . . . , x { , so that 



V = ^2 \ K 1d X ^h 



k,l=l 



where 



dXkdx i/xi = . . . = xf = o 



Since the coordinate system is stationary, the kinetic energy is 

T = ]T) %M k ix k ±i. 

k,l=l 



(12-11) 
(12-12) 

(12-13) 



In general, the coefficients M M may be functions of the coordinates, 
but since the velocities are to be small, to second order in x u . . . , x f ; 
x u . . . , if we may take Mm to be the values of the coefficients at 
Xl = • • • = Xf = 0. 

Equations (12-11) and (12-13) can be written in a suggestive way by 
introducing in the /-dimensional configuration space a configuration 
vector x with components x\, . . . , xf. 



(*i, 



,Xf). 



(12-14) 



1 W. F. Osgood, Advanced Calculus, New York: MacMillan, 1925, p. 179. 




476 THEORY OF SMALL VIBRATIONS [CHAP. 12 

The coefficients Km and Mm become the components of tensors 

fK 11 ---K 1/ \ 
K = 

K ff / 

(12-15) 
M lf \ 
M 

These tensors are symmetric, or can be taken as such, since by Eq. (12-12) 

Kki = K lk> (12-16) 

and in the denning equation (12-13) only the sum i(Mki + Mi k ) is de- 
fined as the coefficient of x k ±i = xi± k . Therefore, we may require that 

M hl = M lk . (12-17) 
The kinetic and potential energies may now be written as 

T = £i • M • x, (12-18) 

V = ix K x. (12-19) 
The Lagrange equations (9-79) may be written as 

M • x + K • x = 0. (12-20) 

This equation bears a formal resemblance to Eq. (2-84) for the simple 
harmonic oscillator. If we write Eq. (12-20) in terms of components, 
we obtain a direct generalization of Eqs. (4-135) and (4-136) for two 
coupled harmonic oscillators. 

We may solve Eq. (12-20) by the same method used to solve Eqs. 
(4-135) and (4-136). We try 

x = Ce pt , (12-21) 

where C = (C x , . . . , C/) is a constant vector whose components C\, . . . ,C/ 
may be complex. We substitute in Eq. (12-20) and divide by e pt : 

p 2 M • C + K • C = 0. (12-22) 

If we write this in terms of components, we obtain 

/ 
2 iV 2 M kl + K kl )Ci = 0, * = 1, ...,/. (12-23) 



12-3] NORMAL MODES OF VIBRATION 477 

If Ci, . . . , Cf are not all zero, the determinant of the coefficients must 
vanish: 



p 2 M fl + X/i • • • v 2 M ff + K ff 



= 0. (12-24) 



This is an equation of order / in p 2 whose / solutions, p 2 = —u 2 , give 
the / normal frequencies of oscillation. We may then substitute any pj 
in Eqs. (12-23) and solve for the components Cjy of the vector Cy (except 
for an arbitrary factor). The solution may then be obtained as a super- 
position of normal vibrations, just as in Section 4-10 for two coupled 
oscillators. In the next section we shall consider an alternative way, 
utilizing the methods of tensor algebra developed in Chapter 10, of de- 
termining the same solution. 

12-3 Normal modes of vibration. If the coordinates x u . . . , x f are 
orthogonal, the tensor M will be in diagonal form: 

In = M k S kl . (12-25) 

If the coordinates are not orthogonal, we can diagonalize M by the method 
of Section 10-4, generalized to / dimensions. (We will use the same 
method below to diagonalize the potential energy.) Let us suppose that 
this has been done, and that the coordinates x u ...,x/ are the com- 
ponents of x along the principal axes of M, so that Eq. (12-25) holds. 
(If xi, . . . , x f are rectangular coordinates of a set of particles, M k is the 
mass of the particle whose coordinate is x k .) 

We now define a new vector y with coordinates yi, ■ ■ ■ ,y/ given by 

y k =(M k x k yi 2 , k=l,...,f. (12-26) 

Note that the configuration of the system is specified now by a vector y 
in a new vector space related to the z-space by a stretch or compression 
along each axis, as given by Eq. (12-26). The kinetic energy in terms 
of y is 

r=iy-y= £&2. (12-27) 

Clearly, the expression for the kinetic energy does not change if we rotate 
the ^/-coordinate system, which is our reason for introducing the vector y. 
The potential energy is given by 

v = y ■ W • y = X) Wkiym, (i 2 -28) 

k,l=l 



478 
where 



THEORY OF SMALL VIBRATIONS 



W kl = 



Ki 



hi 



Mi/zjlfj 1 / 2 



The equations of motion are 



y + Wy 



[chap. 12 



(12-29) 



(12-30) 



The tensor W is symmetric, and can therefore be diagonalized by the 
method given in Section 10-4. Let ey be an eigenvector of W correspond- 
ing to the eigenvalue W ' y. 

W ey = Wfij. (12-31) 



Let aij be the components of ey in the ^-coordinate system: 
ey = (a lh . . . , a f j), j = 1, . . . ,/. 



(12-32) 



Then we may write Eq. (12-31) in terms of components in a form corre- 
sponding to Eqs. (10-108) : 



£ (W ki - Wj S k i)aij = 0, k = 1, 



,/• 



(12-33) 



Again, the condition for a nonzero solution is 

Wu - Wj W 12 

W 21 W 22 - Wj ■■■ 



W n 



W f2 



W 2f 
W ff - Wj 



0. (12-34) 



This is an algebraic equation of order / to be solved for the / roots Wj. 
Note that it is the same as Eq. (12-24) if p 2 = —Wj and we divide the left 
side of Eq. (12-24) by M x • M 2 ■ ■ • M t , remembering that M k i is now 
given by Eq. (12-25). Each root Wj is to be substituted in Eq. (12-33), 
which may then be solved for the ratios aiji a 2i : ■ • • : a/j. The aij can 
then be determined so that ey is a unit vector: 



i=i 



(12-35) 



The proofs given in Section 10-4 can be extended to spaces of any number 
of dimensions, so we know that the roots Wj are real, and therefore the 



12-3] NORMAL MODES OF VIBRATION 479 

coefficients aij are also real. Moreover, the unit vectors ej, e* are orthog- 
onal* for Wj 7* Wi. We have therefore 

e y • e r = d ]r , (12-36) 

or 

/ 
X) ai i aiT = 8 *- (12-37) 

In the case of degeneracy, when two or more roots Wj are equal, we 
can still choose the aij so that the corresponding ej are orthogonal. The 
situation is precisely analogous to that described in Section 10-4, except 
that for / > 3 it cannot be visualized geometrically. The proof of 
lemma (10-125) can be generalized to multiple degeneracies in spaces of 
any number of dimensions. 

Now let the components of the configuration vector y along e u . . . ,e/ 
beq u ..., qf. 

y = £ «*;• < 12 - 38 ) 

In terms of components in the original y-coordinate system, 

/ 
Vk = £ a kj qj. (12-39) 

i=i 

Conversely, by dotting e r into Eq. (12-38) and using Eqs. (12-32) and 
(12-36), we obtain: 

q r = X) a kr y k . (12-40) 

it=i 

These equations are analogous to Eqs. (10-67) and (10-69). 

The potential energy in the coordinate system q\,...,qj, which 
diagonalizes W, is 

V=t, Will d 2 - 41 ) 

i=i 

Since V is a minimum at the origin y = 0, the eigenvalues W u ...,W f 
must all be positive; otherwise for some values of qi, . . . , q/, V would be 
negative. If V were not a minimum, some of the eigenvalues Wj would 
be negative. (The special case Wj = may or may not correspond to 

* It is customary to use the term "orthogonal" rather than "perpendicular" 
in abstract vector algebra when the vectors have only an algebraic, and not 
necessarily a geometric, significance. 



480 THEORY OF SMALL VIBRATIONS [CHAP. 12 

a minimum, depending on higher-order terms which we have neglected.) 
Let us set 

Wj = a) . (12-42) 

The kinetic energy (12-27) in this case is 

T = £ Ml (12-43) 

In view of Eqs. (12-41) and (12-43), the Lagrange equations separate 
into equations for each coordinate qj\ 

Qi + <*hi=0, j=h...,f. (12-44) 

The coordinates q, are called the normal coordinates. The solution is 

q, — A, cos (a jt + Bj sin afi, j = 1, . . . , /, (12-45) 

where Aj, Bj are arbitrary constants. We may write the solution in 
terms of the original coordinates, using Eqs. (12-26) and (12-39) : 

/ 
x h = M k 112 ^ a kj {Aj cos ufl + Bj sin ojjt). (12-46) 

3=1 

The coefficients are 

Aj = qj (0) = X a k jMl l2 x k (0) (12-47) 

and 

/ 
Bj = «r X 4y(0) = 22 o>T 1 akiMl"± k (0). (12-48) 

We therefore have the complete solution for small vibrations about a 
point of stable equilibrium. 

When the number of degrees of freedom is large, solving Eq. (12-34) 
may be a formidable job which, in general, can be done only numerically 
for numerical values of the coefficients. However, in some cases we may 
know some of the roots beforehand (often we know that certain normal 
frequencies are zero), or from symmetry considerations we may know 
that certain roots are equal. Any such information helps in factoring 
Eq. (12-34). 

If V is not a minimum at x\ = • • • = x/ = 0, and some of the co- 
efficients Wj are negative, then we obtain exponential-type solutions. 
This proves that the motion is unstable in this case, since the solution 
(except for very special initial conditions) will contain terms which in- 
crease exponentially with time, at least until the linear approximation 



12-4] FORCED VIBRATIONS 481 

we have made in the equations of motion is no longer valid. The case 
when some Wj is zero will not be discussed in detail here. In the linear 
approximation we are making, the corresponding qj is constant in that 
case, and this corresponds to what was called neutral equilibrium in 
Chapter 2. The motion will proceed at constant fa until qj is large enough 
so that nonlinear terms in qj must be considered. 

It may be noted that in finding the normal coordinates we have found 
a transformation from coordinates x\, . . . , x; to qi, . . . , <?/ which simul- 
taneously diagonalizes two tensors M and K, or more correctly, which 
simultaneously diagonalizes two quadratic forms, T and V. Unless two 
tensors have the same principal axes, it is of course impossible simul- 
taneously to diagonalize them by a rotation of the coordinate system. 
However, if the coordinate system is allowed to stretch or compress along 
chosen axes, as in the transformation (12-26), then we can bring two 
quadratic expressions to diagonal form simultaneously (provided that at 
least one is positive or negative definite). We first find the principal axes 
of the first tensor. By stretching and compressing along the principal 
axes, we can reduce this to a constant tensor (provided that the eigen- 
values are all positive or all negative). In the case above, we reduced 
M to 1 with the transformation (12-26). Since all axes are principal axes 
for a constant tensor, the principal axes of the second tensor, as modified 
by the stretching of coordinates, will reduce both tensors to diagonal form. 
The reader will find it instructive to give a geometrical interpretation of 
this procedure, in the case of tensors in two or three dimensions, by 
representing each tensor by its associated quadric curve or surface, just 
as the inertia tensor was represented in Section 10-5 by the inertia ellip- 
soid. When we are dealing with vectors and tensors in physical space, 
we ordinarily do not consider nonuniform stretching of axes because 
this distorts the geometry of the space. When we deal with an abstract 
vector space, we may consider any transformation which is convenient 

for the algebraic purpose at hand. 

» 

12-4 Forced vibrations. We now wish to determine the motion of the 
system considered in the preceding section when it is subject to prescribed 
external forces F x (0, . . . , F/(t) acting on the coordinates x\, . . . , x f . We 
will again restrict our consideration to motions which remain close enough 
to the equilibrium configuration so that only linear terms in x\, . . . , x/ 
need to be included in the equations of motion. If we introduce the 

vector 

F(0 = (F u ..., F f ), (12-49) 

we may write the equations of motion in the abbreviated form, 

M-x-|-IC-x = F(fl. (12-50) 



482 THEORY OF SMALL VIBRATIONS [CHAP. 12 

where we have simply added the forces F(<) to Eq. (12-20). Note that 
Eq. (12-50) may be obtained from the Lagrangian function 

L = T - V - V, (12-51) 

where T and V are given by Eqs. (12-11) and (12-13), and 

/ 
V = - 22 ****(0- (12-52) 

k=i 

Again suppose that the coordinates xi, . . . , x/ are chosen to be orthog- 
onal so that M is diagonal. If the coordinates x k are not initially orthog- 
onal, and a rotation of the coordinate system is performed to principal 
axes of M, then the components F k (t) must be subject to the same trans- 
formation as the coordinates x k . Since we shall follow this process through 
in the case where we diagonalize the tensor K, we shall not follow it through 
in detail for M, but simply assume that, if necessary, it has been carried 
out and that M is diagonal. 

We transform now to the normal coordinates found in the preceding 
section [Eqs. (12-26), (12-39), and (12-40)]: 

x k - J2 Mk ll2 a kj qj, (12-53) 

3-1 

q 3 - = J2 Ml l2 a kj x k . (12-54) 

k=l 

The generalized forces Qj{t) associated with F k {t) are obtained by using 
Eq. (9-30). 

Qi{t) = X) Mjr lla a kJ F k ®. (12-55) 

The inverse transformation j# 

• / 

F k (t) = J2 MV 2 a kj Qj{t), (12-56) 

;=i 

The reader may also verify Eqs. (12-55) by substituting Eqs. (12-53) in 
Eq. (12-52) and calculating 

dV 
Qi=--^~- (12-57) 

In normal coordinates, 

V'=-J2 «i<?iW. (12-58) 

3=1 



12-4] FORCED VIBRATIONS 483 

so that the equations of motion are 

& + <*hi = Qi(0, 3 = 1, • • • . / • (12-59) 

Each of these equations is identical in form with Eq. (2-86) for the un- 
damped forced harmonic oscillator (b = 0). Therefore the normal modes 
behave like independent forced oscillators, and the solution can be ob- 
tained by the methods described in Chapter 2. 

It is tempting to try to generalize our results to the case when linear 
damping forces are also present. We can easily write down the appropriate 
equations. In the general case when the coordinates are not orthogonal 
and there is frictional coupling between coordinates, the equations of mo- 
tion will be 

X) (M k tfi + B ht ki + K klXl ) = 0, * = 1, ...,/, (12-60) 

or, in vector form, 

Mx+Bx+Kx=0. (12-61) 

Unfortunately, as the reader may perhaps convince himself with some 
experimentation, it is generally not possible simultaneously to diagonalize 
three tensors M, B, K with any linear transformation of coordinates, even 
if stretching is allowed, The method of the preceding section therefore 
fails in this case, and there are no normal coordinates. The situation is 
not improved by assuming that X\, . . . , x/ are orthogonal so that M is 
diagonal, or even by assuming that there is no frictional coupling so that 
B is diagonal. If we apply the transformations (12-53) and (12-54) which 
diagonalize T and V, the coordinates q, are in general still coupled by 
frictional forces: 

9i + L Mr + o>fa = 0> C 12 - 62 ) 

r=l 

where 

b yr = £ M k ll2 Mr ll2 a kJ a lr B hl . (12-63) 

k,l=l 

Note that the matrix bj r is not diagonal even if B k i is. There is a special 
case which sometimes occurs when the frictional forces are proportional 
to the masses, so that B = 2TM. The method of Section 12-3 then works, 
since in the y-coordinate system in which M — ► 1, we have B — > 271 
and the normal coordinates q, along the principal axes of W satisfy the 
separated equations: 

qj + 27ft + o>hi = 0- C 12 " 64 ) 



484 THEORY OF SMALL VIBRATIONS [CHAP. 12 

It may, of course, also happen that in the y-space in which M becomes 1 
the transformed tensors B and K have the same principal axes, but this 
would be an unlikely accident. When the damping forces are very small, 
a perturbation method similar to that which will be developed in the 
next section can be applied to find an approximate solution in terms of 
damped normal modes. 

Except in these special cases, the problem of damped vibrations can 
be handled only by direct substitution of a trial solution like (12-21) in 
the equations of motion (12-60). The secular equation analogous to 
Eq. (12-24) is then of order 2/ in p. Each root allows a solution for the 
vector C. If there are complex roots, they occur in conjugate pairs, p, p*, 
with corresponding conjugate vectors C, C*. The two solutions (12-21) 
can then be combined to yield a real solution which will be damped and 
oscillatory, and can be called a normal mode. If all 2/ solutions are com- 
bined with appropriate arbitrary constants, the general solution to Eqs. 
(12-60) can be written. It is clear on physical grounds, since the frictional 
forces reduce the energy of the system, that the real parts of all roots p 
must be negative if F(x) has a minimum at x = 0; the mathematical 
proof of this statement is a difficult exercise in algebra. 

12-5 Perturbation theory. It may happen that the potential energy is 
given by 

V = V° + V, (12-65) 

where V°(xi, . . . , x/) is a potential energy for which we can solve the 
problem of small vibrations about a minimum point X\ = • • • = x/ = 0, 
and where V'(xi, . . . , xj) is very small for small values of x\, . . . , x/. 
We will call V° the unperturbed potential energy, and V the perturba- 
tion. We expect that the solutions for the potential energy V will approxi- 
mate those for the unperturbed problem. In this section we shall develop 
an approximate method of solution based on this idea. 
We shall assume that V is stationary at Xi = • • • = x; = 0, so that 







f^) = 0. (12-66) 

yXh / x l ^=,...=Xf=0 



If this is not' the case, it is not difficult to find approximately the values 
of x\, ... , Xf for which V is stationary. We leave this as an exercise. 
The origin of coordinates should then be shifted slightly to x\, . . . , x°. 
This will alter slightly the quadratic terms in V°, but these small changes 
can be included in V. In any case, we will therefore have an expansion 



12-5] PERTURBATION THEORY 485 

of V around the equilibrium point of the form (12-65), with 

V° = X) K*Wz, (12-67) 



k,l 

V = X) KiWi. (12-68) 

k,l 

where the coefficients K'u are small. The precise criteria that must be 
satisfied in order that K' k i can be considered small will be developed as we 
proceed. 

We first transform to the normal coordinates g?, . . . , gy for the un- 
perturbed problem. We then have 

V° = £ i^feV, (12-69) 

3=1 

V = £ i^>2°9r°, (12-70) 

T^> = X) M^MT^WmrUu (12-71) 

ft, j 

where TFy are the roots of the secular determinant (12-34) for the un- 
perturbed problem, and where again we assume, for simplicity, that 
Xi, . . . ,'Xf are orthogonal coordinates. The coefficients W'j r are to be 
treated as small. The superscript "°" will remind us that the variables 
q'j are normal coordinates for the unperturbed problem. 
The equations of motion for q u . . . , q, are 

$ + Whl + E W' jr q°r =0, j = 1, • • • , /• (12-72) 

r=l 

We see that the diagonal element of W adds to the coefficient of g°, while 
the off -diagonal elements couple the unperturbed normal modes. We 
expect that if W is small, there will be a normal mode of the perturbed 
problem close to each normal mode of the unperturbed problem, that is, 
a solution with frequency uij near o>° = (W?) 1/2 and for which q° is large 
while the remaining q°, r ^ j, are small. However, if the tensor W° has 
degenerate eigenvalues, so that two or more of the unperturbed fre- 
quencies are equal (or perhaps nearly equal), then we expect that even a 
small amount of coupling can radically change the motion, as in the case 
of two coupled oscillators that we worked out in Chapter 4. This insight 
will help in developing a perturbation method. 



486 THEORY OF SMALL VIBRATIONS [CHAP. 12 

If we try to find a normal mode of oscillation by substituting 

<Z? = (V", j=l,...,f, (12-73) 

in Eqs. (12-72), we obtain 

(p 2 + WfiCj + X) W' jr Cr = 0. (12-74) 

r=l 

Now let us assume that the mode which we seek is close to some un- 
perturbed mode, say j = 1. We then set 

p 2 = -Wl - W'i, 

Ci=l+ C' lt (12-75) 

Cj = Cj, j = 2, . . . ,/, 

where, if W[, C[, . . . , C} are zero, Eq. (12-73) represents a solution of the 
unperturbed problem. Hence for the perturbed problem, we assume that 
W' lt C[,...,C' f are small. We substitute Eqs. (12-75) in Eqs. (12-74), 
and collect second-order terms on the right-hand side: 

-W\ + WU = - X) w 'irC'r + WVJ'u (12-76) 

r=l 
(W> - Wlffl + W'n = - Y, W 'ir C 'r + W'lC'j, j = 2, . . . ,/. 

r-1 (12-77) 

When we neglect second-order terms, the first equation gives W[ : 

Wi = W'n. (12-78) 

Therefore 

w f = -p 2 = W\ + W'n- (12-79) 

Equations (12-77), if we neglect the right members, yield the coefficients 
Cj: 

C '^W^- i = 2 '■ <12 - 80) 

The coefficient C[ is not determined; this corresponds to the fact that the 
normal mode (12-73) may have an arbitrary amplitude (and phase), 
although it must, of course, be near the amplitude (and phase) C? = 1 
which was chosen in Eqs. (12-75) for the unperturbed solution. It will be 
convenient to require that C\, . . . , C/ be the coefficients of a unit vector: 

/ 
J2 C] = 1. (12-81) 

3=1 



12-5] PERTURBATION THEORY 487 

We then obtain the following equation for C{ : 

d = ~i £ (C'j) 2 . (12-82) 

3=1 

To first order in small quantities, 

C\ = 0. (12-83) 

By substituting in Eqs. (12-73), multiplying by an aribitrary constant 
\Ae i *, and superposing the complex conjugate solution, we obtain the 
first-order approximations to the perturbed normal mode: 

ql = A cos (wi< + ff), 

(12-84) 

g° ~ t W ' n o cos ("i* + ®> j = 2,...,f, 

where o>i is given by Eq. (12-79). We see that the first-order effect of the 
perturbation is to shift u>\ by the diagonal perturbation coefficient W[\ 
and to excite the other unperturbed modes weakly with an amplitude that 
is proportional to the perturbation coupling coefficients W'j\ and in- 
versely proportional to the differences in unperturbed normal frequencies 
(squared). This is a physically reasonable result. 

We can now formulate more precisely the requirement that W be small. 
In our derivation, we have assumed that 

W{ « \W°i - TT?|, i = 2, ...,/, . (12-85) 

C'j « 1. (12-86) 

Equations (12-78) and (12-80) show that this is justified if 

W' n « \W°j - W\\, j = 1, ■ • • , /• (12-87) 

This is the condition for the validity of formulas (12-79) and (12-84). 

First-order approximations to the remaining modes are obtained from 
these formulas by interchanging the subscript «i» with any other. 



The astute reader will note that the condition (12-87) in the diagonal co- 
efficient W{\ is necessary only because we have neglected the last term in 
Eqs. (12-77). Equations (12-77) are easily solved for C'j even if the last term 
on the right is included. This allows us to remove the restriction on the size of 
the diagonal coefficients if we wish. This is also obvious because we can always 
include any diagonal coefficient in 7° [Eq. (12-69)]. The normal coordinates 



488 THEORY OF SMALL VIBRATIONS [CHAP. 12 

for the unperturbed problem are still the same; only the (squared) frequencies 
are altered by adding additional diagonal terms. However, in the transformation 
to normal coordinates, diagonal and off-diagonal terms become intermixed, so 
that unless all terms in V (xi, . . . , x/) are small the off-diagonal terms of 
V (<??> • • • > 9/) are unlikely all to be small. 



Conditions (12-87) clearly cannot be satisfied if there is a degeneracy — 
if, for example, W° = W 2 = W%. In that case, as mentioned earlier, we 
expect that even with very small coupling of the unperturbed modes, 
any perturbed mode with w 2 near W\ will show appreciable excitation of 
all three unperturbed modes. We therefore set 

p 2 = -w\ - W, (12-88) 

and assume that only C 4 , . . . , C/ are small, while Ci, C 2 , C 3 may all be of 
order 1. We substitute in Eqs. (12-74) and transpose second-order terms 
to the right members: 

/ 
(W' 11 -W')C 1 + W' 12 C 2 + W' 13 C 3 =-^W' lr Cr, 

r=4 

W' 21 Ci + (W 22 - W')C 2 + w 23 c 3 = - X) w 2r c r , 

r=4 

W'atd + W' 32 C 2 + (W 33 - W')C 3 = - X) W'sfCr, 

r= 4 

(12-89) 
and 

(Wj - Wt)Cj + J2 W 'i£r = - X) W 'ir C * + W ' C i> 3 = 4, . . . , /. 

(12-90) 

If we neglect the right members, then Eqs. (12-89) become a standard 
three-dimensional eigenvalue problem for the eigenvalue W and the 
associated eigenvector (C 1; C 2) C 3 ). There will be three solutions corre- 
sponding to three perturbed normal modes with frequencies w 2 = 
W\ + W near the degenerate unperturbed frequency. In general, the 
three roots W will be different, and so the perturbed modes will no longer 
be degenerate. The remaining coefficients C4, . . . , C/ can be found to a 
first-order approximation from Eqs. (12-90) by neglecting the right mem- 
bers. We may also require that C be a unit vector [Eq. (12-81)]. In 
analogy with Eq. (12-83), this will mean that to first order the three- 
dimensional vector (Ci, C 2 , C 3 ) should be a unit vector. The case of double 



12-5] PERTURBATION THEORY 489 

or multiple degeneracy of any order must be treated in the same way. 
Clearly, if the degeneracy is of high order, the first-order perturbation 
equations [Eqs. (12-89) with right members zero] may be almost as 
difficult to solve as the exact equations (12-74). When / > 4, we may 
have more than one degenerate normal frequency; in that case, the above 
method can be applied separately to each group of degenerate unper- 
turbed modes to find the perturbed modes. 

In cases of approximate degeneracy (W? = W 2 = W 3 ) when condi- 
tions (12-87) fail for a group of neighboring unperturbed modes, the 
method of the previous paragraph can also be applied. Equations (12-89) 
are slightly modified by the addition of small terms, like W% — W°, in 
the diagonal coefficients. The reader can readily formulate the procedure 
for himself. 

When the first-order approximate solution has been found, the ap- 
proximate values of the coefficients W[, C'j may be substituted in the 
right members of Eqs. (12-76), (12-77) [or Eqs. (12-89), (12-90)]. The 
resulting equations are then solved to find a second-order approximation. 
If, for example, we substitute Eqs. (12-80) in Eq. (12-76), we obtain the 
second-order approximation to the frequency correction: 

r=2 W 1 — W r 

where we have used the fact that W is a symmetric tensor. We see that 
the second-order frequency shift in mode 1 contains a contribution due to 
coupling with each of the other modes. The modes tend to repel one 
another in second order; that is, each higher frequency mode (W° T > W°) 
reduces the frequency of mode 1, and each lower frequency mode increases 
it. The same result was observed in the solution to the problem of two 
coupled oscillators in Chapter 4. The procedure can be carried out in a 
straightforward way to successively higher-order approximations, but the 
labor involved rapidly increases. 

To any order of approximation, we may introduce normal coordinates 
qi, . . . , q/ for the perturbed problem by setting 

gP = £ CjrQr, (12"92) 

r=l 

where Cj r , j = 1, ...,/, are the coefficients for the rth perturbed normal 
mode found to any order of approximation by the perturbation theory. 
The vectors C r — {C lr , . . . , C fr ) are, to the given order of approximation, 
orthogonal unit vectors (or can be made so), as we shall see presently. 



490 THEORY OF SMALL VIBRATIONS [CHAP. 12 

We may therefore solve Eqs. (12-92) for 

«r = E Crtl (12-93) 

3=1 

From Eqs. (12-92) and (12-73), if p 2 = -W° — W' r = — a? is the 
approximate value for the frequency, then the approximate solution for 
q r must be 

q r = A r cos o) r t + B r sin u r t. (12-94) 

Comparison with Eqs. (12-45) shows that the q r are (approximate) normal 
coordinates. The Lagrangian, to the given order of approximation, must 
therefore be 

L = E (44? - Wa?), (12-95) 

r=l 

as may also be verified by straightforward substitution of Eqs. (12-92) 
in Eqs. (12-69), (12-70), and (12-43), to any order of approximation 
in Cj r . Alternatively, we may note that Eqs. (12-74) are just the equa- 
tionsjs#e would obtain if we were to look for an eigenvector C of W = 
W° -f- W corresponding to the eigenvalue — p 2 . Hence the approximate 
solutions we have obtained for Eqs. (12-74) are also approximate solu- 
tions to the problem of diagonalizing W. Equations (12-92) must there- 
fore define approximate normal coordinates for the perturbed motion. 

12-6 Small vibrations about steady motion. Let a mechanical system 
be described by coordinates X\, . . . , Xf, and by a Lagrangian function 
L(xi, . . . ,Xf) ±i, . . . , ±f) t). If a solution x\{t), . . . , x%t) is known, 
we may look for solutions close to the known solution by defining new 
coordinates y\, . . . ,yf\ 

x k = x° h (t) + Vk, h = 1, . . . ,/. (12-96) 

We substitute in the Lagrangian L, and expand in powers of y u . . . y t ; 
2/i, ... , yj. Since x\{t), . . . , x%t) satisfy the equations of motion, the 
reader can readily show that no linear terms in y\, . . . , y/; y\, . . . , y/ 
occur in L. Terms in L independent of Vi, ■ . . ,y/',yi, . . • ,y/ do not 
affect the equations of motion and may be ignored. If we assume that 
Vu • • • tVf'iVi) • • ■ tVf are small, and that we may neglect cubic and 
higher powers of small quantities, L becomes a quadratic function of the 
new variables. The equations of motion will then be linear in y i} . . . , y/; 
Vi, ■ ■ ■ , ilf', Vi, ■ ■ ■ , Vf- However, the coefficients in the equations will, 
in general, be functions of the time t, and the methods we have so far de- 
veloped will not suffice to solve them. To develop methods for solving 
equations with time-varying coefficients is beyond the scope of this book. 



12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 491 

We will therefore consider only cases when the coefficients in the linearized 
equations turn out to be constant. 

We can guarantee that the coefficients will be constant by restricting 
ourselves to steady motions. Suppose that some of the coordinates 
Xi, . . . , Xf are ignorable, i.e., do not appear in the Lagrangian function. 
We will also assume that L does not depend explicitly on t. We define a 
steady motion as one in which all of the nonignorable coordinates are 
constant. This definition evidently depends upon the system of co- 
ordinates chosen. We should perhaps define steady motion as motion for 
which, in some coordinate system, the nonignorable coordinates are all 
constant. 

We have seen in Section 9-10 that ignorable coordinates are particularly 
easy to handle in terms of the Hamiltonian equations of motion. Let us 
therefore introduce coordinates x lt . . . , x f , and corresponding momenta 
Pi, ■ • • ,Ps, of which xn+i, . . . , x/ are ignorable. The Hamiltonian 
function is 

H = H(x u . . . ,x N ;vu ■ ■ ■ , Pn, Pn+i, ■ ■ ■ , Pf). (12-97) 

In view of Eqs. (9-198), the momenta Pn+u ■ ■ ■ ,Pf are all constant. 
We have therefore to deal only with 2N equations (9-198), which for 
steady motion reduce to 

For given values Pn+i, . . . , p°, we are to find the solutions, if any, of 
these equations for Xi, . . . , Xn',Pi, • • ■ , Pn- Any such solution a;?, . . . , x%; 
Pi, • • - 1 Pn defines a steady motion. The ignorable coordinates will all 
have constant velocities given by 



*' ~ Uo' 



N+l,...,f, (12-99) 



where the subscript " " implies that the derivative is to be evaluated at 



• £, i> • • • j ■*-#> Pi> • • • , Pf 



Given a steady motion, let us choose the origin of the coordinate system 
so that x\ = • • • — x% = pi = • ■ • '= p% = 0. In order to look for 
motions near this steady motion we hold Pn+i, ■ • ■ ,Pf fixed and expand 
H in powers of x\, . . . , xn, Pi, ... , Pn, which we regard as small. We 
may omit any terms which do not depend on x lt . . . ,pn- Linear terms 
are absent because of Eqs. (12-98). If we neglect cubic terms in small 
quantities, H becomes a quadratic function of x i, ... , xn, Pi, ■ • • , Pn, 
with constant coefficients. It may be that H separates into a positive 



492 THEORY OF SMALL VIBEATIONS [CHAP. 12 

definite "kinetic energy," ' T' {p lt ..., p N ), and a "potential energy," 
'V'(xi, . . . , x N ). In that case, the methods of the preceding sections are 
applicable. In order to apply these methods, we must express the "kinetic 
energy," 'T\ in terms of x u . . . , x N , which may be done by solving for 
Pi, • ■ ■ i Vn the linear equations 

x k = ^ , k=l,...,N. (12-100) 

apk 

The problem is thus reconverted to Lagrangian form, with 'L' {x u . . . , x N ; 
±i, . . . , An) = 'T' — 'V. Note, however, that we cannot obtain the 
correct 'L' simply by substituting x° from Eq. (12-99) in the original L. 
The transition to Hamiltonian form is necessary in order to be able to 
eliminate the ignorable coordinates from the problem by regarding 
Pn+i, ■ ■ ■ , Pf as given constants. The "potential energy" 'V will con- 
tain terms involving p N +i, ■■ ■ ,Pf frcm the original kinetic energy, 
and these will appear with opposite sign in 'L'. 

If l V'(xi, . . . , xn) has a minimum at x\ = • • • = x^ = 0, then 
xi, . . . , xn, if they are small enough, undergo stable oscillations. We 
may then say that the given steady motion is stable in the sense that for 
nearby motions (with the same Pn+i, ■ ■ ■ , Pf), the coordinates oscillate 
about their steady values. These oscillations can be described by normal 
coordinates, which may be found by the method of Section 12-3. 

If the coordinate system we use is a moving one, or if magnetic forces 
are present and must be described by a velocity-dependent potential 
(9-166), or if, as often happens, the ignorable coordinates are not orthog- 
onal to the nonignorable coordinates, then cross product terms XkPi will 
appear in H. Thus, in general, the quadratic terms in H have the form 

JV 

H = ^2 (\a k ipkpi + bkixkpi + ickiXkXi), (12-101) 

k,l=l 

where we may as well assume that 

aik = a k i, cik = c k i. (12-102) 

The coefficients a«, bu, Cki are functions of the constants Pn+i, ■ ■ ■ ,Pf 
and of the particular steady motion whose stability is in question. We 
no longer have a separation into kinetic and potential energies, and the 
methods of the preceding sections can no longer be applied. It may be 
that H as given by Eq. (12-101) is positive (or negative) definite in 
x\, . . . ,xn, Pi, • ■ ■ ,Pn, in which case we may be sure that the steady 
motion is stable. For x lt . . . , X&, p u . . . , Pn must remain on a surface 
of constant H, and if H is positive definite, this surface will be an "ellip- 
soid" in the 2iV-dimensional phase space. 



12-6] 



SMALL VIBRATIONS ABOUT STEADY MOTION 



493 



In any case, we may study the small vibrations about steady motion 
by solving the linearized equations given by the Hamiltonian function 
(12-101) : 



Xk 



N 

i=i 



*k ~ £ ( a kiPi + bi k x{), 



N 



Pk = — ^2 (bkiPi + Ckixi), k = 1, . . . , N. 



(12-103) 



We could return to a Lagrangian formulation involving N second-order 
equations in x 1} . . . , xn, but it is just as easy to deal directly with 
Eqs. (12-103). Let us look for a normal mode in which all quantities have 
the same time dependence: 

x k = X k e pt , p k = P k e pt . (12-104) 

We substitute in Eqs. (12-103) and obtain the 2N linear equations 

N 

X) K&» - V hi)Xi + a k iPi] = 0, 

2=1 

X) [^iXi +(hi + p S k i)Pi] = 0, k = 1, . . . , N. (12-105) 

i=i 

The determinant of the coefficients must vanish: 



0-2N 
blN 

b 2 N 

bNN + P 



11 - 


P 


b 2 i 




■ «ii 


612 




622 — 


p. . 


• <*21 


en 




c 12 




•&11 + P 


C 2 1 




C22 


■ 


2>21 



CiVl 



CJV2 



bNi 



0. (12-106) 



We now note that if p is any root of this equation, so is —p. First, let us 
set p = —p' in the determinant. Now, interchange the upper N rows 
and the lower N rows. Next, interchange the left N columns with the 
right N columns. Finally, interchange rows and columns, i.e., rotate 
about the main diagonal. None of these operations changes the value 
of the determinant (except possibly its sign, which does not matter). 
We now have the same equation for p' that we originally had for p, so 
that if p is a root, so is p'. We see therefore that when we expand the 
determinant (12-106), only even powers of p appear, and we have an 
algebraic equation of degree N in p 2 . If the roots are all negative, as they 



494 THEORY OF SMALL VIBRATIONS [CHAP. 12 

will be if H in Eq. (12-101) is positive definite, then the normal modes 
are all stable. Each root p 2 = — u 2 gives two values p = ±iwy. We 
substitute p = iwj in Eqs. (12-105) and solve for Xij, Pij, which will, 
in general, be complex. There is, of course, an arbitrary constant which 
may be chosen in any convenient way. The solutions for p — —iuj 
will be X*j, P*j. We substitute in Eqs. (12-104), multiply by an arbitrary 
constant Aje i>j and superpose the two complex conjugate solutions to 
obtain the real solution for the normal mode j: 



where 



x k = AjCkj cos (wjt + hi + Oj), 

p k = AjD kj cos (ujt + 7k, + Oj), 

X kj = ^C kj e ih ', 

Pkj = %D kj e iyki , 



(12-107) 



(12-108) 



and Aj and 6j are an arbitrary amplitude and phase. The general solution 
is now a superposition of normal modes : 



N 



Xk 



= J3 A i Ck i cos ( w ^ + Pv + e J')» 



J'=l 



iV 



(12-109) 



Pk = ^2 AjD ki cos {ujt + y k j + dj). 

3=1 



Note that we cannot represent the above result in terms of normal 
coordinates q\, ■ ■ ■ ,qN linearly related to x-y, . . . , xn on account of the 
phase differences f$ k j, 7kj which arise because of the cross terms in co- 
ordinates and momenta. It is possible to find a linear transformation 
connecting the 2N variables x 1; . . . , x^; pi, . . . , p^ with a set of normal 
coordinates and momenta </i, . . . , <?#; pi, . . ■ , Pn, each of which oscil- 
lates at the corresponding normal frequency. Such transformations be- 
long to the theory of canonical transformations of Hamiltonian dynamics, 
and are beyond the scope of this book. 

If any-root p 2 of Eq. (12-106) is positive or complex, the corresponding 
normal mode is unstable, and the coordinates and momenta move ex- 
ponentially away from their steady values. A root p 2 = would corre- 
spond to neutral stability. One common case in which roots p 2 = 
arise occurs when an ignorable coordinate has been included among 
the x i, . . . , xn. If Xj is ignorable, and is included in X\, . . . , x^, then 
Pi is constant and may take any value. If we take pj slightly different 
from p° for the initially given steady motion, there is a new steady 



12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 495 

motion with x j constant and slightly different from xj. This new motion 
is given by 

Xj = Aj + BjXjt, all other x k = Ak, (12-110) 

where Ak may be slightly different from x° k . This motion corresponds to 
a normal mode with p 2 = 0. In some cases, the algebra required to ignore 
explicitly a coordinate Xj is too formidable, and we may prefer to include xj 
among the nonignorable coordinates. This increases the degree of the 
secular equation (12-106) by one, but since the extra root is p 2 = 0, we 
know that p 2 will factor out and the remaining equation will have the 
same degree as if we had ignored Xj. It may also be that we .have chosen 
a coordinate system in which some ignorable coordinate Xj does not ap- 
pear. A root p 2 = of Eq. (12-106) will still occur. Since the coordinates 
actually used will be functions of the ignorable one (among others), in 
the corresponding normal mode, several or all of the coordinates may 
exhibit constant velocities. These may be found by substituting in the 
equations of motion (12-103). 

The case of degeneracy, when a multiple root for p 2 occurs, is more 
complicated for Eqs. (12-103) than for Eqs. (12-20), where the force 
is derivable from a potential energy depending only on x\, . . . , x N . We 
can no longer make use of the diagonalization theory for a symmetric 
tensor to show that for a multiple root p 2 , Eqs. (12-105) have a corre- 
sponding multiplicity of independent solutions, as we did for the analogous 
equations (12-23) or (12-33). Sometimes Eqs. (12-105) may have only 
one independent solution, even when p 2 is a multiple root of Eq. (12-106), 
and we must look for other forms of solution than (12-104). We will not 
carry out the algebraic details here,* but the result is that when 
Eqs. (12-105) do not yield enough independent solutions for X k , Pk for 
a multiple root p 2 , X k , Pk should be replaced in Eqs. (12-104) by poly- 
nomials in t of degree (n — 1), where n is the multiplicity of the root p 2 . 
The resulting expressions must be substituted in Eqs. (12-103), which 
then give 2Nn relations between the 2Nn coefficients in the 2N poly- 
nomials. These relations can be shown to leave just n arbitrary coefficients, 
so that the correct number of arbitrary constants are available. Alter- 
natively, we can slightly alter the coefficients in the Hamiltonian (12-101) 
so that the degeneracy in p 2 is removed, find the solution, and then find 
its limiting form as the coefficients approach their original values. (See 



* For a further discussion of problems of small vibrations, see E. J. Routh, 
Dynamics of a System of Rigid Bodies, Advanced Part. New York: Dover Pub- 
lications, 1955. Chapter 6. A less complete but more elegant treatment using 
matrix methods is given in R. Bellman, Stability Theory of Differential Equations, 
New York: McGraw-Hill, 1953. 



496 THEORY OF SMALL VIBRATIONS [CHAP. 12 

Problem 24, Chapter 2.) When powers of t appear in the solution, it is 
clear that the solution is not stable even when p 2 is real and negative, but 
represents an oscillation whose amplitude after a long time will increase 
as some power of t. Hence degeneracy generally implies instability in the 
case of Eqs. (12-103). It will be found that multiple roots of Eq. (12-106) 
ordinarily mark the boundary between real and complex solutions for 
p 2 in the sense that a small change in some coefficient au, bu, or c k i 
will split the degeneracy and lead on the one hand to two real, or on the 
other hand to two complex, roots for p 2 , depending on the sense of the 
change. The situation is closely analogous mathematically to the problem 
of the damped harmonic oscillator, where a double root for p in Eq. (2-125) 
marks the dividing line between the overdamped and underdamped cases 
and leads to solutions linear in t. In the present case, there is no damp- 
ing; Eq. (12-106) contains only even powers of p, and if complex con- 
jugate roots for p 2 occur, then the corresponding four roots p have the 
form ±7 ± io), and some of the solutions grow exponentially. Thus 
multiple roots can mark the boundary between stable and unstable cases. 
In the case of Eqs. (12-20), where the forces are not velocity-dependent, 
the boundary between stability and instability occurs only when some 
root for p 2 is zero; degenerate negative roots for p 2 always correspond to 
stable solutions. 

We should further remark that even when the roots p 2 are all negative 
and distinct, so that the solutions of Eqs. (12-103) are all stable, we 
cannot guarantee that the exact solutions of the nonlinear equations aris- 
ing from the complete Hamiltonian (12-97) are stable. For vibrations 
about an equilibrium point when the forces are derivable from a potential 
energy, we were able to prove that stable solutions of the linearized equa- 
tions are obtained only around a potential minimum, and that in that 
case, the solutions are absolutely stable if the amplitude is small enough. 
We saw above that if the Hamiltonian H is positive (or negative) definite 
near the steady motion, then we can also show that the solutions are 
stable. But the solutions of Eqs. (12-103) may all be stable even when 
H is not positive or negative definite. In that case, all we can say is 
that, for as long a time as we please, if we start with sufficiently small 
amplitudes, the solutions of the exact problem given by the Hamiltonian 
(12-97) will "approximate those of the linearized problem given by the 
Hamiltonian (12-101). This is true because the nonlinear terms can be 
made as small as we like by making the amplitude sufficiently small, 
and then their effect on the motion can be appreciable, if at all, only if 
they are integrated over a very long time. Nevertheless, if we neglect 
the nonlinear terms, we are prevented from asserting complete stability 
for all time. Cases are indeed known where the linearized solutions are 
stable, and yet no matter how near the steady-state motion we begin, 



12-7] BETATRON OSCILLATIONS IN AN ACCELERATOR 497 

the exact solution eventually deviates from the steady-state motion by 
a large amount. To find the criteria that determine ultimate stability 
in the general case is perhaps the outstanding unsolved problem of 
classical mechanics. 

12-7 Betatron oscillations in an accelerator. In a circular particle ac- 
celerator, for example a cyclotron, betatron, or synchrotron, charged 
particles revolve in a magnetic guide field which holds them within a 
circular vacuum chamber as they are accelerated. Since the particles 
revolve many times while they are being accelerated, it is essential that 
the orbits be stable. Since the particle gains only a small energy incre- 
ment at each revolution, it is permissible to study first the stability of 
the orbits at constant energy, and then to consider separately the ac- 
celeration process itself. We will be concerned here only with the 
stability problem at constant energy E. Let us assume that the mag- 
netic field is symmetrical about a vertical axis, so that we may write, 
using cylindrical polar coordinates (Fig. 3-22), 

B(p, <p, z) = B z (p, z)k + B r {p, z)h. (12-111) 

The magnetic field in a synchrotron or betatron is also a function of time, 
increasing as the energy increases, but since we are treating E as constant, 
we also take B as constant. We will suppose that in the median plane 
z = 0, the field is entirely vertical : 

B(/o, <p, 0) = B l0 (p)k. (12-112) 

A particle of appropriate energy E may travel in a circle of constant radius 
p = a(E). We call this orbit the equilibrium orbit. We are interested in 
the stability of this orbit; that is, we want to know whether particles 
near this orbit execute small vibrations about it. Such vibrations are 
called betatron oscillations because the theory was first worked out for the 
betatron. 

The vector potential (Section 9-8) for a magnetic field with symmetry 
about the z-axis can be taken to be entirely in the ^-direction: 

A = A v {p, z)m, (12-113) 

so that 



B = V X A 



= ^7<"^- h ^- (12 - 114) 



498 THEORY OF SMALL VIBRATIONS [CHAP. 12 

We see from Eq. (12-114) that A is given by 

A,(p, z) = p- 1 fpB z (p, z) dp, (12-115) 

Jo 

since A v must vanish at p = because of the ambiguity in the direction 
of m. Note that 2irpA v is the magnetic flux through a circle of radius p. 
The Lagrangian function is given by Eqs. (9-154) and (9-166) : 

L = im(p 2 + pV + z 2 ) + e - p?A v (p, z), (12-116) 

where e, m are the charge and mass of the particle to be accelerated. If 
the velocity of the particle is comparable with the speed of light, that is, 
if the kinetic energy is comparable with or greater than mc 2 , the relativistic 
form for the Lagrangian should be used (Problem 23, Chapter 9). The 
momenta are 

dL 



Vp = _ = m p, 




p„ = mp 2 <p + - pA v , 


(12-117) 


p„ = mz. 




The Hamiltonian function is given by Eq. (9-196) or (9-200) : 




rr _ Pp 2 , P* , [P*> — (e/c)pA v ] 2 

2m T 2m T 2mp2 


(12-118) 



We see that <p is ignorable and p 9 may be taken to be a given constant. 
The Hamiltonian function then has the form 



with 





h = <r + 'v, 


(12-119) 


trp} 


= ^ = 5^ + ^ 


(12-120) 


'V 


_ [Vv — (e/c)pA v ] 2 

9mn2 


(12-121) 



The problem reduces to an equivalent problem of static equilibrium. 
The steady motions are given by the solutions for p, z of the equations 

d'V e 

— = 1 pvB r = 0, (12-122) 

d'V ( e \ 

— = -pip \m<p + - B A = 0, (12-123) 



12-7] BETATRON OSCILLATIONS IN AN ACCELERATOR 499 

where we have used Eqs. (12-117) and (12-114). The first equation 
above is satisfied in the median plane z = 0, and usually nowhere else 
(p<p 9* 0). The second equation gives 

*=--L B °°( p) ' (12_124) 

7/tC 

which is equivalent to Eq. (3-299). We may solve Eq. (12-124) for p, 
given <p, or alternatively, for </> with p = a, the radius of the equilibrium 
orbit. Note that the energy %ma 2 <p 2 of a particle executing the steady 
motion is less than the total energy H of the particle whose motion we are 
studying, but the difference is of second order in small quantities for 
vibrations near the steady motion. According to the development in the 
preceding section, the two particles should be chosen to have the same p v . 
We now set 

p = a + x, (12-125) 

where a is the radius of the orbit for the steady motion. Next, by expand- 
ing in powers of x, z, x, z, we obtain 

T = \m{x 2 + z 2 ), (12-126) 

'V = 4m« 2 (l - n)x 2 + %ma 2 nz 2 , (12-127) 



where we have set 



co = ID = — 

mc 



ei^a) t (12-128) 

mc 

*=-(irir) ' ( 12 " 129) 



and have used Eqs. (12-114), (12-117), (12-124), and 



dB z 
dp 


dB„ 
dz 


which follows from the fact that 




V X B 


= 



(12-130) 



(12-131) 

as we can see from Eq. (9-162). The quantity n is called the field index. 
We see immediately that the motion is stable only if 

< n < 1. (12-132) 

In a cyclotron, the field is nearly constant at the center, so that n « 1, 
and then falls rapidly near the outside edge of the magnet. In a betatron 
or synchrotron, the magnetic field has a constant value of n and increases 



500 THEORY OF SMALL VIBRATIONS [CHAP. 12 

in magnitude as the particles are accelerated so as to keep a constant. We 
see from Eq. (12-129) that the value of n does not change as B z is increased 
provided the shape of the magnetic field as a function of radius does 
not change, that is, provided dBJdz increases in proportion to B z . 

Since the variables x, z are separated in 'T' and 'V, we can immedi- 
ately write down the betatron oscillation frequencies. It is convenient 
to express them in terms of the numbers of betatron oscillations per 
revolution, v x and v z : 

Wx /-, \l/2 

v x = — = (1 — n) , 

w (12-133) 

w z 1/2 

If there are imperfections in the accelerator, so that B z is not inde- 
pendent of <p, the difference between B z and its average value gives rise 
to a periodic force acting on the coordinate x. The resulting perturbation 
of the orbit can be treated by solving the corresponding forced harmonic 
oscillator equations. If B p is not zero everywhere in the median plane 
(2 = 0), vertical forces act which drive the vertical betatron oscillations. 
In general, such imperfections also lead to variations in the field index n, 
so that n = n(<p), and n for the steady motion becomes a periodic func- 
tion of time. In alternating gradient accelerators, the field index n is 
deliberately made to vary periodically in azimuth <p. The solution of 
this problem is too complex for inclusion here. 

12-8 Stability of Lagrange's three bodies. A particular solution of 
the problem of three bodies moving under their mutual gravitational 
attractions was discovered by Lagrange. This solution is a steady motion 
in which the three masses remain at the corners of an equilateral triangle 
as they revolve around their common center of mass. We wish to in- 
vestigate the stability of this steady motion. This problem is an example 
of a rather general class of problems in celestial mechanics concerned 
with the stability of particular solutions of the equations of motion. 
When the particular solution is a steady motion, the problem can be 
treated by the method of Section 12-6. 

We will simplify the problem by considering only motions confined to 
a single plane. There are then six coordinates, two for each particle. A 
little study shows that there are three ignorable coordinates. Two of them 
represent rigid translations of the three particles, and may be taken as 
the cartesian coordinates of the center of mass. The corresponding con- 
stant momenta are the components of the total linear momentum. The 
third ignorable coordinate will represent a rigid rotation of the three 
particles about the center of mass. The corresponding constant mo- 



12-8] 



STABILITY OF LAGRANGE'S THREE BODIES 



501 



mentum is the total angular momentum. The remaining three non- 
ignorable coordinates will specify the relative positions of the three parti- 
cles with respect to each other. These must be constant in a steady- 
motion; therefore any steady motion must be a rigid translation and 
rotation of the system of three bodies. We may, for example, choose 
the coordinates as in Fig. 12-1. Here, X and Y are coordinates of the 
center of mass, variation of a with the remaining coordinates held fixed 
represents a rotation of the entire system about the center of mass, and 
ri, r 2 , determine the shape and size of the triangle formed by the masses 
mi, m 2 , m 3 . We expect to find three normal modes of vibration of r u r 2 , 6 
about their steady values. If we should happen to overlook any ignorable 
coordinate, we will find only those steady motions in which that co- 
ordinate is constant. The ignorable coordinate will then reveal itself as 
a zero root (p 2 = 0) of the secular equation (12-106). 

According to Eq. (4-127), the kinetic energy will separate into a part 
depending on X and Y, and a part depending on r\, r 2 , 6, and a. The 
potential energy depends only on r u r 2 , and 0. The coordinates X, Y are 
therefore orthogonal to r u r 2 , 6, and a, and the center-of-mass motion 
separates out of the problem. The center-of-mass energy 



T B 



pf (x j + n 



2M 



(Px + Vy), , M = mi + m 2 + m 3 , 

(12-134) 



is constant, and may be omitted from the Hamiltonian. The center of 
mass moves with constant velocity, and we may study separately the 
motion relative to the center of mass. The ignorable coordinate a is 
evidently not orthogonal to 6, since the angular velocity of m x involves 




Fig. 12-1. Coordinates for the three-body problem. 



502 THEORY OF SMALL VIBRATIONS [CHAP. 12 

(6 -\- a) and this appears squared in the kinetic energy. This is not an 
accidental result of our choice of coordinates, but an inherent consequence 
of the fact that rotation of the system as a whole influences the "internal" 
motion described by r\, r 2 , 6. 

It would be a straightforward algebraic exercise to set up the kinetic 
energy in terms of ri, r 2 , 9, a, find the momenta p r iPr2, Pe, Pa, set up the 
Hamiltonian, find the steady motions, and carry through the procedure 
of Section 12-6 to find the secular equation (12-106), which would be a 
third-order equation in p 2 whose roots determine the character of small 
deviations from steady motion. This procedure, however, is extremely 
tedious, as the reader may verify. It turns out, interestingly enough, 
that a less laborious way of finding the actual solution is to abandon the 
Hamilton-Lagrange formalism, and to set up the equations of motion 
from first principles. We will still need the results of the above general 
considerations as a guide to the solution. The algebra is still sufficiently 
involved so that there is a rather high probability of algebraic mistakes. 
It is therefore desirable to replace 0, a by the coordinates «i, a 2 shown in 
Fig. 12-2, so as to introduce an algebraic symmetry between particles 
mi and m 2 . This reduces the amount of algebra needed and provides a 
check on the results, in that our formulas must exhibit the proper sym- 
metry between subscripts "l" and "2". Neither a\ nor a 2 is ignorable now, 
and we see that the ignorable coordinate no longer appears explicitly. 
We could not make use of the ignorable property anyway, since we are 
not going to write the equations in Hamiltonian form. Our secular equa- 
tion will turn out to be of fourth order in p 2 , but we know that one root 
will be p 2 — 0, and can be factored out. We show also in Fig. 12-2 
several auxiliary variables r 3 , 6i, 6 2 , 03 which will be needed. 

We will write the equations of motion of mi in terms of components 
directed radially away from m% and perpendicular to the radius r\. In 
applying Newton's laws of motion directly, we must refer all accelera- 
tions to a coordinate system at rest. The acceleration of Wi is its acceler- 
ation relative to wi 3 plus the acceleration of m 3 . The latter acceleration 
can be found by applying Newton's law of motion to WI3, which is at- 




m 3 
Fig. 12-2. Alternative coordinates for the three-body problem. 



12-8] STABILITY OF LAGRANGE'S THREE BODIES 503 

tracted by m x and m 2 . The force on mi is the gravitational attraction of 
m 2 and m 3 . We have therefore, in the radial direction, 

/.. . 2 , miG . m 2 G . \ m x m 3 G m x m 2 G 
mi I f i — rjai -\ l — + -~ cos 3 J = — cos X . 

(12-135) 
The corresponding equation for the motion of m lt perpendicular to r lt is 

mi La, + 2nd! - ^ sin a ) = - M^ sin X . (12-136) 

Two similar equations can be written for the motion of m 2 . The four 
equations can be conveniently rewritten in the form: 

.2 , (wii + m 3 )(? , m 2 G , m 2 G n 

fi — rial + v ' 1 t- cos 3 H — cos 6i = 0, 

r? r\ r\ 



(12-137) 



.2 , (m 2 + m 3 )G , m-iG „ . mig „ 

r 2 — r 2 a 2 + y ' H 7- cos 3 H — cos 2 = 0, 

r% r\ r% 

, «. . m 2 G . „ . m 2 G . „ n 

nan + 2r i a l t- sin 3 -\ — sin X = 0, 

r 2 r 3 

r 2 a 2 + 2r 2 a 2 H r- sin 3 — sin 2 = 0. 

r\ r 3 

The algebraic symmetry between subscripts "l" and "2" is exhibited in 
the above equations. (Note that 3 = «i — «2 and changes sign if we 
interchange the two particles.) The auxiliary variables X , 2 , 3 , r 3 may 
be expressed in terms of r u r 2 , a u a 2 by using the sine and cosine laws 
for the triangle. 

We now see why it is easier to use Newton's laws directly here. We 
can express the acceleration of m 3 very simply in terms of the gravitational 
forces on m 3 . In the Lagrangian formulation of the corresponding equa- 
tions for r lt r 2 , a lt a 2 (or a, 0), the terms which represent the acceleration 
of m 3 have to be expressed kinematically, i.e., in terms of the coordinates, 
velocities, and accelerations of m x and m 2 , because they are derived by 
differentiation of the kinetic energy T in generalized coordinates. This is 
very complicated, and involves explicitly the position of the center of 
mass relative to m lt m 2 , m 3 , which we do not need in the present formula- 
tion. The resulting equations are equivalent to Eqs. (12-137), but con- 
siderably more complicated in form. One reason for the simplicity of 
Eqs. (12-137) is, of course, the use of the auxiliary variables 6 U 2 , 3 , r 3 ; 



504 THEORY OF SMALL VIBRATIONS [CHAP. 12 

in the Lagrangian formulation, any such auxiliary variables would have 
to be differentiated with respect to r lt r 2 , a lt a 2 in order to write down 
the equations of motion. 

We first look for steady motions. We know from our preliminary dis- 
cussion that a steady motion can only be a rigid rotation about the center 
of mass (plus a uniform translation). We therefore take r\, r 2 , r 3 , 0i, 2 , 03 
to be constant, and set 

a 2 = at, «i = ot + 63- (12-138) 

If we substitute into the last of Eqs. (12-137), we obtain 

\ sin 3 = A sin 02- (12-139) 



From the law of sines, 

rx r 3 



(12-140) 



sin 2 sin 3 
we then have, unless sin 03 = sin 2 = 0, 

r\ = rl. (12-141) 

In the same way, from the third of Eqs. (12-137), we find r 2 = r 3 , unless 
sin 3 = sin 0i = 0. We may therefore set 

r\ = r 2 = r 3 = a, 

- =, - * (12 " 142) 

The only possible steady motion, unless the masses lie in a straight line, 
is one in which the three masses lie at the corners of an equilateral triangle. 
We must still verify that the first two of Eqs. (12-137) are satisfied. 
This is the case if 

a, 2 = ^> (12-143) 

a 3 

where M is the total mass. This is the particular solution of the three- 
body problem found by Lagrange. The case when the three masses lie 
in a straight line is left as an exercise. 

We now seek solutions for motions near the steady motion. Let us set 

r x = a + xi, r 2 = a + x 2 , (12-144) 

«1 = Ut + ^7T + 6l, ol 2 = ut + e 2 , 

where xi, x 2 , e x , e 2 are four new independent variables which we will 



12-8] STABILITY OF LAGRANGE'S THREE BODIES 505 

regard as small. We substitute in Eqs. (12-137), retaining only linear 
terms. We first calculate 

03 = &r + «i - «2, (12-145) 

and, to first order, from the law of cosines, 

r % = a 2 [l + ^^1 + V3 (ei - «„)] • (12-146) 

Now, from the law of sines, 



sin 0i = ^sin0 3 ; (12-147) 

f3 
* = I - \ (€l - €2) - l^ 3 ^ 1 ' ( 12 " M8 > 



Xi 


— 


X2 




a 




X! 


— 


X 2 



hence, to first order, 

and, similarly, 

e 2 = § - \ (6 X - e 2 ) +\ V3^^- (12-149) 

We note as a check that X + 2 + 03 = ir. We are ready to substitute 
in Eqs. (12-137), which, to first order, become 

„ . ( 2 , 2mi + 2m 3 — jm 2 r \ 
xi — 2owei — I w H ^3 O- J xi 

9m 2 G 3\/3 m 2 G 



4a3> ** 4a 2 



(«i - « a ) = 0, 



o • ^ 2 ■ 2m 2 + 2rre 3 — jmi „\ 
x 2 — 2ao>e 2 — I a> H ^ ^ I z 2 



9miC? 3\/3 m x (? , . „ 

- -4^~ Xl 4^~ ^ - e »> = °' 

.. , „ . 3\/3 m 2 G , s 9m 2 G , , _ n 

atx + 2wi;i 4q3 (zi - s 2 ) ^j" («i — «2) = 0, 

oe 2 + 2wi: 2 4q3 (xi - ^2) + -^p - (d — «2) - 0. 

(12-150) 

Note that the second and fourth equations may be obtained from the first 
and third by interchanging the subscripts "l" and "2" and reversing the 
signs of «!, a 2 , and o>; this symmetry follows from the choice of coordi- 
nates in Fig. 12-2. 

A normal mode is to found by setting 



x x = X x e pt , x 2 = X 2 e pt , 
ci = Eie pt , e 2 = E 2 e 



,* . - *.** ( 12 " 151 > 



506 THEORY OF SMALL VIBRATIONS [CHAP. 12 

For convenience we set 



p = (G/a d ) l "P. (12-152) 

Substituting Eqs. (12-151) in Eqs. (12-150) and using Eq. (12-143), 
we obtain 

(t>2 oar , 9 \Xi 9 ^2 

- (^3. m2 + 2M 1/2 p) E,+^ m 2 E 2 = 0, 

9 Xi , ( d2 ,,t , 9 \ X 2 3\/3 _ 
-4 mi ^ + ^ P -3M + i m 1 ) — -- r m 1 E 1 

+ (^m 1 -2M 1 ' 2 P^E 2 =0, 

( 2M ,« P _3Vi m ,)£ + 3vi m ,a 

+ (p 2 ~1 ^2) E x + ? m2 E 2 = 0, 
3\/3 Xi , /, u i/s D , 3\/3 \X 2 

- -r Wl v + \ 2M p + — mi ) t 

+ \ rn 1 E 1 + (p 2 - | m,j E 2 = 0. 

(12-153) 
The secular determinant is 

(pi-nf + lm,) (-?»,) _( 2 M^P + ^ m2 ) (2^1 OT2 ) 

( 2M -p_3V| mj ) (3vf m2 ) ( p2 _| m2 ) (»„,) 

-(?£..,) (»*"'* + ?£.,) (I-,) (P--|«,) 

(12-154) 

We know from previous considerations that this must be a fourth-degree 
equation in P 2 , one root of which is P 2 = 0. This fact, together with the 
symmetries in the array of coefficients, encourages us to try to manipulate 
the above determinant to simplify its expansion and to bring out ex- 
plicitly the factor P 2 . We add the second column to the first, and the 
third to the fourth, then subtract the first row from the second, and the 
third from the fourth : 



= 0. 



12-8] 



STABILITY OF LAGRANGE S THREE BODIES 



507 



(P 2 -3M) (-i m ») -(21f" ! P + ^m ! ) -W U2 P) 

[p 2 - 3M + | (m, + m 2 )l Um 1I2 P -^(m,- m 2 )] 

[2M" 2 P + ^(«n- m,)] - [p 2 - j (m, + m,)] 



= 0. 



(12-155) 



We factor P from the last column, then multiply the last column by 
3M 1/2 /2, and subtract from the first column. We can then factor P from 
the first column, to obtain 



p (-\ m *) ~ (™ ll2 P + ^ m.) -(2M 1 ' 2 ) 

[p 2 - 3M + | (», + m,)l [~2M 1/2 P - ^0- (m, - m a )] 

(1m-) (•£„..) (p 2 -^ 2 ) 

[2Af " 2 P + ^(»,- m a )] - [p 2 - j (mi + m,)] 

(12-156) 

The factor P 2 is now in evidence. To simplify the expansion of the de- 
terminant, we multiply the third row by 2PM~ 112 and subtract from 
the first row: 



= 0. 



* * -(2M 112 + 2P 2 M-" 2 ) 

[p 2 - ZM + I (m, + m„)] [w lla P - ^ (m, - m a )] 



i^ 1 



[2M" 2 P + ^ («., - m,)l - [p 2 - I (m, + m,)] 



= 0. 



(12-157) 



The stars indicate terms that we do not write here, since they will not 
appear in the result. We can now expand in minors of the first column, 
and expand the resulting three-rowed determinant in minors of the last 
column, with the final result: 

P 2 (M + P 2 )^ + MP 2 + ^•(wiim 2 + m 2 m 3 + m a mi)] = 0. 

(12-158) 

Note, as a check on the algebra, that the three masses enter symmetrically 



508 THEORY OF SMALL VIBRATIONS [CHAP. 12 

in this equation, as they must. It is a fortunate accident that an addi- 
tional factor appears explicitly, so that we have only to solve a quadratic 
equation for P 2 . We have, finally, four roots: 

p2 _ n P 2 = — M 

^ ", r m, (12-159) 

p 2 = —\M ± \\M 2 — 27(m 1 m 2 + m 2 m 3 + m 3 mi)] 1/2 . 

As we know, the zero root results from the fact that there is an addi- 
tional ignorable coordinate. The root P 2 = —M yields a stable oscilla- 
tory mode. The last two roots for P 2 will both be real and negative 
provided that 

(mi + m 2 + m 3 ) 2 > 27(raim 2 + m 2 m 3 + m 3 m{). (12-160) 

If this inequality is reversed, the last two roots are complex, and we have 
four complex values of P. Of these roots two give rise to damped and two 
to antidamped oscillatory solutions. In the intermediate case when the 
two members of the inequality (12-160) are equal, it can be shown 
that the amplitude of oscillation grows linearly in time. Hence the La- 
grangian motion of three bodies is unstable when condition (12-160) is 
not satisfied. If one of the bodies, say mi, is much smaller than the other 
two, we have the restricted problem of three bodies studied in Section 
7-6, and the condition for stability reduces to 

(m 2 + m 3 ) 2 > 27m 2 m 3 , (12-161) 

or, if m 3 is the largest, 

m 3 > 24.96m 2 . (12-162) 

If m 3 is the sun and m 2 is the planet Jupiter, this condition is satisfied; 
hence, if we neglect the effects of all other planets, there are stable steady 
motions in which a small body revolves around the sun with the same 
period as Jupiter and at the corner of an equilateral triangle relative to 
the sun and Jupiter. (There are two such positions.) The Trojan aster- 
oids are a group of bodies with the same period as Jupiter which appear 
to be in this position. Since Eq. (12-161) is also satisfied by the earth- 
moon system, the corresponding steady motion of an artificial satellite 
in the earth-moon system is stable. Consideration of motions perpendic- 
ular to the plane of steady motion does not alter these conclusions. How- 
ever, in view of the remarks at the end of Section 12-6, our conclusions 
about the stability are valid only for limited periods of time. 

We leave as an exercise the solution of Eqs. (12-153), to determine 
the ratios of the variables and hence the oscillation pattern for each 
normal mode (see Problem 30). In solving Eqs. (12-153), it may be 



PROBLEMS 509 

helpful to subject them to the same series of manipulations which led 
from the determinant (12-154) to the determinant (12-157). Note that 
adding one row of the determinant to another corresponds to adding the 
corresponding equations. Adding two columns corresponds to grouping 
the corresponding variables, that is, to introducing a new variable which 
is a linear combination of the two original ones. 

Problems 

1. Find the transformation to normal coordinates for the two coupled oscil- 
lators shown in Fig. 4-10. 

2. Solve Problem 26, Chapter 4, by transforming from x\, X2 to normal co- 
ordinates by the method of Section 12-3. 

3. A mass m moving in space is subject to a force whose potential energy is 

V = Vo exp [(5x 2 + 5y 2 + 8z 2 - 8yz - 2Qya - 8za)/a 2 ], 

where the constants Vo and a are positive. Show that V has one minimum 
point. Find the normal frequencies of vibration about the minimum. 

4. A mass m is hung from a fixed support by a spring of constant k whose 
relaxed length is I = 2mg/k. A second equal mass is hung from the first mass 
by an identical spring. Find the six normal coordinates and the corresponding 
frequencies for small vibrations of this system from its equilibrium position. 
Each spring exerts a force only along the line joining its two ends, but may 
pivot freely in any direction at its ends. 

5. An ion of mass m, charge q, is held by a linear attractive force F = — kr 
to a point A, where r is the distance from the ion to the point A. An identical 
ion is similarly bound to a second point B a distance I from A. The two ions 
move (in three-dimensional space) under the action of these forces and their 
mutual electrostatic repulsion. Find the normal modes of vibration, and write 
down the most general solution for small vibrations about the equilibrium point. 

6. The mass W2 in Fig. 4-10 is subject to a force F2 = B sin ut. The sys- 
tem is at rest at t = 0. Find the motion by the method of normal coordinates, 
using the result of Problem 1 . 

7. The pair of ions in Problem 5 is subject to a plane polarized electromagnetic 
wave incident perpendicular to the line AB, whose electric field Eo cos cot is 
directed at 45° to the line IB. Find the steady-state motion. 

8. The mass in Problem 3 is subject to a force 

F x = F y = F s = Be'"'. 

Find a particular solution. 

9. Set up the tensors M, B, K for Problem 25, Chapter 4, show that all 
three can be simultaneously diagonalized, and solve the problem by the method 
of normal coordinates. 

10. The masses wii and rti2 in Fig. 4-10 are subject to frictional forces 
— Tmixi, — TTO2X2, respectively. Find the general solution. 



510 THEORY OF SMALL VIBRATIONS [CHAP. 12 

11. Assume that V°(xi, . . . , x/) has a minimum at x\ = • • • = x; = 0, and 
that V'(xi, . . . ,Xf) is small, but that Eq. (12-66) does not necessarily hold. 
Find approximate expressions to first order in V and its derivatives at x\ = • • • 
= Xf = 0, for the coordinates x\, . . . , x° f of the new equilibrium point for 
V = V° +• V. If the expansion of V° about x\ = • • • = x/ = is given by 
Eq. (12-67) (plus higher-order terms), and if the quadratic terms in the expan- 
sion of V are to be 

V = ]T i(K° kl + K' kl )y k y h 

k,l 

where y k = x k — x° k , find approximate first order expressions for the coeffi- 
cients K' kl . 

12. Find the second-order approximations for the coefficients Cj, C[, which 
are given to first order by Eqs. (12-80) and (12-83). 

13. Find the third-order approximation to the frequency correction given to 
second order by Eq. (12-91). 

14. Formulate the equations to be solved to obtain a first-order approxima- 
tion in the case when conditions (12-87) fail for a group of four nearby modes, 
i.e. the case of approximate degeneracy. 

15. A triple pendulum is formed by suspending a mass M by a string of 
length I from a fixed support. A mass m is hung from M by a string of length I, 
and from this second mass a third mass m is hung by a third string of length I. 
The masses swing in a single vertical plane. Set up the equations for small 
vibrations of the system, using as coordinates the angles 6%, 02, 03 made by each 
string with the vertical. Show that if M )$$> m, the normal coordinates can be 
found if terms of order (m/M) 112 are neglected. Find the approximate normal 
frequencies to order m/M. [Hint: Transform K to a constant tensor, and diag- 
onalize M.] 

16. In Fig. 12-3, the four masses move only along a horizontal straight line 
under the action of four identical springs of constant k, and a weak spring of 
constant k' <<C k. Find, to first order in k', an approximate solution for the 
normal modes of vibration. 

17. Find the approximate solution to Problem 16 for the case when the 
masses are all equal. Does the approximate result suggest a way to solve the 
problem exactly? 

18. A uniform ellipsoid of revolution of mass M, whose axis of symmetry is 
two thirds as long as its equatorial diameter, is modified by placing masses m, 
2m, 3m, m, 2m, 3m in sequence around its equator at points 60° apart. Two 
masses, each 4m, are placed at opposite ends of a diameter, making an angle of 
45° with the axis and at the longitude of the masses m. If m <£. M, find, to 
first order in m/M, the new principal axes. [Hint: The perturbation procedure 



k m 1 k m2 k' m^ k m 4 k 
Fig. 12-3. Four coupled harmonic oscillators. 



PROBLEMS 511 

developed in Section 12-5 for diagonalizing the tensor W may be applied to 
diagonalize approximately any symmetric tensor 1° + I' if the eigenvectors of 1° 
are known and I' is small.] 

19. Apply the perturbation method to the problem of the string with variable 
density considered in the last paragraph of Section 9-9, assuming that a <JC <ro- 
Find the lowest normal frequency to second order in o, and write out the cor- 
responding solution u(x, t) to first order in a. 

20. Formulate a first-order perturbation method of solving Eqs. (12-60), 
treating the friction as a small perturbation, and assuming the solution without 
friction is already known. Show why, even in first order, one cannot introduce 
normal coordinates which include the effects of friction. 

21. Two charges -\-Ze are. located at fixed points a distance 2a apart. An 
electron of mass m, charge — e, moves in the field of these charges. Find the 
steady motions and the small vibrations about the steady motions. 

*22. Two charges -{-Ze and — Ze are located at the fixed points z = a and 
z = —a. An electron of mass m, charge — e, moves in the field of these charges. 
Sketch a graph of z vs. r, where r is the distance from the z-axis, showing the 
values of z, r for which there are steady motions. Investigate the stability of 
these steady motions. 

23. A mass m slides without friction on a smooth horizontal table. It is tied 
to a weightless string of total length I which passes through a hole in the table 
and is tied at its lower end to a mass M which hangs below the table. Set up 
the Hamiltonian function using as coordinates the polar coordinates r, a of the 
mass m relative to the hole, and the spherical angles 9, <p of the mass M relative 
to the hole. Find the steady motions and the normal frequencies of small vibra- 
tions about a steady motion. 

24. Assume that the three masses in Fig. 4-16 are free to move in a plane but 
are constrained to remain in a straight line relative to one another. Choose 
your coordinates so that as many as possible will be ignorable, find the steady 
motions, and find the normal modes of vibration about them. 

25. A symmetrical rigid body is mounted in weightless, frictionless gimbal 
rings. A hairspring is attached to one of the rings so as to exert a restoring 
torque — h}> about the z-axis, where <t> is the Euler angle. Find the steady mo- 
tions and investigate the character of small vibrations about them. 

26. In Problem 13, Chapter 11, a hairspring is connected between the disk 
axle and the rings which exerts a restoring torque — b//', where \[>' is the rela- 
tive angle of rotation between disk and rings. The "gyroscope" moves freely 
in space with no external forces. Find the steady motions and investigate the 
small vibrations about them. 

27. Two masses m are connected by a rigid weightless rod of length 21. One 
mass is connected with the origin by a spring of constant k, the other by a 
spring of constant 2k. The relaxed length of both springs is zero. The masses 
move in a single plane. Choose as coordinates the polar coordinates r, 6 of the 
center of mass relative to the origin, and the angle a which the rod makes with 
the radius from the origin to the center of mass, taking a = when the stronger 
spring is stretched least. Find the steady motions and the conditions under 
which they are stable. 



512 THEORY OF SMALL VIBRATIONS [CHAP. 12 

*28. In Problem 23, a second mass m slides without friction on the table, and 
is connected to the first mass by a rigid weightless rod of length a. (Assume 
the arrangement is ingeniously contrived so that the rod and string do not 
become entangled.) Assume also that M = 2m. Use as an additional coordi- 
nate the angle /? between the rod and the string. Find the steady motions, and 
determine which are stable. Find the normal vibrations about the stable steady 
motions. 

29. The vector potential due to a magnetic dipole of magnetic moment n is, 
in spherical coordinates relative to the dipole axis, 

H sin 6 
A = ffl. 

2*r 2 

Find the steady motions for a charged particle moving in such a field, and 
show that they are unstable. 

30. Find the solution of Eqs. (12-153) for Xi, X 2 , Ei, E 2 , when P 2 = —M, 
and describe the corresponding oscillation. [See the hint in the last paragraph 
of Section 12-8.] 

*31. Analyze the case which was omitted in Section 12-8 when the three 
bodies mi, m 2 , mz lie in a straight line. Show that there are three possible 
steady motions, one for each mass lying between the other two. [Hint: You 
will need Descartes' rule of signs.] Show that motions near each of these steady 
motions are unstable. Compare your results with those of Section 7-6 and 
Problem 17 of Chapter 7. 

32. Find the solution of Eqs. (12-153) for the double root P 2 = —ffl, 
when the inequality (12-160) becomes an equality. Show that in this case, 
Eqs. (12-150) have a second solution in which Xi, X 2 , E\, E 2 are certain linear 
functions of t, say Xi = X{ + Xi't, etc. (You can simplify the algebra a little 
by assuming that one of the additive constants, say X[, is zero. This is allow- 
able, since X{ can always be made zero by subtracting from the second solution 
a suitable multiple of the first solution you found in which Xi is constant. 
The linearity of the equations permits linear superposition of solutions.) 

33. Find the solution of Eqs. (12-153) for X h X 2 , E lt E 2 , for. the root P 2 = 0, 
and show that it corresponds to a new steady motion near the chosen one. 
Since your solution has only one arbitrary constant, there must be another 
solution of Eqs. (12-150) corresponding to P 2 = 0. Guess its form, and verify 
by substitution. 

*34. Show that if motions of Lagrange's three bodies out of the plane of the 
steady motion are considered, at least one of the three additional coordinates 
is ignorable. Choose as two nonignorable coordinates the distances qi = z\ — zz 
and q 2 = z 2 — 23, where z% is the perpendicular distance of to,- from the plane 
of steady motion. Set up the linearized equations of motion by the method 
used in Section 12-8. Solve for the corresponding normal vibrations and show 
that the result can be interpreted as corresponding simply to a small change in 
the orientation of the plane of steady motion. 



BIBLIOGRAPHY 



515 



BIBLIOGRAPHY 



The following is a list, by no means complete, of books related to the 
subject matter of this text which the reader may find helpful. 

Elementary Mechanics Texts 

1. Campbell, J. W., An Introduction to Mechanics. New York: Pitman, 
1947. 

2. Millikan, R. A., Roller, D., and Watson, E. C, Mechanics, Molecular 
Physics, Heat, and Sound. Boston: Ginn and Co., 1937. 

Intermediate Mechanics Texts 

3. Beckeb, R. A., Introduction to Theoretical Mechanics. New York: 
McGraw-Hill, 1954. 

4. Lindsay, Robert Bruce, Physical Mechanics, 2nd ed. New York: 
D. Van Nostrand, 1950. 

5. MacMillan, William D., Theoretical Mechanics. New York: McGraw- 
Hill. Vol. 1: Statics and Dynamics of a Particle, 1927. Vol. 3: Dynamics of Rigid 
Bodies, 1936. 

6. Osgood, William F., Mechanics. New York: Macmillan Co., 1937. 

7. Scott, Merit, Mechanics, Statics and Dynamics. New York: McGraw- 
Hill, 1949. 

8. Stephenson, Reginald J., Mechanics and Properties of Matter. New 
York: John Wiley & Sons, 1952. 

9. Synge, John L., and Griffith, Byron A., Principles of Mechanics, 3rd 
ed. New York: McGraw-Hill, 1959. 

Advanced Mechanics Texts 

10. Corbin, H. C, and Stehle, Philip, Classical Mechanics. New York: 
John Wiley & Sons, 1950. 

11. Goldstein, Herbert, Classical Mechanics. Reading, Mass.: Addison- 
Wesley, 1950. 

12. Lamb, Horace, Hydrodynamics, 6th ed. Cambridge: Cambridge Uni- 
versity Press, 1932. (New York: Dover Publications, 1945.) 

13. Landau, L. D., and Lifshitz, E. M., Mechanics. London: Pergamon 
Press, 1960. (Reading, Mass.: Addison- Wesley, 1960.) 

14. Landau, L. D., and Lifshitz, E. M., Fluid Mechanics. London: Pergamon 
Press, 1959. (Reading, Mass.: Addison- Wesley, 1959.) 

15. Landau, L. D., and Lifshitz, E. M., Theory of Elasticity. London: 
Pergamon Press, 1959. (Reading, Mass.: Addison- Wesley, 1959.) 

16. Lord Rayle