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WAVES
UNIVERSITY MATHEMATICAL TEXTS
GENERAL EDITORS
ALEXANDER C. AITKEN, D.So., F.R.S.
DANIEL E. RUTHERFORD, DR. MATH.
DETERMINANTS AND MATRICES . A. C. Aitken, D.Sc., F.R.S,
STATISTICAL MATHEMATICS . A. C. Aitken, D.Sc., F.R.S.
WAVES C. A. Coulson, M.A., Ph.D.
INTEGRATION . . . . . R. P. Gillespie, Ph.D.
INTEGRATION OF ORDINARY DIFFERENTIAL EQUATIONS
E. L. Inco, D.Sc.
FUNCTIONS OF A COMPLEX VARIABLE
E. G. Phillips, M.A., M.Sc.
VECTOR METHODS . . D. E. Rutherford, Dr. Math.
THEORY OF EQUATIONS . Prof. H. W. Turnbull, F.R.S.
Other volumes in preparation
WAVES
A MATHEMATICAL ACCOUNT OF THE
COMMON TYPES OF WAVE MOTION
BY
C. A. COULSON, M.A., Ph.D.
LECTURER IN MATHEMATICS IN THE UNIVERSITY OF ST ANDREWS,
FORMERLY FELLOW OF TRINITY COLLEGE, CAMBRIDGE
With 29 Figures
OLIVER AND BOYD
EDINBURGH I TWEEDDALE COURT
LONDON : 98 GREAT RUSSELL STREET, W.C.
1941
PBINTED IN GREAT BRITAIN BY
OLIVER AND BOYD LTD., EDINBURGH
PREFACE
THE object of this book is to consider from an elementary
standpoint as many different types of wave motion as
possible. In almost every case the fundamental problem
is the same, since it consists in solving the standard equation
of wave motion ; the various applications differ chiefly
in the conditions imposed upon these solutidn.jj For this
reason it is desirable that the subject of waves should be
treated as one whole, rather than in several distinct parts ;
the present tendency is in this direction.
It is presupposed that the reader is familiar with the
elements of vector analysis, the simpler results of which
are freely quoted. In a sense this present volume may
be regarded as a sequel to Rutherford's Vector Methods,
published in this series^.
In a volume of this size, it is not possible to deal
thoroughly with any one branch of the subject : nor
indeed is this desirable in a book which is intended as
an introduction to the more specialised and elaborate
treatises necessary to the specialist. This book is intended
for University students covering a general course of Applied
Mathematics or Natural Philosophy in the final year of
their honours degree. A few topics, such as elastic waves
in continuous media, or at the common boundary of two
media, and radiation from aerials, have unavoidably had
to be omitted for lack of space. The reader is referred to
any of the standard works on elasticity and wireless for
a discussion of these problems,
vii
viii PREFACE
This book would not be complete without a reference
of gratitude to my friends Dr D. E. Rutherford and
Dr G. S. Rushbrooke, who have read the proofs, checked
most of the examples and contributed in no small way
to the clarity of my arguments. My thanks are also
offered to my wife for her share in the preparation of
the manuscript.
C. A. 0.
January 1941.
CONTENTS
CHAPTER I
THE EQUATION OF WAVE MOTION
PAGE
Introductory 1
General Form of Progressive Waves ..... 1
Harmonic Waves 2
Plane Waves 4
The Equation of Wave Motion 5
Principle of Superposition ....... 5
Special Types of Solution 6
List of Solutions ........ 13
Equation of Telegraphy 15
Exponential Form of Harmonic Waves . . . .16
A Solved Example 17
Examples 19
CHAPTER n
WAVES ON STRINGS
The Differential Equation 21
Kinetic and Potential Energies ...... 23
Inclusion of Initial Conditions ...... 25
Reflection at a Change of Density 25
Reflection at a Concentrated Load 28
Alternative Solutions 29
Strings of Finite Length, Normal Modes .... 30
String Plucked at its Midpoint , . . . , .31
The Energies of the Normal Modes 33
Normal Coordinates ....... 36
String with Load at its Midpoint . . . . .37
Damped Vibrations ........ 39
Method of Reduction to a Steady Wave .... 40
Examples 41
x CONTENTS
CHAPTER III
WAVES IN MEMBRANES
PAGE
The Differential Equation ....... 43
Solution for a Rectangular Membrane .... 44
Normal Coordinates for Rectangular Membrane ... 47
Circular Membrane ........ 48
Examples ......... 50
CHAPTER IV
LONGITUDINAL WAVES IN BARS AND SPRINGS
Differential Equation for Waves along a Bar . . .51
Free Vibrations of a Finite Bar ...... 53
Vibrations of a Clamped Bar . . . . . .53
Normal Coordinates ........ 53
Case of a Bar in a state of Tension ..... 54
Vibrations of a Loaded Spring ...... 55
Examples . . . . . . . . .59
CHAPTER v
WAVES IN LIQUIDS
Summary of Hydrodynamical Formulae .... 60
Tidal Waves and Surface Waves ..... 62
Tidal Waves, General Conditions 63
Tidal Waves in a Straight Channel ..... 64
Tidal Waves on Lakes and Tanks 67
Tidal Waves on Rectangular and Circular Tanks ... 70
Paths of the Particles 70
Method of Reduction to a Steady Wave . . . .71
Surface Waves, the Velocity Potential . . . .72
Surface Waves on a Long Rectangular Tank ... 74
Surface Waves in Two Dimensions . . . . .75
Paths of the Particles 77
The Kinetic and Potential Energies . * . . .78
CONTENTS xi
PAGE
Rate of Transmission of Energy ..... 79
Inclusion of Surface Tension, General Formulae . . .81
Capillary Waves in One Dimension ..... 83
Examples ......... 84
OHAPTEB VI
SOUND WAVES
Relation between Pressure and Density .... 87
Differential Equation in Terms of Condensation ... 87
Boundary Conditions ....... 90
Solutions for a Pipe of Finite Length ..... 90
Normal Modes in a Tube with Movable Boundary . . 91
The Velocity Potential, General Formulae .... 92
The Differential Equation of Wave Motion .... 93
Stationary Waves in a Tube of Finite Length ... 95
Spherical Symmetry ........ 95
The Kinetic and Potential Energies ..... 96
Progressive Waves in a Tube of Varying Section ... 97
Examples 100
CHAPTEK VII
ELECTRIC WAVES
Maxwell's Equations . . . . . . . .102
Nonconducting Media, the Equation of Wave Motion . .105
Electric and Magnetic Potentials . . . . .106
Plane Polarised Waves in a Dielectric Medium . . .109
Rate of Transmission of Energy in Plane Waves . . .111
Reflection and Refraction of Light Waves . . . .113
Internal Reflection . . . . . . . .118
Partially Conducting Media, Plane Waves . . . .119
Reflection from a Metal 122
Radiation Pressure . .124
Skin Effect 125
Examples ......... 125
xii CONTENTS
CHAPTER VIII
GENERAL CONSIDERATIONS
PAGE
Doppler Effect 128
Beats 130
Amplitude Modulation . . . . . . .132
Group Velocity 132
Motion of Wave Packets 135
Kirchhoff's Solution of the Equation of Wave Motion . . 138
Fresnel's Principle ........ 142
Fraunhofer Diffraction Theory . . . . . .145
Retarded Potential Theory 149
Examples ......... 151
Index 154
CHAPTER I
THE EQUATION OF WAVE MOTION
1. We are all familiar with the idea of a wave ; thus,
when a pebble is dropped into a pond, water waves travel
radially outwards ; when a piano is played, the wires
vibrate and sound waves spread through the room ; when
a wireless station is transmitting, electric waves move
through the ether. These are all examples of wave motion,
and they have two important properties in common :
firstly, energy is propagated to distant points ; and
secondly, the disturbance travels through the medium
without giving the medium as a whole any permanent
displacement. Thus the ripples spread outwards over a
pond carrying energy with them, but as we can see by
watching the motion of a small floating body, the water
of the pond itself does not move with the waves. In the
following chapters we shall find that whatever the nature
of the medium which transmits the waves, whether it be
air, a stretched string, a liquid, an electric cable or the
ether, these two properties which are common to all these
types of wave motion, will enable us to relate them
together. They are all governed by a certain differential
equation, the Equation of Wave Motion (see 5), and
the mathematical part of each separate problem merely
consists in solving this equation with the right boundary
conditions, and then interpreting the solution appropriately.
2. Consider a disturbance </> which is propagated
along the x axis with velocity c. There is no need to
l A
2 WAVES
state explicitly what < refers to ; it may be the elevation
of a water wave or the magnitude of a fluctuating electric
field. Then, since the disturbance is moving, <f>.will depend
on x and t. When t = 0, <f> will be some function of x
which we may call f(x). f(x) is the wave profile, since
if we plot the disturbance cf> against x, and " photograph "
the wave at t 0, the curve obtained will be < ~f(x).
If we suppose that the wave is propagated without change
of shape, then a photograph taken at a later time t will
be identical with that at t = 0, except that the wave
profile has moved a distance ct in the positive direction
of the x axis. If we took a new origin at the point x ct,
and let distances measured from this origin be called X,
so that x = X\ct, then the equation of the wave profile
referred to this new origin would be
Referred to the original fixed origin, this means that
<f>=f(xct) . . . . (1)
This equation is the most general expression of a wave
moving with constant velocity c and without change of
shape, along the positive direction of x. If the wave is
travelling in the negative direction its form is given by
(1) with the sign of c changed, i.e.
(2)
3. The simplest example of a wave of this kind is the
harmonic wave, in which the wave profile is a sine or
cosine curve. Thus if the wave profile at t = is
(<f>)t=o = # cos mx,
then at time t y the displacement, or disturbance, is
= a cos m(xct) ... (3)
The maximum value of the disturbance, viz. a, is called
the amplitude. The wave profile repeats itself at regular
THE EQUATION OF WAVE MOTION 3
distances 27r/m. This is known as the wavelength A.
Equation (3) could therefore be written
(f) a cos ~(x~ ct) .... (4)
A
The time taken for one complete wave to pass any point
is called the period r of the wave. It follows from (4) that
2rr
^(xct) must pass through a complete cycle of values
A
as t is increased by r. Thus
27TCT
T = 277 '
i.e. r = A/c . . . (5)
The frequency n of the wave is the number of waves
passing a fixed observer in unit time. Clearly
n=l/r . . . . (6)
so that c = n\, .... (7)
and equation (4) may be written in either of the equivalent
forms,
</) == a cos27r{^ J (8)
eft = a cos 27rl^ nt\ ... (9)
Sometimes it is useful to introduce the wave number k,
which is the number of waves in unit distance. Then
*=1/A, .  (10)
and we may write equation (9)
(f) = a cos 27r(kxnt) . . . (11)
4 WAVES
If wo compare two similar waves
<^ a cos 27r(kxnt),
we see that </>% is the same as <^ except that it is displaced
a distance e/27r&, i.e. eA/277. is called the phase of <f> 2
relative to <f> v If e = 27r, 4?!, ... then the displacement
is exactly one, two, ... wavelengths, and we say that the
waves are in phase ; if e = TT, STT, . . . then the two waves
are exactly out of phase.
Even if a wave is not a harmonic wave, but the wave
profile consists of a regularly repeating pattern, the
definitions of wavelength, period, frequency and wave
number still apply, and equations (5), (6), (7) and (10)
are still valid.
4. It is possible to generalise equation (1) to deal
with the case of plane waves in three dimensions. A
plane wave is one in which the disturbance is constant
over all points of a plane drawn perpendicular to the
direction of propagation. Such a plane is called a wave
front, and the wavefront moves perpendicular to itself
with the velocity of propagation c. If the direction of
propagation is x : y : z = I : m : n, where Z, m, n are the
direction cosines of the normal to each wavefront, then the
equation of the wavefronts is
Ix \my\nz = const., . . . (12)
and at any moment t, <f> is to be constant for all x, y, z
satisfying (12). It is clear that
<f)=^f{lx+my+nzct) . . . (13)
is a function which fulfils all these requirements and
therefore represents a plane wave travelling with velocity
c in the direction I : m : n without change of form.
THE EQUATION OF WAVE MOTION 5
5. The expression (13) is a particular solution of the
equation of wave motion referred to on p. 1. Since
I, m, n are direction cosines, Z 2 +m 2 +i& 2 = 1, and it is
easily verified that <f> satisfies the differential equation *
This is the equation of wave motion. j It is one of the
most important differential equations in the whole of
mathematics, since it represents all types of wave motion
in which the velocity is constant. The expressions in
(1), (2), (8), (9), (11) and (13) are all particular solutions
of this equation. We shall find, as we investigate different
types of wave motion ^subsequent chapters, that equation
(14) invariably appears, and it will be our task to select
the solution that is appropriate to our particular problem.
There are certain, types of solution that occur often, and
we shall discuss some of them in the rest of this chapter,
but before doing so, there is one important property of
the fundamental equation that must be explained.
6. The equation of wave motion is linear. That is
to say, (f> and its differential coefficients never occur in
any form other than that of the first degree. Consequently,
if < x and <f> 2 are any two solutions of (14), a 1 <^ 1 +a 2 <^ 2 is
also a solution, a x and 2 being two arbitrary constants.
This is an illustration of the principle of superposition,
which states that, when all the relevant equations are
linear, we may superpose any number of individual
solutions to form new functions which are themselves also
solutions. We shall often have occasion to do this.
A particular instance of this superposition, which is
important in many problems, comes by adding together
* This equation has a close resemblance to Laplace's Equation
which is discussed in Rutherford, Vector Methods, Chapter VII.
t Sometimes called the wave equation, but we do not use this
phrase to avoid confusion with modern wave mechanics.
6 WAVES
two harmonic waves going in different directions with the
same amplitude and velocity. Thus, with two waves
similar to (11) in opposite directions, we obtain
<f) = a cos k 2n(kxnt)\a cos 2tr(Tcx\nt)
= 2a cos 2nkx cos 2irnt (15)
This is known as a stationary wave, to distinguish it from
the earlier progressive waves. It owes its name to the
fact that the wave profile does not move forward. In fact,
<f) always vanishes at the points for which cos 2irkx = 0,
135
viz. x i~7> T7> dbr> These points are called the
4& 4& 4fc
nodes, and the intermediate points, where the amplitude
of c/> (i.e. 2a cos 2irkx) is greatest, are called antinodes.
The distance between successive nodes, or successive
antinodes, is l/2k, which, by (10), is half a wavelength.
Using harmonic wave functions similar to (13), we find
stationary waves in three dimensions, given by
(f) = a cos ~Y (Ix+my+nzct) + a cos (lx\my+nz+ct)
A A
= 2a cos ~r (Ix+my+nz) cos ct . . . (16)
A A
In this case < always vanishes on the planes Ix \my\nz
A 3A
= "> db 9 > an d these are known as nodal planes.
4 4
7. We shall now obtain some special types of solution
of the equation of wave motion ; we shall then be able to
apply them to specific problems in later chapters. We
may divide our solutions into two main types, representing
stationary and progressive waves.
We have already 'dealt with progressive waves in one
dimension. The equation to be solved is
THE EQUATION OF WAVE MOTION 7
Its most general solution may be obtained by a
method due to JD'Alembert. We change to new variables
u = x ct, and v = x+ct. Then it is easily verified that
d<f> d< ty ty d<(> 3<f>
~ transforms to + , TT transforms to c +c
dx du dv dt du dv
o2 J
so that the equation becomes = ; the most general
dudv
solution of this is
/ and g being arbitrary functions. In the original variables
this is
. . . (17)
The harmonic waves of 2 are special cases of this, in
which / and g are cosine functions. The waves / and g
travel with velocity c, in opposite directions.
In two dimensions the equation of wave motion is
c d
and the most general solution involving only plane * waves
is </> = f(to+my ct) +g(lx+my+ct), . (19)
where, as before, / and g are arbitrary functions and
p+m*  1.
In three dimensions the differential equation is
?V , ^.^ = i^ /9m
dx*^~dy*^ dz 2 c* dt* ' ' l " ;
and the most general solution involving only plane waves is
<f) ~f(lx+my+nz~ct)+g(lx+my+nz+ct) . (21)
in which l*+m*+ri* = 1.
* Strictly these should be called line waves, since at any
moment < is constant along the lines Ix } }ny = const.
8 WAVES
There are, however, other solutions of progressive type,
not involving plane waves. For suppose that we transform
(20) to spherical polar coordinates r, 0, $* The equation
of wave motion becomes
o t 9 __ t\ on i "" * on I '
(22)
If we are interested in solutions possessing spherical
symmetry (i.e. independent of 8 and ifj) we shall have to
solve the simpler equation
This may be written
1 8 2
showing (cf. eq. (17)) that it has solutions
^ =*f(rct)+g(r+ct),
f and g again being arbitrary functions. We see, therefore,
that there are progressive type solutions
<f> = f(r~ct) + *g(r+ct) . . (24)
Let us now turn to solutions of stationary type. These
may all be obtained by the method known as the separation
of variables. In one dimension we have to solve
Let us try to find a solution of the form
= X(x)T(t),
* See e.g. Rutherford, p. 62, equation 20.
THE EQUATION OF WAVE MOTION 9
X and T being functions of x and t respectively, whose
form is still to be discovered. Substituting this value of
</> in the differential equation and dividing both sides
by X(x)T(t) we obtain
(25^
X dx* c 2 T dt* ' ' * V ;
The lefthand side is independent of t, being only a function
of x, and the righthand side is independent of x. Since the
two sides are identically equal, this implies that each is
independent both of a: and t, and must therefore be constant.
Putting this constant equal to p 2 , we find
X'+p^X^Q, T"+c 2 p 2 T = Q. . . (26)
These equations give, apart from arbitrary constants
~ cos m cos A ._,_.
X = . px , T = . opt . . (27)
sm^ sin * '
A typical solution therefore is a cos px cos cpt, in which
p is arbitrary. In this expression we could replace either
or both of the cosines by sines, and by the principle of
superposition the complete solution is the sum of any
number of terms of this kind with different values of p.
The constant p 2 which we introduced, is known as
the separation constant. We were able to introduce it in
(25) because the variables x and t had been completely
separated from each other and were in fact on opposite sides
of the equation. There was no reason why the separation
constant should have had a negative value of p 2 except
that this enabled us to obtain harmonic solutions (27).
If we had put each side of (25) equal to \p 2 , the solutions
would have been
X = ev* , T = e c & . . . (28)
and our complete solution should therefore include terms
of both types (27) and (28). The same distinction between
the harmonic and exponential types of solution will occur
frequently.
10 WAVES
This method of separation of variables can be extended
to any number of dimensions. Thus in two dimensions a
typical solution of (18) is
, cos cos cos
d> = . px . ay . ret , . . (29)
^ sin^ sin sm '
in which p 2j rq 2 = f 2 , p and # being allowed arbitrary
values. An alternative version of (29), in which one of
the functions is hyperbolic, is
= COS pxe G S rct. . . (30)
T sm^ sm '
in which p 2 q z r 2 .
It is easy to see that there is a variety of forms similar
to (30) in which one or more of the functions is altered
from a harmonic to a hyperbolic or exponential term.
In three dimensions we have solutions of the same type,
two typical examples being
. COS COS COS COS , 0,0,0
* = sin * sin W Bin " Hm** ' * +<1 +* = *
^ e rz SC , p'f+r* = ^ (32)
sin sin x a
There are two other examples of solution in three
dimensions that we shall discuss. In the first case we put
x = r cos 0, y = r sin 0, and we use r, # and z as cylindrical
coordinates. The equation of wave motion becomes *
A solution can be found of the form
$ = B(r)@(6)Z(z)T(t), . . . (33)
* See Rutherford, p. 63.
THE EQUATION OF WAVE MOTION II
where, by the method of separation of variables, R, @, Z,
T satisfy the equations
_ o.
r dr r*
2 2 02 (VA\
jp  r > n p q . . (M)
The only difficult equation is the first, and this * is just
Bessel's equation of order m, with solutions J m (nr) and
Y m (nr). J m is finite and Y m is infinite when r = 0, so
that we shall usually require only the J m solutions. The
final form of is therefore
, J m/ .cos n cos cos A /0 _
6 = (nr) . m6 . qz . cpt . . (35)
1 Jr TO sin sm sin
If (f> is to be single valued, m must be an integer ; but n,
q and p may be arbitrary provided that n 2 = p*q 2 .
Hyperbolic modifications of (35) are possible, similar in
all respects to (31) and (32).
Our final solution is one in spherical polar coordinates
r, Q, ifj. The equation of wave motion (22) has a solution
sm
* See Ince, Integration of Ordinary Differential Equations, p. 127.
12 WAVES
ra, n and p are arbitrary constants, but if W(ifj) is to be
single valued, ra must be integral. The first two of these
equations present no difficulties. The 0equation is the
generalised Legendre's Equation * with solution
= P n (cos 0),
and if & is to be finite everywhere, n must be a positive
integer. When m = and n is integral, P n m (cos 9)
reduces to a polynomial in cos of degree n y known
as the Legendre's polynomial P n (cos 9). For other
integral values of m, P n m (cos 9) is defined by the equation
A few values of P n (cos 9) and P n m (cos 9) are given
below, for small integral values of n and m. When m>n,
P n m (cos 9) vanishes identically.
P (cos 9)  1
P! (cos 0) = cos
P 2 (cos0) = i (3 cos 2 01)
P 3 (cos 0) = $ (5 cos 3 03 cos 0)
P 4 (cos 0)  (35 cos 4 030 cos 2 0+3)
P^ (cos 0) = sin
Pa 1 (cos 0) = 3 sin cos
P 3 X (cos 0) = I sin (5 cos 2 01)
P 2 2 (cos 0) = 3 sin 2 0.
To solve the JSequation put R(r) = r"" 1 / 2 /S(r), and we
find that the equation for S(r) is just Bessel's equation
Therefore S(r) = e/ n+1/2 (pr) or
* See Ince, Integration of Ordinary Differential Equations, p. 119,
for the case m = 0.
THE EQUATION OF WAVE MOTION 13
Collecting the various terms, the complete solution, apart
from hyperbolic modifications, is seen to be
^ = f 1/8 +1/2 (:pr) p^ (cos 6} mi/1 cpt . (36)
* n 4 1/2 Slli felli
If ( has axial symmetry, we must only take functions
with ra = 0, and if it has spherical symmetry, terms with
m = n = 0. Now ' J^(z) = \/(2/7rz) sin z, and also
Y l!2 (z) = <\/(2ir/z) cos 2;, so that this becomes
A solution finite at the origin is obtained by omitting the
cos pr term.
8. We shall now gather together for future reference
the solutions obtained in the preceding pages,
Progressive waves
1 dimension
<l>=f(xct)+g(x+ct) . . (17)
2 dimensions
d*<f> dty __ 1 3 2 <f>
fa* + %~ 2 ^ ^ M*
<f> =f(lz+myct)+g(lx+my+ct), l 2 +m* ^ 1 . (19)
3 dimensions
(21)
3 dimensions, spherical symmetry
14
WAVES
Stationary waves
1 dimension
2 dimensions
COS COS
sill & x sin
.
(f)
cos cos cos
sin P x sin ^ sin '*
COS
3 dimensions
(27)
(28)
(29)
= r z
(30)
COS COS COS COS
TJ ^ ^ *<*>
COS
sm * J sin sin
cosh
(31)
cos cos
Plane Polar Coordinates (r, 6)
r dr
Cylindrical polar coordinates (r, 9, z)
(32)
COS COS
sin sin cn '
(35a)
i J w COS COS COS 22<>
V ^ Y m (nr > sin m ^ sin ^sin ^ n = p ~q~
and other hyperbolic modifications.
Spherical Polar Coordinates (r, #, iff)
a 2 ^ 2 a<^ i
n+1 / 2
i a 2 <j i a 2 <^>
' ..9 , ' Q /I 0/9 ^2 P)/2
(36)
. COS . COS
M w (cos C7) . mib . (
sin ^ sm
THE EQUATION OF WAVE MOTION 15
Spherical symmetry
d*J> 2 3d) I d 2 (h ' , cos cos
(37)
In solving problems, we shall more often require progressive
type solutions in cases where the variables x, y, z are allowed
an infinite range of values, and stationary type solutions
when their allowed range is finite.
9. There is an important modification of the equation
of wave motion which arises when friction, or some other
dissipative force, produces a damping. The damping effect
is usually allowed for (see e.g. Chapter II and elsewhere)
O I
by a term of the form k 9 which will arise when the
damping force is proportional to the velocity of the vibra
tions. The revised form of the fundamental equation,
known as the equation of telegraphy, is
/\2 i
If we omit the term ~~ this equation is the same as that
occurring in the flow of heat. If we put < ue~ Jct/2 ) we
obtain an equation for u of the form
(39)
Very often k is so small that we may neglect & 2 , and then
(39) is in the standard form which we have discussed in
8, and the solutions given there will apply. In such a
case the presence of the dissipative term is shown by a
decay factor e~ ktl2 . If this is written in the form e~ tft ,
then t (= 2/k) is called the modulus of decay. When
the term in fc 2 may not be neglected, we have to solve
(38) and the method of separation of variables usually
enables a satisfactory solution to be obtained without
much difficulty.
16 WAVES
There is an alternative solution to the equation of
telegraphy that is sometimes useful. Taking the case of
one dimension, and supposing that k is so small that
k 2 may be neglected, we have shown that the solution of
(38) may be written in the form
ct), .... (40)
where / is any function. Since / is arbitrary, we can put
k
f(xct) = e'*!***' g(xct),
and g is now an arbitrary function. Substituting this in
(40) we get
ct) ..... (41)
This expression resembles (40) except that the exponential
factor varies with x instead of with t.
10. Most of the waves with which we shall be
concerned in later chapters will be harmonic. This is
partly because, as we have seen in 8, harmonic functions
arise very naturally when we try to solve the equation of
wave motion ; it is also due to the fact that by means of
a Fourier analysis, any function may be split into harmonic
components, and hence by the principle of superposition,
any wave may be regarded as the resultant of a set of
harmonic waves.
When dealing with progressive waves of harmonic type
there is one simplification that is often useful and which
is especially important in the electromagnetic theory of
light waves. We have seen in (11) that a progressive
harmonic wave in one dimension can be represented by
= a cos 2m(kxnt). If we allow for a phase e, it
will be written <f> = a cos {%rr(kxnt){}. Now this
latter function may be regarded as the real part of the
complex quantity a e*(2flrtf <)+}. It is most convenient
for our subsequent work if we choose the minus sign and
THE EQUATION OF WAVE MOTION 17
also absorb the phase and the amplitude a into one
complex number A. We shall then write
< = A e^^kx) } A = a e~ i . . . (42)
This complex quantity is itself a solution of the equation
of wave motion, as can easily be seen by substitution, and
consequently both its real and imaginary parts are also
solutions. Since all our equations in <j> are linear, it is
possible to use (42) itself as a solution of the equation
of wave motion, instead of its real part. In any equation
in which (f> appears to the first degree, we can, if we wish,
use the function (42) and assume that we always refer to
the real part, or we can just use (42) as it stands, without
reference to its real or imaginary parts. In such a case the
apparent amplitude A is usually complex, and since
A = a e~~ i , we can say that ^4 is the true amplitude,
and arg A is the true phase. The velocity, of course,
as given by (7) and (10), is n/k.
We can extend this representation of <f> to cover waves
travelling in the opposite direction by using in such a case
<f> = A e W*+W ..... (43)
There is obviously no reason why we should not extend
this to two or three dimensions. For instance, in three
dimensions
& A e 27Ti { nt ~( pX+ w +rz V . . . (44)
would represent a harmonic wave with amplitude A
moving with velocity 7&/y'(# 2 +g 2 +r 2 ) in the direction
x : y : z = p : q : r.
11. We shall conclude this chapter with an example.
Let us find a solution of ~ + ~ = ~ ~~ such that <b
dx 2 dy 2 c 2 dt 2 ^
vanishes on the lines # = 0, x = a, y = 0, y = b. Since
18 WAVES
the lines x 0, a, and y 0, b are nodal lines, our
solution must be of the stationary type. Referring to 8,
equation (29), we see that possible solutions are
. COS COS COS A , 0,0 9
= . px . qy . ret , where 0"+<r = r .
r sin sin sm * *
Since < is identically zero at x 0, and y = 0, we shall
have to take the sine rather than the cosine in the first
two factors. Further, since at x = a, </> for all
values of /, therefore
sin pa 0.
Similarly, sin qb = 0.
Hence p = ?/wr/a, and # = ^77/6, m and n being integers.
A solution satisfying all the conditions is therefore
, . nrnx . mni cos
^ = sin smgi ^frt,
where r 2 = 7r 2 (m 2 /a 2 +n 2 /6 2 ) .
The most general solution is the sum of an arbitrary
number of such terms, e.g.
sin  {C mn cos rct\D mn sin re/}. (45)
m,n &
At = 0, this gives
. mry
sm ,
rr r^
o = 27rcD mn 8m sin .
By suitable choice of the constants C mn and D wn we can
make (f> and ^ have any chosen form at t = 0, The value
at any subsequent time is then given by (45).
THE EQUATION OF WAVE MOTION 19
12. Examples
(1) Show that (/> = f(x cos Q\y sin. ct) represents a
wave in two dimensions, the direction of propagation making
an angle 6 with the axis of x.
(2) Show that <f> = a cos (Ix \myct) is a wave in two
dimensions and find its wavelength.
(3) What is the wavelength and velocity of the system of
plane waves <f> = a sin (Ax+By + CzDt) ?
(4) Show that three equivalent harmonic waves with 120
phase between each pair have zero sum.
(5) Show that (/> r~~ l t' 2 cos %0f(rj~ct) is a progressive type
wave in two dimensions, r and 6 being plane polar coordinates,
and / being an arbitrary function. By superposing two of
these waves in which / is a harmonic function, obtain a
stationary wave, and draw its nodal lines. Note that this is
not a singlevalued function unless we put restrictions upon the
allowed range of 6.
(6) By taking the special case of f(x) = g(x) = sin px in
equation (24), show that it reduces to the result of equation
(36) in which m n = 0. Use the relation
(7) Find a solution of ~4 +  = 0, such that < =
dx 2 c 2 dt*
when t = oo, and <j> = when x = 0.
d 2 <I> 1 2 <
(8) Find a solution of ,^  ^ such that < = when
c)x c ot
x +00 or^= +00.
d*z d*z
(9) Solve the equation  = c 2  given that z is never
ot ox
infinite for real values of x and t, and z = when x = 0, or
when t 0.
dV
(10) Solve ~ ' given that F = when t ~ co and
when x = 0, and when x I.
20 WAVES
(11) x, y, z are given in terms of the three quantities
> *?? by th e equations
x = a sinh sin 77 cos
y = a sinh sin 77 sin
z = a cosh cos 77
3V 3V ^V J 0V
Show that the equation  f 7 + ^ =   is of the
dx 2 dy 2 dz 2 c 2 dt 2
correct form for solution by the method of separation of
variables, when , 77, are used as the independent variables.
Write down the subsidiary equations into which the whole
equation breaks down.
12. Show that the equation of telegraphy (38) in one
dimension has solutions of the form
where m and p are constants satisfying the equation
"[ANSWERS : 2. 2rr/(l* + m 2 )* ; 3. A = 27r/(^. 2 + B* f C a )*,
vel. = AD/27r; 7. ^L sin naje~ cni ; 8. Ae~ n(x + ct ^ ; 9. ^ sin ^
sin cp^ ; 10. u4e~^ 8< sin px , p = *r/l , %ir/l 9 ; 11 Show
that ^ = const., 77 = const., const, form an orthogonal
system of coordinates, and transform y 2 ^ in terms of , 17, f
as in Rutherford, Vector Methods, 47. The result is
$ ~ X(g)Y(7))Z( > )T(t), where m, p and 7 are arbitrary con
stants, and
sinh {  X + p 2
sinh ^ cff dg smh 2 f
_ s i n ~  ^4^23^2 Y = g 2 F,
 2
szn 77 dt] dtj sin 2 ?/
CHAPTER II
WAVES ON STRINGS
13. In this chapter we shall discuss the transverse
vibrations of a heavy string of mass p per unit length. By
transverse vibrations we mean vibrations in which the
displacement of each particle of the string is in a direction
perpendicular to the length. When the displacement is
in the same direction as the string, we call the waves
longitudinal ; these waves will be discussed in Chapter IV.
We shall neglect the effect of gravity ; in practice this
may be achieved by supposing that the whole motion takes
place on a smooth horizontal plane.
In order that a wave may travel along the string,
it is necessary that the string should be at least slightly
extensible ; in our calculations, however, we shall assume
that the tension does not change appreciably from its
normal value F. The condition for this (see 14) is that
the wave disturbance is not too large.
Let us consider the motion of a small element of the
string PQ (fig. 1) of length ds. Suppose that in the
equilibrium state the string lies along the axis of x, and
that PQ is originally at P Q Q . Let the displacement of
PQ from the x axis be denoted by y. Then we shall obtain
an equation for the motion of PQ in terms of the tension
and density of the string. The forces acting on this
element, when the string is vibrating, are merely the two
tensions F acting along the tangents at P and Q as shown
in the figure ; let if/ and \fj\difs be the angles made by these
two tangents with the x axis. We can easily write down
21
22
WAVES
the equation of motion of the element PQ in the y direction ;
for the resultant force acting parallel to the y axis is
F sin (\fj+d\ls)F sin $. Neglecting squares of small
quantities, this is F cos ifj di/j. The equation of motion is
therefore
F cos i/j difj pds
(1)
FIG. 1
Now tan
from (1)
= , so that
ooc
= 7~ 2 dx, and so,
ox
. (2)
f /%\ 2 1 ~ l
But cos 2 A ~ 4 1 f 1  J I ,80 that if the displacements
I \&*y J
WAVES ON STRINGS 23
small enough for us to neglect I  1 compared with
unity, we may write (2) in the standard form for wave
motion * (Chapter I, 5), viz.,
are
It follows from Chapter I, equation (17) that the general
solution of this equation may be put in the form
y=f(xct)+g(x+ct), ... (4)
/ and g being arbitrary functions. f(xct) represents a
progressive wave travelling in the positive direction of
the x axis with velocity c, and g(x\ct) represents a
progressive wave with the same velocity in the negative
direction of x. Thus waves of any shape can travel in
either direction with velocity c = \/(F/p), and without
change of form. A more complete discussion, in which
we did not neglect terms of the second order, would show
us that the velocity was not quite independent of the
shape, and indeed, that the wave profile would change
slowly with the time. These corrections are difficult to
apply, and we shall be content with (4), which is, indeed,
an excellent approximation except where there is a sudden
" kink " in y, in which case we cannot neglect I
14. Since the velocity of any point of the string is y,
we can soon determine the kinetic energy of vibration. It is
T = Jj pfdx .... (5)
* The student who is interested in geometry will be able to
prove that the two tensions at P and Q are equivalent to a single
force of magnitude FdsfR, where JR is the radius of curvature of
the string. This force acts perpendicularly to PQ. Putting
E = l + , and neglecting , we obtain (3).
24 WAVES
The potential energy F is found by considering the increase
of length of the element PQ. This element has increased
its length from dx to ds. We have therefore done an
amount of work F(dsdx). Summing for all the elements
of the string, we obtain the formula
V= V(<foefc) = lF
J J
r/%\ 2
= JF I I J dx, approximately. . (6)
The integrations in (5) and (6) are both taken over the
length of the string.
With a progressive wave y =f(xct), these equations
give
T = J lf**(f')*dx = W /(/')*  (7)
(8)
Thus the kinetic and potential energies are equal. The
same result applies to the progressive wave y = g(x\ct) 9
but it does not, in general, apply to the stationary type
waves y f(xct) +g(x+ct).
We can now decide whether our initial assumption is
correct, that the tension remains effectively constant.
If the string is elastic, the change in tension will be pro
portional to the change in length. We have seen in (6)
that the change in length of an element dx is  I ^ I dx.
2 \dxj
dy
Thus, provided that is of the first order of small
ox
quantities, the change of tension is of the second order,
and may safely be neglected. This assumption is equivalent
to asserting that the wave profile does not have any large
" kinks," but has a relatively gradual variation with x.
WAVES ON STRINGS 25
15. The functions / and g of (4) are arbitrary. But
they may be fixed by a knowledge of the initial conditions.
Thus, with a string of unlimited length, such that
^ =0 = </>(#), y t== Q = 0(aj),* we must have, from (4),
Integrating this last equation we have
and so
The displacement at any subsequent time t is therefore
if i r x ~ ct i r x+ct }
= 5 1 t(xct) + </>(x+ct)   $(x)dx +  i/,(x)dx \
z I C J C J J
I f x + ct }
 t(x)dx\. . . (9)
C J xct J
16. The discussion above applies specifically to
strings of infinite length. Before we discuss strings of
finite length, we shall solve two problems of reflection
of waves from a discontinuity in the string. The first is
when two strings of different densities are joined together,
and the second is when a mass is concentrated at a point
of the string. In each case we shall find that an incident
wave gives rise to a reflected and a transmitted wave.
Consider first, then, the case of two semiinfinite
strings 1 and 2 joined at the origin (fig. 2). Let the
* This function \f/(x) must be distinguished from the angle ^ in
13.
26
WAVES
densities of the two strings be p and p 2 . Denote the dis
placements in the two strings by y and y 2 . Let us suppose
that a train of harmonic waves is incident from the negative
direction of x. When these waves meet the change of
wire, they will suffer partial reflection and partial trans
mission. If we choose the exponential functions of 10
to represent each of these waves, we may write
where
= 2/incidcnt + 2/reflccted
== ^/transmitted
?/ . __ A eZTTidl
^/incident ^^
?V * i 7?
//reflected ^l
... . yJ
transmitted ^2
(10)
(ii)
FIG. 2
A 1 is real, but ^4 2 and B l may be complex. According to
10 equation (42), the arguments of A% and S l will give
their phases relative to the incident wave. All three waves
in (11) must have the same frequency n, but since the
velocities in the two wires are different, they will have
different wavelengths 1/& X and l/& 2 . The reflected wave
must, of course, have the same wavelength as the incident
wave. Since the velocities of the two types of wave are
WAVES ON STRINGS 27
n/kL and n/k 2 (Chapter I, equations (7) and (10)), and we
have shown in (3) that c 2 = F/p, therefore
*lW = ft/ft .... (12)
In order to determine A 2 and B l we use what are known
as the boundary conditions. These are the conditions
which must hold at the boundary point x 0. Since the two
strings are continuous, we must have y = y 2 identically
for all values of t, and also the two slopes must be the same,
so that ~ = ~ for all t. If this latter condition were
ox ox
not satisfied, we should have a finite force acting on an
infinitesimal piece of wire at the common point, thus
giving it infinite acceleration. We shall often meet
boundary conditions in other parts of this book ; their
precise form will depend of course upon the particular
problem under discussion. In the present case, the two
boundary conditions give
A 1 +B 1 = A 2 ,
27ri(k l A l ~\k l B l ) = 27ri(k 2 A 2 ).
These equations have a solution
_ *"" 2 *
A J/t I 1/t A J/> l_ T*
"i A^I~ /I/O "^^1 i / i T'* / 2
Since k lt k 2 and A 1 are real, this shows that B 1 and A 2
are both real. A 2 is positive for all k and & 2 , but jBj is
positive if Tc^>Tc^ and negative if k^k^ Thus the
transmitted wave is always in phase with the incident
wave, but the reflected wave is in phase only when the
incident wave is in the denser medium ; otherwise it is
exactly out of phase.
The coefficient of reflection R is defined to be the
k k
ratio \BtlAi\ 9 i.e. 7^ r 2 , which, by (12), we may write
28
WAVES
Similarly, the coefficient of transmission T is equal
to \A 2 fA 1 1 , i.e.
o. /^
. . . . (15)
17. A similar discussion can be given for the case of
a mass M concentrated at a point of the string. Let us
take the equilibrium position of the mass to be the origin
(fig. 3) and suppose that the string is identical on the two
sides. Then if the incident wave comes from the negative
side of the origin, we may write, just as in (10) and (11) :
2/1 = 2/incident + 2/reflected
2/2 == 2/transmitted
where
ttnddent
Reflected
2/transmitted
The boundary conditions are that for all values of t
(i) foiLo = M*=o ....
(16)
(17)
WAVES ON STRINGS 29
The first equation expresses the continuity of the string
and the second is the equation of motion of the mass M .
We can see this as follows : the net force on M is the
difference of the components of F on either side, so that
if ifj^ and if/ 2 are the angles made with the x axis, we have
~ a .
L ~ _U=o
Since ^ and ifj 2 are small, we may put sin */r 2 =
tan i/r 2 = , sin ^ = , and (18) is then obtained.
ox fix
Substituting from (16) into (17) and (18), and cancelling
the term e 27rint , which is common to both sides, we find
= A,
Let us write Trn 2 M/kF = p . . . . (19)
A solution of the equations then gives
A, l+ip l+p* ' '
. . . (21)
A 2 I lip
A l l+ip
In this problem, unlike the last, B t and A z are complex,
so that there are phase changes. These phases (according
to 10) are given by the arguments of (20) and (21).
They are therefore tan 1 (p) and tan~ 1 ( l/p) respectively.
The coefficient of reflection R is IBJA^, which equals
) 1/2 > and the coefficient of transmission T is
i.e. l/(l+p 2 ) 1/2 . If we write p = tan 0, where
then we find that the phase changes are
and 77/2+0, and also R = sin 0, T = cos 6.
18. The two problems in 16, 17 could be solved
quite easily by taking a real form for each of the waves
30 WAVES
instead of the complex forms (11) and (16). The student
is advised to solve these problems in this way, taking, for
example, in 17, the forms
^incident = a l cos %ir(nt kx)
^reflected = b l GOS{2rrr(nt+kx)+}
2/transmiitcd = 2 cos {^(ntJcx) +rj} . (22)
In most cases of progressive waves, however, the complex
form is the easier to handle ; the reason for this is that
exponentials are simpler than harmonic functions, and
also the amplitude and phase are represented by one
complex quantity rather than by two separate terms.
19. So far we have been dealing with strings of
infinite length. When we deal with strings of finite length
it is easier to use stationary type waves instead of progres
sive type. Let us now consider waves on a string of length
Z, fastened at the ends where x = 0, I. We have to find a
d 2 y 1 3 2 y
solution of the equation (3), viz.  , subject to
dx 2 c* dt*
the boundary conditions T/ 0, at x 0, I, for all t.
Now by Chapter I, 8, we see that suitable solutions are
of the type
cos cos
sin ** sin c ^'
It is clear that the cosine term in x will not satisfy the
boundary condition at x = 0, and we may therefore write
the solution
y = sin px (a cos cpt}b sin cpt).
The constants a, b and p are arbitrary, but we have still
to make y = at x = 1. This implies that sin pi = 0,
i.e. pi = TT, 277, STT. . . . It follows that the solution is
. TTTX t TTTCt . T7TCt\
y = sin la cos   1 b sin r , r=^l,2,3, . . . (23)
I \ I IJ
WAVES ON STRINGS 31
Each of the solutions (23) in which r may have any positive
integral value, is known as a normal mode of vibration.
The most general solution is the sum of any number of
terms similar to (23) and may therefore be written
. TTTX\ TTTCt . T7TCt\
y == Z sin \ a r cos + b r sin \ . (24)
r l \ i l }
The values of a r and b r are determined from the initial
conditions ; thus, when t = 0,
y tssQ = S a r sin ' . . . . (25)
r
T7TC . T7TX
y t=s{) = E b r ysm ... (26)
If we are told the initial velocity and shape of the string,
then each a r and b r is found from (25) and (26), and hence
the full solution is obtained. We shall write down the
results for reference. If we suppose that when t = 0,
y = <f)(x) 9 y = *fj(x) 9 then the Fourier analysis represented
by (25) and (26) gives
2 C J ., . TTTX
a r = ~ (f)(x) sm dx
'Jo I
2 f l
b r MX) si
me J o
TTTY
sin ~dx . . (27)
In particular, if the string is released from rest when
t = 0, every b r 0.
20. As an illustration of the theory of the last section,
let us consider the case of a plucked string of length I
released from rest when the midpoint is drawn aside
through a distance h (fig. 4). In accordance with (25)
and (26) we can assume that
t TTTX met
= L a sin  cos r .
32
WAVES
When t = 0, this reduces to 2a r sin , and the coefficients
r ^
a r have to be chosen so that this is identical with
y = ~x
2h
y= (lx) ,
FIG. 4
TTTX
If we multiply both sides of the equation y = 27a r sin
by sin y , and integrate from x = to a: ?, as in the
I
method of Fourier analysis, all the terms except one will
disappear on the righthand side, and we shall obtain
1 C l l 2 2h . TTTXj C l 2Jl (1 . T7TX J
a r = x sin y dx+ = (lx) sin ax.
2 J o ^ ^ J 112, I *
Sh . TTT
a =  sin when r is odd,
22 2
Whence
= when r is even.
So the full solution, giving the value of y at all subsequent
times, is
Sh 1
; Sin
sin
cos
21
(28)
WAVES ON STRINGS 33
Thus the value of y is the result of superposing certain
normal modes with their appropriate amplitudes. These
are known as the partial amplitudes. The partial
amplitude of any selected normal mode (the rth for
example), is just the coefficient a r . In this example, a r
vanishes except when r is odd, and then a r is proportional
to 1/r 2 , so that the amplitude of the higher modes is
relatively small.
21. The rth normal mode (23) has a frequency re/21.
Also, there are zero values of y (i.e. nodes) at the points
x = 0, Z/r, 21 /r .... (r 1)1 /r, I. If the string is plucked
with the finger lightly resting on the point l/r it will be
found that this mode of vibration is excited. With even
order vibrations (r even) the mid point is a node, and with
odd order vibrations it is an antinode.
We can find the energy associated with this mode of
vibration most conveniently by rewriting (23) in the form
. . TTTX . ,__ x
y = A sin cos 4 + e \ . . (29)
/ { I j
Here A is the amplitude and e is the phase. According
to (5) the kinetic energy is
T = P f dx = A* sin* + .. . (30)
Similarly, by (6) the potential energy is
V* f r&)*te^A^ + <. (31)
2 J o\dxj 4Z \ I j
Now by (3) F/p = c 2 , and so the two coefficients in (30)
and (31) are equal. The total energy of this vibration
is therefore
34 WAVES
The total energy is thus proportional to the square of the
amplitude and also to the square of the frequency. This
is a result that we shall often find as we investigate various
types of wave motion.
As a rule, however, there are several normal modes
present at the same time, and we can then write the dis
placement (24) in the more convenient form
* . TTTX (met {
y = 2M r sin cos^  + r \. . (33
r=i I \ I }
A r is the amplitude, and c r is the phase, of the rth
normal mode. When we evaluate the kinetic energy
as in (30) we find that the " cross terms " vanish, since
sin sin j dx = 0, if r ^ s. Consequently the total
I i
kinetic energy is just
r
J c
and in a precisely similar way tho total potential energy is
By addition we find that the total energy of vibration is
^2>M r *. . . . (34)
This formula is important. It shows that the total energy
is merely the sum of the energies obtained separately for
each normal mode. It is due to this simple fact, which
arises because there are no crossterms involving A r A s ,
that the separate modes of vibration are called normal
modes. It should be observed that this result holds for
both the kinetic and potential energies separately as well
as for their sum.
WAVES ON STRINGS 35
We have already seen that when a string vibrates more
than one mode is usually excited. The lowest frequency,
viz. c/2Z, is called the ground note, or fundamental, and
the others, with frequencies rc/2Z, are harmonics or over
tones. The frequency of the fundamental varies directly
as the square root of the tension and inversely as the
length and square root of the density. This is known as
Mersenne's law. The tone, or quality, of a vibration is
governed by the proportion of energy in each of the
harmonics, and it is this that is characteristic of each
musical instrument. The tone must be carefully distin
guished from the pitch, which is merely the frequency of
the fundamental.
We can use the results of (34) to determine the total
energy in each normal mode of the vibrating string which
we discussed in 20. According to (28) and (33) A 2n = 0,
A A 8A 1 . (271 + 1)77 ^ ^ ^.
and A 2n+l = sin . Consequently, the
7T (^72~p~JL) Zi
total energy of the even modes is zero, and the energy
of the (2n+l)th mode is 16c 2 /^>/(2^+l) 2 7r 2 /. Tlu's shows
us that the main part of the energy is associated with the
normal modes of low order. We can check these formulae
for the energies in this example quite easily. For the total
energy of the whole vibration is the sum of the energies of
each normal mode separately : i.e.
total energy = ~ 2
It is shown in books on algebra that the sum of the series
l/P+l/3 2 +l/5 2 + ... is77 2 /8. Hence the total energy is
2c% 2 p/, i.e. 2Fh 2 /l. But the string was drawn aside and
released from rest in the position of fig. 4, and at that
moment the whole energy was in the form of potential
energy. This potential energy is just F times the increase
in length, i.e. 2jF{(Z 2 /4+/& 2 ) 1 / 2 Z/2}. A simple calculation
shows that if we neglect powers of h above the second,
36 WAVES
as we have already done in our formulation of the equation
of wave motion, this becomes 2Fh 2 /l, thus verifying our
earlier result.
This particular example corresponds quite closely to
the case of a violin string bowed at its midpoint. A listener
would thus hear not only the fundamental, but also a
variety of other frequencies, simply related to the funda
mental numerically. This would not therefore be a pure
note, though the small amount of the higher harmonics
makes it much purer than that of many musical instru
ments, particularly a piano.
If the string had been bowed at some other point than
its centre, the partial amplitudes would have been different,
and thus the tone would be changed. By choosing the
point properly any desired harmonic may be emphasised
or diminished, a fact well known to musicians.
22. We have seen in 21 that it is most convenient
to analyse the motion of a string of finite length in terms
of its normal modes. According to (33) the rth mode is
. . TTTX (met , )
y r = A r sin cos j + e r k
i \ i }
We often write this
TTTX
y r = <f> r sin ... (35)
The expressions (/> r are known as the normal coordinates
for the string. There are an infinite number of these
coordinates, since there are an infinite number of degrees
of freedom in a vibrating string. The advantage of using
these coordinates can be seen from (30) and (31) ; if the
displacement of the string is
y  S<{> r sin . . . (36)
rl *
WAVES ON STRINGS
37
then
T
41
(37)
The reason why we call (f> r a normal coordinate is
now clear ; for in .mechanics the normal coordinates
(?i> #2 $n are suitable combinations of the original
variables so that the kinetic and potential energies can be
written in the form
V 
(38)
The similarity between (37) and (38) is obvious. Further,
it can be shown, though we shall not reproduce the analysis
here, that Lagrange's equations of motion apply with the
set of coordinates <^ r in just the same way as with the
coordinates q r in ordinary mechanics.
23. We shall next discuss the normal modes of a
string of length I when a mass M is tied to its midpoint
(fig. 5). Now we have already seen in 21 that in the
FIG. 5
normal vibrations of an unloaded string the normal modes
of even order have a node at the midpoint. In such a
38 WAVES
vibration there is no motion at this point, and it is clearly
irrelevant whether there is or is not a mass concentrated
there. Accordingly, the normal modes of even order are
unaffected by the presence of the mass, and our discussion
will apply to the odd normal modes.
Just as in the calculations*' of 16, 17, in which there
was a discontinuity in the string, we shall have two
separate expressions y l and y 2 valid in the regions 0^x^.1/2
and 1/2^x^1. It is obvious that the two expressions
must be such that y is symmetrical about the midpoint
of the string. y l must vanish at x and y 2 at x = I.
Consequently, we may try the solutions
y^ = a sin px cos (cpt\e)
y 2 = a sin p(lx) cos (cpt~\) . . (39)
We have already satisfied the boundary condition y l = y 2
at x = 1/2. There is still the other boundary condition
which arises from the motion of M. Just as in (18) we
may write this
Substituting the values of y and y 2 as given by (39) and
using the relation F c 2 p, we find
pi pi pi
The quantity pl/2 is therefore any one of the roots of the
equation x tan x = pl/M. If we draw the curves y = tan x,
y = pl/Mx, we can see that these roots lie in the regions
to 7T/2, TT to 3?7/2, 2?r to 57T/2, etc. If we call the roots
x l9 x 2 ... then the frequencies cp/27T become cx r /7rl. If M
is zero so that the string is unloaded, x r = (r+l/2)7r,
so the presence of M has the effect of decreasing the
frequencies of odd order.
If we write n for the frequency of a normal mode,
then, since n = cpj27r, it follows that (40) can be written
WAVES ON STRINGS 39
in the form of an equation to determine n directly ; viz.,
x tan x = pljM, where x = (irl^n . . (41)
This equation is called the period equation. Its solutions
are the various permitted frequencies (and hence periods)
of the normal modes. Period equations occur very fre
quently, especially when we have stationary type waves,
and we shall often meet them in later chapters. This
particular period equation is a transcendental equation
with an infinite number of roots, ^j
24. In the previous paragraphs we have assumed that
there was no frictional resistance, so that the vibrations were
undamped. In practice, however, the air does provide
a resistance to motion ; this is roughly proportional to the
velocity. Let us therefore discuss the motion of a string
of length I fixed at its ends but subject to a resistance
proportional to the velocity. The fundamental equation
of wave motion (3) has to be supplemented by a term in
and it becomes
dt
ex* c*
A solution by the method of separation of variables (cf . 9)
is easily made, and we find
Since y is to vanish at tho two ends, we must have, as
before, sin pi = 0, and hence p = TTT/I, r = 1, 2, 3 ....
The normal modes of vibration are therefore
y = A r e~ w sin y cos (_#+e r ) . . (43)
where
40
WAVES
The exponential term erW represents a decaying amplitude
with modulus (see 9) equal to 2/k. The frequency
q/27T is slightly less than when there is no frictional resist
ance. However, Jc is usually small, so that this decrease
in frequency is often so small that it may be neglected.
25. There is another interesting method of obtaining
the velocity of propagation of waves along a string, which
we shall now describe and which is known as the method
of reduction to a steady wave. Suppose that a wave is
moving from left to right in fig. 6 with velocity c. Then,
FIG. 6
if we superimpose on the whole motion a uniform velocity
c the wave profile itself will be reduced to rest, and
the string will everywhere be moving with velocity c,
keeping all the time to a fixed curve (the wave profile).
We are thus led to a different problem from our original
one ; for now the string is moving and the wave profile
is at rest, whereas originally the wave profile was moving
and the string as a whole was at rest. Consider the motion
of the small element PQ of length ds situated at the top
of the hump of a wave. If R is the radius of curvature at
this top point, and we suppose, as in 13, that the string
is almost inextensible, then the acceleration of the element
PQ is c 2 /R downwards. Consequently, the forces acting on
it must reduce to (c*/R) pds. But these forces are merely the
two tensions F at P and Q, and just as in 13 (especially
WAVES ON STRINGS 41
note at foot of p. 23), they give a resultant Fds/R downwards.
Equating the two expressions, we have
This is, naturally, the same result as found before. The
disadvantage of this method is that it does not describe
in detail the propagation of the wave, nor does it deal
with stationary waves, so that we cannot use it to get
the equation of wave motion, etc. It is, however, very
useful if we are only concerned with the wave velocity,
and we shall see later that this simple artifice of reducing the
wave to rest can be used in other problems as well.
26. Examples
( 1 ) Find the velocity of waves along a string whoso density
is 4 gms. per cm. when stretched to a tension 90000 dynes.
(2) A string of unlimited length is pulled into a harmonic
shape y = a cos kx, and at time t = it is released. Show
that if F is the tension and p the density of the string, its
shape at any subsequent time t is y ~ a cos kx cos kct, where
c 2 = F/p. Find the mean kinetic and potential energies per
unit length of string.
(3) Find the reflect ion coefficient for two strings which
are joined together and whose densities are 25 gms. per cm.
and 9 gms. per cm.
(4) An infinite string lies along the x axis. At t = that
part of it between x = a i s given a transverse velocity
a z x 2 . Describe, with the help of equation (9) the subsequent
motion of the string, the velocity of wave motion being c.
(5) Investigate the same problem as in question (4) except
that the string is finite and of length 2a, fastened at the
points x = ia.
(6) What is the total energy of the various normal modes
in question (5) ? Verify, by smnmatioii over all the normal
modes, that this is equal to the initial kinetic energy.
(7) The two ends of a uniform stretched string are fastened
to light rings that can slide freely on two fixed parallel wires
a distance I apart. Find the normal modes of vibration.
42 WAVES
(8) A uniform string of length 3Z fastened at its ends, is
plucked a distance a at a point of trisection. It is then
released from rest. Find the energy in each of the normal
modes and verify that the sum is indeed equal to the work
done in plucking the string originally.
(9) Discuss fully the period equation (41) in 23. Show in
particular that successive values of x approximate to rrr,
and that a closer approximation is x = TTT f pl/Mrn.
(10) Show that the total energy of vibration (43) is
%plA r *e kt {q* + kq cos (qt+c r ) sin (#+ f ,)+Jfca cos* (qt+c r )} 9
and hence prove that the rate of dissipation of energy is
%kplA r 2 e~ kt {2q sin (qt+c r )+k cos (qt+ r )}*.
(11) Two uniform wires of densities p L and p a and of equal
length are fastened together at one end and the other two ends
are tied to two fixed points a distance 21 apart. The tension
is F. Find the normal periods of vibration.
(12) The density of a stretched string is m/x*. The end
points are at x = a, 2a, and the tension is F. Show that the
normal vibrations are given by the expression
y=A sin [6 log, WY'" * pt '
' '
Show that the period equation is 6 log e 2 = nir, n ~ 1, 2, ....
(13) A heavy uniform chain of length I hangs freely from
one end, and performs small lateral vibrations. Show that
the normal vibrations are given by the expression
y = A J (2p\/{x/g}) cos (pt+e),
where J represents Bessel's function ( 7) of order zero.
Deduce that the period equation is J^(2p^/{l/g}) = 0, x
being measured from the lower end.
[ANSWERS :
1. 150 cms./sec. ; 2. %Fa 2 k 2 siu*kct, %Fa 2 k* cos 2 kct ; 3. 1/4;
5. y = Sb r cos ^i^ sin ( I*>^, b r = (
Ct d
trirct
6. 8pa 5 /15; 7. y~a r cos cos I + r l ; 8. energy in rth
normal mode = sin 2  ; sum = 3c 2 a 2 />/4Z ; 11. 2n/p
4t7T T O
where c 1 ta>n(pl/c l ) = c 2 tan(pZ/c a ), c 1 2 =j^ T /p 1 , c a 2 = F/ p 2 ]
CHAPTER III
WAVES IN MEMBRANES
27. The vibrations of a plane membrane stretched to a
uniform tension T may be discussed in a manner very
similar to that which we have used in Chapter II for
strings. When we say that the tension is T we mean that
if a line of unit length is drawn in the surface of the
membrane, then the material on one side of this line
TSx
TSy
exerts a force T on the material on the other side, and this
force is perpendicular to the line we have drawn. Let us
consider the vibrations of such a membrane ; we shall
suppose that its thickness may be neglected. If its
equilibrium position is taken as the xy plane, then we are
concerned with displacements z(xy) perpendicular to this
plane. Consider a small rectangular element ABCD
(fig. 7) of sides Sx, y. When this is vibrating the forces
43
44 WAVES
on it are (a) two forces T$x perpendicular to AB and CD,
and (b) two forces "TSy perpendicular to AD and BC.
These four forces act in the four tangent planes through
the edges of the element. An argument precisely similar
to that used in Chapter IT, 13, shows that the forces (a)
d 2 z
give a resultant TSo: . 8y perpendicular to the plate.
Similarly, the forces (b) reduce to a force TSy . $x. Let
c/*c
the mass of the plate be p per unit area ; then, neglecting
gravity, its equation of motion is
c) 2 z r) 2 z f) z
T SxSy+J SxSy = pSxSy ,
dy 2 dx 2 r dt 2
(8*z <Pz\ __ 8%
[dx 2 'dy 2 ) ~~ P ~dt 2 '
This may be put in the standard form
L ^ n \
where c 2 = "T/p .... (2)
Thus we have reduced our problem to the solution of the
standard equation of wave motion, and shown that the
velocity of waves along such membranes is c \S(l"/p).
28. Let us apply these equations to a discussion of the
transverse vibrations of a rectangular membrane ABCD
(fig. 8) of sides a and 6. Take AB and AD as axes of x
and y. Then we have to solve (1) subject to certain
boundary conditions. These are that z = at the boundary
of the membrane, for all t. With our problem this means
that z = when x = 0, x = a, y = 0, y = 6, independent
WAVES IN MEMBRANES
45
of the time. The most suitable solution of the equation
of wave motion is that of 8, equation (29). It is
cos cos cos ,
z = . px . qy . ret ,
sin sin sin
If z is to vanish at x = 0, y = 0, we shall have to reject
the cosines in the first two factors. Further, if z vanishes
FIG. 8
at x = a, then sin pa 0, so that p = m7r/a, and similarly
q = nir/b, m and n being positive integers. Thus the
normal modes of vibration may be written
. . mrrx . mry
z = A sin sin  cos
a b
(3)
where
We may call this the (m, n) normal mode. Its frequency
is cr/27r, i.e.
6 2
(4)
If &>&, the fundamental vibration is the (1, 0) mode, for
which the frequency is c/2a. The overtones (4) are not
related in any simple numerical way to the fundamental,
and for this reason the sound of a vibrating plate, in
which as a rule several modes are excited together, is
much less musical to the ear than a string, where the
harmonics are all simply related to the fundamental.
46
WAVES
In the (m, n) mode of (3) there are nodal lines x = 0,
a/m, 2a/m, a, and y = 0, 6/w, 26/w, ... 6. On opposite
sides of any nodal line the displacement has opposite sign.
A few normal modes are shown in fig. 9, in which the
shaded parts are displaced oppositely to the unshaded.
(0,0)
(1,0)
(2,0)
(2,1)
(2,2)
FIG. 9
(1,3)
The complete solution is the sum of any number of
terms siich as (3), with the constants chosen to give any
assigned shape when t 0. The method of choosing
these constants is very similar to that of 19, except that
there are now two variables x and y instead of one, and
consequently we have double integrations corresponding
to (27).
According to (4) the frequencies of vibration depend
upon the two variables m and n. As a result it may
happen that there are several different modes having the
same frequency. Thus, for a square plate, the (4, 7),
(7, 4), (1, 8) and (8, 1) modes have the same frequency ;
and for a plate for which a = 36, the (3, 3) and (9, 1)
modes have the same frequency. When we have two or
more modes with the same frequency, we call it a
degenerate case. It is clear that any linear combination
of these modes gives another vibration with the same
frequency.
WAVES IN MEMBRANES 47
29. We can introduce normal coordinates as in the
case of a vibrating string (cf. 22). According to (3)
the full expression for z is
A / j , v nir y /K \
z = 27 A mn cos (rct\~ r ) sin sm  . (5)
m,n a b
We write this
'. . WITX . ft7n/
^ 270^ sm sin , . . (6)
m t n a u
where </> mn are the normal coordinates. The kinetic
energy is
. . . (7)
and this is easily shown to be
T  2 \ patymn ... (8)
m, n 8
The potential energy may be calculated in a manner
similar to 14. Referring to fig. 7 we see that in the
displacement to the bent position, the two tensions T8y
have done work T8y . (arcAB 8x). As in 14, this
reduces to approximately  T I I 8x8y. The other two
2 \dxj
tensions TSo; have done work  T I ~ I 8x8 u. The total
2 \dyj
potential energy is therefore
In the case of the rectangular membrane this reduces to
F = Z \ p6cVVl (10)
m,n
It will be seen that T and F are both expressed in the form
of Chapter II, equation (38), typical of normal coordinates
in mechanical problems.
48 WAVES
30. With a circular membrane such as a drum of
radius a, we have to use plane polar coordinates r,9,
instead of Cartesians, and the solution of equation (1),
apart from an arbitrary amplitude, is given in 8,
equation (35a). It is
f*OS
z = J m (nr) g  n m9 cos net.
We have neglected the Y m (nr) term since this is not finite
at r 0. If we choose the origin of 6 properly, this normal
mode may be written
z == J m (nr) cos m9 cos net. . . (11)
If z is to be singlevalued, m must be a positive integer.
The boundary condition at r = a is that for all values of
6 and t, J m (na) cos m9 cos net equals zero, i.e., J m (na) 0.
For any assigned value of m this equation has an infinite
number of real roots, each one of which determines a
corresponding value of n. These roots may be found
from tables of Bessel functions. If we call them n m , ly
n m, 2> n m, fc> > then the frequency of (11) is nc/27r,
i.e. cn m ^ fc/27r, and we may call it the (m, k) mode. The
allowed values of m are 0, 1, 2, ... and of k are 1, 2, 3, ... .
There are nodal lines which consist of circles and radii
vectores. Fig. 10 shows a few of these modes of vibration,
shaded parts being displaced in an opposite direction to
unshaded.
The nodal lines obtained in figs. 9 and 10 are known
as Chladni's figures. A full solution of a vibrating mem
brane is obtained by superposing any number of these
normal modes, and if nodal lines exist at all, they will
not usually be of the simple patterns shown in these
figures. As in the case of the rectangular membrane so
also in the case of the circular membrane, the overtones
bear no simple numerical relation to the fundamental
frequency, and thus the sound of a drum is not very
WAVES IN MEMBRANES
49
musical. A vibrating bell, however, is of very similar
type, but it can be shown,* that some of the more important
overtones bear a simple numerical relation to the funda
mental ; this would explain the pleasant sound of a well
constructed bell. But it is a little difficult to see why
the ear so readily rejects some of the other overtones
(0,2)
(2,2)
whose frequencies are not simply related to the fundamental.
A possible explanation ) is that the mode of striking may
be in some degree unfavourable to these discordant
frequencies. In any case, we can easily understand why
a bell whose shape differs slightly from the conventional,
will usually sound unpleasant.
* See Slater and Frank, Introduction to Theoretical Physics,
1933, p. 161.
t Lamb, Dynamical Theory of Sound (Arnold), 1910, p. 155.
50 WAVES
31. Examples
(1) Find two normal modes which are degenerate (28)
for a rectangular membrane of sides 6 and 3.
(2) Obtain expressions for the kinetic and potential
energies of a vibrating circular membrane. Perform the in
tegrations over the ^coordinate for the case of the normal mode
z = A J m (nr) cos m6 cos net.
(3) A rectangular drum is 10 cm. X 20 cm. It is stretched
to a tension of 5 kgm., and its mass is 20 gm. What is the
fundamental frequency ?
(4) A square membrane bounded by x = 0, a and y = 0, a
is distorted into the shape z = A sin  sin  and then
a a
released. What is the resulting motion ?
(5) A rectangular membrane of sides a and b is stretched
unevenly so that the tension in the x direction is Tj_ and in
the y direction is T 2 . Show that the equation of motion is
d 2 z 8 2 z d 2 z
Ti +T 2 = p . Show that this can be brought into
dx 2 dy 2 dt 2
the standard form by changing to now variables x/ VT~ ,
2//VT" ' anc ^ nence fiftd. the normal modes.
(6) Show that the number of normal modes for the
rectangular membrane of 28 whose frequency is less than N is
approximately equal to the area of a quadrant of the ellipse
vj2 /i2 A *
 1  = =r N 2 . Hence show that the number is roughly
a 2 b 2 T
[ANSWERS: 1. (2, 0) and (0, 1): in general (2m, n) and
(2n, m) ; 2. T = J7rpn 2 c 2 ^ 2 sin 2 net f {J m (nr)}*rdr,
J o
V = %7T P 2 2 A 2 cos 2 net (* [n 2 {J rn / (nr)} 2 +m 2 { J m(^)} 2 / r2 ] r dr >
which becomes, after integration by parts
V = %7rpn 2 c 2 A 2 cos 2 nctj a {J m (nr)} 2 rdr; 3. 1751 ;
4. z = A sin (27rx/a) sin (^nyja) cos (^13rTCt/a) ;
5. z A sin (mnx/a) sin (rnry/b) cos npt,
CHAPTER IV
LONGITUDINAL WAVES IN BARS AND SPRINGS
32. The vibrations which we have so far considered have
all been transverse, so that the displacement has been
perpendicular to the direction of wave propagation. We
must now consider longitudinal waves, in which the
displacement is in the same direction as the wave. Sup
pose that AB (fig. 11) is a bar of uniform section and
P 1 Q 1
FIG. 11
mass p per unit length. The passage of a longitudinal
wave along the bar will be represented by the vibrations
of each element along the rod, instead of perpendicular
to it. Consider a small element PQ of length Sx, such
that AP = x, and let us calculate the forces on this element,
and hence its equation of motion, when it is dis
placed to a new position P'Q' '. If the displacement of
P to P' is f , then that of Q to Q' will be f +8f , so that
P'Q' = 8x +8. We must first evaluate the tension at
P' . We can do this by imagining Sx to shrink to zero.
51
52 WAVES
Then the infinitesimally small element around P' will be
in a state of tension T where, by Hooke's Law,
extension
Tp> A .
orig. length
= A Lim 
= Ag (1)
Returning to the element P'Q', we see that its mass is the
same as that of PQ, i.e. px, and its acceleration is ~.
ut
Therefore
~*~'W ^ T ^'~ T ^'
Thus the equation of motion for these longitudinal waves
reduces to the usual equation of wave motion
The velocity of waves along a rod is therefore \/(A//>),
a result similar in form to the velocity of transverse
oscillations of a string.
The full solution of (2) is soon found if we know the
boundary conditions.
(i) At a free end the tension must vanish, and thus,
r\>
from (1), = 0, but the displacement will not, in general,
dx
vanish as well.
(ii) At a fixed end the displacement must vanish,
but the tension will not, in general, vanish also.
LONGITUDINAL WAVES IN BARS AND SPRINGS 53
33. If wo are interested in the free vibrations of a
bar of length Z, we shall use stationary type solutions of
(2) as in 8, equation (27). Thus
= (a cos px\b sin px) cos {cpt^e}.
If we take the origin at one end, then by (i) fig/dx has to
vanish at x = and x = I. This means that 6 = 0, and
sin pi = 0. i.e. pi mr, where n = 1, 2, ... . The free
modes are therefore described by the functions
co l~T~ +e 4 ' * (3)
This normal mode has frequency nc/2Z, so that the funda
mental frequency is c/2Z, and the harmonics are simply
related to it. There are nodes in (3) at the points x = l/2n,
3l/2n, 5l/2n, .... (2nl)l/2n ; and there are antinodes ( 6)
at x = 0, 2l/2n, 4Z/2/1 .... Z. From (1) it follows that
these positions are interchanged for the tension, nodes of
motion being antinodes of tension and vice versa. We
shall meet this phenomenon again in Chapter VI.
34. The case of a rod rigidly clamped at its two ends
is similarly solved. The boundary conditions are now
that  = at x 0, and at x = Z. The appropriate
solution of (2) is thus
. mrx (mrct }
,sm  cos \   +}. . . (4)
This solution has the same form as that found in Chapter II,
19, for the transverse vibrations of a string.
35. We may introduce normal coordinates for these
vibrations, just as in 22 and 29. Taking, for
example, the case of 34, we should write
, , . nm
g = S n sin , . . . (5)
w = l l
where . (nrrct ,
tn == a n COS j +
54 WAVES
The kinetic energy of the element PQ is pS# .  2 , so
that the total kinetic energy is
f
J
... (6)
1 o n
The potential energy stored up in P'Q' is approximately
equal to onehalf of the tension multiplied by the increase
1 P
in length : i.e.  A . S. Thus the total potential energy is
36. The results of 33, 34 for longitudinal vibrations
of a bar need slight revision if the bar is initially in a state
of tension. We shall discuss the vibrations of a bar of
natural length Z stretched to a length Z, so that its equili
brium tension T is , ,
T =A J p .... (8)
^0
Referring to fig. 11, the unstretched length of P'Q' is not
Sx but 2$x, so that the tension at P f is not given by (1),
i
but by the modified relation
8x+$ l j Sx
J P . = A Lim   
^To+' using (8). . (9)
The mass of PQ is p(l /l)8x where p refers to the unstretched
bar, so the equation of motion is
LONGITUDINAL WAVES IN BARS AND SPRINGS 55
We have again arrived at the standard equation of wave
motion
It follows that c = (Z// )c , where c is the velocity under
no permanent tension. Appropriate solutions of (10) are
soon seen to be
.. . UTTX (nrrct } , rt
 = a n sin y cos J r + e w } , tt=l,2 . . . (11)
I ^ J
The fundamental frequency is c/2l, which, from (10), can
be written c /2 . Thus with a given bar, the frequency is
independent of the amount of stretching.
The normal mode (11) has nodes where x = 0, l/n,
2l/n, . . ,1. A complete solution of (10) is obtained by
superposition of separate solutions of type (11).
37. We shall conclude this chapter with a discussion
of the vibrations of a spring suspended from its top end
and carrying a load M at its bottom end. When we
neglect the mass of the spring it is easy to show that
the lower mass M (fig. 12) executes Simple Harmonic
Motion in a vertical line. Let us, however, consider the
possible vibrations when we allow for the mass m of the
spring. Put m = pi, where p is the unstretched mass per
unit length and I is the unstretched length. We may
consider the spring in three stages. In stage (a) we have
the unstretched spring of length 1. The element PP' of
length Sx is at a distance x from the top point A. In stage
(6) we have the equilibrium position when the spring is
stretched due to its own weight and the load at the bottom.
The element PP' is now displaced to QQ'. P is displaced
a distance X downwards and P' a distance X+SX. Lastly,
in stage (c) we suppose that the spring is vibrating anc
the element QQ' is displaced to RR'. The displace!
of Q and Q' from their equilibrium positions are f ar
56
WAVES
The new length EE' is therefore 8x +SJ?+S. The mass of
the element is the same as the mass of PP' } viz. p8x, and
is of course the same in all three stages.
We are now in a position to determine the equation
of motion of EE' . The forces acting on it are its weight
A 1
(a)
unstretched
(6)
stretched
equilibrium
FIG. 12
stretched
vibrating
downwards and the two tensions at E and E '. The
tension TR may be found from Hooke's Law, by assuming
that 8x is made infinitesimally small. Then, as in 36,
= A.
extension
orig. length
A Lim
Sx
(12)
LONGITUDINAL WAVES IN BARS AND SPRINGS 57
So the equation of motion of RR' is
2
pSx  = resultant force downwards
r '
s , *
== gp + dx
Dividing by p&x and using (12), this becomes
a 2 g A
This last equation must be satisfied by 0, since this is
merely the position of equilibrium (6) . So
.
= 9 +
p
By subtraction we discover once more the standard equation
of wave motion
x c pm
This result is very similar to that of 36. However,
before we can solve (13) we must discuss the boundary
conditions. There are two of these. Firstly, when x = 0,
we must have = for all t. Secondly, when x = I,
(i.e. the position of the mass M ) we must satisfy the law
of motion
Using (12), this becomes
58 WAVES
As before, this equation must be satisfied by = 0, since
this is just the equilibrium stage (6). Thus
So, by subtraction we obtain the final form of the second
boundary condition
A
(14)
The appropriate solution of (13) is
a sin px cos {pc+e}. . . (15)
This gives = when # = 0, and therefore satisfies the
first boundary condition. It also satisfies the other
boundary condition (14) if
plt&npl = mlM. . . . (16)
By plotting the curves y = tan x, y = (m/M)/x, we see
that there are solutions of (16) giving values of 2>Z in the
ranges to vr/2, TT to 37T/2, .... The solutions become
progressively nearer to nrr as n increases.
We are generally interested in the fundamental, or
lowest, frequency, since this represents the natural vibra
tions of M at the end of the spring. The harmonics
represent standing waves in the spring itself, and may be
excited by gently stroking the spring downwards when in
stage (6). If m/M is small, the lowest root of (16) is
small ; writing pi = z, we may expand tan z and get
z(z+z 3 /3+...) =m/M.
Approximately
z 2 (l+z 2 /3) =m/M.
We may put z 2 in the term in brackets equal to the first
order approximation z 2 = m/M 9 and then we find for the
second order approximation
_mA_
. 1+m/BM'
LONGITUDINAL WAVES IN BARS AND SPRINGS 59
The period of the lowest frequency in (15) is 2jTJpc, i.e.,
27rl/cz. Using the fact that c 2 = A?/w, this becomes
277 A/ 1 . If the mass of the spring m had been
' A
neglected we should have obtained the result 2?r\/(Of/A).
It thus appears that the effect of the mass of the spring
is equivalent, in a close approximation, to adding a mass
one third as great to the bottom of the spring.
38. Examples
(1) Find the velocity of longitudinal waves along a bar
whose mass is 225 gms. per cm. and for which the modulus
is 90 . 1C 10 dynes.
(2) Two semi infinite bars are joined to form an infinite
rod. Their moduli are Aj and A 2 and the densities are p
and p 2 . Investigate the reflection coefficient (see 16) and
the phase change on reflection, when harmonic waves in the
first medium meet the join of the bars.
(3) Investigate the normal modes of a bar rigidly fastened
at one end and free to move longitudinally at the other.
(4) A uniform bar of length I is hanging freely from one
end. Show that the frequencies of the normal longitudinal
vibrations are (njJ) c/2Z, where c is the velocity of longi
tudinal waves in the bar.
(5) The modulus of a spring is 72 . 10 3 dynes. Its mass
is 10 gms. and its unstretched length is 12 cms. A mass
40 gms. is hanging on the lowest point, and the top point is
fixed. Calculate to an accuracy of 1 per cent, the periods of
the lowest two vibrations.
(6) Investigate the vertical vibrations of a spring of un
stretched length 21 and mass 2m, supported at its top end
and carrying loads M at the midpoint and the bottom.
[ANSWEKS : 1. 2 km. per sec. ; 2. B =
/(r+i)wc \
( +4 : 6 "
3. e = A r sin v ' ' cos  v ' *'" +e r l ; 5. 1690 sees.,
0252 sees. ; 6. Period = 2ir/nc where k 3 Skcotnl} cot 2 nl = 1,
k = Mln/m.]
CHAPTEB V
WAVES IN LIQUIDS*
39. In this chapter we shall discuss wave motion in
liquids. We shall assume that the liquid is incompressible,
with constant density p. This condition is very nearly
satisfied by most liquids, and the case of a compressible
fluid is dealt with in Chapter VI. We shall further assume
that the motion is irrotational. This is equivalent to
neglecting viscosity and assuming that all the motions
have started from rest due to the influence of natural
forces such as wind, gravity, or pressure of certain bound
aries. If the motion is irrotational, wo may assume
the existence of a velocity potential <j> if we desire it.
It will be convenient to summarise the formulae which
we shall need in this work.
(i) If the vector u j with components (u, v, w) J
represents the velocity of any part of the fluid, then from
the definition of (f>
u = \7< == grad <f>, . . (I)
so that in particular u = ~8<f>/dx, v = d(f)/dy,
w = fy/dz.
(ii) On a fixed boundary the velocity has no normal
* Before reading this chapter the student is advised to read
Rutherford's Vector Methods, Chapter VI, from which several
results will be quoted.
f Using Clarendon type for vectors.
J Many writers use (u x u y u z ) for the velocity components. We
shall find (u, v, w) more convenient for our purposes. It is necessary,
however, to distinguish u, which is a vector representing the velocity
and u, which is just the x component of the velocity.
60
WAVES IN LIQUIDS 61
component, and hence if 8/8 v denotes differentiation along
the normal
= ..... (2)
(iii) Since no liquid will be supposed to be created or
annihilated, the equation of continuity must express the
conservation of mass ; it is
v"r + ? + 7r<> (3)
dx dy dz
Combining (1) and (3), we obtain Laplace's equation
 <
(iv) If H(x, y, z, t) is any property of a particle of the
flff
fluid, such as its velocity, pressure or density, then
Oli
is the variation of H at a particular point in space, and
T)TJ
 is the variation of H when we keep to the same particle
JJt
Dff
of fluid. is known as the total differential coefficient,
jLJt
and it can be shown * that
DH dH ,
fjr  ^T+
Dt dt ,*
DH OH dH , dH , dH ' ( }
{ Q   L u  4 v  U w 
Dt dt ^ dx 8y ^ 8z
(v) If the external forces acting on unit mass of liquid
can be represented by a vector F, then the equation of
motion of the liquid may be expressed in vector form
Du 1
* See Rutherford, 66.
62 WAVES
In Cartesian form this is
8u 8u , 3u 8u _ 1 dp /a .
+ U+V+W =F x f, . (6)
dt dx % dz p ox
with two similar equations for v and w.
(vi) An important integral of the equations of motion
can be found in cases where the external force F has a
potential V, so that F = yF. The integral in question
is known as Bernoulli's Equation :
 + l* + v jl = c > < 7 >
p 2 dt
where C is an arbitrary function of the time. Now
according to (1), addition of a function of t to <j> does
not affect the velocity distribution given by <f> ; it is often
convenient, therefore, to absorb C into the term ~ and
ot
(7) can then be written
A particular illustration of (8) which we shall require later
occurs at the surface of water waves ; here the pressure
must equal the atmospheric pressure and is hence constant.
Thus at the surface of the waves (sometimes called the
free surface)
\ u 2 + V  ^ = constant. . . (9)
ct
40. We may divide the types of wave motion in
liquids into two groups ; the one group has been called
tidal waves,* and arises when the wavelength of the
oscillations is much greater than the depth of the liquid.
Another name for these waves is long waves in shallow
water. With waves of this type the vertical acceleration
* Lamb, Hydrodynamics, Chapter VIII.
WAVES IN LIQUIDS 63
of the liquid is neglected in comparison with the horizontal
acceleration, and we shall be able to show that liquid
originally in a vertical plane remains in a vertical plane
throughout the vibrations ; thus each plane of liquid
moves as a whole. The second group may be called
surface waves, and in these the disturbance does not
extend far below the surface. The vertical acceleration
is no longer negligible and the wavelength is much less
than the depth of the liquid. To this group belong most
wind waves and surface tension waves. We shall consider
the two types separately, though it will be recognised that
Tidal Waves represent an approximation and the results
for these waves may often be obtained from the formulse
of Surface Waves by introducing certain restrictions.
TIDAL WAVES
41. We shall deal with Tidal Waves first. Here we
assume that the vertical accelerations may be neglected.
One important result follows immediately. If we draw
the z axis vertically upwards (as we shall continue to do
throughout this chapter), then the equation of motion in
the z direction as given by (6), is
Dw 1 dp
___ = __0____.
We are to neglect and thus
JLJt
dp
= gp, i.e. p = gpz ( constant.
cz
Let us take our xy plane in the undisturbed free surface,
and write (#, y, t) for the elevation of the water above
the point (x, y, 0). Then, if the atmospheric pressure is
p Q , we must have p = p Q when z = . So the equation
for the pressure becomes
) (10)
WAVES
We can put this value of p into the two equations of
horizontal motion, and we obtain
Du dt> Dv d
~Dt~~ ~ g dx y ~Dt = ~ g dy
(11)
The righthand sides of these equations are independent
of z, and we deduce therefore that in this type of motion
the horizontal acceleration is the same at all depths.
Consequently, as we stated earlier without proof, on still
water the velocity does not vary with the depth, and
the liquid moves as a whole, in such a way that particles
originally in a vertical plane, remain so, although this
vertical plane may move as a whole.
42. Let us now apply the results of the last section
to discuss tidal waves along a straight horizontal channel
whose depth is constant, but whose crosssection A varies
dx
FIG. 13
from place to place. We shall suppose that the waves
move in the x direction only (extension to two dimensions
will come later). Consider the liquid in a small volume
(fig. 13) bounded by the vertical planes x, x~\dx at P
and Q. The liquid in the vertical plane through P is all
moving with the same horizontal velocity u(x) independent
of the depth. We can suppose that A varies sufficiently
slowly for us to neglect motion in the y direction. We
WAVES IN LIQUIDS 65
have two equations with which to obtain the details of
the motion. The first is (11) and may be written
du , du du dt,
+ u +w = 0^.
dt dx 8z dx
Since u is independent of 2, = 0. Further, since we
oz
shall suppose that the velocity of any element of fluid
is small, we may neglect u which is of the second order,
ox
and rewrite this equation
dU a 8 ^ (12)
8* ~ 9 te (")
The second equation is the equation of continuity. Equa
tion (3) is not convenient for this problem, but a suitable
equation can be found by considering the volume of liquid
between the planes at P and Q, in fig. 13. Let b(x) be the
breadth of the water surface at P. Then the area of the
plane P which is covered with water is [A \~b] P ; therefore
the amount of liquid flowing into the volume per unit
time is [(A~}b)u]p. Similarly, the amount flowing out
per unit time at Q is [(A +&)^]Q. The difference between
these is compensated by the rate at which the level is
rising inside the volume, and thus
Therefore
Since bQu, is of the second order of small quantities, we may
neglect this term and the equation of continuity becomes
66 WAVES
Eliminating u between (12) and (13) gives us the equation
In the case in which A is constant, this reduces to the
standard form
This is the familiar equation of wave motion in one
dimension, and we deduce that waves travel with velocity
^/(Ag/b). If the crosssection of the channel is rectangular,
so that A = bh, h being the depth,
c=V(fl*) .... (16)
With an unlimited channel, there are no boundary
conditions involving #, and to our degree of approximation
waves with any profile will travel in either direction.
With a limited channel, there will be boundary conditions.
Thus, if the ends are vertical, u = at each of them.
We may apply this to a rectangular basin of length I,
whose two ends are at x = 0, 1. Possible solutions of (15)
are given in 8, equation (27). They are
= (a cos px+f3 sin px) cos (cpt+e).
Then, using (13) and also the fact that A = bh, we find
8u cp ~
= = (a Gospx+p sinpx) sin (cpt+e).
OX fl
Xi
u  (a sin pxj3 cos px) sin (cpt+c).
ft
The boundary conditions u = at x = 0, Z, imply that
j8 = 0, and sin pi = 0. So
, r == 1,2,3,... (17)
WAVES IN LIQUIDS 67
It will be noticed that nodes of u and do not occur at
the same points.
The vertical velocity may be found from the general
form of the equation of continuity (3). Applied to our
case, this is
8u 8w _
T; h ~T~ == 0
 ox GZ
Now u is independent of z and w = on the bottom of
the liquid where z h. Consequently, on integrating
dx
7rra r c TTTX . /ryrc^ \
(z+A cos sin I 7 +* r f
Ih I \ I )
We may use this last equation to deduce under what
conditions our original assumption that the vertical
acceleration could be neglected, is valid. For the vertical
Dw . dw .
acceleration is effectively , i.e.
The maximum value of this is TrWaJZ 2 , and may be com
pared with the maximum horizontal acceleration Trrc 2 a r /lh.
The ratio of the two is rirh/1, i.e. 277A/A, since, from (17)
A = 2l/r. We have therefore confirmed the condition
which we stated as typical of these long waves, viz. that
the vertical acceleration may be neglected if the wavelength
is much greater than the depth of water.
43. We shall now remove the restriction imposed in
the last section to waves in one dimension. Let us use
the same axes as before and consider the rate of flow of
liquid into a vertical prism bounded by the planes
x, x\dx, y, y+dy. In fig. 14, ABCD is the undisturbed
surface, EFGH is the bottom of the liquid, and PQES is
the moving surface at height (x } y) above ABCD. The
68
WAVES
rate of flow into the prism across the face PEHS is
[u(h+)dy] a , and the rate of flow out across RQFO is
The net result from these two planes is
FIG. 14
o
a gain {ufy^Qftdxdy. Similarly, from the other two
c/x
o
vertical planes there is a gain  {v(h{^)}dxdy. The
total gain is balanced by the rising of the level inside the
prism, and thus
 ~{u(h+)}dxdy  j^{v(h+)}dxdy  ^ . dxdy.
As in 42, we may neglect terms such as ut, and vt, and
write the above equation of continuity
d(hu) d(hv) __ __a^
dx ~ + ~~dy ~8t '
(20)
WAVES IN LIQUIDS 69
We have to combine this equation with the two equations
of motion (11), which yield, after neglecting square terms
in the velocities
8u dv d
Eliminating u and v gives us the standard equation
8
If h is constant (tank of constant depth) this becomes
This is the usual equation of wave motion in two dimensions
and shows that the velocity is \/(gh). If we are concerned
with waves in one dimension, so that is independent of
y (as in 42) we put = and retrieve (15).
We have therefore to solve the equation of wave
motion subject to the boundary conditions
(i) w = at z = h,
(ii) = at a boundary parallel to the y axis, and
ox
o5*
at a boundary parallel to the x axis,
dy
f\Y r\
(iii) at any fixed boundary, where denotes
ov ov
differentiation along the normal to the boundary. This
latter condition, of which (ii) is a particular case, can be
seen as follows. If Ix {my = 1 is the fixed boundary,
then the component of the velocity perpendicular to this
line has to vanish. That is, lu \rnv = 0. By differentiating
partially with respect to t and using (21), the condition (iii)
is obtained.
70 WAVES
44. We shall apply these formulae to two cases ; first,
a rectangular tank, and, second, a circular one, both of
constant depth.
Rectangular tank. Let the sides be x = 0, a and y = 0, 6.
Then a suitable solution of (23) satisfying all the boundary
conditions (i) and (ii) would be
= A cos cos cos (rTTCt+e), . (24)
a b
where p = 0, 1, 2 ... , gr  0, 1, 2, ... , and r 2 = p*/a*+q*lb*.
This solution closely resembles that fora vibrating membrane
in Chapter III, 28, and the nodal lines are of the same
general type. The student will recognise how closely the
solution (24) resembles a " choppy sea."
Circular tank. If the centre of the tank is origin and
its radius is a, then the boundary condition (iii) reduces to
r\Y
at r = a. Suitable solutions of (23) in polar
dr
coordinates have been given in Chapter I, equation (35a).
We have
= A cos mO J m (nr) cos (cnt\e) . (25)
We have rejected the Y m solution since it is infinite at
r = 0, and we have chosen the zero of 6 so that there is
no term in sin m9. This expression satisfies all the condi
tions except the boundary condition (iii) at r ~ a. This
requires that J m '(na) 0. For a given value of m (which
must be integral) this condition determines an infinite
number of values of n, whose magnitudes may be found
from tables of Bessel Functions. The nodal lines are
concentric circles and radii from the origin, very similar
to those in fig. 10 for a vibrating membrane. The period
of this motion is 2?r/ 'en. \
45. It is possible to determine the actual paths of
individual particles in, many of these problems. Thus,
WAVES IN LIQUIDS 71
referring to the rectangular tank of 42, the velocities
u and w are given by (18) and (19). We see that
w ~Trr(z\h) TTTX
= cot 7.
u I I
This quantity is independent of the time and thus any
particle of the liquid executes simple harmonic motion
along a line whose slope is given by the above value of
w/u. For particles at a fixed depth, this direction changes
from purely horizontal beneath the nodes to purely vertical
beneath the antinodes.
46. We shall conclude our discussion of tidal waves
by applying the method of reduction to a steady wave,
already described in 25, to the case of waves in a channel
of constant crosssection A and breadth of waterline b.
This is the problem of 42 with A constant. Let c be the
velocity of propagation of a wave profile. Then super
impose a velocity ~c on the whole system, so that the
wave profile becomes stationary and the liquid flows under
it with mean velocity c. The actual velocity at any point
will differ from c since the crosssectional area of the liquid
is not constant. This area is A\b,, and varies with .
Let the velocity be c~\6 at sections where the elevation
is . Since no liquid is piling up, the volume of liquid
crossing any plane perpendicular to the direction of flow
is constant, i.e.
(A +&) (c +0) = constant = Ac. . (26)
Wo have still to use the fact that the pressure at the free
surface is always atmospheric. In Bcrnouilli's equation
at the free surface (9) we may put d</>/dt = since the
motion is now steady motion ; also V = g at the free
surface. So, neglecting squares of the vertical velocity,
this gives
= const. = c 2 .
72 WAVES
Eliminating between this equation and (26), we have
i.e.
Whence
. _ . (27)
l '
If is small, so that we may neglect compared with 4/6,
then this equation gives the same result as (16), viz.
c 2 = gA/b. We can, however, deduce more than this
simple result. For if >0, the righthand side of (27) is
greater than gA/b, and if <0, it is less than gA/b. Thus
an elevation travels slightly faster than a depression and
so it is impossible for a long wave to be propagated
without change of shape. Further, since the tops of waves
travel faster than the troughs, we have an explanation of
why waves break on the seashore when they reach shallow
water.
SURFACE WAVES
47. We now consider Surface Waves, in which the
restriction is removed that the wavelength is much greater
than the depth. In these waves the disturbance is only
appreciable over a finite depth of the liquid. We shall
solve this problem by means of the velocity potential </>.
<f> must satisfy Laplace's equation (4) and at any fixed
boundary d<f>/dv = 0, by (2). There are, however, two
other conditions imposed on <j> at the free surface. The
first arises from Bernoulli's equation (9). If the velocity
is so small that u 2 may be neglected, and if the only
external forces are the external pressure and gravity, we
WAVES IN LIQUIDS 73
may put u 2 = and F = gr in this equation, which
becomes
free surface
The second condition can be seen as follows. A particle
of fluid originally on the free surface will remain so always.
Now the equation of the free surface, where z = (#, y, t)
may be written
= /(3, y, z, t) = (&, 2/, Q z.
Consequently, / is a function which is always zero for a
particle on the, free surface. We may therefore use (5)
with H put equal to /, and we find
A Df . , 0f
^j^ == ^ + u ^ +v ^~ w '
Dt dt dx 8y
Now from (28)  =   I  I =  TT on the surface.
v } dx g di \dx) g 8i
ftY
Thus ~ is a small quantity of order of magnitude not
ox
f\Y ?\Y
greater than u\ consequently u and v , being of
ex cy
order of magnitude not greater than u 2 , may be neglected.
We are left with the new boundary condition
%  w   * (29)
dt ~ W ~~ 8z (M)
Combining (28) and (29) we obtain an alternative relation
We summarise the conditions satisfied by <f> as follows :
(i) Laplace's equation y 2 < = in the liquid . (2)
(ii) d<f>l&v = on a fixed boundary ... (4)
74 WAVES
1 firh
(iii) =  ? on the free surface . . . (28)
(7 d
oJ" o J
(iv) = on the free surface . . . (29)
vt dz
( v ) .? + g, f!r = on the free surface . . (30)
Gv 0%
Only two of the last three conditions are independent.
48. Let us apply these equations to the case of a
liquid of depth h in an infinitely long rectangular tank,
supposing that the motion takes place along the length
of the tank, which we take as the x direction. The axes
of x and y lie, as usual, in the undisturbed free surface.
Condition (i) above gives an equation which may be
solved by the method of separation of variables (see 7),
and if we want our solution to represent a progressive
wave with velocity c, a suitable form of the solution would
be
<f> = (Ae mz + Be~ mz ) cos m(xct).
A, B, m and c are to be determined from the other condi
tions (ii)(v). At the bottom of the tank (ii) gives d</>/dz~0,
i.e. Ae~ mh Be mh = 0. So Ae~ mh = Be mh = W, say, and
hence
(/) C cosh m(z\h) cos m(xct). . (31)
Condition (v) applies at the free surface where, if the
disturbance is not too large, we may put z = ; after
some reduction it becomes
c 2 = (g/m) tanh mh.
Since m = 27T/A, where A is the wavelength, we can write
this
. . . (32)
WAVES IN LIQUIDS 75
Condition (iii) gives us the appropriate form of ; it is
Y mcO .
 cosh mft sin m(xct).
9
This expression becomes more convenient if we write a
for the amplitude of ; i.e., a =  cosh mh. Then
&
= a sin m(xct), .... (33)
gfl coshw(2+&) , .. /0 , x
A = *  ^ ! i cosm( c). . (34)
me cosh mh
If the water is very deep so that tanh (27rA/A) = 1, then
(32) becomes c 2 0A/277, and if it is very shallow so
that tanh (2?r^/A) = 27rh/X, we retrieve the formula of 42
for long waves in shallow water, viz. c 2 = gh.
We have seen in Chapter I that stationary waves result
from superposition of two opposite progressive harmonic
waves. Thus we could have stationary waves analogous
to (33) and (34) defined by
= a sin mx cos met, .... (35)
. ga . . .__
6 =    sin mx sin met. . (36)
me cosh mh
We could use these last two equations to discuss stationary
waves in a rectangular tank of finite length.
49. We shall now discuss surface waves in two dimen
sions, considering two cases in particular.
Rectangular tank. With a rectangular tank bounded
by the planes x = 0, a and y = 0, 6, it is easily verified
that all the conditions of 47 are satisfied by
PTTX qiry
= A cos  cos ~ cos ret ,
a o
. a A .
A = *  v cos ^L_ cos 1^! sm
rc cosh rA a b
76 WAVES
where
p = 1, 2, ... ; q = 1, 2, .... ; r 2 = 7r 2 (p 2 /a 2 + 2 /& 2 ) and
c 2 = (gr/r) tanh r/L . . . (37)
Circular tank. Suppose that the tank is of radius a
and depth h. Then choosing the centre as origin and
using cylindrical polar coordinates r, 0, z, Laplace's equation
(cf. Chapter I, 7) becomes
$ 2 cA 1 dJ> 1 ffij) d^d)
' .. 1 '  > [ r ___ f\ /QC\
A suitable solution can be found from Chapter I, equation
(35a), which gives us a solution of the similar equation
in the form
. J m cos cos
' ~ Y m ^ nr ' sin sin
In this equation let us make a change of variable, writing
ct = iz> where i 2 = 1. We then get Laplace's equation
(38) and its solutions are therefore
In our problem we must discard the Y solution as Y m (r)
is infinite when r 0. So, choosing our zero of 6 suitably,
we can write
< = J m (nr) cos m9 (A cosh nz+B sinh nz).
At the bottom of the tank condition (ii) gives, as in 48,
A sinh rih = B cosh rih,, so that
</) = (7 J m (nr) cos m0 cosh n(z+A).
The constants m and w are not independent, since we
have to satisfy the boundary condition at r = a. This
gives J m '(na) = 0, so that for any selected m, r& is restricted
WAVES IN LIQUIDS 77
to have one of a certain set of values, determined from
the roots of the above equation. The function C above
will involve the time, and in fact if we are interested in
waves whose frequency is /, we shall try C oc sin 2irft.
Putting C = D sin Sir/I, where D is now a constant
independent of r, 6, z or t, we have
$ = DJ m (nr) cos m6 cosh n(z +h) sin 27rft. (39)
The boundary condition 47 (iii) now enables us to find ;
it is
 i J m (nr) cos m9 cosh nh cos 2rrft . (40)
t7
The remaining boundary condition 47 (iv) gives us the
period equation ; it is
47T 2 / 2 D J m (nr) cos m9 cosh nh sin 2^
{gnD J m (nr) cos m# sinh n& sin 2rrft = 0.
i.e. 4rr 2 / 2 == grn tanh TiA. . . . (41)
For waves with a selected value of m (which must be
integral) n is found and hence, from (41) / is found. We
conclude that only certain frequencies are allowed. Apart
from an arbitrary multiplying constant, the nature of the
waves is now completely determined.
50. In 48 we discussed the progressive wave motion
in an infinite straight channel. It is possible to determine
from (34) the actual paths of the particles of fluid in this
motion. For if X , Z denote the displacements of a particle
whose mean position is (x, z) we have
8<h ga Goshm(z+h)
X = TT =   r T 1
ex c cosh mh
84> qa
Z =   = 
.
dz c cosn mh
78 WAVES
in which we have neglected terms of the second order of
small quantities. Thus
X =       cosra(x~c),
me 2 cosh m h
gasmhm(z+h) .
Z = * _  .  L smm(xct).
me 1 cosh mh
Eliminating t, we find for the required path
* (42)
^ '
These paths are ellipses in a vertical plane with a constant
distance (2ga/mc z ) sech mh between their foci. A similar
discussion could be given for the other types of wave
motion which we have solved in other paragraphs.
51 . The Kinetic and Potential energies of these waves
are easily determined. Thus, if we measure the P.E.
relative to the undisturbed state, then, since (#, y) is the
elevation, the mass of liquid standing above a base dA
in the xy plane is p dA. Its centre of mass is at a height
, and thus the total P.E. is
4, .... (43)
the integral being taken over the undisturbed area of
surface. Likewise the K.E. of a small element is J pu 2 dr,
dr being the element of volume of the liquid, so that the
total K.E. is
... (44)
the integral being taken over the whole liquid, which may,
within our approximation, be taken to be the undisturbed
volume.
With the progressive waves of 48, and <f> are given
WAVES IN LIQUIDS
79
by (33) and (34), and a simple integration shows that the
K.E. and P.E. in one wavelength (27r/m) are equal, and
per unit width of stream, have the value
..... (45)
In evaluating (44) it is often convenient to use Green's
Theorem in the form *
/W
\\dxj l
The latter integral is taken over the surface S which
bounds the original volume, and d/dv represents differen
tiation along the outward normal to this volume. Since
d</)/dv = on a fixed boundary, some of the contributions
to T will generally vanish. Also, on the free surface, if
is small, we may put d0/dz instead of d(f)/dv.
52. We shall next calculate the rate at which energy
is transmitted in one of these surface waves. We can
Q
dz
P
FIG. 15
illustrate the method by considering the problem discussed
in 48, i.e. progressive waves in a rectangular tank of
depth h. Let AA' (fig. 15) be an imaginary plane fixed
in the liquid perpendicular to the direction of wave
* See Rutherford, Chapter VI, p. 66 (ii).
80 WAVES
propagation. We shall calculate the rate at which the
liquid on the left of A A' is doing work upon the liquid on
the right. This will represent the rate at which the energy
is being transmitted. Suppose that the tank is of unit
width and consider that part of A A' which lies between
the two lines z, z+dz (shown as PQ in the figure). At
all points of this area the pressure is p, and the velocity
is u. The rate at which work is being done is therefore
f
pudz. Thus the total rate is I pudz. We use Bernouilli's
J h
equation (8) to give us p ; since u 2 may be neglected,
and F = gz, therefore
p =
Now, according to (1) u = d(f>/dx and from (34),
ga .
6 =    cosm(xct).
me cosh mh
Putting these various values in the required integral we
obtain
f ga cosh m(z+h)
Binm(xct)    (p Q gpz)dz
J ~h c coshwA r
/ A f P0 2 a 2
(x ct)
J ^ h c
2 /
Bin 2 m(
This expression fluctuates with the time, and we are
concerned with its mean value. The mean value of
sin m(xct) is zero, and of sin 2 m(xct) is J. Thus the
mean rate at which work is being done is
/0
sech 2 m& cosh 2 m(z \Ti)dz.
After some reduction this becomes
%gpa?c (I + 2mh cosech 2mh).
WAVES IN LIQUIDS 81
In terms of the wavelength A = 27r/m, this is
1 ^Jl + ^eosech^l . . (46)
4 I A A J
Now from (45) we see that the total energy with a stream
of unit width is %gpa 2 per unit length. Thus the velocity
of energy flow is
C ( , . 4:7Th , 4:7Th} /Am ^
_l + _cosech_ . . (47)
We shall see in a later chapter that this velocity is an
important quantity known as the Group Velocity.
53. In the preceding paragraphs we have assumed
that surface tension could be neglected. However, with
short waves this is not satisfactory and we must now
investigate the effect of allowing for it. When we say
that the surface tension is T, we mean that if a line of
unit length is drawn in the surface of the liquid, then
the liquid on one side of this line exerts a pull on the
liquid on the other side, of magnitude T. Thus the effect
of Surface Tension is similar to that of a membrane
everywhere stretched to a tension T (as in Chapter III,
27), placed on the surface of the liquid. We showed in
Chapter III that when the membrane was bent there was
a downward force per unit area approximately equal to
f
T i
\
g Thus in fig. 16, the pressure p 1 just inside
cy
the liquid does not equal the atmospheric pressure
but rather
(48)
The reader who is familiar with hydrostatics will
recognise that the excess pressure inside a stretched film
(as in a soap bubble) is ^(l/E^+l/B^), where R and B 2
are the radii of curvature in any pair of perpendicular
82 WAVES
planes through the normal to the surface. We may put
E l = d^/dx* and E 2 = d 2 /% 2 to the first order of
small quantities, and then (48) follows immediately.
Thus, instead of making p = p Q at the free surface of
fd* 3 2 )
the liquid, the correct condition is that p j~H +  n r
(dx z cy*)
is constant and equal to p Q . We may combine this with
Bernoulli's equation (9), in which we neglect u 2 and put
V = gz. Then the new boundary condition which replaces
47 (iii) is
(49)
p x y
We still have the boundary condition 47 (iv) holding,
since this is not affected fey any sudden change in pressure
at the surface. By combining (29) and (49) we find the
new condition that replaces 47 (v). It is
8z p
We may collect these formulae together ; thus, with surface
tension
(i) y2^ = o in the body of the liquid . . (4)
(ii) 3<f>/dv = en all fixed boundaries . . (2)
Qt T f P2^* Q"* }"^\
(m)  ^ + ~ + = on the free surface
(49)
(iv) d/dt = d<j>ldz on the free surface . . (29)
( y )  + ~^+^l ^ on the free
3t z dz p \dx* dy*\ 3z ^^
^ ^ y J surface . (50)
Only two of the last three equations are independent.
WAVES IN LIQUIDS 83
54. Waves of the kind in which surface tension is
important are known as capillary waves. We shall
discuss one case which will illustrate the conditions (i)(v).
Let us consider progressive type waves on an unlimited
sheet of water of depth h, assuming that the motion takes
place exclusively in the direction of x. Then, by analogy
with (31) we shall try
C cosh m(z\Ji) cos m(xct). . . (51)
This satisfies (i) and (ii), (iv^ ffives the form of , which is
= (C/c) sinh mh sin m(x >cl). . . (52)
We have only one more condition to satisfy ; if we choose
(v) this gives
m 2 c 2 (7 cosh mil cos m(xct)~{mCg sinh mh cos m(xct)
H m 3 (7 sinh mh cos m(xct) = 0,
P
i.e. c 2 = (g/m}Tm/p) taiih mh. . . . (53)
Tliis equation is really the modified version of (32) when
allowance is made for the surface tension ; if T = 0, it
reduces to (32).
When h is large, tanh. mh = 1, and if we write m = 277/A,
we have
'+ <.
The curve of c against A is shown in fig. 17, from which
it can be seen that there is a minimum velocity which
occurs when A 2 ^Tr^T/gp. Waves shorter than this, in
which surface tension is dominant, are called ripples, and
it is seen that for any velocity greater than the minimum
there are two possible types of progressive wave, one of
which is a ripple. The minimum velocity is (4</T//>) 1/4 ,
and if, as in water, T = 75, p 1*00 and g = 981 c.g.s.
units, this critical velocity is about 23 cms. per sec., and
84
WAVES
the critical wavelength is about 17 cms. Curves of c against
A for other values of the depth h are very similar to fig. 17.
FIG. 17
55.
Examples
(1) Find the Potential and Kinetic energies for tidal
waves in a tank of length I, using the notation of 42.
(2) Find the velocity of any particle of liquid in the
problem of tidal waves in a circular tank of radius a ( 44).
Show that when m = in (25), particles originally on a vertical
cylinder of radius r coaxial with the tank, remain on a coaxial
cylinder whose radius fluctuates ; find an expression for the
amplitude of oscillation of this radius in terms of r.
(3) Tidal waves are occurring in a square tank of depth h
and side a. Find the normal modes, and calculate the Kinetic
and Potential energies for each of them. Show that when
more than one such mode is present, the total energy is just
the sum of the separate energies of each normal mode.
(4) What are the paths of the particles of the fluid in the
preceding question ?
(5) A channel of unit width is of depth h, where h = kx,
k being a constant. Show that tidal waves are possible with
frequency p/27r, for which
f = AJ Q (ax l l 2 ) cos pt,
WAVES IN LIQUIDS 85
where a 2 = 4p z /kg, and J Q is BesseFs function of order zero.
It is known that the distance between successive zeros of
J (x) tends to TT when x is large. Hence show that the wave
length of these stationary waves increases with increasing
values of x (This is the problem of a shelving beach.)
(6) At the end of a shallow tank, we have x = 0, and the
depth of water h is h h x zm . Also the breadth of the tank
b is given by 6 = b x n . Show that tidal waves of frequency
p/2ir are possible, for which
= Ax u J Q (rx s ) cos pt,
where
s ~ 1 m, a 2 = p 2 /gho, r = a/s, 2u l 2m~n arid q =  u/s .
Use the fact that J m (x) satisfies the equation
d*J 1 dJ / m 2 \ __
dx 2 xdx \ x*\
dx
(7) Prove directly from the conditions (i)(v) in 47
without using the results of 48 that the velocity of surface
waves in a rectangular channel of infinite depth is ^(gXjZn).
(8) Find the paths of particles of fluid in the case of surface
waves on an infinitely deep circular tank of radius a.
(9) A tank of depth h is in the form of a sector of a circle
of radius a and angle 72. What are the allowed normal modes
for surface waves ?
(10) If X, Y, Z denotes the displacement of a particle of
fluid from its mean position x, y, z in a rectangular tank of
sides a and b when surface waves given by equation (37) are
occurring, prove that the path of the particle is the straight line
d WTTX b Qirij 1
cot X = cot ^^ Y =  coth r(z+h) Z.
PIT a qir b r
(11) Show that in surface waves on a cylindrical tank of
radius a and depth h, the energies given by the normal modes
(39) are
V = J ~J cosWnh cos z 2<rrft J m 2 (nr) r dr, and
Jo
i r a
T =  n?rpD 2 sin 2 27r/ cosh nh siiih nh I J m 2 (nr) r dr.
2 Jo
86 WAVES
Use the fact that the total energy must be independent of the
time to deduce from this that the period equation is
47T 2 / 2 = gn tarih nh.
(12) Show that when we use cylindrical polar coordinates
to describe the capillary waves of 53, the pressure condition
at the free surface 53 (iii) is

dt p (dr* r'dr ? 2 d0 a J
Use this result to show that waves of this nature on a
circular basin of infinite depth are described by
<f> = C J m (nr) cos mB e nz cos 2irft,
m ,
2nf
where J m '(na) = and 4?r 2 / 2 = gn +~Tn 3 /p.
(13) Show that capillary waves on a rectangular basin of
sides a, b and depth h are given by
cosh r(z{h) mrrx niry
A ~ A  :  cos  cos ^ cos ZTrft,
smh rh a b
v rA mirx nny .
4 = cos  cos   sin 2<nft,
ATTJ a o
where m = 0, 1, 2, ... ; w = 0, 1, 2 ... ; r a = 7r 2 (m 2 /a 2 +n 2 /6 2 ),
and the period equation is
47T 2 / 2 = (gr+Tr 3 //)) tanh rh.
Verify, that when n = 0, this is equivalent to the result of
54, equation (53).
[ANSWERS :
(rnct \ irjTct \
~Y + *r 1, i ffpfar* Sin 8 I y + r 1 ;
(2) radial vol. is (gA/c) cos m6 J m '(nr) sin (cnt + t), trans
verse velocity is (gAm/cnr) sin m^ J m ( nr ) B i n ( cn ^ ~f~ e),
(gA/c) J '(nr) ; (3) J = A cos (pnxfa) cos (qiry/a) cos (met /a),
sin 2 (rnctja), P.E. = 
cos 2 (tTrttf/a) ; (4) = tan cot ; (8) X : F : Z =
r g a a
nrJ m '(nr) : mJ m (nr) tan m^ : nr J m (nr) ; (9) Same as in
a. (39)(41) except that m = 5A;/2, where k = 0, 1, 2  ]
CHAPTER VI
SOUND WAVES
56. Throughout Chapter V we assumed that the liquid
was incompressible. Ail important class of problems is
that of waves in a compressible fluid, such as a gas. In
this chapter we shall discuss such waves, of which sound
waves are particular examples. The passage of a sound
wave through a gas is accompanied by oscillatory motion
of particles of the gas in the direction of wave propagation.
These waves are therefore longitudinal. Since the density
p is not constant, but varies with the pressure p, we require
to know the relation between p and p. If the compressions
and rarefactions that compose the wave succeed each other
so slowly that the temperature remains constant (an
isothermal change) this relation is p kp. But normally
this is not the case and no flow of heat, which would be
needed to preserve the temperature constant, is possible ;
in such cases (adiabatic changes)
P = W, (i)
where Jc and y are constants depending on the particular
gas used. We shall use (1) when it is required, rather
than the isothermal relation.
57. There are several problems in the propagation
of sound waves that can be solved without using the
apparatus of velocity potential <f> in the form in which
we used it in Chapter V, 4754 ; we shall therefore
discuss some of these before giving the general development
of the subject.
87
88 WAVES
Our first problem is that of waves along a uniform
straight tube, or pipe, and we shall be able to solve this
problem in a manner closely akin to that of Chapter IV,
32, where wo discussed the longitudinal vibrations of a
rod. We can suppose that the motion of the gas particles
is entirely in the direction of the tube, and that the velocity
and displacement are the same for all points of the same
crosssection.
Suppose for convenience that the tube is of unit cross
sectional area, and let us consider the motion of that
part of the gas originally confined between parallel planes
at P and Q a distance Ax apart (fig. 18). The plane P
p Q
dx I
+ d
P' Q 1
FIG. 18
is distant x from some fixed origin in the tube. During
the vibration let PQ move to P'Q', in which P is displaced
a distance from its mean position, and Q a distance
g+dg. The length P'Q' is therefore dx+dg. We shall
find the equation of motion of the gas at P'Q' . For this
purpose we shall require to know its mass and the
pressure at its two ends. Its mass is the same as the
mass of the undisturbed element PQ, viz. p dx, where p
is the normal average density. To get the pressure at P'
we imagine the element dx to shrink to zero ; this gives
the local density p, from which, by (1), we calculate the
pressure. We have
p = Lim Pd tx/(dx+dg) = P Jl  ^V
dx^o \ cx)
(2)
SOUND WAVES 89
if we may neglect powers of d/dx higher than the first.
The quantity (p p )//> w ^ often occur in this chapter ;
it is called the condensation s. Thus
s= dfldx, p = Po (l+s). . . (3)
The net force acting on the element P'Q' is p t p /9 and
hence the equation of motion is
We may rewrite (4) in the form
8^__dp8p_ dp8*t
po dt*~ dpte p( >dpdx* ilom (2) '
It appears then that satisfies the familiar equation of
wave motion
g^.'** <>
This equation shows that waves of any shape will be
transmitted in either direction with velocity \/(dp/dp).
In the case of ordinary air at C., using (1) as the relation
between p and p, we find that the velocity is c = 332
metres per sec., which agrees with experiment. Newton,
who made this calculation originally, took the isothermal
relation between p and p and, naturally, obtained an
incorrect value for the velocity of sound.
A more accurate calculation of the equation of motion
can be made, in which powers of d$/dx are not neglected,
as follows. From (2) we have the accurate result
p = p =
KT
90 WAVES
So, now using (4) in which no approximations have been
made,
Equation (5) is found from (6) by neglecting 8g/dx compared
with unity. A complete solution of (6) is, however, beyond
the scope of this book. It is easy to see that, since (6)
is not in the standard form of a wave equation, the velocity
of transmission depends upon the frequency, and hence
that a wave is not, in general, transmitted without change
of shape.
58. We must now discuss the boundary conditions.
With an infinite tube, of course, there are no such condi
tions, but with a tube rigidly closed at x = X Q , we must
have = at x X Q , since at a fixed boundary the gas
particles cannot move.
Another common type of boundary condition occurs
when a tube has one or more ends open to the atmosphere.
At this end, the pressure must have the normal atmospheric
value, and thus, from (1) and (2), dg/dx = 0.
To summarise :
02 I 02
(i) ^ = Qp in the tube, and c 2 = dp/dp . (5)
(ii) = at a closed end (7)
Pt
(iii) ~ = s = at an open end. ... (8)
c/x
59. We shall apply these equations to find the normal
modes of vibration of gas in a tube of length L These
waves will iiaturally be of stationary type.
(a) Closed at both ends x = 0, 1. This problem is the
same mathematically, as the transverse vibrations of a
SOUND WAVES 91
string of length Z, fixed at its ends (cf. Chapter II, 19).
Conditions (i) and (ii) of 58 give for the normal modes
.. . . TTTX (rirct 1
f  4, am T cos { + f  , r = 1, 2, .... (9)
(6) Closed atx = 0, opera aZ # = Z (a " stopped tube ").
Here conditions (ii) and (iii) give = at x 0, and
r\
= at a; = I. The normal modes are
dx
\ "^ I i r\ i 9 nn\
'\' ' 2/ Z W "(V '"2; Z +*rj>rU,l,Z,... l">)
(c) Opera 6o/fc eratfo # = 0, Z. We have to satisfy
the boundary condition (iii) d/dx = at x = 0, Z. So
the normal modes are
r *//* ^
r = l,2,... (11)
In each case the full solution would be the superposition
of any number of terms of the appropriate type with
different r. The fundamental frequencies in the three
cases are 2l/c, 4l/c, and 2Z/c respectively. The harmonics
bear a simple numerical relationship to the fundamental,
which explains the pleasant sound of an organ pipe.
60. We shall now solve a more complicated problem.
We are to find the normal modes of a tube of unit sectional
area, closed at one end by a rigid boundary and at the
FIG. 19
other by a mass M free to move along the tube. Let
the fixed boundary be taken as x 0, and the normal
equilibrium position of the moveable mass be at x I
92 WAVES
(fig. 19). Then we have to solve the standard equation
of wave motion with the boundary conditions that when
x = 0, (ii) gives = 0, and that when x I the excess
pressure inside, pp Q , must be responsible for the
acceleration of the mass M. This implies that
d 2
PPQ = MQp when x = I
The first condition is satisfied by the function
= A sin nx cos (nct\e) . . . (12)
To satisfy the second condition, we observe that
pPo  (dp/dp)(pp ) = ~c* Po d/dx, from (3).
So this condition becomes
M i* = *' **='
Using (12) this gives, after a little reduction,
ril tan ril = IpJM.
The allowed values of n are the roots of this equation.
There is an infinite number of them, and when M = 0, so
that the tube is effectively open to the air at one end,
we obtain equation (10) ; when M oo, so that the tube
is closed at each end, we obtain equation (9).
61. So far we have developed our solutions in terms
of , the displacement of any particle of the gas from its
mean position. It is possible, however, to use the method
of the velocity potential <f>. Many of the conditions which
(f> must satisfy are the same as in Chapter V, but a few
of them are changed to allow for the variation in density.
It is convenient to gather these various formulae together
first.
(i) If the motion is irrotational, as we shall assume,
u = V</>, (cf. Chapter V, equation (1)) . (13)
SOUND WAVES 93
(ii) At any fixed boundary, dfydv = (cf. Chapter V,
equation (2)y ....... (14)
(iii) The equation of Continuity (cf . Chapter V, equation
(3)) is slightly altered, and it is *
(iv) The equations of motion are unchanged ; if F is
the external force on unit mass, in vector form,
they are
Du 1
j~ = F  yp (cf. Chapter V, equation (6)) . (16)
JL/Z p
(v) In cases where the external forces have a potential
F, we obtain Bernouillf s equation (cf. Chapter V,
equation (8))
+ u 2 +F  = const. . . (17)
p ct
in which we have absorbed an arbitrary function of the
time into the term 8(f>/dt (cf. Chapter V, equation (8)).
62. In sound waves we may neglect all external
forces except such as occur at boundaries, and thus we
may put F = in (17). Also we may suppose that the
velocities are small and neglect u 2 in this equation. With
these approximations Bernouilli's equation becomes
dp dcf>
= const.
p ct
We can simplify the first term ; for = I f ] ,
J P J \ d p' P
and if the variations in density are small, dp/dp may be
* Rutherford, 67.
94 WAVES
taken as constant, and equal to c 2 as in (5). Thus
Cdp Cdp
= c 2 M = c*log e p = C 2 {log,(l+s)+log,p }. So
J p J p
P = c 2 s+const., if s is small. If we absorb this constant
P
in <, then Bernouilli's equation takes its final form
Laplace's equation for <f> does not hold because of the
changed equation of continuity. But if u, v, w and s
are small, (15) can be written in a simpler form by the
aid of (13) ; viz.,
This is effectively the same as
 = v*    do)
Now let us eliminate s between (18) and (19), and
we shall find the standard equation of wave motion
wig .  . <
This shows that c is indeed the velocity of wave propaga
tion, but before we can use this technique for solving
problems, we must first obtain the boundary conditions
for (/>. At a fixed boundary, by (ii) d(f)/dv = 0. At an
open end of a tube, the pressure must be atmospheric,
and hence s = 0. Thus, from (18),
8^/0* = (21)
This completes the development of the method of the
velocity potential, and we can choose in any particular
problem whether we solve by means of the displacement
 or the potential <. It is possible to pass from one to
the other, since from (3) and (18)
SOUND WAVES 95
63. We shall illustrate these equations by solving the
problem of stationary waves in a tube of length I, closed
at one end (x 0) and open at the other (x = I). This
is the problem already dealt with in 59 (6), and with
,, , , . . , , . c &<(> 1 &<$>
the same notation, we require a solution of ~ = _ r
H dx* c 2 dt 2
subject to the conditions
d<j>/dx = at x = 0,
It is easily seen that
<f) = a cos mx cos (cmt\e)
satisfies all these conditions provided that cos ml = 0,
i.e. ml 77/2, 877/2, .... (r+l/2)7r/2 .... So the normal
modes are
and from this expression all the other properties of these
waves may easily be obtained. The student is advised
to treat the problems of 59 (a) and (c) in a similar manner.
64. Our next application of the equations of 62
will be to problems where there is spherical symmetry
about the origin. The fundamental equation of wave
motion then becomes (see Chapter I, equation (23))
a 2 ^ 2 ty __ i d^
~3r^ + r Or c 2 ~dt? '
with solutions of progressive type
< = lf( r ct)+lg(r+ct).
There are solutions of stationary type (see Chapter I,
equation (37))
cos cos
<A = (l/r) . mr . cmt.
r v ; ; sm sin
96 WAVES
If the gas is contained inside a fixed sphere of radius a,
then we must have </> finite when r = 0, and dcf>/dr =
when r = a. This means that
with the condition
tan ma = ma .... (23)
This period equation has an infinite number of roots which
approximate to ma = (n+l/tyn when n is large. So for
its higher frequencies the system behaves very like a
uniform pipe of length a open at one end and closed at
the other.
This analysis would evidently equally well apply to
describe waves in a conical pipe.
65. We shall now calculate the energy in a sound
r i
wave. The Kinetic energy is clearly  p u 2 ^F, where
dV is an element of volume. In terms of the velocity
potential this may be written
^ (24)
The last expression follows from Green's theorem just as
in Chapter V, 51, and the surface integral is taken over
the boundary of the gas. There is also Potential energy
because each small volume of gas is compressed or rarified,
and work is stored up in the process. To calculate it,
consider a small volume F , which during the passage
of a wave is changed to F x . If s is the corresponding
value of the condensation, then from (3), we have, to the
first degree in s l9
F^FoU^) . . . (25)
Further, suppose that during the process of compression,
F and s are simultaneous intermediary values, Then we
SOUND WAVES 97
can write the work done in compressing the volume from
' f Fl
F to F! in the form I p dV. But, just as in (25),
J Fo
V = F (l s), and henco
dV = V Q ds.
We may also write p pQ+(dp/dp)(pp )
Thus the potential energy may be written
r*i
r
J o
This is the contribution to the P.E. which arises from the
volume F . The total P.E. may be found by integration.
The first term will vanish in this process since it merely
represents the total change in volume of the gas, which
we may suppose to be zero. We conclude, therefore, that
(26)
f i
the Potential Energy is  c 2 p s 2 dV ...
It can easily be shown that with a progressive wave
the K.E. and P.E. are equal ; this does not hold for
stationary waves, for which their sum remains constant.
66. We conclude this chapter with a discussion of
the propagation of waves along a pipe whose crosssectional
area A varies along its length. Our discussion is similar
in many respects to the analysis in 57.
Consider the pipe shown in fig. 20, and let us measure
distances x along the central line. It will be approximately
true to say that the velocity u is constant across any
section perpendicular to the x axis. Suppose that the
gas originally confined between the two planes P, Q at
G
98
WAVES
distances x, x+dx is displaced during the passage of a
wave, to P'Q', the displacement of P being  and of Q
being g+dg. Consider the motion of a small prism of gas
FIG. 20
such as that shaded in the figure ; its equation of motion
may be found as in 57, and it is
PO "ol? ~oT
dt* dx
(27)
We must therefore find the pressure in terms of . This
may be obtained from the equation of continuity, which
SOUND WAVES 99
expresses the fact that the mass of gas in P'Q' is the same
as that in PQ. Thus, if p is the density,
p Q A(x) dx = p A(x
Neglecting small quantities, this yields
\ dx A dx)
Therefore
Eliminating p between (27) and (28) we find
8'*__zp2e_. 1/I1
P W~ dp dx ~ c PQ dx [A dx
where, as usual, c 2 = .
dp
So the equation of motion is
In the case in which A is constant this reduces to the former
equation (5). An important example when A is not
constant is the socalled exponential horn used on the
best acoustic gramophones ; here the tube is approximately
symmetrical about its central line and the area varies with
the distance according to the law A = A Q e Zax , where a and
A are constants.
With this form of A, (29) reduces to
100 WAVES
A solution is possible by the method of separation of
variables (see 7). We soon find
where m l and m 2 are given by a\/(a 2 n 2 ). In most
exponential horns ri 2 is considerably larger than a 2 in the
range of audible frequencies, so that m 1 and m z may be
written a^in. Thus
+*>} . . (30)
The first term represents a wave going outwards and the
second a wave coming inwards. We conclude from this
that waves can be sent outwards along the horn with a
velocity c which is approximately independent of the
frequency, and with an attenuation factor e~ ax which is
also independent of the frequency. It is this double
independence which allows good reproduction of whatever
waves are generated at the narrow end of the horn, and
which is responsible for this choice of shape in the best
gramophones. Other forms of A will not, in general, give
rise to the same behaviour.
67. Examples
(1) Use the method of 58 to investigate sound waves
in a closed rectangular box of sides a l9 a 2 and a 3 . Show
that the number of such waves for which the frequency is less
than n is approximately equal to oneeighth of the volume
of the quadric x^/a^+y^/a^+z 2 /^ = 4n 2 /c 2 . Hence show
that this number is approximately 47m 3 a 1 a 2 a 3 /3c 3 .
(2) Investigate the reflection and transmission of a train
of harmonic waves in a uniform straight tube at a point
where a smooth piston of mass M just fits into the tube and
is free to move.
(3) Show that the kinetic and potential energies of a plane
progressive wave are equal.
(4) Show that the kinetic and potential energies of
stationary waves in a rectangular box have a constant sum.
(5) Find an equation for the normal modes of a gas which
SOUND WAVES 101
is confined between two rigid concentric spheres of radii
a and b.
(6) Show that a closer approximation to the roots of
equation (23) is ma = (n + l/2)7r l/{(nfl/2)ff}.
(7) Find numerically the fundamental frequency of a
conical pipe of radius 1 metre open at its wide end.
(8) The crosssectional area of a closed tube varies with the
distance along its central line according to the law A = A Q x n .
Show that if its two ends are x = 0, and x = I, then standing
waves can exist in the tube for which the displacement is
given by the formula
f = x^ l ~ n ^J m (qx/c) cos {qct + c},
where m ~ (nf l)/2 and J m (ql/c) = 0.
Use the fact that J m (x) satisfies the equation
<PJ I dJ
[ANSWERS: 2. reflection coefft. = {
transmission coefft. = {l+^^V^Po 2 }" 1 ^ 2 > ^ period = Sir/
where (abp*{I) sin p(ba) ~ p(b a) cos p(b+a) ; 7. 166.]
CHAPTER VII
ELECTRIC WAVES *
68. Before we discuss the propagation of electric waves,
we shall summarise the most important equations that we
shall require. These are known as Maxwell's equations.
Let the vectors E (components E x , E v , E z ) and H (com
ponents H x , H y , Hy) denote the electric and magnetic
field strengths. These are defined  as the forces on a
unit charge or pole respectively when placed inside a
small needleshaped cavity, the direction of the cavity
being the same as the direction in which we wish to measure
the component of E or H. We shall suppose that all our
media are isotropic with no ferromagnetism or permanent
polarisation ; thus, if we write e for the dielectric constant,
and fj, for the permeability, then the related vectors,
viz. the magnetic induction B and the dielectric dis
placement D are given by the equations B = /zH, D = eE.
Further, let j (components j x , j y , j z ) denote the current
density vector, and p the charge density. Then, if we
measure j, B and H in electromagnetic units, E and D
in electrostatic units, writing c for the ratio between the
two sets of units,}; Maxwell's equations may be summarised
in vector form as follows :
div D = 47rp (1)
div B = (2)
* Before reading this chapter, the student is advised to
familiarise himself with the equations of electromagnet ism, as
found in text books such as those by Jeans, Pidduck, or
Abraham  B ecker .
f See, e.g., AbrahamBecker, Chapters IV, VII.
J This system is known as the Mixed System. If we had used
entirely e.s.u., or entirely e.m.u., the powers of c would have been
different. Particular care is required in discussing the units in
(3) and (7). In this chapter c will always denote the ratio of the
two sets of units.
102
ELECTRIC WAVES
curlH
103
(3)
D = cE ..... (5)
B r^ pM ...... (6)
To these equations wo must add the relation between j
and E. If a is the conductivity, which is the inverse of
the specific resistance, this relation is
J = orE ..... (7)
For conductors a is large, and for insulators it is small.
The above equations have been written in vector form ;
until the student has acquired familiarity with the use of
the vector notation and operation, he is advised to verify
the various calculations of this chapter, using the equations
in Cartesian form as well as vector form. This will soon
show how much simpler the vector treatment is, in nearly
every case. If we wish to write these equations in their
full Cartesian form, we have to remember that
__ dD x dD y
dx dy
+
 and that
dz
(orr ) ZJ 3T7 ^ U 3 II 3TJ \
Ctlz Vtly VFLx Cflz Oily Cl2x\
dy dz ' dz dx dx dy /
The preceding equations then become
dx dy
L.*E*.
dy dz
dHx dHz
dx
ldD x }
c ct
1 dD s
d_E,
' dz
dE 2
dx dy
c dt
idB s
(4')
104 WAVES
D x = E X , D y = E V ,D Z = E, . . (5 7 )
B x = nH x , B y = fjiH y , B z = fjiH z . . (6')
j x = o^ , j y = <7^ , j, = aE z . . (7')
Equations (l)(4) are sometimes called Maxwell's
Equations and equations (5) (7) constitutive relations.
Simple physical bases can easily be given for (l)(4).
Thus, (1) represents Gauss' Theorem, and follows from the
law of force between two charges ; (2) represents the fact
that isolated magnetic poles cannot be obtained ; (3) is
Ampfere's Rule that the work done in carrying a unit pole
round a closed circuit equals 4rr times the total current
enclosed in the circuit ; part of this current is the conduc
tion current j and part is Maxwell's displacement current
1 3D
  ~ ; (4) is Lenz's law of induction.
4:7TC Ot
These seven equations represent the basis of our
subsequent work. They need to be supplemented by a
statement of the boundary conditions that hold at a change
of medium. If suffix n denotes the component normal to
the boundary of the two media, and suffix s denotes the
component in any direction in the boundary plane, then
on passing from the one medium to the other
D n , B n , E s and H a are continuous . . (8)
In cases where there is a current sheet (i.e. a finite
current flowing in an indefinitely thin surface layer) some
of these conditions need modification, but we shall not
discuss any such cases in this chapter.
There are two other important results that we shall
use. First, we may suppose that the electromagnetic
field stores energy, and the density of this energy per unit
volume of the medium is
. . . (9)
Second, there is a vector, known as the Poynting
ELECTRIC WAVES 105
vector, which is concerned with the rate at which energy
is flowing. This vector/ whose magnitude and direction
are given by
(E X H), . . . . (10)
represents the amount of energy which flows in unit time
across unit area drawn perpendicular to it. E and H are
generally rapidly varying quantities and in such cases it is
the mean value of (10) that has physical significance.
69. We shall first deal with nonconducting media,
such as glass, so that we may put a = in (7) ; we suppose
that the medium is homogeneous, i.e. and p, are constants.
If, as usually happens, there is no residual charge, we may
also put p = in (1), and with these simplifications,
Maxwell's equations may be written
div E == , div H = 0,
\
ITT
ur H = 
c dt
ir. ITT e l
curl E =  , curl H = 
c dtj
These equations lead immediately to the standard equation
of wave motion, for we know * that
curl curl H = grad div H y 2 H.
Consequently, from the fourth of the equations in (11),
we find
, ,. o,,. . as a .
grad div H \7 2 H =  curl =  curl E.
6 v c dt cdt
Substituting for div H and curl E, we discover the standard
equation
<>
* Rutherford, Vector Methods, p. 59, equation (10).
106 WAVES
Eliminating H instead of E we find the same equation for E :
According to our discussion of this equation in Chapter I,
this shows that waves can be propagated in such a medium,
and that their velocity is c/y^cju,). In free space, where
e = ju, = 1, this velocity is just c. Now c, which was
defined as the ratio of the two sets of electrical units,
has the dimensions of a velocity, and its magnitude can
be obtained experimentally; it is approximately 2*998 . 10 10
cms. per sec. But it is known that the velocity of light
in free space has exactly this same value. We are thus
led to the conviction that light waves are electromagnetic
in nature, a view that has subsequently received complete
verification. Xrays, yrays, ultraviolet waves, infrared
waves and wireless waves are also electromagnetic, and
differ only in the order of magnitude of their wavelengths.
We shall be able to show later, in 71, that these waves
are transverse.
In nonconducting dielectric media, like glass, e is not
equal to unity ; also JJL depends on the frequency of the
waves, but for light waves in the visible region we may
put fj, = 1. The velocity of light is therefore c/\/e. Now
in a medium whose refractive index is K, it is known
experimentally that the velocity of light is c/K. Hence,
if our original assumptions are valid, = K 2 . This is
known as Maxwell's relation. It holds good for many
substances, but fails because it does not take sufficiently
detailed account of the atomic structure of the dielectric.
It applies better for long waves (low frequency) than for
short waves (high frequency).
70. A somewhat different discussion of (11) can be
given in terms of the electric and magnetic potentials.
Since div H = 0, it follows that we can write
H = curl A, .... (14)
ELECTRIC WAVES 107
where A is a vector yet to be determined. This equation
does not define A completely, since if t/r is any scalar,
curl (A+grad $) curl A. Thus A is undefined to the
extent of addition of the gradient of any scalar, and we
may accordingly impose one further condition upon it.
r\Tir
If H curl A, and curl E =  , it follows, by
c ct
elimination of H, that
Integrating,
where (j> is any arbitrary function,
i.e. E=grad^ . . (15)
In cases where there is no variation with the time, this
becomes E = grad </>, showing that <f> is the analogue
of the electrostatic potential.
Eliminating H from the relations H = curl A,
curl H =  , and using (15) to eliminate E, we find
c ct
Let us now introduce the extra allowed condition upon A,
and write g /
divA+^0 . . . (16)
c ct
Then A satisfies the standard equation of wave motion
<"
108 WAVES
Further, taking the divergence of (15), we obtain, by (16)
Thus <f> also satisfies the standard equation
**? " 8 >
A similar analysis can be carried through when p and j
are not put equal to zero, and wo find
H = curl A .... (14')
~ . . (15')
. . . (16')
c dt
<f) and A are known as the electric potential and magnetic
or vector potential respectively. It is open to our
choice whether we solve problems in terms of A and <f>,
or of E and H. The relations (14') (18') enable us to pass
from the one system to the other. The boundary condi
tions for <f> and A may easily be obtained from (8), but
since we shall always adopt the E, H type of solution,
which is usually the simpler, there is no need to write
them down here.
There is one other general deduction that can be made
here. If we use (3), (5) and (7) we can write, for
homogeneous media,
curl H = 47TcrE )  .
c dt
ELECTRIC WAVES 109
Taking the divergence of each side, and noting, from (1),
that div E = 47r/>/e, we find
it
Thus, on integration,
p = /) e~^, where 6 = e/47rcrc . . (19)
6 is called the time of relaxation. It follows from (19)
that any original distribution of charge decays exponentially
at a rate quite independent of any other electromagnetic
disturbances that may be taking place simultaneously,
and it justifies us in putting p = in most of our
problems. With metals such as copper, 6 is of the order
of 10~ 13 sees., and is beyond measurement ; but with
dielectrics such as water is large enough to be deter
mined experimentally.
71. We next discuss plane waves in a uniform non
conducting medium, and show that they are of transverse
type, E and H being perpendicular to the direction of
propagation. Let us consider plane waves travelling with
velocity V in a direction I, ra, n. Then E and H must be
functions of a new variable
u ES Ix+my+nzVt . . . (20)
When we say that a vector such as E is a function of u,
we mean that each of its three components separately
is a function of u, though the three functions need not
be the same. Consider the fourth equation of (11). Its
^component (see (3')) is
en, _ afly __ as*
~dy dz ~ c dt *
HO WAVES
If dashes denote differentiation with respect to u, this is
mH z '~nH y f = ~~~
/
Integrating with respect to u, this becomes
 E
y ^
in which we have put the constant of integration equal
to zero, since we are concerned with fluctuating fields
whose mean value is zero. There are two similar equations
to the above, for E y and E g , and we may write them as
one vector equation. If we let n denote the unit vector
in the direction of propagation, so that n = (I, m, n), we
have
nxH  E (21)
c
Exactly similar treatment is possible for the third equation
of (11) ; we get
c
Equation (21) shows that E is perpendicular to n and H,
and (22) shows that H is perpendicular to n and E. In
other words, both E and H are perpendicular to the direc
tion of propagation, so that the waves are transverse, and
in addition, E and H are themselves perpendicular, E,
H and n forming a righthanded set of axes. If we
eliminate H from (21) and (22) and use the fact that
n x [n x E] = (n . E)n (n . n)E = E,
since n is perpendicular to E and n is a unit vector, we
discover that F 2 = c 2 /e/>t, showing again that the velocity
of these waves is indeed c/\/(eju.).
It is worth while writing down the particular cases of
(21) and (22) that correspond to plane harmonic waves
ELECTRIC WAVES ill
in the direction of the z axis, and with the E vector in the
x or y directions. The solutions are
J0 = H a = V(*lriM ip(t ~ zlV)
E y = aeWW H y = (23)
E 9 = H z = 0.
E x =
E y =:0 H y = 4V(*//*) 6e '' x '~ z/7) (24)
E 8 ==Q H 2 = 0.
In accordance with 10, a and b may be complex, the
arguments giving the two phases. It is the general
convention * to call the plane containing H and n the
plane of polarisation. Thus (23) is a wave polarised
in the xz plane, and (24) a wave polarised in. the yz plane.
By the principle of superposition ( 6) we may superpose
solutions of types (23) and (24). If the two phases are
different, we obtain elliptically polarised light, in which
the endpoint of the vector E describes an ellipse in the
xy plane. If the phases are the same, we obtain plane
polarised light, polarised in the plane y/x b/a. If
the phases differ by ?r/2, and the amplitudes are equal,
we obtain circularly polarised light, which, in real form,
may be written
E x ^=a cos p(tz/V) H x = V( e /j^) a sin Pttz/V)
E y = a sin p(tz/V) H y = +V r ( //*) a cos P(t*IV)
E z = H z = 0. (25)
The endpoints of the vectors E and H each describe
circles in the xy plane.
In all three cases (23) (25), when we are dealing with
free space (e === /z = 1) the magnitudes of E and H are
equal.
72. By the use of (10) we can easily write down the
rate at which energy is transmitted in these waves. Thus,
* To which, unfortunately, not all writers conform.
112 WAVES
(CdP I \
0, 0, TIA/ /
This vector is in the direction of the positive zaxis, showing
that energy is propagated with the waves. According to
(9), the total energy per unit volume is
877 1 ' r J 477
From these two expressions we can deduce the velocity
with which the energy flows ; for this velocity is merely
the ratio of the total flow across unit area in unit time
divided by the energy per unit volume. This is c/v'fc/x),
so that the energy flows with the same velocity as the
wave. This does not hold with all types of wave
motion ; an exception has already occurred in liquids
(52).
When we calculate the Poynting Vector for the waves
(23) and (24), we must remember that ExH is not a
linear function and consequently (see 10) we must choose
either the real or the imaginary parts of E and H. Taking,
for example, the real part of (23), the Poynting Vector
lies in the z direction, with magnitude
c
' 4^
This is a fluctuating quantity whose mean value with
ca 2 /e
respect to the time is A/ The energy density, from
877 \ jj,
0?
(9), is cos 2 jp( z/F), with a corresponding mean value
ea 2 /877. Once again the velocity of transmission of energy
is rA/ r = c/v/(eu), which is the same as the
077 V jU, 077
wave velocity.
ELECTRIC WAVES
113
73. We shall next discuss the reflection and refraction
of plane harmonic light waves. This reflection will be
supposed to take place at a plane surface separating two
nonconducting dielectric media whose refractive indices are
KI and K 2 . Since we may put ^ = {JL 2 1, the velocities
in the two media are c/K^ cjK^. In fig. 21 let Oz be the
FIG. 21
direction of the common normal to the two media, and let
AO, OB, OC be the directions of the incident, reflected
and refracted (or transmitted) waves. We have not yet
shown that these all lie in a plane ; let us suppose that
they make angles 0, 0' and <f> with the z axis, OA being
in the plane of the paper, and let us take the plane of
incidence (i.e. the plane containing OA and Oz) to be the
xz plane. The y axis is then perpendicular to the plane
of the paper.
114 WAVES
Since the angle of incidence is 0, then as in (20), each
of the three components of E and H will be proportional to
gip{ct KI(X sin 6 f z cos 0)}
Let the reflected and transmitted rays move in directions
(I l9 m v n ) and (1 2 , w 2 , n 2 ) respectively. Then the corres
ponding components of E and H for these rays will be
proportional to
Thus, considering the E x components, we may write the
incident, reflected and transmitted values
A $&& ~ R i(% sin f z cos 6)} ^ e ip{ft  Ki(l& f m\y + n^)} an( J
e ip{ct  E z (l s x + m a y + n^
These functions all satisfy the standard equation of wave
motion and they have the same frequency, a condition
which is necessary from the very nature of the problem.
We shall first show that the reflected and transmitted
waves lie in the plane of incidence. This follows from the
boundary condition (8) that E x must be continuous on the
plane z 0, i.e. for all x, y, t,
This identity is only possible if the indices of all three
terms are identical : i.e.
ctKiX sin 9 = ct K^^x^m^) ~ ct K 2 (l 2 x{~m 2 y).
Thus #! sin = K^ =
= K l m 1 = K 2 m 2 .
The second of these relations shows that m = m 2 = 0,
so that the reflected and transmitted rays OB, OG lie in
the plane of incidence xOz. The first relation shows that
Z x sin 0, i.e. that the angle of reflection 0' is equal to
the angle of incidence 0, and also that
KI sin = KZ sin < . . . (26)
ELECTRIC WAVES
115
This wellknown relationship between the angles of
incidence and refraction is known as Snell's law.
Our discussion so far has merely concerned itself with
directions, and we must now pass to the amplitudes of
the waves. There are two main cases to consider, according
as the incident light is polarised in the plane of incidence,
or perpendicular to it.
Incident light polarised in the plane of incidence. The
incident ray AO has its magnetic vector in the xz plane,
directed perpendicular to AO. To express this vector in
terms of x, y, z it is convenient to use intermediary axes
> ??> (see fig 22, where 7? is not shown, however, as it
FIG. 22
is parallel to Oy and perpendicular to the plane of the
paper). is in the direction of propagation, and is in
the plane of incidence. Referred to these new axes, H
116 WAVES
lies entirely in the direction, and E in the 77 direction.
We may use (23) and write
J0f = JSff = , Jff,, = ateW**^
H r) ==H^==0 ) H^ = K^e^K^.
Now = x sin 6{z cos 0, and so it follows that :
incident wave
E x = , H x = #!% COS 9 e M*
E a l e ip ^ ct ~~ :K ^ xaln ^ +zco& ^ , H
E z = , #z = #!% sm 0e ? '^~^
Similar analysis for the reflected and refracted waves
enables us to write
reflected wave
E x = , fls = #A COS eMrftfiteBinflscosfl)^
y 3= & ie &**  *<* sin ^ ^os 19)} } H y = Q,
E z = , # 2  EA sin e^W
refracted wave
E x = , #3. = ^2 cos < e^^
JB y = a^cf^^sin^fzcos^)} ? H y = 0,
E z = , # z = # 2 a 2 sin <^ e*2^c
We may write the boundary conditions in the form that
E x , E y , K 2 Ey, H x> H y and H z are continuous at z = 0.
These six conditions reduce to two independent relations,
which we may take to be those due to E y and H x :
Kfl^ cos 0+K^ cos = K 2 a% cos <f>.
Thus
KI cos 6+E" 2 cos K 1 cos 0~K 2 cos ^ ~~ 2^ cos ^ '
Using Snell's law (26) in the form K l : K 2 = sin ^ : sin 0,
this gives
a "
_
am(6<f>) 2 sin ^ cos 6'
ELECTRIC WAVES 117
Equation (27) gives the ratio of the reflected and refracted
amplitudes. If medium 2 is denser than medium 1 , K 2 > K 1?
so that 6>(f>, and thus b l /a l is negative ; so there is a
phase change of 77 in the electric field when reflection takes
place in the lighter medium. There is no phase change on
reflection in a denser medium, nor in the refracted wave.
Incident light polarised perpendicular to the plane of
incidence. A similar discussion can be given when the
incident light is polarised perpendicular to the plane of
incidence ; in this case the roles of E and H are practically
interchanged, H y for example being the only nonvanishing
component of H. It is not necessary to repeat the analysis
in full. With the same notation for the amplitudes of the
incident, reflected and refracted waves, we have
(2g)
_ . _
sin 29+ sin 2</> sin 20 sin 2<f> 4 cos 9 sin </> '
If reflection takes place in the lighter medium, 7 1
9>(f>, and there is no phase change in E at reflection ;
if K^>K^ then there is a phase change of TT.
It follows from (28) that the reflected ray vanishes if
sin 29 = sin 2<. Since 9 ^ ^, this implies that 9\</) 77/2,
and then Snell's law gives
K l sin 9 = K% sin </> = K 2 cos 0,
So
tan 9 = K^Ki = V(*a/*i)   (29)
With this angle of incidence, known as Brewster's angle,
there is no reflected ray.
In general, of course, the incident light is composed of
waves polarised in all possible directions. Equations (27)
and (28) show that if the original amplitudes in the two
main directions are equal, the reflected amplitudes will
not be equal, so that the light becomes partly polarised
on reflection. When the angle of incidence is given by
(29) it is completely polarised on reflection. This angle is
therefore sometimes known as the polarising angle.
118 WAVES
74. An interesting possibility arises in the discussion
of 73, which gives rise to the phenomenon known as
total or internal reflection. It arises when reflection
takes place in the denser medium so that <>#. If we
suppose 9 to be steadily increased from zero, then <f> also
increases and when sin 6 = K^K^ , <f> = rr/2. If 6 is
increased beyond this critical value, <f> is imaginary.
There is nothing to disturb us in this fact provided that
we interpret the analysis of 73 correctly, for we never
had occasion to suppose that the coefficients were real.
We can easily make the necessary adjustment in this
case. Take for simplicity the case of incident light
polarised in the plane of incidence. Then the incident
and reflected waves are just as in our previous calculations.
The refracted wave has the same form also, but in the
exponential term, K 2 sin c/> = K l sin 0, and is therefore
real, whereas
KZ cos < = <S(K 2 *KJ sin 2
and is imaginary, since we are supposing that internal
reflection is taking place and therefore K t sin 6>K Z . We
may therefore write K 2 cos <f> = jiq, where q is real.
Thus the refracted wave has the form
J5J __ a e
= a
For reasons of finiteness at infinity, we have to choose
the negative sign, so that it appears that the wave is
attenuated as it proceeds into the less dense medium.
For normal light waves it appears that the penetration
is only a few wavelengths, and this justifies the title
of total reflection. The decay factor is
e pqz __ e p^(KSm\*eK&Z f
This factor increases with the frequency so that light of
great frequency hardly penetrates at all. In actual
physical problems, the refractive index does not change
ELECTRIC WAVES 119
from K l to K 2 abruptly, as we have imagined ; however,
Drude has shown that if we suppose that there is a thin
surface layer, of thickness approximately equal to one
atomic diameter, in which the change takes place smoothly,
the results of this and the preceding paragraphs are hardly
affected .
75. In our previous calculations we have assumed
that the medium was nonconducting, so that we could
put a 0. When we remove this restriction, keeping
always to homogeneous media, equations (l)(7) give us
div E = 0,
div H = 0,
curl H = 47TC7E+ ,
c dt
, LJL rH
curl E = ~~*~ .
c 'dt
Now curl curl E = grad div E y 2 E = y 2 E, so that
dt c dt c dt c* dt 2 '
Ua 2 E 477(7LldE
V 2 E=^ + / . . (30)
A similar equation holds for H. Equation (30) is the
wellknown equation of telegraphy (see 9). The first
term on the righthand side may be called the displacement
I )T\
term, since it arises from the displacement current 
477c dt
and the second is the conduction term, since it arises from
the conduction current j. If we are dealing with waves
whose frequency is p/2rr, E will be proportional to e*** ;
the ratio of these two terms is therefore ep/47rccr. Since
e is generally of the order of unity, this means that if
pl^TT is much greater than ccr, only the displacement term
matters (this is the case of light waves in a nonconducting
120 WAVES
dielectric) ; but if p/2?r is much less than cor, only the
conduction term matters (this is the case of long waves
in a good metallic conductor). In the intermediate region
both terms must bo retained. With most metals, if p<10 7
we can neglect the first term, and if >>10 15 we can neglect
the second term.
Let us discuss the solutions of (30) which apply to
plane harmonic waves propagated in the z direction, such
that only E x and H y are non vanishing (as in (24)). We
may suppose that each of these components is pro
portional to
eWrt .... (31)
where pfZir is the frequency and q is still to be determined.
This expression satisfies the equation (30) if
q is therefore complex, and we may write it
q = aifi,
where
The " velocity " of (31) is l/q ; but we have seen in 73
that in a medium of refractive index K the velocity is c/JfiT.
So the effective refractive index is cq which is complex.
Complex refractive indices occur quite frequently and are
associated with absorption of the waves ; for, combining
(31) and (33) we have the result that E x and H v are
proportional to
e ppz e ip(t) m . t (34)
This shows that a plane wave cannot be propagated in
such a medium without absorption. The decay factor may
ELECTRIC WAVES 121
be written e~ kz where k pp. k is called the absorbtion
coefficient. In the case where 4wac/p is small compared
with unity (the case of most metals), k is approximately
equal to 27rcr\/(iJL/). Now the wavelength in (34) is
A = 27T/ap, so that in one wavelength the amplitude
decays by a factor e , approximately ^^a^P AS
we are making the assumption that ccr/ep is small, the
decay is gradual, and can only be noticed after many
wavelengths. The distance travelled before the amplitude
is reduced to l/e times its original value is l/k, which is
of the same order as a.
The velocity of propagation of (34) is I/a, and thus
varies with the frequency. With our usual approximation
that ccr/ep is small, this velocity is
l/2rroc\
We can show that in waves of this character E and H
are out of phase with each other. For if, in accordance
with (31), we write
then the y component of the vector relation
i*. P *
curl E = ^ ,
c dt
gives us the connection between a and b. It is
8E X ___ p dH y
~dz ~~ "~~ c ~dT '
i.e. qa = ^ 6 . . . . (36)
c
Thus b/a is equal to (c///,)g. Now q is complex and hence
there is a phase difference between E 9 and H v equal to
the argument of q. This is tan~ 1 (j3/a), and with the same
approximation as in (35), this is just tan~ 1 (27ro i c/e^>), which
is effectively 27rac/p.
122
WAVES
76. It is interesting to discuss in more detail the
case in which the conductivity is so great that we may
completely neglect the displacement term in (30). Let
us consider the case of a beam of light falling normally
on an infinite metallic conductor bounded by the plane
z 0. Let us suppose (fig. 23) that the incident waves
FREE SPACE
METAL
FIQ. 23
come from the negative direction of z, in free space, for
which e = fju = 1, and are polarised in the yz plane. Then,
according to (24) they are defined by :
incident wave
E x = a l e^ 2 / c > , H v = a l e^^.
reflected wave
E x = b l eW+*M , H v = 
ELECTRIC WAVES 123
In the metal itself we may write, according to (31) and (36),
E x = a 2 e ^ (t ~^ , H y =  q a 2 e^^.
These values will satisfy the equation of telegraphy (30)
in which we have neglected the displacement term, if
02 _
pc
where y 2 = Sira^/pc. Thus
= y(i) ....... (37)
Inside the metal, E and H have a 7r/4 phase difference,
since, as we have shown in (36), this phase difference is
merely the argument of q.
The boundary conditions are that E x and H y are
continuous at z = 0. This gives two equations
Hence
<>
_*
^
2
Since g is complex, all three electric vectors have phase
differences. The ratio R of reflected to incident energy
is !&!/&!  2 , which reduces to
(cy/*) 2 +
In the case of nonferromagnetic metals, cy is much larger
than JLC, so that approximately
124 WAVES
This formula has been checked excellently by the experi
ments of Hagen and Rubens, using wavelengths in the
region of 10~ 5 cms.
It is an easy matter to generalise these results to apply
to the case when we include both the displacement and
conduction terms in (30).
We can use (38) to calculate the loss of energy in the
metal. If we consider unit area of the surface of the metal,
the rate of arrival of energy is given by the Poynting Vector.
This is \a 1  2 . Similarly the rate of reflection of energy is
oTT
So the rate of dissipation is j  a^  2  b l  2 \ .
STT 877
This must be the same as the Joule heat loss. In our
units, this loss is ccrE 2 per unit volume per unit time.
If we take the mean value of E x 2 in the metal, it is an
,00
easy matter to show that I ccrE x 2 dz is indeed exactly
J o
equal to this rate of dissipation.
77. When the radiation falls on the metal of 76, it
exerts a pressure. We may calculate this, if we use the
experimental law that when a current j is in the presence
of a magnetic field H there is a force JLCJ X H acting on
it. In our problem, there is, in the metal, an alternating
field E, and a corresponding current aE. The force on the
current is therefore ju,aE x H, and this force, being perpen
dicular to E and H, lies in the z direction. The force on
the charges that compose the current is transmitted by
them to the metal as a whole. Now both E and H are
proportional to e~ p Y e (see equation 37) so that the force
falls off according to the relation e~~ 2p Y z . To calculate the
total force on unit area of the metal surface, we must
integrate juaExH from z = to z = oo. ExHisa
fluctuating quantity, and so we shall have to take its mean
ELECTRIC WAVES 125
value with respect to the time. The pressure is then
c f 30 1
 Kl 2
P J o z
.e.
Using (38) this may be expressed in the form
78. There is another application of the theory of
76 which is important. Suppose that we have a straight
wire of circular section, and a rapidly alternating e.m.f.
is applied at its two ends. We have seen in 76 that
with an infinite sheet of metal the current falls off as we
penetrate the metal according to the law e~ p Y z . If py
is small, there is little diminution as we go down a distance
equal to the radius of the wire, and clearly the current
will be almost constant for all parts of any section (see,
however, question (12) in 79). But if py is large, then
the current will bo carried mainly near the surface of the
wire, and it will not make a great deal of difference whether
the metal is infinite in extent, as we supposed in 76, or
whether it has a crosssection in the form of a circle ; in
this case the current density falls off approximately
according to the law e~ p Y f as we go down a distance r
from the surface. This phenomenon is known as the
skin effect ; it is more pronounced at very high frequencies.
We could of course solve the problem of the wire quite
rigorously, using cylindrical polar coordinates. The
formulae are rather complicated, but the result is
essentially the same.
79. Examples
(1) Prove the equations (17') and (18') in 70.
(2) Find the value of H when E x = E y = 0, and E z =
A cos nx cos net . It is given that H = when t 0, and also
6 = ^=1, p = cr = 0. Show that there is no mean flux of
energy in this problem.
126 WAVES
(3) Prove the equation (28) in 73 for reflection and
refraction of light polarised perpendicular to the plane of
incidence.
(4) Show that the polarising angle is less than the critical
angle for internal reflection. Calculate the two values if
KI  6, K 2 = 1.
(5) Show that the reflection coefficient from glass to air at
normal incidence is the same as from air to glass, but that
the two phase changes are different.
(6) Light falls normally on the plane face which separates
two media K! , K 2 . Show that a fraction R of the energy is
reflected, and T is transmitted, whore
R = /
\
K t +K t
Hence prove that if light falls normally on a slab of dielectric,
bounded by two parallel faces, the total fraction of energy
reflected is ^ T, and transmitted is r o l * . It is
necessary to take account of the multiple reflections that take
place at each boundary.
(7) Light passes normally through the two parallel faces
of a piece of plate glass, for which K = 15. Find the fraction
of incident energy transmitted, taking account of reflection
at the faces.
(8) Show that when internal reflection ( 74) is taking
place, there is a phase change in the reflected beam. Evaluate
this numerically for the case of a beam falling at an angle
of 60 to the normal when K l = 1*6, K 2 = 1, the light being
polarised in the plane of incidence.
(9) Show that if we assume /A = 1, then the reflection
coefficient with metals (76) may be written in the form
R = 1 2/V(c<7/i>), where v is the frequency. If a is 16 . 10 7
(in our mixed units), calculate R for A = 10~ 3 cms. and
A = 10* cms.
(10) A current flows in a straight wire whose cross section
is a circle of radius a. The conduction current j depends
only on r the radial distance from the centre of the wire,
and the time t. Assuming that the displacement current
can be neglected, prove that H is directed perpendicular to the
ELECTRIC WAVES 127
radius vector. If j(r, t) and H(r, t) represent the magnitudes
of j and H, prove that
^ iTT \ 4 ' ty 1 L<J 3H
(rLr) = 47rn , =
dr v ' J ' dr c ct
(11) Use the results of question (10) to prove that j" satisfies
the differential equation
1 d I dj\ 4m7/z dj
r dr \ dr/ c dt '
Show also that H satisfies the equation
12 tT 1 n ~ET TT A 1 TJT
G H J. (JJLL JLJ. TrTTLtO" ufl
Use the method of separation of variables to prove that
there is a solution of the ^ equation of the form j =/(r)e^,
where
d z f I df
dr 2 r dr
Hence show that / is a combination of Bessel functions of
order zero and complex argument.
(12) If a (in question (1 1)) is small, show that an approximate
solution of the current equation is j = A(l \iar* Ja 2 r 4 )e^,
where A is a constant. Hence show that the total current
fluctuates between iJ", where, neglecting powers of a above
the second, J = 7Ta 2 u4(lja 4 a 2 /24). Use this result to show
that the heat developed in unit length of the wire in unit
time is o (l+aV/12). (Questions (10), (11) arid (12)
are the problem of the skin effect at low frequencies.)
[ ANSWERS : 2. H x H t = , H y A sin nx sin net ;
4. 9 28', 9 36' ; 7. 12/13 of the incident energy is trans
mitted; 8. 100 20'; 9, 0984,0950.]
CHAPTER VIII
GENERAL CONSIDERATIONS
80. The speed at which waves travel in a medium is
usually independent of the velocity of the source ; thus, if a
pebble is thrown into a pond with a horizontal velocity, the
waves travel radially outwards from the centre of disturb
ance in the form of concentric circles, with a speed which is
independent of the velocity of the pebble that caused them.
When we have a moving source, sending out waves
continuously as it moves, the velocity of the waves is
often unchanged,* but the wavelength and frequency, as
noted by a stationary observer, may be altered.
Thus, consider a source of waves moving towards an
observer with velocity u. Then, since the source is moving,
ntX
f\ j^\ ^\. _i^\ uXTV  '
A
'Vi^^' ^^^^ 'V^' ~^^r ^^r
B
SOURCE
OBSRVR
ut ntX 1
U /\ /\ S\ /\
Ib)
A 1 B 1
FIG. 24
(a) Waves when source is stationary.
(b) Waves when source is moving.
the waves which are between the source and the observer
will be crowded into a smaller distance than if the source
had been at rest. This is shown in fig. 24, where the waves
are drawn both for a stationary and a moving source. If
the frequency is n, then in time t the source emits nt waves.
* It is changed slightly when there is dispersion ,* see 83.
128
GENERAL CONSIDERATIONS 129
If the source had been at rest, these waves would have
occupied a length AB. But due to its motion the source
has covered a distance ut, and hence these nt waves are
compressed into a length A'B', where ABA'B' = ut.
Thus
ntXntX' = ut,
i.e. A' = Xufn = A(l w/c), .... (1)
if c is the wave velocity. If the corresponding frequencies
measured by the fixed observer are n and n', then, since
n\ = c = n'X' > therefore
' = ^ . . . . (2)
If the source is moving towards the observer the frequency
is increased ; if it moves away from him, the frequency is
decreased. This explains the sudden change of pitch
noticed by a stationary observer when a motorcar passes
him. The actual change in this case is from hc/(c u) to
nc/(c+u), so that
Arc = 2ncu/(c*u*). ... (3)
This phenomenon of the change of frequency when a source
is moving is known as the Doppler effect. It applies
equally well if the observer is moving instead of the source,
or if both are moving.
For, consider the case of the observer moving with
velocity v away from the source, which is supposed to be
at rest. Let us superimpose upon the whole motion,
observer, source and waves, a velocity v. We shall
then have a situation in which the observer is at rest,
the source has a velocity v y and the waves travel with
a speed cv. We may apply equation (2) which will then
give the appropriate frequency as registered by the observer ;
if this is n*, then
n(cv) n(cv)
    (4)
130 WAVES
To deal with the case in which both source and observer
are moving, with velocities u and v respectively, in the
same direction, we superimpose again a velocity v upon
the whole motion. Then in the new problem 3 the observer
is at rest, the source has a velocity uv, and the waves
travel with velocity c v. Again, we may apply (2) and
if the frequency registered by the observer is n'" 9 we have
(cv) (uv) cu
(5)
These considerations are of importance in acoustic and
optical problems ; it is not difficult to extend them to
deal with cases in which the various velocities are not in
the same line, but we shall not discuss such problems here.
81. We have shown in Chapter I, 6 that we may
superpose any number of separate solutions of the wave
equation. Suppose that we have two harmonic solutions
(Chapter I, equation (11)) with equal amplitudes and nearly
equal frequencies. Then the total disturbance is
(f) = a cos 27r(k 1 xn 1 t)\a cos
[2 2 J [2 2
The first cosine factor represents a wave very similar to
the original waves, whose frequency and wavelength are
the average of the two initial values, and which moves
with a velocity ^ ~. This is practically the same as the
velocity of the original waves, and is indeed exactly the
same if %/ij = 7i 2 /& 2  But ^ e secon d cosine factor, which
changes much more slowly both with respect to x and t,
may be regarded as a varying amplitude. Thus, for the
resultant of the two original waves, we have a wave of
approximately the same wavelength and frequency, but
with an amplitude that changes both with time and distance.
GENERAL CONSIDERATIONS 131
We may represent this graphically, as in fig. 25. The
outer solid profile is the curve
y = 2a cos 2* x  2 *.
The other profile curve is the reflection of this in the x
axis. The actual disturbance (f> lies between these two
boundaries, cutting the axis of x at regular intervals,
and touching alternately the upper and lower profile
curves. If the velocities of the two component waves
are the same, so that n^fk^ = n^k^ then the wave system
shown in fig. 25 moves steadily forward without change
FIG. 25
of shape. The case when njk^ is not equal to n 2 /k z is
dealt with in 83.
Suppose that refers to sound waves. Then we shall
hear a resultant wave whose frequency is the mean of
the two original frequencies, but whose intensity fluctuates
with a frequency twice that of the solid profile curve.
This fluctuating intensity is known as beats ; its frequency,
which is known as the beat frequency, is just n l ^n 29
that is, the difference of the component frequencies. We
can detect beats very easily with a piano slightly out of
tune, or with two equal tuningforks on the prongs of
one of which we have put a little sealing wax to decrease
its frequency. Determination of the beat frequency
between a standard tuningfork and an unknown frequency
132 WAVES
is one of the best methods of determining the unknown
frequency. Beats of low frequency are unpleasant to
the ear.
82. There is another phenomenon closely related to
beats. Let us suppose that we have a harmonic wave
<f> = A cos 27r(nt fcr), with amplitude A and frequency n.
Suppose further that the amplitude A is made to vary
with the time in such a way that A = a\b cos 2irpt.
This is known as amplitude modulation. The result is
<f> = (a +6 cos %7rpt) cos 27r(ntkx)
= acos27r(n kx)
+~cos 27r[(n+p)t kx] + cos
The effect of modulating, or varying, the amplitude, is to
introduce two new frequencies as well as the original one ;
these new frequencies np are known as combination
tones.
83. If the velocities of 81 are not the same (n^k^
not equal to n 2 /k 2 ), then the profile curves in fig. 25 move
with a speed (n l n 2 )/(k l k 2 ) ) which is different from that
of the more rapidly oscillating part, whose speed is
(n 1 +n 2 )l(k 1 +k 2 ). In other words, the individual waves
in fig. 25 advance through the profile, gradually increasing
and then decreasing their amplitude, as they give place to
other succeeding waves. This explains why, on the sea
shore, a wave which looks very large when it is some
distance away from the shore, gradually reduces in height
as it moves in, and may even disappear before it is
sufficiently close to break.
This situation arises whenever the velocity of the
waves, i.e. their wave velocity F, is not constant, but
depends on the frequency. This phenomenon is known
as dispersion. We deduce that in a dispersive system
the only wave profile that can be transmitted without
GENERAL CONSIDERATIONS 133
change of shape is a single harmonic wave train ; any
other wave profile, which may be analysed into two or
more harmonic wave trains, will change as it is propagated.
The actual velocity of the profile curves hi fig. 25 is known
as the group velocity U. We see from (6) that if the
two components are not very different, F = n/k, and
U = (% Wa)/^ fc a ) == dn/dk. . . (7)
In terms of the wavelength A, we have k = I/A, so that
We could equally well write this
dk dk dk dX
We have already met several cases in which the wave
velocity depends on the frequency ; we shall calculate
the group velocity for three of them.
Surface waves on a liquid of depth h :
The analysis of Chapter V, equation (32) shows that the
velocity of surface waves on a liquid of depth h is given by
According to (9) therefore, the group velocity is F \dVjd\,
TT 1 T r f\ , ^Trft , &nh}
i.e. 2 I T" C ~A~ I '
When h is small, the two velocities are almost the same,
but when h is large, U = F/2, so that the group velocity
for deep sea waves is onehalf of the wave velocity.
Equation (10) is the same as the expression obtained hi
52, equation (47), for the rate of transmission of energy
in these surface waves. Thus the energy is transmitted
with the group velocity.
134 WAVES
Electric waves in a dielectric medium :
The analysis in Chapter VII, 69, shows that the wave
velocity in a dielectric medium is given by
F 2 = c a //A.
We may put fju = 1 for waves in the visible region. Now
the dielectric constant 6 is not independent of the frequency,
and so F depends on A. The group velocity follows from
(9) ; it is
<>
In most regions, especially when A is long, e decreases
with A so that U is less than F. For certain wavelengths,
however, particularly those in the neighbourhood of a
natural frequency of the atoms of the dielectric, there is
anomalous dispersion, and U may exceed F. When A
is large, we have the approximate formula
It then appears from (11) that
Electric waves in a conducting medium :
The analysis in Chapter VII, 76, shows that the electric
vector is propagated with an exponential term e ip ^~^ z \
1 pc
where y 2 = 27rcrfJi/pc. Thus F 2 = = ~ . According
y 2 27T0y/.
to (7), the group velocity is
77 _ dP _ fa
~~ d(py) ~
If we suppose that a and JJL remain constant for all
frequencies, then this reduces to
U = 2/y = 2V. . . . (12)
GENERAL CONSIDERATIONS 135
The group velocity here is actually greater than the wave
velocity.
84. We shall now extend this discussion of group
velocity to deal with the case of more than two component
waves. We shall suppose that the wave profile is split
up into an infinite number of harmonic waves of the type
e 2m(kxfU) 9 .... (13)
in which the wave number k has all possible values ; we
can suppose that the wave velocity depends on the
frequency, so that n is a function of k. If the amplitude
of the component wave (13) is a(k) per unit range of k,
then the full disturbance is
(, t) = I a(k) . eW**nt)dk .
k  00
This collection of superposed waves is known as a wave
packet. The most interesting wave packets are those in
which the amplitude is largest for a certain value of &,
say & , and is vanishingly small if k~k Q is large. Then
the component waves mostly resemble e 27r ^ k x ~ not \ and
there are not many waves which differ greatly from this.
We shall discuss in detail the case in which
a(k)=A e^**')'. . . . (15)
This is known as a Gaussian wave packet, after the
mathematician Gauss, who used the exponential function
(15) in many of his investigations of other problems.
A, a and k Q are, of course, constants for any one packet.
Let us first determine the shape of the wave profile at
t = 0. The integral in (14) is much simplified because
the term in n disappears. In fact,
0)
= J
136
WAVES
On account of the term e~ a(k ~ k ^, the only range of k
which contributes significantly to this integral lies around
fc ; since when & & l/V a ^is term becomes er\ and
for larger values of kk it becomes rapidly smaller, this
range of k is of order of magnitude A& l/\/a.
[n order to evaluate the integral, we use the result *
4, (17)
This enables us to integrate at once, and we find that
ITT
fL/ T f\\ A I ajT*z*/cr p 27rikoX (1R\
Y'V* </ j "/ ** A / t* O . \JHJI
V a
The term e 27n '^ represents a harmonic wave, whose
wavelength A = l/& , and the other factors give a varying
mplitude A J e~^^ G . If we take the real part of (18),
FIG. 26
>(#, 0) has the general shape shown in fig. 26. The outer
urves in this figure are the two Gaussian curves
nd </>(%, Q) oscillates between them. Our wave packet
* Gillespie, Integration, p. 88.
GENERAL CONSIDERATIONS 137
(14) represents, at t = 0, one large pulse containing several
oscillations. If we define a half width, as the value of x
that reduces the amplitude to l/e times its maximum
value, then the halfwidth of this pulse is (\/cr)/7T.
At later times, >0, we have to integrate (14) as it
stands. To do this we require a detailed knowledge of n
as a function of k. If we expand according to Taylor's
theorem, we can write
n = n Q +a(kk Q )+p(kk Q )*j2+...
where
a = (dn/dk) , j3 = (d*nldk z ) , ..... (19)
As a rule the first two terms are the most important, and
if we neglect succeeding terms, we may integrate, using
(17). The result is
+ 00
= f
t)
When t 0, it is seen that this does reduce to (18), thus
providing a check upon our calculations. The last term
in (20) shows that the individual waves move with a
wave velocity n /k Qt but their boundary amplitude is given
by the first part of the expression, viz. A ^ e~ 7rl(
Now this expression is exactly the same as in (18), drawn
in fig. 26, except that it is displaced a distance at to the
right. We conclude, therefore, that the group as a whole
moves with velocity a = (dn/dk) Q , but that individual
waves within the group have the wave velocity n /& .
The velocity of the group as a whole is just what we have
previously called the group velocity (7).
If we take one more term in (19) and integrate to
obtain </>(%, t) we find that <f> has the same form as in (20)
138 WAVES
except that a is replaced by or \7rj3it. The effect of this
is twofold ; in the first place it introduces a variable
phase into the term e 27ri ^ k x ~ nQt \ and in the second place it
changes the exponential term in the boundary amplitude
curve to the form
This is still a Gaussian curve, but its half width is increased
t0 {(<j*+ir*p*P)l<m*} l l* . . . . (21)
We notice therefore that the wave packet moves with
the wave velocity ft /& , and group velocity (dn/dk)^
spreading out as it goes in such a way that its half width
at time t is given by (21).
The importance of the group velocity lies mainly in
the fact that in most problems where dispersion occurs,
the group velocity is the velocity with which the energy
is propagated. We have already met this in previous
paragraphs.
85. We shall next give a general discussion of the
standard equation of wave motion y 2 ^ ~ , in which
C Ov
c is constant. We shall show that the value of </> at any
point P (which may, without loss of generality be taken
to be the origin) may be obtained from a knowledge of
of o /
the values of <t, ~~ and ~ on any given closed surface S,
dn dt
which may or may not surround P ; the values of cf> and
its derivatives on S have to be associated with times
which differ somewhat from the time at which wo wish
to determine <f> P .
Let us analyse <j> into components with different
frequencies ; each component itself must satisfy the
equation of wave motion, and by the principle of super
position, which holds when c is constant, we can add the
GENERAL CONSIDERATIONS
139
various components together to obtain the full solution.
Let us consider first that part of </> which is of frequency
p ; we may write it in the form
t(x, y, z) e<** 9 . . . (22)
where k = ZnpIc (23)
i/j is the space part of the disturbance, and it satisfies the
Poisson equation
( v *+*')lA = 0. . . . (24)
This last equation may be solved by using Green's theorem.*
This theorem states that if ifj 1 and t/r 2 are any two functions,
and S is any closed surface, which may consist of two or
more parts, such that i/^ and 2 have no singularities inside
it, then
(25)
The volume integral on the lefthand side is taken over
the whole volume bounded by S, and d/dn denotes
differentiation along the outward normal to dS.
FIG. 27
In this equation ^ 1 and i/r 2 are arbitrary, so we may
/? tfcf
put I/TJ, equal to i/j 9 the solution of (24), and ?/r 2 = ,
* See Rutherford, p. 65, equation (29).
140 WAVES
r being measured radially from the origin P. We take
the volume through which we integrate to be the whole
volume contained between the given closed surface 8
(fig. 27) and a small sphere 2 around the origin. We
have to exclude the origin because *Jj 2 becomes infinite at
that point. Fig. 27 is drawn for the case of P within S j
the analysis holds just as well if P lies outside S.
Now it can easily be verified that V 2 <// 2 ~& 2 <A2>
so that the lefthand side of (25) becomes I ^ 2 (\7 2 +& 2 )0 dr,
and this vanishes, since (V 2 +& 2 )*A by (24). The
righthand side of (25) consists of two parts, representing
integrations over $ and 2. On 2 the outward normal
is directed towards P and hence this part of the full
expression is
When we make the radius of Z tend to 0, only one term
remains ; it is
where da) is an element of solid angle round P. Taking
the limit as r tends to zero, this gives us a contribution
4:7Tift P . Equation (25) may therefore be written
r(e*d<lt . ijcr 3 /1\ , . 7 .e^
~ J  r e ' fcr (  1 +tM 
J\rdn r 8n\rJ Y r
_
dS.
Since by definition <f> = \jj(xyz)e ikct 9 we can write this
last equation in the form
. . . (26)
GENERAL CONSIDERATIONS 141
where
e ik(ctr)
X =
_4 e <rfr>l A\
dn V dn\r)
r
= A B +C, say.
We may rewrite X in a simpler form ; for on account of
the time variation of <, iff e ik(ct ~ r} is the same as <, taken,
not at time t y but at time tr/c. If we write this symboli
r\ /T \
cally [6]tric> ^en # = ^ I 1 [<lr/c I 11 a similar way,
(77i \7y
. 1 r^l i ^ 1 0P f^l i. r
A = I I , and O = I I , wnere, lor
r l^ n \trlc cr dn L ltr/c
example, ~ means that we evaluate dc/>/dn as a
L^ n ]tr{c
function of x, y y z, t and then replace t by tr/c. We
call tr/c the retarded time. We have therefore proved
that
1 f
</>P =  \
X dS, where
So far we have been dealing with waves of one definite
frequency. But there is nothing in (27) which depends
upon the frequency, and hence, by summation over all
the components for each frequency present in our complete
wave, we obtain a result exactly the same as (27) but
without the restriction to a single frequency.
This theorem, which is due to Kirchhoff, is of great
theoretical importance ; for it implies (a) that the value
of < may be regarded as the sum of contributions X/far
from each element of area of S ; this may be called the
law of addition of small elements, and is familiar in a
slightly different form in optics as Huygen's Principle ;
and (6) that the contribution of dS depends on the value
01 <f>, not at time t t but at time trfc. Now r/c is the
142 WAVES
time that a signal would take to get from dS to the point
P, so that the contribution made by dS depends not on
the present value of <f> at dS, but on its value at that
particular previous moment when it was necessary for a
signal to leave dS in order that it should just have arrived
at P. This is the justification for the title of retarded
time, and for this reason also, [^>]^_ r / c is Sometimes known
as a retarded potential.
It is not difficult to verify that we could have obtained
a solution exactly similar to the above, but involving
t\r/c instead of tr/c ; we should have taken if/ 2 in the
previous work to be  instead of  . In this way
we should have obtained advanced potentials, [0]j +r / c ,
and advanced times, instead of retarded potentials and
retarded times. More generally, too, we could have
superposed the two types of solution, but we shall not
discuss this matter further.
In the case in which c = oo, so that signals have an
infinite velocity, the fundamental equation reduces to
Laplace's equation,* y 2 ^ = 0, and the question of time
variation does not arise. Our equation (27) reduces to
the standard solution for problems of electrostatics.
86. We shall apply this theory to the case of a source
sending out spherical harmonic waves, and we shall
take S to be a closed surface surrounding the point P
at which we want to calculate <, as shown in fig. 28.
Consider a small element of dS at Q ; the outward normal
makes angles O l and with QO and PQ, and these two
distances are r x and r. The value of </> at Q is given by
the form appropriate to a spherical wave (see Chapter I,
equation (24)) :
a
(j) Q =  cos m(&r^ . . (28)
r i
* See Rutherford, p. 67, equation (33).
GENERAL CONSIDERATIONS
143
Thus = ~ cos 0.
em >!_
fl m . . 1
a cos t/ t I cosm(c r t ) smm(a r.) V.
(r^ r x J
Now A = 277/m, so that if r x is much greater than A, which
FIG." 28
will almost always happen in practical problems, we may
put
~dn
ma cos 0, .
sm m i c i r \ ^
Also
and
f  = Acos^
a^ \r/
amc
The retarded values are easily found, and in fact, from (27),
ma cos 9, .
v c.i^
_j C os 9 cosm(ci~[r+r 1 ]) cos 9 sin ra(c [r+r,]).
r 2 n cr r n
144 WAVES
We may neglect the second term on the right if r is much
greater than A, and so
ma . * /n .
X = (cos 0+cos X ) smm(d [r+rj) . (29)
i
Combining (29) with (26) it follows that
1 fraa
<h p = I (cos 6 + cos cM sin m(ct[r \rJ\dS
477J rr x
= r (cos 6 f cos 0j) sinm(c^ [r+rj])^ . (30)
2AJ rri
If, instead of a spherical wave, we had had a plane
wave coming from the direction of 0, we should write
r x now being measured from some plane perpendicular to
OQ, and (30) would be changed to
<h p = ^ I  (cos + cos 0J sin m(ct[r +r l ])dS. (31)
2A J r
We may interpret (30) and (31) as follows. The effect
at P is the same as if each element dS sends out a wave
 4 /cos 0+cos 0A
of amplitude y I \dS, A beuig the amplitude
of the incident wave at dS ; further, these waves are
a quarter of a period in advance of the incident wave,
as is shown by the term sin w(c [rffi]) instead of
cQ$m(ct r x ). ~ (cos 0+cos gj { s called the inclination
2
factor and if, as often happens, only small values of
and t occur significantly, it has the value unity. This
interpretation of (30) and (31) is known as Fresnel's
principle.
The presence of this inclination factor removes a
difficulty which was inherent in Huygen's principle ; this
GENERAL CONSIDERATIONS 145
principle is usually stated in the form that each element
of a wave front emits wavelets in all directions, and these
combine to form the observed progressive wavefront. In
such a statement there is nothing to show why the wave
does not progress backwards as well as forwards, since
the wavelets should combine equally in either direction.
The explanation is, of course, that for points behind the
wave front cos 6 is negative with a value either exactly
or approximately equal to cos 6 l9 and so the inclination
factor is small. Each wavelet is therefore propagated
almost entirely in the forward direction.
Now let us suppose that some screens are introduced,
and that they cover part of the surface of S. If we assume
that the distribution of <f> at any point Q near the screens
is the same as it would have been if the screens were not
present, we have merely to integrate (30) or (31) over
those parts of S which are not covered. This approxi
mation, which is known as St Venant's principle, is not
rigorously correct, for there will be distortions in the
value of (f)Q extending over several wavelengths from the
edges of each screen. It is, however, an excellent approxi
mation for most optical problems, where A is small ;
indeed (30) and (31) form the basis of the whole theory
of the diffraction of light. With sound waves, on the other
hand, in which A is often of the same order of magnitude
as the size of the screen, it is only roughly correct.
87. We conclude this discussion with an example of
the analysis summarised in (31). Consider an infinite
screen (fig. 29) which we may take to be the xy plane.
A small part of this screen (large compared with the wave
length of the waves but small compared with other distances
involved) is cut away, leaving a hole through which waves
may pass. We suppose that a set of plane harmonic waves
is travelling in the positive z direction, and falls on the
screen ; we want to find the resulting disturbance at a
point P behind the screen.
K
146
WAVES
In accordance with 86 we take the surface S to be
the infinite xy plane, completed by the infinite hemisphere
on the positive side of the xy plane. We may divide the
P (x, y, z)
Fia. 29
contributions to (31) into three parts. The first part
arises from the aperture, the second part arises from the
rest of the screen, and the third part arises from the
hemisphere.
GENERAL CONSIDERATIONS 147
If the incident harmonic waves are represented by
= a cos m(ct z) this first contribution amounts to
a Cl
2Aj ;
(1+ cos 6) smm(ctr)d8.
We have put 9 l = in this expression since the waves
fall normally on to the xy plane. We shall only be con
cerned here with points P which lie behind, or nearly
behind, the aperture, so that we may also put cos 0=1
without loss of accuracy. This contribution is then
T  smm(ctr)dS . . (32)
The second part, which comes from the remainder of
the xy plane, vanishes, since no waves penetrate the
screen and thus there are no secondary waves starting there.
The third part, from the infinite hemisphere, also
vanishes, because the only waves that can reach this part
of $ are those that came from the aperture, and when
these waves reach the hemisphere their inclination factor
is zero. Thus (32) is in actual fact the only nonzero con
tribution and we may write
 s
Aj r
mim(ctr)d8 . . (33)
Let P be the point (.E, y, z) and consider the contribution
to (33) that arises from a small element of the aperture at
Q (, ??, 0). If OP = /, and QP = r, we have
77 2 . . (34)
Let us make the assumption that the aperture is so small
that  2 and 17 2 may be neglected. Then to this approxi
mation (34) shows us that
K2
148 WAVES
So
J
Again without loss of accuracy, to the approximation to
which we are working, we may put l/r I//, and then
we obtain
</>P = A sin {m(ct /)+},
where
A* = <7 2 + 2 , tan e = S/C,
and
c(x, y) = ^> I c s^(z+^?)d<fy,
. . (35)
Once we know the shape of the aperture it is an easy
matter to evaluate these integrals. Thus, if we consider
the case of a rectangular aperture bounded by the lines
= a, T? = j3, we soon verify that 8 = 0, and that
Fa
= Xf J J C S A?
___ ^ ^ ^
A/ px py
where j9 = 2?r/A/. If we are dealing with light waves,
then the intensity is proportional to O 2 and the diffraction
pattern thus observed in the plane z = / consists of a
grill network, with zero intensity corresponding to the
values of x and y satisfying either sin pax = 0, or
sin pfty = 0.
The theory of this paragraph is known as Fraunhofer
Diffraction Theory.
GENERAL CONSIDERATIONS 149
88. We conclude this chapter with a discussion of
the equation
... (37)
where p is some given function of x, y y z and t. When p =
this is the standard equation of wave motion, whose solution
was discussed in 85. Equation (37) has already occurred
in the propagation of electric waves when charges were
present (Chapter VIII, equations (17') and (18')). We
may solve this equation in a manner very similar to that
used in 85. Thus, suppose that p(x, y, z, t) is expressed
in the form of a Fourier series with respect to t, viz.,
p(*,y,*> t) = Za k (x, y, z)e ikct . . . (38)
k
There may be a finite, or an infinite, number of different
values of &, and instead of a summation over discrete
values of k we could, if we desired, include also an integra
tion over a continuous range of values. We shall discuss
here the case of discrete values of k ; the student will
easily adapt our method of solution to deal with a
continuum.
Suppose that <^(#, y, z, t) is itself analysed into com
ponents similar to (38), and let us write, similarly to (22),
t(x, y, z, t) = 2fa(x, y, z)e*** . . . (39)
k
the values of k being the same as in (38). If we substitute
(38) and (39) into (37), and then equate coefficients of
e ikct , we obtain an equation for \f/ k . It is
This equation may be solved just as in 85. Using
Green's theorem as in (25), we put ^ = *l* k (x 9 y, z),
e ikr
^ a .  1 taking H and 8 to be the same as in fig. 27.
150 WAVES
With these values, it is easily seen that the lefthand side
of (25) no longer vanishes, but has the value
y> ^ttr,* t t e (41)
the integral being taken over the space between S and S.
The righthand side may be treated exactly as in 85,
and gives two terms, one due to integration over , and
the other to integration over S. The first of these is
y p9 Z P ) .... (42)
The second may be calculated just as on p. 140. Gathering
the various terms together, we obtain
_ I J JL* ih k e~ ikr I 1 \ikdfk \dS . (43)
477 J ( r dn dn\r/ r dn]
Combining (38), (39) and (43) we can soon verify that our
solution can be written in the form
^(^p? UP* Z P) = t "~ r!c dr{ X d8, . . (44)
J J
where X is defined by (27). This solution reduces to (27)
in the case where p = 0, while it reduces to the wellknown
solution of electrostatics in the case where c = oo.
We have now obtained the required solution of (37).
Often, however, there will be conditions imposed by the
physical nature of our problem that allow us to simplify
(44). Thus, if p(x, y, z, t) is finite in extent, and has only
had nonzero values for a finite time t>t Q) we can make
X = by taking 8 to be the sphere at infinity. 'This
follows because X is measured at the retarded time trjc,
and if r is large enough, we shall have tr/c<t Q , so that
GENERAL CONSIDERATIONS 151
[(/>]t~ric and its derivatives will be identically zero on 8.
In such a case we have the simple result
%^r, . . . (45)
the integration being taken over the whole of space.
Retarded potentials calculated in this way are very
important in the Classical theory of electrons.
89. Examples
(1) Aii observer who is at rest notices that the frequency
of a car appears to drop from 272 to 256 per second as the
car passes him. Show that the speed of the car is
approximately 20 m.p.h. How fast must he travel in the
direction of the car for the apparent frequency to rise to
280 per second, and what would it drop to in that case ?
(2) Show that in the Dopplor effect, when the source and
observer are not moving in the same direction, the formula)
of 80 are valid to give the various changes in frequency,
provided that u and v denote, not the actual velocities, but
the components of the two velocities along the direction in
which the waves reach the observer.
(3) The amplitude A of a harmonic wave A cos 27r(nt kx)
is modulated so that A a\~b cos 2?rpt{c cos 2 2irpt. Show
that combination tones of frequencies np, n2p appear,
and calculate their partial amplitudes.
(4) The dielectric constant of a certain gas varies with the
wavelength according to the law e = A\B/\ 2 (7A 2 , where
A, B and C are constants. Show that the group velocity U
of electromagnetic waves is given in. terms of the wave velocity
V by the formula
(5) In a region of anomalous dispersion ( 83) the dielectric
A A 2
constant obeys the approximate law e = 1f . A more
A AQ
152 WAVES
j4A 2 (A 2 _ A 2 )
accurate expression is e = 1 H     ~ , where A, B and
(A AQ ) jx>A
A are constants. Find the group velocity of electric waves
in these two cases.
(6) Calculate the group velocity for ripples on an infinitely
deep lake. ( 55, equation (54).)
(7) Investigate the motion of a wavepacket ( 84) for
which the amplitude a is given in terms of the wave number k
by the relation
a(k) = 1 if \kkfl <kt
otherwise,
k and k l being constants. Assume that only the first two
terms of the Taylor expansion of n in terms of k are required.
Show that at time t the disturbance is
=
TT(X at)
where a = (dn/dk) . Verify that the wavepacket moves as
a whole with the velocity a.
(8) Show that when dS is normal to the incident light
( 86), the inclination factor is  . Plot this function
2i
against 6, and thus show that each little element dS of a
wave gives zero amplitude immediately behind the direction
of wave motion. Using the fact that the energy is proportional
to the square of the amplitude of <f> t show that each small
element sends out 7/8 of its energy forwards in front of the
wave, and only 1/8 backwards.
(9) A plane wave falls normally on a small circular
aperture of radius b. Discuss the pattern observed at a large
distance / behind the aperture. Show that with the formulae
of 87, if the incident wave is ^ = a cos m(ct z), then
S = 0, and if P is the point (x, 0, /), then
C I \/(^ 2 "~ 2 ) cos Pfdg where p =
= p f cos (pb cos 0) sin 2 dO.
A / Jo
GENERAL CONSIDERATIONS 153
Expand cos (pb cos 0) in a power series in cos 0, and hence
show that

~ A/ \
where & = pb/2 = nbx/Xf. Since the system is symmetrical
around the z axis, this gives the disturbance at any point
in the plane z =/. It can be shown that the infinite series
is in fact a Bessel function of order unity. It gives rise to
diffraction rings of diminishing intensity for large values of x.
(10) The total charge q on a conducting sphere of radius a
is made to vary so that q = 47ra 2 cr, where a = for <0, and
a = cr sin pt for t>0. Show that if = ^ = 1, ( 70 eq. (18'))
the electric potential ^ at a distance E from the centre of the
sphere is given by
ct<Ra, (f> = 0,
Ra<ct<R+a 9 </> = ^^{lcosp(t 
#/t i \ c
47racor ?a / JR\
R+a<ct, <f> =  sin siiipU  ).
p,R c \ c/
[ANSWERS : 1. c/34, where c = vel. of sound, 2485 ;
3. a+c, 6/2, c/4;
i, A7 iv
where v = wave v oclty;
6. U  JF.]
INDEX
The Numbers refer to pages
Absorption coefficient, 121
Adiabatic, 87
Advanced potentials, 142
times, 142
Ampere's Rule, 104
Amplitude, 2
modulation, 132
Amplitudes, partial, 33
Anomalous dispersion, 134
Antinodes, 6
Bars and springs, longitudinal
waves in, 5159
Basins, tides in, 66, 70, 7476
Beats, 131
frequency, 131
Bell, vibrations of, 49
Boundary conditions, 1, 27, 30,
38, 52, 69, 73, 82, 90, 94
Brewster's angle, 117
Capillary waves, 8184
Chladni's figures, 48
Circularly polarised light, 111
Combination tones, 132
Compressible fluid, 87
Condensation, 89
Conductivity, 103
Conical pipe, sound waves in, 96
Constant of separation, 9
Constitutive relations, 104
Coordinates, normal, 36, 47, 53
D'Alembert, 7
Damping, 15, 39
Decay, modulus of, 15
Degenerate vibrations, 46
Dielectric displacement, 102
Diffraction of light, 145
theory, Fraunhofer, 148
Dispersion, 132
anomalous, 134
154
Displacement current, Maxwell's,
104, 119, 122
Doppler effect, 129
Drude, 119
Electric and magnetic field
strengths, 102
waves, 102127
Elliptically polarised light, 111
Energy, kinetic, 23, 33, 47, 54,
78, 96
loss of, 124
potential, 24, 33, 47, 54, 78, 96
rate transmitted, 79, 111, 133
Equation of telegraphy, 15
wave motion, 120, 5
wave motion, complex solu
tions, 1617
Exponential horn, 99
Field strengths, electric and
magnetic, 102
Fraunhofer diffraction theory,
148
Free surface, 62, 82
Frequency, 3
Fresnol's principle, 144
Fundamental, 35, 49, 58, 91
Gaussian wave packet, 135
Gauss' theorem, 104
General considerations, 128151
Ground note, 35
Group velocity, 81, 133, 135,
137, 138
Hagen and Rubens, 124
Halfwidth, 137
Harmonic wave, 2, 1617
Horn, 96, 99
Huygen's Principle, 141, 144
Inclination factor, 144
Incompressible liquid, 60
INDEX
155
Index, refractive, 106, 118
Induction, magnetic, 102
Intensity, 148
Internal or total reflection, 118
Isothermal, 87
Joule heat, 124
Kinetic energy in bars, f>4
liquids, 78
membranes, 47
sound, 96
strings, 23, 33
Kirchhoff, 141
Lenz's law of induction, 104
Light, velocity of, 106
Liquids, waves in, 6086
Longitudinal waves, 21
in bars and springs, 5159, 87
Long waves in shallow water, 62
Lowest frequency, 35
Magnetic and electric tield
strengths, 102
Maxwell's displacement current,
104, 119, 122
equations, 102
relation, 106
Membranes, waves in, 4350
Mersenne's law, 35
Mode, normal, 30, 33, 35, 37,
39, 45, 48, 55, 91, 95
Modulation, amplitude, 132
Modulus of decay, 15
Nodal planes, 6
Nodes, 6
Nonconducting media, 105
Normal coordinates, 36, 47, 53
Normal modes in bars, 55
circular membranes, 48
rectangular membranes, 48
sound waves in pipes, 91, 95
strings, 3033, 35, 37, 39
Observer, moving, 129
Organ pipe, 91
Overtones, 35, 49
Packet, wave, 135
Partial amplitudes, 33
Paths of particles, 70, 77
Period, 3
equation, 39, 77, 96
Phase, 4, 16, 17
Pipes, sound waves in, 9096
Pitch, 35
Plane of polarisation, 1 1 1
polarised light, 111
nodal, 6
wave, 4
Polarisation, piano of. 111
Polarising angle, 1 1 7
Potential, advanced, 112
electric, 106, 108
energy in bars, 54
oiiorgy in liquids, 78
energy in rnembraiio;, 47
onergy in sound, 96
energy in strings, 24, 33
magnetic or vector, 106, 108
retarded, 142, 151
velocity, 60, 72, 87, 92
Poyiit ing vector, 104, 112, 124
Pressure, radiation, 124
Principle of superposition, 5, 130,
135, 138
Profile, wave, 2
Progressive waves, 6, 13, 23, 26,
28, 30, 40, 66, 71, 74, 77,
83, 95, 100, 109125
Reduction to a steady wave, 40,
71
Reflection coefficient, 27, 117,
123
of light waves, 113
total or internal, 118
Refraction of light waves, 113
Refractive index, 106, 118
complex, 120
Relaxation, time of, 109
Resistance, specific, 103
Retarded potential, 142, 151
time, 141
Ripples, 83
Screen, 145, 146
Separation constant, 9
Skin effect, 125
156
INDEX
Snell's law, 115
Sound, velocity of, 89
waves, 87101
Source, moving, 128
Springs and bars, longitudinal
waves in, 5159
vibration of, 55
Stationary waves, 6, 32, 38, 45,
48, 53, 75, 95
Strings, normal modes, 3 1
waves on, 2142
St Venant's Principle, 145
Superposition, principle of, 5,
130, 135, 138
Surface, free, 62, 82
tension, 63, 81
waves in liquids, 63, 7281
Telegraphy, equation of, 15, 119
Tidal waves, 62, 6372
Time of relaxation, 109
Tone, 35
combination, 132
Total or internal reflection, 118
Transmission coefficient, 28, 1 17,
123
Transverse waves, 21, 44, 109
Vector, Poynting, 104, 112, 124
Velocity, group, 81, 133, 135,
137, 138
of light, 106
Velocity, of sound, 89
potential, 60, 72, 87, 92
wave, 132
Vibrations, degenerate, 46
Wave, capillary, 8184
electric, 102127
harmonic, 2, 1617
in bars and springs, 5159
in liquids, 6086
in membranes, 4350
long, in shallow water, 62
longitudinal, 21, 5159, 87
motion, equation of, 120, <
number, 3
on strings, 2142
packet, 135
plane, 4
profile, 2
progressive,
30, 40, '
45, 4S,
6, 13, 23, 26, 28,
71, 74, 77, 83,
95, 100, 109125
reduction to a steady, 40, 7 1
sound, 87101
stationary, 6, 32,
53, 75/95
surface, 63, 7281
tidal, 62, 6372
transverse, 21, 44,
velocity, 132
Wavefront, 4
Wavelength. 3
109
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