THE LIBRARY
OF
THE UNIVERSITY
OF CALIFORNIA
LOS ANGELES
SOUTHERN BRANUn,
UNIVERSITY OF CALIFORNIA,
LIBRARY,
ANGELES. CALIF.
A TREATISE ON
ANALYTICAL STATICS
CAMBRIDGE UNIVERSITY PRESS
C. F. CLAY, MANAGER
LONDON : FETTER LANE, E.G. 4
NEW YORK : THE MACMILLAN CO.
BOMBAY )
CALCUTTA I MACMILLAN AND CO., LTD.
MADRAS )
TORONTO : THE MACMILLAN CO. OF
CANADA, LTD.
TOKYO : MARUZEN-KABUSHIKI-KAISHA
ALL RIGHTS RESERVED
A TEEATISE ON
ANALYTICAL STATICS
WITH ILLUSTRATIONS TAKEN FROM THE THEORIES OF
ELECTRICITY AND MAGNETISM
BY
EDWARD JOHN ROUTH, Sc.D., LL.D., M.A., F.R.S., ETC.
VOLUME II
CAMBRIDGE:
AT THE UNIVERSITY PRESa
1922
62220
First Edition 1892.
Meprinted 1902, 1908, 1922
Matnematical
Sciences
Library
PREFACE.
IN the first edition of this treatise the subject of Attractions
was presented only in its gravitational aspect. This limitation
was formerly customary, when electricity was less studied than
now, but the result has become somewhat unsatisfactory. When
lecturing on the subject the Author found that some of the most
striking examples of Attraction were those derived from the
theory of electricity. While it was impossible wholly to pass
these over, it appeared that the interest in them was sensibly
diminished if they were discussed without explanations of their
meaning. Examples on the attractions of thin layers of matter,
subject to what appeared to be arbitrary laws, seemed to have
no real applications.
For these reasons a selection has been made of those pro-
positions in Magnetism and Electricity which appeared most
forcibly to illustrate the theory of Attraction. These have been
joined together, with brief introductions, so as to form a con-
tinuous story which could be understood without reference to
any other book.
These illustrations have been so far separated from the rest
of the volume that any portion of them may be omitted by a
reader who desires to confine his attention chiefly to gravitational
problems.
Some theorems, which it was not deemed expedient to include
in the text, have been shortly discussed in the notes at the end
of the volume. These are not always closely connected with the
theory of attractions, yet, being natural developments of the text,
will probably assist the reader in following the argument
VI PREFACE.
The general arrangement of the gravitational part of "At-
traction" has been only slightly altered. New theorems have,
however, been introduced and the demonstrations of some of the
old ones simplified.
The second part of this volume is on the stretching and
bending of rods. The investigation of the stretching, and
consequent thinning, of a rod is founded on Hooke's law. The
fact that (with certain restrictions) the stress couple is pro-
portional to the bending is assumed as an experimental result
and applied to determine the bending of rods and springs under
various circumstances. The problem, when put into this form,
is properly included in a treatise on Statics. Although this
chapter is not a treatise on the theory of Elasticity, it did not
seem proper wholly to omit the theoretical considerations by
which the truth of the fundamental law is confirmed. Accordingly
some simple examples which had been briefly discussed in the
last edition have been retained.
The theory of Astatics occupies the third part of this volume.
It was discussed with sufficient fulness in the first edition and
only very slight alterations have now been made.
A separate index to each of the three chapters has been given.
So many results are included under the head of Attraction that
it was found impossible to mention them all without unduly
lengthening the list. It was also necessary to classify some
theorems only under one heading.
Finally, I desire to express my thanks to Mr J. D. H. Dickson
of Peterhouse for the very great assistance he has given me in
correcting most of the proof-sheets and for his many valuable
suggestions.
EDWARD J. ROUTH.
PETERHOUSE,
December, 1901.
CONTENTS.
ATTRACTIONS.
Introductory remarks.
ARTS.
PAGES
1—6.
Law of attraction, units and magnitude ....
1—3
7—9.
Law of the direct distance
3—4
Attraction of Rods, Discs, &c.
10—13.
Components and resultant attraction
4—6
14.
Attraction of an infinite cylinder
6
15—16.
Examples and theorems on rods
6—8
17—19.
Curvilinear rods. Examples and theorems ....
8—9
20.
Some inverse problems
9—10
21—23.
Circular disc. Infinite plate. Young's rule ....
10—12
24.
Attraction of cylinders
12—13
25—30.
Attraction of a lamina. Playfair's rule
13—16
31—32.
Solid of greatest attraction
16—17
33—38.
Attraction of mountains; density of the earth . . .
17—19
The Potential.
39 — 43.
20—22
44.
Work and potential
22—23
45—48.
Level surfaces, various theorems
23—24
49—51.
Potentials of rods. Eectilinear lamina, with confocal level
24—26
52.
26
53-54.
Discs and cylinders
27
55—57.
Attraction of a homogeneous cylindrical shell at (1) an internal,
27 — 29
58.
29
59—63.
30—33
Vlll
CONTENTS.
Spherical Surface.
ABT8.
PAGES
64—67.
Potential and attraction of a thin shell ....
33—35
68—72.
Ellipsoidal and cylindrical elliptic shells ....
35—37
73—76.
Various theorems
37—38
77—78.
Solid sphere
38—39
79.
39—40
80.
Other laws of force
40—41
81—83.
Eccentric shells
41—42
84.
Gauss and Poisson on mean potential
42
85 — 88.
43 — 44
89—91.
Find the law of force that the attraction may be the same as
that of a particle
44 — 45
92—93.
Method of differentiation
45^6
94.
Similar solids .........
47
Laplace's, Poisson's and Gauss' Theorems.
95 -98.
Laplace's theorem. Jellett's theorem, &c
47—49
99.
Potential constant through a finite space (see Art. 139)
49
100.
Invariant property of Laplace's operator ....
49
101—104.
Potential at internal points (see Art. 146) ....
49—51
105.
Poisson's theorem
51—52
106—107.
Gauss' theorem, jFd<r=4rM
52—54
108—110.
Polar, cylindrical, orthogonal and elliptic coordinates. .
54—56
Theorems on the Potential.
HI H4.
56 — 58
115—117.
Potential constant through any space
58—59
118—121.
Points of equilibrium. Stability. Separating cone
59—61
122—125.
Intersections of a level surface. Eankine's theorem .
61—62
126—128.
Tubes of force, &c. Theorems. Note A, page 356
62—63
129—132.
Bodies with equal potentials ......
63—65
133.
Extensions of these theorems
65—66
134.
To trace level curves and lines of force ....
66
135.
Potential at a distant point. MacCullagh's theorem .
66-67
136.
Theorem on equipotential bodies
67—68
137—138.
Centrobaric bodies
69
139—141.
Potential constant through a finite space (see Art. 99)
69—70
Attraction of a Thin Stratum.
142.
70—72
143.
Attraction of a stratum on an element of itself, X' + X=2F
72—73
144—146.
Changes of force and potential at the surface
73
147.
Green's theorem deduced from Gauss' ....
73—74
148.
Relation between normal force and constant potential
74
CONTENTS.
IX
Green's Theorem.
AKTS.
PAGES
149—153.
A volume integral replaced by a surface integral. Also a
surface integral by a line integral, Note A, page 356
74—77
154—157.
Green's equivalent layer
77—79
158—160.
Points at which V is infinite
79—80
161—162.
Multiple-valued functions
80—81
163.
81 — 82
Given the Potential, find the Body.
164—165.
Solid and superficial matter
82—83
166 — 167.
To find these, see also Art. 299
83 — 84
Method of Inversion.
168—180.
84 — 89
181—183.
Inversion from a line
89—90
184.
Extension of the theory, r'=Arn, 6'=nO
90—92
Circular Rings and Anchor Rings.
185—189.
Potential found by the use of a conic ....
92—93
190—191.
Law of the inverse /cth power. See Note B, page 358 .
93—96
192—193.
Anchor rings
96—97
Attraction of Ellipsoids.
194—195.
Two kinds of sheila
97—98
196—200.
Homoeoid, internal point. The integrals I and <7 .
98—101
201—207.
Thin homoeoids, attractions and lines of force
102—105
208—210.
Homoeoid, external point, F=Mp'ja'b'c', &c. .
105—106
211—215.
Solid homogeneous ellipsoid. Note C, page 358
106—108
216—217.
Level surfaces
108—109
218.
Other laws of force. Note D, page 359 ....
109—110
219—221.
Spheroids. Nearly spherical ellipsoids ....
110—111
222—223.
Ivory's theorem
111—112
224—227.
Maclaurin's theorem, &c
112—114
228.
114 — 115
229.
Mutual attraction ........
115
230.
115
231.
Ellipsoidal quadratic layers. Heterogeneous shells .
116—117
232—238.
Elliptic cylinders, attraction and potential
117—120
239—244.
Heterogeneous ellipsoid, similar strata ....
120—122
245—250.
Ellipsoids with any law of density. Note E, page 360 .
122—128
251—254.
Elliptic discs. Confocal level surfaces, &o.
128—130
CONTENTS.
ABT9.
255—261.
262.
Rectilinear Figures.
Potential of a lamina found in terms of those of the sides
and that of a solid iu terms of those of the faces .
Application to a tetrahedron, &c
130—133
133
Laplace's Functions and Spherical Harmonics.
263—277. Legendre's functions. Elementary theorems. Note F,
page 360 134—138
278—282. The integrals J/(p)Pn<fy>=0, /Pn2dp = 2/(2» + l). See
also Note G, page 361 138—140
283. Expression for the potential of a body .... 140
284 — 287. Laplace's second equation . . . . . . . 110 — 142
288—291. Three fundamental theorems. See Note H, page 362 . 142—145
292 — 293. Examples and theorems 145 — 146
294 — 296. Heterogeneous spherical strata, &c 146 — 148
297—298. Nearly spherical bodies. See Art. 221 . . . 148—150
299. Potential found when given on spheres .... 150 — 151
300—302. Solid of revolution 151—153
303. Potential of a ring, &c. Oblate spheroid . . . 153—154
304 — 309. Clairaut's theorem, first and second approximations . 154 — 159
310. Figure of Saturn acted on by the ring .... 159 — 160
311 — 313. Jacobi's ellipsoid. Nearly spherical earth and Haughton's
problem on a fluid nucleus 160 — 162
Magnetic Attractions.
314 — 322. Potentials, couples, forces and energy .... 162 — 167
323. Lines of force of a rectilinear row of particles and of
parallel rods in a plane 168 — 169
324—325. Examples and applications 169—170
326 — 332. Magnetic body. Elementary rule, rod, sphere, ellipsoid,
cylinder. Note I, page 363 170—174
333. Infinite cylinder in a field of force 174—175
335—338. Terrestrial magnetism 175 — 177
339 — 341. Poisson's general theorem 177—178
342 — 345. Magnetic force and Magnetic induction. See Notes K, L,
page 363 178—180
346—349. Solenoids 180—181
350 — 362. Lamellar shells. Potential force, mutual energy. Circular
shells, &c. See Note M on vector potential, page 364 181 — 187
Electrical Attractions.
363 — 375. Introductory statement and elementary examples . . 187 — 193
376—385. Ellipsoids. Quantity on a portion of the surface. Elliptic
discs, potentials, &c. ....... 193 — 196
CONTENTS. XI
ARTS. PAGES
386—392. Conductor with a cavity. Screens 196—198
393 — 396. Two methods of solution. Green's theorem . . . 198 — 201
397 — 406. Sphere acted on by a point charge. Lines of force, &c.
Quantity on a portion of the sphere and its potential.
See Note N, page 365. Electrified ring. See Art. 422 201—207
407—411. Cylinders with parallel axes 207—210
412 — 416. Plane conductors. Intersecting at right angles or at an
angle ir/n 210—212
417—419. Condensers. Electric cables 212—215
420—421. Nearly spherical bodies, &c. Point charge . . . 215—219
422. Sphere with electrified ring. See Art. 406 ... 219
423 — 432. Two orthogonal spheres. Capacity 219—224
433—437. Spheres intersecting at irfn. Three orthogonal . . 224 — 227
438 — 450. Theory of a system of conductors. Energy. Junction and
introduction of conductors 227—233
451 — 455. Circular disc with point charge 233 — 236
456—458. Spherical bowl 236—238
459—464. Two separate spheres 238—241
Magnetic Induction.
465 — 472. Induced magnetism. The boundary condition . . . 242—245
473 — 486. Specific Inductive Capacity. Dielectrics. Kelvin's theorem.
Problems on Condensers and on Induction . . . 245 — 258
487. Magnetic shells 258
488 — 491. Surface integral of magnetic induction .... 258 — 260
492—494. Gauss' and Poisson's theorems applied to dielectrics . 260—261
495. Potential of an electric system ...... 261 — 262
THE BENDING OF RODS.
Introductory Remarks.
1 — 4. Definition of central line, thickness of rod, extent of defor-
mation, <fcc. 263—264
•
The Stretching of Rods.
5. Hooke's law 264
6. Isotropic and aeolotropic bodies 264
7—9. Theory of a stretched rod. Change of sectional area . 265 — 266
The Bending of Rods.
10 — 12. Equations of equilibrium. Two methods
13. The additional experimental law. Flexural rigidity
267—269
269—270
Xll
CONTENTS.
ARTS. PAGES
14 — 15. Limits to the law of elasticity 270
16. Work of bending an element 270 — 271
17—18. Deflection of a straight heavy rod 271 — 274
19 — 20. Equation of the three moments 274—275
21. General method of solving problems. Yielding . . . 275—277
22. Examples of bent rods. Determination of the constants of
bending by experiment 277 — 278
23. Problems on the Britannia Bridge . . . . . 278
24. Supported rods with weights. Relation of the pressures at
consecutive points of support. Also the deflection . 278 — 279
25. How 2m + 1 supports should be arranged that the pressures
of a heavy beam may be equal 279 — 280
27 — 32. A bent bow. Euler's theorem on the strength of columns.
Greenhill's problems on columns 281 — 285
33—35. Theory of a bent rod. The displacements of the fibres . 285—288
86. Airy's problem on the proper method of supporting stand-
ards of length 288 — 289
37 — 39. Bending of circular rods ; inextensible and extensible . 289 — 291
40 — 42. Work of bending and stretching a circular rod . . . 291 — 292
43—44. Very flexible rods 292 — 294
Rods in Three Dimensions.
45. Measures of twist 294 295
46 — 50. Resolved curvature 295 297
51. Relations of stress to strain 297 — 298
52 — 53. Work of bending 298 299
54. Helical twisted rods 299 301
55 — 56. Spiral springs 301 — 303
57 — 59. Equations of equilibrium 303 — 306
ASIATICS.
1—2.
3.
4—5.
6—7.
8.
A static Couples.
Definitions of the astatic 'arm, the astatic angle. History . 307—308
A couple may be moved parallel to itself .... 308
Resultant of two couples, (1) forces parallel, (2) arms parallel 308 — 309
Working rule 309—310
Three couples cannot in general be in astatic equilibrium . 310 — 311
The Central Ellipsoid.
9. Reduction to a force and three couples
10. Astatic notation. The twelve elements.
11 — 12. Conditions of equilibrium, &c.
See Art. 59
311—312
312—313
313—314
CONTENTS.
Xlll
ARTS.
13.
14—17.
18.
19—20.
21.
22.
If a system be equivalent to three forces their points of
application lie in a fixed plane
The central ellipsoid
Principal axes at any base. The principal couples
The initial positions
There are only four positions in which the forces are equi-
valent to a single resultant passing through the base .
Other ellipsoids
PAGES
314—315
315—317
318
318—319
319—320
320—321
The Central Plane and the Central Point.
23. The central ellipsoids at different points compared . . 321
24. If .8 = 0, they are all similar 321
25 — 26. When the central ellipsoid is a cylinder the locus of the
base is the central plane 321 — 322
27. Eeduction to a force and two couples 323
28 — 29. The central point. The arms of the two couples are perpen-
dicular ; the forces are perpendicular to each other and
to E 323—324
30. Working rule to find the central point .... 324 — 325
31. Equation to the central plane expressed in terms of the
forces and their mutual inclinations .... 325
32. Summary of reductions 325 — 326
33. Analogy to moments of inertia 326 — 327
The Confocals.
34. Arrangement of central ellipsoids about the central point . 327 — 328
35. The principal axes are the normals to the three confocals . 328 — 329
36. The focal conies and the focal lines 329 — 330
37. Theorems on focal lines. Distance of a focal line from the
centre. Four real focal lines can be drawn through a
given point or parallel to a given line. Locus of focal
lines passing through a point on a focal conic is a right
cone. Locus of the base when the central ellipsoid is a
surface of revolution. Other theorems .... 330 331
Arrangement of Poinsot's Central Axes.
38 — 40. When the body is rotated about R, Poinsot's axis is fixed
in space and describes a right cylinder in the body.
To so place the body that, when possible, a given straight
line is a Poinsot's axis 331 — 335
41—42. The couple-moment follows the law F = T0 sin (0 - <f>0) . 335—336
43. Inversion of the body 336—337
44. Equation to Poinsot's axis referred to the principal axes at *
the central point. Minding's theorem. See Art. 48 . 337 — 338
45. To find the axes when the couple-moment is given.
Another construction on Poinsot's axis .... 338 — 339
XIV
CONTENTS.
ABTS.
PAGES
46—47.
Cylinders which intersect the circular cylinder in four
Poinsot's axes of given moment
339—340
48—50.
Minding's theorem. See Art. 44
340—341
51—54.
Relations of Poinsot's axes to the confocals
341—342
Reduction to Three and to Four Forces.
55—56.
Reduction to three forces .......
342—344
57—58.
Astatic triangles lie on the central plane.
Hence a rule to find the central plane ....
344—345
59.
Interpretation of a column of the twelve elements
345—346
60—62.
Beduction to four forces
346—347
63.
347—348
64.
The imaginary focal conic
348
65.
348 — 349
66—69.
Various theorems
349—350
70—71.
Transformation of astatic tetrahedrons ....
350—351
72.
A system cannot in general be reduced to fewer than three
forces .
351
73—77.
Conditions that it can be reduced to two ....
352—354
78.
The invariants of two forces
354—355
79.
Conditions that a system can be reduced to a single force.
355
NOTES.
'
A. A surface integral replaced by a line integral
356—358
B. Potential of a thin circular ring . . .
358
C. Attraction of a solid ellipsoid ; another proof
358—359
D. Potential of a homoeoid for any law of force
359
E. Heterogeneous ellipsoids
360
F. The expansion of the potential for other laws of force
360—361
G. Theorems on Legendre's functions ....
361—362
H. Laplace's equation $YmYndu=Q
362-363
I. A magnetic sphere
363
E. Magnetic forces
363
363—364
M. The vector potential
364—365
N. Figure showing the distribution of electricity on a sphere
365—366
366
ATTRACTIONS.
Introductory remarks.
1. Law of attraction. If two particles of matter are
placed at any sensible distance apart, they attract each other with
a force which is directly proportional to the product of their
masses and inversely proportional to the square of the distance.
Let m, mf be the masses of two particles, r their distance
apart ; if F be the mutual attraction which each exerts upon the
ffl/Yfif
other, then F is given by the equation F = K — — .
If f be the acceleration produced by the attraction of m at the
ffn
distance r, then f=x—.
The quantity K is called the constant of attraction. Its magni-
tude depends on the particular units in which the masses m, m,
the distance r and the force F are measured. To avoid the
continual recurrence of this constant running through every
equation, it is usual to so choose the units that K = 1. When this
is done the units are called theoretical or astronomical units.
Putting K = 1 in the equations, we see that when m and r are
both unity the acceleration / is also unity. We infer that the
astronomical unit of mass is that mass which, when collected into
a particle, produces by its attraction at a unit of distance the unit
of acceleration. The expression for F shows that the unit of force
is the attraction which a particle whose mass is the astronomical
unit of mass exerts on an equal particle at a unit of distance.
To avoid the continual repetition of the same set of words, we
shall use the phrase attraction at a point to mean the attraction
on a unit of mass collected into a particle and placed at that point.
E. s. ii. 1
2 ATTRACTIONS. [ART. 3
It is convenient to use different systems of units for different purposes. The
astronomical units should be used in analytical investigations. In any numerical
applications we may choose such units of space and time as we may find convenient,
and then introduce into our formulae the factor K with its appropriate value.
It may be noticed that in using different units for different purposes we are
following the analogy of other mathematical sciences. In practical trigonometry
we measure angles in degrees, in theoretical trigonometry we adopt that unit by
which our analytical formula; are most simplified. Also in algebra we have one
base in logarithms for use in calculations and another for theoretical investigations ;
and so on through all the sciences.
2. Numerical estimate. To obtain a numerical estimate
of the magnitude of the force of attraction, we must determine by
experiment the mutual attraction of some two bodies. We may
exhibit the result in either of two forms : (1) we may determine
the value of K when the units of space, mass, &c. have been
chosen ; (2) we may determine the magnitude of the astronomical
unit of mass by expressing it as a multiple of some other known
mass.
The two bodies on which the experiment should be tried are
obviously the earth and some body at its surface. Regarding the
earth as a sphere, whose strata of equal density are concentric
spheres, it will be shown further on that its attraction on all
external bodies is the same as if its whole mass were collected
into a particle and placed at its centre. If then ra be the mass
of the earth and a its radius, the acceleration of a body at its
surface is /era/a2. Let g be the acceleration actually produced by
the attraction of the earth on any body placed at its surface. We
thus form the equation /cm/a2 = g.
Several experiments have been made to determine the mean
density of the earth. One of these is the Cavendish experiment,
but there have been others conducted on different plans. The
result is that the mean density has been variously estimated to
be from 5£ to 6 times that of water. According to Baily's
repetition of the Cavendish experiment the ratio is 5'67. Repre-
senting this ratio by /3, we learn that the attraction of a sphere
of water, of the same size as that of the earth, will produce in a
body, placed at its surface, an acceleration equal to
3. To find the value of K when the units of space, mass, and time are the
centimetre, the gramme and the second. Since the mass of a cubic centimetre of
water is one gramme nearly, the mass TO of a sphere of water of the same size as
the earth is fa-a3 grammes, where the radius a is measured in centimetres. By
°on
ART. 7] INTRODUCTORY REMARKS. 3
the experiment just described -j s"C; taking /3=5'67, 0 = 981 (see Vol. i. Art. 11),
Cl O
we find K=
' 4ira/3 1543 x 104 '
If therefore the attracting masses are measured in grammes and the distances
in centimetres, the expression for F with this value of K gives the attraction in
dynes.
Let m be the mass, measured in grammes, of a particle which produces by its
attraction at the distance of one centimetre a unit of acceleration. Then m is the
astronomical unit of mass. The formula /=/ow/r2 gives l=/c?n, .-.m=:1543 x 104
grammes.
Let F be the force measured in dynes which one astronomical unit of mass
exerts on another at the distance of one centimetre. The formula F=Kmm'[r*
gives F=1/K since m=m' and m/c=l. The force F is 1543 x 104 dynes.
4. To find the value of n when the units of space and time are the foot and the
second, and those of mass and force are the pound and the poundal. Since the weight
of a cubic foot of water is the same as that of 7=6! pounds nearly, the mass m of a
sphere of water of the same size as the earth is $irasy, where the radius a is measured
in feet. By the experiment just described ^ = ~ . If we take a =20926000 feet
Qi J3
this gi
If therefore the attracting masses are measured in pounds and the distances in
feet, the expression for F with this value of K gives the attraction in poundals.
The astronomical unit of mass, when the foot and the second are the units of
space and time, is 93 x 107 pounds and the astronomical unit of force is 93 x 107
poundals. A poundal is roughly equal to the weight of half an ounce. See Vol. i.
Art. 11.
6. Dimensions of K and m. When the unit of mass is arbitrarily chosen the
attraction F of a particle of mass m on a particle of equal mass is F=Km2/r2. It
follows that the dimensions of K are the same as FL2/j.~2 or L*fj.~lt~2 where F, L, p, t
stand for force, length, mass, and time. When the factor K is omitted the dimen-
sions of astronomical mass include those of K and become the same as those of
fj,K^ or, which is the same thing, F^L or I/^/x^t"1. This also follows at once from
the formula .F=m2/r2. These dimensions are the same as those of the electrostatic
measure of electricity. See Maxwell's Electricity, Arts. 41, 42.
6. Ex. 1. Prove that the mass of the particle which at the distance of one
centimetre from a particle of equal mass attracts it with the force of one dyne is
3928 grammes. Everett's Units and Physical Constants.
Ex. 2. Show that a cubic foot of water, collected into a particle, attracts an
equal particle placed at the distance of one foot with a force equal to the weight of
l/(8 x 10«) pounds.
7. Law of the direct distance. There are other laws
besides that of the inverse square which may govern the attraction
of bodies in special cases. Some of these will be mentioned as we
proceed. But the most useful is that in which the attraction
» 1—2
4 ATTRACTIONS. [ART. 10
varies as the distance. In this case the attraction of two
particles, each on the other, is represented by F = mm'r, where
m, m' are their masses, and r, the distance between them.
8. When the attraction obeys the law of the direct distance,
the resultant attraction of any body at any point is found at
once by using Art. 51 of Vol. I. Let 0 be any point, Al} A2,
&c. the positions of the attracting particles; let ra^ ra2, &c. be
their masses. The component attractions at 0 are then given by
X = 2ma? = x£m, Y=y"£m, Z=z'£m, where x, y, z are the
coordinates of the centre of gravity of the body or system of
attracting points.
It immediately follows that the resultant attraction at 0 is the
same as if the whole mass 2<m of the attracting system were
collected into a single particle placed at the centre of gravity.
The resultant force on a particle at 0 tends therefore towards the
centre of gravity of the attracting system, and is proportional to the
distance of the attracted point from it.
9. In what follows, when no special law of force is mentioned,
it is to be understood that the law meant is that of the inverse
square. This is often called the Newtonian law.
When the law of attraction is said to be f(r\ it is meant that
the mutual attraction of two particles whose masses are mt m'
placed at a distance apart equal to r is mm'f(r).
Attraction of rods, discs, &c.
10. Attraction of a rod. To find the attraction of a
uniform thin straight rod AB at any external point P.
Let 77i be the mass of a unit of length, then m is called the
line density of the rod. Let p be the length of the perpendicular
PN from P on the rod. Let QQ' be any element of the rod,
NQ = tK', let also the angle NPQ=6, then #=j)tan#.
The attraction at P of the element QQ' is
mdx _ md (p tan 0) _ md&
~PQ?~ <j>sec<9)2 ''~p~'
Let X, T be the resolved attractions at P parallel and perpen-
dicular to the length AB. Let the angles NPA, NPB be a, /3,
then
v f dO . _ m ,
}L — \m — sm 6 ==• — (cos a — cos a) ......... (1),
J p p^
ART. 13]
A STRAIGHT ROD.
~r f dd a m s • a • \ /n\
Y— Im — cos 6 = — (sm 8 — sin a) (2).
J p P
JS Q' Q A
11. Substitute for cos a, cos (3 their values obtained from
the triangles PNA, PNB ; the resolved attraction parallel to the
rod takes the useful form X = ^-7 — ^TFT (3).
PA PB
It should be noticed that this is the attraction at P of the rod
AB resolved in the direction from A towards B.
12. Describe a circle with centre P and radius PN and let
the portion CD included between the distances PA, PB represent
a thin circular rod of the same material and section as the given
rod AB.
The attraction at P of the element RR' of the circular rod is
f m . RR' pdO dd _ , . , . ,
therefore — pp2 - = m±—^- — m, — . But this has just been proved
Jt ,/t jp jp
to be the same as the attraction of the element QQ'. Thus each
element of the rod AB attracts P with the same force as the
corresponding element of the rod CD. The resultant attraction of
the straight rod AB is therefore the same in direction and magnitude
as that of the circular rod CD.
13. The resultant attraction at P of the circular rod CD
must clearly bisect the angle CPD. It
immediately follows that the direction of
the resultant attraction at P of a straight
rod AB bisects the angle APB.
To find the magnitude of the resultant
attraction at P of the circular arc CD, we
draw PE bisecting the angle CPD. Let
the angle any radius PR makes with PE
be A/T. Let 2y be the angle CPD. Since
RR' = pd^r the attraction of the whole
circular arc when resolved along PE is
6 ATTRACTIONS. [ART. 15
I 3 cos ^ = — • 2 sin 7, the limits of the integral being ^ = — 7
and i/r = 7. TAe magnitude F of the resultant attraction at P of
2m . APB
a straight rod AB is given by F= — sin — ^ — .
14. When the rod AB is infinite in both directions the angle
APB is equal to two right angles. The resultant attraction of an
infinite rod at any point P is equal to 2m/p, and it acts along the
direction of the perpendicular p drawn from P to the rod.
This proposition leads to a useful rule which helps us to find
the attraction of any cylindrical surface or solid which is infinitely
extended in both directions. We pass a plane through the
attracted point P perpendicular to the generating lines and
cutting the cylinder in a cross section. If the attracting body
be composed of elementary rods of line density m, each of these
attracts P as if a mass 2m were collected into its cross section and
the law of attraction were changed to the inverse distance. The
attraction of the whole cylinder is then equal to that of this cross
section. If the cylinder be solid and of volume density p, the
cross section is an area of surface density 2p ; if the cylinder is a
surface of surface density a; the cross section is a curve of line
density 2<r. The same rule will apply to a heterogeneous cylinder
provided the density along each generator is uniform.
Three laws of attraction are therefore especially useful. These
are (1) the law of the inverse square, (2) that of the inverse
distance, and (3) that of the direct distance.
15. When the point P moves about and comes to the other side of the
attracting rod AB, crossing AB produced but not passing through any portion of
the attracting rod, the components X, Y remain continuous functions of the
coordinates of P, and will continue to represent the component attractions. When
P lies in AB produced Y takes the singular form 0/0, but it is evident that it
changes sign through zero. The resultant attraction is then given by (3) which
is free from singularity.
When P passes through the material of the rod the case is somewhat different.
When P approaches the thin rod, the angles § and a become ultimately \ir and
-\ir, the Y component becomes infinite while X remains finite. The attraction
is therefore ultimately perpendicular to the rod and finally changes sign through
infinity. When P is inside the indefinitely thin rod the Y component is zero by
symmetry and the X component represents the attraction.
In the preceding analysis we have regarded the linear dimensions of the
transverse section of each element QQ' as infinitesimal when compared with the
distance from P. This however is not true for any material rod when P approaches
ART. 16] EXAMPLES ON RODS. 7
very closely to any point of it. The rod (or at least the portion which is near to P)
must then be regarded as a cylindrical solid.
16. Ex. 1. If two forces be applied at P acting along AP, PB taken in order,
and each equal to m/p, prove that their resultant is equal in magnitude to the
attraction of the rod AB and acts in a direction perpendicular to that attraction.
Ex. 2. The sides of a triangle are formed of three thin uniform rods of equal
density. Prove that a particle attracted by the sides is in equilibrium if placed at
the centre of the inscribed circle.
If one side of the triangle repel while the other two attract the particle, prove
that the centre of an escribed circle is a position of equilibrium. [Math. T.]
This follows at once from Art. 12. Draw straight lines from the centre I of the
inscribed circle to the corners A, B, G of the triangle, cutting the circle in A', B', C'.
The attractions of the sides AB, BG, GA are the same as those of the arcs A'B',
B'G', C'A', that is their resultant attraction is the same as that of the whole circle
on the centre. This attraction is clearly zero.
Ex. 3. Four uniform straight rods of equal density form a quadrilateral, and
their lengths are such that the sum of two opposite sides is equal to the sum of the
other two opposite sides. Find the position of equilibrium of a particle under the
attraction of the four sides.
Ex. 4. Every particle of three similar uniform rods of infinite length, lying in
the same plane, attracts with a force varying inversely as the square of the distance ;
prove that a particle will be in equilibrium if it be placed at the centre of gravity of
the triangle ABC enclosed by the rods. [Math. Tripos, 1859.]
The attractions at P are perpendicular to the sides of the triangle and therefore,
when P is in equilibrium, their magnitudes are proportional to those sides. Hence
by Art. 14 the areas APB, BPG, CPA are equal and therefore P is the centre of
gravity.
Ex. 5. A particle is placed at any point P on the bisector of the angle C of a
triangle. Show that the direction of the resultant attraction of the three sides at P
bisects the angle APB and is equal in magnitude to 2m ( ) sin — -— , where a
\7 aJ 2
and 7 are the perpendiculars from P on the sides BG, AB respectively.
Describe a circle centre P to touch the sides AC, BG. The resultant attraction
of these two sides is equal and opposite to that of the arc of the circle which lies
between the straight lines AP, BP on the side remote from G (Art. 12).
Ex. 6. Two uniform parallel straight rods AB, GD attract each other : show
that the components of their mutual attraction, respectively perpendicular and
parallel to the rods, are
mm' BG' + BG AD'+AD
Y= — (BC-BD-AC + AD), Z,wm' tog _—.__,
where C', D' are the projections of G, D on the rod AB, p the distance between the
rods, and m, m! the masses per unit of length.
Ex. 7. P is a particle in the diagonal AC of a square ABCD, and within the
square ; show that the attraction of the perimeter of the square upon P is equal to
OP
M . -p-j — ; where M is the mass of the perimeter, 0 the centre of the square.
[Trin. Coll., 1882.]
Ex. 8. Let the finite rod AB be produced both ways to infinity and let the
portion beyond A attract and the portion beyond B repel P, the portion between A
8 ATTRACTIONS. [ART. 17
and B exerting no force at P. Prove that the resultant force at P bisects the angle
external to APB and is equal to — cos —5— .
Describe a circle, centre P, to touch AB and intersect PA in C and BP produced
in H. The resultant force at P is therefore equal to the attraction of the arc CH .
Art. 12.
Ex. 9. The law of attraction of a uniform thin straight rod is the inverse
»cth power. Prove that the components of attraction at a point P parallel and
perpendicular to the rod are respectively
the latter integral can be found by a formula of reduction in the usual way.
Ex. 10. The law of attraction of a cylinder infinitely extended in both
directions *is the inverse /cth power. Prove that the attraction at a point P is
equal to that of the cross section provided (1) the law of attraction of the section
is the inverse (K - l)th power and (2) ratio of its density to the cylindrical density
is 2 /(cos 0)"-1 d0, the limits being 0 to %ir, (see Art. 14).
17. Curvilinear rods. The method by which the attraction
of the straight rod AB is replaced by that of the circular arc CD
in Art. 12 may be extended to other curves.
Two curvilinear rods AB, CD are so related that if any two
radii vectores OAC, OBD are drawn, the attractions of the inter-
cepted arcs AB, CD at the origin 0 are the same in direction and
magnitude. It is required to find the relation between the densities
of the rods.
Since the attractions are equal for all arcs, they are equal for
infinitesimal arcs. Let OQR, OQ'R'
be two consecutive radii vectores ; ds,
ds' the arcs Qty, RR' ; m, m' the
masses at Q, R per unit of length.
Then if the law of attraction is the
inverse /cth power of the distance we
mds m'ds'
have — — = - , - ,
r* rK
where r = OQ, r' = OR. If </>, <£' be the angles the radius vector
OQR makes with the tangents at Q and R, this gives
m m'
_
r*-1 sin <f> r'*-1 sin $ '
The densities of the curvilinear rods at corresponding points
must therefore be proportional to r*""1 sin <£.
ART. 20] INVERSE PROBLEMS. 9
If the law of attraction be the inverse square, two curvilinear rods in one plane
equally attract the origin, if the densities at corresponding points in the two rods
are proportional to the perpendiculars from the origin on the tangents.
18. If the two curves are so related that each is the inverse of the other, we
have OQ. OR—OQ' . OR'. A circle can therefore be described about the quadri-
lateral QRR'Q'. In the limit when QQ', RR' become tangents this gives
sin0 = sin0'. If also K=l, we see that m=m'. It follows therefore that when
the law of attraction is the inverse distance, any curvilinear rod and its inverse,
if of equal uniform line density, equally attract the origin.
19. Ex. 1. Let the law of attraction be the inverse distance and let P be any
point attracted by a uniform straight rod AB. Draw PN perpendicular to the rod
and describe a circle on PN as diameter. Prove that the attraction of AB at P is
the same as that of the corresponding arc CD of the circle intercepted between the
straight lines PA, PB, if the line densities are equal. Compare Art. 12.
Ex. 2. Two rigid and equal semicircular arcs of matter with uniform section
and density are hinged together at both extremities. The matter attracts according
to the law of gravitation. If equal and opposite forces applied along the line join-
ing the middle points of the semicircles keep them apart with their planes at right
angles, the magnitude of each force will be 4m2 log (1+^2), where m is the mass of
unit length of arc. [Math. Tripos, 1874.]
20. Some inverse problems. Ex. 1. A uniform rod is bent into the form
of a curve such that the direction of the attraction of any arc PQ at the origin 0
bisects the angle POQ. Show that the curve is either a straight line or a circle
whose centre is 0.
The data lead to the differential equation / — sin 6= tan- I -^ cos 0. The
limits of the integrals being 0 and 0. The equation may be solved by differ-
entiation.
Ex. 2. Find the law of density of a curvilinear rod of given form that the
direction of the attraction at a given point 0 of any arc PQ may bisect the angle
POQ. If the law of attraction be the inverse ccth power of the distance, the result
is that the line density m at P must be proportional to pr*~* where r—OP and p is
the perpendicular on the tangent at P.
Draw any circle, centre 0, intersecting OP, OQ in G, D. The attraction of CD
(regarded as a uniform rod) at 0 is by hypothesis the same in direction as that of
PQ and may (by giving CD the proper density) be made the same in magnitude
also. Include the additional elements QQ', DD'. It is clear that unless their
attractions at 0 are equal the attraction of PQ' cannot coincide in direction with
that of CD'. The attractions at 0 of corresponding elements of the two rods are
therefore equal. Hence as in Art. 17 the density m at every point of PQ varies as
p/-""2. The proposition may also be proved analytically as indicated in the last
example.
Ex. 3. A uniform rod is bent into a curve such that the direction of the
attraction at the origin of any arc PQ passes through the centre of gravity of the
arc. Prove that, either the law of attraction is the direct distance, or the curve is a
straight line which passes through the origin.
Ex. 4. If any uniform arc of an equiangular spiral attract a particle placed at
the pole, prove that the resultant attraction acts along the line joining the pole to
the intersection of the tangents at the extremities of the arc.
10
ATTRACTIONS.
[ART. 21
Prove also that if any other given curve possess this same property, the law
of attraction must be F= —„ -/- ,
p*dr '
where p is the perpendicular drawn
from the attracted particle on the
tangent at the point of which the
radius vector is r.
Reversing the attracting forces,
we may regard the rod as acted
on by a 'centre of repulsive force.
Since the resultant force on any
arc PQ acts along OT, where T
is the intersection of the tangents
at P and Q, we may resolve that
force into two components which act along TP and TQ. It follows that the
resultant force on any arc PQ may be balanced by two forces or tensions acting
along the tangents at P and Q.
To complete the analogy of the force at P to a tension, we must show that that
force is always the same whatever the length of the arc PQ may be. To prove
this let PQ, QB be two contiguous arcs, and let the tangents at P, Q meet in T,
those at Q, E in U, those at P, B in F. Resolving the forces at T, U, V as before,
the components along PT, QT and BU, QU must together be equivalent to the
components along PF, BV. We have to deduce from this that the components
along PT and PV are equal. This follows at once by taking moments about U.
The conditions of equilibrium of the rod are therefore the same as those of
a string acted on by a central force. Referring to Art. 474, Vol. i., the tension is
obviously T=A/p and the force f(r) has the value given above. See the Solutions
of the Senate House problems for the year 1860, page 61. The analytical solution
leads to an interesting differential equation which can be solved without great
difficulty.
21. Attraction of a circular disc. To find the attraction
of a uniform thin circular disc at any point in its axis.
Let 0 be the centre, ABA' the disc seen in perspective;
OZ the axis, i.e. a straight line drawn through 0 perpendicular to
the plane of the disc. Let a be the radius of the disc, m the
mass per unit of area, usually called the surface density. Let P
ART. 22] A CIRCULAR DISC. 11
be the point at which the attraction is required, OP = p, and the
angle OP A = a.
Describe an elementary ammhis, represented in the figure by
QQf. Let x, x + da be its radii, and let 6 be the angle OPQ. The
resultant attraction of the disc at P is F= I— jyp- — cos 8,
J ty-L
where the limits of the integral are x = 0 and x — a. Since
#=jptan# and QP—psec0, we find
F = 2?rm/sin ddd = 2?rm (1 — cos a).
Here a is the acute angle subtended at the attracted point by the
radius of the disc.
It appears from this that the attraction of a uniform thin
circular disc at a point P in its axis depends only on the surface
density and on the angle 2a subtended at P by a diameter of the
disc. It will be presently seen that if a> be the solid angle sub-
tended at P by the disc, the attraction is ma>, (Art. 26).
22. From this we deduce the attraction of an infinite thin
plate or disc by putting a = ^TT. We thus find that the attraction
of an infinite thin plate at any point P is 2-Trm and is therefore
independent of the distance of P from the plate.
We also infer that the attraction of a circular disc of finite
radius a at a point P on the axis very near the disc is ultimately
27TW. The attraction is A =/27r(l —p/r)pdp where r = PA, p is
the density, t the thickness, m = pt, and the limits are p=p to
p + 1. After integration this reduces to 27rw, provided p/a and
t/a are ultimately zero,
At first sight it may appear anomalous that the attraction of an infinite plate
should be independent of the distance of the particle from the plate, but we may
understand how it can happen by considering what elements of the disc are effective
in producing the attraction. Each element of an annulus QQ', whose centre is O,
attracts P with a force acting along the straight line joining P to that element, and
the component of force along PO is obtained by multiplying this attraction by
cos OPQ. When the point P is near 0, this cosine is small and therefore it is only
the portion of the disc near 0 which exerts any sensible attraction in the direction
PO. As P recedes from 0, the cosine for each annulus gets larger and the resolved
attraction becomes greater. Thus the area of effective attraction increases in size
as the particle recedes. At the same time as the particle P recedes from 0 the
actual attraction of each annulus on it decreases. It follows from the analysis in
the last article that the increase of effective area just balances the decrease of
attraction due to increased distance, so that on the whole the attraction is
independent of the distance.
12
ATTRACTIONS.
[AET. 24
23. If g, g' be the attractions due to gravity on two table-
l — -
4 a
lands whose difference of level is x, show that q' = q(
J y \
approximately, where a is the radius of the earth.
To obtain this result, we regard the attraction of the table-land
as sensibly the same as that of an infinite plate, Art. 22. The
attraction is therefore Zirpx, where p is the density of the table-
land or flat mountain. If p' be the mean density of the earth,
its attraction, viz. g', is f Trp'a. There are reasons for believing
that the mean surface density of the earth is about half the mean
density of the whole earth ; when therefore the true density of the
table-land is unknown we may as an approximation put p — ^p'.
The attraction of the table-land is thus approximately $gx/a. The
(d \2 / X\
- =<7 ( 1 — 2 - I
a + xl * \ aj
approximately. Adding this to the attraction of the table-land we
arrive at the result given.
This theorem was first used by Bouguer in his Figure de la Terre. A short
account of this treatise is given in Art. 363 of Todhunter's History of Attractions,
<fec. A similar result is also given by Poisson in Art. 629 of his TraitS de
Mtcanique. See also Clarke's Geodesy. It is often called Dr Young's rule.
According to Nature, Feb. 10, 1898, a good account of the controversy about the
second term of Bouguer's formula is given by G. B. Putnam in the scientific work
of the Boston party on the sixth Peary expedition to Greenland. Eeport A.
Ex. Sir W. Siemens invented an instrument to measure the depth of the sea
under a ship on the principle of balancing gravitation by the force of a spring. If
the mean surface density of the earth be three times that of sea water, and the
mean density of the whole earth five and a half times that of sea water, show that
at a depth h in the sea, the diminution of gravity is fthgja, where a is the radius
of the earth.
24. Attraction of a Cylinder. Ex.1. Find the attraction of a uniform solid
right circular cylinder at a point P on its axis.
Let p be the density of the cylinder, a its
radius. Let 0 be the centre of gravity, OP=p.
Let us take as the element of volume the slice
of the cylinder between two planes drawn per-
pendicular to the axis at distances x and x + dx
from 0, measured positively towards P.
First , let P be outside the cylinder. Let 20 be
the angle subtended at P by any diameter QQ'
of the slice, and let PQ=r. Since the mass per
unit of area of the slice is m=pdx, the attraction
A'
B'
Q
-- But (p-a;)2+aa=r2, /. (x-p)dx=rdr.
The whole attraction of the cylinder at P is therefore F=2vp$(dx + dr), where the
AET. 25] CYLINDERS. 13
limits of integration are x= -%AB to x=%AB and r=PB to r=PA. The resulting
attraction is therefore F=2irp(AB+PA -PB), where AB is any generating line
and A is the extremity nearest to P. We notice that AB is equal to the difference
of the distances from the plane sections passing through A and B.
Next, let P be inside the cylinder, but nearer to the plane section A'A than to B'B.
Since 0 is the acute angle subtended at P by the radius of the attracting slice, we
must equate cos 6 to ± (p - x)/r, the sign being different on opposite sides of P. To
avoid this discontinuity we draw a plane C'C perpendicular to the axis so that P lies
midway between the sections A'A and C'C. The resultant attraction of the matter
between AfA and G'C at P is therefore zero. The resultant attraction of the rest of
the cylinder is given by F= 2irp (CB + PG- PB)
Here CB is equal to the difference of the distances of P from the plane sections
through A and B, measured positively in opposite directions.
Another Solution. We may also find the attraction by dividing the cylinder into
elementary columns or filaments parallel to the axis. We find that the resolved
force parallel to the axis is therefore the difference between the values of the
integral Jpd<r/r for the two plane faces, where r is here the distance of da- from P.
Since dcr=2irrdr, and the limiting values of r for the faces A A', BB' respectively
are PM to PA and PN to PB, we easily arrive at the same result as before.
Ex. 2. Find the ratio of the radius of the base to the height of a right circular
cylinder of given volume so that the attraction at the centre of one of the circular
ends may be the greatest possible. The required ratio is i (9-^17). Playfair's
problem. See Todhunter's History of Attractions, Art. 1585.
Ex. 3. A right circular cylinder is of infinite length in one direction and is
homogeneous, the finite extremity being perpendicular to the generators. Prove
that the attraction at the centre of this end is 2Mja, where M is the mass per unit
of length, and a is the radius.
If the cylinder be elliptic, of the same density and mass per unit of length as
before, and of eccentricity e, then the attraction will be n times the former value,
where n=- (1 - e2)* I* ^ — [St John's Coll., 1887.]
TTV J O^/l-^sin2!?
Ex. 4. A solid right circular cylinder of uniform density p stands on the plane
of xy and is infinite in the positive direction of the axis of z. Show that the z
component of its attraction at a point P of its base is pi, where I is the perimeter of
an ellipse having the base for the auxiliary circle and P for one focus. See Art. 11.
Ex. 5. A vertical solid cylinder of height h, radius a, and density p, bounded by
plane ends perpendicular to the axis, is divided by a plane through the axis into two
parts. Show that the horizontal attraction of either part at the centre of the base is
2ft/, log L±fil{£±j9 . [Coll. Ex., 1888. ]
25. Attraction of a surface. All sections of a uniform
cone which are of the same thickness, and have their plane faces
parallel to a given plane, exert equal attractions at the vertex.
14 ATTRACTIONS. [ART. 26
Let AB, A'B' be two thin parallel laminae of the same
thickness dt. Let p be the density of
the cone. With the same vertex 0 de-
scribe an elementary cone cutting the
laminae in QR, Q'R'. The attractions
of QR, Q'R' at 0 are to each other as
their masses divided by the squares of
the distances. Since the thicknesses are
equal, the masses are proportional to the »'
areas, and these by similar figures are
proportional to the squares of the distances OQ, OQ'. Thus the
attractions of the elements QR, Q'R' at 0 are equal. Hence the
attractions of the laminae AB, A'B' at 0 are the same both in
direction and magnitude.
This being true for all thin laminae must, by integration, be
also true for all thick sections. And in general any two parallel
slices of the same cone, whether thick or thin, attract the vertex in
the same direction with forces proportional to their thicknesses.
26. As the attraction of the element QR at any point 0 is
wanted in several theorems further on, it is convenient to determine
an expression for its magnitude.
Let dcr be the area of the element QR, m its mass per unit of
area, r its distance from 0 ; the attraction at 0 is then mdcr/r3.
To simplify this expression, we use the solid angle subtended
at 0 by the area. Just as in plane trigonometry an angle is
measured by the arc subtended in a circle of unit radius, so the
solid angle contained by any cone is measured by the surface cut
off by the cone from a sphere of unit radius with its centre at the
vertex.
Let the elementary cone whose base is QR intercept on the
unit sphere an elementary area qr, and let this area be da, then da>
measures the solid angle subtended at 0, Let i/r be the angle the
normal to the elementary area QR makes with the radius vector
OQ, then da- cos ty is the area of a section of the cone made by a
plane drawn through Q perpendicular to OQ. Hence by similar
figures dcr cos ty/rz = area qr = dm.
The attraction of the element is therefore m sec i/r . dm. If p be
the perpendicular from 0 on the plane of the element, then
r cos ^=p, and the attraction of the element at 0 may also be
ART. 29] SURFACES PLANE AND CURVED. 15
written in the form mrdw/p. If r, 6, (j> are the Eulerian polar
coordinates of a point referred to any axes with the origin at 0, it
is clear that da) = sin 0ddd<j).
27. It follows from this result that the attraction at 0 of an
element da when resolved perpendicular to its plane is mdco.
Hence we may deduce by integration that the attraction at 0
of a plane uniform lamina of any form when resolved perpendicular
to the plane is mm, where m is the mass of a unit of area of the
lamina, and to is the solid angle subtended at 0 by tlie lamina,
This theorem is due to Playfair, Edin. Trans. Vol. vi., 1812.
Ex. If I, m, n be the direction cosines of the radius
vector of an element of a surface, and if I, m, n can be ex-
pressed in terms of two parameters a and &, show that the
normal attraction of the element on the origin is Adadbdk,
where dk is the thickness of the element and A is the deter-
minant in the margin. [Caius Coll.]
28. The method explained in Art. 17 by which the attraction
at the origin of one thin rod may be replaced by that of another
of a more convenient form may be extended to surfaces.
Let the law of attraction be the inverse /eth power of the
distance. Refer to the figure of Art. 17 and equate the attractions
of the elementary areas QR, Q'R', we have — — = — ^- .
I, m, n
dl dm dn
da da ' da
dl dm dn
db' ~db' db
By Art. 26 da cos •x/r = r*dco, hence — — -=— — ; -ry.
rK~2 cos ty r'K-z cos ^
It follows that, if two curvilinear lamina? are so related that
their masses per unit of area, at points on the same radius vector
drawn from a point 0, are connected by the above equation, then the
attractions at 0 of the portions included within any conical surface
whose vertex is 0 are the same in direction and magnitude.
For example, if the law of attraction be the inverse cube, the
attraction at a point 0 of any portion of a thin plane area is the
same in direction and magnitude as that of the corresponding
portion of a spherical surface having its centre at 0, and touching
the plane, the masses per unit of area of the plane and sphere
being equal. This corresponds to the theorem in Art. 12, which
connects the attraction of a straight rod with that of a circle.
29. If the plane area be bounded by an ellipse (the law of
attraction being the inverse cube) the resultant attraction at any
point 0 acts along the axis of the enveloping cone whose vertex is 0.
16 ATTRACTIONS. [ART. 31
To prove this we notice that the enveloping cone is a quadric
cone and that therefore the portion of the spherical surface (centre
0) enclosed within it is symmetrical about the internal axis of the
cone. The resultant attraction of the spherical surface at 0 must
therefore act along that axis. By a known theorem in geometry
this axis is normal to the ellipsoid which passes through 0 and has
the given ellipse for a focal conic.
30. Ex 1. Show that the attraction at a point 0 of any portion of a thin
plane disc is the same in direction and magnitude as that of the corresponding
portion of a spherical surface having for a diameter the perpendicular ON drawn
from 0 to the plane. The two attracting surfaces are supposed to be homogeneous
and of equal mass per unit of area.
Ex. 2. A tetrahedron is constructed of thin metal, the faces heing of equal and
uniform density. Prove that if the law of attraction were the inverse cube of the
distance, a particle would be in equilibrium if placed at the centre of the inscribed
sphere. See Art. 16, Ex. 2.
Ex. 3. Prove that the ratio of the attractions of a solid right cone at the centre
of the base and at the vertex is — TJT^ 1 the angle at the vertex being a
\/^ — •!•
right angle.
Ex. 4. An infinite lamina is bounded by two parallel straight lines. Prove
that its component attractions X and Y respectively parallel and perpendicular to
its plane are X=2mlogr'lr and Y=2m.6, where r1, r are the distances of the
attracted point P from the two edges, 6 the angle these distances make with each
other and m the surface density. See Art. 14.
Ex. 5. Prove that the resultant attraction of a uniform rectangular plate at a
point P on its axis is 4m sin"1 (sin a sin /3) where a, /3 are the angles subtended at P
by perpendiculars drawn from the centre on the sides and m is the surface density.
Playfair, Edin. Trans. 1812.
Ex. 6. Prove that the attraction of a uniform elliptic disc at the focus is
— {1-^(1 -e2)} where m is the surface density.
The attraction is X=$$mrdOdrcos0lr*. Describe a circle of arbitrary radius c
with its centre at the focus : the attraction of the enclosed area is zero. Integrate
from r=c to the elliptic rim and from 0 = 0 to 2 jr. In this way we avoid the
infinite logr at the origin.
31. Tfce solid of greatest attraction. To find the solid of revolution of given
mass which exerts the greatest attraction at a point 0 situated on the axis.
Let us trace the surface such that the attraction at the given point 0, of a particle
of given mass m placed at any point of the surface, when resolved along the given
axis, is equal to a given constant (7. Taking 0 for origin and the given axis as the
M
axis of reference, the equation of that surface is clearly -^ cos 0=C. By giving C
different values we obtain a system of surfaces. It is evident from the definition
that the surface defined by any value of C lies outside that defined by a greater
value of C. It follows that the resolved attraction of a particle lying on any one
surface is greater than that of an equal particle situated on any external surface.
ART. 33] SOLID OF GREATEST ATTRACTION. 17
It is evident from the equation that all these surfaces are similar and similarly
situated, and that they all touch a plane drawn through 0 perpendicular to the
given axis.
Let us select that surface whose volume would just contain the given mass.
The solid of greatest attraction must coincide with the surface thus selected ; for if
any portion lies outside the selected surface, the attraction would be increased by
moving that portion into the vacant places within the selected surface and thus
filling them up.
The solid of greatest attraction has therefore such a form that the attraction at
the given point of a given particle placed at any point of the surface when resolved
along the given axis is always the same.
The problem of finding the solid of greatest attraction was proposed and solved
by Silvabelle. The principle used above, that the resolved attraction must be
constant over the surface, is due to Playfair, Edin. Trans. 1812. The following
example is also due to him.
32. Ex. Supposing the law of attraction to be the inverse /cth power of the
distance, find the form of an infinitely long cylinder so that the attraction may be a
maximum at an external point.
Take the point for origin; pass a plane through it perpendicular to the
generating lines of the cylinder. Let r be the radius vector of any point on this
section, 0 the angle made by r with the direction of the resultant attraction. The
equation of the curve is included in cos9=CrK~l. When the law of attraction is
the inverse square the required cylinder is right circular.
33. Attraction of mountains. It is a matter of some
importance to determine by direct experiment the effect of the
attraction of a neighbouring mountain on the direction of the
plumb line. This was attempted by Bouguer in Peru but without
any great success. In 1772 Maskelyne, then Astronomer Royal,
proposed to repeat the experiment. He pointed out a mountain
in Yorkshire as suitable for the purpose. He suggested also that
the defect of matter in the valley between Helvellyn and Saddleback
might produce an effect of an opposite character which would be
sensible. The mountain Schehallien in Scotland was finally chosen.
It is a narrow ridge running east and west in a comparatively flat
country and is about 2000 feet above the general level.
Let f, f be the horizontal attractions of the mountain at two
stations north and south. The angular deviations of the plumb
line from the direction of gravity will then be a =flg, and a.' =f'jg.
The meridional distance between the two stations was found by a
survey over the mountain to be 4364*4 feet. By dividing this by
the radius of the earth, the difference of latitude of the two stations
was found to be 42"'9. By observing the zenith distance of the
same star at both stations the difference of the angles which the
R. s. n. 2
18 ATTRACTIONS. [ART. 35
direction of the star made with the directions of the plumb line
at the two stations was found to be 54"'6. The difference between
these two angles, viz. H"'7, is evidently equal to the sum of the
angular deviations a, a.' produced on the plumb lines by the
attraction of the mountain.
34. To find the attraction / at a station A, a contour map of the country was
made. This was divided into twenty rings by circles having A for a common
centre, their radii heing in arithmetical progression. These rings were subdivided
into rectangular spaces by radii vectores drawn from A. The mountain was thus
theoretically divided into elementary columns placed on these rectangular bases.
Let GP be a vertical drawn through the centre of gravity G of any base cutting the
surface of the mountain in P. Let z be the angle PAG subtended by PG at A.
The attraction of this column is nearly equal to 2m sin \z\A G and its direction
bisects the angle PAG, where m is the line density of the column (Art. 13).
Let r, 0 be the polar coordinates of G referred to A as origin and the meridian
AM as axis of x. Let Ar be the difference of two consecutive radii, and A0 the
angle between two consecutive radii vectores. Then m = fj,r . Ar . A0 nearly, where
p. is the density of the column. The resolved attraction of the column along the
meridian is therefore
2?re z z
X— — sin ^cos - cos 6 = /j. sin z . Ar . A sin 6
7* mm
nearly. The constant difference Ar was taken to be 666£ feet. The radii vectores
were drawn according to the following law. The first being directed along the
meridian, the others were drawn making with the meridian angles whose sines were
successively 1/12, 2/12, 3/12, &c. There were therefore 48 columns over each ring.
Also A sin 6 was constant and equal to 1/12. It is now evident that the attraction /
of the mountain may be found by forming the sum
sin 2j + sin z24-sin z3+...
for all the columns and multiplying the result by ^joi . Ar. The twenty rings drawn
round each station included 960 columns. This space was bounded by a circle of
radius 2£ miles. It was assumed that the attraction of the matter beyond this
distance might be neglected.
35. By such processes as these the sum of the two opposite
attractions at the two stations was found as a known multiple of
the density p, of the hill. If R be the radius of the earth, p its
density, we have g = %irpR, Art. 77. We thus have a + a' expressed
as a known multiple of p/p. By equating this result to the
circular measure of 11"'7, we find that the mean density of the
hill is |ths of that of the earth.
A geological survey was subsequently made by Playfair to
discover the average density of the hill. After many corrections
Hutton gave 4-95 as the mean density of the earth, that of water
being unity.
ART. 38] ATTRACTION OF MOUNTAINS. 19
Other mountains also have been used for this purpose. The
observations of James and Clarke on Arthur's Seat gave 5'316,
while those of Mendenhall in Japan led to 5*77 as the mean
density of the earth.
36. There are two other methods of finding the mean density,
one by observations in mines and the other by processes analogous
to the Cavendish experiment. These have been used many times
and lead to results which differ slightly, in excess or defect,
from 5^-.
A short history of the older experiments may be found in Airy's Figure of the
Earth. Much however has been done since 1830, the date of this treatise. An
account of the experiments up to the year 1894 is given in Poynting's essay on the
mean density of the Earth.
37. At the end of a paper on the Schehallien experiment (Phil. Trans. 1821)
Hutton suggested that one of the great pyramids of Egypt might be used instead of
a mountain to find the mean density of the earth. He calls to mind the great size of
one of these, its height being nearly double that of St Paul's Cathedral. Its regular
figure and known composition would, he says, yield facilities in the calculation of
its attraction. Observations could then be made at four stations, one on each face,
and these could be placed much nearer to the centre of gravity of the attracting
mass than was possible in an irregular mountain. Such was his enthusiasm, that
he declared that even his age of eighty years would hardly prevent him from joining
an expedition for this purpose.
38. Ex. 1. The tide in the Bay of Fundy rises 100 feet from low to high water
mark. It has been proposed to find the density of the earth by determining the
attraction of the tide-wave on a plumb-line at high and low tide on the same
principle as Maskelyne's experiment at Schehallien. Supposing the attraction of
the tide-wave at a point 0 on the shore to be represented by that of the water
within a cylinder whose axis is the vertical at 0, whose height I is 100 feet and
31 2r
radius r, show that the deviation of the plumb-line is log — , where R is the
AirRD L
radius of the earth, D its mean density, and r is large compared with I.
Show that this expression increases slowly compared with r, and that if r be
taken between 2 and 4 miles, the deviation to be observed will be about two-fifths of
a second. This is much smaller than the deviation to be observed in Maskelyne's
experiment, which was about eleven seconds. On the other hand the attracting
mass is a homogeneous body instead of a heterogeneous mountain.
Ex. 2. The section of a long wedge-shaped mountain is an equilateral triangle
having BG for base. If P be the point on the face AB at which the horizontal
attraction is greatest, prove that log 3Mffi~a;' =N/3 sin"1 a^ - where x=BP, y = CP
x iy
and a is the length of any side of the section. The equation is nearly satisfied
by z = £a.
2—2
20 ATTRACTIONS. [ART. 40
The Potential.
39. Let AI, AS, &c. be the positions of any number of fixed
attracting particles; m^, m?, &c. their masses. The potential of
these particles* at any proposed point P is defined to be
where rlt r2, &c. are their distances from P regarded as positive
quantities. For the sake of distinction this is sometimes called
the Newtonian Potential. See Art. 9.
This may be called the geometrical definition of the potential.
Another definition founded on the principle of work will be given
a little further on. In discussing the attractions of geometrical
figures the former is the more convenient for use, but in many
physical applications the latter will be found the more satisfactory.
We may notice that as the point P moves in space the potential
is, by the definition, a continuous function of the position of P.
We must however except the case in which any one of the
distances rlt r^, &c. vanishes or changes sign, for then the term
m/r ceases to represent the potential of the particle from which r
is measured. The potential is also a one-valued function of the
coordinates of P.
40. If m be the mass of any one of the attracting particles,
A its position, r its distance from a point P, the potential of m at
P is m/r. Let P' be any point adjacent to P, and let PP' = ds.
The difference of the potentials of m at P and P' is then
d fm\ , mdr ,
ds\rj r2 ds
* The function now called the Potential was used by Legendre in 1784 who
refers to it when discussing the attraction of a solid of revolution. Legendre
however expressly ascribes the introduction of the function to Laplace and quotes
from him the theorem connecting the components of attraction with the differential
coefficients of the function. M. Bianco in the Rivista di Matematica, 1893, gives
quotations from Bist (Institut Paris, 1806) and from Baltzer (Geschichte des Po-
tentials, 1878) showing that Lagrange used the same function in 1777 when
discussing the motion of several bodies mutually attracting each other (Academy
of Berlin, 1777). See also "II problema Meccanico della figura della Terra"
(Torino, 1880) by M. Bianco. The name, Potential, was first used by Green in his
Essay an the application of Mathematical Analysis to the theories of Electricity and
Magnetism, published in 1828. Green gave many of the theorems on this
function now in continual use, which have been since associated with the
names of others who have discovered them a second time. Gauss also uses the
name in Art. 3 of his memoir on Forces acting inversely as the square of the distance,
Leipsic 1840, translated in the third volume of Taylor's Scientific Memoirs. The
reader may also consult Todhunter's History, Arts. 790, 1138, and Thomson and
Tait's Treatise on Natural Philosophy, Art. 483.
ART. 42] THE POTENTIAL. 21
If </> be the angle AP'P, we have cos <£ = drjds. The attraction
of m at P acts in the direction
PA, and is equal to m/r2; hence
its resolved part in the direction
r> n> • ^ D D'
PP is — cos APP = -j- .
r2 r2 cfo
Comparing this with the above
result we see that if P, P/ be
two adjacent points, the excess
of the potential at P' over that at P, divided by the distance PP',
is equal to the resolved attraction in the direction PP'.
This, being true for every particle of an attracting system, is
necessarily true for the whole. We have therefore the following
theorem. If V, V be the potentials of a system at two neighbouring
points P, P', the attraction at P resolved in the direction PP' in
V —V dV
which s is measured is the limit of — p~p7~~ = ~T~ •
So long as the point P is situated outside the attracting mass the potentials F
and V are both finite and this proof is free from ambiguity. The case in which P
lies within the attracting mass will be considered a little further on.
41. By taking the displacement PP' parallel to the axes of
x, y, z in turn, we see that the components of the attraction in the
positive directions of the axes are respectively
y. dV _ dV dV
.A = — =— , I = -j — , Zt = — T— .
dx dy dz
In the same way the components of the attraction in polar
coordinates may be expressed. Let r, 6, <f> be the polar coordinates
of any point P, let F, G, H be the components at P in the
directions in which dr, rdd, r sin 0d(f> are drawn, then
dV dV dV
H -
dr ' rdd ' r sin
In the theory of gravitation the attraction of one particle on another is taken to
be mm'/?*2 (Art. 3), and repulsion is then represented by supposing that the mass
of one of the particles is negative. In other theories, for example in that of
electricity, repulsion is taken as the standard case and then attraction occurs when
the masses have opposite signs. In both cases the geometrical definition of the
potential is F=Sm/r (Art. 39). When therefore repulsion is taken as the standard
the signs of the forces given above must be changed. Thus the force in the positive
direction of the axis of x is X= -dV/dx, and so on.
42. It appears from this proposition that, when the potential
V of a body fixed in space is given, its resolved attractions at any
22 ATTRACTIONS. [ART. 44
point P can be found by simply differentiating the potential with
regard to the coordinates of that point. It follows that, if two
different bodies have equal potentials throughout any space, they
equally attract any particle placed in that space. Thus the at-
traction of a body is determined by the single function V instead
of the three components X, Y, Z,
One chief reason for the use of the potential is that a body, so
far as its quality of attraction is concerned, is analytically given
by a single function without the necessity of stating either the form
or the structure of the attracting body.
When the potential is used merely to find the forces, it is
obvious that we may add an arbitrary constant to its value as
defined in Art. 39. We then have V=^m/r+C, where C is
the constant added. When the attracting bodies are finite, it is
convenient to choose C so that V is zero at an infinite distance ;
this assumption makes (7 = 0. When the attracting bodies extend
to infinity, the potential, as defined in Art. 39, is sometimes found
to contain an infinite constant. It may then be preferable to
keep C arbitrary and to absorb into its value all constants not
immediately required. There is a certain inconvenience in having
different definitions of the potential for finite and infinite bodies,
especially when we wish to proceed from one to the other as a
limit. In stating the results therefore for the Newtonian law of
force we shall adhere to the definition of Art. 39. In special cases
such a constant may then be added as may most simplify the
expression for V.
43. Potential for other laws of force. When the law of
ffrt
force is the inverse «th power, the potential is V = 2 .
K — 1 r*"1
We then find by the same reasoning as in Art. 40 that dV/ds is
the resolved force at P in the direction in which ds is measured.
When the law of force is the inverse distance, the potential is
V = C — 2ra log r. This is sometimes called the logarithmic potential.
44. Work and potential. A definition of the potential may
also be given founded on the principle of work. Referring to the
figure of Art. 40, let a particle of unit mass travel along the
elementary arc PP'. It has been already shown that the resolved
attraction in the direction PP' is dV/ds. The work done by the
attraction is therefore (dV/ds)ds. If the particle continue its
ART. 46] LEVEL SURFACES, &C. 23
journey along any curve, starting from some point P and arriving
at some other point Q, the work done by the attraction is
fdV = Vz — V-i, where Fi and V% are the potentials at P and Q.
Thus the excess of the potential at Q over that at P is the work
done by the attraction on a particle of unit mass as it travels by
any path from P to Q.
If the attracting body is finite in all directions, the potential at
a point P infinitely distant is zero. It follows that, the potential at
any point Q is the work done by the attracting forces on a particle
of unit mass, as it travels from an infinite distance along any path
to the point Q. In the same way the potential at Q is the work
which must be done against the attraction by some external cause
to move a unit particle from Q to an infinite distance.
The several particles of the attracting mass are supposed to
remain fixed in space while the attracted particle makes its
journey from P to Q.
45. Level surfaces. The locus of points at which the
potential has any one given value is called a level surface. It
is also called an equipotential surface.
At any point of a level surface the resultant force acts along the
normal to the surface. To show this, let Pl be a point on a level
surface, and let P2 be any neighbouring point also on the surface.
If Vlt Vz be the potentials at these points, the component force
V — V
in the direction of any tangent PjP2 will be the limit of — ' p *.
PI Pa
This is zero since V1= V2. The resultant force must therefore
act along the normal at Pa.
46. Let two neighbouring level surfaces be drawn at which
the potentials are respectively Fx = c and V2 = c + <5c. The normal
attraction at any point P of either surface is inversely proportional
to the length of the normal at that point intercepted between these
level surfaces. To prove this, let the normal at any point Px on
the first surface intersect the second surface in P2. The normal
Y _ v 8c
force at Px is then ultimately F= * p 1 = p p- . It is therefore
"
evident that F varies inversely as PiP2.
If a rigid surface were constructed having the form of a level
surface and coincident with it, it is clear that a particle, placed at
any point of the surface, would be pulled by the attracting body
24 ATTRACTIONS. [ART. 50
in a direction normal to the surface. The particle, if placed on
the proper side, would therefore be in equilibrium. Level surfaces
are therefore also called surfaces of equilibrium.
47. A Line of force is a curve such that the direction of the
resultant force at any point is a tangent to the curve. It is
evident that the whole system of level surfaces is cut orthogonally
by the system of lines of force.
48. Ex. 1. A free particle placed at rest at any point of a line of force will
move along the curve in such a direction that the potential increases.
Ex. 2. Show that, if attracting matter be arranged so that the direction of
the resultant attraction at any external point P shall always pass through a fixed
point 0, the magnitude of the resultant attraction will be a function only of the
distance OP, and will not depend on the angular coordinates of OP.
To prove this we notice that the level surfaces are spheres, because the normal
at every point P passes through 0. Hence the potential is a function of r only,
Art. 45.
49. Potentials of rods. To find the level surfaces and the
potential of a thin uniform straight rod AB at any point P.
It has already been proved that the direction of the attraction
of a rod AB at any point P
bisects the angle between the
distances PA, PB (Art. 13).
It follows from Art. 45 that
the level surfaces are prolate
spheroids having their foci at B G Q Q' A
the extremities of the rod.
To find the potential we notice that at all points on the same
level surface the potentials are equal. It is therefore sufficient to
find the potential at some convenient point on each spheroid.
Let C be the middle point of the rod, 21 its length, m the line
density. Let r, r be the distances of P from A, B. Let 2a and
e be the major axis and eccentricity of the spheroid, then ae = I,
2a = r + r'. The potential at the extremity of the major axis and
therefore at any point on the spheroid is
T7. f mdx , a + 1 . r + r' + 21
V = \ - = m log . = m log -, — =-, ,
J a-x *>a — l & r + r —21'
the limits of the integral being x = - I to I.
50. When the rod is infinite in both directions the potential
is easily deduced from the attraction already found in Art. 14
ART. 51] POTENTIALS OF RODS, &C. 25
Since the magnitude of the attraction is 2m/p and its direction is
PN, it is evident that the potential must be V = C — 2m log p,
where G is a constant and p is the distance of P from the rod.
We may also deduce this result from the expression for the
potential of a finite rod. Suppose the point P to be situated in
the straight line drawn through G perpendicular to the rod.
z>2
Then r = (I* + p*)* = I + $ -j and we have
T + I
V= m log — - — -, = 2m log 21 — 2m log p.
We thus see that the constant C is really infinite and equal to
2m log 21 when we adhere to the definition of Art. 39.
51. Ex. 1. Let the rod AB be produced both ways to infinite distances. Let
the portion beyond A attract and that beyond B repel P, the part between A and B
exerting no force. Prove that the level surfaces are hyperboloids having A and B for
foci and that the potential at P is m log— — -^ — . Prove also that if the portion A B
£t(/ — T-^~T
is evanescent the level surfaces are right cones and that the potential is 2m log cot ^\f/
where \j/ is the angle of the cone.
Ex. 2. Show that the potential of a thin rod AB at any point P is
V=m log (cot \PAB . cot %PBA).
Ex. 3. A thin uniform rod AB is attracted by a body of any form : show that
the component of the attraction along the length BA of the rod is m(VA-VB),
where V^ and VB are the potentials of the body at A and B, and m is the mass of
the rod per unit of length.
By Art. 11 this theorem is true when the rod is attracted by a single particle ;
it is therefore true by summation when attracted by any body.
Ex. 4. A uniform thin chain AB is enclosed in a smooth curvilinear tube
which it just fits, and is attracted by a body of any form. Show that the force
urging the chain to move in the tube is m(V^-VB). Hence show that the
position of equilibrium may be found by equating the potentials of the body at
the extremities of the chain.
That the force depends only on the positions of the extremities of the chain,
and not on its length or form, may also be shown by another kind of reasoning.
Let the chain be completed into a circuit by uniting two chains in different tubes
at their extremities. If the forces were not equal the chain would begin to move
round the circuit, and thus a perpetual motion would be caused by the mere
presence of an attracting body.
Ex. 5. When the law of attraction is the inverse cube, the potential of a
uniform thin rod AB at any point P is m-y/2p, where y is the angle APB, p the
perpendicular from P on the rod, and m is the line density.
When the law is the inverse fourth power, the potential is m(sin/3-sina)/3p2
where j3, a are the angles p makes with PB and PA.
Ex. 6. A plane lamina is bounded by two parallel straight lines whose distance
apart is 2Z. The surface density at any point Q is /3 (QM . QN)* where QM, QN are
26 ATTRACTIONS. [ART. 52
the distances of Q from the bounding straight lines and 2\ = K-3. The law of
attraction is the inverse *th power of the distance. Prove that the level surfaces
are confocal elliptic cylinders having the foci on the bounding straight lines. Find
also the potential for any cylinder.
First find the attraction at P of an elementary strip Q whose breadth is dx. Put
x=p tan 0 where p=PO is the perpendicular from P to the lamina. The attraction
A /sin MPO . sin NPQ\* /cos e\* Jn
of the strip can then be put into the form - I -: — .,,,?, I dd,
p \sin OMP . sin ONPJ \ p )
where A is a constant and /t=K-3-2X. If then K and X are so related that the
exponent /j,=0 the attraction of the elementary strip at Q is a symmetrical function
of the angles PQ makes with PM , PN. The elements on each side of the bisector
of the angle MPN will then equally attract P. The direction of the attraction
therefore bisects the angle MPN. The magnitude of the attraction is found by
resolving along the bisector and the potential by using the method of Art. 49.
In this proof the plane PMQN is taken to be perpendicular to the boundaries.
52. Ex. A number of infinite straight attracting rods are arranged at equal
distances on the surface of a circular cylinder of radius a. If n be the number of
rods, m the mass of each per unit of length, prove that their potential at any point
P is given by V= G - m log (r271 - 2anrn cos n9 + a2l»),
where r is the distance of P from the axis of the cylinder and 6 the angle r makes
with a plane through the axis and one of the attracting rods.
By making n infinite while the whole mass is given, show that the potential of
a uniform thin cylindrical shell at the point P is C - ±iraM log a or C - &iraM log r
according as P is inside or outside the cylinder, the mass per unit of area being M.
These expressions follow from Art. 50 by using De Moivre's property of the
circle.
These results are of considerable interest because they help us to understand
how the potential of a thin cylindrical shell is a discontinuous function of the
coordinates, being constant at all points within the cylinder and depending on
the logarithm of the distance from the axis at points outside. Supposing the
number of rods to be very great but not infinite, the potential at any point P is
represented by a continuous function of the coordinates of P, i.e., as P travels from
the interior to the exterior through the interstices between the rods the potential is
always the same function of the coordinates. When P is inside the cylinder, rja is
less than unity, and by expanding the logarithm in powers of r/a we see that
V=C-2mnloga+2m(-j cosn0+&c.
It follows that when n is large the potential is sensibly constant throughout the
interior except in the immediate neighbourhood of the surface of the cylinder on
which the rods lie. When P is outside, a/r is less than unity and by expanding
the logarithm in powers of a/r we find V=C-2mnlogr + 2m ( - ) cosn0 + &c. It
appears that, except in the immediate neighbourhood of the surface of the cylinder,
the potential when n is large does not sensibly differ from G - 2wmlog r at any point
outside.
As n increases, the small space within which the potential differs from the first
term of these series gets continually less, and in the limit is zero, so that we may
say that the potential is constant throughout the interior of the cylinder and, except
for C, varies as the logarithm of the distance throughout external space.
ART. 55] POTENTIALS OF CYLINDERS. 27
53. Potentials of discs and cylinders. To find the poten-
tial of a circular disc at any point P situated in its axis.
Referring to the figure of Art. 21, the potential at P of the
annulus QQ' is 27rmxdx(PQ, where a; and x + dec are the radii of
the annulus, and m the mass of the disc per unit of area. If p be
the perpendicular from P on the disc and r the distance PQ, we
have r2 = #2 -f _p2 and rdr = xdx. Substituting, we find that the
potential V of the disc is V = 2?rm fdr= 27rm (r\ — p), where r±
is the distance from P of any point on the perimeter.
If a be the radius of the disc, we may also write this in the
form V = STTW (Va2+j92 - p}.
When the radius a of the disc is infinite we expand the
radical and retain only the lowest power of p/a. We thus find
V=A — 27rmp where A is an infinite constant.
54. Ex. 1. The law of force being the inverse /cth power of the distance,
prove that the potential of an infinite disc at a point distant p from its plane is
C — -. - JY75 - \ where C is infinite or zero according as K < 3 or K> 3. When
(K - i) (6 - K)
K = B the potential is C — irmlogp, where G is infinite.
Ex. 2. Show that the potential of a circular cylinder of density p, radius a,
and small thickness h at an external point P on the axis close to the cylinder is
2irph(a-p), where p is the mean of the distances of P from the two plane faces of
the cylinder. See Art. 9.
55. Infinite Cylinders. An indefinitely thin homogeneous
layer of attracting matter of surface density m is placed on an
infinitely long right circular cylinder. It is required to find the
potential and the attraction at any internal or external point P.
We replace the cylinder by a fine ring of line density m' = 2m
which occupies the position of the cross section through P and
attracts according to the law of the inverse distance, Art. 14.
Let QR, Q'R' be two chords passing through P and making a
small angle dQ with each other. Let PQ = uly PR — u^, QQ' = dst,
RR' = dsn. Let <£ be either of the equal angles OQR, ORQ.
a fm'd0 , f cos0d0 Zm'a[ . r' sin 01
= 2 - —cos 0 = 2m a -f?-^ ,, . ,m = — — sin-1 --- .
J cos<£ J V(* — «" sm20) r |_ a J
28 ATTRACTIONS. [ART. 57
The attractions at P of the elements QQ': RR' are respectively
m'dsju! and m'ds^/u^ But since u-id6jdsl and u2dd/ds2 are each
equal to cos <f>, we see that each of these attractions is equal to
m'dd/cos <£. The attractions are therefore equal.
If P is inside the circle, these attractions balance each other.
The resultant attraction of the whole circle is zero. The potential
is therefore constant and equal to that at the centre.
56. If P is outside the circle- as at P', let 0 = OP'Q, r' = OP' ;
then r sin 6 = a sin <£. The attraction of each of the elements at
Q and R being m'dO/cos <j>, the resolved attraction at P' of the
whole circle along P'O is
X
The limits of 0 are found by drawing two tangents from P' to the
circle ; the integral is to be taken from sin 0 = — a/r to + a/r'.
We therefore find X = M/r' where M = Ziram'.
The attraction therefore of the ring is the same as if its whole
mass were collected into its centre. The attraction of the cylindrical
layer at an external point is the same as if its whole mass were
equally distributed along the axis.
The potential is deduced from the attraction by integrating
dV/dr = — M/r. The potential at an external point is therefore
V=C — M\ogr. We know by Art. 50 that the constant C is
really infinite.
57. The attraction of a solid cylindrical shell bounded internally
and externally by coaxal right circular cylinders may be deduced
from the preceding results.
By dividing the body into elementary cylindrical shells we see
at once that the attraction and potential at an external point are
the same as if the whole mass were equally distributed along the
axis.
At an internal point the attraction is zero and the potential is
an infinite constant.
Lastly, let P be any point in the substance of the shell, r its
distance from the axis. Let us describe a coaxal cylinder passing
through P and dividing the whole attracting body into two shells.
The attraction at P of the outer shell is zero ; the attraction of
the inner is the same as if its mass were arranged along the axis.
ART. 58] HETEROGENEOUS CYLINDERS. 29
The line density is pir (r2 — a2) where a is the radius of the
inner boundary of the attracting cylinder and p the density.
The attraction is by Art. 14
2-Trp (r2 - a2)/r.
58. Heterogeneous cylinder. An indefinitely thin layer of attracting matter
is placed on an infinitely long cylinder of radius a, so that the surface density m is
uniform along any generating line but varies from one generating line to another. It
is required to find the potential at any point P.
We replace the cylinder by a fine ring, of line density m' = 2m, which occupies
the cross section through P, the law of attraction of the ring being the inverse
distance, Art. 14.
Let the plane of the circle be that of xy, the centre 0 being the origin. Let the
polar coordinates of P be r, <f>. Let QQ' be an element of the ring, the angle xOQ
being q ; let q - <t> = ij>.
The line density m' at Q is some given function of q, this we expand (using
Fourier's rule if necessary) in a series of the form
m'=I,(Anco&nq + Bn8innq) .............................. (1),
where S implies summation for integral values of n from ra=0 to oo. We write
this in the form
m'=S(Encosn^ + FnsinnV) .............................. (2),
where En = A n cos n<j> + Bn sin n$, Fn = - An sin n<f> + Bn cos n<j>.
The element of mass at Q being m'ad\J/, and the distance PQ being u, the
potential at P of the whole circle is
V= - \adtym' log u + C = - \ jad^m' log (a2 - 2ar cos ^ + r2) + G,
where the limits are ^=0 to 2?r and C is a constant.
By writing 2 cos ^=£ + 1/£ where £ is an imaginary exponential we have
log (1 - 2ft cos ^ + A2) = log (1 - fc£) + log (1 - A/£)
= - 2 { A cos ^ + $7z2 cos 2^ + £7i3 cos 3^ + <fec. } ..
This series is convergent when h is less than unity.
To obtain a convergent series for V, we expand the logarithm in powers of r/a
or a/r according as P is inside or outside the circle. We therefore write the
potential in the forms
according as P is inside or outside the circle.
Suppose first that m' = EnGosn\j/. Then by remembering that Jcosn^coswi^ty=0
or IT according as m and n are unequal or equal, the limits being 0 and 2n-, we
easily find that V=E- — (-) +G or En — (-) +C according as P is inside or
* n \aj n n \r /
outside, except when w=0.
Next suppose that m'^-F^sinn^, then since Jcosn^sinTtt^di^O, the limits
being 0 and 2w, we find by the same reasoning that the potential at P is constant,
whether P is inside or outside.
When ra=0, we have m' = E0 and the potentials take the form - EQva log a2 + C,
or - Etiira log r2 + C, according as the point P is inside or outside.
30 ATTRACTIONS. [ART. 59
When the line density at Q is given by a single term of the series (1), it is
evident from (2) that En represents the line density of the ring at the point where the
radius vector OP cuts the ring.
Finally, the potential for the whole ring is found by adding together the
potentials for the separate terms of the series (1).
Ex. The density of a thin stratum, on a right circular cylinder of radius a, is
proportional to the distance from a plane through the axis, and its greatest value is
D. Prove that the potential at any point P is 27ra2Z>|/r2 or 2irZ>f according as P
is outside or inside, where | and r are the distances of P from the given plane and
from the axis respectively.
59. Systems of particles. If a particle of mass w/ travel
from a position at which the potential is zero along any path to
any assigned position Slt it is clear from what precedes that the
work done by the attracting forces is Vlm^) where Vl is the
potential at B^ If a second particle m2 travel from a position
of zero potential to the position B2, it is clear that the additional
work is F2ra./, where F2 is the potential at Ba of the same
attracting forces.
Generalizing this, let there be two systems of particles ; let the
masses of the first be m1} m^, &c., and let these be situated at the
points Alt Az, &c. Let the masses of the second be m/, ra2', &c. and
let these be situated at the points Blt B2) &c. Let F1( V2, &c. be
the potentials of the first system at B1} _B2> &c. ; F/, F2', &c. the
potentials of the second system at Alt A2, &c. Let us also suppose
that each particle of either system acts on all the particles of the
other, but does not attract any particle of its own system. The
work done by the attracting forces in moving the particles of the
second system from positions of zero potential to their assigned
positions is W = V1ml' + Tr27na'+ ...
In the same way the work of bringing the particles of the first
system from positions of zero potential to the positions Al} A2, &c.
under the influence of the attracting forces of the second system is
F=F1'm1+F3'm2 + ...
If r12 be the distance between the particles m^, m^', and rn
that between the particles m^, m/, and so on, the values of the
potentials Vlt F/ are
Fi-a + S + fca, Fl'_«L + <+4o.
fu ?*2i ru r12
Substituting, we find that each of the expressions W, W is
equal to
ART. 61] MUTUAL POTENTIAL. 31
If the forces were repulsive instead of attractive this formula
expresses the work the system would do if the particles (under the
influence of their repulsions) retired to infinite distances.
This symmetrical expression is called the mutual potential
energy or the mutual work of the two systems according as the
standard of force is repulsion or attraction (Art. 41).
The work done when either system moves from one given
position to another under the influence of the attractions of the
other system is the difference of their mutual works in the two
positions. If both systems are moved, each from one given position
to another, under the influence of their mutual attractions, it easily
follows, by moving them one at a time, that the work done is the
excess of their mutual work in their final positions over that in
their initial positions.
60. If the particles are elements of a solid body the argument
is still the same. Let dv' be an element of the volume of any
finite mass M', p its density, V the potential of any fixed system
of attracting bodies at the element dv'; the work of collecting
together the mass M' is JVp'dv'.
This formula may be put into the form of a rule. To find the
mutual work of two attracting masses in assigned positions, we
multiply the mass of each element of one body by the potential of the
other at that element, and then integrate the result throughout the
volume of the first body.
61. The particles of a system mutually attract each other and
are in assigned positions. Supposing them to have been originally
at distances so far apart that their mutual attractions were zero, it
is required to find the work done by their attractions as they are
collected together and brought each into its assigned position.
Let us suppose that the particles m1, m2, ... mn-l have been
brought into their proper places. We now bring mn from infinity
into its place under the attraction of ml...mn-.i. The work is
j, y n_1 _,, . , . , , „
mn i -- --- K.. - f. Ihus mn is taken once with each of
j /y* /»» fy>
\! in 'zn * n— i, n)
the masses m^, m^, ... m^^. When we bring in succession mn+1,
mn+2, &c. from infinity we obtain a similar series for each and
therefore mn is taken once with each of these masses as it is
brought in. Thus mn is taken once with every mass except itself.
If m, m' are the masses of any two particles, r their distance apart
32 ATTRACTIONS. [ART. 62
in the final arrangement, the work of the attractions when
collecting the system is W = 2,— - — .
Let Fa be the potential in the given final arrangement at the
particle raj of all the particles except m-^ ; F2 the potential at m%
of all the particles except ra2 and so on. Then
Let us consider how often the mass mn occurs in the expression
+ F2ra2+ .... It occurs once in V^m^ combined with n^,
once in F27/i2 combined with m^ and so on. Again it occurs in
Vnmn combined with every other mass. Thus on the whole mn
occurs twice combined with every other mass. It follows that the
work of collecting the system is
W =£ (FX + F2m2 + ...) = iSFm.
We thus arrive at the following rule. To find the work done
by the attractions of a system of particles brought from infinite
distances to any assigned positions we multiply the mass of each
element by the potential at that element, integrate throughout the
volume, and halve the result.
This rule, when the final sign is reversed, also gives the work
when the particles move from any assigned positions to infinite
distances. To find the work when the particles move from one
assigned arrangement to another, we add together the work
when taken from the first arrangement to infinite distances and
the work when brought from infinite distances to the second
arrangement. If the system be moved, like a rigid body, from one
place to another so that the relative positions of the particles in
the two places are the same, it is clear that no work is done by the
mutual attractions of the particles.
62. In this investigation we have treated the masses m^, wi2, &c. as if their
linear dimensions were infinitely small compared with their distances apart. In
the case of a continuous body the portions of matter not in contact can be divided
into elements so small that the above assumption is correct, but the argument
might be supposed to fail for two elements which finally become contiguous.
We notice however that in finding the potential of any solid mass at a point P
we may omit the matter within any indefinitely small element of volume enclosing
P if its density is finite. For since potential is " mass divided by distance " and
the mass varies as the cube of the linear dimensions, it follows that the potentials
of similar bodies at points similarly situated must vary as the square of the linear
dimensions. The potential must therefore vanish when the mass becomes ele-
ART. 64]
SPHERICAL SURFACE.
33
mentary and the distance indefinitely small. In applying therefore the form
W=%2Vm to a solid body we write pdv for m and take V to be the potential of the
whole mass at the element dv.
In the same way, in finding the potential of a surface at a point P on the
surface we may omit the element contiguous to P if the surface density is finite.
For, the potentials of similar areas at similarly situated points vary as the linear
dimensions, and are zero when the areas become elementary.
63. It appears from the definition of potential that its dimen-
sions are not the same as those of work. The potential of a particle
whose mass is m at 'a point P distant r is m/r. If a particle of
mass m' is situated at the point P, the mutual potential energy
or work of these two particles is mm'/r. The dimensions of
potential are therefore mass divided by distance, those of work are
mass squared divided by distance.
Spherical Surface.
64. To find the potential of a, thin uniform spherical shell at
any point.
Let 0 be the centre of the shell, a the radius of either bound-
ing surface, m the mass per unit of area. Let P be the point at
which the potential is required, OP = r.
Taking on the surface of the shell an annulus QQ' whose axis
is OP, let the angle
POQ = 6, and QP = u.
Since the mass of the an-
nulus is m . add . 2,7ra sin 9
by Pappus' theorem (Vol. I.
Art. 413), the potential at
P of the whole shell is
2-Trma2 sin 6 d9
Since u2 = r* + a2 — 2ar cos 9, we have udu = ar sin 9 d9.
Substituting, we find V-— - fdu. If the point P is external
to the surface as shown in the figure, the limits of u are u = PG to
u = PC' i.e. u = r — a to r + a. In this case V= . If the
r
point P is inside the shell as at P', the limits of u are u = P'C to
u = P'C', i.e. u = a — r to a + r. In this case V = .
a
R. s. ii. 3
34 ATTRACTIONS. [ART. 67
If M be the whole mass of the shell, M=4>7rmaz, and these
M M
expressions take the form V= — or V—— according as the
attracted point P lies outside or inside the shell.
When the point P is at the centre, M is constant and cannot be properly taken
as the independent variable. But since every element of the attracting mass is
equally distant from P, it is evident that the potential at the centre is equal to the
mass divided by the radius, and this agrees with the above result.
65. Since the potential is the same at all points within the
spherical shell, it follows that its differential coefficient with regard
to each of the coordinates is zero. Thus the attraction of a thin
uniform spherical shell at an internal point is zero.
Since a thick shell bounded by concentric spheres may be
regarded as composed of a sufficient number of thin shells, it
follows that the attraction of a thick shell bounded by concentric
spheres at an internal point is zero.
This theorem is also true for a heterogeneous thick shell provided
the strata of equal density are concentric spheres. For in this case
each of the thin shells into which it is analysed is homogeneous.
66. Since the potential at an external point of a uniform thin
shell is Mjr, we see that the force at an external point P resolved
in the direction OP is equal to — Mfr2. The attraction therefore
acts in the direction from P towards the centre, and is the same as if
the whole mass were collected at its centre.
As before, since a thick shell may be analysed into elementary
thin shells, it follows that the attraction of a thick shell bounded by
concentric spheres or of a solid sphere at any external point is the
same as if the whole mass were collected into its centre. Also this
is true for heterogeneous shells provided the strata of equal density
are concentric spheres.
These theorems on the attraction of a spherical shell as well as that of a
spheroid at an internal point are due to Newton.
67. It remains to find the attraction of a thin uniform shell
on an elementary area which is part of itself. Let the attracted
point P be at G, then CQ = u, r = a, and cos QCO = uj2a. Pro-
ceeding as before, we find the attraction X at G is
' 2-Trraa2 sin 6 dd u irm . .
the limits of u being 0 and 2a. This gives X = Zirm = — - .
L a
ART. 68] SPHERICAL SURFACE. 35
The attraction of a thin uniform shell on an element of its surface
is the same as if half the mass of the shell were collected at its
centre.
68. That the attraction of a thin uniform shell bounded by
concentric spheres at an internal point P is zero, may be shown by
an elementary geometrical argument which applies also to the case
of some ellipsoidal shells.
With P as vertex describe an elementary cone cutting the
surfaces of the shell in QQ'qq', RR'rr' respectively. Let PQ = r,
Qq = dr ; PR = r', Rr = dr'. If do> be the solid angle of the
elementary cone, the volumes of the elementary solids at Q and R
will be respectively r^dcodr and r'2d<adr'. Their attractions at P
are therefore pdcodr and pdcadr', where p is the density. These
attractions will balance each other whenever the form of the shell
is such that the intercepted parts Qq, Rr of the chord qQRr are
equal. This being true for all chords through P, the attraction of
every element is balanced by that of the opposite element, and the
resultant attraction on P is zero.
When the shell is bounded by concentric spheres these in-
tercepted parts are evidently equal. The resultant attraction on
any internal point is therefore zero.
When the shell is bounded by similar and similarly situated
concentric ellipsoids the same is also true. To prove this we
notice that, since the chords parallel to QR have in the two
ellipsoids a common diametral plane, the chords QR and qr must
have the same middle point. It follows that the intercepted parts
Qq and Rr are equal.
3—2
36 ATTRACTIONS. [ART. 71
Since a thick shell may be analysed into elementary thin
shells, it follows that the attraction of any homogeneous shell
bounded by similar and similarly situated concentric ellipsoids
at any internal point is zero.
69. If P is on the outside of a thin ellipsoidal shell, bounded
by similar concentric ellipsoids, we may show by similar reasoning
that the enveloping cone whose vertex is P divides the surface into
two portions whose attractions at P are the same in direction and
magnitude.
When P is indefinitely close to the outer margin of the shell,
the infinitely small portion on the nearer side of the polar plane
exerts the same attraction at P as all the rest of the shell. If the
thin shell is spherical, the resultant attraction is known to be the
same as if the whole mass were collected at its centre. Putting m
for the mass per unit of area, the attraction at P of each of the
portions on the two sides of the polar plane is 2?rm.
70. We may apply these results to the solid bounded by two
concentric similar and similarly situated hyperboloids. If one
sheet attract and the other repel, the attraction on P is zero,
provided both sheets are on the same side of P.
Also a paraboloidal shell bounded by two equal paraboloids
having their axes coincident but their vertices separate exerts no
attraction at an internal point.
71. If the thin shell is ellipsoidal and P is very close to the
outer margin, the distance of P from the polar plane is infinitely
smaller than the linear dimensions of the curve of contact. The
attraction at P of the portion on the nearer side of the polar plane
is therefore the same as that of an infinite plate of the same thick-
ness, see Art. 22. The attraction at P of each of the portions on
the two sides of the polar plane is therefore 2irm, where m is the
mass of the shell in the neighbourhood of P per unit of area. The
attraction of the whole shell at a point P, just outside the shell, is
therefore twice that of an infinite plate of the same thickness as
that of the shell at P, i.e. the attraction is 4>7rm. It also follows
that the direction of the attraction is the same as that of the
infinite plate and is normal to the shell. This line of argument
will be more fully considered further on.
Let a, b, c be the semi-axes of the inner boundary of the shell,
ART. 73] ELLIPSOIDAL SHELL. 37
p a perpendicular drawn from the centre to the tangent plane, p
the uniform density of the shell, then m = pdp. The volume v of
the ellipsoid is f^ra3 (6c/a2), and the volume dv of the shell (being
the differential obtained on the supposition that b/a and c/'a are
constants) is 4>7rbcda and the mass M of the shell is pdv. Also
since the bounding surfaces are similar da fa = dp/ p. The resultant
attraction of a thin ellipsoidal shell bounded by similar ellipsoids
Mp
at an external point close to its surface is therefore equal to — r-
and its direction is normal to the surface.
72. Cylindrical elliptic shell. By making one axis of the
ellipsoidal shell infinite, we deduce that the attraction of any
homogeneous shell bounded by similar and similarly situated con-
centric elliptic cylinders at any internal point is zero.
Let (i be the mass per unit of length of a thin cylindrical shell,
and let the infinite axis be c ; then the whole mass of the shell is
p'c. The resultant attraction at any point just outside is equal to
and its direction is normal to the surface.
73. Ex. 1. Prove that, if the attraction of a shell is zero at all internal points
and the inner surface is an ellipsoid, the outer surface is a similar and similarly
situated concentric ellipsoid.
If possible let the outer surface have some other form. Describe a similar
ellipsoid to enclose and touch the outer surface at some point T. The difference
between the ellipsoidal shell thus formed and the given shell possesses also the
property that the attraction is zero. This shell has no thickness at the point T of
contact, and the surface density m at T is zero.
Let P be a point inside this shell very near T, draw a plane through P parallel
to the tangent plane at T. The attraction of the matter on one side of this plane
balances that on the other. But the attraction of the matter on one side is
ultimately zero (being in fact Ivm), hence the attraction of the other is unbalanced
and the particle P cannot be in equilibrium. [Todhunter's History, &c. Art. 1473.]
Ex. 2. If the matter composing a thin shell bounded by concentric spheres
attract according to the inverse /cth power of the distance, prove that the resultant
force on an internal particle P acts towards or from the centre according as K is
less or greater than 2. Cavendish, Phil. Trans. 1771.
The plane section whose centre is P is such that the longer segment PQ of every
chord QPR of the sphere is on the same side of the plane as the centre of the
sphere. Since the masses of the elements Q, E are as PQ2 to P.R2, the attractions
are as PQ2-* to PJJ2~*. The first is greater or less than the second according as
n<2 or >2.
Ex. 3. If matter attracting according to the law of gravitation be uniformly
distributed upon the circumference of a circle, show that the chord of contact
of tangents drawn to the circle from any external point divides the circle into
two arcs, such that the potentials at the point due to each arc are the same.
[Math. Tripos.]
62220
38
ATTRACTIONS.
[ART. 77
74. Potential of an annulus. We may use the method of
Art. 64 to find the potential of an annulus of a thin uniform
spherical shell at a point P on its axis.
Let DD'EE' be the annulus ; let PD = u, , PE = u» OP = r.
The potential of an elementary annulus QQ' being the same as
before, the potential V of the
whole annulus is
TT %7rma f , 27rma , N
F= 1 du = (u2 — MJ),
r J r
since in our case the limits of
integration are u = PD and
u = PE. In the same way the
mass M of the annulus is
2-Trma f 7 Trma,
- \udu=-
The potential of the whole annulus is F=
M
i +
75. If we suppose the annulus to form a c&mplete sphere except for two small
holes DD', EE', we have an expression for the potential which applies equally to
points inside and outside the shell, provided they lie on the axis. Let y be the
radius of either hole. When P is inside the shell the sum of the distances MJ and
wa differs from the diameter only by small quantities of the order y2 and the
potential is therefore sensibly constant. When P passes through the hole DD'
the distance u^ has a minimum value equal to y and then begins to increase
without vanishing or changing sign. When P is outside the shell the sum of
the distances ul and u% differs from twice the distance of P from the centre by
quantities of the order y2, so that the potential sensibly follows the law of the
inverse distance. See Art. 39.
76. Ex. 1. The internal and external radii of a thin spherical shell of density
p are a-t and a. Prove that the difference of the potentials at two points, one
inside and the other outside, both close to the surface, is 2irpt2. We notice that
this difference is of the second order of the small quantity t.
Ex. 2. A thin spherical shell of radius a attracts an internal particle P at a
distance r from the centre. If the shell be divided into two parts by a plane
through P perpendicular to the radius the resultant attraction of each part at P is
• {a- (a2-r2)*} where m is the surface density. [Todhunter's History, 1615.]
77. A solid sphere. To find the attraction of a solid uni-
form sphere at an internal point P.
Describe a sphere concentric with the given surface to pass
through P. The attraction at P of the matter between this
sphere and the given surface is zero ; Art. 65. The attraction
at P of the matter within this sphere is the same as if it were
ART. 79] SPHERICAL SURFACE. 39
collected at the centre, Art. 66. If r be the distance of P
from the centre 0, the attraction is ^Trpr3/^, where p is the
density. It follows that the attraction of a solid homogeneous
sphere at an internal point distant r from the centre is f irpr.
If (x, y, z) be the coordinates of P referred to the centre as
origin, X, Y, Z the components of attraction, we have also
X = - f TT/CXC, Y = - %irpy, Z=- %-rrpz.
These are obtained by resolving the resultant attraction, viz. |-7r/jr,
parallel to the axes.
78. To find the potential of a solid sphere at an internal point P.
If x and x + da are the radii of an elementary shell, taken
within the sphere passing through P, its potential at P is
^irpx^dxjr, Art. 64. In the same way, if y and y + dy are the
radii of an elementary shell outside the same sphere, its potential
at P is kirpy*dyly. The potential at P of the whole sphere is
therefore
_ fr 4tTrpa?dx fa
J» r Jr
If the density p of the sphere is uniform, this integral becomes
If the density is any function of the distance from the centre,
the integration can be effected when the function is given.
79. Ex. 1. A portion of a homogeneous spherical shell is cut off by a cone
whose vertex is at the centre and whose solid angle is dw. Show that the
attraction, per unit of mass, of the rest of the shell on this portion is
. &8
irp (o — a) — rn
^v ' 62
where a and b are the internal and external radii of the shell. Hence show that
when the shell is indefinitely thin the attraction is half that just outside.
Since the resultant attraction of a body on itself is zero, the attraction of the
rest of the shell is the same as that of the whole shell. The attraction on the
portion included is / -£ p2 . — ^- r2drdw ; dividing this by the mass attracted, viz.
/ o T*
\p (r3 — a3) du, we have the result above given.
Ex. 2. Prove that the pressure per unit of length, on any normal section of a
spherical shell of mass M and radius a, due to the mutual gravitation of the particles
tends to the limit M 2/16ira3, as the thickness of the shell is indefinitely diminished.
[Math. Tripos.]
Ex. 3. A solid homogeneous sphere is divided by a plane through its centre
into two hemispheres. These being placed with their plane faces coincident, show
40 ATTRACTIONS. [ART. 80
that the force required to pull them apart is ^J/-/a2, where M is the mass of the
sphere and a its radius.
Ex. 4. The density of a solid sphere varies as the nth power of the distance
from the centre. Show that the potential at an internal point ia
where p is the density at the surface and n + 2 is positive.
Ex. 5. A homogeneous sphere is divided into two parts by a plane QNR
bisecting OP at right angles, P being any point within the sphere and 0 the
centre. If a be the radius of the sphere and c = OP, prove that the attraction
at P of the larger part of the sphere cut off by the plane QNR is n times the
attraction at P of the whole sphere, where n=(3a-c)/4c.
Ex. 6. If I be an external point and C the centre of a sphere, prove that the
sphere on 1C as diameter, the sphere with centre I and radius 1C or the polar plane
of I will divide the sphere into two parts exerting equal attractions at I, according
as the law of attraction is the inverse square, the inverse cube, or the inverse fourth
power of the distance. [St John's Coll., 1885.]
If the law be the inverse nth power, and a radius vector from I as origin cut the
sphere in Q, B and the dividing surface in S, then 2 (IS)3-n=(IQ)3-n + (IR)3~n
except when n=3. The results given follow at once.
Ex. 7. If a homogeneous solid hemisphere of radius a and density p be referred
to the centre of the complete sphere as origin, the bounding plane circle as plane of
xy and the radius of the hemisphere perpendicular to the plane of xy as axis of z,
then the attraction at the origin is along the axis of z and is equal to irpa. Further
show that if V be the potential at a point xyz near the origin, then
V=irpa?+irpaz-%irp {z2 + ya + 4.22} (within the hemisphere),
and V=wpa?+irpaz-$irp {x* + y2-2z2} (without the hemisphere).
[St John's Coll., 1886.]
Ex. 8. The potential of a solid hemisphere of radius a and unit density, at an
external point P situated on the axis at a distance £ from the centre, is
the upper or lower sign being taken according as P is on the convex or plane side
of the body. The potential at an internal point may be found by subtracting from
the potential of the complete sphere, that of the missing half.
Ex. 9. A point P is situated very near to the rim of a thin hemispherical shell
on a prolongation of a radius of the rim. Prove that the component of attraction
at P of the shell in a direction perpendicular to the plane of the rim is ultimately
2m log 8ajx, where a is the radius, x the infinitely small distance of P from the
rim, and m the surface density.
Ex. 10. Two mutually attracting spheres are placed at rest in a vacuum. The
radius of each is one foot and the distance apart (surface to surface) is l/4th inch,
and the density the same as the mean density of the earth. Prove that they will
meet in less than 250 seconds. This problem is due to Newton who gave a wrong
numerical result. [Todhunter, History, &c. Art. 725.]
SO. Other laws of force. Ex. 1. Let the law of force be the inverse
/cth power of the distance. Let the potential of a thin homogeneous spherical
ART. 82] SPHERICAL SURFACE. 41
surface at a point P be V* or VK according as P is external or internal. Prove
the following results
,_ M _ (r-a)a-'-(r + q)8-* _ M _ (a - r)s~K - (a + r)3~K
*~' :~ "
2ar ~(K-1) (/c-3) ' " 2ar
o - , H--r T-
/c(/t-2) or /e(K-2)dr
where Jf is the mass and a the radius.
Ex. 2. Prove that the potential of a homogeneous solid sphere of unit density
at an internal point P distant r from the centre is
(a - r)4-* + (a + r)4~K)
4a- _ ((a
~(K-l)(K-3) \
2(5-«)r
To this we add an infinite constant when K>4. The integral takes another form
when /c=4.
Ex. 3. Let the law of attraction be the inverse cube. Prove that the potential
of a thin spherical shell at a point P distant r from the centre is F8' or F3 according
as P is external or internal, where
^ , M r+a rr M a+r
F.'= -7— log - , F3= -j— log - .
4ar r-a ' 4ar a - r
Prove also that the potential of a solid sphere of unit density at an internal point is
„. IT fa2-?-2, a+r \
V=- \— s— log — — + arV.
r [ 2 6a-r j
81. To /rcd iAe potential of a shell bounded by any two non-
intersecting spheres.
Let A and B be the centres of the spheres, a and b their radii.
Let p be the density of the attracting matter which fills the space
between these spheres.
The potential at any point P is evidently the difference of the
potentials of the spheres each regarded as a solid sphere of density
p. If r, r be the distances of P from A and B respectively, the
potential at P is
according as P is outside or inside both spheres. If P lie between
the spheres
(2/>3\
3a2-r2-=^7- .
r I
82. We may use the same principle to find the attraction of a
shell bounded by two non-intersecting spheres.
Suppose, for example, that the attracted point lies within both
spheres. The force at P is evidently the resultant of two forces,
(1) an attraction equal to ^irp.PA acting along PA, and (2) a
repulsion equal to f -jrp . BP acting along BP. By the triangle of
42 ATTRACTIONS. [ART. 84-
forces, the resultant of these is equal to $7rp . BA acting parallel
to BA. Thus the attraction at all internal points is the same in
direction and magnitude. The attraction at an external point may
be found in the same way.
83. Ex. Two spheres touch at a point O, and the space between is filled with
homogeneous attracting matter. Show that, when the radii differ by an infinitely
small quantity, the attractions at two external points, one at O and the other at the
opposite extremity of the diameter through 0, are as 1 : 5. What is the ratio if the
points are inside both spheres ?
84. A theorem of Gauss. The mean value of the potential
of any attracting system, taken for all points on any spherical sur-
face, is equal to the potential at the centre due to that part of the
attracting system which lies outside the sphere plus the quotient of
the mass inside the sphere by the radius.
Let da- be any element of surface of the sphere, V the potential
of all the attracting mass at this element. Let M be the mass
inside the sphere and M' that outside, and let Fa be the potential
of the latter at the centre C. Let a be the radius of the sphere,
f Vdo- T, M
then we have to prove that J- — - = V- , H .
47ra2 a
Let m be the mass of any particle of the attracting system, and
let it be situated at a point A. Its potential at any point Q of
the sphere is therefore m/AQ. The part of the integral fVda- due
to this mass is therefore $md(r/AQ. The integral jdajAQ is
evidently the potential at A of a thin stratum placed on the
sphere, of unit surface density, and is therefore equal to 4>7ra*/AC
or 4nra*/a according as the point A is situated outside or inside the
sphere.
Taking all the particles of the attracting system, every particle
m outside the sphere contributes a term 4?ra2 . m/ACto the integral
/ Vd<r while every particle m' inside contributes a term 4?ra2 . m'/a.
We therefore have -. — - = S -r~ H . Since Fa is the potential
4-n-a2 AC a
of the external mass at the centre of the sphere, the result follows
at once.
Ex. Prove that the mean value of the potential of a body, taken for all points
equally distributed throughout the volume of a sphere which is external to the
body, is equal to the potential of the body at the centre. This theorem was given
by Poisson for the component of attraction on any given direction. Comptes Rendus,
vol. VH., 1838.
ART. 86] SPHERICAL SURFACE. 43
85. Heterogeneous spherical shells. The potential of a
heterogeneous spherical shell may be found by the help of La-
place's functions more easily than by any other method. Although
there are several cases of heterogeneous shells whose attractions
may be found by special artifices, it does not seem useful to stop
over these when they can all be treated by one comprehensive
method. We must however postpone the discussion of this
method until after we have reached Laplace's equation. In the
meantime there are some general theorems on heterogeneous
shells which are independent of Laplace's functions, and to these
we shall now turn our attention.
86. The potential of a thin heterogeneous spherical shell being
supposed known at all internal points, it is required to find the
potential at all external points.
Let 0 be the centre, a the radius of the sphere. Let P, P' be
two points on the same radius, one inside
and the other outside, such that
OP. OP' = a2.
The points P, P' are called inverse
points. Let OP = r, OP' = r'. 0
Let Q be any point on the surface,
then since OP.OP' = OQ2 the triangles QOP, P'OQ are similar.
It follows that the ratio QP/QP1 is constant for all points on the
sphere, and that this ratio is equal to a/r'.
Let V, V be the potentials of the whole shell at P, P '. If m
be an element of mass at Q, the potentials of m at P and P' are
respectively m/QP and m/QP'. Since these have a constant ratio
for all positions of Q and all values of m, the potentials V, V must
have the same ratio. We therefore have V = V — .
r
If the law of force is the inverse «th power of the distance, the potentials of
the mass m at P and P' respectively are in the ratio 1/(QP)'C-1 to 1/(QP')'C-1. We
(a\K-i
-j I .If the law of force is the inverse distance we find in
the same way that V — F=Afloga/r' where Mia the whole mass of the shell.
We notice that these theorems do not require the shell to be homogeneous or the
sphere to be complete. They apply to any distributions of attracting matter on the
surface of the sphere.
Ex. The potential at an internal point of a thin homogeneous shell of radius a
being V=M/a, prove that the potential V at an external point distant r' from the
centre is V' = M\r'.
44 ATTRACTIONS. [ART. 89
87. A theorem of Stokes. Let X, X' be the radial components of the attrac-
tions at P, P', estimated positively when directed from the centre. Then since
-' ~d' ~~
•when the points P, P7 approach indefinitely near to the surface r'=a, and this
equation reduces to X'-^X—-Vja.
We therefore have the following theorem. The sum of the inward normal attractions
at two points on the same radius, one just inside and the other just outside a thin
heterogeneous spherical shell, is equal to tJie potential at either point divided by the
radius. This theorem is given by Sir G. Stokes in his article on the Figure of the
Earth, and is there proved by the use of Laplace's functions.
88. Let Y, Y' be the components of the attraction at P, P' perpendicular to
OPP'. Let the radius vector OPP' turn round 0 through an angle dO. Then
dF^_dF aL_dV^f^
"Vd-0 ~ de ' r12 ~ rdO \r>
When the points P, P' approach indefinitely near to the surface we have F'=F.
89. A converse problem. To determine the law of force when it is given that
the attraction of every thin uniform spherical shell at every external point is the same
as that of an equal particle placed at the centre. Laplace, Mec. Celeste, vol. i. p. 163.
Let the potential of a particle m at a distance u be mf(u). The potential of the
shell at a point P is — — JM/(M) du, the limits being r-ator+aora-rtoa+r
according as P is external or internal, Art. 64. Since the attractions are equal, the
potentials of the shell and the central point must differ by a quantity independent
ofr. Hence — $uf(u)du=4Trmaif(r) + 2irmaA ........................ (1),
where A may be a function of a but is independent of r. If the potentials of the
shell and the central point are also to be equal we must have A=Q.
Put uf(u)=F'(u)> the equation (1) then becomes
F(r + a)-F(r-a)=2aFJ(r) + Ar ........................... (2),
where r>a. Since the equality is to hold for shells of all radii, we may differentiate
this equation with regard to a. Differentiate twice with regard to r and twice with
regard to o, we then have flT (r+a)=Flv(r-a) ................................. (3).
Since r and a are independent variables this equation cannot hold unless each side
is a constant, for if we write r=a, we have Flv(2a) = & constant. We therefore have
.F(r) = a + /3r+yr2 + 5r3 + er* .............................. (4),
where a, /3, y, S, e are constants. Since (3) has been obtained from (2) by
differentiation, this value of F(r) may not satisfy (2). Substitute in (2) and we
find 5=0, 4 = 8ase. We thus have
(5).
The only laws of force therefore which can make the attraction of every shell equal
to that of its central point are the inverse square, the direct distance and any com-
bination of these. It is unnecessary to include the case in which the potential is
constant, since the attraction is then zero. // the potentials also are to be equal we
must have .4 = 0 and therefore e=0. The potential must then vary as the inverse
distance. If the potential of m is -£mu2 the force varies as the distance. It is
easy to prove by direct integration that the potential of the shell cannot be equal to
that of the central point, but exceeds it by - |Mu2.
ART. 92] METHOD OF DIFFERENTIATION. 45
90. We may also enquire the law offeree when it is given that the attraction of
every thin spherical shell is zero at all internal points. We then find
F(a + r)-F(a-r)=Ar (6).
Differentiate twice with respect to r, we find F" (a + r) = F" (a-r). Since both a
and r are independent variables, this as before requires that each side should be
equal to some constant /3. We then find
/(«)=£+£ (7),
where 2p=A. The only law of force is therefore that of the inverse square.
91. We have assumed in this investigation that the law of force is required to
be independent of the radius of the spherical shell. If we remove this restriction,
there may be other laws of force which make the attraction of a given shell at all
external points equal to that of a central mass*.
To determine these laws we must solve equation (2) without differentiating it
with regard to a, because a is no longer arbitrary. Since (2) is linear, we follow
the rule in differential equations and put F(r) = er*+MePr, where the first term
represents a particular integral introduced to clear (2) of the term Ar. Substituting
this value of F(r) in (2) we arrive at the equation ePa-e-J3a=2jja. This equation
gives all the possible values of p.
This equation has three roots equal to zero and has no other real values of pa.
These lead to the value of F(r) given in (4). To find the imaginary roots we put
pa = a + pi, we then have cos p . sinh a = a, sin p . cosh a = p.
By roughly tracing these curves (regarding a and p as coordinates) we find that
there is an intersection between p = 2mr and (2n + J) ir, where n is any integer
except zero. There is therefore a possible law of potential which however is a
function of the radius of the spherical surface.
We may obtain a simpler result if we enquire when the potential of a thin shell
can be equal to that of a central particle whose mass is ft times that of the shell.
The right-hand side of (2) must then be multiplied by ft and we have .4=0. We
then find eva-e~^a=2ij,pa. This equation determines /* when p has any given real
value. The law of potential isf(r) = (Beffr+Ce~Pf)jr. This law is the same for all
spheres but the ratio of the central mass to that of the shell depends on the radius.
That this law of potential satisfies the conditions given above is easily verified by
actual integration.
92. Method of differentiation. Let the potential of a
homogeneous body of density p at any point P, (£, rj, £), be
V = </> (f, 77, £). If we move the body a small distance dg, the
point P remaining fixed, the potential at P of the body in its new
position is V—(dV/d^)d^. Let us now construct a composite
body whose density at any point Q is the difference of the densities
at Q of the given body in its two positions. Since the boundaries
are not the same, the composite body consists solely of a thin layer
of matter placed on the boundary of the given body. The surface
* It is stated in Nature, No. 1572, Dec. 1899, that Dr Bakker has written a
paper on this subject in the Proceedings of the Royal Academy of Sciences of
Amsterdam. The author has not been able to see this memoir.
46 ATTRACTIONS. [ART. 93
density at any point R is pcos<f)d^, where </> is the angle the
outward normal at R makes with the axis of £. We therefore
arrive at the following rule ; if V = <£ (£, 77, £) is the potential at P
of a solid homogeneous body, the potential at P of a layer on its
boundary of surface density Ap cos <£ is — Ad V/d^, or, which is the
same thing, AX where X is the j~ component of attraction at P.
Here A is a constant for all elements of the attracting body.
If the body is heterogeneous, let its density be p = ty (x', y, z);
the interior of the composite body is not now vacant, its density is
Adp jdx', while the surface density at R is, as before, Ap cos <£,
where p is the density at R of the given body. We notice that
when the density of the given body is zero along the bounding
surface, the potential of a body of density dp'jdx is
93. Ex. 1. As an example consider the case of a homogeneous solid sphere.
The £ components of attraction at P are fira^f/r3 or $wp£ according as P is external
or internal. Hence these are also the potentials of a surface layer of density p cos <f>,
or px'ja if x' is measured from the centre.
Ex. 2. If F be the potential at P of a homogeneous body, prove that the
potential at the same point of a thin layer on its surface of surface density
A (xft - y\) is A ( x -; — y — j where X, p, v are the direction cosines of the
normal. [Turn the body round the axis of z through an angle 30.]
Ex. 3. The surface density at any point Q of an infinitely extended plane is m,
E is a given point distant E0=z from the plane. The potential of the plane at
any point P on the side of the plane opposite to E is F. Let EQ=r", EP=r and
let 0 be the angle EO makes with EP. Assuming the first of the following theorems
deduce the others.
If m=-5z then F=-^,
r 3 zr
_3|t yJ^i I* COS0\
3.5/* „ 2flyi 3 3cos0 3cos20-l)
l— — TT~ » '=— sr \~9.^ + 5 !•»
r17 z*r [z* zr r* I
_Ax'+By'
-7T3- •• -T-«r-
To deduce the second result from the first we perform the operation — on
z dz
both m and F. The third is similarly deduced from the second and so on. To
obtain the fourth we refer E to fixed coordinates x, y, z and operate on the first
with d/dx and d/dy.
The first result for a point P on the axis EO produced is obtained by an easy
integration. It follows by a theorem of Legendre on the attraction of solids of
revolution (to be proved presently) that this result being true for a point P on the
axis is necessarily also true when P does not lie on the axis.
ART. 95] SIMILAR SOLIDS. 47
94. Similar solids. Let dv, dv' be the volumes of two
corresponding elements Q, Q' ; p, p' their densities; r, r' their
distances from two corresponding points P, P'. The lines QP,
Q'P' are parallel and the forces have the ratio pdv/r2 to p'dv'/r'2,
which is the same as the constant ratio pr to p'r'. The resultant
attractions of similar and similarly situated solids at corresponding
points are therefore parallel and have the ratio pr to p'r'.
In the same way the attractions of similar surfaces at corre-
sponding points are in the ratio of their surface densities.
Heterogeneous bodies. Let the density of a solid body at any
point Q be p = ^(x, y, z), where ^ is a homogeneous function of
the coordinates of s dimensions. Let the potential at a point P
be F=<£(£ *?,£>.
Increase the dimensions of the body and the distance of P
from the origin 0 in any given ratio 1 : /3. We thus have two
bodies bounded by similar surfaces 8, S' attracting two points
P, P' similarly situated. Since the potentials at the points P, P'
of corresponding elements at Q, Q' are proportional to the masses
divided by the distances, the potential at P' of the enlarged body
is F'=3*-
The potential at P' of a thin shell bounded by the surfaces ft and fi + djB may be
found by differentiating V with regard to ft on the supposition that the coordinates
of P' (viz. /3£, &c.) are constant. If we finally put /3 = 1, this shell will become a thin
layer placed on the surface S. Since d£/| = - d/3//3, &c. we have for the potential
where F=0(£, tj, f). Since this shell is bounded by similar surfaces, and its
density is \f/ (x, y, z), its surface density a at x, y, z, is <r=p\f/ (x, y, z) dp, where p
is the perpendicular on the tangent plane. Also if M, M' be the masses of the
original body and the stratum, M' = M(s + S) dj3. We may substitute for dp one or
other of these values according as we wish to express the potential in terms of the
surface density or the mass.
Laplace's, Poissons and Gauss' theorems.
95. Laplace's theoremf. Let (£, 77, £) be the coordinates
of any particle A of the attracting matter, and let m be the mass
of that particle. Let (as, y, z) be the coordinates of any point P.
* This formula for the potential of a heterogeneous stratum placed on the
surface of a known body was given by Ferrers in Q. J. vol. xiv. 1877.
t Mecanique Celeste, T. n.
48 ATTRACTIONS. [ART. 96
Taking the particle m apart from the rest of the matter, its
potential at P is Vl = m/r,
where r2 = (x - & + (y - tf + (z - £)2 (1).
dr j.
Since r~dx = X~
^y x — £ d2V^ m 3m (as — £)2
we find -7— = — m— -r- , .'.— j-r- = — ^H -5 •
d?Vl_ m , 3m(y-7y)2 dzV: _ m
' ~df~ i* r* ' dz2= i* r8
Adding these three expressions and remembering equation (1)
tfFi d-Ft efrF,
wefind -dtf+-df+-d*=°'
Let now F be the potential of the whole attracting matter at P.
Then, since F is the sum of the potentials of the several particles,
d*V d*V d*V
it immediately follows that -j-^ + -^ + -r-^ = 0.
In this investigation we have assumed that the point P does
not coincide with any one of the attracting particles. If it did the
meaning of the potential of that particle would require some
further consideration. The theorem has therefore been proved to be
true only for a point external to the attracting matter. It will be
presently shown that the right-hand side is not zero when the
attracted particle forms a part of the attracting mass.
Laplace's equation is a differential equation which must be
satisfied by the potential "of every body at all points not occupied
by attracting matter. If a general solution of the equation could
be found, that solution would comprise within its compass the
potential and therefore the component attractions of all bodies.
<$V d?V d?V
Laplace's function -7— + -j-j + -7-^ is often written in the
(jL&j dy (Lz
abbreviated form V2F.
96. When the law of attraction is the inverse *cth power of the distance we
have VK= — ^-S — -, (Art. 43). We may then prove that
K — 1 —l
When therefore the potentials of a body at an external point P are known
functions of the coordinates of P for the laws of the inverse cube and the inverse
fourth power, this theorem enables us to find by simple differentiation the potentials
of the same body for any higher inverse power. [Jellett, Brit. Assoc., Dublin 1857.]
ART. 101] LAPLACE'S THEOREM. 49
Ex. If the point P is internal and the body is homogeneous and of density p,
prove that the left-hand side of Jellett's equation should be increased by the
constant 4irpQ2-*. [See Arts. 80, 105.]
97. When the attracting body is a heterogeneous spherical surface we find that
where a is the radius and r the distance of the point P from the centre. This
result holds whether P is external or internal.
We have VK= I , u*=a2+r'2-2apr, where da is an element of area at Q,
K-iJ u""1
u = QP and p=cosQOP. To obtain the result, substitute this value of F< on the
right-hand side and eliminate p.
This theorem may be used to find the potential of a circular ring or of any curve
which can be drawn on a sphere.
98. When the attracting body is a lamina of finite extent, not necessarily
homogeneous, and the potentials at points in the plane of the lamina only are
required, the formula takes either of the forms
When the potential of the lamina is required at a point P not in its plane, we
notice that the component of force at P normal to the plane due to any particle m
of the attracting plane is - mz[rK+l where r is the distance of m from P. Summing
1 dV
up for all the particles we find FK+2= -5—. [James Eoberts, Quarterly
K -f* I. zaz
Journal, 1881.]
99. The potential VK of a body when the law of force is the inverse nth power
cannot be constant througlwut any finite space unoccupied by matter unless the law of
force is the inverse square. It is assumed that all the m's have the same sign, that is
every particle must attract or every particle must repel. For if FK=0, we have by
Jellett's theorem either F<+2=0 or *=2. But VK+2 is by definition the sum of a
number of terms all of which have the same sign, and therefore cannot vanish. In
the same way the potential of a lamina cannot be constant throughout any finite area
in its plane unless the law of force is the inverse distance.
100. Another important theorem should be noticed. If we transform the
coordinates from one system of rectangular Cartesian axes x, y, z
to another x', y', z' according to the scheme in the margin, it is
well known that
y'
6,
Thus x, y, z and djdx, dfdy, djdz are transformed by the same
rules. It immediately follows that since x2 + t/2 + zz=x'z + y'2+z'*
<FF d*V d*V_d?V dzV d?V
dx* + ~dy* + dz* ~ dx'* + dy1* + ~dz* '
This is an analytical proof of the invariant property of Laplace's equation. The
result follows more simply from Poisson's theorem (Art. 105), for each side of the
equation is there proved to be equal to - 4.wp.
1O1. Potential at an internal point. The potential at a point P of any
particles situated at the points Alt A2) &c. has already been defined in Art. 39
R. S. II. 4
50 ATTRACTIONS. [AET. 102
to be 2m/r. It is evident from this definition that, if a finite quantity of matter
be situated at any one of the points Alt A2, &c. in a condensed form, the potential
at a point P in the immediate neighbourhood of that point is very great, and at
that point itself this definition would make the potential infinite. But if the
attracting matter is so distributed in space that the mass which occupies any
elementary volume dv is pdv where p is finite, we shall now show that the potential
in this portion of space need not be infinite.
The potential at a point P in the interior of a body of finite density may be
found by taking P as the origin of polar coordinates and integrating all round
throughout the body. In this way we make r positive for every particle, Art. 39.
Describe a small surface S enclosing P and let its equation be r=e/(0, <£), where e
is a small constant factor. An element dv of the volume distant r from P is equal
to r^dudr, where du is the solid angle subtended at P. If then F2 be the potential
at P of the matter filling this surface, we have F8=J— —^pdurdr ......... (1),
where the limits of integration for r are 0 and e/(0, <j>). It is evident therefore that
F2 is of the order e2.
It follows that when e is evanescent the value of F2 is zero. Thus the matter
filling the surface may be removed without altering the potential of the whole
attracting mass. In finding therefore the potential of a body at any internal point
P we may regard P as situated in an infinitely small cavity, and determine the
potential as if P were an external point.
Let us consider next the resolved attraction at the point P of the matter filling
the small surface described above. Let X.2 be the component parallel to the axis of
x, then -X2=J^-^cos0=JJpcos0da>dr ........................... (2),
where 6 is the angle the radius vector r makes with the axis of x. It is evident
that X2 is at least of the order e of small quantities, and therefore vanishes when the
size of the surface is evanescent. Since cos 0 is negative when 0 > ?r/2 the order of the
term may be higher than e.
Lastly let us find the order of dV2jdx. To simplify the integrations let us
suppose that the surface is spherical, so that we may use the formula for the
potential already obtained in Art. 78. Let the radius of the sphere be e, let the
coordinates of its centre be (a, b, c) and those of P be (x, y, z). Then
*-«)2-(y-&)8-(*-«)2} ..................... (3).
It follows at once that =-^.(x-a), J=_|P (4)
dx 3 dx- 3
Since x - a is less than e, it is clear that dV^dx is a small quantity of at least the
order e, and vanishes when e is evanescent. In the same way the first differential
coefficients of F2 with regard to y and z are evanescent with e. The second
differential coefficients of F2 with regard to x, y or z are however not small.
We have supposed the density of the matter within the evanescent sphere to be
aniform. It is however clear that, if we substituted for p an expression of the form
p=p0+A (x-a)+&c.
we should merely add to the expression for F2 terms of the order e3.
1O2. To prove that the relation X=dVjdx which has been established for an
external point also holds for an internal point. Let Vlt F2 and Xlt X% be the
potentials and components of force at P due respectively to the matter outside and
inside a small spherical surface S. Then F= Fj + F2 and X=Xl + Xi. Since P is
ART. 105] POTENTIAL AT AN INTERNAL POINT. 51
external to that part of the body which is outside S, X1 = dVlfdx. "We have just
proved that dV^dx and X2 are each equal to zero when the size of the surface is
evanescent. Hence X=dV/dx. . ..
Ex. 1. Let the point P be situated at the middle point of the axis of a right
circular cylindrical cavity of altitude 2h, and let x be measured along the axis.
Prove that -j— = - ±irp ( 1 - y j where I is the distance of P from any point of either
rim. Thence show that in a flat cylindrical cavity dX^dx is - 4trp and in a long
cylinder is zero.
Ex. 2. Let the law of force be the inverse Acth power of the distance. Prove
that for a homogeneous body the relation X=dVjdx holds for an internal point P.
It is sufficient to prove this for a sphere enclosing the point P. Take P for
origin, we then find by an easy polar integration the value of X. The value of F at
the same point has been given in Art. 80. The integrations are shortened by taking
P near the centre.
103. We shall now prove that, when a point P passes from the interior of a
body of finite density into external space, both the potential and the attraction
undergo no sudden change of magnitude, but the second differential coefficients of the
potential are discontinuous in value.
Describe round the point A of emergence a small surface S of any convenient
form. Since both the potential and the attraction due to the matter within S are
zero, the points near A may be regarded as both external and internal.
All that is meant is that there is a numerical continuity in the potential. The
potentials of a solid sphere, for example, are represented by different analytical
expressions at points inside and outside, but at the surface both these have the
same numerical value, viz. Mja, Art. 78.
104. When P traverses an infinitely thin stratum whose surface density is finite
the volume density is not finite. It will be shown further on that the potential is
continuous, but that the attraction does undergo a sudden change of value, and an
expression will be found for the change.
It is at once evident from Arts. 15 &c. that when P arrives at an infinitely thin
line of finite line density, both the attraction and the potential are infinite.
105. Poisson's theorem*. If Fbe the potential of a body
at an internal point P at which the density p is finite, then
d*V d*V d?V
Describe a spherical surface of radius e enclosing the point P,
let (a, b, c) be the coordinates of its centre, (x, y, z) those of P.
Let the radius e be so small that the matter enclosed by the
sphere may be regarded as of uniform density.
Let Va be the potential at P of the matter within the sphere,
* This theorem is given by Poisson in the third volume, page 388, of the
Nouveau Bulletin des Sciences par la Soc-Hite Philomathique de Paris, 5e Anne'e 1812.
He proceeds very nearly as in Art. 105.
4—2
52
ATTRACTIONS.
[ART. 106
F! that of the rest of the body, then F=F1+F2. But
Laplace's theorem, V2Fi = 0, hence
where X2, F2, Z2 are the resolved attractions at P of the matter
within the sphere. But by Art. 77
X3 — — ±irp(x — a\ F2 = — 47rp(y — b), &c.
It easily follows by substitution that V2F=-4<7rp. Another
proof of this theorem founded on Gauss' theorem is given a
little further on.
We may notice that the centre of the sphere, though arbitrary in position, must
not be taken coincident with P. The reason is that we differentiate F2 with regard
to the coordinates of P, i.e. we make P travel from the point (a;, y, z) to a neigh-
bouring point (x + dx, &c.). But since the centre of the sphere is fixed, it cannot be
made to coincide with both the positions of P.
Ex. When, the law of force is the inverse distance and the attracting body
is a lamina attracting particles of its own substance prove that -3-5 + -7-5- = — 2rp.
ax ay
[Deduce this from the attraction of a cylinder (Art. 14) or from that of a circular
area (Art. 57) by the method of Art. 105.]
106. Gauss' theorem. Let S be any closed surface, and let
Ml be tJie sum of the attracting masses which lie within the surface,
M2 the mm of the masses outside. Let da- be any element of area of
this surface, F the normal resolute at this element of the attraction
of the whole mass both internal and external. Then fFda- = ± 4!'7rMl
where the integration extends over the whole surface of S and the
upper or lower sign is taken according as F is estimated positive or
negative when the normal force acts inwards*.
* This theorem was given by Gauss in 1839, his paper is translated in Vol. m.
of Taylor's Scientific Memoirs. It was also given by Sir W. Thomson in 1842 in
his papers on Electrostatics and Magnetism. The demonstration given by Sir G.
Stokes in 1849 has been followed here. He also deduces the Cartesian form of
Poisson'a equation from Gauss' theorem. See his Mathematical and Physical
Papers.
ART. 106] GAUSS' THEOREM. 53
Let m be the mass of any particle of the attracting system, and
let it be situated at the point A. A straight line drawn through
A to intersect the surface S in any point will also intersect it in
some other point, but, if the surface is re-entrant, it may enter and
issue from the surface any even number of times. Let the points
of intersection, taken in order, be Plt P2, &c., and let the direction
PiP2, &c. be called the positive direction of the straight line.
Let 0lt B3, &c. be the angles the positive direction of PiP2, &c.
makes with the normals PiNi, P&Nz, &c. drawn outwards. It is
evident that where the line enters the surface cos 9 is negative,
and where it issues from the surface cos 6 is positive, thus the
angles B1} 02, &c. are alternately obtuse and acute.
With A for vertex describe about this straight line an elemen-
tary cone whose solid angle is dw, and let it intersect the surface S
in the elementary areas dcr1} da-^, &c. If the distances J.P1 = r1,
APz = r2, &c., these elementary areas by Art. 26 are
d&i = ri2dw sec (vr — 6^), d<r2 = rz2dco sec 02, &c ...... (1).
If the point A is external to the surface as in the upper part
of the figure, the normal resolutes taken positively when acting
Wi 777
outwards are Ft= — cos (TT — #a), F2 = -- - cos $2> &c ...... (2).
Since the signs of these terms are alternately positive and
negative, it follows that when A is external
F1da-1 + F2do-.2 + &c. = 0 ..................... (3).
If the point A is internal and lies between Pj and P2, as
represented in the lower part of the figure, the sign of the force
F! must be changed. We therefore have
Fjd^ + F2da-z + &c. = - '2mda> ............... (4).
If the point A lie between P2 and P3, the signs of the first two
terms in the series (2) are changed, and the equation (4) resumes
the form (3), and so on.
If we now let the straight line AP^P^ &c. revolve round A into
all positions, all the elements of the surface will be included in the
integration. We therefore find for an external point fFda = 0 (5).
For an internal point the integration of the right-hand side of
(4) is limited to a hemisphere of the unit sphere, Art. 26. We
therefore have fFda = — 4>7rm ........................... (6).
Let now the system consist of any number of particles mt, m^,
54 ATTRACTIONS. [ART- 108
&c. inside, and m/, wi2', &c. outside the .surface S. The particles
outside contribute nothing to the integral $Fd<r, while the particles
inside contribute respectively — 47r-m1, — 4nrm2, &c. On the whole,
when F is measured positively outwards, we have
fFda- = -4>7rM1 ........................... (7),
where M^ stands for the sum of the internal particles m^, m.2, &c.
The truth of the theorem is not affected if some of the matter,
instead of being attractive, be repulsive. Such matter must however
be regarded as having a negative mass.
107. The product Fda- represents the product of the normal
resolute of the attraction at an element multiplied by the area of
the element across which it is supposed to act. This product is
sometimes called the flux or flow of the attraction across the
elementary area da- in the direction in which the component F is
measured. When the particles of the body attract, the proposition
asserts that the whole inward flux across any closed surface is equal
to 4?r multiplied by the mass inside. The product Fda- is also
called the induction through the element ; see Maxwell's Electricity.
We sometimes require the flux or induction across a portion only of the surface
S instead of across the whole. Let this portion subtend a finite solid angle w at
any one attracting point m. Then by what precedes the flux or induction across
this portion due to the attraction of m is mu>. If there are several attracting points
we may find the flux due to each and add the results together.
1O8. To deduce Laplace's and Poisson's theorems from Gauss1 theorem. To
effect this we take as the closed space to which we apply Gauss' theorem the
element suited to the coordinates we intend to use. Let P be any point of space
and let df , dij, df be the lengths of the three edges which intersect at P.
In Cartesian coordinates the element has its edges parallel to the coordinate
axes and therefore d£=dx, drj = dy, d£=dz. The sides of the polar element are
d£ = dr, dt) = rd&, df=rsin0d0, while those for cylindrical coordinates are d% = dR,
It should be noticed that in all these cases the three edges which meet at any
corner of the element are at right angles. The mass inside the element is
M=pd£d-r)d£ in every case.
Let V be the potential at P. Consider first the two faces perpendicular to the
edge at d£ ; the inward flux for the one and the outward flux for the other are
f
The total outward flux for these two is therefore — df. Treat the two other pairs
of faces in the same way and equate the whole flux to - 4irM. We then have
ART. 109] GENERALISED COORDINATES. 55
If we now substitute for d£, <fr/, df their values in Cartesian coordinates and
divide by the product dxdydz, this becomes
<PV cPV d?V_
dx2 + dy* + dz* ~ p-
If we substitute the polar values of d£, di\, d'f and divide by r2dr sin 0ddd<f> we
find
1 & . dV\ 1 d2F
1 d (»dV
*to\ W
If we substitute the cylindrical values we have
1 d /dF\ 1 &V d"-V_
~*-
Poisson's equation in oblique Cartesian coordinates takes the following form.
Let o, p, 7 be the sines of the angles between the axes ; A, B, G the angles between
the coordinate planes, then
dxdy dydz
where D
1O9. Orthogonal and elliptic coordinates. Let the equations of three
surfaces which intersect at right angles be
a=f1(x,y,z), P=f3(x,y,z), y=f,(x, y, z) .................. (1),
where a, /S, 7 are three parameters whose values determine which surface of each
system is taken. These parameters may be regarded as the coordinates of the point
P of intersection of the three surfaces.
Let P£, Pi), Pf be normals to the surfaces a, /3, 7 at the point P of intersection.
Let the direction cosines of these normals be (\ /Xj v^), (\2 ^ P2), (\3 /j.3 v3). We
therefore have \1 = ~, /j^^^, Vl = —z- h^ — axz + ay2 + az2 where sumxes denote
fin li-i ftn
partial differential coefficients.
Let PQ = d£ be an element of the normal P| and let (xyz), (x + dx, &c.) be the
coordinates of the extremities of d£. Then
da 1
-;- = -r-(axdx + aydy + aidz) = \-idx + fjL1dy + vldz.
"T. "i
The right-hand side represents the sum of the projections of dx, dy, dz on the
normal and this is d£. Hence
da. dp dy
d*=v d"=v d^'
The general equation of flux for the orthogonal element d^drid^ is by Art. 108
~ df +&c.= -
Substitute the values of d£, dr), df, and we find after division by dadfidy
d_(Jh_dV\ d_(.Jh_dV\ d f hs dV\ _ - 4wp
da \h^h3 da) + dp \h^hlJp) + dy \V^^7/ ~ h^h^
The quantities h^h^n^ are given by
V = axa + a,3 +«,»», h^=p^ + p^ + p^ V = 7z2 + 7/ + 7/,
and are supposed to be expressed in terms of the orthogonal coordinates 0^7, the
Cartesian coordinates xyz being eliminated by using the equations of the orthogonal
surfaces. This equation is sometimes called Lame's transformation of Poisson's
equation.
56 ATTRACTIONS. [ART. Ill
11O. Since F is regarded as a function of a, |8, y, we have
dV _dV dV dV
d^~daa'+d^fi* + dyy"
with similar expressions for dV/dy and dF/dz. These we differentiate again and
substitute in Poisson's equation. Since the surfaces a, /3, y are orthogonal the
coefficients of tPF/dad^ &c. are zero. We therefore have
d?V dV
- 4*p - - (az2 + a,2 + a,2) + - (azx + *„, + a J + &C.
Let the arbitrary functions o, /3, 7 be so chosen that they satisfy Laplace's
equation. The Poisson equation then becomes
Let a be the potential of a thin ellipsoidal shell of unit mass, such as that
described in Art. 68. Let (a&c) be its semiaxes. It will be shown in the chapter
on the attraction of ellipsoids that the level surfaces of the shell are the confocal
ellipsoids. Let (a'b'cr), (a" &c.), (a'" &c.) be the semiaxes of the three confocais
which pass through any external point P.
Since V = ax3 + aj,2 + a,2, it is evident that h^ is the component of force at P due
to the shell in a direction normal to the ellipsoid (a'b'c1). It will also be shown
that this force is h, = -^-, = — ^n , where p' is the perpendicular from the centre
dp a o c
on the tangent plane. Similar expressions must hold for the hyperbolic confocais
by the principle of continuity.
If Dlf J>2, are the semi-diameters of the confocal ellipsoid respectively parallel
to the normals at P to the confocal hyperboloids we know that p'D1D3 = a'b'c' by
the properties of conjugate diameters. Also by the properties of confocal quadrics
D1*=a'2-a"!>, D22=a'2-a'"a and p'dp' = a'da'. By using these expressions, we put
the equation (1) into the form
(o"« - a"*) + (a"* - a") + (a* - a"*) = **p (a"* - a'"*) (a'""- - a") (a" - a'*) .
Sincej/dp'=a'da' the potentials a, /3, y are to be found from
da 1 d|8 1 dy 1
da'~ b'c" da" b"c"' da'"" b'"c'"'
This form of Poisson's equation agrees with that given by Lame.
Theorems on Hie Potential.
111. The potential of any attracting system cannot be an absolute
maximum or minimum at any point unoccupied by matter*.
If V be the value of the potential at any point P whose
* The theorems in this section may for the most part be found in Gauss'
memoir on Forces varying inversely as the square of the distance, 1840. In the
Cambridge and Dublin Mathematical Journal, Vol. iv. 1849, there is an interesting
collection of theorems on the potential by Sir G. Stokes. Most of these were already
known, but the proofs were much improved and put into new and better forms.
This paper is. reprinted in his collected works Vol. i. p. 104. The reader may also
refer to papers by Lord Kelvin in various volumes of the Cambridge and Dublin
Mathematical Journal, 1842 and 1843, reprinted in his Electricity and Magnetism.
There is also a memoir by Chasles in the additions to the Connaissances des Temps
for 1845.
ART. 113] THEOREMS ON THE POTENTIAL. 57
coordinates are as, y, z, the value V of the potential at any
neighbouring point P' whose coordinates are #+£, y + f]> 2+ £
will be given by
F«8) + &c.,
where partial differential coefficients are represented as usual by
suffixes.
If F were a maximum or minimum at the point as, y, z, the
first differential coefficients Vx> Vy, Vz would each be zero, and the
three second differential coefficients Vxx, Vyy, V& (besides fulfilling
some other conditions) would have the same sign. But since the
point P is unoccupied by matter, they must satisfy Laplace's
equations, Art. 95. Their sum must therefore be zero. It is
therefore impossible that all three should have the same sign.
It has not been assumed that the masses of all the particles
have the same sign. The theorem is still true if the forces due to
some particles are attractive, and those due to others are repulsive.
When the law of force is the inverse distance and the attracting body is a
lamina, we have at all points in that plane F^.^- Vyy = Q, Art. 105. It follows that
in this case also the potential cannot be an absolute maximum or minimum at any
point in the plane of the lamina unoccupied by matter. For other laws of force in
which the sum of F^, Vyy, Va is not zero, the argument does not apply.
We have here assumed that we may apply Taylor's theorem to the potential.
That we may do so follows from the definition given in Art. 39. It is clear that
the potential at P of a single particle and therefore of a system of particles whose
total mass is finite is a function of the coordinates of P which is continuous and
finite as long as P does not traverse any attracting matter. We may however put
the argument into another form which has the advantage of avoiding the use of
series.
112. Another proof. With P as centre describe a sphere of
small radius. If the potential F were an absolute maximum at
P, the potential at any point Q of the sphere must be less than
that at P. Thus F is decreasing for a displacement along every
radius of the sphere. It follows from Art. 41 that the outward
normal force F at Q is negative at every point of the sphere. But
by Gauss' theorem fFda- = 0 (Art. 106), which requires that F
should be positive for some elements of the sphere and negative for
others. In the same way it may be shown that the potential
cannot be an absolute minimum at P.
113. If the point P be situated within the substance of a continuous attracting
body of finite positive density p, the potential may be a maximum but cannot be a
minimum at P.
58 ATTRACTIONS. [ART. 116
To prove this we observe that the potential function F here satisfies Poisson's
equation instead of Laplace's. Since the sum of the three differential coefficients
FJUJ, Vn, Vn is negative, it ifl possible that each may be negative. In that case V
is in general a maximum.
If we adopt the second proof, we notice that Gauss' theorem requires jFJcr to be
equal to — 4irlf, where M is the mass inside the sphere of small radius. It follows
that F may be negative, and therefore F be decreasing, for a displacement along
every radius. The quantity V may therefore be a maximum at P.
114. If any arbitrary curve is drawn in space not intersecting
any portion of the attracting matter, the potential may vary from
point to point of the curve. At some points the potential may be
a maximum and at others a minimum for displacements restricted
to that curve. For example, if the curve touch a level surface the
space differential coefficient of the potential is zero at the point
of contact and the potential may be either a maximum or a
minimum. What we have proved in Art. Ill is that the potential
cannot be a maximum or minimum at any point for displacements
in every direction.
If the curve is a line of force, it cuts the level surfaces at right
angles and the space differential coefficient of the potential cannot
vanish unless the resultant force is zero, (Art. 47). The potential
at a point P which travels along a line of force always in the
same direction must therefore continually increase or continually
decrease until P arrives at a point of equilibrium.
At a point of equilibrium there are some directions in which
the potential increases and others in which it decreases (see
Art. 120). The point P may therefore resume its journey (though
not necessarily in the same direction as before) so that the
potential at P continues to increase or decrease. The journey
can be continued to an infinite distance unless stopped by arrival at
a point of the attracting mass.
115. If the potential is equal to any given constant quantity A
at all points of a closed surface S which does not contain any portion
of the attracting mass, it must be constant and equal to A at all
points of the space contained within the surface S.
For if it were not constant, there would be some point at which
either it is greater than at all the other points or less than at all
other points. But this has just been proved to be impossible.
116. Ex. 1. As an example of this theorem consider the case of a spherical
shell of uniform thickness and density. Describe a concentric sphere within the
shell. By symmetry the potential must be the same at all points of its surface.
ART. 119] POINTS OF EQUILIBRIUM. 59
Since there is no attracting matter within this sphere, it follows that the potential
is constant throughout its interior.
Ex. 2. If the potential is not constant throughout the superficies of any closed
surface S, let A be the greatest and B the least value. Prove that the potential at
all points within S lies between A and B. [Stokes.]
Ex. 3. A level surface S completely encloses all the attracting matter of a
Bystem. If the consecutive level surfaces extending from S to infinity be drawn,
prove that the potential continually decreases outwards from each to the next until
it vanishes at an infinite distance.
117. If the potential is constant throughout any finite space, it
is also constant throughout all external space which can be reached
without passing through any portion of the attracting mass. [Stokes.]
The external boundary of the space is necessarily a level surface. If possible
let A be a point outside the space at which the potential is a little greater than
within the space. Since the level surface through A cannot cut the boundary, the
potential at all points in the neighbourhood of A is greater than within the space.
We can therefore describe an indefinitely small sphere, passing through A and
having its centre 0 within the space, such that the potential is increasing outwards
along every radius drawn from O to any point on the sphere outside the space and
is constant along every radius which lies wholly within the space. It follows that
the normal force has the same sign at every element of this sphere. This however
by Gauss' theorem is impossible. In the same way it may be shown that no point
A can exist in the neighbourhood of the space at which the potential is less than
within the space.
Another Proof. It has already been pointed out in Art. 39 that the potential at
P is a continuous function of the coordinates of P. It follows that when an
expression has been found which represents the potential throughout any finite
empty space that expression must also represent the potential throughout all
external space which can be reached without passing through any portion of the
attracting mass.
118. Points of equilibrium. If an isolated particle placed
at any point P be in equilibrium under the attraction of any
system, that point is called a point of equilibrium. When every
point of a curve is a point of equilibrium, the curve is called a line
or curve of equilibrium.
When the potential of the attracting mass is known, the
positions of the points of equilibrium are found by equating the
first differential coefficients of the potential to zero, viz. dV/dx,
dV/dy, «Scc. ; for these represent the resolved parts of the forces
parallel to the axes.
119. The equilibrium of a free isolated particle attracted by
fixed bodies cannot be stable for all displacements or unstable for
all displacements, but must be stable with reference to some
displacements and unstable with reference to others. Earnshaw's
theorem. Camb. Transac., 1839.
(JO ATTR ACTIONS. [ART. 121
If the equilibrium were stable when the particle occupied a
position P, the potential must decrease in all directions from P,
i.e. the potential would be an absolute maximum at P, which has
been proved impossible. In the same way the equilibrium could
not be unstable for all displacements.
120. A particle is in equilibrium at a point P. It is required
to find the equation of the cone which, having its vertex at P,
separates the displacements for which the equilibrium is stable from
those for which it is unstable.
The level surface which passes through any given point has in
general a tangent plane at that point, but when the given point is
a point of equilibrium, such as P, the first differential coefficients
Vx, Vy and Fz are zero, and the equation of the plane is nugatory.
Resuming the expression for the potential V at any point
(# + £, &c.) neighbouring to (x, y, z), we have, (Art. Ill)
V - V = \VXX? + &c. + Vxy&i + &c- + cubes (!)•
For any small displacement from P which makes V greater
than V, the force on the particle will act from P, and the equili-
brium will therefore be unstable (Art. 41). For any displacement
from P which makes V less than V, the equilibrium at P will be
stable. To find the directions which separate the stable and
unstable displacements, we put V = V. The equation of the
separating cone is therefore found by equating to zero the terms
of the lowest order on the right side of equation (1).
The separating cone is therefore a quadric cone, unless all the
differential coefficients of the second order are also zero. It is a
real cone, since by Laplace's theorem Vxx, Vyy and Vzz cannot all
have the same sign whatever rectangular axes it is referred to.
The level surfaces in the immediate neighbourhood of a point
P unoccupied by matter are in general planes, but if P be a
position of equilibrium, they are hyperboloids with the separating
cone for a common asymptotic cone. If PQ be any radius vector
of one of these hyperboloids, the force of restitution for a given
small displacement along PQ varies inversely as PQ.
121. Ex. 1. Show that three straight lines at right angles can always be
drawn through the vertex on the surface of the separating cone. There is an
infinite number of such systems of straight lines.
Ex. 2. If the attracting body is symmetrical about an axis and the point of
equilibrium lie on the axis, prove that the separating cone is a right circular cone
ART. 124] LEVEL SURFACES. 61
of semi-vertical angle tan"1 ^2. [This follows at once from Laplace's theorem,
Art. 95.]
Ex. 3. The lines of force in the immediate neighbourhood of a point of equili-
brium, when referred to the principal diameters of the separating cone as axes, are
ze=Mxa=Nyb, where a, b, c are the reciprocals of Vxx, Vyy, Vtt at the point of
equilibrium, and M, N are two arbitrary constants.
Ex. 4. If a number of mutually repelling particles are enclosed in a rigid
boundary, show that when in stable equilibrium they all reside on the surface.
[If any one were not on the surface, that particle would be in unstable equilibrium,
the remaining particles being held at rest.] Kelvin, see Papers on Electrostatics,
&c., p. 100.
Ex. 5. Three uniform thin rods AB, BC, CA, which form a triangle, attract a
particle P placed at the centre of the inscribed circle. The particle is therefore in
equilibrium. Show that the equilibrium is unstable for all displacements in the
plane of the triangle.
122. If two sheets of a level surface intersect along a line,
every point of that line is a point of equilibrium.
Let P be such a point, then at least three tangents can be
drawn to the sheets of the level surface not all lying in one plane
and making finite angles with each other. Since the force along
each of these is zero, it follows that the particle is in equilibrium.
123. At every point of the curve of intersection of two sheets
of a level surface, the tangent cone becomes two planes which are
the tangent planes to the two sheets. The tangent cone may
therefore be written in the form
(a£ + brj + cO (a'£ + b'rj + c'£) = 0.
Comparing this with the form already found (Art. 120), we have
aa' + W + cc' = Vxx + Vyy + Vzz.
This is zero by Laplace^s theorem; the tangent planes are therefore
at right angles. We therefore infer that, if two sheets of a level
surface intersect, they intersect at right angles.
124. Ex. 1. The tangent cone becomes two planes whenever its discriminant
is zero. Prove that in a level surface these planes cannot be imaginary. [If it were
possible, the cone could be reduced to the form (a% + b-r) + cf)*+(a'i- + b'r) + c'£')z = ().
This would make a2 + a'2 + 62 + &c. = 0, by Laplace's theorem, which is impossible.]
Ex. 2. Show that an isolated line in free space cannot form part of a level surface.
If the potential at a point P were greater than that at some neighbouring point
Q and less than that at E, it would follow from the principle of continuity that there
must be some point between Q and R on every path from one to the other at which
the potential is equal to that at P. If then an isolated line form part of a level
surface, the potential must be either greater than at all neighbouring points not on
the line or less than at all such points. On either alternative the second proof, by
which it is shown that the potential cannot be an absolute maximum or minimum,
is contradicted, Art. 112.
62 ATTRACTIONS. [ART. 127
125. Rankine's theorem. If at any point of a level surface all the differential
coefficients of V up to the ?ith inclusive with regard to x, y and z are zero, we know
from solid geometry that there is a tangent cone of the (rc+l)th order at that point.
If (n + 1) sheets intersect along a line, the same thing will be true at every point of
that line, and the tangent cone will be the product of the (n + 1) tangent planes.
Let us suppose that the level surface is such that at two consecutive points P, P'
all the differential coefficients of F up to the nth are zero ; let us examine the form
of the surface in the immediate neighbourhood of those two points.
Taking P for origin and PP for the axis of z, we have at the origin all the
following differential coefficients equal to zero :
d*V d»F d«F dnV d»V dnV
_ _ .-, _ _ , _ ._. _ — Jfrrt
dxn' dxn-1dy "" dyn ' dxn~ldz' dyn~ldz' dae*-*d&'
These are also zero when z receives an increment dz ; hence their differential coeffi-
cients with regard to z are all zero. It therefore follows that every differential
coefficient of F of the (ra+l)th order which has dz, dz-, &c. in the denominator is
zero at the origin. If therefore F' be the value of the potential at a point £, 17, f,
we find on making the expansion by Taylor's theorem
F' - r=A0?+i + Aj»i + ... + An+lr,™ )
+ powers of £, 77, f of (n + 2)th order )
where A0, Al, &c. are constants. It follows that the terms of the lowest order in
the expansion do not contain f.
The level surface which passes through the origin is given by V ' - F=0. This
level surface has therefore (n + 1) tangent planes at the origin given by
U=A£** + A1pi+...+Art.ilf"*1 = 0 ........................ (3).
All these tangent planes pass through the two given consecutive points P, P'.
We shall now "prove that all these tangent planes are real, and that each makes the
same angle with the next in order. The expression for V given in (2) must satisfy
Laplace's equation, hence the expression for U given in (3) must also satisfy that
equation. Transforming to cylindrical coordinates, U becomes Z7=Pr™+I, where P
is some function of <f>. By Art. 108, since z is absent from U, we have
<PU 1 dU JL <PU
dr* +r dr + r2 d<t?
d?P
Substitute, and we find {(n + l)n + n + l} P+ —— = 0.
The equation (3) therefore reduces to cos {(n + 1) <p + a} = 0, which gives n+1 planes,
making equal angles, each with the next in order.
126. Tubes of force. If we draw a line of force through
every point of a closed curve, we construct a tube which is called
a tube offeree. By choosing the closed curve properly we can
make the section of the tube indefinitely small ; it is then called a
filament. It is evident that the resultant attraction at any point
P of a filament acts in the direction of the tangent to the length
of the filament.
127. The magnitude of the attractive force at any point of the
same filament is inversely proportional to the area of the normal
section of the filament at that point.
ART. 129] TUBES OF FORCE. 63
Let cr, a-' be the areas of the normal sections of the filament at
any two points P, P'. Let F, F' be the attractive forces at the
same points. These forces act along the tangents at P, P' to the
length of the filament and are supposed to be measured positively
in the same direction along the arc.
Let us apply Gauss' theorem to the space enclosed by the
filament and the two normal sections. Since the filament contains
no attracting matter the total flow of the attraction across the
whole surface is zero. The flow across the sides of the tube is
zero, because at each point the resultant force acts along the
length of the tube. The flow across the two normal sections must
therefore be zero, hence Fa- — F'a-' = 0, that is Fa- is constant for
the same tube.
128. Ex. 1. Let the attracting body be a sphere. The lines of force are by
symmetry normals to the surface ; the filaments are therefore conical surfaces of
small angle. If r be the distance of P from the centre, (? = r2du; hence Fr1 is
constant along any line of force. Thus it follows at once that the force of attraction
at any external point varies inversely as the square of its distance from the centre.
Ex. 2. If jPbe the normal force at any point P of a level surface; p, p' the radii
of curvature of the principal sections and ds an element of the arc of a line of force
at the same point P. then will — ^ 1 1- — =0.
cu p p
Construct on the level surface an elementary rectangle PQSR such that the sides
PQ, PR are elements of the lines of curvature at P. Let the tube of force having
this rectangle as base intersect a neighbouring level surface in P'Q'S'E'. If a, <r'
are the areas of these rectangles and ds = PP', we have by the properties of similar
ff' (p + ds)(p' + ds) . /I 1\
figures — =— '-¥• -=! + ( - + -; }ds.
<r pp \P pj
If F, F+dF be the forces at P, P7, we know that Fff = (F+dF) ff', Art. 127.
This immediately reduces to the required result. See Bertrand on isothermal
surfaces, Liouville's J. 1844, vol. ix.
129. If two different bodies have equal potentials over the
surface of any space not including any attracting matter, they
have equal potentials throughout that space, and also at all external
space which can be reached without passing through any of the
attracting matter of either body.
For let the attraction of one of the bodies be changed into
repulsion. Then the potential due to both bodies is zero over
the surface of the given space. That is, the united potential
is constant over the surface ; it is therefore also constant and zero
throughout the enclosed space, and at all points of external space
which can be reached without crossing any attracting matter;
64 ATTRACTIONS. [ART. 131
Arts. 115 and 117. Returning then to the original supposition
that both the bodies attract, it easily follows that their potentials
are equal.
130. If two different bodies have equal potentials over the whole
boundary of any surface enclosing both, they have equal potentials
throughout all external space.
As before, changing the attraction of one body into repulsion,
let us consider the potential of both bodies regarded as one system.
Their united potential is therefore zero over the whole boundary of
the surface. It is also zero over the boundary of an infinite sphere.
Since the space between the surface and the sphere contains no
attracting matter, the potential is also zero throughout that space,
Art. 115. Returning to the original supposition, that both bodies
attract, we see that their potentials must be equal.
Ex. An unknown body is surrounded by a sphere of radius a.- The direction
of the attraction at all points of this sphere is normal to the sphere and its
magnitude is equal to a given constant F. Prove that the attraction at any
external point is FcPjr*.
The sphere is a level surface because the force is normal. The potential of the
body at any point of the sphere is therefore equal to that of a particle whose mass
is Fa? placed at the centre.
131. If two different bodies have the same level surfaces throughout any empty
space, tJieir potentials throughout that space are connected by a linear relation.
Let F and V be the two potentials. Since when F is constant, F' is also
constant, it follows that F' is some function of F, say F'=/(7). Then by differen-
tiation we easily find
d*V' dW d^F' _ df_ <)d*V dW d-V) cPf_ \(dV\z (dZ^\ (*L\{
dx*~ + ~dy* + ~az* ~ dV (da? + dy* + dPj + dV~z \\~dx) + \dy) + \~dz ) ]'
Since the space is external to both bodies, this, by Laplace's equation, reduces to
d2f
0 = j~2 » unless F is constant throughout the space considered. If Fis constant, the
level surfaces for both bodies are indeterminate and therefore V also is constant.
We therefore have in both cases V=AV+B, where A and B are two constants.
Suppose the space considered includes the points at infinity, then when the
attracting masses are finite in size and density both F and F' vanish at such
points. We then have B = 0. Again F and V must vanish at infinity in the ratio
of the attracting masses ; we therefore find V'IV=M'IM if M, M' be the masses of
the attracting systems. We thus have the theorem ; if two finite bodies have the
same external level surfaces and have equal masses, their attractions at all external
points are the same in magnitude and direction. Quarterly Journal of Mathematics,
1867.
When the space in which the two bodies have the same level surfaces encloses
both bodies, this theorem follows at once from that proved in Art. 130. Since the
two bodies have the innermost level surface common, we can by altering the mass
of one of them make their potentials equal over that surface. The potentials of the
ART. 133] BODIES WITH EQUAL POTENTIALS. 65
changed bodies are then equal over all external space and the potentials of the ori-
ginal bodies have a constant ratio.
132. As an example of this theorem, consider the case of a spherical shell.
The external level surfaces of such a shell and those of an equal mass placed at its
centre are both spheres. Hence the attraction of a spherical shell at any external
point is the same as that of an equal mass placed at its centre.
Again, the level surfaces of two equal and parallel infinite plates are both planes.
Hence their attractions at any point are in a constant ratio. But at an infinite
distance the attractions of two such plates when separated by a finite interval tend
to equality, hence the ratio of the attractions is unity. It follows that the attraction
of an infinite plate at an external point is independent of its distance. In the same
way the attraction of an infinite circular cylinder is the same as if the whole mass
were uniformly distributed along the axis.
133. The theorems in this section have been enunciated with
special reference to the potential of an attracting system, but a
little consideration will show that they have a more extended
application.
If V be any continuous one-valued function which satisfies
Laplace's equation and is not infinite within any given space, it
follows from the argument in Art. Ill that F cannot be an
absolute maximum or minimum at any point within that space.
Most of the other theorems are simple corollaries from this one
general principle, and apply therefore to any finite continuous
function which satisfies Laplace's equation.
For example, if such a function be constant over the boundary
of any space and not infinite within that space, it must be constant
throughout that space.
To take another example, let V be a finite continuous function
which satisfies Laplace's equation, then V = c is a system of
surfaces. If any member of this system intersects itself in a
singular line, the two sheets are at right angles. If several sheets
intersect in a singular line, each tangent plane makes the same
angle with the next in order.
Let V, V be two continuous solutions which are both finite
and one- valued at all points of space bounded by a surface S and
are equal at every point of that surface, then they are equal
throughout that space. The space considered may be external
to 8 provided the functions are also equal at all points on the
surface of some sphere of infinite radius enclosing S. This
theorem shows that when the values of a function V are known at
all points of the boundary of a space, it is determinate throughout
R. 8. II. 5
66 ATTRACTIONS. [ART. 135
that space, provided it is known to satisfy Laplace's equation and to
be finite throughout that space.
134. To trace level curves and lines of force. Ex. 1. Three equal
particles are placed at the corners ABC of an equilateral triangle. Trace the level
curves and lines of force in the plane of the triangle.
We first search for the points of equilibrium. The centre of gravity G is evidently
one such point. The level curves near G are conies
(Art. 120) which must have GA, GB, GG for three
principal diameters. The conies are therefore
circles. The equilibrium is clearly stable for a
displacement perpendicular to the plane and is
therefore unstable for some (and therefore also for
all) displacements in the plane (Art. 111). The
potential is therefore a minimum at G for displace-
ments in the plane ABC.
Let D, E, F be the feet of the perpendiculars from
the corners. Since the force at F tends towards G and G is a point of minimum
potential, there must be a point of equilibrium between G and F. There are
therefore three points of equilibrium which are H, K, L. The level curve which
passes through these points governs the whole sketch and is exhibited in the figure.
Some of the other level curves fill up the four vacant areas and others surround the
three loops.
To sketch the lines of force. It will be found convenient to mark the level
curves or surfaces with small arrows to indicate the direction of the normal
force. We then have the following rule ; no line of force can pass from a point A
on one level surface to a point B on another unless either the arrows at both A and
B tend in the same direction along the line of force or the line of force passes
through a point of equilibrium which lies between A and B.
The arrows on the sides of the curvilinear triangle HKL all tend outwards from
G, while those on the three curvilinear triangles which surround A, B, C tend
inwards towards those points. Hence a line of force beginning at A must either
proceed to an infinite distance or cross KL. If it enter the triangle HKL it cannot
emerge without passing through G. It must then proceed onwards to either B or C.
There are conical points at H, K and L. The level surfaces near G are not
closed but bend over A, B, C, and surround the conical points.
Ex. 2. Two particles whose masses are m, ml are placed at A and B, both being
attractive. Trace their level surfaces.
Ex. 3. Three equal particles are placed at three points A , B, C in a straight
line. The particles A and C attract while B repels. Trace the level surfaces.
135. Potential at a distant point. To find the potential of
a body finite in all directions at any distant external point*.
Let the origin 0 be a point not far from the body. Let Q be
* The expansion of the potential at a distant point is originally due to Poisson,
but was put into a convenient form by MacCullagh, jR. Irish Trans. 1855. Some
of the following theorems were given by the author in the Quarterly J. 1857. The
name centrobaric is due to Lord Kelvin, who gave several theorems on these bodies
in the Proc. R. S. E. 1864. The results in Arts. 140, 141 are taken from Thomson
and Tait, 1883.
ART. 136] POTENTIAL AT A DISTANT POINT. 67
the position of any particle of the body, m its mass, (x, y, z) its
coordinates, r its distance from the origin. Let (£, t], £) be the
coordinates of the point P, OP = r', and the angle POQ = 9.
To generalize the investigation we shall assume that the law of
attraction is the inverse /cth power of the distance. We then have
F-.-V =*
(r/2-2rr/cos^ + r2)^~
__, m f 1 rcos# (/e + l)cos20 — 1 /r\a
-*7^{^1*'~^~ ~2~ ~Vr7+'
^* t
2^ffft
The first term of the series is -^— ^ =- . Hence the attraction at
I K ~" " J-
a very distant point is ultimately the same as if the whole mass
were collected into a single particle and placed at 0.
To find a closer approximation to the true attraction, let
the point 0 be such that the second term of the series vanishes.
This requires that 2mr cos 0 = 0. Since rr' cos 6 = x% + yrj + z%,
this gives ^mx + rfemy + ^mz = 0 for all values of f , 77, £ The
point 0 will therefore be the centre of gravity of the body.
We have now to consider the third term of the series. Let
A, B, G be the moments of inertia of the body about any three
straight lines at right angles meeting in 0, I the moment of
inertia about the straight line OP, then
Writing 1 — sin2 6 for cos2 6 and making these substitutions we
find for the third term
4 r'«+i ...............
When the law of force is the inverse square and the centre of
gravity is the origin we arrive at MacCullagh's expression for the
. .. , ^ M . A + B + C-3I . _
potential, viz. y=—-\ -- — -- [• ............... (3),
where M is the mass of the body.
Ex. When the law of attraction is the inverse distance, the potential of a
single particle takes the form G - m log r7. Prove that the potential of a body at a
distant point is V=C-M logr/+ ,, "4I+ ........................ (4).
' *
136. If two bodies have equal potentials at all external points,
their centres of gravity must coincide and their masses must be
equal. If the law of force is the inverse icth power the bodies are
5—2
68 ATTRACTIONS. [ART. 136
equimomental, unless K. — — 1. If the law is the inverse square, the
difference of their moments of inertia about every straight line must
be constant.
The potential of each body can be represented by the series
described in Art. 135 and these series must be equal, term to
term. The equality of the first terms requires that the masses
should be equal. Taking the origin 0 at the centre of gravity of
one body, the second term must be missing for both series and
therefore the centre of gravity of the second body must also be at.
the origin.
Comparing the third terms of the series we have
K (A + B+ C)- 2(* + 1)7= K (A' + B' + (7) - 2 (« + 1) /'...(o),
where unaccented and accented letters refer to corresponding
quantities in the two bodies. It follows that (unless K = — 1)
/ — /'is the same for all axes passing through the common centre
of gravity. The axes of maximum and minimum moments of
inertia in the two bodies are therefore the same. Since these
are the principal axes of inertia, the two bodies must have the
directions of their principal axes coincident. Since / — /' is the
same for every axis, it follows that the four differences A — A',
B — R, 0 — C', and /— /' are equal. The equation (5) then
becomes (K- 2) (/-/') = 0 (6).
Unless K — 2, we have /=/' and therefore the moments of
inertia of the two bodies about every axis are equal, each to each.
If however K = — 1 these conditions are not necessary. When
K has this value the law of attraction is the direct distance. In
this case it has already been proved that a body, whatever be its
form, attracts any particle as if it were collected into its centre of
gravity (Art. 8).
These are necessary conditions that two bodies should be
equipotential (unless # = — 1), but they are not sufficient. It is
also necessary that all the subsequent terms of the potential series
should be equal, each to each.
We have assumed here that the law of attraction is some one integral inverse
power of the distance. If the law he represented by a series of inverse powers such
as pliJC + (i'lrK+1 + &c., it is evident that so far as the series (1) in Art. 135 is
concerned we need only consider the three lowest powers of r in the law of
attraction. The remaining powers enter only into the terms of that series not
included in our approximation. Proceeding in the same way we again arrive at the
results stated in the enunciation.
ART. 139] CENTROBARIC BODIES. 69
137. Centrobaric bodies. When a body is such that its
potential at every point is equal to that of a particle of mass M
situated at some fixed point 0, the body is said to be centrobaric.
In other words, the body is equipotential to a particle. We infer
immediately from Art. 136 that M is equal to the mass of the body
and that 0 is its centre of gravity. Since for a particle /' = 0, it
follows that the moment of inertia / of the body about every axis
is the same. The body therefore cannot be centrobaric unless every
axis at the centre of gravity is a principal axis.
The condition (6) now becomes (K — 2) 7 = 0. It appears
therefore that the series (1) of Art. 135 cannot reduce to its
first term unless K = 2 or 7=0. The latter condition cannot be
satisfied unless the masses of some of the particles are negative,
that is unless some of the particles attract and others repel P.
Assuming that all the particles attract P, according to some inverse
power of the distance, we see that the attraction of a body cannot
be the same as if the whole mass were collected into its centre of
gravity unless the law of force be either the direct distance or the
inverse square of the distance.
138. Ex. If the law of force be the inverse square, the potential of a body at
all external points cannot be the same as that of two masses M^ and Jf, placed at
two points A , B fixed in the body unless (1) the body and masses have their centres
of gravity coincident, (2) the moments of inertia of the body about every axis
through the centre of gravity perpendicular to AB are equal.
139. Potential constant in a cavity. In a similar manner,
when a body has a cavity within its substance we may determine the
necessary conditions that the potential should be constant throughout
the cavity. Taking the origin within the cavity, we have at all
points close to the origin
^ m ( 1 /cosfl (K + 1) cos2 B - 1
V — Z ^^7 1 T "i r
"-1 \K - 1 r 2
the expansion is in powers of r'/r because r' is less than r.
This cannot be independent of r' unless the coefficient of each
power of r' is zero. Equating the coefficient of r'* to zero, we have
„, .,. Q „ ^ v ^ , . , . D
Writing a, ft, 7 for 2 — ^, 5-jJji ^^+3 and putting the point P
in succession on the axes of x, y, z we have KOL = ft + 7, /c/3 = 7 + a,
/cry = a + /3. These give K = 2, or K = — 1 and a + ft + 7 = 0, or
70 ATTRACTIONS. [ART. 142
a, $, 7 each zero. The two latter alternatives require that all the
m's should not have the same sign. Hence if every particle of the
body be attractive, the potential cannot be constant throughout any
cavity unless the law of attraction is the inverse square, (Art. 99).
140. Assuming that a body attracts all points in external space as if the whole
mass were collected into its centre of gravity, prove that (1) the centre of gravity is
inside the external boundary, (2) the external boundary is a single closed surface.
If the centre of gravity O were in the same external space as the attracted point
P, we could surround it by a small sphere, centre 0, radius c, which does not enclose
any particle of the attracting mass. The flux across this sphere is therefore zero,
Art. 106. But since the force on P tends always to 0, the flux is also ^irM. These
results contradict each other unless the whole mass is equal to zero.
Again, if the attracting system consist of two separate portions, the centre of
gravity 0 must lie inside one of them. Enclosing the other portion in a sphere, the
flux across the surface is 4irM', if M' be the mass of this portion. But since 0 lies
outside the sphere, it is also zero. These results cannot coexist unless the mass of
that portion is zero.
141. A body B is such that the resultant attraction between it and a given
body A is a force which always passes through the centre of gravity 0 of B, in
whatever position A is placed. Prove that the resultant attraction between B and
every body is a force which passes through the centre of gravity of B.
Let the body A be turned about a fixed point P sufficiently distant from B, that
the body A in its motion never meets the fixed body B. In all these positions the
resultant attraction of A on B is & force which passes through the centre of gravity
of B. Hence if every particle of the mass of A be uniformly distributed over the
surface of the sphere which that particle describes in its motions, the resultant
attraction of the mass thus obtained is also a force which passes through the centre
of gravity of B. The mass thus obtained is a spherical shell whose resultant attraction
at any point of B is the same as if it were collected at the centre P. The resultant
action between the body B and a particle placed at P is a force which passes both
through P and the centre of gravity of B. The body B is therefore centrobaric for
all points P beyond a certain distance and therefore for all points of space which can
be reached from P without passing over any of the attracting mass, Art. 129.
Attraction 'of a thin stratum.
142. A theorem due to Green *. Let a thin heterogeneous
stratum of attracting matter be placed on a surface which has no
* The theorem X' -X=4:irm is of great importance in the theory of attraction.
The principle of the demonstration given in Art. 142 was used for a spherical shell
by Lagrange in 1759 and was afterwards applied by Coulomb to the case of a thin
electrical film of any form (Paris Memoires, 1788). Poisson gives a generalization
of the theorem to any film (Mem. de.,.VInstitut, 1811, Connaissance des Temps for
1829, p. 375). Cauchy deduces the same result for any film from the general
formulae of attractions (Bulletin. ..Soc. Philornathique, 1815, p. 53). The theorem is
commonly called Green's theorem (Essay... on Electricity and Magnetism, 1828). It
was afterwards re-discovered by Gauss, 1840. A proof on the same general principle
as that in Art. 142 was given by Kelvin in 1842, see the reprint of his papers on
Electrostatics and Magnetism, 1842, and Thomson and Tait, Art. 478.
ART. 142]
ATTRACTION OF A THIN STRATUM.
71
conical points or other singularities. Let p be the density and t
the thickness at any point A of the surface, and let m = pt, so that
m is the surface density at the point A. In what follows we shall
regard m as finite and t as indefinitely small, so that p is very
large.
Let P, P' be two points situated on the normal at A, one
inside the surface and the other outside,
both close to the stratum; it is required
to find the attractions at P and P'.
With centre A, and a small geodesic
radius a, describe on the surface a circle
whose circumference is DE, and let DN
be a perpendicular on the normal at A
drawn from any point D. The radius a
of this circle is infinitely greater than
either AN or the thickness t but in-
finitely less than either radius of curva-
ture of the surface.
This circle divides the whole attract-
ing stratum into two parts whose attractions at P and P' will be
separately considered. Let us first find the attraction of the
small portion DAE which we may suppose to lie in the tangent
plane at A.
We take the attracted point P for origin and the normal PA
for the axis of x, let PA =p. By Arts. 21, 22, the attraction is
Zirpjdx (l - * ) = 27r/> [* - [a? + (p + *)f - {a2 +^>f ],
\ V \a + x )'
the limits being p to p + 1. Now ultimately p/a and tja are zero,
while pt = m. We have therefore for the attraction
The attractions at P, P' are therefore each equal to 2-Trw. They
are directed along the normal in opposite directions, and their
difference is 4t7rm.
We have supposed the stratum DAE to lie in the tangent plane at A. But the
effect of the curvature would be simply to change the attraction 2irm of a plane disc
into 27T/W (l-/3/r), where /3 is a quantity of the order a or p. These additional
terms are zero because both a and p are infinitely smaller than r. That this is so
may be made clearer by considering the case in which the surface is spherical. The
disc DAE is then bounded by a right cone whose vertex is at the centre and whose
72 ATTRACTIONS. [ART. 143
semi-angle is a/r. By using Art. 74 and retaining the first powers of pja, tja and
a/rwefind p = %(a + 2p).
Consider next the attraction of the portion of the stratum
remote from A. Let F, F' be the ^-components of attraction at
P, P'. Since these depend on the attracting mass, each contains
the factor pt or m. Also, since the distance PP is infinitely
smaller than the distance of either P or P' from the nearest
attracting element, F' differs from F by (dF/dx) t. The difference
is therefore of the order pt* or mt. We may therefore regard
F, F' as equal.
Taking both portions of the attracting stratum into the account
and representing by X, X' the normal attractions of the whole
system at P, P' we have
X = F-27rm, X' = F' + 27rm (1),
where X, X' are measured positively from P' to P. Since F, F'
are ultimately equal, these give
X'-X^iirm, F=$(X' + X) (2).
The equation X' — X = kirm shows that when attraction is taken
as the standard case, 47rm. is equal to the sum of the normal
attractions at each side of the stratum, the attractions being
measured towards the stratum. When repulsion is the standard
case, 4-Trm is equal to the sum of the normal repulsions, the
repulsion being measured on each side from the stratum.
If there are any other attracting bodies in the field which are
at finite distances from the points P and P', their attractions at
these points are ultimately equal. It follows that in both the
formulas (2) we may suppose X, X', and F to mean the normal
components due to all causes.
143. The equation F=%(X + X') enables us to find the
normal attraction of a thin heterogeneous stratum on an elemen-
tary portion of itself.
Let the element be a small cylinder whose base is the area
da situated at A and whose altitude is the thickness t of the
stratum. The normal attraction of the adjacent portion DAE on
the cylindrical element is ultimately zero because it is the same as
the normal attraction of an infinite plate on a portion of itself.
The attraction of the remote portion of the stratum is Fmda: It
follows therefore from (2) that the whole normal force per unit of
mass acting on the element is the arithmetic mean of the normal
ART. 147] ATTRACTION OF A THIN STRATUM. 73
attractions just inside and just outside the stratum. The normal
force on the matter mdar which covers an element of area da- is
•y-/2 TP-2
Fmda- and is therefore equal to — 5 dor, where X, X' are the
O7T
normal forces at each side of the element.
144. We may also show that the parallel tangential compo-
nents of attraction just inside and just outside the stratum are
equal. Let the axis of y be parallel to a tangent at A to either
boundary of the stratum. Let Y, Y' be the components of attrac-
tion at P, P. Considering first the adjacent portion DE of the
stratum, it has already been shown that the resultant attractions
at P, Pr are each directed along the normal PP' ; hence this
portion contributes nothing to Y or Y. Considering next the
remote portion of the stratum, it may be shown as in Art. 142
that the components F, Y' differ by terms of the order mt. In
the limit therefore when t is very thin, we have Y' = Y.
145. We shall now show that the potentials at P, P' are also
equal. The potentials due to the remote portion of the stratum
for the same reasons as before can differ only by terms of the
order mt. Consider next the portion of the stratum adjacent to A',
the potentials at two points equally distant from the two faces of
the stratum evidently differ by terms of an order higher than mt.
See also Art. 76, Ex. 1. Taking both portions of the stratum, we
see that the potentials at P and P' are ultimately equal.
146. It follows from this proposition that if a point travel
from a position P just within a thin stratum to another P' just
outside, both on the same normal, the normal component of the
attraction is increased by the quantity 4mm, where m is the surface
density. At the same time the tangential components of the attrac-
tion and the potential are unaltered.
147. We may also deduce Green's theorem from the propo-
sition, due to Gauss, that the flux of the attraction over a closed
surface is 4>jr multiplied by the mass inside. See Art. 106.
Let the axis of a; be a normal to the stratum, measured
positively inwards, and let it cut the boundaries in the points
A, A'. Let us consider the flux of the attraction across an
element of volume whose edges parallel to the axes as, y, z are
respectively A A' = t, dy and dz.
74 ATTRACTIONS. [ART. 149
Proceeding as in Art. 108 we have
(Xr — X) dydz + -T- tdydz + -r~ tdydz = 4t7rptdydz,
where X1 — X has not been equated to (dXfdx) dx because there
is attracting matter on one side only of each of the two faces
perpendicular to the axis of x. Substituting m=pt in the equation,
dividing by dydz and taking the limit, we find X' — X = 4-Trm.
148. Ex. 1. A thin layer of heterogeneous attracting matter is placed on a
sphere of radius a. If F be the potential and m the surface density at any point A,
show that the normal attractions on each side of the stratum are F/2a ± 2irm, Art. 87.
Ex. 2. Prove that, if matter attracting according to the law of the inverse
square be so distributed over a closed surface that the resultant attraction on every
external particle in the immediate neighbourhood is in the direction of the normal,
the resultant attraction on every internal point is zero.
The outer boundary of the stratum is by definition a level surface. The inner
boundary is therefore also a level surface. The result then follows from Art. 115
because there is no attracting matter within that surface.
Green's Theorem,
149. Let a portion of space be enclosed by a surface which
we shall call S. Let V, P, Q, R be any one- valued finite functions
of x, y, z, and let dv = dxdydz. Let us integrate
throughout the given space S. The first term becomes by an
integration by parts
j[[PV]dydz-jfjV~dxdydz ............ (2).
We have here integrated all the elements which lie in a column
parallel to the axis of x. Let AB be one of these columns and
let it intersect the surface S at A and B in the elementary areas
d<r, da'. If (\'fjfv) be the direction cosines of the outward normal
at the upper limit B we have dydz = \'dcr'. In the same way if
(A./ii>) be the direction cosines of the outward normal at the limit A,
we have dydz = — \da, since V is positive and A, negative. The
quantity in the square brackets in the first term of (2) is to be
taken between the limits A and B and is therefore
(PV)BXd<r'-(PV)A(-\d<r) ............... (3),
where the suffix indicates the place at which the value of the
quantity in brackets is to be taken. The two terms in (3) have
ART. 149]
GREEN'S THEOREM.
75
now to be integrated, the first for all elements such as B on the
right-hand side of the bounding curve CD and the second for all
>
elements such as A on the left. These together are the same as
fPV\d<r taken for all elements of the surface, where X now stands
for the cosine of the angle the outward normal at da- makes with
the axis of x.
Treating the other terms of (1) in the same way we have
Let P, Q, R be the components of a vector / and let / cos i be
the normal component at the element d<r. The equation (4) then
becomes
TT f-rrr -j F-rrfdP dQ dR\ 7
U= F/costcZo-- |F(-r- +-j^ +T- dv ......... (5).
J / \dx dy dz )
In this way the volume integral (1) has been replaced by a
surface integral, when the vector is such that
dP + dQ+dR = Q
dx dy dz
Let this vector be the attractive force of some system whose
potential is V. To be more general, let P = dV'/dx, Q = dV/dy,
R = dV'jdz, then
Let dn be an
where F, F' are two arbitrary functions of xyz.
element of the outward normal at da, then
dV dV dV _dV
~~j A. H j U, -J = V — 7
dx dy dz dn
Also let p, p' be such functions of xyz that
- 47T/0 = V2 F, - 4>Trp = V2 V.
.(7).
76 ATTRACTIONS. [ART. 153
Then by (4) the symmetrical expression U takes either of the
forms U~jV$J£-d<r+4arjVp'dv (9)
=/F/ d^d(T+4>7rSY'pdv (io)>
The equality of the expressions (6), (9) and (10) is usually called
Green's theorem.
150. If the functions V, V satisfy Laplace's equation we
have p = 0, p = 0, the equality then becomes
.(11).
151. Let F, V be the potentials of two attracting systems. Let W be the
mutual work of the first and that portion of the second system which is internal to
S ; let W be the mutual work of the second and that portion of the first which is
internal to S. Then, by Art. 59, Green's equation becomes
U=$VFdff + ±irW=$V'Fda + ±irW' (13),
where F, F' are the outward normal components of force at the element dy.
The expression for U admits also of interpretation. Let (XYZ), (X'Y'Z') be the
components of force due to the two systems at any point (xyz) within S. Let
R, R' be the resultant forces, <f> the angle between the directions of E, R'. Then,
by(l), U=$(XX' + YY' + ZZ')dv=$RR'co8<t,dv (14).
If the two systems are the same, i.e. if the particles occupy the same positions
in the two systems and have equal masses, Green's equation becomes
U= \R*dv = \VFdff + 4:Tr$Vpdv,
where F is the outward normal force at the element do-.
152. Instead of considering the space internal to S we may integrate through
the space between S and a sphere of infinite radius enclosing S and having its centre
at a finite distance from S. We must then of course include this sphere in the
surface integration over S. Let V, V be the potentials of some masses M, M' respec-
tively, then for points on the surface of the sphere V=H/a and dV'\dn= -M'/a?
also dff=a?du>, where a is the radius and du is the elementary solid angle subtended
at the centre by da: We therefore have for the sphere
' [^dV , ,,, fdu . MM'
I V d<r = -MM' I — =-4ir ,
J an J a a
and this is zero when a is infinite. We may therefore in this case apply the equality
(9) and (10) without further change to the space outside S. We notice that an
is always to be measured outwards from the space over which the integration
extends.
153. To deduce Gauss' theorem. Let us put unity for V.
Since this value satisfies Laplace's equation, we have p = 0. The
equality (9) and (10) takes the form
ART. 155] GREEN'S THEOREM. 77
Let the space of integration be the finite space enclosed by a
surface S. We thus avoid the integration over the surface of a
sphere of infinite radius. Supposing V to be the potential of any
attracting mass, the function of x, y, z represented by p becomes,
by Poisson's theorem, the density of the mass at the element dv.
The right-hand side of this equation is therefore —4nrM, where
M is that portion of the mass which is inside S. Also dV/dn
represents the outward normal force. The equation therefore
asserts that the whole outward flux across any surface 8 is
— 4>7rM. This is Gauss' theorem.
154. Green's equivalent layer. Let V = l/r' where r' is
the distance of any point within the space of integration from some
given point P. Let the integration extend throughout the space
internal or external to S according as P is external or internal
In this way we make l/r' finite throughout the integration.
Since ^irp = — V2 V, p' is now zero, and Green's equation
/•_ d fl\ [dVda- [pdv
becomes j V dn(r') d" ~ j d^ / =4?rjV ......... (16)'
Here the r' on the right-hand side is the distance of P from dv
and on the left-hand side r is the distance of the same point
from the element da of the surface.
We shall now suppose that V is the potential of some
attracting system, part of which may be inside S and part outside.
The right-hand side of the equation is evidently ^irVl where Fj is
the potential at P of that part of the attracting mass which is on
the side of S opposite to P.
The equation asserts that the potential at P of that part of the system on the
opposite side is equal to that of a thin layer placed on the surface S whose surface
density D at any point Q, (PQ=r') is given by
-
dn \r'/ dn r1 dn
where V is the potential at Q of the whole system. To make D independent of the
position of P we shall get rid of the terms which contain r'.
155. Let the surface 8 be such that the potential V of the
whole attracting system is constant and equal to Fg over its area.
Then S is a level surface, or a closed portion of a level surface, of
the whole system. Since l/r' is the potential at d<r of a unit
mass placed at P, we have by Gauss' theorem
{v^
J dn
78 ATTRACTIONS. [ART. 156
according as P is within or without the finite space enclosed by
the surface S. The plus sign is given to 4?rF, because we are
integrating throughout the space on the side of S opposite to P,
and dn is therefore measured towards P.
Lastly let us place on the surface S a thin layer of matter
1 dV
whose surface density p" is given by p" = — ^— -p where dn' is
measured outwards from the finite space enclosed by S. Let V"
be the potential of this layer at P, then
iirjdn* r"
In the equation (16), when P is internal dn is measured inwards
and therefore dn = — dn' ; when P is external dn = dn'. That
equation therefore becomes
F,-F" = y;, or F'^F, ............... (17),
according as P is internal or external. We deduce the three
theorems enunciated in the next article.
156. Let S be a level surface of an attracting system. Let a
thin layer of attracting matter be placed on the surface S such
that its surface density p" at any point Q is given by the equation
where F is the potential at Q due to the attracting system, and
dn' is measured positively outwards from the finite enclosed space.
(1) The potential of the layer at any point P, external to the
level surface S, is equal to the potential at the same point of that
portion of the attracting system which is within S.
. (2) The potential of the layer at any point P internal to S,
increased by the potential at the same point of that portion of the
attracting system which is external to 8, is constant for all
positions of P, and is equal to the potential V, of the whole
attracting system at the level surface S.
(3) The whole mass of the stratum is fp"d<r, and by Gauss'
theorem, this is equal to the mass of that portion of the attracting
system which is inside 8.
If the surface S encloses all the attracting system, the second
theorem asserts that the potential of the layer at all internal
points is constant and equal to that of the attracting system at
the level surface S.
ART. 159] GREEN'S THEOREM. 79
This form of the theorem will enable us to find the law of
distribution of a charge of electricity, given to any solid insulated
conductor whose boundary is a level surface of some known
attracting system.
157. Ex. It is known that a prolate spheroid is a level surface of a uniform thin
attracting rod whose extremities are at the foci S, H of the spheroid, Art. 49. Find
the surface density of the thin stratum which, when placed on the spheroid, has
the same attraction at all external points as the rod.
The surface density p" at any point Q of the spheroid is given by ±irp"=F,
where F is the resultant attraction at Q. Also F= 2m sin ^SPHjy, where y is the
distance of Q from the rod. By using some geometrical properties of conies this
leads to the result that p" is proportional to the perpendicular p from the centre on
the tangent plane at Q. The whole mass of the stratum is equal to that of the rod.
158. Points at which V is infinite. If P be any arbitrary
point taken in the interior of the space bounded by the surface 8,
it is evident that one of the columns of integration parallel to each
coordinate axis will pass through P. It is necessary that in each
of these three columns the subject of integration should be finite.
We have therefore assumed in the proof given in Art. 149 that
(1) both the functions V, V are finite and continuous, (2) that
their first and second differential coefficients with regard to x, y, z
are each finite throughout the space considered. If any of the
functions be infinite at some point A within S, we must sur-
round that point by an infinitesimal sphere, and integrate only
over the space between the sphere and the surface S.
159. Green's equation is
jv^-dr + iw (vp'dv = /V'^d(r+4ir f V'pdv ............... (I.).
Let us suppose that one term of V is I//, where r1 is a distance measured
from P. We shall substitute this term in Green's equation, and the space of
integration shall be that between a small sphere, centre P, radius e, and the
surface S.
Consider first the integrals taken over the surface of the sphere. Since
da=e'2d<a, we have by changing to polar coordinates
[IdV, fldV».
I — -j- da- = I - — e2dw=0,
J r an J e an
where dn has been measured from the space of integration, that is inwards on the
sphere, and VP has been written for the value of V at P.
Consider next the volume integrals. Since / is finite throughout the space of
integration, p'=0 and the term $Vp'dv disappears. The integral Jpdv/r' is to be
taken only for the space outside the sphere, but since dw=r'2dw if we include the
integral for the space within the sphere we have only added zero (see Art. 101).
80 ATTRACTIONS. [ART. 162
Green's equation far the term V = 1/r' takes the form
where the surface integrals are taken only over the surface S and the volume
integrals throughout the space S, ignoring the sphere altogether.
Since VP is the potential at an internal point P of the whole mass and Jpdy/r'
the potential of the mass inside S, this equation becomes identical with (16) of
Art. 154 when we change the sign of dn.
Let F be the potential of some attracting system, part of which may be
inside S and part outside. Also let
give the surface density D at any point Q of a thin layer placed on S, where r' = PQ
and V is the potential at Q of the whole mass. The equation (II.) then asserts that
the potential, at a point P inside S, of that part of the attracting system which is
also inside S, exceeds the potential of the stratum by the potential Vf of the whole
mass at P.
160. We may notice that if F or V be the potential of a system of bodies of
finite density, neither V nor its first differential coefficients are infinite at any point
of the mass, see Art. 101.
If one term of F' were m/r' we may regard the particle m as the limit of a small
sphere of radius e and density pQ, where $irp0e3 = m. The integrations in (I.) can
then be made throughout the space enclosed by S without reference to the sphere.
The integral 4v$Vp'dv will supply an additional term equal to ±irVPm. In this
way we arrive at once at the final equation (II.).
161. Multiple-valued functions. It has been supposed in these theorems
that the functions F, F' have only one value at the same point of space. If they
are potentials of attracting masses, they are each of the form Sm/r and can have
only one value. But if they are obtained as solutions of Laplace's equations, as in
hydrodynamics, they may be many-valued functions. Thus let a fluid be running
round in a ring-like vessel. If F be the velocity potential at any point P, we know
by the principles of hydrodynamics that dV/ds=u, where « is the arc described,
and « is the velocity at P. Since the velocity is always positive, the velocity
potential F must always increase as P travels round the ring. When P has made
a complete turn, it comes to the point it started from, and F has a different value.
If we put Laplace's equation into cylindrical coordinates (Art. 108), we easily see
that V=ta.u~lylx=<f> satisfies the equation and represents such a motion.
163. In order to apply Green's equation to a multiple-valued function by
integrating throughout the space enclosed in a ring-shaped surface we must deprive
the function of its multiple values by placing a barrier at any point and including
this barrier as one of the boundaries. In this way the point P is prevented from
making a complete circuit and the function is reduced to a single-valued form. It
may be that the surface has several ring-like passages interlacing, and it may then
be necessary to insert several barriers before the function is reduced to a single-
valued form.
Taking the simpler case of a single ring-like surface, let us suppose that the
potential F is always increased by the same quantity c when the point P starting
from any position has made a complete circuit and has returned to the same position
ART. 163] GREEN'S THEOREM. 81
again. Similarly let V be increased by c?. Let da be an element of the area of a
barrier placed anywhere across the ring-like cavity. Let s be an arc measured from
the barrier round the ring to the barrier again, say from *=0 to s = l. Consider the
part of the boundary formed by the two sides of the barrier; remembering that dn is
measured outwards, we have dn= -ds for the side defined by 8=0, and dn = ds
for the side s = I. We thus have, when we integrate over both sides of the barrier,
(v^da = -lvd^
J dn J ds
.
ds
Supposing F and V to be solutions of Laplace's equation, Green's theorem becomes
[^dV' fdV' [v,dV. ,[dVJ
U- I F-r-dff+c rda= I V' — dv + c' \ -^- da,
J dn J ds J dn J ds
where along the surface S, dn is measured outwards, and across the barrier ds is
measured in the positive direction round the ring.
163. Ex. 1. Let F, F' represent as before any two functions of (x, y, z), and
let a be a third finite function of the same variables. Beginning with
fff*fdVdV dVdV dVdV'\ ,
U= I I I a2 -j- -j- + -r- -j— + -r- -J- ) dxdydz,
J J J \ dx dx dy dy dz dz /
show, by the same succession of integrations as in Art. 149, that
U= ( a?V ^-dff + iw lv'pdv = f a*V ^ d<r + 4w tvp'dv,
d ( ,dF\ d ( , dF\ d / , dF\
where - 4irp = -r- ( a2 — + -j- a2 -j- ) + — ( a2 -r- I ,
dx \ dx J dy \ dy J dz \ dz )
and - 4wy>' represents a similar expression with F' written for F. This is Kelvin's
extension of Green's theorem. See Thomson and Tait, Part i., p. 167.
Ex. 2. If F, V be two solutions of the differential equation
dx \ dx J dy \ dy J dz\ dz J
and if also F= F' at all points of a closed surface S, prove that F= F' throughout
the enclosed space.
Let u—V-V, then u also is a solution of the differential equation. Writing u
for both F, F' in the general theorem of Ex. 1, we have
The right-hand side is zero since u vanishes at.all points of the surface S. But the
left-hand side is the sum of a number of positive quantities and cannot be zero
unless each vanishes. Thus dujdx, du/dy, dujdz are each zero at all points inside
S, i.e. the function u is a constant. Since it is given equal to zero at the surface S,
it must be zero at all points within 8. Lejeune Dirichlet uses a similar argument
in Crelle, xxxn. 1844.
This differential equation is of great importance in the analytical theory
of heat.
Ex. 3. Show in the same way that if dV\dn=AV'\dn at all points of the surface
S, then F= V throughout the space enclosed. [Here du/dn = 0.]
Ex. 4. If both F and F', besides being solutions of the differential equation,
also satisfy the equation dV/dn= —kV at all points of S, where k is a function of
the coordinates which is always positive, prove that F= F'. [Here the right-hand
side of Ex. 2 would otherwise be negative.]
R. S. II. 6
82 ATTRACTIONS. [ART. 165
Ex. 5. If V be one solution of the differential equation in Ex. 2 such that
dV/dn=-kV at all points of a surface S, where fc is always positive, prove that
there is no other solution of that differential equation which satisfies this condition.
[Use a proof similar to that in Art. 133.]
Given the potential, find the body.
164. Poisson's equation 4?rp = — V2F supplies a partial solution
to this question. The potential V being given throughout all
space we find p by differentiation. This value of p, if finite
throughout space, determines the only body which could have the
given potential. If the potential is given as a discontinuous
function of the coordinates difficulties may arise in applying
Poisson's equation at the points or surfaces of discontinuity. The
following theorem will therefore be necessary.
165. Let the potential V throughout a given space S be the
given function <£ (x, y, z), throughout a neighbouring space 8', let
the potential be ty (x, y, z\ and so on. In this way we regard all
space as divided into compartments within each of which the
potential is a different function of the coordinates. We suppose
in the first instance that the given potentials are nowhere infinite.
As a point P moves in space, passing from one compartment
to the next, we know by Art. 145 that there should be no sudden
change in the numerical value of the potential. "We therefore
suppose that the given potentials (f>, ty have equal values at all
points of the common boundary. This implies that the space rates
of the potential tangential to the common boundary are equal.
The tangential components of force must therefore be equal.
If the normal forces at the boundary are not also equal, there
will be a film of attracting matter at the boundary (Art. 146)
whose surface density <r is given by Green's equation
4 _ d<f> d-fr
dn dri *
where dn, dn' are measured in directions outwards from the spaces
S, 8', and therefore, at points inside each space, towards the
boundary.
We have now proved that the only arrangement of matter
which could produce the given system of potential values is one
consisting partly of solid matter given by Poisson's equation filling
the compartments and partly of films on the boundaries. It
ART. 167] GIVEN THE POTENTIAL, FIND THE BODY. 83
remains to prove by integration that this arrangement does
actually fulfil the given conditions. The results of these inte-
grations are supplied by Green's theorem.
166. Let us write <j> for the arbitrary function V in Green's theorem and let
D=-r- \r'(j>-r- ( -j } --r\ as denned in Arts. 154 and 159. Then the potential at
4?r ( dn \rj dn\
P when P is outside S is equal to that of a stratum of surface density D placed on
S, and when P is inside S, the potential at P exceeds that of the stratum by
<t> (x, y, z). Let this stratum be included (with the sign of D changed) as part of
the attracting system, the potential at a point outside S is then zero and at a point
inside S the potential is <f>. The proposed conditions are satisfied for the space S.
Treating the neighbouring space Sf in the same way, we obtain an internal
density determined as before by Poisson's equation and a superficial density which,
when its sign is changed, is the same as that given by D except that the function <f>
is replaced by if/ and the element dn of the normal is measured in the opposite
direction.
Adding together the two superficial densities and remembering that </> and ^ are
equal at those points of the boundary which are common to S and S', we observe
that the first terms of each destroy each other. We therefore find for the density
of the superficial stratum
' = -, — i~r-
4ir (dn
z) + T-7
' '
dn'
where dn and dn' inside each compartment are measured towards the boundary, so
that dn= -dn'. We notice that this law of density is independent of the position
of P.
// the given potential 0 is infinite at any point A within the space S we must
suppose that a finite quantity Q of attracting matter is situated at A (Art. 101).
The quantity Q may be found by enclosing A within a small sphere and using
Gauss' theorem, 4irQ=$Fdtr. If the potential is infinite along a curve, the line
density may be found by enclosing an elementary arc within the sphere.
167. Ex.1. The potential at a point Q is 0=2ir(62- a2), f = f*-(3&2-r!-2a3/r)
or X=|T (b3- as)/r, according as the distance r of Q from the origin is less than a,
lies between a and b, or is greater than 6. Find the attracting system.
Considering the space in which r is less than a, we see that both the volume
density and the part of the surface density <£0/4irdra are zero.
Considering the space in which r lies between a and b, the volume density is
found by substituting in P=-T -- -p^-, Art. 108, and is equal to unity. The
part of the superficial density found by substituting in d\l//4wdn is zero at the inner
boundary and - (b3 - a3)/362 at the outer.
Lastly in the space in which r is greater than b, the volume density is zero and
the part of the superficial density dx/^irdn = (b3- a3)/362.
Joining these together, we find that each of the two surface densities is zero and
that the attracting body is a spherical shell of radii a and b and unit density.
Ex. 2. Find the attracting system whose potential V is equal to
at all points within the ellipsoid Lxa + My'-i + Nza=l and zero at all external points.
The system is a homogeneous ellipsoid whose density is fi(L + M+N)l2*,
6—2
84 ATTRACTIONS. [ART. 168
together with a superficial stratum whose surface density at Q is - w/2irp, where p
is the perpendicular on the tangent plane at Q.
Since this stratum is equivalent to a thin homogeneous confocal shell (see Vol. i.
Art. 430), this result supplies a simple relation between the potential of a homo-
geneous solid ellipsoid and that of a homogeneous confocal shell. See Art. 224.
Method of Inversion.
168. Inversion from a point*. Let 0 be any assumed
origin, and let Q be a point moving in any given manner. If on
the radius vector OQ we take a point Q so that OQ . OQ = &*,
then Q and Q are called inverse points. If Q trace out a curve,
Q' traces out the inverse curve ; if Q trace out a surface or solid,
Q traces out the inverse surface or solid. The points Q, Q are
sometimes said to be inverse with regard to a sphere whose centre
is 0 and radius k.
Let F, Q' be the inverse points of P, Q, then since the products
OP . OF, OQ. OQ are equal and the
angles POQ, P'OQ' are the same, the
triangles POQ, P'OQ' are similar.
We therefore have
J- .-1 W- m
P'Q'PQ'OF
Let m, m be the masses of two particles placed respectively at
k
Q, Q', and let the densities be such that m' = m^ (2).
Multiplying equations (1) and (2) together, we see that the
potential at F of m is equal to that at P of m, after multiplication
by a quantity k/OP' which is independent of the position of Q.
Let any number of particles of given masses m^, 7??2, &c. be
placed at different points Qlt Q2, &c., and let the corresponding
masses m^, m?', &c., be placed at the inverse points Q/, Q2', &c.
Then since an equation similar to (2) holds for each pair of masses,
we have by addition
/ Potential at P' \ _ / Potential at P \ k
\of the inverse system/ \of the given system/ OP'
k
which may be compendiously written V = "
* The Method of Inversion is due to Sir W. Thomson, now Lord Kelvin. In a
letter addressed to M. Liouville and published in Liouville's Journal, 1845, a short
history and a brief account of some of its applications are given. This letter may
also be found in the Reprint of papers on Electrostatics and Magnetism.
ART. 171] METHOD OF INVERSION. 85
169. If the given masses ml} m2, &c. are arranged so as to
form an arc, surface or solid, the inverse masses will also be
arranged in the same way. It will therefore be necessary to
discover some rule by which we can compare the density at any
point of the given system with that at the corresponding point of
the inverse system.
Using the same figure as before but changing the meaning of
P, let PQ now represent any elementary arc of the locus of Q,
then P'Q represents the corresponding inverse arc. If the locus
of Q is a curve, we infer from the similarity of the triangles POQ,
P'OQ' that the lengths of the elementary arcs P'Q', PQ are in the
ratio OQ/OP, i.e. OQ/OQ ultimately. Hence by (2) the ratio of
the line densities of the arcs P'Q', PQ is equal to k/OQ.
If the locus of Q is a surface, the elementary areas P'Q', PQ are
in the ratio of the squares of the homologous sides, i.e. as OQ'2 to
OQ1. Hence by (2) the ratio of the surface densities at Q' and Q
is equal to (k/OQ)3.
If Q travel over all points of space enclosed by a surface, the
elementary volumes at Q', Q are ultimately in the ratio OQ*
dco.d(OQ') to OQ2da>.d(OQ). Since OQ.OQ = k*, this ratio is
equal to OQ3/OQ3. Hence by (2) the ratio of the densities at Q
and Q is equal to (k/OQ)6.
Summing these results, we see that
/ density at Q \ _ f density at Q \ / k \zd~1
\of the inverse system/ \of the given system/ ' \OQ')
where d represents the dimensions of the system, i.e. d = 1, 2, or 3
according as the system is an arc, a surface or a volume. When
the system is a point, d = 0 ; the equation (4) then agrees with (2)
and gives the relation between ra and ra'.
170. The mass of any portion of the inverse body is equal to
the potential at the centre of inversion of the corresponding portion
of the primitive body multiplied by the radius k of inversion. By
Art. 168, we have ra' =mk/OQ, i.e. m' is equal to the potential of
ra at 0, multiplied by k. The theorem being true for each ele-
ment of mass is necessarily true for any finite portion of the body.
171. Ex. If the law of force be the inverse nth power of the distance, the
potential of a particle m takes the form - — ^ ^—^ . Prove that the equations
corresponding to (2), (3), and (4) become
,1-n
gg ATTRACTIONS. [ART. 174
When the law of force is the inverse distance n=l, and the potential of the
attracting mass takes a different form. In this case the quantity here called
becomes Sm/(n-l), and is therefore proportional to the mass of the body.
theorems therefore of inversion, though they no longer apply to the attractions of
hodies, will still enable us to find their masses when their densities vary as some
power of the distance from a point. See Quarterly J., 1857.
172. Some Geometrical properties. It is convenient to notice that if the
points P, Q invert into P', Q>, then ^ = gJ^, where PQ, P'Q' are the linear
. PQ.RS .
distances between P, Q and P7, Q' respectively. For example the ratio ^R sp is
unaltered by inversion; because each letter occurs the same number of times in the
numerator and denominator.
173. To find the inverse of a sphere. Let Q describe a sphere whose centre is
C, and let OQ.OQ'=k*. Let OQQ' cut the primitive sphere in R, then since
OQ . OR is constant, it follows that OQ'jOR is constant. The locus of Q' is there-
fore similar to that of M, that is, the inverse is a sphere and 0 is a centre of
similitude.
T'
The centre D of the inverse sphere lies in 00 produced, and by the properties of
similar figures, is at such a distance from 0 that OD/OC is equal to the constant ratio
OQ'/OR. The centre C of the primitive sphere does not invert into the centre D of
the inverse sphere, but into some point C' such that OC . OC'= fc2. It is easy to see,
by similar triangles, that C' lies on the polar line of the centre of inversion O with
regard to the inverse sphere.
A sphere inverts into a plane when the centre of inversion O is on the surface of
the primitive sphere. The inverse of a plane with regard to any centre 0 of inver-
sion is a sphere which passes through 0.
A circle is the intersection of two spheres and in general inverts into a circle,
but when the centre of inversion lies on the circle, the inverse is a straight line.
Ex. Let P, P' be two inverse points with regard to a sphere S ; prove that
every sphere passing through P, P' cuts S orthogonally. Conversely, if a sphere
S' cuts S orthogonally and GPP' is any chord through the centre of S, then P, P'
are inverse points with regard to S. See figure of Art. 86.
174. An angle is not altered by inversion. Let PQ, PR be elementary arcs of
two curves which meet in P and are not necessarily in the same plane with the
centre 0 of inversion. Let P'Q', P'R' be the inverse arcs, we have to prove that
the angles QPR, Q'P'R' are ultimately equal. Describe a sphere through the four
points P, Q, R and P"; then since the products OP . OP1, OQ . OQ' and OR . OR' are
equal, the sphere also passes through Q', R'. The planes OPQP'Q.' and OPRP'R'
cut the sphere in two circles whose planes intersect in OPP'. The opposite angles
QPR, Q'P'R' contained by the tangents to these circles are evidently equal by
symmetry. It is also evident that the planes of the angle and its inverse, viz. QPR
and Q'P'R', make equal angles with the opposite directions of OPP'.
ART. 176]
METHOD OF INVERSION.
87
175. It follows at once, from the theorem, that two given orthogonal spheres
invert into orthogonal spheres.
176. Ex. 1. The potential of a homogeneous spherical surface at a point P
is 4:irap or 47ra'2/5/CP according as P is inside or outside the surface, where C is the
centre and a is equal to the radius. It is required to invert this theorem with regard
to an external point 0.
Since the product of the segments OQ . OQ' is constant in a sphere, it is clear
that if we take ft equal to the length of the tangent OT, the sphere will be its own
inverse. When only one sphere occurs in the system this choice of the value of k
will simplify the process, but when there are several spheres it will be more
convenient to keep the value of ft indeterminate.
If P is within the sphere, the inverse point P' is also within the sphere. By (4)
the density of the inverse sphere at Q' is equal to p(ft/OQ')3> ax1^ its potential at P
is 4irapfc/OP'.
If P is without the sphere, P' is also without. The density at Q' of the inverse
system is the same as before, but the potential at P' is . ~-^, . Let €' be the
CP (JJ:
point on the straight line 00 such that C and C" are inverse points. Then by the
similar triangles COP, C'OP' we have CP . OP'=OC. C'P'. The potential at P' is
471-ffl2/) ft
~~
therefore
~OC~
If M' is the mass of the inverse system, the relation between M' and p may be
easily deduced from either of these expressions for the potential. Take the first,
where P' is inside the sphere, we notice that since every element of the sphere is
equally distant from the centre, the potential at the centre is M'/a. Hence putting
P' at the centre and comparing the two values of the potential, we have
M' = ±irpa"klOC. Take the second case, when P' is without the sphere, we notice
that the potential at a very distant point must be mass divided by distance. By
equating these two values of the potential, we arrive at the same value of M' as
before. This value of M' may also be easily deduced from Art. 170.
Taking both these results, we arrive at the following inverse theorem.
Let a mass M' be distributed over a spherical surface, centre C, so that its
density at any point Q' is p (k/OQ')3, where 0 is an external point, and ft is the
length of the tangent from O. Then p=M'c/47ra2ft, where c = OC; and the potential
c 1 M'
at any point P1 is M' - _ or
a UJr
according as P' lies within or without the
sphere. The points C' and C are inverse points with regard to 0, and it is easy to
see that C' lies on the polar line of 0.
The potential of this heterogeneous spherical stratum at all external points is the
same as if its whole mass M' were collected at C', and at all internal points is the
same as if a mass M'c]a were collected at 0,
88 ATTRACTIONS. [ART. 179
It follows from Art. 136 that the centre of gravity of the heterogeneous stratum
is at C" and that every straight line through C" is a principal axis.
Ex. 2. If the density of a spherical surface vary as the inverse cube of its
distance from an internal point O, find its potential at any point.
If the centre of inversion O is inside the primitive sphere we can still make
the sphere its own inverse by drawing OQ' from 0 in the direction opposite to OQ,
and taking k2 equal to the product of the segments of all chords through O. With
these changes we may show that the potential at all external points is the same as
if its whole mass M' were collected at 0, and at all internal points is the same as if
the mass .M'c/a were collected at C',
Ex. 3. The potential of a homogeneous solid sphere at an external point P is
$vpa?/CP, where C is the centre and a the radius. Invert this theorem with regard
to an external point 0.
The result is that the potential at an external point of a heterogeneous sphere,
whose density at any point Q' is p(k/OQ')s, is the same as if its whole mass M' were
collected into a fixed point C". This point C' is the inverse of the centre with regard to
0 and is also the centre of gravity of the sphere. The constant p may be found from
the relation M'c = %irpa?k, where c=OC, and k is the length of the tangent from 0.
Ex. 4. A heterogeneous spherical shell is bounded by eccentric spheres whose
radii are a, b, and its density at any point Q is m/OQ5, where m is a constant and
0 a given external point. Show that its potential at any internal point P is
r (a? &2\ 1 (AP* BP*\ 1 -|
™ L \74 ~g*JOP~ \OA* ~ OB*J Ol*J '
where A and B are the points where the polar planes of 0 intersect the diameters
drawn through 0, and /, g are the tangents from O.
Ex. 5. An infinitely thin layer of matter is placed on the surface of elasticity
esr4=a2x2 + 62y2 + c222, so that the surface density at any point distant r from the
centre varies as pjr6, where p is the perpendicular from the origin on the tangent
plane. Show that the potential at any external point is the same as if the whole
mass were collected at its centre of gravity.
177. If $ is a level surface of any attracting points, the
inverse of S is not in general a level surface of the inverse of the
attracting points, because the ratio of the potentials (being given
by V = Vkjr') is not constant. But if S is a level surface of zero
potential, the inverse of S is also a level surface of zero potential of
the inverse attracting points.
178. Let P, P' be inverse points with regard to a sphere S.
If Q be any point on the surface, the ratio PQ/P'Q is constant by
the similar triangles OPQ, OP'Q, (Art. 86). Let this ratio be
a//3, then a/PQ — /3/P'Q — 0, that is the sphere is a level surface
of zero potential of two particles placed at P, P', whose masses are
measured by a and — /3.
179. Two points P, P are inverse to each other with regard
to a sphere £ Let the inverse of this system, taken with regard
ART. 182] INVERSION FROM A LINE. 89
to a new origin 0, be the points Q, Q' and the sphere S'. Then
the points Q, Q' are inverse points with regard to the sphere 8'.
By putting particles of proper masses at P, Pf, the sphere can
"be made a level surface of zero potential. The inverse of 8 with
regard to the new origin 0 is therefore also a level surface of zero
potential of the inverse masses at Q, Q'. Hence Q, Q' are inverse
points with regard to 8'.
A purely geometrical proof"* of this theorem is given in Lachlan's Modern
Geometry.
ISO. If a particle of finite mass m is at the centre of inversion 0, the inverse
is a distribution of matter at infinitely great distances from 0. The theory of
inversion gives the potential of the whole inverse system including the infinitely
•distant matter. If we wish to remove the latter from the field under consideration
we must subtract its potential. Now by equation (3) of Art. 168 its potential at any
point P' is V = V -=-=% — _ p = -T. We may therefore disregard this infinitely
OP (JP . (JP K
distant matter if we subtract from the potential of the inverse body as given by the
theory, the constant m/k.
If the mass at O merely forms part of a stratum passing through 0, the mass
actually at 0 is zero and the constant to be subtracted is also zero.
181. Inversion from a line. Instead of inverting the
attracting system with regard to a point 0 we may invert it
with regard to some straight line Oz. Let a point Q move in any
manner, and let QN be a perpendicular on the axis Oz. If on NQ
we take a point Q' so that NQ.NQf = k*, where A; is a given
constant, then Q' is the inverse of Q with regard to the axis of z.
With this definition it is clear that any cylindrical surface
with its generators parallel to Oz inverts into another cylindrical
surface also having its generators parallel to that axis. This
method of inversion will therefore help us to deduce the potential
of one cylindrical surface or solid from that of its inverse. We
shall suppose that the density of the cylindrical body is uniform
along any generating line but varies from one generator to
another.
182. If an infinite rod is parallel to the axis of z, its attraction at any point P
on the plane of xy is known to be 2m/QP, where Q is the intersection of the rod
with the plane of xy and m is the line density. The potential of such a rod at P is
therefore V= C-2m log QP, where C is some constant, Art. 50. Let us invert
this rod with regard to the axis of z into a parallel rod, and P into another point
P. Supposing the inverse rod to have the same line density as the primitive rod, its
OP'
potential at F is V = C - 2m log Q'P. But by Art. 168 P'Q' = PQ, . — ^ . Hence
F' + 2mlogOP'=F+2mlogCK> : (1).
90 ATTRACTIONS. [ART. 184
Let there be a system of rods intersecting the plane of xy in the points Qlt
Qz, &e., and let the inverse rods intersect the same plane in Qi'» Qs', &c. Let mlt
m2, &c. be the line densities of the several pairs. Then for each pair we have an
equation similar to (1) ; adding all these together we find
(Potential at P' of inverse system)
- (Potential at P' of the whole mass collected at the axis)
= (Potential at P of given system) - (Potential at O of given system).
183. If the primitive system of rods intersect the plane of xy in an arc or an
area, the inverse system will also be arranged in the same way. To compare the
densities we observe that the masse* of the given system and the inverse are the same
but differently distributed. If the loons of Q is an arc, the ratio of the elementary
arcs at Q', Q is equal to OQ'/OQ, and the ratio of the line densities is therefore
equal to OQ/OQ', i.e. (k/OQ1)*. If the locus of Q is an area, the ratio of the surface
densities is equal to (&/OQ')4.
We should notice that m is the mass per unit of length of a rod. Hence when
the attracting rods form a cylindrical surface whose surface density is p, we have
m=pds, where ds is an element of arc of the section of the cylinder by a plane
perpendicular to the axis. For example, in the case of a right circular cylinder of
radius a we have 2m=25ro/). If the rods form a cylindrical volume of density p, we
have m=pdA, where dA is an element of area of the curve of section.
Ex. 1. A heterogeneous stratum is placed on a right circular cylinder, the denxity
being uniform along any generator. It is required to compare the potentials at an
internal and an external inverse point. If we invert the system with regard to the
axis and the radius k of inversion be the radius of the cylinder, the stratum inverts
into itself. If P, P be the internal and external points, V, V the potentials, we
have by Art. 182 V - (€' - 2Sm. log OP) = V-V0. Collecting all the constant terms
into one, we have V ' - V=A -2Swlog OP. The corresponding proposition for a
sphere is given in Art. 86.
Ex. 2. Invert the following theorem with regard to an eccentric internal
straight line. The potential of a homogeneous right circular cylindrical surface at
any internal point is constant and equal to that along the axis.
The resulting theorem is as follows. If matter be distributed in a thin stratum
over a right circular cylinder so that the surface density at any point Q1 is
proportional to the inverse square of the distance of Q' from an internal straight
line OZ parallel to the generators, the potential at any external point is the same
as if the whole mass were evenly distributed over the straight line OZ.
184. Extended theory. Let Qlt Q2,...Qn, be n points arranged at equal
distances on the circumference of a circle of radius p. Taking the centre 0 as
origin, let the polar coordinates of these points be (p, <f>), (p, <f> + a), (p, 0+2a) &c.,
where na=2ir. Let P be any point and let (r, 6) be i'ts coordinates. By De
Moivre's property of the circle we have
r2»-2rnp»cos»(0-0)+/>2n=PQ1s.PQ22...P()ll2 (1).
Let us now take two other points Q', P' whose coordinates (/>', 0') and (r1, 6') are
such that p' = c(p/c)n, <f>' = n<p; r' = c(rlc)n, $'=nO, where c is any constant. It
immediately follows that the left side of (1) is equal to c2!1*"1) . (P'Q')2. Taking the
logarithm of both sides, we find
logP'Q' + (n-l)logc=logPQ1 + logPQ2 + &c.+logPQn (2).
Let us now suppose that two infinite thin rods, each of uniform line density m,
ART. 184] EXTENDED THEORY OF INVERSION. 91
are placed perpendicularly to the plane of the circle at P and P' respectively. It
follows at once from equation (2) that the potential of the second rod at Q' differs
by a constant from the sum of the potentials of the first rod at the points Qlt Q2, &c.
In the same way, by properly placing pairs of corresponding rods we may build
up two corresponding cylindrical bodies, which have the property that the potential
of the second body at Q' differs by a constant from the sum of the potentials of the
first at Qi...Qn.
We may express this result in the form of a theorem. An infinitely long
cylindrical body has its density uniform along any generating line and attracts
according to the law of nature. The body, being referred to cylindrical coordinates
with the axis of z parallel to the generators, is transformed into another cylindrical
body by moving each cylindrical element (r, 6) into the position (r', ff), where
r'=c(rjc)n, 0'=n0, without altering the mass of element. If the potentials of the
original body at the n points (p, <f>), (p, 0 + a), (p, </> + 2a) &c. be Vlt F2, V3 &c. then
the potential of the transformed body at (p't <j>'), where p' = c(pjc)n, tp' = n<f>, differs by
a constant from the sum Vl + V2 + (&c. + Vn.
If one be a continuous cylindrical solid, the other body may be made also
continuous by altering the areas of the sections of the transformed elements,
keeping the mass unchanged. Since the elementary areas at P, P' are respectively
rd0dr and r'dd'dr' we easily see that the volume densities at P, P' must be in the
ratio of (rzr')2 to r2.
If one body be a continuous surface, the other may be made also a continuous
surface. Since the masses on the corresponding arcs ds, ds' are equal, the surface
densities a, <r', must be such that <rds = <r'ds'. This ratio may be put into other
forms. Let \ff, \j/ be the angles these arcs make with their respective radii
vectores, then since r'=rnlcn~lt 0'=n0,
dtf de
It appears that the radial angle \f/ is unaltered by the transformation. Since
sin \j/ = rd0jds, sin^'^r'dd'/ds', we see that dslds'=r/nr', and therefore ro- = nr'ff'.
Since the coordinates of the corresponding points of the two figures are connected
by the relations r/=rn/c™~1, 0' = n0. it is clear that when 6' has increased from 0 to
2ir, 6 has varied from 0 to 27r/n, and thus an arc only and not a closed curve is
obtained. If P' travel n times round its curve, the curve traced out by P will
consist of n equal and similar arcs, fitting together and forming a closed curve.
Since a = 27r/n, it is also evident that these n arcs are similarly placed with regard
to the n points Qlt Q2, &c., and that therefore the potential of the whole closed
curve at each of these points is the same.
The potential therefore at Q' of the n coincident cylindrical strata generated by
the rod P' in n revolutions (which of course is n times that of the cylinder taken
once) is equal to n times the potential of the complete cylindrical stratum
generated by the rod P at any one of the points Qlt Qa, &o. It follows that the
potentials of the two closed cylinders (each taken once) are equal at the corresponding
points Ql and Q'. If one stratum (like an electrical stratum) is equipotential
throughout all space on one side of the surface, the other is also equipotential on
the corresponding side.
Ex. Thin layers of attracting matter are placed on the cylinders
Ax* + 3 (3B - 2A) «y + 3 (3 A - 2B) x*y* +By«~l;
92
ATTRACTIONS.
[ART. 185
if the surface densities are proportional respectively to r2? and r*p, where r is the
distance from the axis and p is the perpendicular on the tangent plane, prove that
the potentials are constant at all internal points.
If a thin stratum is placed on the cylinder Lx'2 + My'2=l whose surface density
is ff'=icp', the potential inside is constant, Art. 72. Transform this theorem by
writing r' = r2/c, ff=2ff. The elliptic cylinder becomes the first of the two cylinders
in the example. The surface density a follows at once from <rr=n<r'r', if we
remember that pjr—p'lr' at corresponding points.
Circular rings and anchor rings.
185. When the potential of a thin heterogeneous circular
ring for any law of attraction is known at all points in its plane
within the circle, the potential at every point of space may be
deduced by inversion.
Let the plane of the circle be the plane of xy, the centre 0
being the origin. Let the
plane of xz contain the G'
point R at which the
potential is required and
let it cut the circle in
A', A. In the figure the
attracting circle is repre-
sented by the dotted line.
Describe a circle through
the points A, A' and R,
then
CP.CR = CO.CC'=CA*.
The points P and R are therefore inverse with regard to C. If
then F, V" are the potentials of the ring at P and R, when the
law of force is the inverse /rth power, we have F" = F(4,
\r
where c = OA and r" = OR.
When the law of attraction is the inverse distance, the
potential takes a logarithmic form, Art. 43. Let ra be the mass
of a particle of the ring situated at Q. Its potentials at P and R
areC-m log QP and (7 - m. log QR. But since the triangles QCP,
QCR are similar (Art. 168) QP/QR = c/CR. If then F and V"
are the potentials of the whole ring at P and R, we have
V"= V+M log ^7, where M is the mass of the heterogeneous ring.
AET. 190] CIRCULAR RINGS AND ANCHOR RINGS. 93
186. The geometrical relations between the positions ofP and R
are most easily obtained by describing an ellipse or hyperbola
whose foci are A, A' and which passes through R.
Since the angles ARC, A'RG stand on equal arcs, these angles
are equal and RPG is therefore a normal to the ellipse. We thus
have r = e*x where r — OP, x is the distance of R from the axis Oz
of the ring, and e is the eccentricity. We also have GA/CR or
c/r" = e. If p, p are the least and greatest distances of R from
the ring, the focal distances A R = p, A'R = />'. Hence, a being the
radius of the ring, ex = \ (p — p) and a/e = %(p' + p).
The semi-major axis of the ellipse is a' = a/e.
187. The result may be stated as follows. Let the potential
at an internal point P in the plane of the ring be V =f(r). Then,
if the law of force is the inverse icth power of the distance, the
potential at R is V" = eK~1f(e2si). If the law of force is the
inverse distance V" = V + M log e. We may use any of the preceding
geometrical results to express e in terms of the coordinates of R.
The points R and P' are inverse points with regard to a sphere whose centre is
C" and radius C'A. These may be used to deduce by the same rule the potential at
R from that at the external point P'. Instead of the ellipse we then use the
hyperbola which has its foci at A', A and which passes through R.
188. Ex. Prove that the component forces at R along the tangents to the
ellipse and hyperbola are e*+lF sin RPA and e'K+lF' sin RP'A, where F, F1 are the
forces at P and P' resolved in the directions OP, OP respectively ; and e, e' are the
eccentricities of the ellipse and hyperbola. Prove also that ee'=a[x, and that
P, P' are inverse points with regard to the ring.
189. Ex. The potential of a uniform circular ring, when the law of force is
the inverse distance, is known to be constant at all points within the circle
and in its plane, Art. 55. The potential at any point R of space is therefore
V"=C+M loge. It follows that the level surfaces are oblate spheroids having the
circle for a focal conic.
Prove that the resultant force at R takes either of the forms
F= M . n/pp'= M.(l- e2)/w,
where p, p' are the focal distances and n is the length of the normal UP.
If the ring is heterogeneous, let its law of density be given as described in
ira fe*x\n
Art. 58. The potential at any point R of space is then V"='2En—(—-} +C
(except when n=0), where a is the radius of the ring.
190. To find the potential of a thin uniform circular ring of
line density m, the law of attraction being the inverse icth power of
the distance.
First, let the point P at which the potential is required be in
94 ATTRACTIONS. [ART. 190
the plane of the circle. Taking the figure of Art. 55, let OA = a,
OP = r, we then have
a sin 0 = r sin 9, w2 — 2ur cos 9 + r2 — a2 = 0 (1 ),
where 9 = OPR, u = PR. When P is outside the circle both the
values of u given by the quadratic are positive and represent
geometrically the distances PQ, PR; these we distinguish as
v^, u*. When P is inside the circle the geometrical distances are
— MJ and it*.
The elementary masses at Q, R being mu^ d6/cos </> and
muz dd/cos <f>, the potential V of the whole ring at an external
v_ m [ dQ ~ a 1 1 x»v €
when P is inside we write — ut for ut ; we notice that when K
is an even integer, the same formula represents the potential
whether P is internal or external. The value of SK^ may be
deduced from the quadratic, thus S0 = 2, S-L = 2r cos d/(r* — a2).
The limits of the integral are different according as P is
outside or inside the circle. When P is outside, <£ varies from
0 to %TT and sin# from 0 to a/r, the final result being doubled.
When P is inside, 9 varies from 0 to \TT and sin <f> from 0 to r/a,
the final result being doubled. To simplify the limits we express
the integral V in terms of <£ or 9 according as P is outside or
inside.
Representing these potentials by VKf and VK, we have after
using (1)
<SK_a ad(f> „ _ 2m f S^ adO
the limits in both integrals being 0 to |TT. When the law of force
is the inverse square,
4iinad(f> ~ f 4<madd
(r2- a2 sin2 </>)*' *= J (a2 - r2 sin2 0)* *
When the law of force is the inverse cube we find for an external
point (using (1))
V — [mad<t> 2rcos#
2
r cos 9 r3 — a2 r2 — a2 '
The potential at an internal point may be deduced from the
general expression for VK, if we remember to write -M! for u^.
It follows however at once from the expression for F3' by using
the rule of inversion (Art. 171). We write o2/^ for r and
ART. 191] CIRCULAR RINGS AND ANCHOR RINGS. 95
mult
„
* » =
multiply by (a/r^)"-1, where rx = OP and « = 3. We thus find
mira
a a ff> &
I J
attraction of the ring at any external point P' in the plane
of the ring is the sum of resolved attractions of the elements at Q
M r
and R. In this way we find X' ' — — S^dd), where the limits are
TrrJ
0 to |TT.
The potentials for these two laws of force being known, the
corresponding potentials for any other inverse law may be deduced
from the theorem (Art. 97)
The potentials at points in the plane of the ring being known,
the potential at any point R of space may be found by the rule
of Art. 187. If (x, z) be the coordinates of R, we write e*x for r
and multiply by e*"1.
For example, when the law of force is the inverse square, the
potential at R is -^—, - r- — -r-, - ., . ,.n,, the limits being
J {(p + pf - (p - p)2 sin2 0}*'
0 to \TT.
Instead of using the angles OPR = 0, ORP — <f>, we may use the third angle
= \{/ of the triangle OPR, or the angle x subtended by OR at A, so that
Supposing the law of force to be that of the inverse square of the distance, the
potential at P is then
_ f mds _ f 2mad\f/ _ f 4tmadx
2~J PR ~ J (a2 + r2-
2-
where the limits are ^ = 0 to IT, and x = 0 to £TT, and ds is an element of arc. These
results hold whether P is internal or external, provided it is in the plane of the circle.
191. Ex. 1. Investigate Plana's theorem that the attraction of a uniform
circular ring at an external point in the plane of the ring is
^
**>
the limits being 0 to \v. [Turin Memoirs, 1820.]
Ex. 2. Prove that the potential of a thin uniform attracting ring at an internal
point P in its plane very close to the circumference is ultimately 2m log 8a/|, where
£ is the distance of P from the ring.
We take the general expression for the potential given as an elliptic integral in
Art. 190 and put r/a = k, where k is to be ultimately put equal to unity. We have
» = f __ *?. _ = [
m /l-fc»ain»0)? J
4m /(l-fc»ain»0)? \l + fcsin0/ (1 - A;2 sin2
The last integral can be found, and is equal to -log{(l- &2) */(! + &)}. The other
96
ATTRACTIONS.
[ART. 193
integral presents no singularities and we may put k = 1 before integration ; its
value is then log 2.
This value of the potential agrees with that given by H. Poincare in his TMorie
du Potentiel Newtonien, 1899, p. 132. The use of the first integral on the right-
hand side of (1) to find the elliptic integral K when k = l was suggested in a College
examination paper, Dec. 1896.
A plane drawn through the axis of the ring will cut the level surfaces in a series
of curves. By using the theorem V"=eV of Art. 187 we may prove that these are
circles in the immediate neighbourhood of the ring.
Ex. 3. Prove that the level surfaces of a thin circular ring, when "the law of
attraction is the inverse cube, are given by pp' = /u2, where p, p' are the greatest and
least distances of any point R from the ring, and the constant p. is given by 2fj? = M[ V".
192. Anchor rings. An anchor ring is generated by the revolution of a circle
of radius a about an axis Oz in the plane of the circle, the centre C describing a
circle of radius c. A thin homogeneous layer is placed on its surface. Prove that
the potential of the layer at any point P of the axis is
nfjr
where R = CP, and M is the whole superficial mass. If m be the surface density,
M=2ira . 2irc . m by Guldin's theorem.
Let QQ' be an arc of the generating circle ; let PQ make an angle <j> with the
outward normal CQ to the anchor ring.
Let the angles CPQ=B, CPO=ft, and
PQ=p. Since the arc QQ' = pdd sec <p,
the potential at P of the annulus gene-
rated by the revolution of QQ' about
Oz is
V=m J sec <pd& . 2irp sin (0 + j3)
= 2irmJ sec <f>d0p (cos 0 sin ft + sin 9 cos ft).
Since the integration extends over the
whole circumference of the generating
circle, the last term is zero. Also
0
.: cosed0=cos(f)d(p(afR).
.: F= Iwm sin ft (a/ R) J pd<p.
The limits are 0 to w if we double the result. Produce PQ to cat the circle again
in R and let PR=p'. Then
Since p+p'=2.PN=2^/(Ry -a2 sin4 <p) this reduces to the result given above
without difficulty.
193. Ex. 1. The potential of a solid homogeneous anchor ring at any point P
of the axis may be expressed in either of the following forms
sin2 \pd$
The limits for <p are 0 to \v, and for ^, 0 to *-. If /x be the density the whole mass
ART. 194] ELLIPSOIDS. 97
Let r, 6 be the polar coordinates of an element of area at S of the generating
circle referred to P as origin and PC as axis of x. The potential is then
" T/j.r sin (6 + /3) = 2?r/i sin /3 \\rd6dr cos 8,
'-II"-
r
since, as in the last example, one term of the integral is zero. Let us integrate this
first with regard to r. Then using the geometrical relations connecting 0, <t>, p, p'
given in the last article, we find
The limits of sin 0 are - aj R to a/B and those of <f> are — JTT to £TT.
Let us integrate the double integral first with regard to 0. Let the circle whose
centre is P and radius r=PS cut the generating circle in T, 2". Let the angles
CPT=e1, PGT=\(>. Then
We now find ^ : — 3 = J rdr . 2 sin 0, = faJJ sin \f/c
27T/U, sm |8 J l J
the limits of ^ are evidently 0 to ir. This is equivalent to the second expression
given above. The second expression for F agrees with that given by Dyson, Phil.
Trans. 1893, p. 55.
Ex. 2. Express the potential of a solid anchor ring in elliptic functions. Let
^2=l-&2sin20, k = ajR, then
1= J cos2 <j>Xd<p=fX cos 4>d sin <f>
= f sin2 <pXd<f>+%P J sin2 0 cos2 0 (d^>jX)
by integrating by parts. The integrated part is zero at each limit.
- 21+ J(l - sin2
Substituting for sin2 0 its value in terms of X, we find
where the limits throughout are 0=0 to ^TT.
Attraction of Ellipsoids.
194. For the sake of brevity we shall adopt in this section two
new terms taken from Thomson and Tait's Natural Philosophy.
A homoeoid is a shell bounded by two surfaces similar and
similarly situated with regard to each other. In what follows we
shall somewhat restrict this definition and use the term only
when the shell is bounded by concentric ellipsoids.
A.focaloid is a shell bounded by two coufocal ellipsoids.
Thomson and Tait restrict these terms to infinitely thin shells,
but it will be convenient for us to use them in a more general
sense, distinguishing the shells as thick or thin according as the
thickness is finite or infinitely small.
A shell bounded by two similar and similarly situated surfaces
R. s. ii. 7
98 ATTRACTIONS. [ART. 196
has been called a homothetic shell by Chasles in the Jour. Pol.,
Tome XV., 1837. This is a convenient term when the surfaces are
either not concentric or not ellipsoids.
195. Let (a, 6, c) be the semi-axes of the internal surface of a
thin ellipsoidal shell, (a + da, &c.) those of the external surface.
Let OPQ be any radius vector drawn from the common centre 0
cutting the ellipsoids in P and Q, let OP = r. Let p be the
perpendicular from 0 on the tangent plane at P, p + dp the
perpendicular on a parallel tangent plane to the outer ellipsoid.
Then dp is equal to the thickness at P.
When the thin shell is a homoeoid we have by the properties
da db dc dp dr 77
of similar figures — = T~ = — = — = — — ak.
a o c p r
Since the volume of a solid ellipsoid is fodbc, we find by differen-
tiation that the volume v of the shell is v = 4nrabcdk. Two thin
homoeoids are said to be confocal when their inner boundaries are
confocal conicoids.
When the shell is a focaloid, we have a'2 = a2 + X, 6'2 = 62 + X,
&c., where (a', &', c') are the semi-axes of the external surface.
These give for a thin shell ada = bdb = cdc = pdp = ^d\. The
volume v of the shell may be shown by differentiation to be
_ 4?r 62c2 + c2a8 + a262 d\
= T' aba ~ 2 '
If we regard either shell as a thin stratum placed on an
ellipsoidal surface the mass on any elementary area d A is pdp . dA
where p is the density. The surface density is therefore pdp and
it varies directly or inversely as p according as the stratum is the
limit of a homoeoid or a focaloid.
196. Thick homoeoid, internal point. To find the poten-
tial of a thick homogeneous homoeoid at an internal point.
It has been shown in Art. 68 that the attraction of such a shell
at all internal points is zero. The potential is therefore constant
throughout the interior, and it will be sufficient to find the potential
at the centre.
Taking polar coordinates with the centre as origin, the mass of
any element is pr'drdw, where p is the density of the element.
The potential V of the whole solid at the centre is therefore
V=pffrdrda>. If rlt r-2 be the radii vectores of the two surfaces
of the shell, we have F= ^pfr22d(a — ^pjr^dco.
ART. 197] ELLIPSOIDS. 99
The determination of the potential at the centre of a thick
shell, bounded by any concentric ellipsoids, depends therefore on
the evaluation of the integral fr^da) taken over the superficies of
an ellipsoidal surface.
When the shell is a homoeoid these surfaces are similar. Let
(a, b, c), (ma, mb, me) be the semi-axes of the external and
internal surfaces. We then find V= Jp(l — m2)/r2d&>, where r is
the radius vector of the external boundary.
When the shell is a thin homoeoid m is nearly equal to unity.
The surface density is pdk.p and the potential is pdkfr^da} where
dk = 1 — m. When the surface density is pp the potential is /j,fr2dca
and the whole mass is ^Trabcp>.
It easily follows that the potential of a thin homoeoid is two-
thirds of the potential at the centre of a solid homogeneous
ellipsoid of equal mass and having the same external boundary.
197. To find the integral fr*dw we write dco = sin 6d6d$.
Substituting for r2 its value found from the equation to the
ellipsoid, we have
ff sin 0d0d(i>
Jr>da> = 1 I— — j , — £?—
T- + sin2 e Pi + -
c2 \ a2 o2
where the integration extends over the whole surface of the
ellipsoid. Taking only an octant, the limits are 6 — 0 to 6 = \TT,
<f> = 0 to <f> = \TT. The order of integration is immaterial.
Let us integrate first with regard to <£. Dividing both
numerator and denominator by cos2<£, we find
sin OdOd tan <ft
cosa0 sin20 /cos2 6 sin20
c2 a2
By obvious processes in the integral calculus
tan-
[
Jo
o
It therefore follows that
2 ,
_ TT f
~2'J
sin 6d9
//c
V (
ff
To interpret this expression, let us produce the radius vector OP
or r to cut the tangent plane drawn at the extremity C of the axis
of z. Let R be the point of intersection and let CR = u, then
7—2
100 ATTRACTIONS. [ART. 200
u = c tan 0. Since the limits of 6 are 0 and I-TT, those of u are
0 and oo . Substituting, we find
where the integration on the left side extends over the whole surface
of the ellipsoid.
/•tfi J
198. If we write 2 = I j-r- — rryrr - > . , . , — rr f we find
J0 (a3 + »)* (&2 + w)* (c2 + «)*
that the potential of a thick homoeoid is
i _ wa
F=p(l-m2)7m&c. / = «/*-— -,.
1 — m3
The potential of a thin homoeoid is V=\M.I, where, in each
case, M is the mass of the attracting body.
It follows that the integral I may also be defined as the ratio of
the internal potential of a thin homoeoid to half its mass. If the
homoeoid represent an electrical stratum 2/7 is the capacity.
199. Since the first integration in Art. 197 has been made
with regard to <£ it is evident that we may introduce any function
of 0 as a factor without disturbing the argument. We therefore
have
, , „/}, 0 7 r cos20cfo2
Jr2 cos2 6 aw = 27raoc j -7—n - ^-^ - ^y- - ^ .
Jo («2 + u*)* (62 + w2)* (c2 + w2)*
Since u = c tan 6 and z = r cos 6, this gives JWa> = — 27rabc -c-^-y
ac
/. fix?da> = -2Trabc.a-j-, JWo> = — STTO&C . b -jr ,
aa do
where the integrations extend over the whole surface of the
ellipsoid.
The polar equation of the ellipsoid is
Differentiating fr2dw = 2jra&cl with regard to c, we have fr*n2dw = irabc* d^-
do
with similar expressions for JV*Z2dw and
200. Ex. 1. Prove $\Vfi?ov™du=-^ ^ (f> ^ (ff> L (h> , where (X, 0, r) are
the direction cosines of a radius vector, n=f+g + h, and L (/) stands for the
quotient of all the natural numbers up to 2/ by the product of the same numbers
up to /, both included. A short proof is given in the author's Rigid Dynamics,
VoL i. Art. 9.'
Ex. 2. Show that the integral I, and therefore the potential F, may be
expressed as an elliptic integral. Thus j— 2 f f
*/(c2 — b'2} I v/(l — '
ART. 200] THIN ELLIPSOIDAL SHELLS. 101
c" — a2 /c2 — 6*
where X = 8 and the limits are ^=0 to sin \f/= I — - — . The integral is real
if the axis of c is the longest of the three.
To prove this we revert to the value of 7 (Art. 197) after the integration with
regard to <t> has been performed. Putting cos 0=0, the integral takes a known
form. This is reduced to the standard form given in the example by putting
Ex.3. Show that
(-*-„ + -^ + £ \I = - 4~ ,
\cia2 db* dc2J abc
Let Q2=(a2+M)(62 + u) (c2 + w), then I=Jdu/Q;
. d£= [ du 2 dQ _ 1
" da?~ ~ * J Q(a2 +
u)' Q du ~ a2 + «
The results follow by simple substitution.
By writing b=ma, c=na, u = va?, we see that J is a homogeneous function of
a, b, c of - 1 dimensions. The second result then follows from Euler's theorems on
homogeneous functions.
Ex. 4. If $r*mdw=abcRm, prove that
^ 2
The first result follows from Euler's theorem on homogeneous functions. By
differentiating fr*mdu=abcRm with regard to c2 we obtain Jr:im+2n2dw as in Art. 199.
The two other results follow easily.
Ex. 5. Instead of the standard integral represented by I we may use the
/"" abcdu
integral J = I
Jo «
i - : - r .
(a3 + M)* (62 + M)* (c2 + »)*
dJ bcdl dJ ca dl
We then have -T-= -- -T-, -3r=-ir^i:'&c-
aa a aa ao b do
If we write a, /3, <y for the reciprocals of a2, fc2, c2 we easily find
^, fr2do,= -4»- f A + ^ + ^.) j=2irabcl,
dy ' \da dp dy/
where the integrations extend over the surface of the ellipsoid.
Differentiating Jr2n2dw= - kirdJIdy, i tunes with regard to o, j times with regard
to /3 and A - 1 times with regard to 7 we arrive at the last result.
Ex. 6. If f(P, m8, n2) be a homogeneous function of J2, m2, n2 of s dimensions,
provethat fr2/(F> m2, n2) <*W=*/
Prove alBo that
£ ^ ^ )
da d/3 ^7/7 o
102 ATTRACTIONS. [ABT. 202
201. Theorems on thin homoeoids. The potential at any
internal point of a thin homoeoid being known it is required to find
the potential at any external point.
Let two ellipsoids have for their semi-axes (a, b, c), (a', b', c') ;
points on these are said to correspond when their coordinates
are connected by the relations
a~a" b~b" c c'"
Let do; da be two triangular elements of area at P, P' such that
the corners are corresponding points; let p, p' be the perpen-
diculars from the centre 0 on the tangent planes. The volumes of
the tetrahedra whose bases are do-, do' and common vertex 0 are
respectively %pda and ^p'da-'. The first of these volumes is ex-
pressed by one sixth of the determinant in the margin, x y z
where the several rows express the coordinates of the ac^ yt z±
corners. The second volume is expressed in the same ac3 y2 z%
way with accented letters to represent the corresponding points
on the second ellipsoid. It is evident from the relations (1) that
these determinants are in the ratio abc : a'b'c'. We therefore infer
that the elements of surface of the two ellipsoids are connected by
the equation pd* abc_
p'da-' a'Vc'
Since any elementary areas at P and P' may be subdivided into
triangles, it is evident that this relation holds for elementary areas
da; d<r' of any shape, provided only their boundaries are formed by
corresponding points.
Since the thickness of a thin homoeoid is represented by kp,
it follows that the volumes of corresponding elements of two thin
homoeoids are in a constant ratio. Adding these elementary volumes
together, it is easily seen that this constant ratio is equal to that
of the whole volumes of the two shells. If the shells are of such
thicknesses that their whole volumes are equal, then the volumes
of all corresponding elements are equal. See Vol. I. Art. 428.
202. We shall now require the following geometrical theo-
rem :— the distance between two points one on each of two confocal
ellipsoids is equal to the distance between their corresponding points.
A proof may be found in Smith's Solid Geometry, Art. 166. This
theorem is usually called Ivory's theorem after its discoverer, who
ART. 204] THIN ELLIPSOIDAL SHELLS. 103
also applied it to determine the potential of an ellipsoid at an
external point.
Let P, P' be two corresponding points, one on each of two
confocal thin homoeoids of equal volume ; let also Q, Q' be any
two corresponding elementary volumes each equal to dv. Let the
equal distances PQ', P'Q be represented by R. Iff (R) represent
the law of attraction, the potentials at P and P' of these ele-
mentary volumes are each f(R) dv. Integrating over the whole
surfaces of the shells, we see that the potential of the inner thin
homoeoid at the external point P' is equal to that of the outer thin
homoeoid at the corresponding internal point P, provided the
densities are equal at corresponding points*,
Thus when the potentials of thin homoeoids at all internal
points are known, their potentials at all external points are also
known.
203. It is evident that the potentials of these shells are equal
whatever be the law of attraction provided the potential is a function
of the distance only.
The potentials are also equal if the shells are heterogeneous, and
the density at any point is a function of (x/a, y/b, zfc). In this
case it is evident that the densities of the shells are equal at
corresponding points. The equality of the potentials is also true
when the shells are incomplete, provided only the existing parts
" correspond " to each other.
204. The theorem may also be used (though not so simply)
to compare the potentials even when the density is any function
of the coordinates. It will be convenient to express this result in
an analytical form.
Let the density p of a thin homoeoid (semi-axes a, b, c) be
f(x, y, z), and let v be the volume of the shell. It is required to
find its potential at any external point (£', 17', £'). Let a confocal
ellipsoid be described passing through the point (£', if, £') so that
* Chasles in his Nouvelle solution du probleme de I 'attraction d'un ellipsolde
hgterogene sur un point exterieur, Liouville, vol. v. 1840, shows that thin confocal
homoeoids have potentials at corresponding points proportional to their masses,
but considers only the case in which they are homogeneous. Knowing that the
potential of the outer at an internal point is constant, he deduces several theorems
on the attractions of the inner shell at external points. He finds the attraction of
a solid heterogeneous ellipsoid by dividing it into thin elementary homoeoids, the
strata of equal density being the elementary homoeoids. The case in which the
homoeoid is heterogeneous is not discussed.
104 ATTRACTIONS. [ART. 206
its semi-axes a, V, c' are given by a'2 - a2 = X, b'2 - 62 = X, c'2 - c2 = X,
where X is a root of the equation
= i
Let this ellipsoid be the inner boundary of a second thin homoeoid
whose volume is equal to that of the former. Let its density at
any point (x', i/, z) be p' =f(ax\d, by/b', czjc'). The potential of
this second homoeoid at the internal point (af '/a' ','brj' '/&' ', c£"/c') is
equal to the potential required.
We shall in general take a2, &2, c2 to be in descending order of
magnitude. If X is either positive, or negative and numerically
greater than c2, the surface (1) is an ellipsoid. If X is negative
and numerically greater than c2 the surface is one of the hyper-
boloids or is imaginary. The root of the cubic to be chosen must
therefore be the algebraically greatest root. Since the attracted
point is external to the ellipsoid a2 + X is necessarily greater than
aa, the greatest root is therefore positive.
205. Taking the case in which the two thin homoeoids are
homogeneous, the potential of the outer has been proved constant
for all internal points, Art. 68. It immediately follows that the
potential of the inner is the same at all external points which lie
on the same confocal. We therefore infer that the level, surfaces
of any thin homogeneous homoeoid are confocal ellipsoids.
It follows from, this proposition that the direction of the
attraction of a thin homoeoid at any external point P' is normal
to the confocal ellipsoid which passes through that point.
It is proved in treatises on solid geometry that this normal is
also the axis of the cone which has its vertex at P' and envelopes
the ellipsoid.
This result was given by Poisson (M€m. de I'Institut, 1835). There is an
elementary demonstration by Steiner in Grelle's Journal, voL xii.
206. Since two thin confocal homoeoids have the same level
surfaces, their potentials can be made equal over any level surface
enclosing both by properly adjusting their masses. It immediately
follows that their potentials are also equal throughout all external
space, Art. 130. Since the potentials of finite bodies vanish at
infinity in the ratio of their masses, it is evident that the masses
of the two homoeoids must be equal. We have therefore the
following theorem, the potentials, and therefore also the resolved
ART. 209] THIN ELLIPSOIDAL SHELLS. 105
attractions, of two confocal thin homoeoids of equal masses are
equal throughout all space external to both.
207. Lines of force. The lines of force of a homogeneous
thin homoeoid are the orthogonal trajectories of all the confocal
ellipsoids. Let (a', b', c'), (a", b", c"), (a", V", c'") be the semi-
axes of the confocal ellipsoid and hyperboloids which pass through
any external point (f, 17', f ). Then by a theorem in solid
geometry
a' a" a'" b'b"b'" c'c"c'"
~ V(a2 - &2) (a2 - c2) ' V(&2 - a2) (&2 - c2) ' ~ V(c2 - a2) (c2 - 62) *
see Salmon's Solid Geometry, Art. 160. Since these conicoids inter-
sect at right angles, the curve of intersection of the two hyperbo-
loids is an orthogonal trajectory of all the confocal ellipsoids. The
required trajectories are therefore found by regarding (a", b", c")
and (a'", V", c'") as constants. It follows that g'/a', y'/b', f/c' are
constant for the same orthogonal. Thus it appears that any line
of force of a homogeneous thin homoeoid intersects all the confocal
ellipsoids in corresponding points.
208. Thin homoeoid, external point. To find the potential
and the attraction of a homogeneous thin homoeoid at an external
point P'. The potential V of the given homoeoid at P' has been
proved equal to that of a confocal homoeoid of equal mass having
P' just outside (Art. 206). This again is equal to the potential at
a point just inside (Art. 145). It follows from Art. 198 that the
potential at P7 is
r* du'
= '
where M is the mass of the homoeoid and (a', b', c') the semi-axes
of the confocal which passes through P'.
This integral may be put into another form which contains the
semi-axes a, b, c of the given homoeoid instead of those a', b', c'
of the confocal. Putting a/a = a2 + X, 6'2 = 6a + X, c'2 = c2 + A. and
u' = u — X, we have
v, _ M T00 du
=
where \ is defined in Art. 204.
209. To deduce the resultant attraction, we notice that the
level surfaces of the given homoeoid are confocal quadrics. The
resultant force F therefore acts parallel to the perpendicular p'
106 ATTRACTIONS. [ART. 211
drawn from the centre 0 on the tangent plane at F. Hence
F=dV'/dp', and
dV d\ _ M 2/ = Mp'_
= ~d\dp'~ ~ 2 (a2 + X)i (62 + X)* (c2 + \)» a'6V '
since, by Art. 195, d\ = 2p'dp'.
The expression for the attraction F may be obtained inde-
pendently. The attraction of the homoeoid at F is equal to that
of a confocal homoeoid having F just outside (Art. 206) and
Mv'
therefore, by Art. 71, F=~^fj*'
Again, assuming this value of F, the potential V follows at
once by integration.
21O. Ex. 1. If an attracting body has an ellipsoid enclosing the whole
attracting mass for one of its level surfaces, prove that all the external level
surfaces are confocal ellipsoids. See Art. 130.
Ex. 2. The attractions of a given thin homoeoid on two corresponding elementary
areas taken on any two confocal ellipsoids are equal. [Chasles, Journal Polytech-
nique, 1837, Tome xv.]
Ex. 3. The attraction of a thin homoeoid at any point situated on its external
surface is proportional to the thickness of the shell at that point. [Chasles.]
Ex. 4. A thin prolate spheroidal shell of mass M is divided into two portions
by a diametral plane perpendicular to its axis. Prove that the pressure per unit of
length on the line of separation, due to the mutual attraction of the parts, is
M2 log a - log 6 .
8rt a*-b* ' [Math. Tnpos.]
The resultant force on any element of the shell is half the force just outside,
Arts. 68 and 143. If da be an elementary area, I the cosine of the angle p makes
with the axis, the resultant pressure is ^^Fldedp, where F has the value given in
Art. 209. Putting da =Qiryds, dpjp — da/a, the integration can be effected.
Ex. 5. The mutual potential of a thin homoeoid (mass M, semi-axes a, b, c)
and any internal mass M' is %MM'I, where I is the integral denned in Art. 198.
The mutual potential of the same homoeoid and any external mass M' placed as
a stratum on a confocal quadric (semi-axes a', b', c') is \MM'T, where I* is the same
integral with a', b', c1 written for a, 6, c. [See Arts. 61, 208.]
211. Solid homogeneous ellipsoid. To find the potential
at an internal point P whose coordinates are (f, rj, £).
Describe a double cone with vertex P cutting the surface
in two opposite elementary areas Qlt Qa. If R1} Rz are the
distances PQl} PQ2, the potential of the double cone at P is
^pfffi+Rf^da) (Art. 196). It is evident that if we integrate
this expression all round the point P every element of volume of
* This expression for the resultant force is given by Chasles in the Journal
Poly technique, 1837, Tome xv. See also the Quarterly Journal, 1867.
ART. 212] SOLID ELLIPSOID. 107
the ellipsoid will be taken twice over; we must therefore halve
the result. The potential at P is then
V=lpf(R1* + R2*)d<t> ..................... (1).
The distances R1} R2 are the roots of the quadratic
b
where (I, m, n) are the direction cosines of Q2PQi- This quadratic
7?2
may be written shortly — + 2FR-E = 0 .................. (3),
where r is the semi-diameter parallel to Q2PQi- Hence
ICO.
It is obvious that the term containing the product Im disappears
on integration, for the elements corresponding to (I, m) and (I, — m)
destroy each other. The terms containing mn and nl are also zero.
n f ( /2Z2r* r2\ )
Hence F= g Pf-- -J f + &c. + &c. + A da> ...... (4).
"J \\ Q> Q> J )
We have J^dco = Mirabel ........................... (5).
By differentiating (5) with regard to a, as in Art. 199, we have
ro lSr> j « i d(/a) , fdl I\
f 2r — r- da> = 27roc j ' = 2-7T abc -y- + - ) ....... (6).
a8 da \da a/
A
After substituting from (5) and (6) in (4), we find
V =I+dLp+dLrf+M (7)
= -» T __ 7 ? T 7 77 '/ T 7,, fe ............ V«>
irpabo ada bdb cdc
212. If we substitute I=fdu(Q this becomes
F
fee XT., f £2 OT2
— I 1 S ^
Jo Q 1 ~ a? + u b* + u
where Q2 = (a2 + u) (b2 + u) (c2 + u). These two important expres-
sions are often written, for brevity, in the form
(9).
Here D - 1 - (^ _A_ *L- f" du (10)
" ~ ~J Q ' 27rabc~ ada~J<> (a? + u)Q"
with similar values for B and G.
The component forces at any internal point P are then
...... (11).
108 ATTRACTIONS. [ART. 216
213. By putting f, 17, £ equal to zero, we see that %pD is the
potential at the centre of the solid ellipsoid. We also notice that
by Art. 199
<fe> ...... (12),
where the integrations extend over the whole surface of the body,
The first of the results (13) follows also from Poisson's theorem
since d^Vjdx^ — — Ap &c. The second may also be deduced from
(4) ; for the sum of the coefficients of £2, rj2, £2 after multiplication
by a2, 62, c2 is evidently — r1.
1 Z2 ra2 n2 , ... ,. .,, . ,.^
Since — = -; + -rr + -; , we see by substitution either in (4) or
r2 a2 62 c2
(12) that the constants A, B, G are functions of the ratios of the
axes and are therefore the same for all similar ellipsoids.
214. The four integrals A, B, C, D have here been expressed in terms of the
integral I and its differential coefficients with regard to a, b, c. Other standard
integrals might also have been taken. Thus we might use the integral called J in
Art. 200, Ex. 5. We might also express the components X, Y, Z in terms of any
one of the four integrals A, B, C, D. We deduce from the third part of Ex. 3,
BV-Aa* ,dA C#-Aa* dA ....
Art. 200,. — rs - 5- = &-jr» —a - *-=c^~ ........................ (14)-
o2 - a? do c2 - oa dc
These relations enable us to deduce the formulas for X, Y, Z given by Laplace in the
Mecanique Celeste, vol. n. p. 12.
210. Ex. Prove that the three numerical constants A, B, C lie between
«/a8 and v/c8 where v is the volume and a, c are the greatest and least axes of the
ellipsoid. Prove also that D lies between lira? and 41TC3.
To prove the first theorem we notice that the integral (10) is decreased by
writing a for b and c ; the integration can then be effected. A superior limit is
found by writing c for a and 6. The second theorem follows from the equations
(13) Art. 213 by eliminating first A and then C.
216. To find the level surfaces inside the attracting ellipsoid.
These surfaces are given by A£a + Erf + C^ = K, where K is a
constant. Since A,B,G are necessarily positive, the level surfaces
are similar and similarly situated ellipsoids.
To trace their forms, we must consider the magnitudes of the coefficients
A, B, C. We have (Art. 212)
A-B _ <*> Zirabc.du Aa?-Bb* *> 2irabc.udu
Q(a2+w) (b2+u)*
ART. 218] SOLID ELLIPSOID. 109
Both these integrals are essentially positive. It follows that when a, b, e are in
descending order of magnitude, both I/A, 1JB, I/O and Aa?, Bb-, Cc2 are also in
descending order.
A level surface so far resembles the attracting ellipsoid that both quadrics have
their longest, their shortest and their mean axes respectively in the same directions.
The axes of a level quadric are (K/A)*, (K/B)*, (£/C)*. If c be the least axis of
the attracting ellipsoid and K= Ccs, the level quadric touches the ellipsoid at the
extremities of the least axis, while the other two axes are less than the corre-
sponding axes of the ellipsoid. The level quadric therefore lies wholly within the
ellipsoid, for if not it would cut the ellipsoid in two curves one on each side of the
plane of xy and also touch it at the extremities of the axis of z. This of course is
impossible. Any level quadric therefore lies wholly within the attracting ellipsoid or
intersects its surface according as K is less or greater than <7ca.
217. Ex. 1. Prove that the level quadrics are more spherical than the
bounding surface of the attracting ellipsoid.
The eccentricities of the sections of the two quadrics by the plane of xy are
respectively given by eJ2=I-AjB and e2=l-62/aa. It follows immediately that
e"*-e* is negative.
Ex. 2. If a concentric ellipsoidal cavity be cut out of a solid homogeneous
sphere, show that within the cavity the equipotential surfaces are given by
(2A - B - G) *a + (25 - C - A ) y* + (2(7 - A - B) *2= constant,
where A, B, C are constants depending on the shape of the cavity.
[St John's Coll. 1887.]
218. Other laws of force. The potential of a solid homogeneous ellipsoid at
an internal point P when the law of force is the inverse nth power of the distance
may be found by the method used in Art. 211.
By describing a double cone with the vertex at P as before, we find that the
potential is F= J _ ISdu + C, S= Rf-* + (-Ez)*~K.
When K is even, S is a symmetrical function of the roots of the quadric (2) of
Art. 211. The double integrals take forms similar to that in equation (4) and may
be reduced to single integrals by differentiations of fr*d<a=2irabcl.
We notice that when *c>4 the expression S is an integral rational function of
the direction cosines (I, m, n) and the final integrals can be evaluated without
difficulty (Art. 200, Ex. 1). The potential for these laws of force can therefore be
found in finite terms free from all signs of integration.
When the law of force is the inverse fourth power of the distance we have for
? if* f2
the potential at an internal point £, 17, £ , F4 = £ irp log E+Ct .E = 1 — 5 ~ p — s •
From this result the potential for the inverse sixth, &c., powers may be deduced
free from integrals by using Jellett's theorem (Art. 96).
The component attractions at an external point may be deduced by Ivory's
theorem (to be presently proved). Thence by integration the potential for the
inverse fourth power of the distance is found to be
The potential of a thin homogeneous homoeoid may be found in a similar
manner, but it may also be deduced from that of a solid ellipsoid by taking the
total differential with regard to a, 6, c on the supposition that the ratios a : b : 6
HO ATTRACTIONS. [ART. 221
are unaltered. See Arts. 195 and 92. The potentials at an internal and external
point are for the law of the inverse fourth power,
219. Spheroids. To find ike potential and attraction of the
solid spheroid whose semi-axes are a, a, c at an internal point.
To find the constants A, C we use the equations
The limits of integration are 6 = 0 to TT and <f> = 0 to £77. If we
put cos 6 = z, the second equation becomes
dz
where the limits are z = 1 to z = — 1.
If the spheroid is oblate, a is greater than c, and
. V(a2 — c2) /i \
tar' -^ ....... a).
If the spheroid is prolate, a is less than c, and
We also have ZTrabcI = fr2da) = D. Thus the values of J. and (7
may be found either by solving these equations or by using the
formulae A= — 27rabcdI/ada &c. The potential at any internal
point is then F= $p [D - A (£2 + if) - C?}. We notice that
tan"1 V(«2 — c2)/c in an oblate spheroid is equal to the angle
subtended at the extremity of the axis of revolution by the distance
between the centre and either focus.
32O. Ex. 1. The earth being regarded as an oblate homogeneous spheroid
the ratio c/a=l-e where e is the elliptic! ty. Since the value of e is 1/300 nearly,
it is generally sufficient to retain only the first powers. Prove that
[z* , a2 ( f cos2 6 sin 6d6 dd>
[We have <7=|-dw=^M i + '2ecos20 Expand the subject of integration
in powers of e.]
Ex. 2. Show that an attracting homogeneous oblate spheroid of eccentricity },
in the centre of which there acts a repulsive force /tr, will have its own surface for
one of its level surfaces if 3fi.=8irp (SirJB - 27). [Coll. Ex. 1888.]
221. Nearly spherical ellipsoids. Ex. 1. The axes of an ellipsoid are so
nearly equal that the square of the difference can be neglected. Prove that
ART. 222]
IVORY'S THEOREM.
Ill
Put &/a=l — /3, c/a = l — 7; then since A is a function of &/a and c/a we have
A = |TT (Z/ + M/3 + ^7) nearly where L, M, N are independent of the axes. Now A
cannot be affected by interchanging 6 and c, hence M=N. Also when /3=0 and
7=e the ellipsoid becomes a spheroid and the expression for A must become
identical with that found in Art. 220. Hence L = l, N=-%. This proof is
commonly ascribed to D'Alembert.
Ex. 2. Prove that to a second approximation the constants of the attraction
(Art. 212) are
=|TT {1-f (-2/3 + 7) _^(_ 18^ + 407 + 97') + ...},
when &/«=!-£ and c/a — 1-7. We notice that since A+B + C=iir for all
values of 0 and 7, the sum of the coefficients of any power in the three first
expansions must be zero.
222. Ivory's theorem. To find the attraction of a solid
homogeneous ellipsoid at an external point P' whose coordinates
are %', 77', £".
Let R be the distance of any element QQ' of the ellipsoid from
P', and let <f> be the angle this
distance makes with the axis of
x. Thus R = QPf, <f> = P'QQ'.
Iff (R) be the law of attraction,
the x component of the attraction
of this element at P' is
pdxdydzf (R) cos <£.
Draw Q'n perpendicular to
P'0, then cos<f> = -^7 = --7-.
fi W
Vyvaj \Juw
The x attraction at the element at P', measured positively in the
positive direction of x, is therefore pdydzf (R)dR. Let LM be a
column having its length LM parallel to the axis of x and the
elementary area dydz for base. Integrating with regard to R we
find that the x component of its attraction at P' is
pdydzff (R) dR = pdydz {f(P'M) -f(P'L)}.
Let us now describe an ellipsoid through P' confocal to the
external surface of the attracting solid. Let a', b', c' be the semi-
axes of this new ellipsoid. If L', M', P be points corresponding to
L, M, P', the column LM' will have for its base the elementary
area dy'dz', where y'jb' = y/b and z'/c' = zjc. The coordinates
112 ATTRACTIONS. [ART. 224
£, i), £ of P are known in terms of those of P' by similar relations ;
see Art. 201. The attraction of the column L'M' at P is
pdy'dz*{f(PM')-f(PL%
By Ivory's theorem, P'M = PM', FL = PL', Art. 202; the x
attractions of the columns LM, L'M' are therefore in the ratio
of the areas dydz, dy'dz' of their bases, i.e. the x attractions are
in the constant ratio be to b'c'.
If we fill one ellipsoid with columns like LM, the other ellip-
soid is filled by the corresponding columns, and the x attractions
of the corresponding columns are in the same ratio. We therefore
x attn of inner ellipd at P' be
infer that — — — 5 -- ^—A — —77 = TT~, .
x attn of outer ellipd at P be
Similar theorems apply to the y and z components of the
attractions of the two ellipsoids.
This theorem was enunciated and proved by Ivory in the Phil. Trans, for 1809.
We ought perhaps to speak of it as Ivory's demonstration of Laplace's Theorem.
But Ivory's own proof is not now exactly followed, as further simplifications have
been introduced. The extension of the theorem to any law of force is due to Poisson,
Bulletin... la Societg Philomathique 1812, 1813.
223. When the law of attraction is the inverse square, the
axial components of the attraction of the outer ellipsoid at the
internal point P or (£, 17, £) are
The axial components of the inner ellipsoid at the external
point P' or (f, T/, ^) are therefore given by
Here a, b', c' are the semi-axes of the confocal drawn through the
attracted point P', and A', B', C' are the same functions of the
ratios of the axes a, b', c' that A, B, C in Art. 213 are of the ratios
of a, b, c.
224. From these values of X', Y', Z' we may at once deduce
a theorem often called Maclaurin's theorem. If we compare the
attractions at the same point of two different ellipsoids bounded
by confocals, we notice that a', b', c are the same for each, so that
each of the components X', Y', Z' is proportional to abc, i.e. to
the product of the axes. The attractions therefore at the same
external point of different homogeneous ellipsoidal bodies bounded
by confocals are the same in direction and their magnitudes are
ART. 225] SOLID ELLIPSOIDS.
113
proportional to their masses. The law of attraction is that of the
inverse square of the distance.
Let V, V be the potentials of the two ellipsoids at any
external point P', then since the component attractions are pro-
portional to the masses M, M't the ratios V/M and V'/M' can
differ only by a quantity which is independent of the coordinates
of P'. Since both potentials are zero at an infinite distance this
constant must also be zero.
Hence the potentials of two confocal solid homogeneous ellipsoids
at any point external to both are proportional to their masses.
Since a focaloid is the difference of two confocal ellipsoids, it
follows that the potentials of thick focaloids are also proportional
to their masses.
225. To find the potential V of a solid homogeneous ellipsoid
at an external point P' whose coordinates are %', ?/, £"*.
Through the external point P' describe an ellipsoid confocal
with the given ellipsoid. If the matter composing the given
ellipsoid be made to fill the confocal (by changing the density
from p to p) the attraction, and the potential, are unaltered at
all external points. Let a', b', c' be the semi-axes of the confocal
ellipsoid, then p'a'b'c = pabc.
Since the point P' is on the surface of the confocal ellipsoid
the potential is the same as that found in Art. 212 for an internal
point. We therefore have by (9)
v' =
where A', B', C', D' are the same functions of a, b', c' that A, B,
C, D are of a, b, c. The potential may also be written in either of
the two other forms given in Art. 211.
- v' *
* irpabc
~Jo
o
where Q'* = (a'8 + u') (6/2 + u'} (c/2 + u').
* The expressions for the potentials of a homogeneous ellipsoid at an external
and internal point were given by Rodrigues as early as 1815 (Correspondance sur
I'Ecole Roy ale Poly technique, vol. IIL). An analysis of his method is given by
Cayley in the Quarterly Journal, vol. n. 1858. There is a memoir by Poisson on the
attraction of a homogeneous ellipsoid (Memoires de I'Institut de France, 1835) in
which he gives a history. He finds the component attractions of the ellipsoid.
R. S. II. 8
114 ATTRACTIONS. [ART. 228
226. If we put a/2 = a2 + X, b'2 = V + \, c/2 = c2 + X, and
u' = u — \, the last expression becomes
where Q2 = (a2 + M) (&2 + u) (c2 + u) and X is to be found from the
cubic of Art. 204.
To deduce the component forces X', 7', Z' we differentiate V
with regard to f , 17', £' respectively. Since the lower limit X is a
function of the coordinates we shall here require the value of
dV'fd\. But the subject of integration is zero when we write A,
for u, hence d V'/d\ = 0. We therefore have at once
X' °°du-2? = ,™ du'
*
__
Qa? + u o (af* + u')Q -rrafVc"
by Art. 212. This agrees with the result in Art. 223.
227. Ex. Let p, q, r be the lengths of the axial intercepts of any external level
surface of a solid homogeneous ellipsoid (a>b>c). Prove (1) that p is greater than q
but less than qa/b, (2) that p2- a2 is greater than g2-o2 but less than g2-^8.
Putting V=vpabcK, the intercepts are given by
Q~
See Art. 226. If the inequalities to be proved were reversed it may be shown that
these equations could not be true.
228. Some special cases. Ex. 1. The attraction of a thin homoeoid at any
external point is the same as that of a thin disc bounded by its elliptic focal conic
/ x- y-\ $
and having the surface density at any point P inversely proportional to ( 1 — 2 - ^ j ,
where (x, y) are the coordinates of the point P and 2a, 26 the axes of the focal conic.
Prove also that the level surfaces of the disc are confocal quadrics.
This follows from the theorem in Art. 206, since the disc may be regarded as a
confocal homoeoid in which the axis c is evanescent. To find its law of density we
notice that the mass on any elementary area dxdy is 2p — dxdydc. Now — = -
c2 c2
because za=c2 — 5 x2 - jj,y*, and, the surfaces of the disc being similar, c/a and cjb are
constants. The masses being made equal, the result follows.
Ex. 2. The attraction of a solid ellipsoid at any external point is the same as
that of a thin disc, of equal mass, bounded by its elliptic focal conic, axes 2a, 26,
and having its density at any point directly proportional to (l-^-_^I) . Use
j
Maclaurin's theorem, Art. 224.
Ex. 3. The attraction of a thin prolate spheroidal homoeoid at any external
point is the same as that of a thin homogeneous straight rod joining the foci.
This result may be deduced from that given in Art. 224, but it follows more easily
from Art. 131. The thin shell and the straight line have the same level surfaces
(viz. confocal conicoids) and masses, hence their attractions are also the same.
ART. 230] SOLID ELLIPSOIDS. 115
Ex. 4. The attraction of a solid prolate spheroid at any external point is the
same as that of a straight rod joining the foci, and having its line density at any
point P proportional to SP . PH.
Ex. 5. If Vt be the potential of a thin focaloid at an internal point P, prove
where v is the volume enclosed by the shell, dv that of the shell itself, F is the
potential at the same point of the enclosed volume supposed to be of the same
density as the shell itself, and X is the difference of the squares of the semi-axes of
the two boundaries of the shell. See Art. 195.
Y di
For a solid ellipsoid we have - — =I+2-r-5|2+&c., as in Art. 211. To deduce
irpabc da?
the potential of a thin focaloid we find 5F on the supposition that a?, 62, c2 are each
increased by the same quantity X. This is evidently effected by performing on both
sides of the equation, as it stands, the operation ^=^(j~2 + 372 + ^-2) • The
result follows at once from Ex. 3, Art. 200.
Sv
Ex. 6. Show that the potential of a thin focaloid at an external point is — V.
229. Mutual attraction. Ex. 1. A homogeneous ellipsoid attracts a body
M according to the law of the inverse square ; prove that if M be a spherical or
cubical portion of the mass of the ellipsoid itself, the resultant attraction will be
the same as if the mass M were collected at its centre of gravity. Prove also that if
M be a segment of a thin exterior confocal ellipsoidal shell, and if its principal axes
at its centre of gravity be parallel to the axes of the ellipsoid, the attraction of the
ellipsoid on it will reduce to a single force through its centre of gravity.
[Math. Tripos.]
Ex. 2. A solid homogeneous ellipsoid is divided into two parts by a plane
perpendicular to an axis. Prove that the mutual attraction of the parts for varying
positions of the plane varies as the square of the area of section.
[May Exam. 1881.]
Ex. 3. Show that any plane divides a solid homogeneous ellipsoid into two
parts such that the attraction between them reduces to a single force.
[Em. Coll. 1891.]
230. Elliptic coordinates. We may express the potential of an attracting
ellipsoid at any internal or external point P in terms of its elliptic coordinates by
using a geometrical theorem usually ascribed to Chasles.
Let a', a", a'", be the semi-major axes of the three confocal quadrics which pass
through the point |, 17, £ ; let A, B, G be the semi-axes of any arbitrary confocal, then
To apply this theorem, we put 42=a2 + u, Bz=V*+u, CP=:c?+u and substitute in
the formulas already found for the potential in Arts. 212 and 225. We thus find
that the potential of a solid homogeneous ellipsoid at the point a', a", a'", is given by
At an external point the limits are u=a"2-a* to u=oo ; at an internal point «=0
to u=<x> .
8—2
116
ATTRACTIONS. [ART. 231
231. Linear and quadratic layers. Ex. 1. If a thin layer of attracting
•matter, distributed over the surface of an ellipsoid, be such that the surface density
p at any point (x, y, z) is p (Lx + My + Nz), where p is the perpendicular on the
tangent plane, prove (1) that the axial components of the attraction at any internal
point are constant and respectively equal to La'A, MbW, Nc*C, where A, B, C, have
the meaning given in Art. 212 and (2) that the potential is a linear function of the
coordinates.
To prove this we refer to Art. 92. Since the component attractions of a
homogeneous ellipsoid at an internal point are Ap%, Spy, Cp£, the potential of a
thin superficial layer of surface density pcos<f> is Ap£. Since co&<p=px/a2, the
potential of a layer of surface density pLx is La?A(. The x component of attraction
is therefore La* A, while the y and z components are zero. It is evident from
the symmetry of the law of density that the mass is zero. The potential is
La? A£ + MWB-n + Nc* Of.
This example has an electrical meaning. An uncharged ellipsoid is placed in a
field of uniform force, the direction cosines of the constant force being proportional
to the arbitrary quantities La?A, Mb2B, Nc2C. Since the resultant force due to the
electricity and to the field must be zero at all internal points, the electrical density
must be represented by -p. The result shows that the ratio pip is a linear function
of the coordinates.
In the same way we enquire in the next example what must be the field of force
that pjp may be a quadratic function of the coordinates.
If the ellipsoid is charged with a quantity E of electricity, this quantity is to be
so distributed over the surface that its attraction at any internal point is zero (Art.
68). The additional electrical surface density is therefore icp, where K is such that
the whole quantity is equal to E. By Art. 71 or 195 this gives /c =
Ex. 2. If a thin layer of attracting matter, distributed on the surface of an
ellipsoid, be such that the surface density at any point (x, y, z) is pf(x, y, z), where
/ is a homogeneous quadratic function of (x, y, z), prove (1) that the potential at
any internal point is also a quadratic function of the coordinates of that point
together with a constant, and (2) that the axial components of the attraction at any
internal point are linear functions of the coordinates of that point.
Let us regard the layer as occupying the space between two concentric ellipsoids
having their axes nearly coincident in direction. The second ellipsoid is derived
from the first by small rotations 50, d<f>, 5$ round the axes and a change of the axes
a, b, c into a + 8a <fec. By choosing 50, d<f>, 5\j/ and 5a <fec. properly, this thin layer
may be made to represent the given quadratic distribution over the surface.
Consider first the rotation 8\f/. The component displacements of a point Q are
dx= -yS\f/, dy=xd\f/, Sz = Q; the direction cosines of the normal at Q are \=pxja2&c.
The thickness of the layer is the sum of the projections on the normal. Omitting
a2-62
the factor 5^, the surface density becomes p — ^- xy, and the potential of the shell
at an internal point ~Sj,~x~^ -- y~j~=(A-B)sey.
When the surface density is pxylab the potential becomes Zirabc / g° '/^ " — ,
by substituting for A, B their values, Art. 212. The upper limit is oo and the lower
limit is zero or X (Art. 226) according as the point is internal or external. See
Art. 93, Ex. 2.
ART. 233] ELLIPTIC CYLINDERS. 117
Consider next the change of a into a + Sa. Let r be the radius vector measured
from the centre. The thickness of the shell at Q, being the projection of dr on
the normal, is j>8r/r. But, since l/r2=F/a2 + &c., where I, m, n are the direction
cosines of r, we have 5r/r=a;23a/a3. After omitting the factor Saja the surface density
of the shell becomes pz2/a2 and its potential at P is adVjda, where V is the potential
of the solid ellipsoid at the same point. After substituting for V the potential
becomes
a+u a+u +u c + w/ a2+wj
where the limits are 0 to oo or X to oo according as the point P is inside or outside
the shell.
232. Elliptic cylinders. To find the attraction at an in-
ternal point of a solid homogeneous cylinder whose cross section is
an ellipse and whose length is infinite in both directions.
The axial components of this attraction may be immediately
deduced from those of an ellipsoid by making one of the axes
infinite. Let us make c = oo , so that the infinite cylinder stands
on an ellipse whose axes are along the axes of x and y. The axial
components of the attraction at any internal point (£, r), £) are
X = - Ap%, Y=- Bpr], Z=Q, where A=Kdu and B = K dw.
Since in a cylinder (x, y) may be regarded as the coordinates
of any point on the elliptic section, we have obviously
A + B = 47r, Aa* + Bb* = fr'*da>,
where i* is the radius vector of the cross section in the plane of
xy. Putting for dco its usual polar value sin 8ddd(f> we have
//2rfw = /sin Ode . //2ety,
where the limits are 0 = 0 to tr and <£ = 0 to 2?r. The first
integral is obviously equal to 2 and the second integral is twice
the area of the ellipse, i.e. ^irab. We thus have Aa* + Bb* — 4nrab.
The axial components are therefore
ab % v ab v)
T-, ^ = -4l7rp-TTr-
+ba r a+bb
233. The potential also may be deduced from that for an
ellipsoid, Art. 212. After substituting the values of A, B, and
putting (7 = 0, we find
Since the potential of the cylinder is infinite at points on its
axis, Art. 50, it is evident that D is an infinite constant which
may be omitted when the axes a, b are not varied. This ex-
118 ATTRACTIONS. [ART. 236
pression for V may also be obtained by integrating the expressions
The level surfaces inside the attracting cylinder are similar and similarly
situated concentric cylinders. Considering a cross section we deduce from (1) that
the longest axis of a level surface is in the direction of the longest axis of the
attracting cylinder. See Art. 216.
234. Ex. If in a spheroid the axis of revolution c is very great, the spheroid
becomes a cylinder. Prove that C and Cc are ultimately zero, while Cc2 is infinite,
Art. 219.
235. To find the attraction at an external point of a solid
homogeneous elliptic cylinder.
The attraction at an external point may be deduced from
that at an internal point by an application of either Ivory's or
Maclaurin's theorem. Let a'b'c' be the semi-axes of a confocal
through the attracted point P. Then a'2 - a? = V* - 62 = c'2 - c2.
Since a'2 — a2 is finite it follows that when c and c' are both
infinite their ratio is unity. Since the components of attraction
of the given cylinder and the confocal are proportional to their
masses, we have (as in Art. 224)
, abc fc/ V ab fc,
X = — A 0 -7T7-, | = — 4-7T/3 , 7/-7T-/ S »
r a b c a + b a b
by substituting for A' its value found in Art. 232.
In this way we find that the axial components X', T', Z' of
the attraction of a solid cylinder at an external point (f ', T?', £") are
ab r „,
where (a', 6') are the semi-axes of a cross section of a confocal
cylinder drawn through the attracted point.
236. Ex. 1. Show that the resultant attraction of an infinite cylinder is the
same in magnitude at all internal points situated on a coaxial cylinder similar and
similarly situated to the boundary. Show also that the direction of the attraction
at any point on the surface of such a cylinder is parallel to the eccentric line of
that point.
Ex. 2. Show that the resultant attraction of an infinite cylinder is the same in
magnitude at all external points situated on a cylinder confocal with the boundary.
Show also that its direction at any point on a confocal is parallel to the eccentric
line of that point.
Ex. 3. If a thin stratum of attracting matter distributed on the surface of an
infinite elliptic cylinder be such that the surface density p at any point (x, y, z) is
p(L- + Mj-+N], prove that the axial components of the attraction at an
internal point (£, i), f) are X=L- - ^, Y=M -- r , Z=Q, where the coordinate
(t T~ o a -i* o
axes are the principal diameters of a cross section and the axis of the cylinder.
ART. 237] ELLIPTIC CYLINDERS. 119
This result has an electrical meaning. If the electrical density on the surface of
the elliptic cylinder be represented by - p, the electricity will be in equilibrium
when the system is placed in a field of uniform force whose components are
X, Y, Z; see also Art. 231, Ex. 1.
Ex. 4. If the surface density p of a thin stratum of attracting matter placed on
the surface of an infinite elliptic cylinder be given by p=p ( L-. • + M~ + N^-\
\ a2 ab b2J '
prove that the x component of the attraction at any internal point (£, 77) is
X= - ----- -TJ {{L-N)% + Mri}, with a similar expression for the y component.
(a -J- oj
Ex. 5. Show that the potential at an internal point of an infinite cylindrical
mass bounded by two coaxial cylinders is infinite. Art. 50.
Ex. 6. The components of the attraction of a right elliptic cylinder whose
section is (o;/a)2 + (2//&)2=l, and whose ends are any two planes perpendicular to the
axis, at an external point £', if, f , are X', Y', Z'. A confocal cylinder having the
same ends is described through £', 17', f , and attracts an internal point £, TJ, f, with
components X, Y, Z. Show that if |/a =£'/»', i?/6 = ??7&'> f=f » then X'/A~ =&/&',
F/r=a/a'. [Math. T. 1879.]
237. To find the potential at an ^external point of an elliptic
cylinder we use Maclaurin's theorem.
Let F, V be the potentials of two confocal cylinders whose
semi-axes are respectively a, b and a', b'. Since their component
attractions at all external points are proportional to their masses,
we must have F = -777 V + E,
a 0
where the constant E is independent of the coordinates f, 17' of the
attracted point but may be a function of the axes of either cylinder.
Let the external cylinder (a', b') pass through the attracted
point P', then by Art. 233
YL- a>b' (%"* ^ !L _ZL ab ft"1 'jLl\ F'
2-rrp ~ a' + b' (a' + V J + ^ ' ''' ZTTP ~ a' + b' (a' + b' ) 4
where D and E' are independent of %', w' but are functions of a', b'.
Let 2/ be the distance between the foci of the given elliptic
cylinder, then a/2-o/2=/2. ,
To find E', we place the attracted point on the axis of x, then
I' = a! and TJ = 0. By Art. 235 we have
dV „ . ab ab ( , ,/v
_7 = Z = -47rp^-^ = -4^-a(a -6),
after substituting b' = V(«'2 ~/2), we find by an easy integration
F= - 27^ {a'2- a' V(a"-/') +/" log (a' + b')} + G,
where the constant G of integration is independent of a', V.
120 ATTRACTIONS. [ART. 239
Comparing the two values of V we see that E' — — ab log (a + b').
Hence F= - 2*rp -j-, + - ^pab log (a' + V) + 6?.
a + o \ a o /
338. The four variables £', if, a', b' in the expression for V are connected by
the relations |/2/a/2 + ij'2/&'2=l and a'*-b''i=fi. It is often convenient to reduce
these to two independent coordinates. Let
The value of <f> determines the particular confocal elliptic cylinder on which the
attracted particle P lies, and 6 (being the eccentric angle) determines the position
of P on that cylinder. Substituting we find*
V= - irpab (e~2* cos 26 + 20) + G,
where some constant terms, functions of a, b, have been included in the infinite
constant G.
239. Heterogeneous ellipsoid, similar strata. To find
the potential of an ellipsoidal shell whose strata of equal density
are similar to, and concentric with, a given ellipsoid.
Let a, b, c be the semi-axes of the given ellipsoid ; ma, mb, me,
(m + dm)a &c. those of the inner and outer boundaries of an
elementary homoeoid. If (x, y, z) be any point on this homoeoid,
o? if z*
the value of m is given by — + ^ + -; = m2 .................. (1).
a o c
Let the density at any point (xyz) of this homoeoid be p =f(m-).
The mass of the element is therefore f(m2) . 4r7rabcm?dm.
The potential of this homoeoid at any point P is (by Art. 208)
f°° dv
2,7rabcm?dmf(m2) I - — — - ^-7 — -r-0 - ... , „ - r, ...(2),
22 22- *22 v '
where the lower limit X' is 0 when the point P is internal and is
the greatest root of the cubic
2 2 £2
, 'I __ i ___ b _ -I /q\
' ' ,w,2j>2 i TL ' ~ ^2^,2 L ^' ............ V°/'
when P is external (Art. 204). The potential of the heterogeneous
shell may be obtained by integrating (2). To simplify the integra-
tion we put v = mhi, A/ = m% The potential V of the shell is then
V f f™ du
givenby
where Q2 = (a2 + u) (b2 + u) (c8 + u), and p is zero, or the greatest
f2 rj2 f2
root of — ^ -- H _ — '- -- f- - b — = m2. (^
^
according as P is internal or external. The limits for m depend on
the internal and external boundaries of the shell.
* See a paper by Prof. Lamb in the Messenger of Mathematics, 1878.
ART. 241] HETEROGENEOUS ELLIPSOIDS. 121
First, let us find the potential Fa at an external point of the
mass enclosed by the ellipsoid defined by m = n. To effect this we
change the order of integration in (4).
To make this process clear we trace the curve AB whose ordinate u is given as
a function of the abscissa m by the equation
A M
say $ (w)=m2. When m=0, M is infinite, and when
m=n, u has the value e given by
£2 ,,-2 5-2
* ' I * — * lr
Q
In the order of integration indicated in (4) we integrate I m
along a column LM from u=ft to u= oo and then sum "
the columns from m=0 to m=OC=n. In the reversed order we integrate first
along a row PQ from m2=0(w) to m2 = n2 and then sum the rows from u—CB = e
to M=OO .
The equation (4) then becomes
F,
where Q2 = (a3 + w) (&2 + M) (c2 + M), m2 is given as a function of u by
(6) and e by (7). This formula gives the potential of the mass
enclosed by the ellipsoid defined in equation (1) by putting m=n.
If the attracted point P is on the surface of this ellipsoid, we have,
by (7), e = 0. If the potential of the whole mass enclosed by the
ellipsoid a?fa2 + <&c. = 1 be required, we have n = 1 and e = X, where
\ is denned in Art. 204.
240. Secondly, let us find the potential V2 at an internal point
of the mass between the ellipsoids defined by m — n and m = n. The
limits for the integral (4) are now u = 0 to oo and m = n to n'.
These are constants and the order can be immediately reversed.
The potential is therefore given by
m-abc
241. Lastly, let us find the potential V8 at a point P situated
in the substance of a solid ellipsoid bounded by the surface m = n.
Let the point P be situated on the ellipsoid defined by m = n.
The potentials at P of the two portions of the solid separated by
this ellipsoid are given by the values Fi, Va found above. We find
F3 by adding (8) and (9) together.
122 ATTRACTIONS. [ART. 245
where Q2 = (a2 + «) (62 + u) (c2 + u), and m2 is given by (6) as a
function of u.
One result may be briefly stated, as follows. The potential V at
P of the solid ellipsoid a?ja* + &c. = I is given by
where the limits are 0 to <x> when P is internal and \ to oo when P
is external. The value of A, is given in Art. 204, and
+ J^- (12).
242. The component forces X, Y, Z due to the attraction of the
solid ellipsoid may be found by differentiating the expression for
V just found. Since both w2 and X are functions of the coordinates
we must find dVfd\ and dV/dm*. When u = \m2 becomes unity
and the subject of integration vanishes. Hence dV/d\ is zero,
The corresponding values of F, ^ follow at once. For an internal
point \ = 0.
243. The expression (11) for the potential of a solid ellipsoid
may be put into another form in which the limits are constant by
putting u = v + X, a2 + X = a'2 &c. Writing the formula at length
we have at an external point
dv
o
The axes of the attracting ellipsoid have disappeared from the
right-hand side and are replaced by the axes a', b't c' of the
confocal which passes through the attracted point.
244. Ex. The density at any internal point T of an ellipsoid is k . OR/OT,
where OR is the semidiameter which passes through T and fc is a constant. Prove
that the integrations to find X, Y, Z can be effected in finite terms.
Prove also that the axial components are the same, at all internal points on
any given radius vector.
The last result is proved by noticing that before integration each component is
a homogeneous function of £ , 17, f of zero dimensions. It should also be remarked
that though the density at the centre is infinite the components X, Y, Z are finite.
Poisson, Connaissance <&c., 1837.
245. If we write /(m2) = A(l- w2)n, the density p at (a?, y, z)
of the solid ellipsoid, and the potential V at (£ , if, f), become
ART. 246] HETEROGENEOUS ELLIPSOIDS. 123
where the limits are X to oo for an external point and 0 to oo for
an internal point, Art. 204.
Since the density is zero at the surface of the ellipsoid, it
follows from Art. 92 that we may differentiate each of these
expressions, so that if the density is p' = dp/dx, the potential is
V = dV/d%. This enables us to find the potential of a hetero-
geneous ellipsoid when the density is as, any, op, &c.
T -i,
Let
then x=
The potentials for these three laws of density are
,, „, .„ f Rdu
p=Ax, V=ira3bcA£ I 7-5 — r-,
sj (a2 + w) Q
Rdu
The proper forms for the three following laws of density may be found by
differentiating E3. We then have
p=Axyz, V=-,
p=Ax3, V
The case in which p=sAxfy<>zh is considered by Ferrers in the Quarterly Journal,
1877, vol. xrv.
It will presently he proved that the potentials of a homoeoid, whose surface
density 0- is numerically equal to pp (where p is the perpendicular on the tangent
plane), may be deduced from that of the solid ellipsoid by differentiating with
regard to R and doubling the result, Art. 249.
The potentials for a homoeoid are therefore
The limits are 0 to oo for an internal point and X to oo for an external point.
See Art. 204.
These agree with the results obtained in Art. 231 by an elementary method.
246. We may use the method of Art. 211 to find the potential at any internal
point P, (£, 7j, £), of a heterogeneous ellipsoid whose density at any point Q is
We describe as before a double cone with its vertex at P cutting the ellipsoid in
two elementary areas Qlt Qz. The distances PQ1=JJ1, PQ3= -R^ are given by the
quadratic (2) of Art. 211. Let PQ=R.
124 ATTRACTIONS. [ART. 246
The density p at any point Q of the cone is
nB)
.............................. (1),
where 8=ld[d£ + Tndldii + ndld£ and L(i) = l .2 ... i.
The potential at P of the positive side of the double cone is S J.R*'+1d.Rda>5*0/Z/ (i),
where the limits are R=0 to R=R1 and S implies summation for all existing values
of t. To find the potential of the negative side of the double cone we let R be the
distance of one of its elements Q' taken positively. The density at Q' is found by
writing -JR for R in the series for p. The potential at P of the,negative side of the
cone is therefore 2 $Ri+1dR(-l)idw8itj>IL(i), and the limits are R=0 toR=-R<i.
Taking the two conical elements together we find for the potential of the ellipsoid
.................. (2).
As we shall integrate all round the point P, every element is taken twice over, we
must therefore halve the value of V thus found.
The quantity Si is a symmetrical function of the roots of the quadratic (2) of
Art. 211. We therefore have Si+z + 2Fr^Si+l-Er^Si=0 ........................ (3).
The initial terms are /S0=2, St= -2J?'ra. Assume that Sf and Si+1 contain only
terms of the form HKrK+i and HK'rK+i+1, where HK represents a homogeneous
function of (I, m, n) of K dimensions. It follows that S,-+2 contains terms of the same
forms, viz. HKrK+i+2, H'K+i'rK+i+3. Again 5*0 is a function of I, m, n of i dimensions,
hence Si+2 8»0 is of the form HK+irK+i~™. The determination of V is therefore reduced
to the integration of expressions in which the index of r exceeds by 2 the sum of the
indices of I, m, n.
The terms of (2) which contain any odd exponents of I, m, n give zero after
integration, as in Art. 211. Omitting these it is clear that every term of V is of the
where
f f
«= J J (
and o, ft, y are the reciprocals of a2, 62, c2. Now by Art. 197
du . fv'^dv
Qi '
where Q2=(a2 + w) (&3 + u) (c2+«), Ql2=(a+t?) (£ + v) (-y + r) and t?=l/u and the
limits for u and z; are 0 to oo . Differentiating this, we find
if
/_d
\dy
'/<
1
where the limits are v=0 to oo .
The remaining integrations cannot be effected. The potential has thus been
found expanded in powers of the coordinates f , 17, f of the attracted point, with
single integrals with regard to v for coefficients. The several powers of £, 17, f may
be collected together, once for all. We then arrive at the formula given in Art. 247.
The algebraic process of collection is however tedious when the density p contains
high powers of x, y, z. It is given at length in the Phil. Trans. 1895, vol. CLXXXVI.
Ex. 1. The density of a solid ellipsoid at any point Q is a homogeneous
function of i dimensions of the coordinates of Q. Prove that the potential of the
ellipsoid at an internal point P is the sum of a series of homogeneous functions of
the coordinates of P of the dimensions i + 2, i, i — 2, &c.
ART. 247] HETEROGENEOUS ELLIPSOIDS. 125
[This may be deduced from the equation (3) by noticing that S0 and S1 are
respectively of zero and one dimension. Thence by induction the dimensions of St
in terms of the coordinates can be found.]
Ex. 2. The density of a homoeoid is Ax and the law of force is the inverse
fourth power of the distance. Prove that the potertial at an internal point is
A da 2x
^A^'E'
247. Two theorems. Let the density of a solid hetero-
geneous ellipsoid (when K> 0) be
, / a? if
Let *-l f • '•>
a2 + u b* + u c2 + u '
-L __
K~ K + «(K+1)'Z(1).22+ +K(K + l)...(K+n)' L(n)Z» +
Q2 _ (0« + w) (& + w) (c2 + w) and Z (w) = 1 . 2 . 3 . . . n.
The potential of the ellipsoid at any point (£, 77, f) is
TT- i ^ T^M Ttf j / a^ ^7 c£ \
F=7ra6cJ. f-^Jf.6 -r-^- , ,. , . b ...... (I),
J Q ^ \a2 + u 62 + M c2 + W
where the limits are u = 0 to oo for an internal point, and u = X
to oo for an external point, Art. 204.
Let the surface density of a thin homoeoid be
x y z
\, -
b c
,,, dMl RuD RnunDn
= = 1 + + -
The potential at any point (£, rj, ^) is
£ brj ct
where the limits are 0 to oo or \ to oo , according as the attracted
point is inside or outside.
The advantages of these formulae* are (1) that the only
differentiations to be performed are those on the expression for
* These formulae were first given in this form by Dyson in the Quarterly
Journal, 1891, vol. xxv. By computing the potentials of a homoeoid for several
laws of density he discovered by induction the forms assumed by the potential when
the density is Axfy»zh. Assuming the potentials to be known he deduces the
attracting body by reasoning similar to that given in Art. 164. He gives the
necessary differentiations at length.
126 ATTRACTIONS. [ART. 248
the density, and (2) that most of the terms containing £, 77, £
have been collected together and expressed in powers of the
function R.
The component attractions at any point may be deduced from these potentials
by differentiation with regard to £, i>, f. When the attracted point is internal the
limits are absolute constants and we merely differentiate the subject of integration.
When the point is external, the lower limit X is a function of £ , i), f (Art. 204) which
makes JS=0 when u=X. Hence (as in Art. 226) we may treat X as a constant
during the differentiation, except in the first term of M'.
248. To prove the two theorems* in Art. 247 we shall adopt
the method described in Art. 164. We assume as given the two
expressions for the potential of a homoeoid at an internal and
external point, and we shall deduce the attracting body.
Let the potentials at an internal and external point be
distinguished as F and V. Then since X = 0 at the surface of
the ellipsoid we have F = V at all points of the separating
ellipsoid. It is also evident by inspection that V is zero at
infinitely distant points.
The expressions for F, V are found by actual differentiation
to satisfy Laplace's equation, Art. 95. As these differentiations
with regard to f, 17, f, present no peculiar difficulties but lead to
long algebraical processes they will not be reproduced here. We
infer however from the result that the attracting matter resides
solely on the separating ellipsoid, Art. 164.
Let a- be the surface density of the separating stratum. If dn
* The potentials of heterogeneous ellipsoids and shells have been investigated in
several ways. First there is Green's paper, Camb. Phil. Soc. 1833, where the law
of attraction is the inverse /cth and the density Enf(x, y, z) where S = l-x2/<z2-&c.
Green uses Cartesian coordinates, but a solution by means of Lame's functions is
given in Ferrers' Spherical Harmonics. The values of X, Y, Z given in Art. '242 are
due to Poisson, Connaissance des Temps, 1837 (published three years earlier). He
begins with the formulas for the component attractions of a homogeneous ellipsoid
which he had obtained in 1835 (M6m. de I'Institut de France). Cayley gives a
formula for the potential of certain heterogeneous ellipsoids in his memoir on
Prepotentials, Phil. Trans. 1875, which is really an extension of the theorem of
Art. 239 to the case in which the force varies as the inverse ccth power of the
distance. In the Quarterly Journal, 1877, Ferrers applies these results to determine
the potential of a solid elh'psoid whose density is xfy»zh, see Art. 245. He also first
discovered the rule to find the potential of a shell by differentiating with regard to
R. His proof is different from that in Art. 249. In vol. xxv. of the Quarterly
Journal Dyson gave the important formulae mentioned in Art. 247. There are
other valuable papers by W. D. Niven, Phil. Trans. 1879, and by Hobson, London
Math. Soc. 1893. There is also a paper by the author on these potentials, Phil.
Trans. 1895. A second memoir is given by Hobson in the Lond. Math. Soc. 1896.
There is also a paper by G. Prasad in the Messenger of Math. 1900. Most of these
give the applications to discs and lamina and assume that the law of force is the
inverse /cth power.
ART. 250] HETEROGENEOUS ELLIPSOIDS. 127
be an element of the normal to the ellipsoid drawn outwards we
have (since F, V differ only in the limit X)
dV _dV =dV d\
dn dn d\ dn '" ' ^
Now X = 0 at every point of the ellipsoid and £ = x, &c., hence
dV , A fx v z^
—j — = — "Traoc — ;— Y" — , r i -
aX abc \a b c>
Again X and p are given by the equations
Differentiate the former and put X = 0, we have
l_ d\ _ 2ao dx „ _ 2
p* dn a? dn ~ p*
since the direction cosines dx[dn, &c., of the normal are pxfd2, &c.
Substitute these values in (1) and we find that <r = Apty (x/a, &c.).
This is therefore the density of the stratum which produces the
potentials F, V.
249. The potential of a solid ellipsoid whose density is p=Axfy°zh being known
we can immediately deduce that of a thin homoeoid having the same law of density.
We write a=mal, b=mb1, c=mc1, u=m^u1 and then differentiate with regard to m.
The thickness dp of the homoeoid thus obtained is given by dpjp = d (ma^jma^ = dm/m,
Art. 195. The surface density a-=pdp=ppdm/m. After the differentiation has
been effected it is convenient to put m=l, so that alt blt clt MJ become again
a, b, c, u. We may also omit the factor dm and regard the homoeoid as a layer
of finite density <r=pp. It is supposed that A is independent of the axes a, b, c.
The potential of the solid elh'psoid becomes after these changes have been made
/°° An n 2/h ^Qr 2ft
?? (Bittf ,) $Vf » .— = - 77 A a \a , 2 - xi •
\, Qi (a12 + w1K(V + Wi)"(ci2 + ui)A
The operator D is unaltered, X=X1m2, and
(t2 \n+l
m2 -- ^ -- &c. ) w1»=/S'l+X"»
Oi^+ll! /
where S represents the quantity in brackets. Since JR=0 when u=\ and therefore
S=0 when M1=X1, we may as before treat Xa as constant when differentiating with
regard to m. We now see that m enters into the expression for V only implicitly
through S, The differential coefficient of F is therefore 2mdF/d5. When m=l,
S=R and exactly replaces B in the formula for F, hence dV/dm=2dVldR. The
potential of the homoeoid is therefore found by differentiating that of the solid
ellipsoid with regard to R and doubling the result.
250. The potential of a heterogeneous homoeoid whose surface density is
ff=Apxfy°zk being known, that of a solid ellipsoid whose density is
p = (l- o;2/a2 - (fee.)"-1 xfy°zh
can be deduced by integration. Let a, 6, c be the semi-axes of the ellipsoid whose
potential is required, ma, &c., (m + dm)a, &c. those of an elementary homoeoid as
128 ATTRACTIONS. [ART. 251
described in Art. 239. If p' be the density of the homoeoid, its surface density is
p'pdmlm. The potential of this element may be found by writing ma, &c., m?u, m2\t
for a, &c., u, X in the general expression (II.) of Art. 247, and multiplying the
density by a/6" ch dm/m. The potential is thus seen to be
where M' =
We shall write m?-l + R for the quantity in square brackets, this being its value
when expressed in terms of R.
To find the potential of the ellipsoid we multiply by (1 - m2)*"1 (see equation (1)
of Art. 239) and integrate from m=0 to m=l. The potential of the solid
heterogeneous ellipsoid at an external point is therefore
V = S / 1 dm? I du (1 - m2)"-1 (m2 - 1 + R)» F (u),
where F (u) contains all the factors which are independent of m, thus
An-
W~ Q (Lrc)222»
Since the attracted point is external to the ellipsoid, Xj is not zero and it is
necessary to change the order of integration by the process described in Art. 239.
The # (u) of that article is here called 1- R and since n = l, we have e=X. The
new limits are at once seen to be m2=l -R to 1 and u=\ to oo . The potential V
' = | du T
dm2 (1 - m4)"-1 (ms - 1 + B)» f (u).
The integration with regard to m2 is effected by putting l-m2=JZw so that
dm2=-.Rdt> and the limits become v = l to 0. We then find by using Euler'a
gamma function, V = S / duR*+* rMr(TC+*) F (u).
J A 1 (W + K + 1)
Substitute for J1 (u) its value given above and this at once reduces to the expression
for V given in Art. 247.
251. To find the potential at an external point of an elliptic
(a? y2\n
1 -- a — j- 1
a o /
where n is not necessarily integral.
We regard the disc as the limit of a solid ellipsoid whose axis
of c is zero. By Art. 245, corresponding values of the density p
and potential V of the ellipsoid are
TTO&C f f _ ^ 0 }Kdu
F-^— Ji1-^-*0-} ^
where the limits are A, to oo .
The mass enclosed by a prism standing on the base dxdy and
extending both ways to the surface of the ellipsoid is
(p*c* - s2)*-1 dz,
ART. 253] ELLIPTIC DISCS. 129
where _p2 = 1 — a^/a2 — y2/b*. Put z = pc sin 6 and this when K is
•*• J ^ ^ Yn *2 3fY-*r(*)r(i) .
positive reduces to dxdy 1 -- - — y- _\ x — V-7 _4c.
17 \ a2 6V F (/c + £)
We now put K = n + % and Ac = F (« + |-)/F (/c) F (|), and the
mass of the prism is a-dxdy. Since F (£) = A/TT, the potential at an
external point of the heterogeneous elliptic disc is
F,
r(n + |) Jx Q V
where Q2 = (a2 4- w) (&2 + u) u, n > — \, and \ is defined in Art. 204.
To find the potential of a homogeneous elliptic disc at any
external point, we put n = 0. The potential is therefore
v/ r«2abdu( r _£__j?\*
Jx Q V a2 + w &' + w «;•
By using Chasles' geometrical theorem (Art. 230) we may express the potential
of the disc in elliptic coordinates. We find at once for a homogeneous disc
^'=f^?[(a2 + «-O(«2 + «-o"s)(«2+«-a'"2)]*
where the limits are a'2 - a2 to oo .
252. Ex. 1. The surface density of an elliptic disc is
ff = (1 - x2/a2 - yzlbz)K ~i 0 (a;, y),
where K is positive. Prove that the potential at an external point is
where Q2=(a2 + «) (62 + w) M, the limits are X to oo , MK has the same meaning as in
f2
Art. 247 and R=l-~
[Proceed as in Art. 251, using Art. 247.]
Ex. 2. The line density of an elliptic ring is p'=p<j> (x, y). Prove that its
potential at (£r;f) may be deduced from that of the elliptic disc in Ex. 1 by putting
/c=£, differentiating V with regard to B and doubling the result.
Ex. 3 . The density of a solid elliptic cylinder is p = A (I - x2/a2 - j/2/*)2)*-1 . xfy«z*.
Prove that the potential at an external point is
1
where Q12 = (a2 + w) (&2 + «), and the limits are X to oo . If the attracted point is
internal the limits are 0 to oo .
The potentials of an elliptic cylindrical shell follow by the rule of differen-
tiation.
[Put (p = (ax)f(by)o(cz)k and c=oo in the formula of Art. 247.]
253. Confocal level surfaces. Ex. Let the law of force be the inverse «th
power. Prove that the level surfaces of an elliptic disc whose surface density is
ff — (\-!L- ^L\n , where n = £ (K - 3), are confocal quadrics of the disc. Conversely
R. S. II.
130 ATTRACTIONS. [ART. 255
prove that if the level surfaces are confocal quadrics the surface density is that
given by the expression for a.
Thence find the potential for any level surface by placing the attracted point on
the axis of the disc. The proofs may be found in the Phil. Trans. 1895.
254. The component attraction at P of a uniform elliptic disc in a direction
perpendicular to its plane may be found by using Playfair's theorem (Art. 27). We
describe a cone whose vertex is P and base the elliptic area. The normal attraction
of the disc is equal to the solid angle of the cone multiplied by the surface density
of the thin disc.
When the. attracted point lies on the hyperbolic focal conic of the attracting
ellipse, the cone is known to be a right cone and the solid angle may be found by
elementary solid geometry.
If in this last case the distance of P from the plane be f, and the major axis of
the confocal ellipsoid through P be a', we have
Z
'
2ir V2-W (aT + &4)*
X _b f a' -,} _<>b If f2 + ft2 \fr 1]
2^~h \(a'2-hrf~ ) h iVa^+W a]'
2ir a *
where 2ft is the distance between the foci of the attracting disc, and the surface
density is unity.
Ex. 1. Prove that the solid angle of the cone Ax*+Byz
-Cz2=0 is I
Q
where Q^=(C-u) (A + u) (B+u). The limits are u = 0 to C, where A, B, C are
positive quantities.
Ex. 2. Prove that the solid angle of the cone
Ax2 + By2 + Cz* + 2Dyz + 2Ezx + 2Fxy = 0
[2Judu A+u, F , E
18 I ~T7 — rr» wnere A is the determinant in the margin.
J v(-A)
The limits are M=0 to that root of the equation A=0 which
has a different sign to the other roots.
PR n
E , D , C + u
Potentials of rectilinear figures.
255. Potential of a lamina. To find the potential at any point P of a plane
lamina of unit surface density.
Let P^V be the perpendicular from P on the plane. Let the plane of the lamina
be the plane of xy, N the origin and NP the axis of z. Let NP=£. Let (r, 6) be
the polar coordinates of a point on the plane of xy.
If QQ' be any elementary arc of the curvilinear boundary, the potential of the
triangular area NQQ' is / ^ J"a , where the limits of integration are r=0 and
r=r. If R=PQ, this reduces to (
Integrating this again for all the elements of the boundary, we see that the
potential V at P of the area of any closed plane curve is J(JJ - f) dO. In this
expression the limits are determined by making the point Q (whose coordinates are
r, 0) travel completely round the curve in the positive direction, the elementary
ART. 256]
RECTILINEAR FIGURES.
131
angle dO having its proper sign according as the radial angle 0 is increasing or
decreasing when Q passes over each element of the perimeter.
When the perpendicular PJV falls within the lamina, the limits of 0 are 0 and
2w, the expression for the potential is then $Rd0 - 2irf. When the perpendicular
falls outside the lamina the upper and lower limits of 0 are the same, so that
Jfd# = 0 and the expression for the potential is simply [RdQ.
256. We may put the expression just found for the potential into another
form which is sometimes more useful.
If rdOdr is any element of the area of the triangle NQQ', u its distance from P
and if> the angle u makes with the normal to the plane, the solid angle du
subtended at P by the triangle is
rd0dr
the limits of u being f and B.
The potential of the triangular area NQQ' at P is, by Art. 255, equal to
i*dO
In fig. 1, the perpendicular PN falls within the attracting area. We then find, by
integrating all round the perimeter of the area, that the potential at P is
where ta is now the solid angle subtended at P by the area.
P
P
Fig. 1.
Fig. 2.
In fig. 2, the perpendicular PN falls without the area. In this case we must
subtract from the potential of NQQ' that of NSS'. Since dO is positive for QQ' and
negative for S'S when a point travels round the curve in the positive direction, the
form of the result is unaltered.
Let ds be the length of any elementary arc QQ' of the perimeter, p the perpen-
dicular from N on the tangent at Q. Then since rzd6=pds, the potential at P of
the area takes the form V = I -^ - f w, where the integration extends all round
the perimeter, and w is the solid angle subtended by the lamina at P.
Ex. If the law of force be the inverse fifth power of the distance, show that
1 / pds ,
the potential of a plane lamina of unit density at a point P is ^r^ I -pj » where
the integration extends all round the perimeter and the letters have the same
meaning as in Art. 255.
9—2
132 ATTRACTIONS. [ART. 260
257. When the lamina is bounded by rectilinear sides, p is constant for each
side and may therefore be brought outside the integral sign. The integral \dsjE is
then the potential of that side at P. We therefore have the following theorem.
If V be the potential at any point P, of the area contained by any plane rectilinear
figure regarded as of unit surface density; Vlt F2, <&c. the potentials at the same point
of its sides each regarded as of unit line density, a the solid angle subtended at P by
the area, then V'= -^<a+p^+p^ + (&c., where f is the length of the perpendicular
PN on the area, and plt p2, dbc. are the perpendiculars from N on the sides taken
with their proper signs. •"
The signs of the perpendiculars are determined by the following rule. If the
point Q travel round the perimeter in the direction of the motion of the hands of a
watch, the perpendicular p is positive or negative according as the origin N lies on
the right or left-hand side of the tangent at Q.
258. Potential of a solid. If V" be the potential at any point P of a solid, of
unit density, and bounded by plane rectilinear faces; Fa', F2', dkc. the potentials at
the same point of its faces each regarded a» of unit surface density, then
where fj, f2, d~c. are the perpendiculars from P on the faces taken with their proper
signs.
Describe an elementary cone whose vertex is P and whose base is any element
of area of the boundary of the solid. Let dw be its solid angle. The volume of an
element of the cone being r^dwdr, the potential of the cone at P is
_ »d<r
r r
where r is now the radius vector drawn from P to the elementary area da- and p is
the perpendicular from P on the tangent plane. The potential of the whole solid
body at P is therefore \
I -
When the boundaries of the solid are planes, p is constant for each plane and
fpdffjr is the potential of that plane face at P. We have at once V"=%2pV.
259. The solid angle subtended at any point P by any triangle ABC is the area
of the unit sphere enclosed by the planes PAB, PP.C, PC A. This area is the same
as that of the spherical triangle traced on the sphere by these planes, and a finite
expression for its value is given in books on spherical trigonometry. Since any
polygonal area can be divided into triangles it follows that the solid angle subtended
at P by any rectilinear figure can always be found. The result may be complicated
but it involves no integrations which cannot be effected.
It immediately follows from Arts. 257, 258 that the potentials of all rectilinear
figures and the potentials of all solids bounded by plane rectilinear faces can be
found. Thus the three integrals which express the components of the attraction of a
rectilinear lamina or solid can be found infinite terms.
260. Components of Attraction. Some simple expressions may be found for
the components of the attraction of the lamina. We know by Playfair's theorem,
that the component along the perpendicular PN on the lamina is equal to the solid
angle subtended at P by the lamina, see Art. 27.
We may obtain an expression for the resolved part of the attraction along a
straight line drawn in the plane. If this straight line be called the axis of x and
the boundary of the lamina be a closed curve in the plane of xy, the x component
ART. 262] RECTILINEAR FIGURES. 133
of the attraction is X'= I -^ , where R is the distance of an element of the boundary
from the attracted \ oint P.
Divide the lamina into elementary rectangles having their lengths parallel to
the axis of x, and let the breadth of each be dy. If AB be any one of these
(regarded as of unit surface density), its x attraction on P in the direction AB is
(— - — ) dy, see Art. 11. The attraction of the whole lamina is therefore $dy/R,
PA PBJ
where R stands for either PA or PB, and dy is taken positive or negative according
as the ordinate y is increasing or decreasing when a point Q travelling round the
curve passes A or B.
261. A solid body of unit density is bounded by plane faces: it is required to
find the resolved part of its attraction at a given point P in a given direction Px.
Whatever the form of the solid may be, its component of attraction in the
direction Px is X"= I *-^-« where d<r is an element of the surface, <f> the angle
J M
the normal at da makes with the given direction Px and R is the distance of dff
from P.
When the solid is bounded by plane faces, cos 0 is the same for all the elements
of the same face. It may therefore be brought outside the integral sign. Since the
integral Jd<r/JJ is obviously the potential at P of the face, we have at once
X" = YI cos 0X + F2' cos 02 + = 2 Fcos 0,
where F/, F2', &c. are the potentials at P of the plane faces regarded as of unit
surface density, and 01} 02, &c. are the angles the normals measured inwards make
with the direction in which X is measured.
262. Ex. 1. If a, £, y, d be the quadriplanar coordinates of a point P referred
to the faces of a tetrahedron, show that the potential of the solid contained by the
tetrahedron regarded as of unit density is ^ (Fjo-f F2/J+ F87+F48) where F,, Fa,
F3, F4 are the potentials at the same point of the several faces regarded as of unit
surface density.
Ex. 2. Show that the solid angle w subtended at any point P by a triangular
area ABC is given by
/ w\a
I ft; cosec - ) =
2J ~ 4 4 4
where v is the volume of the tetrahedron ABGP and p, q, r are the distances of P
from the angular points of the triangle.
Ex. 3. The triangle OBC is right-angled at B, and at 0 a straight line OP ia
drawn perpendicular to its plane. If the triangle be of unit surface density, prove
that its attractions at P resolved parallel to OP, OB, and BC respectively are
tan-1 — (a2 + 62 + c2)* - tan-1 -
ac ^ c
o *• ^
.
log - i- - — '— -- f log
'
where a=OP, 6= OB, c=BC. Since any rectilinear figure in the plane of xy may
be divided into right-angled triangles having a common corner 0 by dropping
perpendiculars from 0 on the sides and joining 0 to the corners, these results give
the three resolved attractions of any plane rectilinear figure. [Knight's problem.
Todhunter's History, p. 474.]
134 ATTRACTIONS. [ART. 264
Laplace's Functions and Spherical, Harmonics.
263. In many parts of the theory of Attractions, the integrations are shortened
and made more comprehensive by the use of Laplace's functions. In other parts
the necessary processes could not be effected without their help. There are several
treatises on these functions from which the reader may acquire a knowledge of this
important branch of Pure Mathematics. The propositions however which are
wanted in Attractions are not very numerous and these books, contain much more
than is here required. At the same time the subject of Attractions is generally
approached by the student at a period of his course when he has not yet reached
the proper study of these functions. For these reasons it seems proper to make
a preliminary statement of a few elementary theorems which the reader acquainted
with Laplace's functions may pass over.
264. Expansion of the inverse distance. Let P, P' be
two points, one of which will afterwards be taken as a point of the
attracting mass and the other as the point at which the attraction
is required. Let (x, y, z), (x, y', /) be their Cartesian coordinates
referred to any rectangular axes, (r, 6, <£), (r', 6', <£') their cor-
responding polar coordinates. Let R be the distance between the
points and let p = cos POP'. We therefore have
R *J{(x-x
It will be found convenient to expand I/R in a convergent
series of ascending powers of either r/r' or r'/r. Supposing first
r < r', we write h = r/r'. We then have by the binomial theorem
Expanding these terms and writing Plf P2, &c. for the
coefficients of the several powers of h
(I-2ph+h*)-* = I + P1h + P2h* + ............ (2).
The terms containing hn are evidently the first in (2ph — A2)w,
the second in (2ph - /i2)""1, and so on. It is therefore clear that
Pn is a rational integral function of p, whose highest power is pn
and whose powers descend two at a time, the terms being alternately
positive and negative. Thus Pn is of the form
Pn = Anpn + A^pn-* + ..................... (3),
where An, -A«_2, &c. are constants.
These constants are easily found when n is a small integer by
the use of the binomial theorem in the manner shown above, thus
, &c.
ART. 268] LEGENDRE'S FUNCTIONS. 135
265. The function Pn is usually called a Legendre's function
of the nth order. It is sometimes written in the form Pn (p)
when it is desired to call attention to the independent variable.
Regarding one of the two radii vectores OP, OP' as a fixed axis
and the other as capable of moving into all positions round the
origin, Pn is a function of the inclination of the latter to the fixed
axis. The fixed radius vector is called the axis of reference of the
function or more shortly the axis of the function.
266. If (a', @', 7') are the direction cosines of OP', we have by
projecting OP on OP pr = ax + J3'y
Regarding OP' as fixed in space and OP as moving about 0 we
see that Pnrn is a homogeneous rational and integral function of the
coordinates of P.
The quantity 1/.R, regarded as a function of the variables (x, y, z),
is known to satisfy Laplace's equation, Art. 95. Since this is true
whatever (xr, y', z') may be, provided they are fixed, it follows that
the coefficient of every power of 1/r' in the expansion
1 1 P^r P2r2
S = ? + 7T + ^ + ..................... (4)
satisfies Laplace's equation.
267. Any homogeneous function of (x, y, z) which satisfies
Laplace's equation is called a spherical harmonic function. Its
degree may be any positive or negative integer, it may be
fractional or imaginary.
When the function is such that it may be written in the form
rnf(9) where 6 is the inclination of the radius vector to a fixed
straight line, it is called a zonal spherical harmonic. We therefore
see that Pnrn is a zonal spherical harmonic of the nth order.
268. The expansion (4) has been made in powers of r/r' on
the supposition that r is less than r1. If the contrary be the case
we must make the expansion in powers of r'/r in order that the
series may be convergent. We then have
1 l + ^ + *£ + .................. (5).
R r r3 r3
It follows in the same way that the coefficient of r/n, viz.
Pnr~(n+1)> is a homogeneous function of the -(w + l)th order
136 ATTRACTIONS. [ART. 27 1
which satisfies Laplace's equation. Thus both Pnrn and Pnr~(n+V
are zonal harmonics of different orders.
269. It is useful to notice that the values of Pn when p = + 1
and p = 0 follow at once from the series (2). Thus when p= ± 1,
Pn is the coefficient of hn in the expansion of (1 + h)~\ When
p = 1, Pn — 1 and when p = — 1, Pn — + 1 or — 1 according as n is
even or odd. Both cases are included in the statement thatPn —pn
when p=±1.
It follows that the sum of the coefficients of the several powers
of p in the expansion of Pn is unity.
After differentiating the series (2) K times we find
dp* dp* dp*
where ft.= 1 . 8 . 5. . . (2*c - 1). It follows that when p = ± 1, ¥%*
' dp* L(n-K)L(K) 2* '
The value when p=0 is somewhat more complicated.
270. Any integral rational function of p of the nth degree,
say F (p), can be expanded in a series of the form
F(p) = BnPn + B^P^ + ...+£0P0.
Since the highest power of Pn is Anpn, we can, by properly
choosing the constant Bn> make F (p) - BnPn = F2 (p) contain
pn~l as the highest power. Choosing again the constant Br^_1
properly we can make F2 (p) - Bn^.1Pn_l contain pn~* as the
highest power and so on until we arrive at zero. In this way,
we find
^ = i (2P2 + P0), tf = \ (2P8 + 3P,), ? = A (8P4 + 20P2 + 7P0), &c.
It follows from Art. 269 that the sum of the coefficients of the
functions Pn in any one of these expressions is unity.
271. To prove that Pn = 1 1 ^ (? - 1)»
Let u = p + % (uz — 1) h, then by Lagrange's theorem
By solving the quadratic and differentiating we find
(itJ
(B).
The coefficient of hn in the expansion (B) is by definition Pn. By
differentiating (A) and comparing the two expansions the theorem
ART. 276] LEGENDRE'S FUNCTIONS. 137
follows at once. The positive sign in (B) must be taken, because
when n = I, Pn =p. This expression for Pn is due to Rodrigues.
272. COR. Since all the roots of (p* - l)n = 0 are real and lie
between + 1 inclusively, those of -r-(£>2 — l)n = 0 are also real and
lie between those of (p* — l)n = 0. By continuing this process we
see that all the roots of Pn = 0 are real and lie between ± 1.
273. The two following equations are important
1 = 0 ......... (2).
The first is usually called the differential equation and the second
the equation of differences and sometimes the scale of relation.
To prove these we notice that, if u = JLPnhn, the left-hand sides
of the equations are the coefficients of hn in the expressions
By substituting u = (1 — 2ph + A2)~i these expressions are found to
be zero.
The following theorems are also useful
These may be proved by substituting in
274. The equation -Pn=0 has no equal roots, for if Pn and dPJdp were zero
simultaneously it would follow from the differential equation (Art. 273) that either
p= ±1 or d2Pn/dp2=0. The first alternative is impossible since these values of p
make PB= ±1. Differentiating again we prove in the same way that <PPn/dps=0
and so on. But this would make dnPJdpn=0 which it is not, for by Art. 264 it is
equal to An [n.
275. The roots of Pn=0 lie between those of Pn+1 = 0. Let alt Oj, ... c^ be the
n roots of Pn=0 in increasing order of magnitude. Then dPJdp is alternately -fr-
aud - when we give these values to p ; it has the same sign as Pn when p > an and
is therefore positive when p = an. But by (4) of Art. 273 Pn+1 and dPJdp have
opposite signs when Pn=0 and jp<l. Hence PB+1 is alternately - and + when we
put^ = 0!, &c. an, and is negative when p — an. Again PB+1 is positive when p = l
(being in fact unity), hence one root of Pn+1 = 0 is >on, n - 1 roots lie between those
of Pn = 0, and the (n+l)th root must be ^oj.
276. The reader is recommended to trace the polar curve r = a + bPn for the
values n=l, 2, 3, &c. where Plf Pa, &c. have the values given in Art. 264 and
138 ATTRACTIONS. [ART. 279
p=cos$. The- constant & should be regarded as much smaller than a. The two
theorems of Arts. 274 and 275 will be found useful in tracing the relations which
exist between the several functions.
377. It is important to notice that the function Pn is not numerically greater
than uuity for any value of p less than unity. For the proof of this we have here
no room.
Supposing h to be less than unity, the series
is convergent even when we replace every coefficient by its greatest numerical value
and make every term positive. The series is therefore absolutely convergent when
both p and h are less than unity.
r+l
278. To prove that I f(p)Pndp = 0, where f(p) is any
integral rational function of p of less than n dimensions. It
follows from this that when m and n are unequal (so that one is
r+i
less than the other) I PmPndp = 0.
J -i
By a theorem in the integral calculus we have
fudv = uv-u'vl + u"vn-...+(-l)nfundvn ......... (1),
where accents denote differentiations and suffixes denote in-
tegrations. Let Q be finite between the limits and let
dn-iQn d"-2^ dn~3Qn
V = ^-> Vl = ^p^' "»=^'&C-
Each of the terms v, v^... vn^ contains the factor Q at least once
and therefore vanishes when p is put equal to any root of Q = 0.
If we also put u=f(p) the series terminates before we arrive at
the final integral. It follows that the integral (1) is zero when
the limits are any two unequal roots of Q — 0. Let Q =p* — ] , the
integral is then zero when the limits are p=±l. See Art. 271.
279. If f(p) is of n or higher dimensions, the only term on the right-hand
side of (1) (Art. 278) which is not zero is the final integral. This is also true if
f(p)=p* (where K is a positive quantity = or >n) and the limits are p=0 to p = l.
In this case all the terms up to U*VK are zero because the first factor vanishes when
p=0 and the second when p = l. The final integral is made one of the standard
forms in the integral calculus by putting p= cos 6 and its value can be written
down. As these integrals are not required here, it is sufficient to state the result
in the form
This result is also true when the limits are - 1 to +1 and K is integral. For if
K+H is even each side is then doubled and if odd each side becomes zero.
Ex. Prove / p*Pndp = — — - \pK~lPn-^dp where the limits are 0 to 1, and
K is a positive quantity greater than or equal to n.
ART. 282] LEGENDRE'S FUNCTIONS. 139
28O. To find $PmPndp between any limits. The functions Pm, Pn satisfy
Multiply the first by Pn and the second by Pm and integrate each product by parts.
We then obtain by subtraction
where the right-hand side is to be taken between the given limits.
It immediately follows that where the limits are - 1 to + 1 the integral is zero.
When m is even and n odd, we deduce from Art. 269
Jo
- .....- .....-
o m - (n-m)(n+m + l)' 2.4.6...m "2.4.6 ... (n- 1) '
When m and n are both even or both odd, the integral is half that of the same
integral with the limits ±1 and is therefore zero.
Since P0=l we find m(m + l) I Pmdp is equal to the value of dPm/dp when p=0.
Also / lPmdp = 0.
f+i 2
281. To prove that
This important result may be deduced from Art. 278 by putting
f(p) = Pn, but the following method is of more general application.
We multiply the equation of differences, viz.
nPn - (2n - l)pPn-, + (»-!) P«-2 = 0,
by Pn and integrate between the limits p= ±1. We then have
nfPn*dp - (2n - l)$pPnP^dp = 0.
In the same way, if we multiply by Pn_2 and integrate between
the same limits, we find
- (2n - VfpPn-iPn-zdp + (n- l)fP\-2dp = 0.
We now write n + 1 for n in the last equation and eliminate
fpPnPn^dp. We thus arrive at
(27i + l)fPnzdp = (2n - 1) JPVidp,
provided n is not zero. By continued reduction we find that each
of these is equal to fP<?dp = 2. The result follows at once.
where p = cos ff.
To prove the first, integrate by parts and notice that since d?PJdp*=P"
is of lower dimensions than Pn, \PnP"dp=Q. To prove the second, write
dd= -dp/,J(l-p*), integrate by parts and use the differential equation.
Ex. 2. Prove that / — ^ — -dp=m(m + l) if n>m, and m + n is even. It
/ -i dp dp '
is evidently zero if m + n is odd.
140 ATTRACTIONS. [ART. 284
Ex.3. Let (1-i
Prove
(n + 1) Q,^! -p (2n+K- 1) Qn+ (n + K- 2) Qn_1=0,
where tf= (2n + /c - 3) (n + K - 2)/n (2n + K - 1) and the limits are - 1 to + 1.
Ex. 4. Prove that / +1 (1 -.p2)* - ro KP"=0, if m and n are unequal. [It
J -1 dp" dp"
follows at once from Ex. 3 by using Art. 269.]
283. Potential of a body*. To apply these expansions to
find the potential of a body, we regard (as', y', z') as the coordinates
of any particle m of the attracting mass. We now multiply l/E
by m and sum or integrate the result for all the attracting
particles. At some points of the body we may have r > r, at
others r>r' ; we may therefore have to use both the expansions
in Arts. 266 and 268 each for the appropriate portion of the
attracting mass. In this way we find
...+++ ...... (6),
where F« = 2 and Zn = 2mr'"Pn.
These summations cannot be effected until the form and law
of density of the heterogeneous body are known. We notice
however that both Fn and Zn are the sums of a number of
Legendre's functions with coefficients and axes depending on the
given structure and shape of the body. Regarded as a function
of (x, y, z) both Ynrn and Znrn are integral rational spherical
harmonics. When therefore we use Cartesian coordinates we
write the series in the form
T T T
V=S0 + $ + &+... + ±-° + £1 + l2 + ...
r r3 r6
where $„, Tn are spherical harmonic functions of x, y, z of n
dimensions.
284. Laplace's equations. In this way we have been led
to an expansion of V in powers of r which must hold for all
attracting masses. Let this be written F=2Fnrn, where n may
be either a positive or a negative integer. Substituting this
* This expression for the potential V is given by Sir G. Stokes in his memoir on
the Variation of Gravity, &c. Camb. Trans. 1849. He obtains the expression by
solving Laplace's equation.
ART. 285] POTENTIAL OF A BODY. 141
series for V in Laplace's equation as expressed in polar co-
ordinates (Art. 108) and equating the coefficient of rn to zero,
we have
where /A = COS#.
The corresponding equation for Tm is found by writing m for
n. If we choose m so that m (m + 1) = n (n + 1) we have m = n or
m = — (n + 1). It follows that there are two powers of r, and only
two, viz. rn and r~(n+1), such that their coefficients in the series
(6), viz. Tn and Zn, satisfy the differential equation (7). It
appears therefore that Yn and Zn are both solutions of the
differential equation (7) and differ only in the arbitrary functions
or constants which occur in the solution.
Any function of two independent angular coordinates (such as
the direction angles 6, <£ of the radius vector) which satisfies
equation (7) is called a Laplace's function. Thus Yn is a Laplace's
function of the order n. The corresponding function Tnrn when
expressed in terms of (x, y, z) satisfies Laplace's equation and is a
spherical harmonic, Art. 267. A Laplace's function when expressed
as a function of the Cartesian coordinates of the point at which
the radius vector intersects some given sphere with its centre at
the origin is called a spherical surface harmonic.
285. If 6', <f>' be the direction angles of a fixed radius vector OF and
cos POP' =p, we have p = cos 6 cos 0' + sin 6 sin 6' cos (0 - <£').
The Legendre's function Pn is therefore a symmetrical function of 0, <f> and
0', <tf. Eegarded as a function of 0, <j>, we see, by comparing the series (4) and
(5) of Arts. 266, 268 with (6) of Art. 283, that Pn is a special case of Fn. It follows
that Pn must also satisfy Laplace's equation (7).
If the axis of the function Pn, i.e. OP', be taken as the axis of reference, we
have n=p and dPJd<j> = 0. The differential equation then becomes
The general solution of the differential equation (8) has two arbitrary constants.
To find the general solution when a partial solution has been found we use a rule
given in the theory of differential equations (see Forsyth's Diff. Eq. Art. 58).
The general solution is thus found to be
where A and B are the two arbitrary constants. Since Pn is an integral rational
function of p we may by using partial fractions effect this integration. The process
is rather long and the results will not be required. It will be sufficient to notice
that the part of the solution derived from the integral is not an integral rational
function of p. It follows that the only integral rational solution is APn,
142 ATTRACTIONS. [ART. 288
In the same way the general solution of the equation of differences
(n + 2) Un+a - (2n + 3) pu^ + (n + 1) nn = 0
is un=APn+BQn where
Both these partial solutions are integral rational functions of p. This result is
easily verified by substitution: if we remember that the equation is satisfied by
wn=PB, we find that the coefficient of every PH is zero.
286. We have seen in Art. 283 that the potential of any body can be expanded
iu a series of spherical harmonics of integral orders. In this expansion Ynrn and
Znr* are both integral and rational functions of x, y, z of a positive integral order.
Changing to polar coordinates we find that Yn is an integral function of cos0,
sin 0 cos 0, sin 0 sin <p. Expanding the powers of sin<f>, cosip in multiple angles,
we have
where A0,A1...An,B1...Bn are all integral and rational functions of sin 9 and cos 6.
Substituting this value of Yn in (7), we see that both Ak and Bk satisfy
where ji=cos0.
Since the equation (10) reduces to the form (8) when fc = 0, we have A0=a0Pn (n),
where a0 is an arbitrary constant.
The values of Alt Bl &c. will not be required; it will therefore be sufficient to
mention that their values found from equation (10) are
where ak and bk are arbitrary constants.
The function (at cos ktf> + bk sin k<j>) (sin 0)* — -^M is called a tesseral surface
dfi"
harmonic of degree n and order ft. In the particular case in which k=n, the
function is called a sectorial surface harmonic of degree n.
287. The case in which Yn=Pn(p) is sometimes useful in the theory of
attractions. Since p is the cosine of the angle between the directions (6, 0),
(6', $>'), Pn is a symmetrical function of (0, <f>), (ff, 0'). We therefore have
•P» (P) = «o*W + So* (sin 6 sin ^)* ^£2 l£n_ cog ft (0 _ ^
where Pn=Pn(n), Pn'=Pn(fj.'), n=cosO, /tt'=cos0' and S implies summation from
* = 1 to n. By putting 0 = 0, 0' = 0 we see that a0 = 1. In a similar way by putting
0=ir, 6'=%r we deduce that at=2* . When ft=o, we take half this value.
288. Three theorems. The great utility of Laplace's
functions depends on three theorems. To these we now turn
our attention.
Theorem I. If 7m, Tn be two Laplace s functions of different
orders then fYmYnd<o = 0, where dco is an elementary solid angle
and the integration extends over the whole surface of the unit sphere.
ART. 289] LAPLACE'S FUNCTIONS. 143
The following is Kelvin's proof. Put V= Ymrm, V = Tnrn and
apply Green's theorem (Art. 150) to the surface of a sphere of
radius a, whose centre is at the origin, then
-.
dr J dr
Substitute for V, V and we have
am+n+1nfYm¥nda> = am+n+lmfYmYndco ;
hence unless m and n are equal, fYmYndo) = 0.
When m and n are positive these values of V and V are both
finite throughout the sphere. If however m, or n, is negative it is
necessary to integrate over the two surfaces of a spherical shell, to
avoid the infinity at the centre. If a and b be the radii we then
have (am+n+1 - bm+n+1) nfYmYnd<o = (am+n+1 - bm+n+1) mfYmYnda>.
It follows that JYmYnd(o = 0 unless m = n or m + n + l=0.
We have also /YmPnda> = 0 and since P0 = l, fYmdo) = 0, where
the integration extends over the whole unit sphere.
289. Theorem II. Let Yn be a Laplace's function of the
angular coordinates (0, <£) and Pn a Legendre's function of the
same coordinates having (6', <f>')for its axis. Let both these be of
4-7T
the same order, viz. n, then fYnPndca = — - - Yn', where the
Z. ft "f~ JL
integration extends over the whole unit sphere, and Yn' is the value
of Yn when (6', <£') have been substituted for (0, <£).
To find the value of fYnPnda>, let us take as the axis of z, the
axis of Pn, (Art. 265) so that Pn =Pn(yu,), where /u-= cos 0. Also
dm = sin 6d0d<f> becomes — dfidfy. The limits of integration are
p = 1 tO - 1, </> = 0 tO 27T.
Taking the value of Yn given in Art. 286, viz.
Yn = aoPn (fji) + 2 ( At cos k<f> + Bk sin k<j>),
we notice that /cos kfydfy = 0 and /sin k<f)d(f> = 0 when the limits
of <£ are 0 to 2-Tr. Hence
2
fYnPnd(o = — a0ffPnzd/j,d<f> = a0.2ir. , - .
It remains to find the value of a0. Referring to equation (10)
of Art. 286, we see Ak = 0 and Bk = 0 when /j,= l except when
k = 0. Also Pn (/j.) = 1 when p, = 1. Thus a0 is the value of Yn
at the point where the positive direction of the axis of z cuts the
unit sphere. Since the axis of Pn has been taken as the axis of z
144 ATTRACTIONS. [ART. 291
it follows that a0 is the value of Yn at the positive extremity or
pole of the axis of Pn, and this value has been represented in the
enunciation by Yn'.
290. Theorem III. Any function of the two angular co-
ordinates of the radius vector can be expanded in a series of
Laplace s functions, and the expansion can be made in only one way.
For a discussion of this important theorem we must refer the reader to the
treatises on these functions. It will suffice here if we consider how we may
practically use the theorem in those simpler cases which generally occur in the
theory of attraction.
Let us first suppose that the given function is an integral rational function of
the direction cosines of the radius vector, i.e. of sin 0 cos <f>, sin 0 sin <p, and cos 0.
On transforming to Cartesian coordinates and multiplying each term by the proper
power of r the function becomes an integral rational function of x, y, z, which we
can arrange in a series of homogeneous functions. Taking any one of these,
say fn (x, y, z), we shall show how it may be expanded in a series of spherical
harmonics combined with powers of r. Thence (if it be necessary) we deduce the
expansion in Laplace's functions by giving r any constant value.
Subtract from /n the expression (a5a+j/2 + z2)/n_2, where /n_a is an arbitrary
integral and rational function of (x, y, z) of the (n-2)th degree, viz. •
Substituting V=fn- (z2+J/2 + 22)/n-2 ^ v"2F, there results a homogeneous function
of (x, y, z) of the (n - 2)th degree, which therefore contains as many terms as there
are ways of making homogeneous products of x, y, z of that degree. But /B_2 is an
arbitrary homogeneous function of the same degree and contains an equal number
of terms. There are therefore just enough arbitrary constants A0, Alt B1 &c. to
enable us to make the coefficients of every term in V2F equal to zero. Assuming
that the linear equations thus formed to find A0, A± &c. are not inconsistent with
each other, the expression /n - (x2+j/2+22)/n_2=SB satisfies Laplace's equation and
is therefore a spherical harmonic.
Repeating this process with the function /n_2 , we have
/»-2-(s2 + !/2+*2)/B_4=<Sn_2,
and so on. We finally end with a constant or an expression of the first degree
according as n is an even or odd integer.
Writing r2 for x2+t/2+«2 we have /n=-SB+r2Sn_2 + r4SB_4+..., where Sn, Sn_^
&c. are all spherical harmonics. It should be noticed that this equality is a mere
algebraical transformation, and involves no assumptions as to the meaning of the
letters.
If we now regard r as the radius of the unit sphere or any suitable sphere, Sn,
SB_2 &c. become Laplace's functions, and the required expansion has been made.
When the function does not contain powers of x, y, z above the cube, this
process will be unnecessary, for the arrangement in harmonics can then be generally
performed at sight.
291. When the Cartesian equivalent of the given function is not an integral
rational function of the coordinates, an expansion in a finite number of terms cannot
be obtained. We then proceed in another way. Assume that the expansion can be
effected in a convergent series, say f(6, <j>) = F0+ Y1 + Fa + ..., where Yn is a Laplace's
ART. 293] LAPLACE'S FUNCTIONS. 145
function of the nth order. Let Pn be the Legendre's function having (&', <f>') for its
axis, so that Pn is a symmetrical function of (0, <f>) and (d1, </>') ; Art. 285. Multiply
both sides of the equation by Pn and integrate over the whole surface of the unit
sphere ; then by Art. 289 JJ/(0, <£) Pndnd<f>= ^^ Fn',
where Yn' is the value of Yn when (0', tp') have been written for (0, <j>). When the
integration on the left-hand side has been effected, the result will be a known
function of 6', <J>' only. Since 6', <j>' are arbitrary we can replace them by 6, 0 and
thus the form of Yn has been found.
Laplace's expansion is an extension to two independent variables of Fourier's
expansion of a function of one variable in a series of sines and cosines of its
multiples, and like that theorem is subject to limitations. The process of expansion
given above is not in any way a proof, it is to be regarded as merely a convenient
method of applying Laplace's theorem to special cases. It fails to give the limita-
tions and must be used with caution when the function to be expanded is not single
valued.
292. Ex. 1. What are the conditions that
(1) ax + by + cz, (2) Aa? + By*+Cz* + 2Dyz+2Ezx + 2Fxy
may be spherical harmonics? The first is always so, the second when A +B+ (7=0.
Ex. 2. Expand sin3 0 cos3 0 in Legendre's functions.
This is the same as ps if the axis of x be taken as the axis of reference. Now
PS = 4 (5P3 - 3p)» hence P8 - ips =lP- Th-e result isp3= |P3 + f Pr
Ex. 3. Expand sin2 8 sin $ cos <f> + cos3 d in Laplace's functions.
The result is Fa + F2 + F3, where Y1=^ cos 0, F2 = sin2 d sin <p cos tf>, F3 = | (5 cos* 6
-3cos0).
Ex.4. Expand log (l+cosec|0) in Legendre's functions. [Coll. Ex.]
The result is P0 + £Pj + $Pa + £P3 + . . ..
Ex. 5. Prove by successive induction or otherwise the equalities
Ex. 6. If ~^= SorPr and —f = 26.P., prove that
o/p dp
ar=2r+l and b,=£ (2s + l) (n-s) (n+l + s).
Multiply the series by Pr and Pg respectively and integrate by parts between the
limits ± 1. The expansion of the with differential coefficient of Pn is investigated in
the Proceedings of the London Math. Soc. 1894.
Ex. 7. ItpK = aKPK + ... + anPn+... prove that
q-2ra-
_ _
an ~2n+lK-n + 2' n~2 .4. 6...(/c-n) . 1 . 3 .
293. Ex. 1. The polar equation of a nearly spherical surface is
where /3 is a small quantity whose square can be neglected. Prove the following
results,
(1) The volume is |ira3 (1 + 3£yo) and the surface is 4*ra2 (1 + 2/3F0).
(2) If rY1=Ax + By + Cz, the coordinates of the centre of gravity of the volume
B. 8. II. 10
146 ATTRACTIONS. [ART. 294
are x=pAa, ff=pBa, t=pCa. The centre of gravity of the surface coincides with
that of the volume.
(3) If r'r, = Ax* + By* + Cz* + 2Dyz + 2Ezx + 2Fxy, the moment of inertia
about the axis of * is *-^?£ (l-pC+5pY0), and the product of inertia about the
axes of x, y is — =- . -=- pF.
It follows from this example that when the origin is placed at the centre of
gravity of the volume the term Yl is absent from the equation. When the constant a
is so chosen that it is equal to the radius of the sphere of equal volume, the term F0
is absent.
To obtain any of these results, we proceed as follows. Let M be the volume,
P =cos0, &c., then Mz = ]$r*drdu.z = l±r4dtj)P1. Substitute for r, expand and use
Art. 289. The result is $ira4pYl', where Y,' is the value of Y^ at the extremity of
the axis of z and in the small terms this is C. Similarly the moment of inertia is
tfr*drdw . r2 sin8 6= \^d<a . $ (1 - P2). We then proceed as above.
Ex. 2. The polar equation of a nearly spherical surface is r=a (1 + /3PJ where
p is a small quantity whose powers above the second may be neglected. Prove that
the area of the surface exceeds the area of a sphere of radius a by 2jra2£- . -,
except when n=0. [Math. T.]
Ex.3. Prove that the surfaces r=a(l + /3Y1), r=a{l+p (Y0+Y1 + Y^}, where
the square of p can be neglected, are respectively a sphere and a conicoid. The
coordinates of the centre are the same as those of the centre of gravity already
found.
294. Attraction of a spherical stratum. A thin hetero-
geneous stratum of attracting matter is placed on a sphere of
radius a. It is required to find its potential at any internal or
external point.
Let p be the surface density at any point Q of the sphere, do-
an element of area at Q ; 6, $ the polar coordinates of Q. then
d<r = sin ddddfy. Let P be the point at which the attraction is
required, and let the coordinates of P be (r, 6', </>').
If R be the distance between the points Q and P, the potential
of the whole stratum at P is F = jpda-jR. Let p be the cosine
of the angle between the positive directions of the radii vectores
OQ and OP, then R* = a2 + r2 - 2apr'.
If the point P is inside the sphere, r' is less than a, and we may
expand l/R in a convergent series of ascending powers of r'ja. If
the point attracted is outside the sphere, we must expand in powers
of a/r'. Since R is a symmetrical function of a and r we have
-
ART. 295] SPHERICAL STRATA. 14-7
The surface density p is a given function of the coordinates of
Q ; let it be expanded in a series of Laplace's functions or surface
harmonics, thus p = F0 + Yl + Fa + .........
Substituting these values of p and l/R in the expression for
F, we have by the theorems I. and II. in Arts. 288, 289,
( 1 r' 1 /r'\s 1 /r'\n
F=47ra F0' + ±F/- + * F/( -) + ... _i_ Fn'(- ) + .
( Sao \al 2n + l \aj
Y, _*••<* \ y, 1 yr/« . 1 y / / '«V * F ' ( "* 1 -
3/5 V// '" 5"r"1 "*
according as r' is less or greater than a. Tfte yirs£ o/ these two
expansions gives the potential at any internal point, the second at
any external point.
If Yn is expressed as a function of the angular coordinates
(0, <f>) of Q, then as already explained (Art. 289) Tn' is the value
of Yn when the polar coordinates 6', </>' of the attracted point P
have been written for (0, <£). If however Yn is expressed as a
homogeneous function of the Cartesian coordinates (x, y, z) of Q,
then Yn is obtained from Yn by writing the Cartesian coordinates
of P for (x, y, z) and multiplying the result by (a/r')n.
We notice that by Art. 86, the potentials at two inverse points
are connected by the equation V = Fa//. It follows that either
of the series in the brackets must change into the other when we
write a2// for r'.
295. Ex. 1. The surface density at any point Q of a sphere is a quadratic
function of the Cartesian coordinates of Q. Find the potential at any point whose
coordinates are (x1, y', z1).
Let the surface density p be given by p=Ax2+By* + Cz2 + 2Dyz + 2Ezx + '2Fxy.
Let us represent this function by/(a;, y, z).
As this function would be a spherical harmonic if A + B + C=0, we make the
necessary expansion in surface harmonics by subtracting and adding G (x* + y'2+zz),
where 3G = A + B + C. We therefore have p = Y0 + Y2 , where
Y0=Ga\ Y2=f(x, y, z)-
The required potential at the point P is therefore
according as P is inside or outside the sphere. Here Y2'={f(x't y', z') - GV2}! - ) ,
and F0' = Ga2. Substituting these values for Y0' and Y2' in the formulae for V and
V the required potentials have been found.
Ex. 2. The surface density at any point of a sphere is p=mxy : show that its
. 4iram . 4tiram . ,/a\5 _,.
potential at any point (a/, y1, z') IB — — x'y' or — ^— x y ( -7 ) » according as the
point is within or without the sphere.
10—2
ATTRACTIONS.
[ART- 2
Ex. 3. The surface density at any point of a sphere is r^yz, show that the
potential at an internal point is f iramdy'z'.
Ex 4 Matter of mass If is distributed on a spherical surface whose centre is
at 0 and'radias a, so that its density at any point is proportional to the square of
its distance from a point C outside the sphere where 00 = b ; prove jhat fce
potential at an external point P distant r from the centre is M |- - 3^^ Jlf •
[Caius Coll. 1897.]
where x=rcosPOC.
Ex 5 If the surface density at any point Q be an integral rational function of
the Cartesian coordinates of Q of a degree not higher than the nth, prove that the
potential at any internal point P is an integral rational function of the Cartesian
coordinates of P also of a degree not higher than the nth.
296. Attraction of a solid sphere. To find the potential
of a solid heterogeneous shell bounded by concentric spheres when
the density p at any point is a homogeneous function of the
coordinates of the kth degree.
Let the density p be expanded in a series of the form
where Fn is a Laplace's function of the angular coordinates. The
potentials of an elementary shell whose radii are r and r + dr at
an internal and external point respectively are
The potentials of the solid sphere are found by integrating
these expansions between the limits a and b, where a, b are the
internal and external radii of the given shell
Ex. 1. The density of a shell bounded by concentric spheres of radii a
and b is given by p=mxy. Show that the potential at an internal point is
| mv (62 - a2) x'y'.
Ex. 2. The density of a solid sphere of radius a is given by p=mxyz. Show
that its potential at an external point is v\ irma9 x'y'z'lr17.
297. Nearly spherical bodies. The strata of equal density
of a solid are nearly spherical and both its internal and external
boundaries are surfaces of equal density. Find to a first approxi-
mation its potential at an internal and an external point*.
Let any surface of equal density be r = a + af(Q, (f>, a), where
a is a constant and f a function whose square can be neglected.
* The formulas here given are those used by Laplace to find the potential of the
earth regarded as a stratified heterogeneous body, M€c. Celeste, vol. n. p. 44.
When the strata are not so nearly spherical that the square of f(0, <f>) can be
neglected the algebraical processes become very complicated. For these the reader
is referred to memoirs by Poisson in the Connaissance des Temps for 1829 and 1831.
ART. 297] NEARLY SPHERICAL BODIES. 149
The quantity a is the parameter of the strata, i.e. by its variation
we pass from one stratum to another. Let the internal and
external boundaries be denned by a=ao and a = aa. Let the
density of any stratum be p = F(d).
Let the equation of the stratum be expanded in a series of
Laplace's functions, viz. r = a(l +2Fn) ..................... (1).
The solid bounded by this surface may be regarded as a sphere of
radius a, together with a stratum of surface density «2 Yn placed
on its external boundary.
The potentials of this solid, regarded as homogeneous and oj
unit density, at an internal and an external point are respectively
If we differentiate each of these with regard to a, we obtain the
potentials of a stratum of unit density bounded by the surfaces
whose parameters are a and a + da. The actual density of the
stratum is p = F (a); if then we multiply the differential coefficients
by p and integrate between the limits a = a0 and a = al, the
required potentials at an internal and external point are found
to be TT.wp. + S*, ......... (4),
a ......... (5),
4-7T f
' = ~
r J
da 2n + 1 r
the limits of the integrals being a0 and a^
We may also find the potential at any point of the solid
defined by the value a = a of the parameter. In this case the
point is external to the strata between a0 and a' and internal to
those between a and a^ The required potential V" is therefore
the sum of the two expressions for F and V ', the first between
the limits a0 and a and the second between a! and £4. The result
s
where 2 implies summation for all the values of n which occur in
the equation (1), r', 6', $>' are the coordinates of the attracted
point P, Y^ is a known function of 6', <j>', a, and p is a function of a.
150 ATTRACTIONS. [ART. 299
After the integration has been effected, the potential V" is expressed as a
function of /, ff, <f>', and a'. In the terms which contain the small factor Yn' we
may put a'=r'. In the first term of the second line where there is no small factor,
we use the equation / — a' (1 + SFn').
To obtain the component attractions at P it is necessary to differentiate the
potential with regard to the coordinates of P. If no substitution has been made
for a' we must remember that a' is a function of /, 6', </>'. We shall however
immediately prove that the partial differential coefficient dV"lda'=0, so that the
first differential coefficients of V" with respect to r7, ff, <j>' may be correctly found by
treating a' as a constant.
We have by differentiating (6)
We now put a^Jr' = a' (1 - Yn') and in the remaining terms r' = a'. It is then easily
seen that the terms independent of Yn' cancel, while the coefficients of both Yn' and
dY'/da' are zero. There are some remarks of Poisson on this point in the memoir
already referred to.
Another proof. The change of a' into a' + dar transfers an element from one
integral of (6) to the other and this is equivalent to moving the stratum bounded by
the surfaces a' and a' + da' from one side of the point P to the other. But this
change does not alter the potential of that stratum at a point on its surface,
(Art. 145), that is dV"lda'=Q. The potential at P'is therefore only altered by the
direct change of the coordinates of P.
298. Ex. There is some reason to suppose that the strata of the earth are
elliptical and that the density decreases from the centre to the surface. Assuming
then that r=a (1 + Y%) and that p=gam, where m is greater than - 2, prove that the
potential at any internal point is
J a2-Hn as+m - as+n» 1 F2' &s+m - a54™!
9 (2 + m + 3+ro ? + 7* 5+m } •
where a is the value of a at the boundary, and /=a (1 + F2'J.
299. Let the potential be given at every point of the surfaces of
two concentric spheres, radii a and b, there being no attracting
matter between the spheres. Find the potential throughout the
intervening space.
The potentials, being given functions of 0, <£ when r = a and
r=b, may be expanded in one way only in a series of surface
harmonics, Art. 290. Let these expansions be respectively
F=2Sn and F' = 2Sn', where 8n and Sn' are known functions of
B, <f>. The general expression for the potential is
The conditions of the question are satisfied if we take
Yna» + Zn/a»+i = Sn, Yn &» + Zn/b™ = Sn'.
Thus Fn and Zn are found. We know by Art. 133 that there is
but one value of F which satisfies the given conditions.
ART. 301] SOLID OF REVOLUTION. 151
If the inner sphere (radius a) include all the attracting matter
we may put b = oo , and then Yn = 0. The potential V takes the
form V = S$n (a/r)n+1 and has only the inverse powers of r.
If all the attracting matter is outside the sphere r = b we may
put a = 0. We then have Zn = 0 and the potential has only the
direct powers given by V = 2$n (r/a)n.
300. Solid of revolution. To find the potential of a solid of
revolution at any point P not occupied by matter.
Let the axis of the solid be taken as the axis of z with any
suitable origin. We have then by Art. 283,
+ ............... (i).
Since the attracting body is symmetrical about the axis of z it is
evident that V cannot be a function of the angular coordinate 6.
Hence by Art. 286, Y0 = c0P0, Z0 = c0'P0, Y± = c^P^, &c., where c0, c0'
&c. are as yet undetermined constants. To find these we put the
attracted point on the axis; we then have P0= 1, P1= I, &c. The
equation (1) thus becomes
+ ............... (2).
Suppose then we know the potential of the solid at all points
of its axis in a convergent series, then (2) is a known series, and
therefore the coefficients c0, c0', &c. are also known. The series (1)
for the potential at P then becomes
\+ (3).
Thus the potential has been found.
In this way we arrive at a theorem of Legendre, viz. if the
attraction of a solid of revolution is known for every external point
which is on the prolongation of its axis, it is known for every
external point. See Todhunter's History, Arts. 782, 791.
301. It may happen that the expansion (2) giving the
potential at points on the axis takes different forms at different
points. Thus when r is less than some quantity a there may be
only positive powers of r, and when r is greater than a there may
be only negative powers. Again, if the solid of revolution have a
cavity extending to the axis, (2) may assume one form within the
cavity and another outside the solid.
152 ATTRACTIONS. [ART. 302
If the solid have a ring-like hollow symmetrically placed about
the axis of revolution but not extending to it, it is clear that
a point P situated in this hollow has no corresponding point Q on
the axis from which the potential may be derived. In such a case
the values of some of the constants c0, GJ., &c. may be determined
when we know the values of V along some line passing through
the cavity and making an angle 0 = a with the axis. It should
however be noticed that one of Legendre's functions may vanish
when 6 = a and the unknown constant which accompanies that
function would remain undetermined. Since each Legendre's
function is unity when 8 = 0 this does not occur when the values
of the potential along the axis are given.
r* d-jf TJ-
302. By integration I - r^ - r = -j— — =-r . We write
Jo a + bcos^ >J(a -&2)
a = 1 — hp, b = h V(/>2 — 1) and expand both sides in powers of h.
Since only the first power of h occurs in the denominator on the
left-hand side, the general term is easily found. Comparing the
coefficients of hn we have
l)cOST/r}»cty = PB ............ (4).
0
This formula is given by Laplace, Mecanique Celeste, Tome v.,
page 40.
Since p is less than unity, this integral appears to be imaginary.
If however we expand the wth power, the integrals of the odd
powers of cos ty will vanish between the limits, and a real
expression for Pn will remain. We may therefore take either of
the signs before the radical. There is another integral which may
be deduced from (1), viz.
p ± Vpa - 1 cos w**1
Suppose that for any portion of the axis the potential is given
ty V=f(r), where /(r) is such an expansion as (2) Art. 300 with
either positive or negative powers of r or both. Substituting
for Pn in (3), the integral (4) in the terms with positive powers
of r, and the integral (5) in those with negative powers, we have
1 C*
P = - J f(rp ± r \/y - 1 cos -«/r) dty ............ (6).
Thus when the potential is known along the axis in the form
ART. 303] SOLID OF REVOLUTION. 153
V =f(r), the potential at other points is known in the form of the
definite integral (6).
Other forms for Pn and therefore for V may be obtained by other substitutions.
/ir d\l> TT
— r - irr - - o , = ~?7~2 — ^ — or an(i Put
0 o + *-*-3
<i=l-ph, b=ph, c==ph^/(-l) we find
2" fn
P = — / (si
* JO
This result is due to Catalan, Bulletin de Soc. Math, de France, 1888, vol. xvi.,
p. 129.
3O3. Ex.1. To find the potential of a uniform circular ring cf infinitely small
section at any point not on the axis.
Let the origin be the centre of the ring and let the axis of the ring be the axis of
z. Let a be the radius of the ring, M its mass.
The potential at any point Q on the axis distant r from the origin is evidently
M/N/a2+r2. We shall expand this in powers of rja or a/r according as r is less or
greater than a. Taking the first supposition, we have
When r is greater than a the expression may be deduced from that just written
•down by interchanging a and r.
The potential of the ring at any point P not on the axis is therefore
according as r is less or greater than a.
Ex. 2. A solid ring is generated by the revolution of a closed curve about an
axis Oz and is symmetrical about the equatorial plane. Prove that the level
surfaces in the immediate neighbourhood of the intersection 0 of the axis with that
plane are given by 2z2 - x2 - y2=/3 where j8 is a constant.
Since the potential at a point on the axis is of the form A + JBr2, the result
follows from Legendre's rule, Art. 300.
Ex. 3. A solid anchor ring is generated by the revolution of a circle of small
radius a, the centre describing a circle of radius c. Prove that in the neighbourhood
M f a2 2z2 — a;2 — •u2)
of the origin the potential at the point xyz is V= — jl— 53 -- 772 — — f •
Ex. 4. Prove that the potential V of a homogeneous oblate spheroid of mass M
at an external point P is
T_, ML 3.P2/oe\» 3.P4/ae\4 (-!)». 3. P^ /aA2" , I
r-7f-r#(7j +iT7 (T) -&c-+(L+i)(2«+3)U) +&cr
where r, 0 are the polar coordinates of P referred to the centre and axis of
revolution, and e is the eccentricity of the generating ellipse.
To prove this we first find the potential V at an external point on the axis and
then use Legendre's rule.
By using Laplace's rule, Art. 297, we at once deduce that the potential of a
ATTRACTIONS. [ART. 304
heterogeneous spheroid whose strata of equal density are co-axial spheroids and
f dV
whose boundary is a surface of equal density ^ Jf-^da, the limits
to a. Here a is the semi-axis major of any spheroid, P=f(a), e = f(a) are the
corresponding density and eccentricity and a=a at the surface.
If this body represent the earth, we notice that e is very small and a few terms
only of the series are necessary to find the potential even at points near the surface.
304. Clairaut's theorem. To investigate the law according
to which gravity at any point on the surface of the earth varies with
the position of that point*.
Without making any hypothesis respecting the distribution of
matter in the interior of the earth, we assume the principle that
the surface of the earth is a level surface of the attraction of the
earth and of the centrifugal forces. If to be the angular velocity
of the earth, the centrifugal acceleration at a distance p from the
axis is a>?p and the potential is ^p\ At all points of the surface
we have therefore F+^<w2rasin2^ = /ic (1),
where 6 is the co-latitude of the point, r the radius vector and K a
constant.
The potential V is therefore such that at all points of the
surface its value is given by (1), and at all points infinitely
distant F=0. It follows by Art. 133 that the potential V is
determinate at all points of space external to the surface.
Let the equation of the surface of the earth be
r = (?(! + «! + «,+ ...) (2),
where Ui, u^, &c. are Laplace's functions of the first and higher
* This famous theorem was given by Clairaut in his Thforie de la figure de la
terre, 1743. No assumption was made about the law of density in the interior
except that the strata of equal density are spheroids of small ellipticity, and that
the external surface is one of equilibrium. The theorem was extended by Laplace
who, assuming only that the strata are nearly spherical and the surface stratum
one of equilibrium, established a connexion between the form of the surface and the
variation of gravity which in the particular case of an oblate spheroid gives directly
Clairaut's theorem. Stokes, without making any hypothesis respecting the state of
the interior of the earth but assuming that the surface is one of equilibrium and
nearly spherical, obtained Laplace's equations. Camb. Phil. Trans. 1849. O'Brien
in his Mathematical Tracts, 1840, remarks that if the surface of the earth and also
the law of variation of gravity are known the effects of the earth's attraction on the
moon follows as a natural consequence independently of any theory except that of
universal gravitation. These effects may also be deduced from MacCullagh's theorem
on the potential of a body given in Art. 135. See also the author's treatise on Rigid
Dynamics, vol. n. chap. xn.
The extension of Clairaut's theorems to include terms of the second order of
small quantities was first effected by Airy, Phil. Trans. 1826, part m. This is also
investigated by Callandreau, Annales de VObservatoire, Paris, 1889. There is also
a paper by Gr. H. Dai-win in the Monthly Notices of the Astronomical Society, London,
1899, who gives a short summary of the works of Helmert, Callandreau, Wiecherj
on the terms of the second order.
ART. 305] CLAIRAUT'S THEOREM. 155
orders. We shall assume as the result of observation that the
surface is so nearly spherical that all the terms after the first are
small quantities. The origin of coordinates is either on the axis
or distant from it by small quantities of the first order. In the
latter case the term o>2rz sin2 6 in (1), which already contains the
small factor &>2, is altered only by terms of the second order. The
constant c is the radius of the sphere of equal volume and the
term u0 has therefore been omitted, Art. 293. The term u^ would
also be zero if the origin were taken at the centre of gravity of the
volume.
The potential at all points external to the earth is given by
where the constants in Y0, T1} &c. depend on those in ult u2, &c.
Since w2 is small, it follows from (1) that V is nearly constant
over the surface of the earth. Hence when we put r = c, the
expression (3) for V must differ from its first term only by small
quantities. It follows that the functions Ylt Fa, &c. are small.
Using (1) and (3) we find
where sin2 6 has been arranged as the sum of two Laplace's
functions. This equation gives r as a function of 6, <£> and must
therefore reduce to an identity if we substitute for r from (2). In
this substitution we write the value of r true to a first approxi-
mation in the term F0/r, but in the subsequent small terms it is
sufficient to put r = c. We therefore have
F F F
-° (1 - M, - ws - &c.) + — * + — 2 + &c. + i »2c2 (| + £ - cos2 0) = /c.
c c c
Equating to zero the functions of the same order, we deduce that
Z'-rf^O, F-cFtfc F=c2Fu-io,V(i-cos20), &c.
c 3
.-. F= F0 (- + ^ + ^p + &c.) - ^4- (j. - cos2 6>) ...(4).
This formula expresses the potential of the attraction at any point
of external space when the form of the surface is known. It is
evident that F0 is here the mass of the earth.
305. The force of gravity at a point on the earth's surface is
the resultant of the attraction of the earth and the centrifugal
156 ATTRACTIONS. [ART. 306
force due to the rotation. If v be the angle between the vertical
and the radius vector, g cos v is the component along the radius
vector. Since v is very small, we have
after substituting for r from (2) and rejecting the squares of small
quantities we find
a = _tt1-«s
c
- f o>'c (I - cos2 0) - o>2c (§ + i - cos4 0).
Let G be the mean value of g taken over the whole surface of the
earth, then (Art. 288)
G=ffg sin 0ded<f>j4,7r = ^ - §o>2c.
c
Let m represent ofc/G, we then have
g = G {1 _ |m (£ _ cos2 0) + M2 + 2w8 + 3u4 + &c.} ...... (5).
The law of variation of gravity is therefore found, when the form
of the surface is given.
306. The surface of the earth is known to be very nearly an
oblate spheroid of such small ellipticity that the difference of the
polar and equatorial semi-diameters is only 1 /300th part of either.
We may therefore write its equation in the form
r = a(l-ecos2(9) ..................... (6).
Putting 6 = \ir and 0 = 0 in turn we see that the equatorial
and polar semi-diameters are a and a (1 — e). In order to make a
comparison between the equations (6) and (2) we write (6) in the
form r = a (1 - £e + € (£ - cos2 0)} = c [I + € (£ - cos* 0)}.
We have therefore
c = a (1 — ^ e), v^ = e (£ — cos2 0), u± — 0, us = 0, &c.
The expression for g therefore becomes
g = G {I - (f m - e) (^ -cos2 0)} = £'{!+ (f m - e)cos2 0}...(7),
where £'= G (1 - £(f m-e)}. Putting (9 = ^7T we see that G'
represents the acceleration due to gravity at the equator.
The centrifugal force at the equator is a>2a and the time of
rotation of the earth (viz. 27r/o>) is 24 hours. Taking a to be
ART. 309] CLAIRAUT'S THEOREM. 157
about 3963 miles, and mean gravity to be 32'18, we find that
afa/G = 1/289. Since this ratio contains the small factor <&*, we
may put a = c and G = G'. We may therefore define the quantity
<m = afc/G to be the ratio of the centrifugal force at the equator to
equatorial gravity.
307. The potential of the earth at any external point follows
from equation (4). If we put E for the mass of the earth, we
have Y0 = E, o>'ic = mG = mE/c*. The potential is therefore
1? T?(&
r-f + ^-4££(«**-t) ............... (8).
If P, Q be the polar components of the attraction at any external
point, say the moon, we have
dV E(?
-e) — sin0cos0.
308. By comparing Laplace's expressions for the potential,
(4) or (8), with that given by MacCullagh (Art. 135) we may
obtain some information respecting the distribution of matter in
the interior of the earth. If the origin in (2) be taken at the
centre of gravity of the volume, the term u^ becomes zero. Since
the term containing 1/r2 in the potential is then absent the origin
is also at the centre of gravity of the mass (Art. 135). The centres
of gravity of the volume and mass must therefore coincide.
Since by (8) the potential is independent of the longitude, the
same must be true in the expression
This requires that the axis of rotation should be a principal axis of
the mass. Again writing B = A, and /= A sin2 6 + (7 cos2 6, we see
that
C-A _ 2 / ra\
EC* ~3V6 2/'
3O9. Clalrant's theorem to a second approximation. It is not difficult to
carry the approximation to the second order of small quantities if we follow the
same reasoning. We make no assumption about the law of density of the earth
except that the potential is symmetrical about the axis of rotation and on each side
of the plane of the equator. As a trial solution, we omit the even powers of 1/r
and take instead of (1) and (3) of Art. 304 the equations
/c (1), |r»-+£&+2?! +.- (3),
* °
158 ATTRACTIONS. [ART. 309
•where E ia the mass of the earth ; P2, P4, &c. are Legendre's functions and /3, y
are two constants. We shall also suppose that the surface of the earth has the
form r=o(l-ecosa0-.p28in:!0co820) ........................... (2),
where a is the semiaxis major and if the form be a spheroid, p2=f e2.
If we substitute from (2) and (3) in (1) as in Art. 304 the result should be an
identity. This will be found to be true if /3 and y are small quantities respectively
of the first and second orders, and the expression for V in (3) is restricted to the
first three terms. Equating to zero the coefficients of cos20 and cos40, (all the
higher powers having coefficients of at least the third order), we thus obtain two
equations to determine /} and y.
Let m be the ratio of the centrifugal force at the equator to equatorial gravity,
then w2 a = m ( — - — w2a ) ,
\ dr J
where a is to be written for r after the differentiation has been performed, and
cos 6 put equal to zero.
In this way we obtain the three results
(1),
**«-»{l-M-*»} ......... . ....... . ........................... (6).
After substituting these values of /3 and y in (3) we have an expression for the
potential of the earth at all external points.
To find gravity g at the surface, we have
where F'=F+ Jw2r2sina0. On substituting this value of V we soon see that the
expression for g contains terms which are constant multiples of cos20 and cos40.
We may therefore write
(8).
To find the three constants G', X, fj. we notice that g = 0' when 6 = \ir. Hence G' is
the value of equatorial gravity, and may be found from (7) by putting r=a and
0=i*r after the differentiations have been performed. We observe next that XG' is
the difference between the values of gravity at the pole and the equator and that
both these may be deduced from (7). Lastly we notice that -/*G' is the coefficient
of cos4 0 in the value of g ; and this may be very shortly deduced from (7). In this
way we find G'= l-f m + e-^me+
The angle 0 is the angle the radius vector r makes with the axis of rotation. If e'
be the angle the direction of gravity makes with the axis of rotation we have
0=0/ + 26sin0'cos0'.
We then find by an easy substitution
g = G' {1 + X cos2 ff + (n - 4\e) sin2 ff cos2 0*}.
We may extend Clairaut's theorem to a third approximation by proceeding in
ART. 310J FIGURE OF SATURN. 159
the same way. We then include a fourth term 5P6/r7 in equation (3) in which 5 is
a small quantity of the third order. We have also an additional term in ('2). The
numerical calculations are troublesome and the additional terms too small to be of
any interest.
310. Figure of Saturn. To find, to a first approximation,
the effect on the figure of Saturn of the attraction of the ring. We
suppose the form of Saturn to be nearly spherical, the ring to be
circular, concentric, homogeneous, of small section and situated in
the plane of the planet's equator. The planet rotates with a small
angular velocity. The principle of the investigation is that the
surface of Saturn is a level surface of the attractions of the planet,
ring and the centrifugal forces.
Let the polar equation of the surface of Saturn be
r = c(l+F1+F2 + &c.) .................. (1).
Since the surface is nearly spherical, all the harmonics F1} F2,
&c. are small quantities whose squares and products are to be
neglected. By omitting the term F0, we have made c to be the
radius of the sphere of equal volume. Also the mass M = %7rpc3,
where p is the density. By Art. 294 the potential of Saturn at
an external point is
+ r + F + &c. ......... (2).
We now substitute from (1) in the first term of (2) and put r = c
in the small terms. We thus find
The centrifugal force at any point is aPx, where a> is the
angular velocity of the planet and x the distance from the axis
of rotation. Putting x = r sin 6, the potential of the centrifugal
forces becomes
Since eo2 is small, we put r = c in this formula.
Lastly if M/n is the mass of the ring, supposed to be condensed
into a circle of radius a, the potential of the ring is, by Art. 303,
Since l/n is small, we again put r = c in the small terms.
We now substitute these three potentials in the equation
F,+ F0+Fr = /t ........................ (6),
160 ATTRACTIONS. [ART. oil
where K is a constant. Since there can be but one expansion of
the potential in harmonic functions, the sums of the several
potentials of each order must separately vanish.
The potentials Ve and Vr contain no harmonics of an odd
order; hence those in F, must also vanish. We therefore have
yi = 0, 78 = 0, &c. After substituting for Vg, Vc, Vr and
equating to zero the sums of the harmonics of the second and
fourth orders, we have
«* 9
The remaining terms contain higher powers of c/a. Since this
fraction is nearly £, these terms may be disregarded in a first
approximation.
Representing these results by F2 = — /9P2 and Y4 = >yP4, we see
that a near approximation to the form of Saturn is given by
r = c{l-/9P2(cos0) + 7P4(cos0)} ....... .....(7),
where 6 is the angle the radius vector makes with the axis of
rotation.
If the last term of (7) were omitted the surface would be an
oblate spheroid, Art. 306. The effect of the small term jPt is to
lengthen slightly both the polar and equatorial diameters and to
shorten those in middle latitudes.
The real shape of Saturn was at one time a matter of great controversy. The
first observations were made by Sir W. Herschel who found that the deviation of
the figure from that of an oblate spheroid was eo great that the longest diameter
was in latitude 43° 2(X. Herschel believed that this peculiarity was due to the
attraction of the ring. But it was soon discovered that this opinion was not
confirmed by a theoretical examination of the effect of the ring. Bessel however
afterwards proved by direct measurements of several diameters that the true form
was very nearly that of an oblate spheroid. Probably the discrepancy was due to
an optical distortion of the planet when seen through its atmosphere. These
measurements of Bessel are given in a memoir On the dimensions and position
of the ring of Saturn and those of the planet. See a translation in the Additions a
la Connaiitance des Temps for the year 1838, page 47.
311. Ex. 1. If the free surface of equilibrium of the earth is an ellipsoid, and
if e is the mean ellipticity of the meridians, 17 the ellipticity of the equator, and I
the longitude reckoned from the meridian of greatest elliptieity, and X the latitude,
prove that g=G {l-($TO-e)(J-8in2X) + ^cossXcos2J}. [Math. T. 1867.]
Ex. 2. Jacobi's ellipsoid. An ellipsoid revolves about a principal diameter with
an angular velocity which is not necessarily small. Prove that the internal level
surfaces due to the attraction and the centrifugal forces are similar ellipsoids.
Prove also that the resultant force at any point P on a given level surface is
ART. 313] FIGURE OF SATURN. 161
proportional to the length of the normal intercepted between P and the principal
plane perpendicular to the axis of revolution. If the boundary of the ellipsoid is
itself a level surface and the angular velocity is small, prove by comparing this
result with Clairaut's formula for gravity that e=5m/4.
By adding to the value of V in Art. 212 the terms due to the centrifugal forces,
viz. kP(x2 + y2), we see at once that the level surfaces are similar ellipsoids. By
Art. 46, the force at any point P on a given level surface is inversely proportional
to the distance dp between two neighbouring level surfaces. In our case dp is
proportional to p (Art. 195) and therefore inversely proportional to the length of
the normal. For points on the axis of rotation but on different level surfaces, the
force is Cpz, (Art. 213).
312. Ex. Let the earth be a solid heterogeneous nearly spherical nucleus
completely covered by a homogeneous ocean. If the system is made to rotate, with
equal angular velocities, about the principal axes at the centre of gravity of the
nucleus in succession, the ocean will assume three different forms. Prove that the
mean of the three radii vectores in any given direction is the same as the radius
vector of the ocean when supposed to be in equilibrium on the nucleus without
rotation.
Let r=o(l + SMn), /=Z>(l + Si>n) be the equations of the surfaces of the nucleus
and ocean as in Art. 304. Then since the nucleus and the mass of the ocean are
given, a, b and un are known and we have to find vn. The potential of a homo-
geneous mass of fluid extending from the centre to the surface of the ocean is given
in (3) of Art. 297. The potential of the excess of the nucleus above that of an
equal volume of fluid, and the potential of the centrifugal forces are given in Art. 304.
The sum of these three potentials is constant along the surface. By equating to
zero the sum of functions of the same order, we notice that vn is independent of w
except when n=2. We find that vz=Z2+A (£ - cos2 0) where Z% is independent of u,
and A is a multiple of w. Since the sum of the squar.es of the direction cosines of
a radius vector is unity, the mean of the three values of v2 is independent of w.
313. Ex. Let the earth consist of a spheroidal homogeneous fluid nucleus
surrounded by a consolidated crust whose external surface is also a spheroid, the
two spheroids being level surfaces of the attractions and centrifugal forces. If e', e
be the ellipticities ; a', a the mean radii of the inner and outer spheroids ; p't p the
densities of the two substances, prove that
«/> + e' 0>' - P) = f (2e - m) A,
where the mean density A is given by the last equation. The whole mass is
supposed to rotate about a principal axis at the centre of gravity with a small
angular velocity w.
To obtain the first two equations we use the formulae (2) and (3) of Art. 297 to
find the potentials of the two portions of the earth. The sum of these together
with that of the centrifugal forces is constant along each spheroid.
In the case of the earth A = 2p, m=l/289, e=l/300, and a=3958 miles. With
these numbers the Rev. S. Haughton deduced from these equations that the
thickness of the crust is 768 miles. Trans. Royal Irish Academy, 1851, vol. xxn.
dated 1855. It is remarkable that the thickness should be so great. The first
R. S. II. 11
162 MAGNETIC ATTRACTIONS. [ART. 315
attempt to discover the thickness of the crust was made by W. Hopkins, who
estimated the minimum thickness to be not less than one-fourth or one-fifth of
the earth's radius, Phil. Trans. 1842. Much has been written on the subject
since then.
Magnetic Attractions.
314. Potentials of Magnets. Two equal particles, each of
mass ra, are placed at two points A, B, whose distance apart is 2a.
Any particle being placed at P one of these repels the particle at P,
while the other attracts it. Such a combination may be called a
simple magnet*. See the figure of Art. 316.
It will be convenient to take repulsion as the standard case.
Let the mass of the particle at A be called positive, then that at
B is the negative mass. The particle at P, if of positive mass,
will then be repelled by the particle at A and attracted by that
at B. The ends A and B are called respectively the positive and
negative poles of the magnet.
Since the particle at each end of a magnet repels a particle of
the same sign, it is a matter of convention to call one positive
and the other negative. The convention adopted in Maxwell's
Electricity is that when used as a compass the positive pole points
north (Art. 394). It follows that the north pole of the earth
attracts the positive pole of the magnet. The south pole is
therefore the positive pole of the earth.
315. The line BGA is called the axis, and the distance BA
the length ; the positive direction is BA. The middle point C is
called the centre. The quantity m is called the strength and the
product of the length by the strength, viz. 2am or M, is called the
magnetic moment.
If the point P lie in the axis, the magnet is said to be end on.
If the axis is perpendicular to the distance CP, the magnet is
broadside on.
The strengths are so measured that the force exerted by m on
m' at a distance r is mm'jr1*. As explained in Art. 5 the dimensions
* The latin treatise of W. Gilbert of Colchester, De Magnete &c., 1600,
(translated by F. Mottelay), 1893, is generally referred to as one of the earliest. The
book discusses in general terms, and without Mathematics, the magnetic theory of
the Earth. The mathematics of Magnetism was first properly discussed by Poisson,
and he was soon followed by other great mathematicians. In 1849 Kelvin gave a
complete theory which, without assuming any hypothetical magnetic fluid, is
founded on facts generally known, see the Reprint of papers on Electrostatics and
Magnetism. The student of Magnetism will find the treatise of J. J. Thomson of
great assistance, and also that of Maxwell when more advanced in the subject.
ART. 316] SIMPLE MAGNETS. 163
of strength are LF$ where L represents length and F force. The
dimensions of magnetic moment are L2F*.
In all that is here said (unless when otherwise specified) the magnets are supposed
to be used in air. The effects of the medium are not included.
316. To find the potential of a simple magnet at any point P.
Let r be the distance of P from the middle point C of AE and
let 6 be the angle PGA. We notice that the angle 6 is measured
from the positive end towards P. We have in a field without
induction
-„. m —m m m
AP BP V(r2 + a2-2arcos0) V(^ + a2 + 2 ar cos 6)
2am
When the length 2a of the magnet is small compared with the
distance r, it is often a sufficient approximation to reject all but
the first term of this series. Put M = 2am, the potential of the
magnet as given by its principal term is then V= - - — . The
order of the first term rejected is the fraction (a/r)1 of the term
retained. Magnets in which it is sufficient to take account of the
principal term only are sometimes called small magnets.
Since repulsion has been taken as the standard case the
component forces (Art. 41) at P in the direction CP and perpen-
dicular to CP are respectively
dV_2Mcos0 p_ dV _Msm6
~~d7 = r3 ' rd8~ r3 '
the latter being measured positively in the direction which makes
6 increase. In the figure the arrow-heads
indicate the directions of the forces at
P due to the repulsion of A and the
attraction of B ; while the double arrows
indicate the positive directions of the
components F and G.
It appears from the investigation that both the potential and
the force at any point P are not altered by changing the length
2a and the strength m provided the product M = 2am is kept
unchanged. A small magnet is therefore given when we know
(1) the position of its centre C, (2) the positive direction of its axis
and (3) the magnetic moment M.
11—2
164 MAGNETIC ATTRACTIONS. [ART. 318
317. Resolution of Magnets. When a small magnet of
moment M' is end on to P so that 0 = 0, it follows from Art. 316
that the resultant force at P is directed along CP and is equal to
ZM'lr3. When a small magnet of moment M" is broadside on to
P so that 6 — £TT, the resultant force at P is perpendicular to CP
and is equal to M "/r*. If we take M' = M cos 6, M" = M sin 0,
we notice that the component forces at P due to the magnet M
are the same in direction and magnitude as those due to two
magnets M', M". It therefore follows, that the small magnet M
may be resolved into two components McosO, Msind. This rule
being true for a rectangular resolution may be extended to include
all cases. Hence small magnetic moments may be compounded and
resolved by the parallelogram law.
One advantage of the resolution into components "end on"
and " broadside on " is that the direction of the force due to each
component is at once evident, the direction being in every case
parallel to the axis of the component magnet. The force at P due
to a magnet " end on " acts in the positive direction of its axis ;
the force due to a magnet "broadside on" acts parallel to the
negative direction of the axis.
318. Mutual action of two small magnets. Let the two
small magnets BOA, B'C'A' be in one plane and let their moments
be M, M'. Let CC' = r, and let r be measured positively from
C to C'. Let 6, & be the angles the positive directions of the
axes make with the positive direction of r, that is with CO'
produced, and let B'C'A' = 2a'.
We resolve the acting magnet M into M cos 6, M sin Q. These
produce forces F and G at the
centre C" of the magnet B'C'A'
respectively where
£<
n
and G = Msin0Jr*. The former
acts along CC' and the latter perpendicularly to CC' in a direction
tending to increase 0.
These may also be regarded as the forces at any point in the
neighbourhood of C", provided the magnets are so small that we
can reject Fa'/r and Ga'/r. We therefore apply them without
alteration of magnitude or direction to the pole A' and also with
ART. 319] MUTUAL ACTION OF MAGNETS. 165
their signs reversed to the pole B'. The action of the one magnet
on the other is therefore a couple. See Art. 320.
To find the magnitude of the couple, we take the moment about
the centre C' of the force which acts at the positive pole A' only and
double the result. The couple tending to increase 6' is therefore
F = - 2wi'a' (.F sin 0f+Gcos 6')
MM'
= -- — — (2 cos 6 sin & + sin 6 cos 0').
319. When the two magnets are not in one plane we proceed in
the same way. Let CC' be taken as the axis of x, and let (X/xy),
(\'fi'v) be the direction cosines of the positive directions of the
two magnetic axes. We resolve the acting magnet into M\ Mp,
Mv. The former being "end on" produces a force at G' which acts
in the positive direction of its axis and is therefore X = 2 M. X/r3.
The two others being "broadside on" produce forces which act in
the negative direction of their axes and are Y= — Mfj,/r3 and
Z = — Mv/r3. These forces are transferred to act at the positive
pole A' whose coordinates are x' = a'\', y' = a'fjf, z' = a'v. Twice
their moments about any axes having C' for origin give the
couples which represent the action of one magnet on the other.
The couples about the axes of x, y, z are (by Art. 257, vol. I.)
To simplify the results, let the plane containing CC' and the
magnetic axis B'C'A' be the plane of xz. Let 0, & be the angles
the magnetic axes make with the axis of x and let <£ be the angle
between the planes in which 6, & are measured. The coordinates
of A' are then x' = a'cos0', 2/ = 0, ^ = a' stuff. The forces
X = 2M cos 0/r* and Z= — Msin0 cos 0/r8 act in the plane xz and
produce a couple
MM'
P = -- -jj— (2 cos 0 sin 6' + sin 0 cos & cos <£}.
This couple when positive tends to increase &. The force
Y= — M sin 0 sin (fr/r3 produces a couple A' in the plane yC'A'
A, MM' .
where A = sin 0 sin <£.
When positive this couple tends to increase <f> and acts from A' to y.
When the plane xz contains the axis B'C'A', X=cos0, /i=sin 0 sin <f>,
i» = sin0cos0, X' = cos0', /t'=0, p'=sin0'. The couples A'= -Kxsind'+Ktco& 6',
and r'= -Kv may then be at once deduced from those of Kx', Kv', Kz'.
166 MAGNETIC ATTRACTIONS. [ART. 322
32O. The component forces at the poles A', B' have been regarded as equal in
magnitude but opposite in sign. To this degree of approximation the forces which
tend to move the centre of gravity of the magnet B'A' are zero. This means that
the expressions for their magnitudes contain an additional factor r in the
denominator so that the forces vary as the inverse fourth power of the distance.
These forces are very small and are generally neglected. We must however
notice that, though the moment about C" of the forces in Art. 318 which act between
the poles of the magnets is f, the moment P about C of the same forces differs
from I" by the moments of the forces which act at the centre C'. Though these
forces are very small, yet the arm r is here very great and the resulting couple is of
the order 1/r8.
It is sufficient to indicate the method of finding these forces and to state their
magnitudes. Let (x, y, z), (x1, y', z?) be the coordinates of the positive poles A, A'
of the two magnets referred to origins G, C' respectively. The distance D between
-4.^'is D2
The forces Z, Y, Z are then
We now expand these expressions in inverse powers of r and effect the summation
of each term for positive and negative values of m, m'. Finally we write x = a\,
af=a'\' &c. We then find
Ex. Two small magnets float horizontally on the surface of water, one along
the direction of the straight line joining their centres and the other at right angles
to it. Prove that the action of each magnet on the other reduces to a single force
at right angles to the straight line joining the centres and meeting that line at one-
third of its length from the longitudinal magnet. [Coll. Ex. 1900.]
321. Potential energy. A small magnet, whose moment is
M', is acted on by a number of given magnets; it is required to find
the potential energy. Let m' be the strength, 2a' the length of the
small magnet B'C'A', then M' = 2a'm'. Let V, V be the potentials
of the field at the negative and positive poles of the small magnet,
then I" = - ( V — V)/2a is the component of force, due to the
field, at the small magnet in the positive direction of the axis,
Art. 40. The mutual potential energy is, by Art. 59,
The required potential energy is therefore found by multiplying the
moment M' of the small magnet by the axial component of force F'
and changing the sign.
322. To find the potential energy of two small magnets. We
use the same notation as in Art. 319. The component forces due
AKT. 322] POTENTIAL ENERGY. 167
to the magnet M are X = 2JlfX/r8, F= - M^r*, Z = - Mv/r*. The
resolved part F' of these along the axis of the magnet M' is
The required potential energy is, by Art. 321,
MM'
W -- (SXX'-^'-wO ............... (!)•
If -\/r is the angle which the positive directions of the magnetic
axes make with each other
cos ty = XX' + fji/jf + vv ;
(2).
If the magnetic axes EGA, KG' A' make angles B, B' with GG'
and if the planes AGG' , A'GG' make an angle </> with each other
we may put, as in Art. 319,
X = cos 8, p = sin B sin <J>, v = sin B cos <j>,
X' = cos 0', p' = 0, v = sin & ' ,
MM'
.-. W= -- — {2cos0cos0y-sin0sin0'cos<£} ...... (3).
The potential energy W being known, we deduce without
difficulty the couples which represent the action of the magnet
M on M'. Referring to the figure of Art. 318 we see that
T" = — d W/dO' is the moment of the couple in the plane in which
B' is measured (Art. 41). The couple in the perpendicular plane
(that is the plane yG'A') is A' = - dW /sin B' d<j>.
Ex. 1. If the law of force be the inverse /cth power of the distance, prove
(1) that the potential of a small magnet at any point P is V=M cos 0/r* and
(2) that the potential energy of two small magnets is
W_MM_ {cosy_(K+y cos 0 cos B'},
flC+l
where the notation is the same as in Art. 322.
To prove the first part we proceed as in Art. 316. To obtain the second result
we follow the method of Art. 322, using the rule in Art. 321.
Ex. 2. A small magnet free to move about its centre is acted on by another
fixed magnet and the law of force between the poles is the inverse /cth power of the
distance. The magnets are placed with the axis of one along and that of the other
perpendicular to the straight line joining the centres. Prove that the couple
tending to produce rotation in the free magnet when the fixed magnet is " end on "
is K times that when "broadside on."
By making experiments on the magnitudes of these couples Gauss determined
the value of K and thus proved that the law of force is the inverse square. The
experiments are shortly described in J. J. Thomson's Electricity and Magnetism.
168 MAGNETIC ATTRACTIONS. [ART. 323
323. A series of particles whose masses (positive or negative}
are m^, mt, &c. are placed in a straight line Ox at given points
Alt A%, &G. Find the equations of the lines of force.
Let rlt r2, &c. be the distances of any point P from A1} Az. &c.;
61} 03, &c. the angles these distances make with Ox. Let <£>!, $2, &c.
be the angles the tangent to the line of force through P makes
with the radii vectores rlt ra, &c. ; then taking any one of these
sin <f> = rdOjds.
Since the resultant force at P acts along the line of force, we
have 2 — sin <£ = 0, .'. 2m — = 0.
When the points Alt A^, &c. lie in the axis of x,
T! sin #! = rv, sin 02 = &c.
Hence 2ra sin 0d0 = Q, .: 2m cos B = K.
The equations of the lines of force and the level surfaces written
at length, are therefore
m^ cos 01 + m2 cos 0a + &c. = K,
m^jr-i + m^/rz 4- &c. = K',
where K and K' are arbitrary constants.
In a magnet rm^ = — <m^, the lines of force and the level
surfaces reduce to
cos 6 — cos & = K!, 1/7*! — l/r2 = .KY.
Line of force from one particle to another. When a line of
force passes through one of the attracting or repelling particles,
the radius vector at that particle becomes a tangent and 9 is then
the angle that tangent makes with the positive direction of the
axis of x. Let a line of force pass between the points At, A^.
Then, equating the values of K at these two points, we have
m^ + &c. + mi cos 04 — mi+l — &c. = 77^ + &c. + mk cos 0k — mk+1 — &c.
.'.mi sin2 £ 0t + mf+j + &c. + mt-i + ™<k cos2 £ 0k = 0.
If all the masses have the same sign the only line of force
which can pass from one particle to another is the straight line Ox
on which all the particles are situated.
Line of force from a particle to an infinite distance. Let a line
of force pass from the particle m* to a point at an infinite distance
in a direction which ultimately makes an angle ft with the axis
of x. We then have in the same way
M! + m? + &c. + mt cos 0t — mi+l — &c. = (2m) cos ft.
AET. 324] LINES OF FORCE. 169
In a magnet where Sm = 0, no line of force can pass to an
infinite distance except the one along Ox.
Parallel rods. We may obtain a corresponding theorem for a
series of thin parallel attracting rods. Let the rods be cut by
a perpendicular plane in the points A1} Az, &c. and let (rlt 6^),
(r2> $2), &c. be the polar coordinates of any point P in this plane
referred to Alt Az, &c. as origins. If m^, m2> &c. are the line
densities of the rods, the lines of force and level curves in this
plane are respectively 2ra0 = K, 2m log r = K'.
324. Ex. 1. Prove that the lines of force of a simple magnet BOA (not
necessarily small) are symmetrical curves concave to the magnet and passing
through its poles. If P be the middle point of one of the lines of force, prove that
the curvature at P is three halves that of the circle BPA, and that the curvatures
at B and A are zero. If BPA be an equilateral triangle prove that the line of
force meets the magnet at right angles. [Math. T. 1871.]
Ex. 2. A small fixed magnet BCA acts on a small magnet B'G'A' free to turn
about its centre. Prove that when the free magnet is in equilibrium its axis lies in
the plane AGO' and that tan 8'= -£tan0.
Let the magnetic forces of the earth be represented by those of a small magnet
placed at the centre with its positive pole pointing south. The north-seeking pole
of the compass needle is then its positive pole. It follows that in north magnetic
latitude X, the dip D below the horizon of a small magnet free to turn about its
centre of gravity (usually called a dipping needle) is found by writing $ir + \ for 6
and f7r-D for 6'. Hence the tangent of the dip is twice the tangent of the magnetic
latitude.
Ex. 3. A small fixed magnet EGA acts on a small magnet B'G'A' free to turn
about its centre in the plane ACG'. Prove that the two positions of B'G'A' in
which the couple I", tending to produce rotation, is greatest and zero are at right
angles. Prove also that the maximum couple is E (1 + 3 cos2 0)2 where E = MM'Jr3,
and that when the magnet B'C'A' makes an angle <f> with its position in equilibrium
the couple is proportional to sin <f>.
Ex. 4. A compass needle B'C'A' is free to turn about its centre G' in a
horizontal plane and is acted on by a small vertical magnet whose centre G lies on
the circumference of a horizontal circle having its centre in the vertical C'Z.
Prove that, if </>, <f>' be the angles the planes ZC'C, ZC'A' make with the magnetic
meridian, sin (<f> - <£')/sin <f>' is approximately the same for all positions of the
disturbing magnet.
Ex. 5. Three small magnets are placed with their centres at the angular points
of an equilateral triangle ABC and being free to move about their centres rest in
the following positions. The magnet at A is parallel to BC whilst those at B and
C are at right angles to AB and AC respectively. Prove that the magnetic
moments are in the ratios *JZ : 4 : 4. [Math. T. 1880.]
(Use Art. 318.)
Ex. 6. Two small magnets of moments M, M' are fixed at two corners of an
equilateral triangle with their axes bisecting the angles. A third small magnet is
170 MAGNETIC ATTRACTIONS. [ART. 326
free to move at the other angular point. Prove that its axis makes with the
bisector of the third angle an angle whose tangent is ^3 (M~M')/7 (M+M').
[Math. T. 1882.]
Ez. 7. Point charges e, - e', - e' are placed at 0, A, B respectively which are
in a straight line and OA = OB. Prove that, if e>2e', the greatest angle a line of
force leaving 0 and entering A can make with OA is a, where esinz^a = e'.
[Coll. Ex. 1900.]
[If the line of force pass from 0 to an infinite distance we must have
ecos0<e-2e' ; if it arrive at A, we have esin2£0=e'sin2$0', where 6, ff are the
angles the tangents at 0, A make with OA and AO respectively. If 0 is greater than
the value of a given above, the line cannot go to A ; if less it cannot go to an
infinite distance. See Art. 323.]
325. To determine by experiment the numerical values of (1) the horizontal
force H due to the earth's magnetism and (2) the magnetic moment M of a given
magnet. There are several ways of effecting this, but in general two experiments
have to be made, one to determine the ratio H/M and the other the product HM.
The two following examples will explain the process without details. A minute
account of the methods of conducting these and other experiments for the same
purpose is given in Maxwell's Electricity, vol. n. chap. vn. The quantity H
represents the horizontal component of force on a unit pole and is directed towards
magnetic north.
Ex. 1. A small compass needle free to turn round its centre in a horizontal
plane is acted on by a fixed magnet of moment M whose length is perpendicular to
the magnetic meridian and whose centre is in the horizontal plane. If the
deviation of the compass needle from the magnetic meridian be <j>, prove that
ta.n<f>=2M/Hrs. This determines M/H when <f> has been observed. It also gives
the value of M or H when the other is known.
Ex. 2. A magnet of moment M is suspended by two fine threads of length Z
from two points D, E of a horizontal bar. The strings are attached to two points
D', E' of the magnet which are equally distant from the centre. The magnet
being acted on by the earth's horizontal force assumes a position of equilibrium. Let
the bar be turned round a vertical axis until the magnet, when again in equilibrium,
is perpendicular to the magnetic meridian. In this position let the bar make an
angle e with the magnet. Prove that (P - 4fe2 sin2 £0)i= Wb* sin Q\H.M, where W is
the weight of the magnet and 26 the length of either DE or D'E'. This experiment
determines the product HM.
326. Potential of a magnetic body. We have hitherto
supposed that the attracting and repelling particles of a magnet
were situated at two definite points of the axis, called the poles.
But there are no such ideal magnets in nature. When a real
magnet is broken into pieces the fragments continue to exhibit
polarity. We must therefore suppose that the magnetism (what-
ever that may be) is distributed throughout the body. We shall
here assume as a working hypothesis that each element of volume
of a magnetic body acts on an external magnetic element as if
it were occupied by a small simple magnet whose strength and
ART. 328] MAGNETIC BODIES. 171
length are indefinitely small. Let m and 2a be the strength and
length of the small magnet which occupies the element dv of
volume, and let M= 2am be its moment. The moment per unit of
volume is 2amfdv. Representing this ratio by I, we have the
relation Idv = 2am = M. The positive direction of the axis of this
ideal magnet represents the positive direction of magnetisation of
the body at the element dv, and the intensity of the magnetisation
is measured by /. The potential of any element of a magnetic body
at a point P which is at a finite distance r from the element is
— — cos 6 where 6 is the anqle which the distance r makes with the
i
positive direction of magnetisation.
327. Elementary rule. The potential Idv cos 6/r* is the
same as the repulsion of the element dv, supposed to be of density
/, when resolved in the direction of magnetisation. It immediately
follows that when the direction of magnetisation is uniform
throughout the body the potential at a point P is the same as the
repulsion at P of that body, supposed to be of density I, when
resolved in the direction of magnetisation. If the intensity / is
not also uniform, the body is supposed to be heterogeneous. This
simple rule frequently enables us to write down the potential of
a magnetic body.
328. Magnetic rod. The potential of a thin uniformly mag-
netised rod AB of volume v and length I at any external point P
Ivf 1 1 \ Iv, . . .
T(Ap-Bp)°r-lp-(am/3-Sina)>
by Arts. 10, 11, according as the direction of magnetisation is
along, or perpendicular, to the length. In the former case we see
that the magnetic rod acts as if it were a simple magnet of equal
length whose strength is Iv/l.
This result may also be arrived at by d priori reasoning. The
effect of the elementary magnet in any element dv of volume is
not altered if its length is increased (without changing the
moment Idv) so that the magnet occupies the full length of each
element. The positive and negative ends of the successive
magnets then destroy each other, leaving a positive element of
magnetism at one end of the rod and a negative element at the
other.
172 MAGNETIC ATTRACTIONS. [ART. 330
It follows from Art. 27 that the potential at P of a thin
circular disc, of volume v, area A, uniformly magnetised per-
pendicularly to its plane is Ivv/A where G> is the solid angle
subtended by the disc at P.
329. Magnetic sphere. Since the attraction or repulsion
of a homogeneous solid sphere of volume v and unit density is
v/r*, it follows immediately that the potential at P of the same
sphere when uniformly magnetised is Iv cos 0/r2, where r is the
distance of P from the centre and 6 the angle r makes with the
direction of magnetisation. The potential of a uniformly magnetised
solid sphere is therefore the same as that of a small concentric simple
magnet, (called the equivalent magnet), whose moment is M = Iv
and whose axis is in the direction of magnetisation.
When equivalent magnets can be determined for two bodies we
can at once deduce from Art. 322 their potential energy. In this
way we see that the mutual potential energy of two spheres
uniformly magnetised in different directions is
— ^— (cos i/r — 2 cos 6 cos &),
where r is the distance between the centres and ^r, 6, & have the
same meaning as in the Art. just referred to.
330. Magnetic ellipsoid. The potential of an ellipsoid
uniformly magnetised in a given direction can be obtained at
once by using the rule. The component repulsions of a homo-
geneous ellipsoid at an internal point are lAx, IBy, IGz. By
resolving these in the direction of magnetisation (I, m, n) we find
that the magnetic potential at an internal point (f, 77, f) is
where A, B, C are the quantities denned in Art. 212. The
components of magnetic force at any internal point are therefore
X = — IAl, Y — — IBm, Z = — ICn: Art. 41. These are constant
in magnitude and direction at all internal points.
At an external point, the magnetic potential is
' >
where A', B', C' and a, b', c' have the meanings defined in Art. 223.
ART. 332] MAGNETIC ELLIPSOID. 173
331. An ellipsoid is placed in a field of uniform magnetic
force, it is required to find the magnetism induced in the ellipsoid.
The theory of induced magnetism is discussed in the section on
magnetic induction. It is enough for our present purpose to say
that when certain neutral bodies are acted on by magnetic forces
each element dv of volume becomes magnetised in the direction
of the resultant force F which acts on that element and that the
intensity I = kF. The constant k is called the magnetic sus-
ceptibility ; another constant /j, = 1 + 4t7rk afterwards introduced
is called the magnetic permeability.
Let I, m, n be the direction cosines of the direction of the
induced magnetisation at any point P of the ellipsoid. Let
X, Y, Z be the components of force at P due to the field,
X'y Y', Z' those of the force due to the ellipsoid now become
magnetic. The force F is the resultant of X, Y, Z and X', T, Z'.
Since the intensity I at P is given by / = kF, we have
Il = kFl = k(X + Xf), Im = k(Y+Y'), In = k(Z+Z').
Let us assume as a trial solution that the ellipsoid becomes
uniformly magnetised in direction and magnitude. We then
have X' = — IAl, &c. while X, Y, Z are given constants. The
equations give at once
kX kY kZ
in =
1 + kB' 1 + kO'
Since these equations give constant values for the components
of magnetisation the trial solution satisfies the conditions of the
problem. This therefore is one solution. If we use the constant
fj. instead of k, these equations become
n
~
332. Ex. 1. A sphere and a circular cylinder, constructed of the same kind
of material, are placed in succession in a uniform magnetic field, the axis of the
cylinder being perpendicular to the force. Prove that the intensities of the
induced magnetisms are in the ratio 3 (ytt + 1) to 2 (/*+2). [In a sphere A, B, C are
each equal to 4jr/3. Their values for a cylinder are given in Art. 232.]
Ex. 2. An elliptic cylinder, which has one transverse axis very much longer
than the other, is placed in a uniform magnetic field with its infinite axis
perpendicular to the direction of the force. Prove that the intensity of the
induced magnetism when the transverse longest axis is in the direction of the force
is approximately /* tunes that when the same axis is perpendicular to the force.
174 MAGNETIC ATTRACTIONS. [ART. 333
Ex. 3. Prove that the potential of a thin plane lamina uniformly magnetised
perpendicularly to its plane at a distant point (&£) is
•where the axes of coordinates are the principal axes of inertia at the centre of
gravity, va2, vb2 the moments of inertia about the axes of x and y, and r is the
distance of the point from the origin. [To prove this we differentiate with regard
to f MacCullagh's expression for F, Art. 135.]
333. Magnetic cylinder. Prop. 1. The density at any point of an infinite
right circular cylinder (radius a) is <p (x, y)t the axis of the cylinder being the axis
of z. Prove that, if tf> (x, y) satisfy Laplace's equation and be of t dimensions, the
potential of the cylinder at an internal point (£, rj) is
We obtain this result by making c infinite in the first theorem of Art. 247,
noticing that Q[c=aa + u when a =b. The potential is therefore
du L Rl+"u»Dn } [ a? *
(«)H- (a^
The operator D = — 5— ( ^r= + -r-^ ) , and 0 satisfies Laplace's equation, hence all
a \of at) J
the terms except that given by 71 =0 are zero. The potential becomes
+u
At an internal point, the limits are 0 to oo ,
At an external point, the limits are X to oo ,
««
Prop. 2. The x and y components of magnetisation of a right circular cylinder
are Il=df/dx and Im=df]dy, where f(x, y) is a homogeneous function of x and y of
i dimensions which satisfies Laplace's equation. Prove that the potential of the
cylinder at an internal point is 2ir/(|, 17).
The potential of a magnetic cylinder whose intensity is H is equal to the
resolved repulsion of a cylinder whose density is IZ (Art. 327). The potential of
the cylinder due to both components of magnetisation is therefore
F=
' *-!
since d//d£ and df/dri are both of i - 1 dimensions.
The potential at an external point is found in the same way. Since a2 + X = £2 + ij2,
the result is F'=2,
Prop. 3. A right circular cylinder is placed in a field of force whose potential is
/(£> if)- Prove that, if /(£, ij) is a homogeneous function of i dimensions which
satisfies Laplace's equation, the magnetic potential inside the cylinder it
ART. 335] HETEROGENEOUS CYLINDER. 175
Assume as a trial solution that the x, y components of magnetisation are
Ldf/dx and Ldfldy. The equation of condition (Art. 331) Il = k(X+X') becomes
). Hence L (1 + 2dfc) = - k. The other equation of
a£ /
condition leads to the same result. The potential inside the cylinder is therefore
The potential at a point outside the cylinder is
Ex. A right circular cylinder is placed in a field of magnetic force whose
potential is A (^-i?2). Prove that the potential of the magnetic force within the
cylinder is A' (£2 - i?2), where A' (1 + M) = 24. [Coll. Ex. 1899.]
In the same way, if the potential of the field were Axy, the magnetic potential
would be A'xy, where A' has the same value. This result follows at once from the
former because £2 - ij2 becomes - 2£V when the axes are turned round OZ through
half a right angle.
334. To find the mutual potential energy of two magnetic todies. By Art. 321
the potential energy of a magnetic body and an elementary magnet of moment M'
is —M'F\ where F' is the component of force due to the magnetic body in the
direction of the axis of the elementary magnet. If the elementary magnet
represent the magnetism of an element dv' of a second magnetic body, we have
M'=I'dv'. The potential energy of the two bodies is therefore W= — J.F'1'di/ where
the integral extends throughout the volume of the second body.
If V be the potential of one magnetic body, X', /*', "' the direction cosines of the
direction of magnetisation at any point of the other, the expression for W takes
the form ^=///(£X' + % * + 5? "') IW^''
This integral is the same as that considered in Green's theorem (Art. 149), and is
equivalent to W= fvi' cos i'dS - fv (^j-
If the magnetisation I' is such that its components I'\'=dfldx't and where /is
a function which satisfies Laplace's equation, the expression for W is reduced to a
surface integral.
335. Terrestrial magnetism. The phenomena of terrestrial
magnetism can be roughly represented by the action of a powerful
small magnet placed near the centre of the earth (Biot, Traite de
Physique, 1816). This supposition is equivalent to treating the
earth as a sphere uniformly magnetised in direction and magnitude
(Art. 329). The theory altogether fails in accuracy when applied
to explain the irregularities at special places. An attempt was
therefore made by a Norwegian observer, Hansteen, to explain
the observed facts by the action of two large magnets within the
earth, both being excentric. But the results, though superior to
those derived from a single magnet, were not satisfactory.
176 MAGNETIC ATTRACTIONS. [ART. 337
336. Gauss* investigated the potential of the magnetism at
a point P on the supposition that it was distributed irregularly
throughout the earth. To effect this he used a formula equivalent
to that given in Art. 283, viz.
&c ......... (1),
where a is the radius and r the distance of P from the centre of
the earth. If the causes of magnetism are inside the earth the
second of these series alone is to be retained. When P is at a
great distance from the attracting mass, this reduces to Z0afr.
It follows that Z0a is the attracting mass and is therefore zero.
After some preliminary trials Gauss decided that it would be
sufficient for a first approximation to retain only the terms up to
and including (afr)4. This is to be regarded as a trial solution
to be accepted or rejected after a comparison of its results with
the observed facts of magnetism. With this limited value of V
the theoretical components of force in three rectangular directions
can be found by differentiation. Let the directions be, one parallel,
a second perpendicular to the meridian, and a third vertical.
Kepresenting these components by X, Y, Z, the declination S of
the needle and the dip i are given by (X2+ Yz) tanH' = Zz and
X tan 8 = F. The values of the declination, dip, and intensity were
known in Gauss' time at nearly 100 places. The observations
at 12 of these (properly chosen) were used to determine the 24
unknown constants which occurred in the functions Zr &c. Gauss
then tabulated side by side the observed and computed values of
the declination, dip and intensity at 91 places on the surface of
the earth, so that an easy comparison could be made.
337. In general the agreement was so accurate as to leave
no doubt on the fundamental correctness of the theory. The
observations made since Gauss' time are also in sufficient accord-
ance with the theory. The small discordances which remain are
ascribed by Gauss partly to errors in the observations and partly
to the fact that all the observations used do not correspond to
the same year. The terms beyond the fourth order in (1) may
have sensible effects and possibly other less influential causes of
magnetism may exist.
* Gauss' paper is translated in Taylor's Scientific Memoirs, vol. n., 1841.
ART. 339] POISSON'S THEOREM. 177
338. The causes of magnetism have been assumed to be
inside the earth. If there are any external causes, their effects
could be represented by including sorrfe terms of the first series
in (1). If the causes were wholly external to the earth the
potential would be represented solely by the first series in (1).
The vertical force would then be — SnFn/a instead of S (n + 1) Zn/a.
Since the observed vertical force does closely satisfy the latter
of these two very different expressions Gauss considers it proved
that only a small part of the terrestrial magnetic force can be due
to causes external to the earth. This argument does not apply
to the periodical changes of the needle which have not been
considered by Gauss.
339. Poisson's theorem. To investigate a general formula
for the potential of a magnetic body. We resolve the intensity /
into three components A = I\ B = ///,, G = Iv. Let us find the
potential due to the first of these. Let QQ'= dx be an element of a
column LM parallel to x (figure of Art. 222). Let QP = r and let
Q'n be perpendicular to QP, then Qn = — dr and cosPQQ'=—dr/dx.
The potential of the column at P is then
os pQQ, _
and the potential of the whole magnetic body at P is
dy
Following the same reasoning as in Green's theorem (Art. 149)
we put this into the form
TT (, .dS [fdA dB dC\ dv
V= II cos ^ -- -5~ + ^-+ -j- }~>
J r J\dx dy dz) r
Avhere dS is an element of the boundary and dv of the volume of
the magnetic body.
It follows from this equation that the magnetic potential at P
is the same as that of a quantity of matter distributed partly
internally and partly superficially. The volume density p of the
internal distribution, and the surface density a- of the superficial
distribution, are
(dA dB dC\ T
p m— I —- + __+ -,_ ) o- = / COS I.
\da dy dz)
Here i is the angle the direction of magnetisation at any point
of the surface makes with the outward normal at that point.
R. s. n. 12
178 MAGNETIC ATTRACTIONS. [ART. 342
340. Since the total quantity of attracting matter in each
elementary magnet is zero, it is evident that the sum of the
internal and superficial distributions in Poisson's theorem is also
zero.
The mass distributed over the surface S of the magnetic body, being Jlcosid/S,
is evidently the flux of the vector I across the boundary of S.
So also, if the surface S' is the boundary of any portion of the body, the mass
distributed internally is equal and opposite to the flux across the surface S'. Thus
-pdxdydz is equal to the outward flux across the six faces of the Cartesian
element dxdydz. We may therefore deduce the value of p for polar, cylindrical, or
other orthogonal coordinates by finding as in Art. 108 the flux across the faces of
the corresponding element.
If Ij , I2 , I3 are the polar components of I in the directions in which r, 6, $
are measured, then (Art. 108)
_1 d&r2) 1 d(I8sinfl) 1 dls
~P~r* dr rsintf d6 + r sin 0 ~d(j> '
In cylindrical coordinates E, 0, z
IdftB) ldI2 dlt
f E dE "^RdQ dz '
341. Ex. A magnetic shell is bounded by spheres of radii a, 6. The
direction of the magnetisation at any distance r from the centre is radial and its
magnitude is icrn. Find the potential at an internal and external point.
The internal distribution is p= - -^ — (KIM+Z) and the superficial distribution
ff1=-Kan and <r2=+ic6* on the two boundaries. The potential of all these
at an internal point (Art. 64) is
_ / 4irr2dr 4?r6a 4?ra2 K (bn+L — an+1)
V= I p + 0-2 — T OS = - 45T — i '- .
J f b a n+l
The potential at an external point is zero.
342. The force of induction*. The magnetic force at a
point P of space void of magnetised matter is the force on the
positive pole of a magnet of unit strength, the positive pole being
placed at P. To find the force at a point P situated within a
magnetic body we imagine an infinitely small space round P to be
removed and a positive unit pole placed at P in the cavity.
Consider the effect of this removal ; the attraction of the solid
distribution of Poisson which once filled the space has disappeared,
and there is now a superficial distribution on the inside of the
cavity. Since the attractions of similar and similarly situated
bodies on the same point vary as their linear dimensions (Art. 94),
* The distinction between the magnetic force and the force of induction is due
to Kelvin and is fully explained in his Theory of Magnetism, Arts. 479 &c. The
former name is due to him, and the latter to Faraday and Maxwell. See the
treatise on Electricity and Magnetism, Art. 428.
ART. 343] THE FORCE INSIDE A BODY. 179
the solid distribution is not affected by the removal of the infinitely
small quantity of matter (Art. 101). But the superficial distribu-
tion on the inside of the cavity does affect the force in a manner
which depends on the form of the cavity.
Thus the resultant force at a point inside a magnetic body is
made up of two components. One of these is due to (1) all external
causes, (2) the whole solid distribution, (3) the superficial distribu-
tion on the external boundary. The other is due to the superficial
distribution on the inside of the cavity alone. The former com-
ponent is defined to be the magnetic force at a point within the
magnetic substance.
343. Let the cavity have the form of an infinitely small
cylinder whose length is 26 and radius a, and let the generating
lines be in the direction of magnetisation. Let P be at the
central point of the cylinder. The superficial surface density,
being / cos i, is zero along the generating lines and + / at the
two circular ends. The " outward " normal for the cavity tends
towards the point P and therefore the surface density is + 1 for
the negative end of the cavity and — / for the positive end. The
repulsion of the two ends at P is 47r/jl — jj—^ — •?-:> by
acting in the direction of the magnetisation of the body in the
neighbourhood of P. It appears that the force depends not on
the absolute dimensions of the cavity but on the ratio of the
length to the breadth. Hence however small the cavity may be
made, the force due to the superficial distribution on its walls will
in general remain finite. If the radius a is infinitely smaller than
the length 6, the force due to the superficial distribution is zero.
If the radius is infinitely greater than the length the force is 4nrl.
The actual force, due to all causes, on a positive unit pole
situated at the central point P of a cylindrical cavity, whose
length is in the direction of magnetisation and is infinitely greater
than the breadth, is the same as the force already called the
magnetic force at P. The actual force, due to all causes, on the
pole when situated at the central point of a thin disc-like
cylindrical cavity, whose plane is perpendicular to the direction
of magnetisation, is called the force of induction at P.
By taking cavities of different forms we may contrast the two
forces in other ways. Let the cavity be of a thin disc-like form,
the normal to the plane making an angle i with the direction of
12—2
180 MAGNETIC ATTRACTIONS. [ART. 347
magnetisation. The distribution on the curved side is ultimately
zero. The distributions ±1 cosi on the two plane faces act on P
as if they were distributed on infinite planes ; the repulsion at P
is therefore 47r/cosi. Thus the actual force on P is the magnetic
force or the force of induction according as the plane of the cavity
contains or is perpendicular to the direction of magnetisation.
344. Ex. Prove that when the cavity is spherical the force at the centre due
to the superficial distribution is |irl (see Art. 93).
345. It appears from what precedes that the force of induction
at P is the resultant of the magnetic force at P and a force 4?r/
acting at P in the direction of magnetisation of the body in the
neighbourhood of P.
Let A = I\, B = Ifj,, G = Iv be the Cartesian components of the
vector /; X, Y, Z and Xlf Yl} Z^ the components of the magnetic
force and the force of induction. Let F be the potential of the
whole magnetic body at any internal point P, as given by Poisson's
theorem, Art. 339. Then
Y=-dV/dy, Y
Z = -dV/dz, Z
346. Bodies not uniformly magnetised. When the mag-
netism of a body is not uniform, either in direction or intensity,
it becomes necessary to choose special forms for the elements.
The magnetic lines are curves such that the direction of
magnetisation at any point is a tangent to the curve at that
point. In a line of force the direction of the force is a tangent,
Art. 47. If we draw a magnetic line through every point of a
closed curve we construct a tube which is called a magnetic tube.
When the section of the tube is very small it is sometimes called
a filament By analysing a magnetic body into elementary tubes
or filaments we may often find its magnetic potential at any external
point P with great ease.
347. Solenoids. Let da- be the area of a section of a
magnetic filament at any point Q, ds an element of length
measured in the direction of magnetisation and / the magnetic
intensity. Using the same notation as before (Art. 316) we notice
that d is the angle in front of the radius vector QP and that
therefore cos 6 = — dr/ds. Hence since dv = da-. ds the potential
of the filament at P is F = f — cos 0 = - (ida — .
] r2 r3
AET. 350] SOLENOIDS. 181
When the magnetism of the body is so distributed that Ida- = d/ju
is constant for each magnetic filament, the body is called a solenoid.
The integration can then be effected at sight. If R, S be the
intersections of the filament with the surface of the body, RS
being the positive direction of magnetisation, the potential at any
external point P is V= ~^ —
^
The potential of a solenoidal filament is independent of its
form and depends solely on the magnetism Ida- of a cross section
and on the positions of its extremities. A closed solenoid exerts 110
action on any external magnet.
348. The potential of a solenoid, or of any portion of a
solenoid, may be found by summing up the potentials of the
filaments which compose the body. Let any filament intersect
the boundary in an area dS and let the direction of magnetisation
make an angle i with the outward normal. Then since
Ida- = IdS cos i,
the potential of the body is the same as that of a thin superficial
stratum on the boundary, and this stratum is the same as that given
in Poisson's theorem (Art. 339).
349. Since this must be also true for every element of volume
of the body, it follows that the solid distribution of Poisson must
dA dB dC
be zero. We have then —p= -7 — r- -5— + -r = 0,
ax ay dz
where A, B, G are the components of / at the point x, y, z. This
is a necessary condition that the magnetism is solenoidal.
To prove that this condition is sufficient. By Poisson's theorem the potential
of every portion of a body is equivalent to that of a surface distribution Jcos 6
and a volume distribution p. Let this portion be an arbitrary length I of &
magnetic filament. The potential of the filament is
1^- a
ro J r
.
ds
where the suffixes refer to the ends of the filament. The potential of the fila-
ment is therefore the same as that of a surface distribution I cos 6 and a volume
distribution p'=^ -- —, — -. Since the surface distribution of the arbitrary filament
d<r ds
is the same as that given by Poisson, the density p' of the volume distribution
must also be the same as p. Hence when p = 0 we must have Ida constant for any
filament.
350. Lamellar shells. If the magnetic lines can be cut
orthogonally by a system of surfaces we can conveniently analyse
182 MAGNETIC ATTRACTIONS. [ART. 352
the body into elementary shells. Let the equation of these
surfaces be f(x,y,z) = c. Consider the shell bounded by the
surfaces c and c + do. Let dcr be an element of area at any point
Q of the first surface, t the thickness of the shell ; the volume of
the corresponding element of the shell is then dv = tdcr. Let t be
measured in the direction of magnetisation and let / be the
intensity. The magnetism of the element dv is equivalent to that
of a small magnet whose moment is Idv and whose axis is normal
to the surface. The potential at any point P is Idv cos #/r2, where
r and 6 have the same meanings as in Art. 347.
Let dco be the solid angle subtended by do1 at P, then
da- cos 8/r* = dto (Art. 26). The potential of the shell at P is
therefore V=fldv cos 0/r2 = /It da).
Here the sign of dco follows that of cos 6. Let that side of da-
be called the positive side to which the direction of magnetisation
points. Since 6 is the angle QP makes with that axis, the solid
angle dco is positive or negative according as P lies on the positive
or negative side of the elementary area da:
Let P travel from a position P1 close to d<r on its positive side to a position P2
also close to d<r on its negative side, the journey being made outside the elementary
area. When P crosses the tangent plane to da at some external point, the solid
angle subtended at P changes from positive to negative. The solid angles at Px
and P3 are 2ir and - 2w, hence if we suppose P to travel from P3 to Px through the
element of area the solid angle is increased by 4?r.
351. The product It is called the strength of the elementary
shell at Q. When the shell is such that the strength is every-
where the same the shell is called a simple magnetic shell and the
distribution of magnetism is said to be lamellar. If the strength
varies from point to point, the shell is called a complex magnetic
shell and the distribution of magnetism is said to be complex
lamellar. Let It = <£.
352. When the distribution of magnetism is lamellar the
potential takes a simple form. Putting eo for the whole solid
angle subtended at P by any portion of the elementary shell, we
find that the potential at P of that portion is
It follows from this result that if two thin lamellar shells have
the same rim and the positive sides are turned the same way, the
potentials and therefore the forces at any point P are equal each to
each. The dimensions of <j> are those of potential.
ART. 354] LAMELLAR SHELLS. 183
Let a thin lamellar shell enclose a space. The potential at P
of any portion has been shown to be Itco. Let this portion
increase and finally cover the shell. If P be inside the empty
space the solid angle &> subtended at P increases and is finally ^TT.
If P be outside, the angle <w will presently begin to decrease and
will be finally zero. It follows that the potential of a closed lamellar
shell at an internal point is 4nrlt, at an external point the potential
is zero.
353. If a thin lamellar shell is in the presence of a number
of magnets, the mutual potential energy of any element dv and
the field is — FIdv by Art. 321, where F is the axial component of
force at the element dv. Since Idv = <f>d<r and <£ is constant when
the distribution of magnetism is lamellar, the mutual potential
energy of the whole shell and the field is — <j)fFd(r. The integral
fFda- represents the flux of the force due to the field entering the
negative side of the shell.
354. To determine the conditions that the distribution of magnetism is lamellar.
Let X, fj., v, expressed as functions of x, y, z, be the direction cosines of the
tangent at any point R of a magnetic line. The analytical condition that the
magnetic lines can be cut orthogonally by some system of surfaces is that
Xdx+ftdy + vdz can be made a perfect differential of some function f(x, y, z) by
multiplication by a factor, and the orthogonal surfaces are then f(x, y, z)=c. Let
p be one of these factors, the three equations
, df df df
,A=-, „»=_, pv=- ........................ (1)
must then be satisfied by simultaneous values of p and /. If A, B, C be any
quantities proportional to X, /t, v, say A=m\ &c., we find, by eliminating pfm and/,
. ............... .
dy ) \dx dz J \dy dx J
Let a shell be formed by the two surfaces /= c, /= c + dc. Let x, y, z; x + dx, &c.
be the coordinates of two adjacent points E, S, one on each surface. The thickness
t of the shell at R is the sum of the projections of dx, dy, dz on the normal at R,
hence pt=p(\dx + fj.dy+vdz) = df.
The product pt is therefore constant and equal to dc for the shell. Now two
quantities (say pt and It) cannot both be constant for the same shell, unless I bear
a constant ratio to p. Thus It will be constant only if 1= Pp, where P is a function
of x, y, z which is constant all over the surface f=c. Hence P=F(f), and it is
evident that p'=Pp is another factor which also makes Xda; + &c. a perfect
differential, viz. the differential F (/) df. It is therefore necessary that I should
be equal to some one of the values of p which satisfy the conditions (1).
Let the magnetism of the body be given by the components A, B, C of the vector
I expressed as known functions of x, y, z. The necessary and sufficient conditions
that the distribution of magnetism should be lamellar are that
A = f, B=«, C=f,
dx dy' dz
184 MAGNETIC ATTRACTIONS. [ART. 357
where f is an arbitrary function of x, y, z. The condition (2) that the magnetic
lines can be cut orthogonally by some system of surfaces is satisfied by these values
of A, B, C. The function / is called the potential of magnetisation. It must be
distinguished from the magnetic potential F.
356. Ex. Each element Q of a thin spherical shell is magnetised along the
direction OQ, where O is a given point on the surface, with an intensity Z which
varies as the distance OQ. Prove that the potential at any external point P is
proportional to cos 0/r2, where r is the distance of P from the centre C and 0 is the
angle r makes with OC.
Eesolve the magnetism at Q into the two directions CQ and OC. Taking the
former alone, the distribution is lamellar and the external potential is zero.
Taking the latter, the distribution is uniform and the potential is known.
356. To find the magnetic force exerted by a lamellar shell of strength <f> on a
unit pole at P.
Describe a cone whose vertex is P and whose generators pass through the rim
or margin AQQ' of the shell,
and let this cone be cut by a
sphere, whose centre is P and
whose radius c is very great,
in the spherical segment
BRR'. We replace the given
shell by another shell with
the same rim, but having for
its surface the spherical seg-
ment and that portion of the
cone which lies between the
rim and the segment (Art.
352).
The small magnet equiva-
lent to the magnetism at TT on any elementary area dS of the cone is " broadside
on " to P and the force exerted at P is therefore tpdS/r3, where r=PT. When P is
on the positive side of the given shell the positive pole of the small magnet at T is
directed inwards towards the given shell and the force at P tends outwards in the
direction indicated by the arrow (Art. 317). Let the angle QPQ' = d\ft, then
dS=rd\f/dr. The force at P due to the magnetism on the strip QQ'BR'
becomes by integration f ) (pd\f/, where r is now PQ. This reduces to its first
term when c is very great.
The equivalent magnets which represent the magnetism on the spherical
segment BRR' are "end on" to P. The force on P due to any elementary area
dS is 20diS/c3. Since dS=c2dw (where dw is the solid angle subtended at P) this
force is zero when c is very great.
To find the force at P due to a thin lamellar shell we divide the rim into elements.
The force due to the element QQ' is equal to <t>d\f//r and also to QdA/r3, where r = PQ,
QPQ' = d\j/ and dA is twice the area QPQ'. This force acts at P perpendicularly to
the plane QPQ'. The resultant of these elementary forces is the force on a unit pole
at P.
367. We notice that the magnitude of the force due to the elementary arc QQ'
and its direction relatively to the plane PQQ' are not changed by rotating that
ART. 359] LAMELLAR SHELLS. 185
plane about QQ' as axis. The side of the plane to which the force tends (when
not already obvious) is therefore easily found. If P is brought by the rotation from
the positive to the negative side of the shell, the force on a positive unit pole at P
acts in the direction of motion.
When the rim is a plane curve and the point P lies in that plane, the force at P
is normal to the plane and equal to
358. To find the Cartesian components of the force at P due to a lamellar shell.
Let (£, i), f) be the coordinates of P, (x, y, z) those of a point Q on the rim.
Let (A, (JL, v) be the direction cosines of the normal to the plane PQQ'. The z
component of force is therefore v^dA/r3. By projecting the area dA on the plane
of xy we have (x-i-)dy -(y-i))dx=vdA.
f (x — £ ) dy — (y — r>) dx , .
Hence Z—<j> I S£- a 3'g — and similarly
x= [(y-r,)dz-(z-$)dy
r-,/!
r3
The integration in each case is conducted round the rim in the positive direction.
The left-handed system of coordinates being used (vol. i. Arts. 97 and 272) ; the
positive direction is clockwise as seen by an observer with his feet on the shell and
his head on the positive side.
We may put these expressions for the forces into another form. Let
Then since rz=(x-£)* + &c., we have - -) = , &c.,
r
_ .
Snmlarly X=<f> - ,
359. To find the potential of a lamellar magnetic body. Let the components
of I be A = df/dx, &c., and let r be the distance of a point P from any element of
volume dv. The potential at P is (by Art. 339)
dzj r J \dx ax dy dy dz dz
by Green's theorem. Now -^ = — = ~ = -= cos 0, where Q is the angle the
on r2 dn r2
distance r makes with the outward normal (Art. 347). Also v2l/r=0 or ~^ir
according as P is without or within the substance of the body. Hence
"/>
(i,,
where (/) is the value of / at the point P.
Let the surface integral I / ^ — be represented by 0.
Then F=O + 4ir (/) (2).
The components of magnetic force at an internal point P are given by
X=-dV]dx &c., and the components of magnetisation by A = d(f)ldx &G. It
186 MAGNETIC ATTRACTIONS. [ART. 362
follows at once from Art. 345 that the components of magnetic induction are related
to fi by the equations Za s= - dti/dx <fec.
Let Oj, 02 be the values of O at points Plt P2 respectively just outside and just
inside the surface S. Now cos 6 dS/rz is the solid angle subtended by dS at P and
hence, by the same reasoning as in Art. 142, 01-02 = (/) (wi~ wa)> where olt w2 are
the solid angles subtended at Pj , P2 ^ the elementary area dS. The difference of
these solid angles is 4ir by Art. 350. We therefore have fy - fl2=4ir (/). If V1 , F2
are the values of V at the same points, Pa and P2, we have by (2) F^fy and
F2=O2 + 47r(/). Hence Fa=F2. The magnetic potential is therefore continuous
when the point P enters the substance of the magnetic body, but the potential ft is
not continuous.
36O. The mutual potential energy of two thin lamellar shells may be found by
r r cog £
the formula W= -0$' I — dads', which involves only integrations round the two
rims. Here ds, ds' are elements of the two rims, e is the angle their positive
directions make with each other, and <f>, <f>' are the strengths of the shells.
Let as first find the mutual potential energy of the shell whose strength is <f>
and a portion of the other shell which is so small that we may regard its rim as a
plane curve. Let this plane be taken as the plane of xy and let (£', ??') be the
coordinates of any element da' of the area. The potential energy is by Art. 321
By the application of Green's method to plane curves (Art. 149) this surface
integral is replaced by an integration round the rim which in our case gives
$(Fd£'+GdT)'), We now substitute for F and G their values from Art. 358 and
remember that df'=0. The expression for the potential energy is then
? + *»**+***? '
To find the energy when the second shell is of finite size we integrate the
expression just found. Let two adjacent elements touch along the arc AB. When
integrating round these two elements, we pass over the arc AB in opposite directions
and therefore for this arc the angle e (being the angle the direction of integration
makes with an arc ds of the first shell) has supplementary values in the two
elements. The sum of the integrals for both elements may therefore be found by
integrating round both as if they were one, omitting the common arc AB. The
game reasoning applies to all adjacent elements, hence the total energy for two
shells of finite size may be found by integrating round their perimeters.
361. The theory of thin lamellar shells derives additional importance from its
connection with electric currents. According to Ampere's theory the forces on a
magnetic pole due to a closed electric circuit are the same as those of a thin
lamellar magnetic shell, of proper strength, whose rim is the closed circuit.
The direction of these currents may be usefully remembered thus : if the earth's
magnetism were due to currents round the axis, their general direction would follow
the sun, that is, would be from east to west. Now the south pole of the earth has
been taken as its positive pole (Art. 314). Hence the currents equivalent to a small
magnet flow round it anti-clockwise when viewed by a person with his head at the
positive and feet at the negative pole.
362. Examples. Ex. 1. Prove that the force due to a thin lamellar shell,
•whose rim is a circle of radius a, (1) at the centre is Z0=2w(pla and (2) at a point
ART. 363] LAMELLAR SHELLS. 187
P situated in its plane at a small distance £ from the centre is Z=
[Use the formula Z=<f>jd\f//r.']
Ex. 2. Prove that the potential of a thin lamellar shell whose rim is a circle of
radius a at a point P distant f from the plane and at a small distance £ from the
axs s =,r-
•where t;2=a2 + f2. Prove also that the component forces at P are
It is evident from the symmetry of the figure on each side of the plane TI$ that the
potential can contain only even powers of £. The expression for V is therefore
correct up to the cuhes of the small quantity £.
Ex. 3. Prove that the mutual potential energy W of a thin circular lamellar
shell and a small magnet whose centre C is on the axis of the circle is
W= - ^— /sin e + 4^aV Bija 0 (2 sin2 0 - 3 cos' 6) + &c.| ,
where a is the radius of the shell, h, I the distances of C from the centre and rim of
the shell respectively, p the half-length of the magnet, 6 the inclination of its axis
to the plane of the shell, and M its moment.
If the magnet C be placed on the axis of the shell at a distance from its plane
equal to half the radius, the terms of the order p2 are zero. It follows that the action
of the shell on the small magnet C is equal to that of another small magnet whose
moment is M'=ira*<j>l5>J5 placed end on at the centre of the shell. A similar theorem
for electric currents is given by M. Gaugain in the Comptes Rendus, 1853.
We may obtain this result by expanding the expression for the potential found
in Ex. 2 in powers of p, then twice the sum of the odd powers after multiplication
by the strength of the small magnet is the energy required.
Ex 4. If the law of force be the inverse /cth power of the distance, prove that
the mutual potential energy W of two thin lamellar shells of elementary areas is
where R is the distance between two points one in each area. It appears that we
cannot extend the theorem to shells of finite area unless the law of force be either
the inverse square or the direct distance.
Electrical Attractions.
363. Introductory statement*. When certain bodies are
electrified one evidence of the presence of electricity is the
* Many theorems in Attractions are important because they are used in the
theory of Electricity and would seem almost purposeless without some notice of
these applications. To enable these to be properly understood it is necessary to
give a brief introductory account of the principles to be afterwards assumed. In
this statement only so much of these principles is given as is required in what
follows. For example, the experimental proofs of the laws of electricity are not
described. These may be found fully discussed in Everett's edition of Deschanel's
Natural Philosophy, 1901, in J. J. Thomson's Treatise on Electricity, and in that
of Maxwell.
188 ELECTRICAL ATTRACTIONS. [ART. 365
attraction exerted on other electrified bodies. For the purpose
of illustrating the theory of attractions we shall here replace the
electricity by that system of repelling particles which exerts an
equal force at every external point.
364. If ra, mf be the masses of two particles we assume here
that the force of repulsion exerted by one on the other is mm'/r3
(Art. 1). But the masses are not restricted to be positive, two
particles whose masses are of opposite signs attract each other.
The electricity is called positive or negative according as it is
represented by particles of positive or negative mass. Since like
particles repel each other the fundamental formulae connecting
potential and force are X= — dV/dx &c. (Art. 41).
It is obvious from this definition that either kind of electricity
may be called positive, provided the other is called negative.
A convention is therefore necessary. Let a piece of glass and a
piece of resin be chosen which exhibit no signs of electricity. Let
these be rubbed together and separated. Each body is now found
to attract the other. The electrification on the glass is called
positive, that on the resin negative. If a body electrified in any
manner repels the glass and attracts the resin, it is positively
electrified. If it attract the glass and repel the resin, its
electricity is negative.
365. When a particle A, say positively electrified, is brought
into the presence of certain bodies it is found that electricity is
immediately developed in them. Some positive electricity is
repelled and driven to the parts of the body most remote from A
and some negative electricity is attracted to the nearer parts. If
a second and then a third particle be made to act on the body
more positive and negative electricity are developed and separated
as before. In all these cases the arrangement of the electricity
when in equilibrium is altered by the approach of a new particle.
This result is interpreted to mean that when electricity is in
equilibrium the force acting on each element of the volume of the
body is zero. If it were not zero, more electricity would be
developed and the existing arrangement would not be in equi-
librium. It follows that the electric potential due to its own charge
and to the external electrical particles is constant throughout tke
body.
ART. 369] INTRODUCTORY STATEMENT. 189
366. We have here supposed that the electricity is able to
move without constraint from one element of volume to another.
A body which permits this transference is called a perfect con-
ductor. There are also bodies in which the electricity cannot
travel through the volume but is forced to remain in the place
where it has been developed. These are called perfect non-con-
ductors. There are in nature no perfect conductors and no perfect
non-conductors, but in some bodies the developed electricity
travels so easily and in others with such difficulty that they are
usually distinguished as conductors and non-conductors. Metals,
fluids and living bodies are conductors, while dense dry gases,
glass, silk are non-conductors.
367. If we represent the electricity by repelling points we
must be able to apply Poisson's theorem to a body which is
without constraints, Art. 105. We then have 4nrp = — V2F, where
pdv is the excess of the positive over the negative electricity in
the elementary volume dv. Since X, Y, Z are zero in equilibrium,
Art. 365, this equation gives p = 0. The element therefore contains
equal quantities of positive and negative electricity.
This holds at every internal point but not at the boundary of
the solid, for here the surface constraint comes into play. The
conductor is supposed to be surrounded by a non-conducting
medium through which the electricity cannot pass. This medium
by its pressure constrains the electricity to remain in the con-
ductor. There may therefore be an indefinitely thin layer of
attracting or repelling particles on the boundary.
When equal quantities of positive and negative electricity
occupy an element, that element is said to be neutral. It exerts
no force at any external point. When there is an excess of either
kind in any element, that excess is said to \>efree. In a conductor
the free electricity resides on the surface.
368. The potential of the electricity is the same as that
of an indefinitely thin layer of repelling matter placed on the
surface of each electrified conductor. We measure the, amount of
the electricity at any point by the surface density of this equivalent
layer. The whole quantity of electricity is measured by the mass
of the layer.
369. Surface density. Since the potential is constant through-
out the interior of a conductor, the theorem of Green proved in
190 ELECTRICAL ATTRACTIONS. [ART. 371
Art. 142 takes a simpler form. Let p be the surface density at
any point P, let X be the normal force in the non-conducting
medium at a point infinitely close to P and let the positive
direction of X be from the conductor into the non-conductor, then
4<7Tp = X. If dn be an element of the normal drawn outwards,
V the potential due to all causes, this equation may be written in
the form 4t7rp = — dV/dn.
The repulsive force on the electricity which covers any element
da- of the surface is ^Xpda; (Art. 143). This may be written in
either of the equivalent forms 27rp2dcr or (Xz/87r)da: Since this
is always positive, the direction of the force is necessarily outwards.
370. If an electrified conductor is joined .by a wire of con-
ducting material to the earth its potential must become the same
as that of the earth (Art. 365). At the same time the potential
of so large a body as the earth is not affected by any transference
of electricity to or from it. Supposing the earth to be in its
ordinary neutral state, the potential then becomes zero. When a,
body is thus joined to the earth, it is said to be uninsulated.
Electrified bodies are in general supposed to be insulated
by air, unless otherwise stated. When the density of the air is
diminished its resistance to the escape of the electricity also
decreases. When the pressure is still further reduced its resistance
increases again. A vacuum, that is to say, that which remains in
a vessel after we have removed everything which we can remove
from it, is a strong insulator. Maxwell's Electricity &c. Art. 51.
371. Capacity. If one conductor is insulated, while all the
other conductors in the field are kept at zero potential by being
put into communication with the earth, and if the conductor,
when charged with a quantity E of electricity, has a potential V,
the ratio of E to V is called the capacity of the conductor.
In this definition the capacity is supposed to be independent of the special
nature of the non-conducting medium which surrounds and separates the con-
ductors. But the medium is itself acted on and reacts on the conductors. To
take account of these actions it is necessary to introduce into the definition a new
factor called the Specific Inductive Capacity of the medium. The effect of this
factor is a subject for separate discussion. In what follows (until otherwise
stated) it will be supposed that the medium is such that the effects of the induction
on it can be disregarded. This is the case if the medium is air or generally any
dry gas. In these media the specific inductive capacity is nearly equal to unity.
Ex. Let a sphere of radius a be at a great distance from all other bodies
and let it be charged with a quantity E of electricity. The potential is E/a,
(Art. 64). The capacity of an insulated sphere is therefore equal to its radius.
ART. 373] ELEMENTARY PROBLEMS. 191
372. The electrical problem. Let a conductor have a
charge M of electricity and be acted on by an external charge M'.
Then in equilibrium a mass M + p of free positive electricity and
a mass — yu- of negative electricity (where //. is one of the unknown
quantities to be found) will be arranged on the surface of the
conductor in such a manner that the sum of the potentials of
M', M + fju and — p is constant throughout the interior. The
electrical problem is to find the superficial density at every point.
Conversely, if the electricity be thus arranged it will be in
equilibrium. First, we notice that the component forces at every
internal point are zero. Next, the tangential component just
inside the surface is zero and therefore by Art. 146 the tangential
force at any point of the stratum is zero. The resultant force on
any superficial element of the electricity is therefore normal and
by Art. 369 acts outwards. It is therefore balanced by the
reaction at the boundary, (Art. 367).
373. Ex. 1. The potentials of an electrical system at the corners of a small
tetrahedron are F15 F2, F3, F4; prove that the potential at the point which is the
centre of mass of particles m^, m2, T%3, mt placed at the corners is SmVjZm.
This follows at once from Taylor's theorem. [Trin. Coll. 1897.]
Ex. 2. An insulated conductor of finite volume is charged ; prove that the
electrical layer completely covers the conductor.
If there were any finite area on the surface unoccupied by electricity, the
potential must also be constant throughout all external space which can be reached
without passing in the immediate neighbourhood of repelling matter, Art. 117.
Hence X=0 both on the inside and on the outside and the surface density would
be everywhere zero.
Ex. 3. A conductor ia charged by repeated contacts with a plate which after
each contact is re-charged so that it always carries the same charge E. Prove
that, if e is the charge of the conductor after the first operation, the ultimate
charge is Eej(E-e). [Coll. Ex.]
When the plate touches the conductor the whole charge on both is divided
between the two bodies, so that their potentials become equal. If the whole charge
were increased in any ratio the potentials would be increased in the same ratio and
equilibrium would still exist. It follows that just after each contact the quantities
of electricity on the plate and the conductor are in a ratio /3 : 1 which only depends
on their forms.
Let £„ be the quantity on the conductor after n contacts. After the next
contact, xn+l - xn is taken from the plate. Hence the ratio of E- xn±1 + xn to
xn+l is /3 : 1. After the first operation the quantities on the plate and conductor
are E - e and e, and this ratio also is /8 : 1. To find the ultimate ratio (when n is
very great) we put xn+l=xn. We then find xn+1 by eliminating /3.
Ex. 4. A soap bubble is electrified ; prove that the difference between the
pressures of the air inside and outside is 2Tjr — 2irp*, where T is the surface
tension, r the radius and p the surface density of the electricity.
192 ELECTRICAL ATTRACTIONS. [ART. 375
374. Two spheres joined by a wire. Two small spheres
of radii a, b, placed at a considerable distance from each other, are
joined by a thin wire and the system is insulated from all other
attracting bodies. The spheres and the wire are made of some
conducting material. A charge E of electricity is given to the
system, determine approximately how it is distributed over the
several bodies.
Let I be the distance between the centres A, B of the spheres ;
x, y the quantities of electricity on their surfaces, z the quantity
on the wire. The electricity is so distributed that the potential
is constant throughout the interior and therefore the potentials at
A, B are equal Since the centre A is equally distant from every
point of that sphere, the potential at A of the electricity on its
surface is x/a. Since A is very distant from every point of the
other sphere, the potential at A of the electricity on the sphere B
is very nearly yjl. Neglecting the electricity on the wire for the
moment we have the two equations
x y as y
---
when the radii are very small compared with their distance we
can reject the terms with I in the denominator. The electricity is
then distributed over the surfaces of the two spheres nearly in the
ratio of the radii.
We shall now prove that the quantity of electricity cm the cylindrical wire may
be neglected if the radius c is sufficiently small. Let D be the average surface
density on the wire, then z = 2irclD. The potential of a cylinder of length I and
of uniform surface density at the middle point of its axis is
very nearly, since cjl is very small. Since F is nearly equal to x\a or y[b it is
clear that z can be made as small as we please by using a wire sufficiently thin
compared with its length.
376. Ex. 1. A conducting sphere, of radius a, is joined to the earth by a
fine wire and is acted on by an electrical point Q at a distance 6 from the centre
of the sphere. Prove that the electricity on the sphere is - Qajb. Prove also that
the mutual attraction between the sphere and the point approximately varies
inversely as the cube of the distance and as the square of the charge.
Ex. 2. Two conducting spheres (radii a, b) are joined by a long thin conducting
wire, and the total charge is zero. A cloud charged with a quantity E of electricity
passes much nearer to one sphere than the other, but at a considerable distance
from both. Prove that the transference of electricity from one sphere to the
other is nearly abEJ(a + b) I', where I' is the distance of the cloud from the nearest
sphere.
ART. 378] AN ELLIPSOIDAL CONDUCTOR. 193
Ex. 3. Two conducting spheres of capacities c, c' are at a great distance I from
each other and are connected by a long fine wire. Prove that the capacity of the
conductor so formed is approximately c + c'- 2cc'/L [Coll. Ex. 1900.]
376. An ellipsoidal Conductor. Let a solid ellipsoid be
charged with a quantity M of electricity and be not acted on by
any external forces. We know by Art. 68 that the stratum
enclosed between the given ellipsoid and a similar and similarly
situated concentric ellipsoid exerts no attraction at any internal
point. We therefore infer that the infinitely thin stratum of
electricity will be in equilibrium, when distributed on the given
ellipsoid so that the surface density p at any point P is proportional
to the thickness of the thin shell. By Art. 71 the surface density
M.f>
at P is p = j — -Hr- , where p is the perpendicular from the centre
on the tangent plane at P.
377. Let points on two ellipsoids be said to "correspond" when their
coordinates referred to the axes are in the ratio of the parallel axes, thus xja=x'ja'
&c. Let two curves be drawn, one on each ellipsoid, such that the points on
one correspond to those on the other and let the spaces enclosed be called
corresponding spaces. The quantities of electricity on corresponding spaces are
in the ratio of the whole charges given to the ellipsoids.
This theorem follows at once from the proof in Art. 201, if we regard each
electrical stratum as a thin homoeoid. It may also be proved by the method of
projecting one ellipsoid into the other as explained in vol. i. Art. 428.
Ex. 1. Prove that the quantity of electricity on the portion of the ellipsoid
bounded by any two parallel planes is the same fraction of the whole electricity
that the portion of the diametral line between the planes it of the whole diameter.
If a portion of an indefinitely thin shell formed by two concentric spheres
be cut off by any two parallel planes we know that the intercepted volume is
proportional to the distance between the planes, (vol. i. Art. 420). Projecting the
spherical shell into an ellipsoidal one, the plane sections project into planes and
the theorem enunciated follows at once.
Ex. 2. Two planes drawn through any diameter POP' of the ellipsoid intersect
the diametral plane of POP' in OR, OR', and D is the diameter parallel to the
chord RR'. Prove that the electricity E on the lune included by these two
planes is given by E = (M/TT) sin"1 (RR'jD) where M is the whole electricity on the
ellipsoid.
378. The constant potential inside the electrified ellipsoid
can be found only as a definite integral, (Art. 197). When the
ellipsoid is one of revolution so that a = b, the integration can be
effected without difficulty.
Let the axis of z be the axis of revolution. The quantity of
E. s. n. 13
194 ELECTRICAL ATTRACTIONS. [ART. 381
electricity between the planes z and z + dz is dz.M/2c, Art. 377.
The potential at the centre is therefore
F- * ,•-*
2cJ r c2
Effecting the integrations we find
V- M cin-V(a2-c2) F= -J-- oa
-VCa'-c2) a V(c2-a2) a
according as the spheroid is oblate (a > c) or prolate (c > a). The
internal potential of the prolate spheroid is found more easily by
taking V to be the potential at the focus, for then r = c + ez.
The potential of the oblate spheroid is also V=M(£>/f, where /is
half the distance between the foci of the generating ellipse and <j>
is half the angle subtended by 2/ at the extremity of the axis of
revolution.
379. We know by Art. 205 that all the external level surfaces of the ellipsoidal
conductor are confocal ellipsoids. If P be any point situated on the confocal
whose semi-azes are a', V, c' the potential at P is
where the limits are X=a'2-a2 to ao , (Art. 208).
380. The external surface of a conductor charged with a
given quantity of electricity is not acted on by any external
body. Prove that the electricity at every point has the same sign.
Let V0 be the potential and first let this be positive. If there
be any point P on the surface at which there is negative
electricity, dV/dn must there be positive because the surface
density m is given by 4?rm = — dV/dn (Art. 369). Hence the
potential increases outwards along the line of force at P. But
this is impossible since the potential at every point between the
surface of the conductor and the surface of a sphere of infinite
radius must lie between F0 and zero, (Art. 116, Ex. 2). A similar
proof applies if V9 is negative.
381. A conductor, charged with a given quantity of electricity,
is acted on by given forces. Prove that there is but one arrange-
ment of the electricity which could be in equilibrium.
If possible let there be two distributions of the electricity,
either of which could be in equilibrium, though the potentials
are not necessarily the same. By subtracting one of these from
the other, as in Art. 129, we obtain a distribution of electricity in
equilibrium in which the external forces are absent and the total
ART. 384] ELLIPTIC DISC. 195
mass is zero. This distribution must have both positive and
negative electricity on the surface, but this has just been proved
to be impossible.
382. Elliptic Disc. To deduce the distribution of electricity
on an insulated elliptic conducting disc from that on an ellipsoid
we put c = 0. Then
The surface density p at any point (x, y} of the disc is then given
, Mp M __ 1 _
P ~ ~
where M is the whole charge on the disc. This value of p gives
the surface density on either side.
The internal potential due to the electricity on both sides is
cos2 sn2
~ ~
where the limits are r = 0 to l/q and 6 = 0 to 2?r. Effecting the
integration with regard to r and writing # = ^TT — 0, we find
T. Mr** d<j>
For a circular disc we have, if p be the surface density at any
distance r from the centre,
M 1 v_ [2frrdr2p _ MTT
(a2-r2)'a ~J F~ '~^a~'
The capacity M/V is therefore 2a/?r.
383. The quantity of electricity on any elementary area dxdy of the disc is
T — r -77= - ^~- — 57^5: . If then two elliptic discs (say an ellipse and its auxiliary
4-n-a - 22 - 22
circle) have equal charges, the quantities of electricity on corresponding elements
are also equal, for in these elements xja=x'la' &c., dx/a=dx'/a' &c. The quantities
on any corresponding finite areas are therefore also equal. This result follows at
once from Art. 377.
384. The potential V of the elliptic conductor at an external
point P follows from the result stated for an ellipsoid in Art. 379.
Let the confocal on which P is situated cut the axis of the disc in
(7, C' and let its semi-axes be a, b', c'. We find after putting
c = 0, X = p? and p = b tan 6,
dJL- Mf d0
~6« +>* ~~ a J I -
_ -
{(a2 + ^)~(6« +>)}* ~~ a J [I - e* sin2 d}* '
13—2
196 ELECTRICAL ATTRACTIONS. [ART. 387
where the limits are //, = c' to yu, = oo and 0 = ^TT — <£ to 6 = \TT
where <j> is half the angle subtended by the minor axis of the disc
at C or C'.
For a circular disc the potential at an external point is M(f>/a
where <f> is half the angle subtended at C by any diameter of the
disc.
385. Ex. 1. A chord is drawn on an elliptic insulated disc, prove that the
quantities of electricity on each side of the chord are in the ratio of the segments
of the conjugate diameter, (Art. 377).
Ex. 2. Prove that the quantity of electricity on an elliptic sector bounded
by the semi-diameters CP, Cf is M (<j>' - 0)/2?r, where <j>, <f>' are the eccentric angles
of P, P' and M is the whole quantity of electricity (Art. 383).
Ex. 3. A similar coaxial ellipse whose semi-axes are na, rib is described on the
electrified disc. Prove that the quantity of electricity between this ellipse and the
rim of the disc is M>J(1 - n2).
Ex. 4. Prove that the line density of a thin electrified insulated rod is constant.
Regard the rod as an evanescent ellipsoid in which a and b are zero and c
finite. The line density />' is such that p'dz represents the mass between two
planes whose abscissae are z and z + dz. This we know is Mdzj2c, Art. 377.
386. Conductor with a cavity. A body is bounded by
two surfaces Sly S2 which do not intersect. It is charged with a
given quantity M of electricity and is acted on by a given system
of electrical points (mass Mj) situated within Si.
Let x be the quantity of electricity on S1} M — x that on S.2.
These are so distributed that the sum of the potentials of Mlf
a; and M— <x is constant throughout the space between Sl} S3.
Describe a surface cr between S1} S2 and enclosing Slt then by
Gauss' theorem (Art. 106) 4-vr (Ml + x) is equal to the flux of force
across <r, and this is zero. Hence x = — H^ ; the charge on $a is
therefore — Ml and that on Sa is M + M^.
387. If the charge M= - Ml} the total quantity on S2 is zero.
It immediately follows from the argument in Art. 380 that the
charge on each element is also zero. For by that article, the
electricity on every element of S2 must have the same sign, which
is impossible when the whole quantity is zero. The whole of the
free electricity is therefore concentrated on S^
The sum of the potentials of the system MI and of the
electricity x = — M1 is constant throughout the space between Sj.
and $2 and therefore throughout all space which can be reached
without passing in the immediate neighbourhood of attracting
matter, Art. 117. It is therefore constant throughout all space
ART. 391] CONDUCTOR WITH A CAVITY. 197
external to 81 and is the same as that at an infinitely distant
point. The sum of the potentials of M1 and x is therefore zero.
It appears that the form of the surface S2 may be changed
without in any way disturbing the electricity on the surface S^
388. Let a solid conductor whose boundary is the surface $2
be acted on by a given external system of electrical points
(mass M2) and let the charge given to the conductor be y. The
condition of equilibrium is that the sum of the potentials of
Mz and y is constant throughout the interior. Since this condition
is not affected by removing any portion of the inside matter the
equilibrium of the electricity on S2 will not be disturbed when the
surface 81 is made the internal boundary.
389. If we now superimpose these two conductors, we have a
conductor bounded by the surfaces 8lt $2 with a given electrical
mass MI inside S1 and a mass Mz outside S.2. Let the charge
given to the conductor be M.
There will now be a charge x — — M1 on Si so arranged that
the sum of the potentials of MI and a; is zero at all points external
to Si. There will be charge y = M+Ml on $2 so arranged that
the sum of the potentials of M2 and y is constant at all points
internal to S2. The condition that the equilibrium should remain
undisturbed is that the sum of the four potentials should be
constant at all points bet ween SltSa and this condition is evidently
satisfied.
We observe that the distributions of electricity on the two
boundaries are independent of each other.
390. Screens. We notice how completely the electricity on
the surface Si screens the repelling masses MI from observation by
an external spectator. If the masses forming the system MI be
moved about in any way within the cavity the electricity on Si
rearranges itself continually so that in equilibrium the resultant
force at every external point is zero.
In the same way if the external masses M2 be moved about,
the electricity on the surface S2 is rearranged and the motion is
imperceptible to an observer within the cavity.
391. We shall now prove that there is but one possible distribution in equi-
librium on the two surfaces Slt S3 when the charge M and the electrical masses
MI, Mz are given.
If possible let there be another arrangement. Then subtracting one of these
198 ELECTRICAL ATTRACTIONS. [ART. 393
from the other, we have an arrangement of electricity in equilibrium, in which
there are no internal or external masses, while there are charges x and y on /Sx and
S2 , whose sum x + y is zero.
The sum of the potentials of x and y is constant over Sj and therefore over
a surface just inside 5X (Art. 145). This surface contains no attracting matter
and therefore the sum of the potentials of x and y is constant throughout all space
within S1. The forces X, X' just inside and just outside Sa are therefore zero.
It follows that the density of the electricity at every point of S1 is zero (Art. 142).
Consider next the surface Sa. The total charge on it is zero. Hence, as in
Art. 381, the charge on each element is zero.
The two possible distributions must therefore have been the same.
393. Ex. 1. A solid sphere, radius a, is concentric with a spherical shell,
radii b and c, both being perfect conductors, and charged with quantities E, E' of
electricity. To find the potential of the sphere.
The quantities of electricity on the three surfaces whose radii are a, b, c are
respectively E, -E, and E' + E (Art. 389). By Art. 64, the potentials at any
points inside the substance of the sphere and shell are respectively
E E E' + E E' + E
y — — — ~i~ T » ' — •
a b c c
If the shell is joined to earth, or if the radius c is very great, in either case
V=0 and the capacity of the sphere, viz. E/V, becomes ab/(b-a). If a and b are
also very nearly equal to each other, the capacity is very great. Supposing
the potential of the sphere to be finite, the charge on the sphere and the equal
opposite charge on the inner surface of the shell are very large.
When two conductors, insulated from each other, are placed very near each
other the system is called a condenser.
Ex. 2. Three insulated conductors A, B, C, are in the form of thin concentric
spherical shells of radii a, b, c; and are so charged that their potentials are
F1, F3, F8. Prove that the charge on the intermediate shell B is
(a-b)(b-c) • (a (6 - c) Fi + 6 (c - «) F2 + c (a - b) F8}. [Coll. Ex. 1897.]
Ex. 3. A condenser is formed of two concentric spherical thin conducting
sheets, the radius of the inner being b, that of the outer a. A small hole exists
in the outer sheet through which an insulated wire passes connecting the inner
sheet with a third conductor, of capacity c, at a large distance r from the
condenser. The outer sheet of the condenser is put to earth, and the charge on
the two connected conductors is E. Prove that approximately the force on the
third conductor is ac*E*j (-^ +cjr». [Trin. Coll. 1897.]
393. Green's Method. The law of distribution of a given
quantity of electricity on a given surface under the influence of
given forces cannot always be discovered. Two methods of finding
a solution in certain limited cases are in general use. The method
of inversion, by which when one problem has been solved the
solution of another can be deduced, has been explained in Arts.
168 &c. We shall now proceed to Greens Method, by which the
ART. 395] GREEN'S METHOD. 199
law of distribution of a certain quantity of electricity can be found
when the boundary of the conductor is a level surface of a known
system of repelling bodies. This method has been already dis-
cussed in Art. 156. Before however proceeding to its application
we shall give an elementary proof in small type which more fully
illustrates the principles of attraction.
394. Let Ml , M2 be two given systems of attracting or repelling particles
and let S be a surface enclosing M1 only within its finite space. If the masses M2
were removed and S made the inner boundary of a conductor, a quantity of
electricity, equal to -Afj, would be found on the surface S and its potential
together with that of the system Ml would be zero throughout all space external
to S (Art. 387). Let this distribution of electricity on S be called E^.
Let us now change the sign of every element of the electricity E1 and constrain
it to remain, otherwise unaltered, on the surface S. Let this new distribution
be called E2. The quantity of the electricity Ez is then equal to +Ml and the
potential of E2 is the same as that of M1 throughout all space external to S.
Let us next suppose that S is a level surface of M1 and M2 and let the potential
at any point of S be Vt. It must therefore also be a level surface of Ez and l/2.
The sum of the potentials of E2 and M2 is therefore constant and equal to Fg at
all points of a surface just inside S. Since no particle of either E2 or M2 lies
within this surface, the sum of the potentials of £8 and M3 is constant and equal to Vt
throughout the interior of S (Art. 115).
If S be made the external boundary of a conductor and the system M-^ removed,
the distribution of electricity Ez would be in equilibrium under the influence
of its own attraction and that of M% (Art. 372). We also know that no other
distribution of the same quantity M1 of electricity will be in equilibrium (Art. 381).
Briefly, El , when acted on by M j , is in equilibrium if S is the inner surface
of a conductor, and E2, when acted on by M2 , is in equilibrium if S is the outer
surface. Also E1 and Ez differ only in sign.
The surface density p2 at any point P of the stratum Ez when placed on the
external surface of a conductor follows at once from Green's theorem, (Art. 142).
By that theorem 4irp2=X where X is the sum of the normal forces due to M2 and
E2 at a point just outside the substance of the conductor. But the normal force
due to Ez has been proved equal to that of M^. Hence X is the sum of the
normal forces at P, due to M1 and Mz, measured positively from the conductor
towards the non-conductor.
The density pl at any point P of the stratum E^ when placed on the inner
surface of a conductor has the sign opposite to p2. Since the non-conductor is
now on the opposite side of Sl the density p± is given by the same rule, 4irp1=X,
where X is the sum of the normal forces due to M1 and Ma measured towards
the non-conductor.
395. We arrive at the following rules.
1. Let 8 be any closed portion of a level surface (potential FO)
of a given electrical system, Ml being inside and Mz outside. We.
may remove either the mass Mg and regard S as the internal
boundary of a solid conductor acted on by the internal mass Mlt or
200 ELECTRICAL ATTRACTIONS. [ART. 396
we may remove Ml and regard 8 as the external boundary of a
conductor acted on by the external mass M2. In either case the
density p at P is given by the rule 4nrp = X. where X is the
normal force at P due to both J/x and Mz measured positively
towards the non-conductor.
2. The quantity of electricity on S is — Ml or + Mt according
as S is an internal or external boundary.
3. The potential of the electrical stratum at any point R on
the side opposite to M^ is numerically equal to that of Mt. It
follows that the stratum and M^ have not only the same mass, but
also the same centre of gravity, (Art. 136). Their principal axes
at the centre of gravity also coincide in direction and the difference
of their moments of inertia about every straight line is the same.
4. The potential of the stratum at any point R' on the same
side as Ml is equal to V8 — Vz where Va is the potential of Mt at
R', when S is an external boundary. It follows that when S is
an internal boundary, the potential at R' is given by the same
expression with its sign changed.
396. ' If the quantity Q of electricity which covers any given
portion a- of the surface S is required we use Gauss' theorem,
(Art. 106). We describe a surface closely enveloping the given
portion of S both inside and out, then 4nrQ=j'Fda: Just inside
the conductor S the force F is zero and all this portion of the
integral may be omitted. The required quantity Q is therefore
given by the above integral, where F is the normal (or resultant)
force due to the given system M1} Mz measured towards the
non-conductor at the element da- of the given area, and the
integral extends over that area.
When the systems Mlt Mz consist of isolated particles of
masses m1,mi&c. the integral can be put into a more convenient
form. For any one particle m we have, as in Art. 106, Fda=mdco
where dco is the solid angle subtended at m by da: This elementary
solid angle is to be estimated as positive when a repulsive force
emanating from m traverses da- outwards into the non-conducting
medium. Adding up the corresponding portions for all the
particles, we see that tfie required quantity Q of electricity is given
by 47rQ = Sraw.
Image. "When an imaginary system of points, if properly placed on one side
of a surface, would produce at points on the other side of that surface the same
ART. 398] SPHERICAL CONDUCTOR. 201
attraction or repulsion which the actual electricity on the surface produces that
system of points is called an image. Thus, when the surface S is the external
surface of a conductor, M1 is the image of Ht, because the attraction which it
exerts at points on the other side of S is the same as that due to the electrification
on S when acted on by Ma .
397. Electricity on a sphere. To find the distribution of
electricity on a sphere of conducting matter when acted on by an
electrical point.
Let A be the centre of the sphere, B any external point, BD a
tangent and DC a perpendicular
on AB. Let the distances of
A, B, G from D be respectively
a, b, c, so that a is the radius.
Let the distances of B and G
from the centre A be f and f.
Since d?=ff, the points B
and G are inverse with regard to
the sphere. If Q be any point
on the sphere, the ratio CQ/BQ
is constant and therefore (by
putting Q at D) equal to c/b (Art. 86). We also have, by the
similar triangles BDG, BAD, c/b = a/f.
If then we place at B and C respectively quantities of
electricity E = fjb, E' = — /AC *, where /j, is any constant, we have
at any point Q of the surface
E E'
BQ+GQ = Q (1)'
i.e. the sphere is a level surface of zero potential. We may
therefore at once apply Green's principles (Art. 395).
398. If the conducting mass is to be a solid sphere sur-
rounded by a non-conducting medium, we remove the electrical
point G and distribute the mass E' over the surface so that the
surface density p at any point Q is given by 4nrp = F, where F is
the normal force at Q due to E and E' tending towards the side
of the sphere on which the non-conducting medium lies. The
* Since CQ2/BQ2= a2//2 =/'//, it follows that the squares of the mass particles
which, placed at the inverse points B, C, make the sphere a level surface of zero
potential are proportional to the distances of those points from the centre. The
centre oj gravity of masses proportional to "L/E2, - 1/.E'2, placed at B, C respectively,
is at the centre A of the sphere. This result enables us to apply any of the theorems
relating to the centre of gravity given in vol. i.
202 ELECTRICAL ATTRACTIONS. [ART. 401
stratum thus formed is in equilibrium under its own repulsion
and the action of the electricity E at the external electrical
point B.
The potential of the stratum at any point R outside the
sphere is equal to that of the electrical point E' and is therefore
E'fCR. At a point R inside the sphere, the potential of the
stratum is equal and opposite to that of the external electrical
point E, because the sum of these two potentials is zero at all
points inside the conductor.
399. If the sphere is the boundary of a cavity in the conductor,
we remove the electrical point B. The surface density p, when
the electricity is acted on by the electrical point E' = — Ea/f
situated at G, is the same as that just found for a solid sphere,
but with the sign changed (Art. 395).
The potential of the stratum at any point within the sphere is
therefore equal to that of E at the same point. The potential of
the stratum at a point outside the sphere is equal and opposite to
that of E'.
Another proof. Let us surround the system by a sphere of infinite radius with
its centre near C. The point B is now included within the boundary formed by the
given sphere and the infinite sphere. We remove the point B and spread its
electricity over the double-sheeted boundary. By Art. 386 the quantity on the
given sphere is equal and opposite to that of C and is therefore - E' ; the quantity
on the infinite sphere is therefore E + E'. Since this is a finite quantity spread on
a sphere of infinite radius, both its potential and attraction at points near B or G
are zero. This electricity may therefore be removed from the field.
4OO. We may at once deduce either of the results given in Arts. 398, 399 from
the other by an easy invertion with regard to the centre A of the sphere, the radius
being the radius of inversion (Art. 86).
When the sphere is an outer boundary the potential of a charge E at B together
with that of a surface distribution p on the sphere is zero throughout the interior.
When we invert this system with regard to the centre, the distribution on the
sphere is unaltered while the charge at B is moved to C and becomes Eajf
(Art. 168). It follows, from Art. 177, that the potential of the same distribution
on the surface of a spherical hollow together with that of a charge Eaff at C is zero
at all points outside the sphere. This distribution is therefore in equilibrium when
placed on the inner surface of a conductor and acted on by the charge at G. This
last result is the same as that obtained in Art. 399 except that the signs of both E'
and p have been changed.
401. To find the surface density p at any point Q in terms of
the distance of Q from either electrical point.
The outward normal force F due to the repulsions of the
points at B and G is the resultant of E/BQ* and E'/CQ2, see
ART. 402] SPHERICAL CONDUCTOR. 203
figure of Art. 397. Hence resolving perpendicularly to CQ, we
have -rr^ s:
But BQsinBQC : AQsin AQC is equal to the ratio of the per-
pendiculars from B and A on CQ and is therefore equal to
„ E BG . a
BC-.GA. ...r—j^-gj-.
By similar triangles we have BG/CA = fr/a?, hence (Art. 369)
Eb*
4<irpa = aF=-jjQ-3 (1).
In the same way we find by resolving perpendicularly to BQ,
ar- (2).
Either of these results may be deduced from the other by
using the known relations E/b + E'/c = 0, b/BQ = c/CQ. Art. 397.
If the sphere is the boundary of a solid conductor, F is to be
measured outwards from the sphere into the non-conductor, and
these expressions give the density at any point. If the sphere is
the boundary of a cavity, the force F must be taken positively
inwards and the signs on the right-hand sides of (1) and (2) must
be changed.
In both cases let the point (B or C) at which the acting charge is situated be
called 0 and let the charge (E or E') be called E1. If &2 be the product of the
segments of a chord drawn through 0, the surface density p at any point Q on the
sphere is given by ±trpa = - — T .
402. In the case of a solid conducting sphere we may super-
impose a uniform stratum of any surface density p0. This addition
changes the potential to V0, where V0 = 4nrp0d*la. If p' be the
resulting surface density at any point Q, we have
*xTTQ = 7>/~»Q = I /-»/->» • \'*/'
a a BQ3 a a CQ3
The quantities of electricity on the sphere due to the two
strata respectively are V0a and E', and the total quantity is
E" = V0a + E' where E'/a = — E/f. The potential at any external
point R is the sum of the potentials of two electrical points, one of
mass E' placed at C, the other of mass F0a = E" — E' placed at the
centre A of the sphere.
204
ELECTRICAL ATTRACTIONS.
[ART. 405
403. The surface density may also be easily found by the method of inversion *.
Proceeding as in Art. 176 we invert the theorem "the potential of a thin spherical
stratum of density p0, radius a, at an internal point P is F0=4irp0a2/a." Let k'2 be
the product of chords through the centre 0 of inversion, so that the sphere inverts
into itself. We immediately arrive at the theorem "the potential at a point P'
of a thin spherical stratum of density Po' = Pv W OQ)3 is V0' = V0 (fc/OP')," where P
lies on the side of the sphere opposite to 0.
Since the expression just found for F0' is clearly the potential at P' of a particle
F k placed at 0, it follows that " the sum of the potentials at P' of the electrical
stratum and of the particle ( - F0fc) placed at 0 is equal to zero." Let the arbitrary
density p0 be such that - V0k = E, then the sum of the potentials at F of a stratum
whose density p= — (777; ) . and of the particle E placed at 0, is zero, and is
Qira yc/^/y
therefore constant throughout the space on the side of the sphere opposite to 0.
If the sphere is to be a solid sphere of conducting matter we place 0 outside, say
at B in the figure of Art. 397. If the sphere is to be a cavity in a conducting
medium we place 0 inside, say at C in the same figure. In either case the density
of the stratum at any point Q when acted on by an electrical point of mass E at 0 is
— Ek^
given by ±irp= — -$ , and the conducting matter is at zero potential.
U> • *J\g
404. In the case of a solid conducting sphere the potential of this stratum
alone at any point R' within the sphere (being equal and opposite to that of E)
is - EjOE'. Placing E' at the centre, we see that the quantity E' of electricity on
the sphere is given by E'/a= -Ejf where /= OA. We also find the potential F at
any external point B by inverting the stratum with regard to its centre A as in
Art. 399. The stratum is unaltered by this inversion. Its potential at R is
therefore F= ( - TTHT ) ~rU. If 0' be a point such that AO . A0'= a2 the triangles
\ OH / AM
OAR', O'AR are similar and OR' . AR= O'R . OA. The potential of the stratum at
any point R is therefore E'lO'R and is equal to that of a particle of mass E' placed
at 0'.
405. Lines of force and level surfaces. To simplify the discussion of these
curves, let us consider the case in which the sphere is at potential zero. We may
* The first determination of the law of distribution of electricity on a sphere
when influenced by an electric point was made by Poisson whose method required
the use of Laplace's functions. Sir W. Thomson discovered the powerful method
of inversion and used it to obtain an easy geometrical solution of this problem.
He also expressed the surface density in a much simpler form than that given
by Poisson. See Section v. of the Reprint of his papers on Electrostatics and
Magnetism. The first application of Green's theorem to this problem is to be found
in Maxwell's Treatise on Electricity, &c.
ART. 405] LINES OF FORCE AND LEVEL SURFACES. 205
then represent the attractions by two electrical points E, E' situated at B, G ; in
our case E'= -Ea/f is negative. We shall put a/f=n for brevity, and we notice
that 71 <1. See figure on page 206.
The equations of the curves have been found in Art. 323. If (r, 0), (/, 0') are
the polar coordinates of a point P referred to B and G as origins, EGA being the
axis of reference, the lines of force and the level curves are given by
E cos e+E' cos d'=K, E/r + E'lr'=K'.
It is clear that the lines of force emanating from one electrical point must either
pass to the other or proceed to an infinite distance, (Art. 114).
When a line of force proceeds from B to an infinite distance we equate the
values of K at B and at infinity. Since the radius vector at the origin B coincides
with the tangent and the angle &' is there equal to it we have
E cos 00 -E'=(E + E') cos /3,
where 00 and j8 are the angles the tangent at B and the asymptote make with the
axis of reference B CA. Since cosjS must lie between ±1, we see that cos00 must
lie between - 1 and 1 - 2n.
When a line of force proceeds from C to an infinite distance we have
E+E' cos 00' = (E + E') cosp,
hence cos 00' must lie between 1 and (2 - n)/n. Since the latter fraction is greater
than unity, no line of force can pass from C to an infinite distance, except that
which coincides with the straight line BCA*.
When a line of force proceeds from B to G we have
E cos 01 - E' = E + E' cos #/,
where 0lt 0/ are the angles the tangents at B, C make with BCA. As cos0j
decreases from unity to 1 - n, the sign of cos #/ is negative and the lines of force
arrive at C on the side nearest B. When cos 61 = 1 - n the line of force at C is
perpendicular to EGA. When cos 8i lies between 1 - n and 1 - 2n the sign of cos 0^
is positive and the lines of force enter G on the side furthest from B. When
cos 01 < 1 - 2n the line of force goes to an infinite distance.
To trace the level surfaces we proceed as in Art. 134. The level surface which
passes through the point of equilibrium X governs the whole figure. This point
lies in BG produced so that CX=BX*Jn. There is a conical point at X, the
tangents making an angle ±tan~1N/2 with BCA produced (Art. 121, Ex. 2). This
surface when complete consists of two sheets ; one of these passes between B and C
because its potential is less than the infinite positive potential near B and
algebraically greater than the infinite negative potential near C. The other
sheet cuts AGE beyond B because its potential is less than that near B and
greater than that at an infinite distance. The two sheets therefore turn from X
towards B and C, one enveloping C only and the other both B and (7. This level
surface is represented by the thick line in the left-hand figure. Its potential is
E(l-,Jri)*ll, where BC=l.
The other level surfaces fill up the enclosed spaces and surround the two sheets.
A few of these are represented by the dotted lines. The level surfaces within the
smaller sheet and those outside both are at potentials less than that at A', while
* Since the sphere of zero potential surrounds C (but not B), it is clear that no
line of force (except CA produced) could proceed from a point on that sphere to a
point at an infinite distance at which the potential is also zero (Art. 114). It is
also clear that there can be no points of equilibrium except on the axis BCA.
206
ELECTRICAL ATTRACTIONS.
[ART. 40 G
those between the two sheets are at greater potential. The complete level surfaces
whose potentials are less than that at X and greater than zero are therefore two-
sheeted surfaces. The two sheets of the level surface of zero potential are a sphere
inside the smaller sheet and a sphere of infinite radius.
Since X lies on the axis of reference, any line of force which passes from B to X
is defined by E cos Ol - E' = E + E' or cos Bl = 1 - 2n. If the potential is to continue
to decrease, this line must make a sharp turn at X, Art. 114. By doing this it
could either reach C or proceed to an infinite distance in the direction BOA. The
line of force from B to X is represented by the thick line in the figure on the right-
hand side. The closed surface formed by all the lines of force which proceed from
B to X separates the other lines of force into two systems, those inside this space
pass from B to C, those outside proceed from B to an infinite distance. A few of
these are represented by the dotted lines.
The figures represent the level surfaces and lines of force due to two particles
placed at B and C. When C is surrounded by a spherical conductor the lines of
force are cut off by the sphere, and exist only outside the sphere. The potential
being constant within the conductor the level surfaces become indeterminate.
The figures are only roughly drawn. The outer sheet, for example, of the level
surface which passes through X should be very much larger.
4O6. Ex. 1. A sphere charged with a given quantity of electricity is acted on
by an external electrified point jB and a tangent from B touches the sphere at D.
Prove that the potential at D of the heterogeneous stratum of electricity is the
same as if it were homogeneous and its density equal to that of the heterogeneous
stratum at D.
Ex. 2. A conducting sphere (radius a) is at potential zero under the action of
a quantity E of electricity at a point B distant / from the centre A. The sphere is
cut by a plane perpendicular to the diameter BA. Prove that the quantity of
electricity on the side remote from B is \^ (*- -J^) where Q is any point
on the curve of section.
Problems of this kind may be solved in three ways : (1) by Gauss' theorem the
quantity Q on any area is given by 4irQ=Ew+E'u' as explained in Art. 396;
(2) the heterogeneous stratum is known to be inverse of a homogeneous stratum'
hence Q/k is equal to the potential at the centre B of inversion of the corresponding
portion of the homogeneous stratum ; (3) the result may be obtained by direct
integration.
Ex. 3. Prove that the potential at any point R on the diameter BA of
the electricity cut off by the plane section as described in the last question is
ART. 407] CYLINDRICAL CONDUCTORS. 207
Eb* 1 /RQ RH\
~ W ' BR OR \Bd~ BH) where S 1S the P°mt of mtersection of BA produced
to cut the sphere.
Ex. 4. A spherical insulated conductor, charged with a given quantity E" of
electricity, is in a uniform field of force defined by the potential Fx. Prove that the
surface density at Q is given by 4irpa? = E" -3aFx where x is the abscissa of Q
referred to the centre. [In Art. 401 make B very distant and E/J3Q2= --F.]
Ex. 5. A solid sphere being charged with a given quantity E" of electricity is
acted on by an electrical particle of mass E situated at a distance / from the centre
A of the sphere. Prove that the mutual repulsion between the sphere and particle
77t// JJT 77*2^
_ £j £/ Ji*tt
/2 f3 (/2-a2)2'
Thence show that if the sphere be close enough to the particle, the mutual force
is attractive ; and if the sphere is uncharged the force is attractive at all distances.
If the sphere be allowed to fall from rest towards the particle find the velocity in
any position.
Ex. 6. A unit charge is brought to a point B, at a distance / from the centre
of an insulated sphere, of radius a and charge E ; prove that the total work done is
7-a/'(*-aV [Coll. Ex. 1897.]
Ex. 7. Outside a spherical charged conductor there is a concentric insulated
but uncharged conducting spherical shell which consists of two segments : prove
that the two segments will not separate if the distance of the separating plane from
the centre is <a6/(a2 + 62)^, where a, 6 are the internal and external radii of the
shell. [Coll. Ex. 1897.]
Ex. 8. If a uniform circular wire charged with electricity of line density - e is
presented to an uninsulated sphere of radius a, the centre of which is in the line
through the centre and perpendicular to the plane of the circular wire, prove that
the electrical density induced at any point on the sphere, whose angular distance
from the axis of the ring is 8, is
/2-a2 Eefsina
ira (a2 - 2af cos (0 + a) +/2}* {a2 - 2af cos (0 - a) +/2} '
where /is the distance of any point of the ring from the centre of the sphere, a is
the angle subtended at the centre by any radius of the ring, and
aa-2a/cos(0 + a)
[Math. Tripos, 1879.]
The density at any point Q of the sphere due to an element of electricity
m=e/sinod(20) at a point B on the ring is given in Art. 401 and is a known
multiple of m/BQ?. To effect the integration between the limits 0 and %ir we first
prove by geometry that BQ is a known multiple of A=^/(l - fc2cos20) and then use
the theorem (1 - fez)jA~3d<^=jA~1d0. This analytical result may be obtained by
differentiating sin0cos0/A and then integrating the result between the limits
0 and %ir.
407. Electricity on cylinders. We may apply either
Green's theorem or the method of inversion to find the distri-
208 ELECTRICAL ATTRACTIONS. [ART. 408
bution of electricity on an infinite circular conducting cylinder
when acted on by a thin uniformly electrified non-conducting rod
placed parallel to the axis either inside or outside.
Referring to the figure of Art. 397 we see that since CQ/BQ
is constant, log CQ — log BQ is also constant for all points on
the circle. Let two non-conducting thin rods infinite in both
directions be placed at B and G perpendicularly to the plane of
the paper ; let these rods be uniformly and equally electrified but
with opposite signs. The infinite cylinder whose cross section is
the circle is then a level surface of the two rods (Art. 43).
If the cylinder is the boundary of a solid conductor, we remove
the electrical rod C and distribute its electricity over the cylinder.
The repulsion of the stratum at any external point R is the same
in direction and magnitude as that of the rod C. Its magnitude
is therefore 2m/CR, where m is the line density of the rod. At
any internal point R' the repulsion is equal and opposite to that
of B, Art. 365.
If the cylinder is the boundary of a hollow in a conductor we
remove the rod B. The distribution of the electricity on the
cylinder is the same as that found for the solid cylinder but
opposite in sign.
To find the surface density p at any point Q we follow the
analysis in Art. 401. We notice that the attractions are 2m/BQ
and - 2mfCQ instead of EjBQ* and E'/CQ2. Making the corre-
sponding changes in the result we find that for a solid cylinder
The external rod has here the positive line density m. If the
cylinder is hollow and the internal rod has a negative line density
— m, the sign of p must be changed.
4O8. The same results follow from the method of inversion. Thus let the
rod be inside the cylindrical hollow as at C. We know, by Art. 183, Ex. 2, that if
the surface density at Q is proportional to 1/OQ2 the attraction at all external
points is the same in magnitude and direction as if the attracting mass were
equally distributed over the rod C. The condition of equilibrium is that the
attraction due to both the surface density and the rod should be zero at all
external points. This is satisfied if the surface density have a sign opposite to that
of the line density of the rod.
The result for the case in which the rod is outside a solid cylinder may be
deduced from this by an inversion with regard to the axis of the cylinder, see
Art. 399.
ART. 410] CYLINDRICAL CONDUCTORS. 209
409. Let the positions of the rods B, G be given; let 0
bisect BC, and let BG = 2t. Let us describe the system of
co-axial circles whose radical axis is perpendicular to BG and
passes through 0. Let the length of the tangent drawn from 0
to any circle be t, then obviously B, G are the evanescent circles
of the system. Let A be the centre of any circle, then
The points B and G are therefore inverse points with regard to
any co-axial circle. The cross section of the cylinder may be any
of these circles.
Since the line densities of the two rods are equal and opposite,
it follows from Art. 323 that the lines of force are defined by
01 — 62=K and the level curves by r1/rz = Kf, where (rlt 0j), (r2, 02)
are the polar coordinates of any point P referred to B and C
respectively as origins and BGA as the axis of reference. The
lines of force are therefore the circles which pass through B and G
and the level curves are the co-axial circles.
410. We may also find the law of distribution on two circular
non-intersecting cylinders (radii a, a) having their axes parallel to
each other and their charges equal and opposite.
Let A, A' be the centres of the two circles made by a
perpendicular cross section of the cylinders. Then two points
B, G can be found (and only two) which are inverse to each
other with regard to both circles. Each cylinder is a level
surface of two parallel rods passing through B and G equally
electrified but with opposite signs.
Let the cylinders be solid conductors, each external to the
other, and let them be separated by a non-conducting medium.
We remove each rod and spread its electricity over the cylinder
within which it lies, according to the law found in Art. 407.
Since the attraction of one electrified cylinder (say A) at all
external points is the same as that of the rod which was inside
its conducting matter, the attraction of the other cylinder (A1)
is in equilibrium when acted on by the electrified cylinder (A).
The electricity on each cylinder is therefore in equilibrium when
acted on by the other.
Several arrangements of the cylinders may be made. First, each cylinder may
be external to the other as just explained, or one cylinder may contain the other
and be separated from it by the non-conducting medium. In both these cases the
E. S. II. 14
210 ELECTRICAL ATTRACTIONS. [ART. 413
rods are removed and each cylinder is occupied by the electricity of the rod which
was within its conducting matter. Secondly, the cylinders may be separate hollows
in an infinite conducting medium, or one cylinder may contain the other with the
conducting medium between their surfaces. In these two cases the rods are not
removed ; each cylinder is occupied by electricity equal in quantity but opposite in
sign to that of the rod within the nearest non-conductor.
411. Ex. An infinite conducting cylinder of radius a is placed with its axis
parallel to an uninsulated conducting plane and at a distance c from it. The
cylinder is maintained at potential V, prove that the charge (m) per unit of length
V v —
is given by — = log — — - - . Prove also that the surface density at any point
— //£ Cl
of the cylinder is proportional to the distance from the plane. [Coll. Ex. 1880.]
Prove also that the mutual attraction between the cylinder and the plane is
ms/(c2-a2)i. [Math. T. 1888.]
[Let the points A, B (through which the rods pass as described in Art. 407) be
BO placed that the plane bisects their distance apart at right angles. Both the
plane and the cylinder are then level surfaces of the two rods.]
412. Electricity on planes. To find the distribution of
electricity on an uninsulated infinite plate when acted on by a
quantity E of electricity collected into a point B at a distance h
from the plate.
Draw BM perpendicular to the plate and produce it to C so
that MG = BM. The surface of the plate is
then a level surface of zero potential of E
placed at B and —EaiC.
The surface of the plate may be regarded
as a sphere of infinite radius enclosing con-
ducting matter on the side C. The elec-
tricity (-E) will then be in equilibrium if distributed on the
surface so that 4?rp is equal to the normal force at Q due
to the electrical points measured towards the non-conductor. We
%E 9 T?h
therefore have 4?rp = -- sin BOM = --
r* rs >
where r = BQ. The total quantity of electricity on the surface is
— E. We obtain the same results by inverting the sphere described
on BM as diameter.
413. Ex. 1. Show that half the whole electricity on the infinite plate is
comprised within any right cone whose vertex is at the influencing point B and
whose semi-angle is 60°.
Use the theorem 4wQ = 2Ew, Art. 396. It also follows that all areas on the
plate which subtend the same solid angle at the influencing point contain equal
quantities of electricity.
Ex. 2. Prove that the quantity of electricity, on one side of any straight line
X drawn on the plate, is Q= -Edjr, where 0 is the angle a plane drawn through X
ART. 414]
PLANE CONDUCTORS.
211
and the influencing point makes with the plate. The angle & is measured
on that side of X which makes Q and 0 numerically increase together, when X is
moved parallel to itself.
The solid angle w subtended at the influencing point is here enclosed by two
planes. These form a lune on the unit sphere whose area is 26.
Ex. 3. A spherical body with an electric charge E is at a height h above the
surface of the earth, the height being large compared with the dimensions of
the body. Prove that the body is attracted downwards with a force approximately
equal to £2/4ft2, in addition to its weight.
Prove also that its capacity is increased by the presence of the ground in the
ratio l + a/2ft: 1 approximately, where a is the radius. [Coll. Ex. 1900.]
414. The planes xOy, yOz intersecting in Oy are the boundaries
of a conductor ; the non-conducting medium being in the positive
quadrant. The system is acted on by an electrical point at A
whose coordinates are %, £ To find the distribution of electricity
on the planes (1) ivhen the angle xOz is a right angle, and (2) when
that angle is Trjn where n is an integer.
(1) Let us try to find a system of electrical points such that
the two planes xy, yz form part of one level surface. One of these
points must be at A, all the others will be inside the conductor.
Describe a circle centre 0, radius OA and let ABA'B' be a rect-
angle. Place at A' a quantity E of electricity equal to that at
the given point A and at B, B' quantities each equal to — E.
The planes xyt yz are then evidently level surfaces of zero
potential.
The surface density p at any point Q on the plane xy is then
given (Art. 412) by
The quantity Q of electricity on the plane xOy is given by
47rQ = ^Ea> = -4>E(OAxf- OB'x) = - 4>E0,
where 6 — angle AOB' and B'Ax' is a straight line parallel to Ox.
The quantity Q' on Oz is - Efffir where &=AQB.
14—2
212 ELECTRICAL ATTRACTIONS. [ART. 417
Ex. A straight line Y is drawn on the plane xy parallel to Oy. Prove that
the quantity of electricity on the side of Y remote from 0 is - E<f>lir where <f> is the
angle AYB'.
415. (2) If the angle xOz — 7r/3 we divide the circle into six
parts by three diameters and place A, A', A"; B, B', B" just
before and just after the alternate divisions. If we suppose each
A to be occupied by + E and each B by — E, it is obvious that
both the planes xy, yz are level surfaces of zero potential. In the
same way we may sketch the figure for the angle xOz = TT/H and
in all these cases the surface density at any point Q on either
boundary can be written down by Green's rule, (Art. 395).
Ex. Prove that the quantity Q of electricity on the plane xOy is - E (ir - 36)jir
where 0 is the angle AOx.
416. Ex. 1. A long rod uniformly charged with electricity is placed perpen-
dicular to a large conducting plane and with an end nearly in contact with the
plane; show that if the plane be put in connexion with the earth, the density
of the electricity induced on the plane will vary inversely as the distance from
the rod. [Caius Coll. 1880.]
Place a similar oppositely electrified non-conducting rod on the other side of
the plane. The plane is then a level surface of zero potential of the two rods and
the electricity can be found by Green's method.
Ex. 2. A uniformly attracting rod is placed parallel to a large conducting
plane. Prove that, if the plane is put in connexion with the earth, the density of
the electricity at any point of the plane will vary inversely as the square of the
distance from the rod.
Ex. 3. A conductor is bounded by the surface of a sphere, whose centre is at
the origin, and by the rectangular planes xy, yz ; the non-conducting medium
being the portion of the positive quadrant inside the sphere. The system is acted
on by an electrical point of given intensity, situated in the non-conducting medium,
whose coordinates are x, y, z. Find the surface density at any point of the
boundary. [Use seven other electrical points situated in the conducting me dium.] .
417. A simple condenser. Let a portion S of the surface
of a conductor A be so near the surface of another conductor A'
that the distance 6 between them at any point is a very small
fraction of the radii of curvature of each surface, and let /3, ft' be
the potentials of the conductors. It is required to find, to a first
approximation, the distribution of electricity on the neighbouring
surfaces.
Let P, Pf be two points on the conductors on the same line of
force ; p, p the surface densities at these points ; F, F' the forces
just outside the conductors at P,P' measured in the direction PP'.
Then 4>7rp = F, 4-rrp' = - F'. By Taylor's theorem
ART. 418] CONDENSERS. 213
8 — 3'
As a first approximation, we have F = F', and 4-Trp = p ;
$
o' _ /g
similarly 4nrpf = 2—-. — . Hence, 4?r times the surface density on
either conductor is ultimately equal to the fall of the potential
from that conductor to the other divided by the distance.
We notice that when the potentials ft, ft' are given the electrical
densities on the neighbouring surfaces can be made very great
by diminishing the distance 6.
If dcr be an element of the area S, the quantity of electricity
on S is ^-P^ (-£. This is (/3 - ft') SIM when the distance 6
47T J O
is constant.
If the conductor A' is joined to the earth, its potential ft' = 0,
and by the definition in Art. 371 the capacity of S becomes S/^-rrd.
To obtain a nearer approximation we take a second term in Taylor's theorem.
We then have p - /3= -j- 6 + \ -—5 -^ + &c.
dn dn2 2
Here, as before, dV/dn= -F, and in the small additional term we write for d2V/dn2
its mean value, viz. -(I*- F)/0. Substituting for F and F' their values iirp and
-4arpf, we find ^^ = - pd + ^^ B .................................... (1).
To obtain another equation connecting the nearly equal quantities p and - p', we
construct a tube of force joining P, P'. Let the areas at P, P' be d<r, da-', then
Fd<r=F'd<r', (Art. 127) and therefore pdff + p'd<r'=Q .................................. (2).
Let R, R' be the principal radii of curvature at P measured positively in the
direction P'P. Then, as in Art. 128, Ex. 2,
Solving these equations we have
These two approximations were given by Green in his Essay on Electricity and
Magnetism, pages 43, 45.
418. Ex. 1. A condenser is formed of two flat rectangular plates, each of
area A, which are very near together but not quite parallel, one pair of parallel
edges being at distance c and the opposite pair at distance c'. Prove that the
capacity is approximately ,. log -> .
The lower part of the condenser is fixed in a horizontal position and the other
is free to turn about a horizontal axis through the centre of its under face. Show
that a slight tilt which draws one pair of opposite edges together and the other apart
through I/nth of their distance will increase the capacity approximately by the
214 ELECTRICAL ATTRACTIONS. [ART. 419
fraction l/3ns of its value. Prove that, when the upper plate is delicately balanced
on its axis, charging the condenser will make its equilibrium unstable.
[St John's Coll. 1897.1
Ex. 2. A conducting plate A is inserted between two conductors B, B' terminated
by plane faces parallel to those of A. Let 0, B' be the distances of B, B' from the
nearest face of A, and let S be the area of either face of A. If B, B' be maintained
at equal potentials F2 and the potential of A be V1 , prove that the ratio of the
quantity Q of electricity on both faces of A to the difference Vl - F~2 is — ( - + — j .
[Coll. Ex.]
419. Cylindrical Condenser. A long straight electric cable,
consisting of a conducting cylindrical core surrounded by a shell
of non-conducting matter whose external surface is a co-axial
cylinder, is placed in deep water. The perpendicular sections of
the two cylinders are concentric circles whose external and internal
radii are a', a. To find the capacity of the cable.
Let mlt m2 be the charges per unit of length on the outer
and inner surfaces of the shell; a, /8 the potentials of the outer
and inner conducting media.
When a non-conducting shell separates two conductors the
sum of the potentials of the charges on the two surfaces of the
shell is constant (and therefore zero) at all great distances. It
follows from Art. 136 that the charges are equal and opposite.
The proof for the special case of cylinders is nearly the same as
for the general case. The potentials of the two cylinders at a
point P in the external conducting medium distant r from the
axis differ only by constants from 27/ijlogr and 2ra2logr (Art. 56).
The sum of these cannot be constant when r varies unless w-j = — w^
(Art. 365).
The potential of the inner cylinder at a point R in the non-
conducting shell differs only by a constant from 2m2 log r while
the potential of the outer is constant ; (Arts. 55, 56). The potential
at R of both cylinders is therefore
V = 2m.j log r + A
where A is a constant. The difference of the potentials at the
two surfaces of the shell is
a — fi = 27W2 (log a' — log a).
The capacity C of a length I of the core, measured by the ratio of
the quantity of electricity to the difference of potentials, is
~ a- (3~* log a! '/a*
ART. 420] CYLINDRICAL CONDENSER. 215
When the radii a, a' are nearly equal, the thickness of the
shell is very small and the capacity is very great. Putting
x = (a' — a)/a, the capacity becomes
n — i ^ ^a
^ ~ 2
x a — a
We may deduce this result from Art. 417. The capacity is
there proved to be approximately S/4<7r0, where the area S = Ziral
and the thickness 6 = a — a.
When the axes of the cylinders bounding the shell are parallel but not coincident,
we proceed in the same way. Let A, A' be the centres of a cross section of the two
cylinders ; a, a' the radii, a'>a. Let B, C be the points in which this cross section
is intersected by the two rods described in Art. 407, C being inside the core and B
in the water. Let m1, m^= -m^ be the line densities of the rods B, G respectively;
rls rs the distances from B, C of any point E between the cylinders. The potential
at E of the electric cylinders is (by Art. 407)
F= 2/»2 log ra + 2/Wj log r^ + A
When E is on the circle whose radius is a, we have r2lr1=alf, where/ is the distance
of B from the centre A. A similar result holds for the other circle. The difference
of potentials at the two surfaces is therefore
a - /3 = 2m2 (log a'// - log a//).
The capacity C' of a length Z of the core is therefore
(V
_ _
o-/3 * log of /a -log /' //'
where /, /' are the distances of the axes of the two cylinders from the external rod
B. Since /'>/, we see that the capacity is least when the two cylinders are co-axial.
420. Nearly spherical surface. To find to a first approxi-
mation the distribution of electricity on the surface of an insulated
conductor which is nearly spherical.
Let the given equation of the surface be expanded in a series
of Laplace's functions
r = a{l + Tl+T9+...} ..................... (1).
The term F0 has been omitted because all constants may be
included in the factor a. The terms Y1} Y2 &c. are so small that
their squares can be neglected. Let the required distribution of
electricity be p = D {1 + Z1 + Z2+ ...} .................. (2).
If the surface were strictly spherical, the distribution of
electricity would be uniform and every Z would be zero. It follows
that when the surface is nearly spherical each Z is of the first
order of small quantities.
Let (r, 6, </>) be the coordinates of any elementary area da- of
216 ELECTRICAL ATTRACTIONS. [ART. 420
the surface ; (r', ff, <f>) the coordinates of any internal point P.
Let dco = sin dddd<f>. The potential V of the electricity at P is
v- 1 pd<r
-]j(r*
where the limits of integration are 0 = 0 to TT, and <f> = 0 to 2?r.
This series is convergent for all positions of P which are at a
distance r' from the origin less than the least radius vector of the
surface. Let ty be the angle the radius vector r makes with the
normal to the element dcr, then r*da) = da- cos i/r. Since o/r is a
quantity whose square can be neglected, we have r2d(o = dcr.
The electricity is so distributed that the potential V is
constant throughout the interior, we therefore equate to zero the
coefficients of the several powers of r' in the series (3). Hence
for all values of n>0, -P» = 0 ........................ (4).
We now substitute for r and p their values given by (1) and (2)
and reject the squares and products of the small quantities Ylt F2,
&c., Zl} Zz, &c. We then have by Art. 290
J{-(n-l)Yn + Zn}Pnda> = 0 ............... (5).
4(7T 4tTT
Now fYnPndco = g^L. F'n, JZnPnda> = ^n Z'"> where F«> Z'n
are the values of Tn, Zn when 6', </>' are written for 6, $ ; Art. 289.
We thus find Z'n=Y'n(n-l) ........................ (6).
The conclusion is that the surface density of the electricity
on the surface (1) is
It may be noticed that the term Fj is absent from the expression
for p. The reason is that the surface r = a (1 + Fa) is approximately
a sphere when Fx is small, Art. 293, Ex. 3. The surface density
is then uniform.
If E be the quantity of electricity on the surface, we have,
since fYndco — 0 and the squares of Yn are neglected,
E = fpr*da> = 4,-jra?D.
This equation determines D when E is given. The potential
at the origin is V=fprda)
The capacity is therefore equal to a.
ART. 421] NEARLY SPHERICAL CONDUCTORS. 217
To find the potential of the stratum at an external point we
make the expansion (3) in powers of r/r. We then have
r«+2
+ &c. + Zm + (n + 2) Ym + &c.}
r
r/n+1 '
After substituting for Z'n its value, this reduces to
421. Ex. 1. The surface r=a (i + ficos-O), where /3 is very small, is charged
with a quantity E of electricity. Prove that the surface density is
(l-0sin20)E/47ra2.
Ex. 2. A nearly spherical conductor whose equation is r=a(l + 2wn) is enclosed
in a nearly spherical shell, the equation of whose inner surface is r=b(l + 'Zvn)
where un, vn are Laplace's functions of (0, <p). If the potentials of the solid and
shell are respectively a and p, find the potential at any point P between the conductor
find shell. See Art. 392, Ex. 1. *
The potential at P is, by Art. 283,
(1).
If the surfaces were truly spherical, the distribution of electricity on each would
be uniform and the expression for V would take the form A+B/r where A, B are
constants. It follows that Fx, F2, &c., Zlt Z2, &c. are in our problem small
quantities. Proceeding as in Art. 299 and rejecting the squares of small quantities
Z Z
we have ro + i^o + Y.2a" + &o. + — $• (1 - u^ - u2 - &c.) + -£ + <fec. =a,
F0 + Yl0 + F262+ &c. + (1 - wj - v2 - &c.) + + &c. -p.
Equating the functions of like order, we find
These give
Substituting these values in (1) the potential at any point P is obtained, the
equations of the two surfaces being given. The surface density p at any point P
of the internal conductor is found from ±irp= -dV/dr, where after differentiation
we put r—a in the small terms and r=a(l + S«n) in the large term. We then find
(n + 2) a2"*1 + (n - 1) 62»+1} un
218 ELECTRICAL ATTRACTIONS. [ART. 421
where l/ju=52n+1-asn+1 and the summation 2 begins at n=l. The surface density
p' at any point of the external conductor is found by interchanging (a, b) and (u, v),
the sign of the first term being also changed. This problem is discussed in a.
nearly similar manner in Maxwell's Electricity.
Ex. 3. A shell is bounded internally by a nearly spherical surface whose
equation is r=fe(l + Svn) and is acted on by an electrical point situated at its
approximate centre. Prove that the electrical density p at any point P of the
surface is given by 4irb2p= - J3{1-S (n + 2)vn}, where the origin is at the electrical
point, E is the quantity of electricity at that point and the summation S begins
at n=l.
Ex. 4. A nearly spherical conductor, which is also a solid of revolution with
the approximate centre near the axis, is placed in a uniform field of force whose
potential is MX where the axis of x is the axis of the solid conductor. Find the law
of distribution of electricity on the surface when the charge is given.
The surface being one of revolution about the axis of reference and also-
nearly spherical, its equation referred to an origin on the axis can be expressed
in the form
r=a(l + .41P1 + .42P2 + &c.) (1),
where all the coefficients Alt Aa, &o. are small. Similarly we may express the
surface density in the form
p=D (l+B1P1+B2P.i+&c.) (2).
If the conductor were accurately spherical, the expression for p would be of the form
D(l+B1coB0) (Art. 406, Ex. 4). It follows that when the surface is nearly
spherical the coefficients P.2 , B3 &c. are small, but B1 is not necessarily small.
Proceeding as in Art. 420, we make the potential at an internal point H
whose coordinates are (/, 6') equal to a constant K.
+ Mr'cosO'=K (3),
where q is the cosine of the angle between the radii vectores r, r1. Expanding and
equating the several powers of r1 to zero, we find
^1 Qn=0 or -McoaO' (4)
according as n>l or =1. Here Qn is a Legendre's function of q.
To find the constants B^ , B2, &c. it will be sufficient to put the point E in some
convenient positions. Let us place R on the axis, then q=p, the Legendre's
function Qn becomes PB, and cos0'=l. We then have when n=l
+B2P.i+. ..)Piaw=-M ........................ (5).
Since $PmP1dp = 0, this gives B1 = - SM/iirD ................................. (6).
When n>l we have, since B2 &c., A^ &c. are small
a + &G. - (n - 1) ,4^ - &c.} Pndp
-Bl(n-l)l{A1Pl + AtP9 + &e.}PJPndp = 0 ...... (7).
The first line presents no peculiarity and reduces to {Bn-(n-l)An}2/(2n + l).
Since P^=p the integral in the second line may be written 2AKfPKPnpdp. Now
by Art. 273 '(n + l)-P^fi-(2n + l)^PB+»Pw_1=0,
.-. (2n + 1) JP« P^ dp = (n+ 1) JP^P^ dp + n /P* Pn^ dp.
ART. 423] SPHERE WITH A RING. 219
This is zero except when K = n ± 1. We then have
The latter of these results follows also from the first by writing n - 1 for n. The
second line of the equation (7) becomes
_B n~l [A 2(M+1)u 2n
*a2n + l ' p*1 2n + 3 +^n-12w-
Finally we have, when n>l,
If E' be the quantity of electricity on the surface we have
Substituting in (2) the values Slt Bn and D given in (6), (8) and (9) we find the
value of the surface density p when the surface of the conductor is given.
The potential at the origin is K=&irDa (1+ ^A^j).
422. Sphere with a ring. Ex. 1. A uniform circular wire (radius 6),
charged with electricity of line density - e, surrounds an uninsulated concentric
spherical conductor (radius a). Prove that the electrical density at any point
of the surface of the conductor is
Ex. 2. A uniform circular wire (centre C), charged with electricity of line
density -e, influences an uninsulated spherical conductor (centre 0), the plane
of the wire being perpendicular to 00. Prove that the electrical density at any
point E of the surface of the conductor is
-£r- s (2n + *) p» (cos a) p» (cos *) ( ?
where S implies summation from n = 0 to 71=00. Also a is the radius of the
sphere, 6 the distance of any point M on the rim of the ring from 0, a. the angle
subtended at 0 by any radius of the ring and & the angle OR makes with the axis
OC of the ring.
The potential of the ring at any point Q on the axis referred to O as origin is
FI = -TTTX-oT^— T^n = T 2 Pn (cos a) ( I
and M =• - 2irbe sin a. The potential at any point S not on the axis is found by
introducing the factor Pn (cos 0) into the general term, where 0 is the angle COS.
The potential F2 of the spherical layer is given in Art. 294. The sum of the two
potentials being zero, the value of Yn follows at once.
423. Orthogonal spheres. The boundary of an insulated
conductor is formed by two orthogonal spheres. Find the law of
distribution of a charge of electricity*.
* The problem of finding the law of distribution of electricity on two orthogonal
spheres when acted on by an electrical point is solved in Maxwell's Treatise on
Electricity dbc. He also gives the solution for spheres intersecting at an angle ir/n,
for three and also four orthogonal spheres.
220
ELECTRICAL ATTRACTIONS.
[ART. 424
Let A, B be the centres, a, b the radii and let AB cut the
plane DD' of intersection of the
spheres in G. Then, as before,
the distances of A, B, 0 from D
are a, b, c. Let mass particles
E, H, E" be placed at A, B, G
such that
E E' -E"
Since the sphere A is a level
surface of zero potential of the particles at the inverse points B, G
(Art. 397), it is a level surface of potential e of all the three particles.
In the same way the other sphere is also a level surface of the
same three particles and is at the same potential.
Using Green's theorem, we see that the quantity of electricity
(a + b — c) e, if distributed properly over the whole surface, will be
in equilibrium at potential e, (Art. 395).
The normal force at any point Q on the sphere A due to both
the points B, C, has been proved to be proportional to 1/CQ3 and
also to I/BQ3 (Art. 401). The normal force due to the particle
at A is E/a*. We have therefore
e . H e K
where H, K are some constants. Since two sheets of a level
surface intersect in the circle DZX, the normal force and therefore
p vanishes when Q is at D (Art. 122), that is when CQ = c or
BQ = 6. The density may therefore be written in either of the
forms
1 3) o ( / I)
The density at any point Q' on the other sphere is given by
424. We may also consider the solid bounded by the convex
portion of the sphere A and the concave portion DND' of the other
sphere. The quantity on the solid is then ea, the potential is e,
and the electricity is acted on by the external electrified points
E' =• eb, E" = — ec. The densities are given by the same formulae
as before, except that the sign of that on the concave portion must
ART. 426]
ORTHOGONAL SPHERES.
221
be changed, because the normal force outwards into the non-
conductor (Art. 369) tends towards the centre of the sphere B
instead of from the centre, as on the convex portion of that sphere.
425. Ex. When both portions are convex the quantities Q, Q' of electricity
on the spheres A, B respectively are
Q = J« {a - c + 6 + c (a2 - 62)/aZ>},
When one portion, as DND', is concave, the electricity on that portion is
426. To find the law of distribution of electricity on a conductor
bounded by the convex portions of two orthogonal spheres when acted
on by an external electrical point.
The two orthogonal planes xOy, yOz in the left-hand figure
are the planes of zero potential of four equal particles Alt Az,
B1} 52; Alf A2 being of positive and Blt B2 of negative mass,
see Art. 414. Let us invert this with regard to any point D.
Consider first the section by the plane xOz. The straight lines
Ox, Oz invert into orthogonal circles which intersect in D and in
another point D' lying in DO produced. The radii a, b of these
circles are arbitrary because D is any point. Let their centres
be A and B as represented in the right-hand figure. The circle
A1B1A2B% inverts into another circle cutting the two former
orthogonally and (being symmetrical about DOD') has its centre
K in DD'. The radius of this circle is such that the perimeter
passes through the inverse point of the arbitrary point Aj_.
Let the points Aly A.2 invert into F1} Fz and the points Blt _Z?2
into (?!, $3; all these four points lie on the circle whose centre is
K. Since the plane xy is a level surface of zero potential of Alf
222 ELECTRICAL ATTRACTIONS. [ART. 427
Bi} the inverse sphere (say the sphere whose centre is A) is a level
surface of zero potential of F^ Gl} Art. 179. It thence follows
that Flt G! are inverse points with regard to that sphere. In the
same way Pa, G^ are inverse with regard to the same sphere, while
F1} Gz and GI, Fz are inverse with regard to the sphere B. Thus
F&A, F^B, GsF2A, G^^B are straight lines. It also follows
that if Fl is external, the other three points Fz, Gly Gz are inside
one or other of the two spheres A and B.
The ratio of the masses m, m' at any two inverse points Q, Q'
is known by the rule m'/m = DQ'/k, Art. 168. The quantities of
electricity at F1} Fit Glt G2 are therefore proportional to their
distances from D. Let these be
The potential at D of each electrical point is therefore
numerically the same. We may also use the rule (proved in
the footnote to Art. 397) that the squares of the quantities of
electricity which occupy points inverse to a sphere, and make the
sphere to be of zero potential, are proportional to the distances
of those points from the centre. Thus E^/E1f2 = AFJAG^,
E^/El'2 = BF2/BG1 and so on.
If we take the convex portions of the spheres A, B to be the
boundary of a solid conductor, that boundary will be a level surface
of zero potential of the four particles at F1} G1} F2, G2. Hence
the quantity of electricity Q = e (DF2 — DG± — DG^) will be in
equilibrium under the influence of a quantity E1 = e.DF1 placed
at jF\ if distributed according to Green's law.
427. The surface density at any point P on the sphere whose
centre is A is found by considering the two doublets Flt Gj. and
Ft, <?,. We have by Art. 401
where a2 and /S2 are the products of the segments of chords of that
sphere drawn from Fl and G2. Since p must vanish when P is
any point D of the intersection of two sheets of a level surface,
we see that
Since D lies on the sphere B with regard to which Flt G2 are
inverse points, we may write G^DjF^D = b/F^, Art. 397. Also
ART. 431] ORTHOGONAL SPHERES. 223
a2 = (F1A)Z — a2 where a and b are the radii of the spheres whose
centres are A, B.
Let the position of the influencing point J^ be at an infinite
distance from the sphere. The electricity at Fl is then infinite
but its potential, viz. EJDFi, becomes the constant e. The
conductor being at zero potential, the sum of the potentials
of the electricities at the three remaining points 6r1} G2, Fz is
therefore — e. The positions of these points are evidently A, B
and G, where C is the intersection of AB and DD'. We thus fall
back on the case of a solid conductor charged with a quantity of
electricity e ( - DA - DB + DC) and at potential - e ; (Art. 423).
428. If we insulate the conductor and give it such a charge
that the potential becomes e, we have, by superimposing the
density found in Art. 423,
= . l - - Rf
429. The rule to find the distribution of electricity on two orthogonal spheres
a,t zero potential may be summed up in the following manner. The point Fl being
given, we seek (1) the inverse points of ^ with regard to the two spheres A and B,
let these be Glt G2; (2) the inverse point of Gx with regard to the sphere B or the
inverse point of (?2 with regard to the sphere A, let this be F%. These four points,
any F being taken with any G, form two doublets. The sphere is a level surface
of zero potential of each doublet. The ratios of the quantities of electricity at the
points of each doublet, and the resulting surface density due to each, follow from
the elementary rules given in Arts. 397, 401. The electricity at any G has an
opposite sign to that at any .F.
430. Ex. An uninsulated conductor consists of a sphere and an infinitely
large and infinitely thin plane passing through the centre B of the sphere. If it
be exposed to the influence of a given charge of electricity at the point F1 where
F-J3 is perpendicular to the plane, prove that G1 being a point on F^ produced
such that BG1 is equal to BFlt the superficial density at any point P on the
hemispherical surface nearest to ^ is proportional to 3 - .
[Math. Tripos, 1877.]
The infinite plane may be regarded as the limiting case of an orthogonal sphere.
"We then follow the rule in Art. 429. The inverse point of F-^ with regard to the
plane is Gfj, the inverse points of Flt G1 with regard to the sphere are GZ,FZ. The
given system of sphere and plane is a level surface of zero potential of these four
points. We use Green's method as explained in Art. 401.
431. Geometrical properties. Ex. 1. Prove (1) that the centre of each of
the three orthogonal circles lies in the radical axis of the other two, and that the
orthocentre of the triangle ABK formed by joining the centres is the radical centre
of the circles. Prove (2) that the diagonals of the quadrilateral .F^GjFjGj intersect
in the orthocentre of ABK.
224 ELECTRICAL ATTRACTIONS. [ART. 434
The first results have been proved in Art. 426 for the centre K and the two
circles whose centres are at A and B, and are therefore true for all the circles. The
diagonals intersect on the polar lines of A and B, and since the circles are orthogonal
this is also the intersection of the radical axes.
Ex. 2. Prove (1) that -L + -* *_ ~ = 0. Prove (2) that, if particles
Sf -C/s" ™*I 2
whose masses are proportional to l/^2, l/£28, - 1/-B?, -W are placed at the
points Fj, Fa, Gx, G2, the sum of their moments about every straight line is zero.
Prove (3) that the centre of gravity of IjEf and 1/-E22 coincides with that of 1/E'j2
and 1/-E'22 and also with the orthocentre of ABK.
We notice that the centre of gravity of each of the doublets l/^2, - 1[E\* and
1/.E22, l/.E'a3 is at A, Art. 397. Thus the centre of gravity of all four particles is
at A. Similarly it is at B and this is impossible unless the results (1) and (2) are
true. To prove the third result we take moments about the diagonals of the
quadrilateral F^FyG^.
432. Ex. A conductor is formed by the outer surfaces of two equal spheres,
the angle between the radii at a point of intersection being 2a-/3. Prove that the
cap'acity of the conductor is • ^ ~ a, where a is the radius. [Coll. Ex. 1899.]
2i *J o
This result follows by inverting with regard to A the second figure of Art. 414.
The inverse of the electrical point A contributes only the constant potential Ejk to
the inverse figure (Art. 180). Omitting this point, the inverse of the rest of the
system is in equilibrium at potential - E[k. By Art. 170 the mass of any portion
of either system is equal to k times the potential at A of the corresponding portion
of the other system. In this way without drawing the inverse figure we find both
the quantity of electricity on the spheres, and its potential. The ratio is the
capacity required.
The capacity of the inverse system is therefore k2V/Q where Q is the quantity
of electricity on the original system and V its potential at the centre of inversion.
In our case the point A in Art. 414 bisects the arc xz and k=a. Also Q= — E and
V is twice the potential at A of B plus twice that of A' plus that of B'.
433. The boundary of a conductor is formed by tlie external boundary of three
spheres which have a common circular intersection, each sphere making an angle ir/3
with the next in order. To find the law of distribution on this conductor we invert
the right-hand figure in Art. 414 just as we inverted the left-hand figure of that
article when we required the distribution on two orthogonal spheres (Art. 426).
Let the plane of the paper contain the centre of inversion D and be perpendicular
to the common intersection Oy of the three planes. These planes invert into
spheres whose centres Clt Cs, C2 lie on a straight line perpendicular to DO. Let
the planes Ox, Oz which bound the conductor invert into the spheres whose centres
are Clt <72, the third plane, which is entirely in the conductor, inverting into the
sphere whose centre is Gt. In the inverse figure therefore the centres of the
outer spheres are Clf <7S. Since these centres lie on the perpendiculars drawn
from D to the planes, the angles C1DGS, C8D(72 are each ir/3. These spheres have
a common circle of intersection and D is any point on that circle.
434. If the position of the centre of inversion D is arbitrary the six electrical
points in the figure invert into six Flf Olt F2, G3, Ft, G3 which lie on the circle
inverse to that containing A, B, &o. and the general results are very similar to
ART. 435]
SPHERICAL CONDUCTORS.
225
those obtained in Art. 426. If we place D on the circle ABA'B'A"B" (say between
A and z) our results will correspond to those found in Art. 423 by the use of Green's
method. Let us consider concisely this last case as presenting some novelty.
Since D lies on the circle ABA'B' &c. the arcs AA', A' A" &c. subtend at D angles
each equal to ?r/3; hence in the inverse figure also the angles subtended by F^F^,
F2F8, G^, G2G8, C^Cg and C8(72 at D are each equal to ir/3. So again F1G1,
F2G%, F3Ga subtend equal angles at D. The six electrical points Flt Glt &c. now
lie on the diameter CjCtC9. Let a radius vector starting from DA turn round D,
it evidently passes in order through the points Flt Qlt C^; F2, G2, C3; F3, G3, C2.
The electrical points and the centres in the inverse figure are therefore arranged
from right to left in this order. By considering the triangles C1DC3, C8DC2 we see
that the three radii are connected by the equation l/rs=l/rj + l/r2. In the same
way if (&, £2, £8), (r^, i)2, ij8) are the distances of (Flt F2, F9), (Glt G,, Gs) from D we
have l/£2 = 1/& + l/£8 , !/% = l/i^ + 1/ijg . The perpendicular p from Don the straight
line CjCjC, is given by # is/fa8 + rza+r82)
G
F
F,
The points (Flt GJ, (F2, G3), (F3, G2) are inverse points with regard to the
point Ca; (Fs, G8), (Flt G2), (F2, GJ are inverse with regard to C3, and (.F2, G2),
(.FI, G8), (1^, <?j) are inverse with regard to (72. The arrangement of the suffixes
suggests an obvious rule to find the inverse of any point with regard to any sphere.
The point Fl being arbitrarily taken outside the spheres (7,, Cs, all the other five
are within the boundary.
The quantities of electricity at the points -F\, F2, G^ Ac. are respectively
#!=«. DFlt Ea=e.DFa, E\ = -e.DG^ &c. by Art. 169; the potentials at D of the
six electrified points are therefore numerically equal.
Since each sphere is a surface of zero potential of the six points Fl , Glt &c. we
may apply Green's theorem. In this way we can find the law of distribution on the
surface formed (say) by the two spheres whose centres are Clt C3 when acted on by an
electrical point situated at any external point .Fj on the diameter C^C^C^ .
435. Let us place the point F1 at an infinite distance from the spheres. Since
the attraction of JPj is then zero (though the potential is finite) we may remove this
point from the system. We now have the case of an insulated conductor bounded
as before by the spheres Clt C2 and charged with a given quantity of electricity.
The points G19 (?2, G, now coincide with Clt C3, C2 respectively. Also since
F^ , F^FS each subtend an angle ir/3 at D, the triangle F^FSD is equilateral. The
potential at D of F: is e . DF^DF-^ and is therefore equal to e. When the point F j
is removed from the system (which was at zero potential) the potential F0 of the
remaining five is -e. The quantity of electricity on the two external spheres is
the sum of the electricities at the five points and is therefore the sum of -erl%
-ert, -era, 2e.D.F2. The capacity is therefore
E. 8. II.
226
ELECTRICAL ATTRACTIONS.
[ART. 437
436. Three orthogonal spheres. Let ABC be any triangle ; a, 6, c the
lengths of its sides. Let AF, BG, CH be the perpendiculars drawn from A, B, C
on the opposite sides. Let O be the orthocentre. Let us describe three spheres
with centres A, B, C and radii a, /3, 7 such that the spheres taken two and two are
orthogonal. Then since the square of the distance between the centres of two
orthogonal spheres is the sum of the squares of the radii, we have
Let the chord of intersection of the circles whose centres are A, B made by the plane
of the paper intersect AB in S. Then
AS3 - BS*= a? - pP= 62 - a*=AH2 - BH*.
The point S therefore coincides with H.
The three chords of intersection of the
circles, taken two and two, are there-
fore the three perpendiculars AF, BG,
CH. If the lengths of these chords are
respectively 2/, 2g, 2h, we have af=fiy,
each of these being twice the area of the
triangle whose base is BC and altitude/.
Similarly bg=ya, cA=o/3.
A circle can be drawn about CFOG,
hence
AO. AF=AG. AC=bccosA
=4(&2+ca-a2)=a>.
Thus the systems of points (0, F), (G, C), (H, B) are each inverse with regard to
the sphere A. Similar results hold for the points in each of the lines through
B and (7.
Let us place at the points A, B, O ; F, G, H, quantities of electricity re-
spectively equal to ea, ep, ey; -ef, -eg, -eh, as explained in Art. 423. Also,
since F and 0 are inverse points with regard to the sphere A, we place at 0 a
quantity of electricity fl = ef.a/AF, (Art. 397). Since AF.a=2A. where A is the
area of the triangle ABC, we find 0=ea/}7/2A. It appears that O is a symmetrical
function of the radii of the spheres.
It follows that any sphere, as At is a level surface of zero potential of the
particles pkced respectively at (B, H), (F, 0), (C, G) while its potential due to the
particle placed at its centre is e. Each of three spheres is a level surface of potential
e of the seven particles placed at A, B, G, F, G, H and 0.
Let the external surfaces of the three orthogonal spheres be the boundary of an
insulated conductor charged with a quantity E of electricity, then the law of
distribution may be found by Green's method. If p be the surface density at any
point Q on the external surface of the sphere A, we have by Art. 401
ft I 0 \9 ( y \* f^ 1
> J1 . \ I _^
* 1* I TW^ I 1 7T7\ I T1 ^TToj
where t is the tangent drawn from F to the sphere A. If we wish to express our
results in terms of the radii a, j8, y, we may prove that
The potential is e and the quantity E is
E = e{a + 18 + 7 -/- g - h + a07/2A}.
437. The law of distribution on three orthogonal spheres may also be deter-
mined very simply by inversion. The three coordinate planes xOy, yOz, zOx are
level surfaces at zero potential of eight points, four of which are represented by
ART. 438] A SYSTEM OF CONDUCTORS. 227
A1B1A^B2 in the figure of Art. 426 and the other four are on the opposite side of
the plane of xz. The coordinates of these are ±«, ±y, ±z and the charges
numerically equal. After inversion with regard to any point D the planes become
orthogonal spheres. We may thus find the law of distribution on the external
surface of three orthogonal spheres at potential zero when acted on by an external
electrical point .Fj .
The three coordinate planes and a sphere whose centre is the origin are level
surfaces at zero potential of sixteen points, viz., the eight described above and
their inverse points with regard to the sphere. By inverting this system with
regard to any point D we find the distribution on four orthogonal surfaces at
potential zero when acted on by an external electrical point J\.
By proceeding as in Art. 427, we deduce the law of distribution when the
conductor is insulated and not acted on by an external electrical point. Finally,
by superimposing the two distributions thus arrived at, we obtain the law of
distribution when the conductor is insulated and acted on by the external electrical
point Fv
438. Theory of a system of conductors. Let Alt
A2,...An be a system of insulated conductors, each being ex-
ternal to all the others. Let pu, p^, ... be the potentials due
to a charge unity given to A1} the others being uncharged. In
the same way let pa, p-^y ... be the potentials when a charge unity
is given to A2 alone, and so on. If we give to A^ alone a charge
E! or to Az alone a charge E2, &c. these potentials will be respec-
tively multiplied by Elt Ez, &c. Superimposing these states of
equilibrium, we see that the potentials inside Alt A9, &c. when
charged with Elt E2, &c. are respectively
(1).
&c. = &c.
If we now solve these equations we have a second set of linear
equations which we represent by
El = quV1 + q2iV2 + ... ]
Ea = q1,V1 + q22Va + ... I (2).
&c. =&c. J
The coefficients pn, p12> &c. and qn, q^, &c. depend only on the
forms and relative positions of the conductors in the field and are
independent of the charges given to them.
The coefficients qu, q^, &c. (in which the two numbers forming
the suffix are the same) are called the electric capacities of the
bodies Alt A2, &c. The capacity of a conductor may be defined to
be its charge when its own potential is wnity and that of every other
conductor in the field is zero.
15—2
228 ELECTRICAL ATTRACTIONS. [ART. 439
The coefficients q^, qw, &c. (in which the numbers in the suffix
are different) are called the coefficients of induction. Any one of
them, as qrs, denotes the charge on Ag when Ar is raised to
potential unity, the potentials of all the conductors except Ar
being zero.
The coefficients pn, plz, P&, &c. are called the coefficients of the
potential. Any one of them as pra denotes the potential of A,
when a charge unity is given to Ar, the charges on all the other
conductors being zero.
Since the dimensions of potential are quantity/distance, it
follows that every coefficient of potential is the reciprocal of a
length. For the same reason every coefficient of induction has
the dimensions of a length.
439. To prove that prg = pw and qrs = qgr. Let the conductors
Ai...An when the charges are E^...En and the potentials V^..Vn
be called system I. Let the same conductors when the charges
are E^...En and the potentials F/...Fn' be called system II. Let
us treat these as independent coexistent systems.
The mutual work between two systems has been proved in
Art. 59 to be equal to the sum of the products of each element of
mass of either system by the potential of the other system at that
element. In the body Ar each element of electricity in one system
is to be multiplied by the potential of the other system at that
body, and the product is either ErVr' or Er'Vr. We may therefore
form the equation
ElVl' + EtVt'+...-El'Vl + E,'Vi + (3)
which may be shortly written ^EV = ^E'V.
Let us now put each of the electricities El,EZy &c., E^, E2f, &c.,
except Er and Eg, equal to zero. Then by equations (1), V8 =prsEr,
Vr'=pgrE8'. The equation (3) then gives prg = p8r.
In the same way if we put each of the potentials F1? F2, &c.,
F/, F2' &c. except Vr and V,' equal to zero we deduce from (2) and
(3)2ra = ?,r.
Ex. 1. Three small conducting spheres, whose radii are TJ, ra, rg, are placed
with their centres at the corners of a triangle whose sides a, b, c are very much
greater than the radii. Prove the following approximate relations
a? - r2r8 _ - (ab - crs) _ - (ac - fc;-2) _ 1 1112
a*r*ri9u~ abcr.^q^ " afccr2g18 ~ T-J r2 rt ~ o5^ ~ 6Vj ~ cV8 + abo '
Proceeding as in Art. 374 we find that the potentials Flf F2, F8, at the centres
of the spheres are given by three linear equations of the form V1=£1lr1 + E
ART. 443] A SYSTEM OF CONDUCTORS. 229
These correspond to equations (1) of Art. 438. Solving these we find E1 expressed
as a linear function of VltV^,Vt, the three coefficients are respectively gn, g12, q^.
Ex. 2. Two insulated electrified spheres (radii rlt r2) are at a considerable
distance c from each other ; prove that the coefficients of potential and induction
are approximately given by
440. The lines of force. Consider the lines of force which
intersect the surface of a conductor. Since at any point of the
surface 4nrp = — dV/dnt it is clear that the potential decreases or
increases outwards along these lines according as they intersect
the conductor at a point of positive or negative electricity, (Art. 114).
Let a point P travel along a line of force in such a direction
that the potential at P continually decreases. The line of force is
said to issue from or terminate at a conductor according as the
point P crosses its surface in an outward or inward direction.
It follows that a line of force can issue from a conductor only
at a point of positive electricity and will then either proceed to an
infinite distance or terminate at a point of negative electricity on
some conductor of lower potential.
If a line of force proceed from one conductor to another, it
joins points A, B on the two conductors which are oppositely
electrified.
441. If a tube of force intersect two conductors, the quantities
of electricity at the two ends are equal and of opposite signs.
Divide the given tube into elementary tubes ; let the areas at
the extremities A, B of any one of these be do; da-'. Let the
forces at A, B measured outwards from the conductors be F, F\
then Fd<T = -F'd<r', (Art. 127). Since kirp = F, kirp' = F', we
have pd<r = — p'dcr'.
442. The conductor of greatest positive potential can have only
positive electricity on its surface. For, if any element of its surface
were negatively electrified, a line of force could terminate at that
element. Such a line must have issued from a conductor of greater
positive potential. Similarly the conductor of greatest negative
potential can have only negative electricity on its surface. See
Art. 380.
443. To prove that all the coefficients of the potential (pn,
Pw> &C.) are positive and that the coefficient pr» is less than either
Prr Or psa.
Let the body Ar be charged with a positive unit of electricity
230 ELECTRICAL ATTRACTIONS. [ART. 444
and let all the others be uncharged. Then Vr=prr and Vg = prs,
by Art. 438. The body Ar cannot be entirely covered with
negative electricity and is therefore not the body of greatest
negative potential, Art. 442. Any other conductor A8 has both
positive and negative electricity on its surface and cannot be the
body of greatest positive or greatest negative potential. The
charged body Ar must therefore be the conductor of greatest
positive potential, and there is no conductor of greatest negative
potential. Hence all the conductors are at positive potential and
Prr >Pr$>
Let the body At be placed in a hollow excavated in Ar and completely surrounded
by it, then, since At is uncharged, there is no development of electricity either on
its surface or on the inside of the shell Art Art. 389. The potential throughout the
interior of Ar is p^ and hence in our present notation prt=ptr* In the same way,
if As is enclosed by a shell At, then prt=Prf
The case in which A, is enclosed by one of the other bodies is thus only a
limiting case of the theorem and is not an exception.
444. To prove that q^ is positive and qn negative, and that the
sum of the series S = qlr + q-zr + • • • + <lrr + • • • + qnr is positive.
Let the body Ar be charged to potential unity, all the others
being at zero potential. The charges given to the conductors
A1} AS, &c. are therefore qlr, <?•»•» &c« (Art. 438). The body
Ar is the conductor of greatest positive potential, its charge gw is
therefore positive, (Art. 442).
The body As is at zero potential. If there were a point of
positive electricity on its surface a line of force could issue from it
and must terminate at some point of lower potential, but there are
no such points. The body A, is therefore covered with negative
electricity, that is qrs is negative.
The unoccupied space outside the system is bounded by the
surfaces of the conductors and by a sphere of infinite radius.
Hence the potential at every point of this space lies between the
greatest and least potential on the boundary, (Art. 116). These
potentials are respectively unity and zero. The potential of the
system at a very distant point is the same as if the whole quantity
of electricity were collected into its centre of gravity (Art. 109)
and its sign is therefore the same as that of the series S. The
sum of this series must therefore be positive.
If At is enclosed by any body At and both are at potential zero, no line of force
can pass between At and the shell At. There is therefore no electricity on the
body Att (Art. 440), and in this case the charge gr,=0.
AET. 447] A SYSTEM OF CONDUCTORS. 231
If At is enclosed by Ar and Ar be at potential unity, At at potential zero, all the
lines of force between Ar and At must issue from Ar and arrive at A . The body
Aa is therefore charged only with negative electricity (Art. 440) and qrt is negative.
445. Ex. Prove that when r and « are unequal
and when r=s, the sum is unity. Thence show that the series represented by 8
in Art. 444 lies between 0 and !/#„..
The first two results follow from Art. 438, by putting Er=l, and every other
E = Q. The third follows from the first two, since prr>prt.
446. To find the mutual potential energy W of a system of
conductors. It has been proved in Art. 61 that W is equal to half
the sum of the products of each element of mass by the potential
at that element. As in Art. 439 this product for the body Ar is
ErVr. We therefore have
(4).
By substituting from equations (1) and (2) of Art. 438 we see
that this may be written in either of the forms
447. Ex. 1. Prove analytically that the expression for W is always positive.
Since qn is negative, let qrt= -pn. Hence by Art. 444 qrr>plr + p!ir+&o. It
follows from the expression (5) in Art. 446 that
...)F22 - 2/312F1F2±&c.
Ex. 2. A given charge is distributed over a number of conductors so that the
potential energy of the system when in electrical equilibrium is least. Prove that
the conductors are at the same potential. [Math. T. 1897.]
Make the expression (5) for W in Art. 446 a minimum with the condition that
Z.E is given.
Ex. 3. Energy of condensers. Two conducting surfaces are separated from
each other by a plate of some non-conducting substance so as to form a condenser; as
described in Art. 417. Find the potential energy.
Let p, p' be the potentials of the conductors ; p, p' the surface densities. Let
dS be an element of area of either surface, 6 the thickness of the conductor at this
element. The potential energy due to this element is (by Art. 446)
dW=$ppdS+%p'p'dS ................................. (1).
Since 4ir/> is equal to the fall of the potential divided by the thickness, we have
4^= (/3 -/JO/0, 4irp'=(p'-p)/e ........................ (2).
The capacity per unit of area, if measured by the ratio of the quantity of electricity
on either conductor to the difference of the potentials, is
C1=PI(P-P') ....................................... (3).
Using the equations (1) and (2) we can express dW in terms of either /3-£' or p.
We find dW=(p-pT"
232 ELECTRICAL ATTRACTIONS. [ART. 448
The value of W may then be found by integration. If 0 is constant at all points of
an area 5, and Q the quantity of electricity on that area, we have
In the case of a spherical conductor separated from a concentric conducting
shell by a thin non-conductor (Art. 392) we have S = 4ira2. The potential V at the
centre is - -- — = %, the capacity C is Q/K=a2/0. The potential energy is
a a + & a2
therefore TF=g=g = ^p.
As a second example, let the condenser be formed by a cylindrical conductor
separated from a concentric cylindrical shell by a thin non-conductor, (Art. 419).
The area of a unit of length is S=%ira. The capacity C' per unit of length is
?) which by (2) reduces to a/20. The energy per unit of length is
__
~~^ 2C"'
448. Junction of conductors. Ex. 1. Two conductors Alt A2, of a system
are joined together by a fine wire. Prove that the capacity of the united bodies is
qu + 2qM+qK. Prove also that this is less than the sum of the capacities before
the junction. [ColL Ex.]
Let the conductors be charged with such quantities of electricity ^ , £2 , &c.
that the potentials of Alt At are equal. By joining these no change is made in the
distribution of the electricity. The total quantity on the united bodies is £1+.E2,
and the n equations of Art. 438 become the following n - 1 equations
El + Ea = (qu + 2qlz + ?22) Vl + (gu + 228) F, + ...
<fec. =&c.
The results follow at once, since qia is negative.
Ex. 2. Five equal uncharged and insulated conducting spheres are placed with
their centres at the angular points of a regular pentagon. Another charged sphere
is moved so as to touch each in succession at the point nearest the centre of the
pentagon. Prove, that if ev..e6 are the charges on the spheres when they have been
each touched once
CJ-CL «!, 0
et-e1, e2, e1
€4 ~~ &\ i £ 3 , £j 4" £3
= 0,
0
= 0.
[Coll. Ex. 1901.]
Let A1...AS be the fixed spheres, At the moveable one. When A6 is close to Alt
but not touching it, we have six equations expressing FJ...FJ in terms of any
charges E^...E6 which may be given to them, (Art. 438). When Al and A6 touch,
El and Ee are so modified that F1=F6, but the sum E1 + E9 remains unaltered.
Equating the potentials Vl and Vt we see that E^ is a linear function of E3...Et.
Let this linear relation be
E1 = aE6 + p(E2 + Es) +y(Es + Et).
Since the five spheres are equal and arranged in a regular figure, this relation will
hold at each successive contact, provided El always represents the electricity on the
sphere which is being touched. We therefore have just after the contacts in order
have occurred,
e1=aEe, ea = a(E6
«4 = « (Ee ~ei - «3 -
ART. 451] A CIRCULAR DISC. 233
Eliminating o, p, y from these five equations we obtain the two results to be
proved.
449. Introduction of a conductor. An insulated uncharged conductor B is
introduced into the system of conductors Alt Aa, &c. Prove that the coefficient
of potential #„. of any one of the others on itself is diminished.
Let the body B be brought into its place as an uncharged non-conductor and
let it suddenly become a conductor. At this instant the potential energy of the
system, viz. £Z.EF, is not altered, because the E of the new body is zero. The
electricity is not now in equilibrium and must tend to assume a new arrangement.
It is a dynamical principle that when a system is in stable equilibrium the potential
energy is a minimum. It follows that in the new position of equilibrium the
energy is less than before.
To separate the effect on p^. from that on the other coefficients, let the
conductor Ar alone have a charge, all the others, as well as the new body B, being
uncharged. The energy before the introduction of B was \Erpn, and after that
event became ^-Ej/p'rr- ^e new value of the coefficient of the potential, viz. p'^.,
is therefore less than #„..
450. Potential Energy. Ex. 1. A conductor having a charge Q and being
at potential V0 is acted on by a quantity E of electricity situated at an external
point B; in this state the potential at an external point B' is V^,. The same
conductor with a charge Q' and at a potential F0' when acted on by E' placed at B'
has apotential VBr at B. Prove that Q'V0 + E'VB, = QV0' + EVB'.
This is the mutual work of the two states described above when regarded as
different systems, see Art. 439.
Ex. 2. An uncharged insulated conductor is acted on by a quantity E of
electricity situated at an external point B. Prove that the potential at any
external point B' is a symmetrical function of the coordinates of B and B'.
This theorem is also true if the conductor is uninsulated, for we may join it to
earth by a fine wire and include the earth as part of the system.
The first result follows from Ex. 1 by putting Q=0, Q'=0, E=E'.
Ex. 3. The locus of a point B at which a given quantity E of electricity must
be placed to develop a given quantity Q of electricity in an uninsulated conductor,
is that level surface of the same conductor (when insulated, charged to potential F0'
and not acted on by any external point) at which the potential is - QV0'IE.
451. A circular disc. To find the distribution of electricity
on a circular disc when acted on by an external electrical point
B situated in its plane*.
The electric density at any point Q on either side of an
insulated circular disc is p = ~-\ /no QR'\k wnere ^> ^ are
* The problem of finding the law of distribution of electricity on a circular
disc and spherical bowl when influenced by an electrical point was first solved by
Sir W. Thomson, see section xv. of the reprint of his papers. In the Quarterly
Journal for 1882 Ferrers found the potential due to the bowl at any point of space.
He uses the method of spherical harmonics. In the same Journal 1886, Gallop
applied Bessel's functions to find the distribution on a circular disc. He also
investigates the distribution on a spherical bowl and finds the capacity of the bowl ;
for this purpose he uses the method of inversion.
234
ELECTRICAL ATTRACTIONS.
[ART. 452
The internal
the intersections of a chord BQ with the circle,
potential is F0 and the quantity
M of electricity is Jf=2aF0/7r,
(Art. 382).
If we invert this with regard
to an external point B, with a
radius of inversion k equal to the
tangent BD, we shall obtain the
law of distribution of electricity
on the same disc when acted on
by an electrical point at B.
Let Q' be the point inverse to Q, then since R, R also are
inverse points,
QR QR' _ fr k* k*
QR' ' QR ~ BQ . BR BQ . BR " BQ*
by Art. 172. The surface density p' at Q' is given by
R'
_
~P
,BQJ 27r2 BQ'*'(QR'.QR)*'
The potential at any point P' within the disc is V0k/BPf. Put
F0fc = E, then the potential of a quantity — E situated at B
together with that of the distribution
o'-^.-i
" ~ 9^-2 z?n
is zero at all points within the disc. Here we have written
k2 = AB* - a?, Q'R' . Q'R = a? - AQ* where A is the centre of the
disc and a the radius.
The expression (1) gives the required surface density at any
point Q on one side of the disc when the internal potential is
zero, and the electricity at B is — E.
452. To find the quantity M' of electricity on the inverse disc
we use the rule M' = kV^ where Fi is the potential of the original
disc at the centre of inversion, Art. 170. This gives by Art. 384
M' = kM<j>/a, where <£ is the angle subtended by any radius of the
disc at the apex of the confocal spheroid through B. Since
M = 2aF0/7r and E = V0k, we have M = 2#</>/7r. Let a', c' be the
semi-axes of the confocal which passes through B, then tan <f> = a/c't
c'2 = a/a — a2 and a = AB. Hence (f> is also half the angle subtended
by the disc at the electrified point B, i.e. 0 = DBA.
ART. 454] A CIRCULAR DISC. 235
453. To Jind the potential at any external point P' in the
plane of the disc. The potential at P of the original disc is
F = — — = - — ° sin"1 — where a' = AP is the semi-major axis of the
a TT a
confocal through P, (Art. 384). The potential V at P' of the
inverted disc is therefore
F/ = 2Fo . _J_a BC.BP'\ k
IT Sm \pP' &2
where k is the length of the tangent BD, and C, the inverse point
of A, is the foot of the ordinate of D, see Art. 172.
454. To find the distribution of electricity on a plane circular disc, centre A,
when acted on by a quantity -E of electricity situated at a point 0 on the axis.
Let us cover the area of the plane outside the disc (regarded as a non-conductor)
7?h 1
•with a layer of electricity whose surface density at any point B is p= --z-^
and let this layer be fixed in the plane (Art. 412). Then if Q be any point on the
conducting disc, the induced density at Q is (by Art. 451)
, Eh f fxdOdx 1 1 fx*-a^
OB*
where x=AB, r=AQ, 6 is the angle QAB. We now substitute
where h= OA. We first integrate with regard to 6 between the limits 0 and 2ir,
using the integral / - 3 = .. T „ . To effect the integration with regard to x,
J i — e cos " v ( ~" ^ /
write a5=fetan^ and express the result in terms of cos^. The ordinary rules of
the integral calculus then show that we should put (&2+a2) cos2^=&2-j/2. The
limits for x being a to oo, those for y are 0 to h. We thus find
Eh t - tan-1 1 2_ft2+ra
"'
The result is that the potential due to the forced distribution p outside the disc
together with that due to the distribution p' on each side of the disc is zero at all
internal points.
Now by Art. 412 an electrical point -E situated at 0 and an infinite plane
whose density is that represented above by p (but with its sign changed) exert no
attraction at all points on the side of the plane opposite to 0, and the sum of their
potentials at all such points is zero.
Superimpose this second electrical system on the first ; then the forced
distributions outside the disc cancel each other. The sum of the potential due to
-E situated at 0 and that due to the electricity on the two sides of the disc is zero
at all points within the conducting substance.
The densities on the sides most remote from and nearest to 0 are respectively
mt-tair-1* „_ , Eh 1
> P ~P
These formulae represent the density at any point Q when the internal potential
is zero and the disc is acted on by an electrified point - E situated at 0.
236 ELECTRICAL ATTRACTIONS. [ART. 456
Here ta_ft +r\ n is easily seen that t=cotil/, where ^ is half the angle
a*-r*
subtended at 0 by the chord drawn at Q perpendicular to the diameter AQ.
455. To find the potential of the electrified disc at any external point P, and
also the quantity of electricity on the disc.
Consider two discs whose surface densities are respectively
By differentiation we find that -£* = -£± a* + h* and that at the rims where r=a and
tan-1t=^ir, the densities also have the ratio 2m to a2+/i2. Let Fx and F2 be the
potentials of the discs at external points similarly situated. Now dV^da is the sum
of the potentials of a disc whose density is dp^da and of an annulus round its rim.
It immediately f ollows that ^J = *£ _*!_ .
Now by Art. 384, Fj is the potential of the electricity on one side of a circular
disc charged with a quantity M=4ira, hence V1 = 2ir<f>. If we put m = Ehj'2ir*, F2
becomes the potential of a circular area whose density is the sum of the densities on
the two sides of the disc. We therefore have
_ 2Eh fa 1 d<f> ,
Fa= / —5 — rs-r- da,
2 r Jo a? + hzda
where <f> is the angle subtended by any radius of the disc at the apex of the confocal
spheroid drawn through P.
When the point P lies in the plane of the disc, the integration is easy. Let
x be the abscissa of P, then x is also the semi-axis major of the confocal through
P and sin0=a/z. We therefore have
= 2Eh f
r J
_x
When x is infinite, this takes the simple form F2=- . — tan-1 -. Now at a great
X 7T fl
distance, potential is mass divided by distance ; the quantity of electricity on the
O Tp „
disc is therefore — tan"1 - . This is the same as 2-E^/ir where ^ is half the angle
subtended at the electrified point 0 by any diameter of the disc.
When the point P is on the axis, we have tan 0= a/2 where z is the ordinate of
P. The potential is then
When e=h, this expression takes the form 0/0. We easily find however that the
potential at the electrified point 0 is
a 1 A
+ tan
, a\
"1 - ) .
hj
T \az+h* ^h
When the point P has a position defined by any values of x, e, both the process
of integration and the final result are somewhat complicated. The whole of the
work is given by Gallop in the Quarterly Journal, vol. xxi.
456. Spherical bowl. To find the distribution of electricity on an insulated
spherical segment with a plane rim.
ART. 458] A SPHERICAL BOWL. 237
The electrical distribution on the bowl may be deduced by inversion from that
on a circular disc at zero potential with a quantity
of electricity - E at a point 0 on the axis.
Let MN\>e the rim of the disc, O the centre
and OM=k the radius of inversion. We regard
O as the centre of a sphere of small radius e.
This sphere inverts into a large sphere of radius
fc2/e. The quantity M' on the inverted sphere is
given by M'=kV1 (Art. 180) and is evidently equal
to - Ek/e. The attraction at any internal point is
therefore zero but the potential is - £/&.
The disc inverts into the segment MA'N, the
sides nearest to or farthest from O corresponding
to the convex and concave sides of the bowl.
To deduce the density at Q' from that at Q we use the formula of Art. 169 as applied
to surfaces. Since fc/OQ'=OQ/A; and OQ2=/i2+r2we deduce that the surface densities
at any point Q' on the concave and convex sides are respectively
E h /* i w . Eh
The sum of the potentials of the electricity on the bowl and of that on the
sphere of infinite radius being zero, the internal potential F0 of the electricity on
the bowl alone is Ejk.
Let A'Q^r1, A'M=a', OM=k, and let the diameter OA' of the sphere be/.
We then have since hf=Jt?
The densities at any point on the concave and convex sides of the insulated
bowl then take the forms
where V0 is the internal potential.
457. To find the quantity M' of the electricity on the bowl, we use the rule
M'=kVlt Art. 170. We have therefore merely to write E = kV0 in the expression
for the potential of the disc at 0 (Art. 455) and to multiply the result by k. Let 2a
be the angle subtended at the centre of the sphere by any radius of the rim, then
a=Atana and h =f cos2 a. The quantity M' is therefore given by
The potential V of the bowl at any point P may be deduced from that of the
disc at the inverse point P. The result takes a simple form when P* lies on the
unoccupied part of the sphere. We then have
-n __ —
v (P'M.P'N)^'
where a'=A'M=fsina.
468. Ex. Prove that the density at any point Q' of a spherical bowl at zero
potential when acted on by a quantity -E' of electricity at any point B' on the
unoccupied part of the sphere is
,_ E' 1 /OM2-QB'2\»
238 ELECTRICAL ATTRACTIONS. [ART. 460
See the figure of Art. 456. This follows from the result in Art. 451 by inversion
with regard to a point 0 on the axis.
459. Electricity on two spheres*. Two electrified conducting spheres are
in presence of each other; it is required to find the resultant force due to their
mutual action. The spheres may be either insulated or maintained at a constant
potential, say, by being joined to a distant large reservoir of electricity by a fine
wire. The investigation depends on the following theorem.
The electricity on a sphere (radius a) is maintained at constant potential and is
in equilibrium under the action of any number of electrical points. Another
electrical point A, charged with a quantity E of electricity, is placed at a distance x
from the centre of the sphere. The new distribution of electricity may be repre-
sented by the addition of a layer on the sphere such that its potential plus that of
the electrical point A is zero throughout the interior. The potential of such a
layer at all external points is the same as that of an electric particle E'= -Eajx
placed at the image or inverse point of A. The increase in the quantity of
electricity on the sphere is then E', Art. 402.
If the sphere is insulated, the additional layer representing the change produced
by A must be such that its mass is zero. The potential of this layer at any external
point is equal to the sum of the potentials of two electric particles. One of these
has a mass E'= -Eajx and is to be placed at the image of A, the other has an
equal and opposite mass and is to be placed at the centre. The potential inside the
sphere is then increased by — E'/a or Ejx.
It is evident that the former case is less complicated than the latter. We shall
therefore in the first instance suppose that both the spheres are maintained at
constant potential, and finally deduce the case of insulation from the former.
460. Let the radii of the spheres be a, 6, and let the distance between the
centres A0, B0 be c. Since c is necessarily greater than either a or 6, we can
express the force in a convergent series by regarding a/c and b/c as small quantities
of the first order. Let the given potentials inside the spheres A, B be u, v. If the
distance c were very great the quantities of electricity on the spheres would be
ua=E and vb=F, and the mutual force of repulsion would be EF/c*.
The electrical point E placed at A0 will disturb the electricity on the sphere B.
The external effect of this disturbance is represented by a mass particle placed at
the image AQ' of A0 with regard to the sphere B. Since this mass is proportional
to E we represent it by Ep0'. In the same way the effect of the electrical point F
is represented by a mass particle Fq0' placed at B0' the image of B0 with regard to
the sphere A. For another approximation we seek the images of A0', B0' and so on
continually.
To fix our ideas, let 1, p0', p^ , p^, &c. denote the masses of the series of which
the first term is a mass unity placed at the centre A0. Then plt p3, &c. are within
* The problem of determining the distribution of electricity over two spheres in
presence of each other was attacked by Poisson in 1811, who expressed the results
by definite integrals, see M6m. de VInstitut. There is a solution founded on the
method of successive images by Kelvin, Phil. Mag. 1853, reproduced in his Papers on
Electrostatics and Magnetism, page 86. In Maxwell's Electricity, edition of 1892,
page 281, there is a short discussion of Kirchhoff's results by J. J. Thomson. He
also gives references to other papers on this subject. The principle of successive
influences was first enunciated by Murphy in his treatise on Electricity, 1833. In
the case of two equal spheres whose distance apart is 100 times either radius he
finds the difference of densities at the ends of the symmetrical diameter.
ART. 461] TWO SEPARATE SPHERES. 239
the sphere A ; p0', jp/, <fco. are within the sphere B. Let f0, /,, /2, &c. denote the
distances of 1, plt pa, &c. from A0; f0', //, &c. denote the distances of p0', p^, &c.
from B9. Then /0=0, /0'=62/c and so on. In the same way, if a unit of mass is
placed at B0, let 1, q0', qlt &o, denote the masses, and g0, ga', glt g^t &o. the
distances of the successive points of the corresponding series from B0 and A
alternately. Then 00=0, gQ'=a2/c, &c. We obviously have the following equations
n — nn
/ _ aPn
Jn+l—,_ f i » Pn+1— ._ f '
c /n c /n
The corresponding relations for the points of the other series are obtained by
interchanging a, /, p with b, g, q.
We notice that all the masses pn, qn are independent of the electrical conditions
of the spheres and are functions of o, 6, c only. If we regard a/c, bjc as small
quantities of the first order, pn and qn are small quantities of the order 2n, while
pn', qn' are of the order 2n + l. The distances /„, /n', gn, gn' are all of the second
order. We also notice that the distance between the masses ps, pt is /,-/«, the
distance between ps, qt is c-ft-gt, and so on.
The whole repulsion between each sphere and the other is equal to the repulsive
force exerted by the fictitious masses inside one sphere on those within the other
sphere. It is therefore represented by
&P.Pt **g.g/ 1 /<»
77^ T ; - ^ - TTTo ~T" T ~ 7\f>i ..... l^/>
ff (c-/.-//)2 (?-gt-9t?\
where the summations extend from s=0 to oo and t=0 to oo , and p0 — l, q0—i-
The total quantities of electricity on the spheres are
E'=E2pn + F2qn', F=ESpn' + F2qn ..................... (3),
where the summations extend from n=0 to oo .
It follows that (2) also represents the mutual force X between the spheres when
insulated and charged with quantities E', f of electricity.
461. When the spheres are not very close to each other it is sufficient to take
a few terms only of this doubly infinite series. Let us reject quantities of the order
EF/cz when multiplied by (a/c)* or (6/c)4. In this case we require only the repulsive
forces between the points E, Fq0', Eplt Fq± inside the sphere A and the points
F, JEp0', Fql , Epi within the sphere B. Taking any two of these we see (since all
the /'s and g'a are of the second order) that their distance apart may be regarded as
equal to c, except in the case of the particles E and Ep0' and the particles F and
Fq0'. The force between E and Ep0' is
with a similar expression for that between F and Fq0'. The whole repulsion is
therefore J+jCtf*£ltf ........................... (5),
where E'=E + Fq0'
F'=F+Ep0'+Fq1
It is evident that E' and F' are the quantities of electricity on the spheres.
240 ELECTRICAL ATTRACTIONS. [ART. 464
Since the masses 1, p0', plt Pi occupy successive inverse points we have when
terms of the fourth order are neglected
ja _ a2 _a2 ,_ &2 _ft2
/o'=-, /i=^r^,-7» ^1-^3^-7
:^, J,i'==rM=_
.(7);
ao- I
^fi~
these results follow directly from (1). Substituting these expressions in (5) and
again rejecting all terms of the fourth order, we find
EFf. Zab\ . E2 |b . 262(a + 6)) F* (a , 2a2(a + 6)]
~~ — "
where u=E/a, v = Fjb are the given potentials of the two spheres.
462. To find the force of repulsion when both the spheres are insulated we
notice that the expression (5) gives the force between the spheres when charged
with the quantities E', F' of electricity and that their potentials are respectively
u=Eja, v=.F/&. It follows immediately from (5) that
E'F' E'2 2V> _ F'* 2a8
" "" ~~ ........................... "•
463. When the spheres are close to each other the method of finding the
functions pn, qn, &c. by continued approximation becomes laborious. If we put
pn = 1/PB and eliminate /„' and pn' from the equations (1) we arrive at the equation
c2 — a2 — fc2
of differences Pn+i + P»-i = -- r - P».
The solution is obviously
1 c2 - a2 - 6s
-m, where ft+-r=
-r .
h ao
We shall suppose that h is the root which is less than unity. To find the constants
c2 — 6*
A, B we have by (2) the conditions P0=l, Pj= — r— . In the same way we find
that Pn' satisfies the same equation of differences, with the conditions
The reader will find methods of reducing the doubly infinite series for the force
X to a single series, and also a discussion of the case in which the two spheres are
in contact in Kelvin's Papers on Electrostatics, <6c., page 89.
464. Ex. 1. Two conducting spheres touch each other externally and are
charged with electricity. Prove that the density at the point of contact is zero.
[Use Art. 142.] [Murphy.]
Ex. 2. A conducting sphere, of radius a, having an electric charge E, is in
front of a large plane conducting surface connected to earth, its centre being at a
distance c from this surface, which is large compared with o. Prove that the
sphere experiences an attraction towards the plane which is approximately equal to
EP f a»\
4^ (l + 2?) ' [St John'8 C°1L 1897'J
Place on the other side of the plane at the same distance a second sphere of
equal radius and let its charge be - E. The required attraction is the force X, given
by (9), which one sphere exerts on the other (Art. 461).
Ex. 3. Two equal conducting insulated spheres of radius a are placed with their
centres at a distance c apart in a uniform field of force, of intensity F, and whose
ART. 464] TWO SEPARATE SPHERES. 241
direction is at right angles to the line joining the centres of the spheres. Show that,
if the spheres are initially uncharged, their mutual repulsion will be
]• [Math. Tripos, 1900.]
The force F acting alone would cause a distribution, of electricity on each sphere
whose surface density is p=ky, where 4:irak=3F (Art. 406, Ex. 4), and the straight
line joining the centres of the two spheres is the axis of x. The potential at any
external point of a thin layer of surface density ky placed on a sphere is the
y component of the repulsion of a solid sphere, of density ka, (Art. 92). This
potential, again, is the same as that of a small doublet, or quasi-magnet, whose
moment is a?F placed at the centre with its axis parallel to the axis of y, (Art. 316).
Such a doublet (strength m, length I), if placed at the centre A of one sphere,
would change the distribution of electricity on the other sphere. By Art. 459, the
changes produced by each mass m acts at external points like a second mass
particle m'= -ma/c, placed at the inverse point of the particle m. These form an
inverse doublet of strength m' and length Z'=Za2/c2. This inverse doublet has
therefore a moment — a*F(a/c)3 and is placed at the inverse point of the centre A,
with its axis parallel to that of the first doublet.
To find approximately the mutual action of the two spheres, we consider each to
be occupied by two doublets. The force exerted by one broadside doublet on
another is proved in Art. 320 to be X=3MM'jr*. The force exerted by one of the
larger doublets on the other is therefore 3a6.F2/e4. The force exerted by each large
doublet on the opposite small one is 3a3JF { - a?F (a/c)3}/r*, where r=c - a2/c. This
when doubled reduces to -6a9^2(c2 + 4a2)/c9. The force exerted by one small
doublet on the other is of an order higher than the terms given in the enunciation.
Adding these together we arrive at the result to be proved.
Ex. 4. Two spheres (centres A, B; AB=c; radii a, b) are charged with
electricity and mutually influence each other. Let / ( - \ and -/ ( - j be the
potentials of the sphere A at any internal and external point respectively, the point
being situated on the line AB (Art. 294). Prove that / must satisfy the equation
where h and k are the potentials of the two spheres.
If the spheres are in contact, deduce
i mz \ - — __ Jimz
L~m+z)~ a a{m+(l-m)z}>
where m=6/(a + 6) and r[a=l-z. Prove also that a solution of this functional
is ,/1.s)=
Deduce the potential of the sphere A at any point P not on the axis. See Art. 178.
To prove these results, let F be the function corresponding to / for the other
sphere. Equate the potentials inside the spheres to h and k. Then eliminate F.
See Poisson's two memoirs, M6m. de VInstitut, dc., 1811, pages 1 and 163. Also
Plana, Mem. de <&c. Torino, ser. n. vol. vn., 1845, and vol. xvi., 1854.
K. 6. II. 16
242 MAGNETIC INDUCTION. [ART. 466
Magnetic Induction.
465. When magnetism is induced in a neutral body A by the
influence of a magnetised body B, it is supposed that each element
dv of the volume of A becomes magnetic*. Let R be the resultant
magnetic force (Art. 342) on a positive unit pole situated in the
element, due to the influencing body B and the induced magnetism
in A. In an isotropic body the axis of magnetisation of the
element dv is in the line of action of the force R. The intensity
/ is nearly proportional to the force R provided that force is not
very large. We therefore put / = kR. When k is positive the body
is said to be paramagnetic and the direction of magnetisation
coincides with that of the force F, when k is negative the body is
diamagnetic and these directions are opposite. The value of k for
soft iron is positive and great, but for bismuth it is negative
and very small. Thus for soft iron k may vary under different
circumstances from 10 to nearly 200, but for bismuth (which is
one of the most highly diamagnetic substances known) k is about
1/400000. The coefficient k is called the magnetic susceptibility,
it is also called Neumann's coefficient.
466. Let U be the magnetic potential of the magnetism of the
influencing body B and ft that of the induced magnetism in A.
Let V= U + ft be the potential due to all causes. Let (I, m, n)
be the direction cosines of the direction of magnetisation of any
element dv of the body A. It immediately follows that
, T j
- -, Im = — k , - -, In= — k—±—; - - ...(1).
dx dy dz
We may in Poisson's manner represent the potential due to
the induced magnetism by that of a distribution of fictitious
matter throughout the volume and over the surface of the body A.
The density p of the former is given by
( dx dy dz )
Here we have introduced the condition that k is constant for
* The mathematical theory of induced magnetism was first given by Poisson,
Memoires de I'Institut, 1824. The difference between his theory and that of Weber
cannot be discussed here. The reader will find the fundamental principles of
induced magnetism explained in the reprint of Kelvin's papers. The theory of
Faraday and Maxwell, that the dielectric is the seat of a peculiar kind of stress,
does not come within the limits of a treatise on attractions.
ART. 469] MAGNETIC INDUCTION. 243
the body A. Since the element dv is outside the body B and
inside A we have
V2 U= 0, V'fl = - 47rp ; .-.(! + 47T/j) /? = 0.
It follows that when the magnetic susceptibility is constant, the
volume density p is zero. The potential of the magnetism induced
in the homogeneous isotropic body A at any internal or external
point may therefore be represented by that of an imaginary layer of
matter on the surface of that body. The surface density cr of this
layer is known by Poisson's proposition to be cr = I cos 0, see
Art. 339.
If F is the normal component of force at any point P close
to the surface but in the substance of the body, the surface
density at P is cr = JcR cos 6 = ± kF, the upper or lower sign being
used according as F is measured positively from P in direction
pointing outwards or inwards from the boundary.
467. The actual distribution of induced magnetism is both
solenoidal and lamellar. Since p = 0 the condition that the
magnetism is solenoidal is satisfied, (Art. 349).
The level surfaces due to the acting forces are defined by
U+ £l = c. Each element of the body is magnetised at right angles
to the level surface which passes through it, and, since / = JcR, the
intensity is inversely proportional to the normal distance between
two consecutive level surfaces c and c + dc (Art. 46). The dis-
tribution is therefore lamellar, Art. 851.
468. The boundary condition. Let -F1( F2 be the normal
components of the magnetic force due to all causes at points
P!, P2 respectively just inside and just outside the stratum but
situated on the same normal. Let these forces be measured
positively in each medium from the stratum on its boundary.
Then by Arts. 142, 466,
F1 + Ff = 4ira; <r = - kF,. ............... (2).
From these we deduce the equation
+ ^ = 0 .................. (3).
469. When two substances, both of which are susceptible of
induced magnetism, are separated by a surface S the conditions at
the boundary are slightly altered. Let fo, ka be their respective
susceptibilities, F^ Fa the normal components of the magnetic
16—2
244
MAGNETIC INDUCTION. [ART. 470
force in the two substances at any point P of the boundary 8, each
being measured positively from S.
The surface of each substance is bounded by a fictitious layer
whose surface densities are respectively
ff^-k.F,, o-2 = -k,F2 .................. (4).
The minus sign is used because each F is measured inwards as
explained above.
We then have by Green's theorem (Art. 142)
Eliminating <rlt o-2 we deduce the condition
(1 + 47T&,) F1+(l + 47T&2) F2 = 0.
The coefficient 1+4-TrHs called the magnetic permeability and
is often represented by the letter /x-. The equation then takes the
form piFi + f**Fa = 0 ........................ (6).
It is often convenient to measure the normal forces F1} F2 in
the same direction. Let either direction of the common normal to
the separating surface be chosen as the positive direction, we deduce
from (6) the following theorem. The normal forces just within the
two substances at any point of the boundary (when there is no
charge on the boundary) are inversely as the permeabilities of the
substances.
When the body is not susceptible of magnetisation k = 0 and
therefore /* = 1. In a paramagnetic body k is positive and p is
greater than unity. In a diamagnetic body k is negative and /* is
less than unity.
470. In some applications of this theory to electricity the
separating surface S is also occupied by a thin layer of matter
capable by its repulsion of inducing polarisation in the two media.
This layer is to be regarded as part of the influencing body. Let
p be its surface density.
We then have by Green's. theorem (Art. 142)
47r(p + o-1 + o-2) = ^1+F2 ............... (7),
as in Art. 469. Eliminating trlf vz by using the fundamental
equations <rl = — k1F1, a-2 = — k2F2 we arrive at the generalised
equation f^F1 + fj^F2 = 4>7rp ..................... (8).
All the conditions are included in the two statements briefly
expressed by a = — kF and the equation (8).
ART. 473] SPECIFIC INDUCTIVE CAPACITY. 245
471. Let H represent the resultant magnetic force and B
the magnetic induction at any point P respectively. When the
magnetism of the body is due solely to induction, the direction of
magnetisation coincides in direction with that of the magnetic
force H, (Art. 465). It follows that the force of induction B (being
the resultant of the magnetic force H and a magnetic force 4>7rl,
Art. 342) must also coincide in direction with that of the magnetic
force. We therefore have £ = H + 4?r/, and since / = kU, this
gives B = fiH.
The equation (6) of Art. 469 then asserts that the normal
component of the magnetic induction at P is unaltered in magnitude
when P passes from one medium into another, the components
being measured in the same direction along the normal.
472. We know by Art. 144, that the tangential component
of the magnetic force is unaltered in magnitude when P passes from
one medium into another, the components being measured in the
same direction along the tangent. The magnetic potential at P
is also unaltered, (Art. 145).
Let Hlt HZ be the resultant magnetic forces in the two media
at any point P of the boundary ; 6lf #3 the angles their directions
make with the normal at P, then
i sn ! = 2 sn 2, fai cos l = ^z cos 2,
.'. tan #,//*! = tan B2//j^.
When therefore a line of magnetic force passes from one
medium into another in which the permeability is greater than
in the first its direction is bent away from the normal.
473. Specific inductive capacity. In the problems on
electricity which have been hitherto solved in this treatise the
non-conducting medium or dielectric which surrounds the con-
ductors has been supposed to be air or some other gas. But the
capacities thus determined do not agree with experiment when
some solid non-conductor is substituted for the air. In this case
the elements of the solid become excited in such a manner that
each assumes a pularity analogous to the magnetic polarity induced
in the substance of a piece of soft iron under the influence of a
magnet. To take account of this state, called polarisation, we
apply the same analysis as that used for induced magnetism.
We suppose each element dv of the dielectric to become an
246 DIELECTEICS. [ART. 474
elementary doublet whose poles are occupied by equal and opposite
quantities of electricity. The direction of polarisation is that of
the electric force F due to all causes and the intensity is I = kF.
This polarity is then replaced by a fictitious stratum of electricity
on the surface of the dielectric whose repulsive force at any point
is equal to that of the polarised dielectric. One effect of the
repulsion of this stratum is to alter the potential of each conductor
and therefore to change its capacity.
The coefficient 1 + 4<Trk is called the specific inductive capacity
of the dielectric and is generally represented by the letter K. It
is evidently analogous to the permeability p in the theory of
magnetism. The two however differ in this particular; the
specific inductive capacity of a dielectric is very approximately
independent of the intensity of the electric force, while the
permeability is not an absolute constant but varies with the
magnetic force when that force is not small. The reader will
find in J. J. Thomson's Electricity and Magnetism (Art. 154) a
diagram which clearly exhibits the variations of p produced by
changes in the magnitude of the magnetic force.
A short table is given in DeschanePs treatise (edited by Everett) Art. 158 of
the corresponding values of the magnetic force H, intensity I, and permeability /j.,
for a specimen of soft iron.
H=0-3, 1-4, 3-5, 4-9, 10'2, 78, 585.
1= 3, 32, 574, 917, 1173, 1337, 1530.
M=128, 299, 2070, 2350, 1450, 215, 34.
The values differ in different specimens. We notice that as the magnetic fore?
increases, ft is at first nearly constant, then rapidly increases and arrives at a
maximum and again decreases. The value of /* depends also on the temperature.
At first it increases slowly with the temperature but at such high temperatures as
600° to 800° the rate of increase is very rapid. It then begins to decrease as
rapidly as it rose.
The specific inductive capacities of the following substances are taken from
J. J. Thomson's treatise, (Art. 67). Solid paraffin 2-29, sulphur 3-97, flint glass
6-7 to 7-4, distilled water 76, alcohol 26.
474. Effect of the substitution of a solid dielectric
for air. Let there be any number of closed conductors Alt Az,
&c. separated from each other by air as the non-conductor. Let
Elt E2, &c. be the charges on the conductors, Z7lf Z72, &c. the
constant internal potentials, plf pz, &c. the surface densities at any
points Qi, Q2> &c. on the several conductors.
When the conductors are separated by a dielectric of specific
ART. 475] DIELECTRICS. 247
inductive capacity K we represent the repulsions and attractions
of the dielectric by equivalent strata placed on the boundaries of
the conductors. Let their surface densities be respectively o-j , er.,
&c. We suppose that the dielectric is uniform so that there are
no equivalent strata in the field except those on the surfaces of
the conductors. If the dielectric have a boundary at an infinite
distance both the force and the density vanish at that boundary.
Let us assume as a trial solution that o-I = Xp1, o-2 = Xp2, &c.
where X is an unknown constant multiplier which is the same at
every point of every conductor. The sum of the potentials of all
the conductors at any point P in the field will then be changed
by the introduction of the dielectric in the constant ratio 1 to 1 + X.
The potentials U1} U^ &c. will also be changed in the same ratio
and will remain constant. The conditions of equilibrium will
therefore not be disturbed (Art. 372).
The test that the trial assumption leads to a correct solution
is that all the boundary conditions can be satisfied by the same
constant value of X. The conditions at the boundary of any
conductor A are given in Art. 470. These are
a = - kF, KF + K'F' = 4,-irp,
where K has been written for p. In our case, F and F' are the
normal forces respectively just inside the dielectric and just inside
the conductor. The latter being zero, we have
<r = -JcF, KF=4>7rp, <r = \p.
Eliminating F and cr, and remembering that 1 + 4?r& = K, we
have at once 1 + X= I/K. Similar equations apply at the boundary
of each conductor and give the same value of X.
The result is that the distribution of real electricity on the
surfaces remains unaltered, but the potential inside each conductor
is changed by the attractions and repulsions of the dielectric and
reduced to l/Kth part of what it was when the separating medium
was air.
475. To find the change of force at any point. Since the
surface density of each equivalent stratum is X times that of the
real electricity at the same point, the force X' at any point P in
the field, due to both the equivalent strata and the real electricity,
must coincide in direction with the force X at the same point P
due to the real electricity alone, and the magnitudes are such that
X'=(l+ X) X. We therefore have X' = X/K
248 DIELECTRICS. [ART. 477
If, when the dielectric is introduced to replace the air, the
potentials of the conductors are kept unaltered, the charges of real
electricity are increased in the ratio 1 to K and the force at
any point will then be unaltered. The force which one conductor
exerts on another will be increased in the ratio 1 to K.
The potential energy is W=^Vm (Art. 61) where m is the
quantity of electricity on the conductor whose potential is V. It
follows that the energy will be divided or multiplied by K
according as the charges or the potentials are kept unaltered.
476. The case in which one conductor A is entirely surrounded by a shell
formed by another conductor B needs some special attention. We suppose at first
that there are no other conductors in the field. The separating medium being in
the first instance air there is a distribution of electricity on the external surface S
of A and the internal surface S7 of B. Let the surface densities at any points Q, Q'
be respectively p and p'. If the conductor B has no external boundary, but extends
to infinite distances, the distributions on S and S' are such that the sum of their
potentials is constant throughout all space external to 6" and is the same as at an
infinite distance. The potential at every point external to S' is therefore zero and
the charges on S, S' are equal and opposite. We may now remove any portion we
please of the neutral matter outside the surface S' and reduce the conductor £ to a
finite size.
In this state of the system, there is no electricity on the external boundary of
the shell B. The potential of the system is zero within the substance of the con-
ducting shell B and equal to some constant a within the conductor A. See Art. 386.
When the whole space between A and B is filled with a dielectric, we represent
its repulsions by those of equivalent strata placed on the surfaces S, S'. Assuming,
as before, that their densities are <r=\p, <r'=\p', where X is some constant, we find
that the conditions at the boundary of A (viz. <r= - kF, KF=&irp) give immediately
1+\=1/K. The conditions at the other boundary give the same value of X.
The result is that the distributions of real electricity on S and S' remain
unaltered, but the potentials inside A and B are reduced to l/J5Tth part of what they
were when the medium was air. The potential inside B was zero and remains zero.
The potential inside A becomes a/K.
The capacity of the conductor A (being measured by the ratio of the charge to
the potential, when the conductor B is at potential zero, Art. 371) is therefore K
times as great as when the two conductors were separated by air.
477. Effect of external conductors. Let us next suppose that the external
surface S" of the shell B is charged with electricity and that other conductors are
placed in the field outside S". These additions to the system will not disturb the
equilibrium of the charges on the surfaces S, S', but will increase the potential
throughout the interior of S" by some constant /3. Supposing the conductors A and
B to be separated by air, the potentials inside B and A become /3 and a+p=a'.
The system thus formed (as explained in Arts. 389, 390) consists of two parts
which are independent of each other. Let us therefore fill the space between the
shell B and the conductor A with a dielectric of inductive capacity K, leaving the
conductors outside the shell still separated by air. The distributions of electricity
ART. 47 8 J PLANE DIELECTRICS. 249
on S and S' are not disturbed by this change, the potential inside B remains equal
to j3, but that within A becomes a" = /3 + a/^T where a = a'-/3. The difference between
the potentials of A and B is therefore decreased in the ratio 1 to K.
The capacity of the conductor A (if measured by the ratio of the charge on A to
the difference of the potentials of A and B) is therefore K times as great as when A
and B were separated by air.
478. A plane dielectric. Two conducting plates of infinite extent are placed
with their nearest plane faces A and B parallel to each other and at a distance Q.
A plate C, of specific inductive capacity K and thickness t, is introduced into the
intervening space with its two faces parallel to the planes A and B, the space on
each side of G being occupied by air. Find the effect of the introduction of the
dielectric G on the capacity and potential energy of the system.
Let a, 6 be the distances of the faces L, L' of the dielectric C from the planes
A, B, L being the nearest to A and L' to B, then d=a + t + b.
Let p and //= - p be the surface densities of the charges on the planes A and B.
Let <r and <r'= -<r be the surface densities of the strata on L and L' which are
equivalent to the polarity of the dielectric.
At a point P between the planes A and L the force F, measured from A towards
B, is constant and equal to 4vp (Art. 22). The constant force F', measured in the
same direction, at a point R between L and L' is found from the condition that the
induction is unchanged when P crosses the boundary of the dielectric (Art. 469),
hence KF'=F. At a point Q between L' and B the force is again F=£irp.
Let a, /3 be the potentials at the planes A, B, and X, X' those at L, L'. The
force at a point P distant x from A is -dVjdx=F, .: V=a-Fx. Similar reasoning
applies to the points Q and R. We have therefore
\=a-Fa, \'=\-F't, fi = \'-Fb.
Adding these three equations together and substituting for F, F' their values, we
find /3-a=-47rp (a + b + t[K)
= -4irp(0-* + t/Z) ................................. (1).
The capacity C (when measured by the ratio of the charge on either of the
conductors A, B to the difference of their potentials) is given by
We notice that this is independent of the position of the dielectric G.
If the whole space between the plates A, B is filled with air, we have t=0 and
the capacity is 1/47T0. The capacity is therefore increased by the introduction of
the dielectric C. When the dielectric G fills the whole space between the plates
A, B, we have t = 0 and the capacity is K times as great as when the separating
medium was air.
The potential energy per unit of area due to the charges ±/> on the plates is by
Art. 61, W=$'2EV=$pa,-bpp=hp(a.-p)'
We may express this result either in terms of p or a -0. We have by (1),
It follows that the introduction of the dielectric decreases or increases the potential
energy according as the charge p or the difference of potentials is kept unaltered.
The force per unit of area which one conductor A exerts on the other B is %Fp
{Art. 143). Since F=4irp this becomes 2?rp2. The force is therefore not changed
by the introduction of the dielectric C provided the charges are kept unaltered.
250 DIELECTRICS. [ART. 480
If the difference of the potentials is kept unaltered, we substitute for p from
equation (1). The force per unit of area is then (j3 - a)2/8w (6 - t + t/K)2.
47O. A cylindrical dielectric. The outer and inner boundaries of two-
conductors A, B are infinite co-axial circular cylinders whose radii are a, b. A
co-axial circular cylindrical dielectric C of specific inductive capacity K is introduced
into the space between A and B, the rest of the space being filled with air. To find
the effect of the shell G on the capacity and potential energy.
Let p, p' be the densities of the charges on the surfaces A, B of the conductors ;
<7, ff' those of the strata on G whose repulsions represent the forces due to the
dielectric. Let the radii of the two surfaces L, L' of the shell C be a', b'; L being
nearer A than B. Let a, /3 be the potentials at the conductors, X, X' those at the
surfaces L, L'.
The repulsion of any one of these cylinders at an internal point is zero. At an
external point the force varies inversely as the distance r from the axis and is equal
to 2m/r where m is the charge per unit of length (Arts. 55, 56). For the cylinder A,
m = 2irpa.
The force at any point P between A and L is 4^-pa/r. Putting r=a', and using
the rule that the product of the force and K is unaltered when P passes into the
dielectric (Art. 469), we see that the force just outside L is ^irpaja'K. The force at
any point R between L and L' is therefore iirpalr'K where r' is the distance of R
from the axis. Similarly the force at a point Q between L' and B is 47r/>a/r"
where r" is the distance of Q from the axis.
We now find by easy integrations
X-o= -47rpalog-, X'-X= - - '
-,.
o
Adding these together we have
-, ..................... (1).
The capacity C per unit of length (measured by the ratio of the charge on A to the
difference of potentials) is given by — ;=log — ( 1 - — ) log — .
2(7 a, \ 2t/ a,
Since the whole quantity of matter given by Poisson's equivalent strata is zero
(Art. 340), we have o-a' + (/fc' = 0. Also since the potential of the whole system at
any point external to B is constant, the quantity ( pa + p'b + eaf -tV6') log r is
independent of r, and this is impossible unless pa + p'b — 0. The charges on the
conductors A and B are therefore equal and opposite.
The potential energy per unit of length (Art. 61) is given by
W= | (27rpa a + 2irp'bp) = trpa (a - /3),
which can be expressed in terms of either the charge or the difference of potentials,
by substituting from (1).
480. A repelling point of mass E is placed at a point A in a
medium of inductive capacity K ; prove that the potential at any
point P distant r from A is EjKr.
The point may be regarded as the limit of a small sphere of
equal mass, radius a, whose specific inductive capacity is unity.
ART. 481] PROBLEMS ON DIELECTRICS. 251
This sphere is then the inner boundary of the dielectric and the
equivalent distribution on its surface must be taken into the
account. The force due to the charge E at all points just inside
the surface is E\d? and, a being small, all other forces in the field
may be neglected. Since there is no real electricity on the
sphere, the normal forces on each side are inversely as the specific
inductive capacities (Art. 469). The force at all points just out-
side the sphere but within the substance of the dielectric is
therefore F — EjKa?. The surface density a- of the stratum on
the sphere is cr = — kF, and is therefore uniform. The resultant
repulsion of the charge E together with that of the uniform stratum
is therefore EjKr* at all points external to the sphere (Art. 64).
If another charge of mass E' be at a point B distant r from A,
we replace it by a small sphere of mass E' and radius b. The
force on the sphere E1 due to a uniform distribution of attracting
matter on this sphere is zero, (Art. 65). When therefore two point-
charges, separated by a uniform dielectric, repel each other, the
force is EE'j'Kr*.
481. Problems on dielectrics. To find the effect of in-
duction on a dielectric we have generally to begin with a trial
solution. Sometimes we assume the density of the equivalent
stratum on the boundary S of the dielectric to be an unknown
constant multiple (say X) of some quantity suggested by a corre-
sponding problem when the dielectric is air (Art. 474). We can
then deduce the potentials on each side of the equivalent stratum
and determine the constant A, by using some one of the forms of
the boundary condition.
In other cases it is more convenient to assume some expressions
for the potentials O, O' due to the repulsions of the dielectrics;
these must be suggested by the circumstances of the case. They
must obviously satisfy the following conditions, (1) the functions
fl, H' must satisfy Laplace's equation at all points not occupied
by attracting matter, and be finite and continuous each on its own
side. If the medium on one side of S extend to infinity, the
potential corresponding to that side must be zero at an infinite
distance. (2) The two functions H, O' must be such that at
every point of S,
252 DIELECTRICS. [ART. 482
where dv, dv are elements of the normal to S, measured positively
from S, and U is the potential of the influencing body. These
are called Poisson's conditions. When XI has been found, the
magnitude and direction of the induced polarity follow from
equations (1) of Art. 466. The surface density of the equivalent
strata can be found by (4) of Art. 469.
482. Ex. Find the polarisation induced in two media of capacities Klt Ks
separated by a plane and acted on by an electric charge E situated at a point B which
is in the first medium at a distance BM=hfrom the separating plane.
First solution. Produce BM to C and make MG=h, see the figure of Art. 412.
Let (rlt r/), (r2, r2') be the distances of any two points P1, P2 in the two media
from B, C respectively. Assume as a trial solution that the potentials due to all
causes at Pj, P2 are ^1 = — — I — /» ^a = ~ ........................... W»
Airi ri rz
where M, N are two unknown constants. These potentials are finite at all points
unoccupied by matter, zero at infinity, and satisfy Laplace's equation. They
must also satisfy the boundary conditions F1 = F2 and K^ + K^F^Q, at all points
which make r1=r1'=r2. We find by resolution
E \h Nh .
' *» 5" ........................... ( )f
/. ~+M=N,
Ki
These equations give M and N, we therefore have
= 2JE__1
2
From these the values of the components of polarisation U, Im, In follow at
once, Art. 466.
The density <r of the equivalent layer on the boundary plane at a point P distant
r from either B or C is given by
In forming the trial solution (1) we may assume that the potential at a point Pj
in the first medium is the sum of the potentials due to the electric point B and any
imaginary electric points properly placed in the other medium. No electric point
(other than the real point JB) in the first medium can be used, because the potential
would then be infinite at that point. Similarly in forming a trial potential at P2
in the second medium, any suitable imaginary points situated in the first medium,
but none in the second, may be used.
Second solution. Instead of assuming some values for the potentials Vlt F2, we
may take as our trial assumption some form for the density a of the equivalent
layer on the plane. By referring to Art. 412 we are led to the assumption
2.E/&
<r = X— -3. The repulsion due to the stratum at any point P on either side is the
same as that of a charge XE situated at a point (B or C) on the side opposite to P.
The normal forces on each side of the separating plane (measured from that plane)
due to the electric point at B and the stratum are therefore
1 .\ Eh
ART. 483] KELVIN'S THEOREM. 253
Jf _jr
The boundary condition K^ +K^F2=0 gives at once X= - Tv * * . Since
(£j ~f~ KTJ -^i
this value of X is constant the trial solution is verified.
The value of <r thus found obviously agrees with that found in the first solution.
The potentials are evidently F, = — 1 7. F0= 1- -—
1 ^i »"i' ' ' Kft ra '
483. Effect of the substitution of a dielectric shell for
some of the air. A conductor A is surrounded by another B at
zero potential, the space between being occupied by air. Charges
E and — E being given to these bodies respectively, let F0 be the
potential inside A. Let a shell G in the space between A and B
be bounded by two equipotential surfaces L, L' of the charges on
A and B, L being the nearest to A. Let U, U' be the potentials
at these surfaces. If a dielectric of capacity K be substituted for
the air in the shell G (the rest of the space between A and B being
still occupied by air) the whole effect of the dielectric is to diminish
the potential in the interior of A by ( 1 — -^ j (U—U'), see Art. 476.
This theorem is due to Kelvin [reprint, &c. Art. 45].
When the separating medium is air the potential of the
system at the interior surface S' of the conductor B and at every
point without its surface is zero (as explained in Art. 476) while
the potential at the surface and within the interior of A has some
constant value a.
Let us place on the surfaces L, L' indefinitely thin layers
whose surface densities cr, a-' are respectively given by 4nra- = \F
and 47TO-' = — \F' where Fy F' are the normal components of
force due to the charges on A and B, both forces being measured
from A towards B. The total masses of these layers are re-
spectively \E and — \E (Art. 156).
Since L is a level surface, the potential due to the charge on
it at any external point P is XFi, where V± is the potential at P
due to the charge E on A ; and at an internal point Q its potential
is \ ( U— F2) where F2 is the potential at Q due to the charge
— E on B (Art. 156). Similar remarks apply to the layer on the
surface L' except that the sign of \ is altered.
Hence at any point P external to both L and L', the effect of
the introduction of one stratum is to increase the potential by
\F! and the effect of the other is to decrease the potential by the
same amount. The potential at any external point is therefore
unaltered.
254 DIELECTRICS. [ART. 484
At any point Q internal to both L and I! the potential is
increased by the sum of \(U-Va) and - X ( U' - F2). The
potential is therefore increased by the constant quantity \(U— U').
At any point R between L and L' the potential is increased
by the sum of XFj and —\(Uf— F2). The potential is therefore
increased by X(F— U'), where F is the potential at R due to the
given charges on A and B, i.e. V=V1+ F2.
The introduction of these strata therefore increases the
potential inside A by a constant quantity and does not alter
the potential within the substance of B. The electric equilibrium
of the two conductors is therefore not disturbed.
The layers placed on L and L' will be the equivalent strata of
the dielectric G if the densities <r, a-' are respectively equal to — k
times the normal components of force due to all causes at points
just within the two boundaries of the dielectric each measured
from its own stratum. The potential at a point R just outside
L being F-f X(F— U') the outward normal force (obtained by
differentiation) is (1 +\)F. We therefore have the two equations
Hence 1 + X = l/K. The conditions at the other boundary give
the same value of \.
The effect of the introduction of the dielectric is not to alter the
level surfaces, but to decrease the potential a in the interior of A by
a known quantity.
Since no restriction has been placed on the size of the external
conductor B, we may replace it by a sphere of infinite radius.
The charge on its surface being finite, we may then eliminate
that conductor altogether from the field. Kelvin's theorem may
therefore be applied when the shell G surrounds a single conductor
A, provided the boundaries of G are equipotential surfaces.
484. Let us now suppose that the shell B and the conductor A are placed
in a field of constant potential (see Art. 477), so that the potential at every point is
increased by the same quantity /3. The electrical equilibrium is not disturbed, but
the potentials inside the conducting matter of B and A (when separated by air)
become respectively p and /3 + a = o'. Each of the potentials U, U' is also increased
by fi, but their difference is not altered.
After the introduction of the dielectric shell (7, the potential inside B remains /3,
•while that inside A becomes <*'-(!- — j(U- U'). The capacity C (if measured by
ART. 485] PROBLEMS ON CONDENSERS. 255
the ratio of the charge Q on A to the difference of the potentials of A, B) is then
given by §=«'-£- (l~Jt)(u- u')-
When the dielectric fills the whole space between A and B, we have U=a!, C7'=/3 ;
so that the effect of introducing the dielectric is to multiply the capacity by K.
The potential energy of the system is by Art. 61 equal to ^EV. The effect of
introducing the dielectric shell G is to diminish the internal potential of the shell
A, leaving that of B unaltered. The potential energy is therefore diminished by
485. Ex. 1. A spherical shell (whose inner radius is c) and a solid concentric
conducting sphere (radius a) are charged with quantities ±23 of electricity. The
space between is filled with two dielectrics separated by a third sphere of radius b.
Prove that the capacity 7 is given by - = ( — ?: ) "^ + ( r — ) »=• • [Coll. Ex.]
y \a */ •"•! \* c ) -"-z
This result follows at once from Kelvin's theorem (Art. 483). Let Q be the
charge on the sphere of radius a. When the separating medium is air, the
potentials F and U at the surfaces of the spheres a and b are
V=Qla-Qlc, U=Q/b-Q/c.
The effect of the dielectric is to reduce the potential within the sphere a to the
value |^.F-
The capacity required is QjV.
Ex. 2. A spherical conductor of radius a is surrounded by a concentric
spherical conducting shell of radius b and the space between is filled with a
dielectric of specific inductive capacity fj.e~^/ps (where p=r]a) at a distance r from
the centre. Prove that the capacity of the condenser so formed is 2^a/(ec-e)
where c=62/a2. [Coll. Ex. 1896.]
Ex. 3. Prove that the capacity of two parallel plates, separated by air and
placed at a distance apart equal to 6, will be increased n-fold by introducing between
them a slab of substance whose specific inductive capacity is K and thickness
2ll J2L where n <K. [Coll. Ex. 1900.]
n xt — 1
Ex. 4. A condenser is formed of two parallel plates, whose distance apart is h,
one of which is at zero potential. The space between the plates is filled with a
dielectric whose specific inductive capacity K increases uniformly from one plate to
the other. Prove that the capacity per unit area of the condenser is
where K^ and Kz are the values of K at the surfaces of the plates, the inequality of
the distribution at the edges of the plates being neglected. [Math. Tripos, 1899.]
Ex. 5. Three closed surfaces 1, 2, 3 are equipotentials of an electric field; if an
air condenser is constructed with faces 1, 2, its capacity is A ; with faces 2, 3 the
capacity is B ; if with 1, 3, the capacity is C. Prove - = — + •=. .
O A Jo
If a dielectric K fill the space 1, 2 and one K' fill 2, 3, prove that the capacity of
the condenser having 1, 3 for faces is i = -^ + — . [St John's Coll. 1898.]
C Ah. X> A
256 INDUCTION PROBLEMS. [ART. 486
Ex. 6. A condenser consists of two confocal ellipsoids, the squares of whose
semi-axes are respectively a2, b'2, c2 and a2 + «, &c. If the dielectric be air and /S
the capacity for electricity, prove that
2 fu du
If the dielectric be a solid arranged in ellipsoidal shells confocal with the
conductors, and such that the specific inductive capacity of each shell is inversely
proportional to the volume of the enclosed ellipsoid, prove that the capacity is
ZKabcju, where K is the specific inductive capacity of the innermost layer.
[St John's Coll. 1879.]
486. Ex. 1. A charge E is placed at a distance / from the centre of a sphere
of s.i. c. K and outside the sphere. Prove that the potential at any point inside the
E 2n + l /r\n
sphere at a distance r from the centre is — S -= - - ( - ) Pnt where the sum-
/ Kn+n + 1 \fj
mation extends from n=0 to GO. [Coll. Ex. 1897.]
Let <T=2AnPn represent the surface density of the charge equivalent to the
polarity of the dielectric. We write Pn instead of Yn because the system is
symmetrical about the straight line joining the charged point to the centre of
the sphere. The potentials due to this stratum are given in Art. 294 at points
inside and outside the sphere. Adding to these the potentials of the external charge
and using the equation (6) of Art. 469 we obtain the result to be proved.
Ex. 2. A sphere of B.I.C. K is placed in air, in a field of force due to a potential
Xn (before the introduction of the sphere) referred to rectangular axes through the
centre of the sphere, where Xn is a solid harmonic of degree n. Prove that the
potential inside the sphere is — — — Xn. [Coll. Ex. 1898.]
71 ~J~ J- -j* -t^ft
Ex. 3. Find the potential at any point when a sphere of specific inductive
capacity K is placed in air in a field of uniform force.
A circle has its centre on the line of force which passes through the centre of
the sphere and its plane perpendicular to this line of force. Prove that if the plane
of the circle does not cut the sphere, the presence of the sphere increases the
JL — 1
induction through the circle in the ratio 1 + 2 — — s sin3 a to 1, where 2a is the
K. + ^
angle of the enveloping cone drawn from any point on the circumference of the
circle to the sphere. [Coll. Ex. 1896.}
Proceeding as in Ex. 1 we find that the potential due to all causes at any point
/ J£— 1 ascos#\
outside the sphere is V'=F( x- - ) where F is the given force of the
\ JL + A r* )
r jy
field. The flux of force through the circle is then - I --- 2-irudu, Art. 107.
J dx
Ex. 4. A circular wire is situated in a uniform magnetic field, with its plane at
right angles to the lines of force; prove that the effect of introducing into the
middle of it a sphere of soft iron of permeability fj., which exactly fits its section, is
to increase the induction through it in the ratio of 3 to 1 + 2//x.
[By Art. 471 the induction is /* times the flux.] [St John's Coll. 1896.]
Ex. 5. A spherical shell of radii a, 6 (fe>a) and specific inductive capacity K is
placed in a field of uniform force F; prove that, if Ft is the force in the space
within both spheres, = l + l- [Coll. Ex. 1899.]
ART. 486] INDUCTION PROBLEMS. 257
Ex. 6. An infinite solid with a plane face is acted on by a small magnet, of
unit moment, situated at a point E outside the solid, the axis of the magnet
being perpendicular to the plane face. Prove that the magnetic potential at any
point P within the solid is . ™*<\ z where r=EP, 0 is the angle EP makes with
the axis of the magnet and p. is the permeability of the solid. [Coll. Ex. 1897.]
We represent the repulsion of the solid by that of a thin stratum of variable
density <r on its surface. The normal force at a point Q close to that surface is due
ultimately to the repelling matter in the neighbourhood of Q and is therefore lira.
If Z be the normal force due to the magnet, the condition at the boundary is
(27rcr + Z)ij.+ (2a-<r - Z) = 0.
This gives by Art. 316 ff=^T" » wliere r' = EQ and 2 is the distance of
E from the plane. The potential due to a stratum z/r'8 is given in Art. 412, that
due to <r is then deduced by differentiation as explained in Art. 93, Ex. 3. Finally
the given result is obtained by adding the potential of the magnet itself.
Ex. 7. A sphere of specific inductive capacity K and of radius a is held in air
with its centre O at a distance c from a point A where there is a positive charge E.
Prove that the resultant attraction on the sphere is
where /3= (K- 1)I(K+ 1). [Math. Tripos, Part n. 1897.]
The potential at an internal point is given in Art. 486, Ex. 1, thence the surface
density y of the stratum equivalent to the polarity of the dielectric may be found by
an obvious differentiation, Art. 466. If R be the distance of any elementary area
dS of the sphere from A, the resultant force on the sphere is X= I ^ — ^ <rdS.
J H' H
The expansion of — ~- = -§2 (ra + l)Pnftn, where ft=o/c, is found by differentiating
-£V C
that for c/R (Art. 264) with regard to h. The integrations can be effected at sight
by using Arts. 288 and 289. The series thus found for X agrees with that obtained
by expanding in powers of a/c the result given in the enunciation.
Ex. 8. The space between two concentric conducting spheres is filled on one
side of a diametral plane with dielectric of specific capacity K, and on the other
side with dielectric of specific capacity K'. The inner sphere is of radius a and has
a charge E. Prove that the force on it perpendicular to this diametral plane is
r 7*2
. [Coll. Ex. 1901.]
The potential V in either dielectric is S /^nr»+-~ jPn, but since Fmust be
independent of 0 both when r=a and r=fc we find F=J + B/r. Since F has the
same value on both sides of the diametral plane (Art. 481) for all values of r
between r=a and r=b, this formula, with the same values of A and B, gives the
potential in both dielectrics. By Art. 470, we find that the real densities p, p' on
the two halves of the sphere are given by 4irp=if£/a2, ±irf!=K'B[a?. Since
27T (p + p')a* = E, we find £ = £ = ~ ^^-, . The pulling force on an element
pdS is^pdS.(- dV/dr), which reduces to 2vp*dSIK. We now write dS=2wa? sin 0 . de
and multiply by cos 0 to resolve the force parallel to *. The integral from 6 =0 to
R. S. II.
258 MAGNETIC SHELLS. [ART. 488
Ifir gives the resolved force on half the sphere. Interchanging K and K' we have
the resolved force on the other half. The difference is the force required.
Ex. 9. A dielectric hemisphere of radius a and inductive capacity K is placed
•with its base in contact with the plane boundary of an otherwise unlimited
conductor. Prove that the potential at any point of the field outside both the
conductor and the dielectric is F'= -4inr cos 6 [r—-= — ~ -5 ) , where the origin is
\ K + zr* )
at the centre of the hemisphere, and <r is the surface density of the charge on the
plane conductor at a great distance from the hemisphere. [Coll. Ex.]
487. Magnetic «in»ii«- Ex. 1. An iron shell (radii a, b, a>l) is placed in a
field of uniform magnetic force /. Find the induced magnetism and the force X
inside the hollow.
Put p=2Y*, p' = ~SZn for the surface densities on the spheres. Their potential
within the material is
The boundary conditions to be satisfied are
where a and b are to be written for r respectively after the differentiations have
been effected. These show that l"n=0, Zn=0 except when n=l. We find
3fc/cos0 f / /6\«M -9A/C030
Yl=—w— |3 + 8r^l-^-J ft , Z,- —^ - ,
where N=9(l+4rk) + 2(4irk)* -jl- (?} j- . The potential V and force X inside
the hollow, due to all causes, are
V=*(Y1 + Z1)r-fx, =
Ex. 2. A solid uniform sphere (radius a) is placed in a uniform field of force
whose potential is -fx, say the magnetic force of the earth. Prove that the
potential of the induced magnetism at all external points is the same as that of a
concentric simple magnet whose moment is a3/ - 5 .
/*+ ^
Ex. 3. A small magnet of moment M is placed at the centre of an iron shell,
radii o, 6. Prove that the potential at any point external to the shell, due to all
. M cos 6 9u
causes, is — -^ — ^ - . _1>2/i - 1\ wnere P = b]a and p is the permeability of
the shell. Thence show that if p. is great and p not nearly equal to unity, the
potential is zero. In this case the induced magnetism on the shell neutralises that
of the magnet at all external points.
488. To find the surface integral of the magnetic induction
through any closed surface 8.
To find the component of the magnetic induction at any point
P in a direction PN, we construct a disc-like cavity at P which
has its plane normal to PN. The normal component of the
induction is then the same as the actual normal component of
force at P due to all causes, (Art. 343).
ART. 490] SURFACE INTEGRAL OF INDUCTION. 259
To find the surface integral of the magnetic induction we
remove a thin layer of matter all over the surface S or, at least,
over that part of 8 which lies within the magnetic body. We
shall now apply Gauss' theorem to the repelling matter situated
within the internal boundary of this empty shell, i.e. within the
surface 8.
Since each magnetic molecule has two equal and opposite poles,
and no magnet lies partly within and partly without the empty
shell, the algebraic sum of the magnetic matter within the surface
$ is zero. For the sake of generality, let us suppose that there
may be other repelling particles (besides the magnetism) situated
within S. Let M be their total mass.
Let H be the magnetic force, B the magnetic induction and
/ the intensity of magnetisation at any point P of 8. Let 6, 0'
and i be the angles their directions respectively make with the
outward normal at P. Then
£003 6' ' = H cos 6 + 4?r7 cost (1).
Applying Gauss' theorem to the surface 8, we have
4irM = fBco80'dS (2),
= f(Hcos0+4i7rIcosi)d8 (3).
489. Another proof. We may also arrive at these results very easily, if we
first replace the magnetism by Poisson's solid and superficial distributions. Let p'
be the density of the solid distribution, I cos i the surface density. If the surface
S lie wholly within the magnetic body, the superficial distribution on the body will
be outside S. We then have by Gauss' theorem
4ir(M+$p'dv)=$HoosedS (a).
Since Poisson's rule applies also to any portion of a magnetic body (Art. 340) we
have also $p'dv+$IcoaidS=0 (£),
where the surface integral extends over the surface S. Eliminating p' we have
47rlf=J(Hcos« + 4irIcosi)d5 (3).
If the surface S intersect the boundary of the magnetic body, we suppose 7=0 at
all points of S which are outside the body.
We must also include on the left-hand side of (a) that portion of the superficial
density on the body which lies within S ; let this portion be called /. At the same
time we must add J to the left-hand side of (/3), since JJcos idS only extends over
that portion of the surface S which lies within the body. When therefore we
eliminate $p'du, the quantity J also disappears and we again arrive at (3).
490. If there is no repelling matter besides the magnetism,
M = 0. We then find that the surface integral of the magnetic
induction across any closed surface 8 is zero.
17—2
260 DIELECTRICS. [ART. 493
491. If the magnetism is wholly induced we have B =
and & = i, (Art. 471). We then have
4arM = / pH cos i dS ..................... (4).
We infer that m a dielectric of specific capacity K the outward
flux across a closed surface S of K times the normal force is equal
to 4-Tr times the repelling mass inside. This is also called the
outward induction across the surface S.
492. Let us apply the modified form (4) of Gauss' theorem
to a Cartesian element of volume of a dielectric. The value of
the right-hand side of (4) for the two faces perpendicular to x is
(as explained in article 108) d(KX)dydz. Treating the other
faces in the same way and writing M=pdxdydz, we find
d(KX) d(KY) d(KZ)
4777) =-S — + ~^7 — - + - -3 — '•
ax dy da
If we use the potential V, this becomes
where p is the density of any real repelling matter which may
occupy the space S independently of the Poisson volume density
p' due to the presence of the heterogeneous dielectric.
If we write I=kH, (Art. 471), the equation (/3) takes the form
Jp'dw + J kH cos 6dS=Q.
Applying this also to a Cartesian element we have
d /, dV\ d t, dV\ d /, dV
p=T* (t^j+Ty^dy-j+Tz (k *
where p' is the density of Poisson's solid distribution.
The equation (a) becomes in the same way
+4F+4? ........................... (7).
Since E=l + 4irk, any one of these three equations follows from the other two.
493. To deduce the condition at the common boundary of two dielectrics from
the modified Gauss equation (4).
Let a thin stratum of repelling matter of surface density m separate two
dielectric media of capacities K, K' ; see Art. 470. We follow the same reasoning
as in Art. 147, but writing KX, KY, KZ for Z, Y, Z. If we take x normal to the
separating surface we then have
(E'X'-KX)dydz+
In the limit this becomes K'X' - KX=iirm,
ART. 495] DIELECTRICS. 261
where X, X' are the normal forces on each side of the separating stratum, both
measured in the same direction, viz. from the medium K towards the medium K'.
This of course is the same as the result arrived at in Art. 470.
We may put this argument in another way. Let us enquire what form the
equation (5), viz. jL(KX)+(KY) + L(KZ) = 4*p ........................ (5),
assumes when the specific inductive capacity changes from K to K' at any surface.
Taking x normal to the surface we notice that dVjdx increases rapidly on crossing
the surface, while dVjdy, dV/dz do not. The left-hand side of (5) is therefore
ultimately reduced to its first term. Integrating from x=0 to x=t, we have
494. As an example, let us consider the problem solved in Art.. 482. At all
points in the medium which contains the point charge E, the density />=0, except
at that charge, while in the other medium p=0 at all points. We may therefore
take as the trial values of the potential
Fl=r+V' FS=^S'
since these satisfy equation (5) of Art. 492 at all points at a finite distance from E.
To find L we apply (5) (or equation (4) of Art. 491 from which (5) was derived) to
the points of space near the charge E. To avoid the difficulties of infinite terms,
we shall choose the equation (4). Taking as the surface S a sphere whose centre
is at E and whose radius is a, we have
where i' is the angle r/ makes with the normal to the sphere. In the limit, when
a is very small, we reject the term containing M. We immediately have L=:E/K,
and the solution may then be continued as in Art. 482.
We notice that, when a is not very small, the term containing M is zero by
Gauss' theorem (Art. 106) because the point C from which r/ is measured lies
outside the surface S.
495. To deduce from the extended form of Gauss' theorem an
expression for the potential of an electric system.
By Art. 61 the potential energy of a system of repelling
particles is W = ^Vm = ^fVpdv,
where V is the potential and p the density at the element of
volume dv. If there be no repelling particles within the element,
then for that element p = 0. The integration extends throughout
the volume of some closed surface S within which all the repelling
particles lie. Substituting for p its value given in Art. 492, we
have
F * fff r&(
STrJJJ (dx\
where K is the specific inductive capacity of the medium which
occupies the element dxdydz.
262 POTENTIAL ENERGY. [ART. 495
We now integrate each term by parts, following Green's method,
(Art. 149). We have
where the square brackets imply that the term is to be taken
between the limits of integration. These are represented by A to
B in the figure of Art. 149. Treating all the terms in the same
way, we have
dyi ^ f r ttdV\* ^ fdV\* ^ (dV\*\j
-T- d<r + I K](-j-) + [-r-) + l-j- )[dv.
dn J (\dx J \dy / \ dz J J
If the integration extend throughout a sphere of large radius
R, the product VdVjdn is of the order I/RS while d<r is of the
order B?. The surface integration therefore vanishes when the
integration extends throughout all space. We thus find
w ! f
W= Q- I
8irJ
^^
i -j— ) + I -j- ) + -J
(\da;/ \dy J \ dz
Ex. Find the potential energy of the system described in Art. 478.
We have 8vW=$KFt'dx, where F= -dVJdx and W is the energy per unit of
area.
Between A and L, F=4irp, K=l and the limits of integration are a; = 0 to o.
Between L and L', F=4rplK, and the limits are x = a to a+t. Between L' and B,
F=4trp, K=\ and the limits are x=a+t to 9. Outside A and B, F=0. Effecting
these integrations and adding the results, we arrive at the result given in Art. 478.
In the same way the energy of the cylindrical condenser described in Art. 479,
is given by 8vW= | F3dv + [bf(F*IK)dv+ I F2dv,
J a J a J V
where F=4r/>a/r and dv = 2vrdr. This evidently reduces to the result given in the
article just referred to.
THE BENDING OF RODS.
Introductory Remarks.
1. OUR object in this chapter is to discuss the stretching,
bending, and torsion of a thin rod or wire. We may define a rod
as a body whose boundary is a tubular surface, of small section.
The surface is therefore generated by the motion of a small plane
area whose centre of gravity describes a certain curve and whose
plane is always normal to the curve. The curve is generally called
the central axis or central line of the rod.
The rod or wire is to be so thin, that, so far as the geometry of
the figure is concerned, it may be regarded as a curved line having
a tangent and an osculating plane. Although this limitation will
be generally assumed it will be seen in the sequel that some of the
theorems apply to rods of considerable thickness. It is not pro-
posed to enter into the general theory of the elasticity of solid bodies,
except where it is necessary for the elucidation of the point under
discussion, and even then the reference will be restricted as far as
possible to the most elementary considerations.
2. In general the deformation of the body will be regarded as
very small, so that each element of the body is only slightly
strained from its natural shape. It will therefore be assumed that
the whole effect, when properly measured, of any number of dis-
turbing causes may be obtained by superimposing their separate
effects.
3. By reference to Art. 142 of the first volume of this treatise,
it will be seen that the action across any section G of a thin rod
AB consists of a force and a couple. On this is founded the
mathematical distinction between a string and a rod. The action
across any section of the former is a force, called its tension, which
acts along the tangent to the string, Vol. I., Art. 442. In the case
of a rod the force may act at any angle to the tangent and there is
in addition a couple.
264 BENDING OF RODS. [ART. 6
4. Let P be any point of a body, let a closed plane curve be
described round P of indefinitely small area, and let this area be
to. If the body is a fluid it is the fundamental principle of hydro-
statics that the action between the fluid on one side and the fluid
on the other side of the area o> consists of a force whose direction
is perpendicular to the plane of the area. It is thence deduced
that the magnitude of this force or pressure is the same for all
inclinations to the horizon of the elementary curve provided its
area remains unaltered. If the body is an elastic solid, the action
across the plane is also a force, but its direction is not necessarily
perpendicular to the plane of the area and its magnitude is not
necessarily the same for all inclinations of the plane.
In discussing the mechanics of a rod, its cross section, though
very small, is not to be regarded as infinitely small. If we divide
any section into elementary areas, the action across each element
will be an elementary force, and the resultant of all these will be,
in general, a force and a couple, Vol. I., Art. 142.
The Stretching of Rods.
5. To determine the simple stretching of a straight rod by a
force applied at one extremity, the other being held fast.
The relation which exists between the force and the extension
of the rod has already been discussed in the first volume of this
treatise under the name of Hooke's law. Ifllfl be the unstretched
and stretched lengths of the rod, o> the area of the section of the
I — I T
unstretched rod, T<o the tension, then 1 = •=, , where E is a
tj A
constant depending on the material of the rod and is usually
called Young's modulus.
When a rod is stretched we know by common experience that
its breadth and thickness are also altered. These lateral changes
follow a law similar to Hooke's law except that the modulus E is
not necessarily the same as that for extension. The study of these
lateral contractions belongs properly to the theory of elasticity and
only a simple case will be considered here.
6. The substance of a homogeneous body is called isotropic when the properties
of a solid of any given form and dimensions cut from it are the same whatever
directions its sides may have in the body. The substance is called ceolotropic when
the properties of the solid depend on the directions which its sides have in the
body. We shall suppose that the material of which the rod is composed is
isotropic.
ART. 7]
STRETCHING OF RODS.
265
7. Theory of a stretched rod. Let the unstretched rod form a cylinder with
a cross section of any form and size. When stretched the rod becomes thinner, so
that the several particles undergo lateral as well as longitudinal displacements.
There is one fibre or line of particles which is undisturbed by the lateral contraction.
Let this straight line, which we may regard as the central line, be taken as the axis
of x, and let the origin be at the fixed extremity of the rod. We suppose that the
stretching forces at the two ends are distributed over the extreme cross sections in
such a manner that after the rod is stretched these sections continue to be plane
and perpendicular to the central axis. It will appear from the result that the force
at each end should be equally distributed over the area.
Let x, y, z be the coordinates of any particle P in the unstrained solid, x+u,
y + v, z + w the coordinates of the same particle P1 of matter in the deformed body.
Then u, v, w are such functions of x, y, z that the equations of equilibrium of all
the elements of the solid are satisfied. We shall now prove that if we take u=Ax,
v=-£y, w= -Bz all the equations of equilibrium may be satisfied by properly
choosing the constants A and B. According to this supposition the external
boundary of the stretched rod will be a cylinder and the particles of matter which
occupy any normal cross section of the unstrained rod will continue to lie in a
plane perpendicular to the axis when the rod is stretched.
Let PQES be any rectangular element of the unstrained solid having the faces
PQ and RS perpendicular to the central axis. By the given conditions of the
question this element assumes hi the strained solid a form P'Q'R'S' in which all the
angles are still right angles and the sides parallel to their original directions. The
direction of the stress across each face of the strained element is therefore perpen-
dicular to that face. To measure these forces we refer each to a unit of area. Let
Nx, Nv, Nt be the forces, so referred; let these act on the three faces which meet
at the corner P7 and are respectively perpendicular to the axes of x, y, z; we shall
regard these forces as positive when (like the tension of a string) they pull the
matter on which they act, and as negative when (like a fluid pressure) they push.
Let a, b, c and a (1 + a), b (1+/3), c (1+7) be the sides of the element before and
after the deformation. Then Nx, Ny, Nt are functions of a, /3, 7, see Art. 489,
Vol. i. We shall expand these functions in ascending powers of a, /3, y and since
we here confine our attention to a first approximation, we shall neglect all the
higher powers of a, p, 7. Assuming the lowest powers in the expansion to be the
first, we have Nx = no. + \ (£ + 7) ,
the coefficients of £ and 7 being the same because the medium is isotropic. For
the same reason the stress Nv must be the same function of /3 and 7, a, that Nx is
of a and /3, 7. Thus Nv= icfi + X (a + 7).
In the same way Nt may be derived from Nx by interchanging a and 7. To make
these more symmetrical, it is usual to write them in the form
(a + p+y).
266 BENDING OF RODS. [ART. 9
The constants X and ft are the same as those chosen by Lame to measure the
elastic properties of a solid; see his Lemons aur la thtorie mathematique de I'elasticite
des corps solides.
8. In the problem under consideration the sides dx, dy, dz of the unstrained
du dv dw
element become dx + du, dy + dv, dz + dw. It follows that a = ^> ^ = ~ ' ^=dz'
Substituting the assumed values of u, v, w, we have
Nx=2pA+\(A-2B), Nv=-2(iB + \(A-2B), N,= -2»B + \(A -2B).
These values are independent of x, y, z, so that the opposite faces of any element
wholly internal are acted on by equal and opposite forces. It follows that every
internal element is in equilibrium.
Consider next the elements which have one or more of their faces on the
boundary of the rod. Such faces must be parallel to the central axis and in a
vacuum are not acted on by any pressure. It is therefore necessary for their
equilibrium that the constant forces represented by Nv and Nt should be zero.
B \ (3\ + 2/j.)/j. A
We therefore have J- ^ A'
Since Ax is the extension, By the contraction of a rod of length x and breadth y
and Nx is the stretching force per unit of area of the section, it follows that
increase of length _ X+M w decrease of breadth _ X _
original length ~~ fi (3X + 2/x) *' original breadth ~ 2jt (3\ + 2fj.) **
Comparing the first of these with the statement of Hooke's law given in Vol. i.
Art. 489, we see that the constant E, usually called Young's modulus, is the
reciprocal of the coefficient of Nx. If E' be the corresponding coefficient for the
decrease of breadth we have .E=M (3X + 2At), .E'=2 (X,+ M) JB.
X
It follows from this solution that when a rod has been stretched, each fibre (or
column of particles parallel to the central axis) is stretched and contracted indepen-
dently of the others and exerts no action on the neighbouring fibres. The total
force required to produce a given extension is therefore independent of the form of
the cross section provided its area remains unaltered.
In this investigation the action across each of the six faces of the element is
normal to that face. In many problems in elastic solids this simplicity does not
exist and there are tangential actions also across the faces. For the discussion of
these questions the reader is referred to A Treatise on the Mathematical Theory of
Elasticity, by A. E. H. Love, 1892.
9. Ex. 1. Show that E and $E' are the forces which would stretch a rod of
unit section to twice its original length and half its original breadth respectively.
Show also that E' is greater than 2E.
If the stretching tension be Nx, v the volume, Sv the increase of volume, prove
*-^*..
Ex. 2. If the side faces of the rod are exposed to a uniform normal pressure
equal to p per unit of area, prove that the force required to produce a given
extension is less than that in a vacuum by Xp/(X-f/t) per unit of area of cross
section.
ART. 11] EQUATIONS OF EQUILIBRIUM. 267
The Bending of Rods.
10. To form the equations of equilibrium of a thin inextensible
rod bent in one plane.
First Method. In this method we consider the conditions of
equilibrium of a finite portion of the rod or wire. The method
has been used in Vol. I. Arts. 142 — 147 to determine the stress at
any point of a rod naturally straight and slightly bent by the
action of given forces, and the same reasoning may be applied to
rods whose natural forms are curved.
Let P be any section of a thin rod APB regarded as a curved
line. Let T and U be the resolved parts of the stress force along
the tangent and normal at P, and let L be the stress couple.
These represent the mutual action of the two parts AP, PB of the
rod on each other. These stresses are then obtained by considering
the conditions of equilibrium of the portions AP, PB separately.
Let Flf Fz &c. be forces acting at the points Dit D2 &c. of the
portion PB in directions making angles 8,, 8a &c. with the tangent
at P. Taking any directions along the tangent and normal at P
as positive, let T and Z7act on the portion PB in these directions;
we then have by resolution
T + 2^ cos 8 = 0, U+2FsmB = Q.
In the same way if plt pz &c. be the perpendiculars from P on
the lines of action of the forces, we have by moments L + ^Fp = 0.
These three equations determine Tt U and L when the form of the
curve is known.
11. Second Method. In this method we form the equations of
equilibrium of an elementary portion of the rod or wire.
Let PQ be any element of the rod and let the arc s be measured
from some fixed point D on the rod up to P in the direction AB,
so that s = DP. Let the stress forces of AP on PB be represented
268 BENDING OF RODS. [ART. 11
by a tension T acting, when positive, in the direction PA and a
shear U acting in the direction opposite to the radius of curvature
PC. Then the stress forces of QB on QA are represented by
T+dTin the direction QB and U + dUin the direction QC, these
directions being represented in the figure by the double arrow
heads. Let the stress couple at P on PB be represented by L, the
positive direction being indicated by the arrow head on the circle
at P; then the stress couple at Q on AQ is represented by L + dL
acting in the opposite direction, i.e. in that indicated by the double
arrow heads on the circle at Q. Let Fds, Gds be the impressed
forces on the element PQ resolved in the direction of the tangent
PQ and normal PC, taken positively when acting respectively
in the directions in which the arc s and the radius of curvature p
are measured. Let dty be the angle between the tangents at P
and Q, and let ty be so measured that -^ and s increase together.
Resolving the forces in the direction of the tangent and normal
at P, we have
In the limits these become
0 ..................... (1),
0 ..................... (2).
Also taking moments about P
.\dL+ Uds = 0 ...... (3).
ART. 13] EQUATIONS OF EQUILIBRIUM. 269
Writing dty = ds/p, these equations take the form
.(4).
+U =0
as
If each element of the rod is acted on by an impressed couple,
as well as by the impressed forces Fds, Ods, it must be taken
account of in the equation of moments. Let Jo's be its moment
taken positively when the couple acts on the element PQ in the
opposite direction to the couple L. We then add Ids to Uds in
equation (3) and therefore add / to the left-hand side of the last
of equations (4).
The positive directions. The positive direction of the couple L
at P on that part of the rod towards which the arc s is measured
is opposite to that in which the angle dty = ds/p is measured. The
positive direction of the shear U on the same part of the rod is
opposite to that in which the radius of curvature is measured.
The positive direction of the tension at P on the same part of the
rod is opposite to that in which the arc s is measured.
12. When we compare the advantages of the two methods of solution we notice
that the second gives differential equations which must be integrated, and the
constants must be determined by the conditions at the extremities. On the other
hand the first method, though it gives expressions for T, C7, and L, introduces into
the equations the action of all the forces on the finite arc PB. When, therefore,
the form of the strained curve is so well known that we can calculate the resolved
parts and the moments of the impressed forces the first method gives the required
stresses at once. When however the form of the strained curve is very different
from that when unstrained, and is itself unknown, the second method presents
several advantages over the first.
13. Experimental Results. When a thin rod or wire is
bent under the action of forces we have to determine not merely
the components of stress, i.e. T, U and L, but also the form of the
strained rod. The equations of equilibrium found above supply
three equations, so that a fourth is required to make up the
necessary number. For this purpose we have recourse to experiment,
Vol. I., Art. 148. If pl} p are the radii of curvature at any point P
before and after the deformation, the stress couple L is given by
270 BENDING OF RODS. [ART. 16
where K is some constant depending on the material of which the
rod is made and on the section at P. It is usually called the
fiexural rigidity of the rod. This expression for L agrees very
well with the results of experiment when the change of curvature
is not very great.
Since the moment L represents the product of a force and a
length, it is evident that the dimensions of K are represented by
a force multiplied by the square of a length. If E be Young's
modulus for the material of the rod and co the area of the section,
Em will represent a force, so that the constant K is often written
in the form K = Ewk2, where & is a length.
It will be shown further on that in certain cases wk2 is the
moment of inertia of the area of the normal section about a straight
line drawn through its centre of gravity perpendicular to the plane
of bending. This result does not agree so well with experiment as
that represented by (5).
14. It is hardly necessary to remind the reader of the remarks
made in Vol. I. Art. 490, on the limits to the laws of elasticity.
When the stretching or bending of the rod exceeds a certain limit,
the rod does not tend to return to its original form, but assumes a
new natural state different from that which it had at first. In
all the reasoning in which the equation (5) is used, it is assumed
that the bending is not so great that the limit of elasticity has
been passed.
16. The theoretical considerations which tend to prove the truth of the
equation (5) depend on the theory of elasticity and therefore lie somewhat outside
the scope of the present chapter. As however this theory clears up some of the
difficulties which belong to the bending of rods, it does not seem proper wholly to
pass it over. One case can be presented in a simple form, and that case will be
discussed a little farther on after the use of the equation (5) has been explained.
16. The work of bending an element. To find the work
done by the stress couple L when the curvature of an element of the
rod is increased from its natural value l/p1 to the value l/pa.
Let PQ be an element of the central line and let ds be its
length. As PQ is being bent, let ty be the angle between the
tangents at its extremities; let p be its radius of curvature. If ^
be the value of ^ when the rod has its natural form, the stress
couple L is L = K(-~^\ = K^~^1
\p PI/ ds
ART. 17] WORK OF BENDING. 271
The work done by the couple L when ty is increased by d^r is
- Ld^fr, (see Vol. I. Art. 292). The negative sign is given to the
expression because, as explained in Art. 11, L is measured in the
direction opposite to that in which ty is measured. The whole
work done by the couple when ty is increased from fa to fa is
therefore equal to - \K . ^2~^
as
Replacing fa, fa by their values in terms of />2, p1} we see that
the work Wds done by the couple L may be written in either of
the forms Wds = - \K ( - - 1Y ds = -^ .
\p» pj %K
If the change of curvature at every point of the rod is known,
the whole work done by the stress couples in the rod may be found
by integrating the first of these expressions along the length of the
rod. If however the change of curvature is unknown, and the
couple is given, the work is found by integrating the latter
expression.
Resilience. Resilience denotes the quantity of work that a spring, or elastic
body, gives back when strained to some stated limit and then allowed to return to
the condition in which it rests when free from strain. The word " resilience " used
without special qualifications may be understood to mean extreme resilience or the
work given back by the spring after being strained to the extreme limit within
which it can be strained again and again without breaking or taking a permanent
set. See Kelvin's article on " Elasticity " in the Encyc. Brit. 1878.
17. Deflection of a straight rod. A heavy rod, originally
straight, rests on several points of support A, B, 0 &c. arranged,
very nearly in a horizontal straight line, and is slightly deflected
both by its own weight, and by a weight W attached to a point H
between B and G. It is required to explain the method of finding
the deflection at any point of the rod and to determine the relations
which exist between the stresses at successive points of support.
Let A, B, C be three successive points of support. These are
so nearly on the same level that the distances AB = a, B0=b,
may be regarded as equal to their projections on a horizontal
straight line. To simplify the process of taking moments the
order of the letters used is exhibited in the upper figure, as if
they were all strictly in a horizontal line, instead of being only
very nearly so.
Let x be measured horizontally from B in the direction BG.
The rod, when bent by its weight, will assume the form of some
272
BENDING OF RODS.
[ART. 17
curve which differs very slightly from the nearly straight line
ABG. Let y be the ordinate at any point Q, between B and C,
measured positively upwards, from a horizontal straight line drawn
through B and let the radius of curvature be positive when the
concavity is upwards. The stress couple at the point Q is Kfp ;
when p is positive the fibres of the under part of the rod are
stretched while those above are compressed, hence the stress couple
at Q acts on QC in the clock direction and on BQ in the opposite
direction. Let the shear at Q be U and let its positive direction
when acting on QC be downwards.
p D' D H Q
(1).
Let Lz and U2 be the couple and shear at a point D indefinitely
near to B on its right-hand side. Let w be the weight of the rod
per unit of length, then the weight of DQ is wx, and this weight
acts at the centre of gravity of DQ. Let BH= f. Taking
moments about Q for the finite portion DQ of the rod, we have
-=Z2
The term containing W is to be omitted when Q is on the left-
hand side of H, i.e. when x < j~ .
In forming the right-hand side of this equation the rod has
been supposed to be straight and horizontal, because the deflections
are so small that only a very small error is made by neglecting the
curvature. If this were not so, the shear would not be vertical,
and the arm of its moment would be different from that used in
the equation. In the same way the thickness of the rod has been
neglected, and in all its geometrical relations the rod is regarded
as if it were a line coincident with its central axis, Art. 1.
ART. 18] DEFLECTION OF A STRAIGHT ROD. 273
The rod is supposed to be of such material that a considerable
effort is required to produce a slight curvature; the coefficient K
is therefore large. On the left-hand side of the equation all the
small terms cannot be rejected because these are multiplied by K.
It is however sufficient, in a first approximation, to retain only the
largest of these small terms. We therefore put
p Oi
The upper sign must be taken because p is measured positively
when the concavity is upwards, and in this case dyjdx is increasing
and therefore d^yjdxz is positive.
The general rule followed in these problems is, (1) that all
terms not containing K are formed on the supposition that the rod
has its natural shape, (2) that in all terms containing K as a factor
only the first power of the deflection y is retained. The natural
shape in our case is a horizontal straight line.
18. The equation (1) now takes the form
(2)
where x is restricted to lie between x = 0 and x = b and the term
containing W is to be omitted when a? < £. Let Z2' and t7"2' be the
stress couple and shear at a point D' indefinitely near B on its
left-hand side, and let R2 be the pressure of the point of support
B on the rod upwards. Consider the equilibrium of the small
portion D'D of the rod. The stress couples and the stress forces
at the extremities act on this element in the directions opposite
to those represented in the figure, the weight wds acts downwards
and the pressure jR2 upwards. We have, by taking moments about
J)', and resolving,
Z2' - Z2 = Uzds - %w (ds)z + Rz . BD' = 0)
// * / / __ ff — . _^ 411/1 Q — — II 1
\J a ™~ *— ' O ^^ -*-**2 ^^ M/W/O — — "I
Hence in the limit
Xa' = ia, U»'~ UZ = R* (3).
If we take a point P between A and B so that BP represents a
negative value of x, we have K -^ = L* — U*x — \wa? . . .(4),
where OB is restricted to lie between a?=0 and x = — a. Since
Zj' = Z2 this equation differs from (2) only in having Un' written
for U2, the term in W disappearing naturally.
R. s. ii. 18
274 BENDING OF RODS. [ART. 1.9
Lastly, if U be the shear at any point of the rod we have by
dL d?v
equation (3) of Art. 11 U= ~~fa= ~ ^dx* ^
It is evident that the two arcs AB, BO of the rod must have
the same tangent at B and therefore the same value of dyjdx. It
follows from the first of equations (3) that the stress couples on
each side of B are equal ; the two arcs have therefore the same
curvature. But the shears on each side of B differ by the pressure
.Ra, and therefore there is an abrupt change in the value ofd?y/da?
at a point of support.
These equations are sufficient to determine the stresses when
the terminal conditions are known. But the integrations must be
effected for each span separately and the conditions at the points
of junction allowed for. To shorten the mathematical labour we
require some method of passing over a point of discontinuity. This
is effected by the theorem or equation of the three moments, by
which a relation is found between the stress couples at any three
successive points of support.
19. Equation of the three moments*. Let us integrate
(2) over the length BQ of the rod. The limits for every term,
except the last, are x = 0 to x, and for the last term x — £ to x.
We thus have
-(^=L0-\U>x*-$wx*-lsW(x-Zr (6)
where $ is the inclination of the rod at B to the horizon.
Integrate again,
K(y-px) = %Lj?-1sU,x*-&wx<-1sW(x-tf (7).
Eliminating U, between (2) and (7) and writing L = Kd*y/da?,
we have
6K(y-/3a;) = (L + 2L2)a* + $wx* + W% (x - % ) (20 - f) ...(8).
This equation holds at any point Q between H and G. When
Q lies between B and H, the term with W is to be omitted.
Since C/2 does not appear in the equation, it also holds when Q
* The theorem of the three moments in its first form is due to Clapeyron,
Comptes Rendus, 1857, Tome XLV. ; but it has been greatly extended since then.
A sketch of these changes is given by Heppel in the Proceedings of the Royal Society,
1870, vol. xix. The extension to include the case of variable flexural rigidity is due
to Webb, Proceedings of the Camb. Phil. Soc. 1886, vol. vi. The allowance for the
yielding of the supports is given by K. Pearson, Messenger of Mathematics, 1890,
vol. xix.
ART. 21] EQUATION OF THE THREE MOMENTS. 275
lies between A and B, provided x is then regarded as negative.
Let y-i and ys be the altitudes of the points of support A and C
above B. The equation (8) becomes when x = 6 and x = — a
67T (y, - /%) = (L3 + 2£2) 62 + {wfr +Wg(b- f ) (26 - f )
6ffa + /8a) = (A + 2Z2)
Eliminate $ and we find
Here w is the weight of a unit of length of the rod. If the
spans AB, BG have unequal values of w, say w-i and ws, we write
^ (w:az + w368) for the fourth term on the right-hand side.
This important relation between the stress couples at any three
successive points of support is usually called the equation of the
three moments. By help of this equation, when the stress couples
at two of the points of support are known, the stress couples at all
the points may be found.
To find the shear at the point B of support, we take moments
about either G or A. We then have
L, = L2- U^b-^wb"- W(b-& ............ (10)
A = Z2 + Ut'a-$wa* ........................... (11)
which may also be derived from (2) by putting x = b and x = — a.
The pressure R2 on the point of support may then be found by (3).
If the point H at which the weight W is attached lie between
A and B instead of B and G, we reverse the positive direction of
x. Let the distance BH = £' measured positively from B towards
A. The last term of (9) must then be replaced by
vrY«-n<te-ry*
This may also be derived from the last term of (9) by writing — f
for £ and — a for 6.
20. The equation of the three moments when written in the
form (9) may be regarded as the relation between the ordinates
y\> 2/s of any two points and the stress couples L1} L3 at those points.
It may be used, for example, to find the deflection y8 at the free
end of a rod where La = 0.
21. If the rod rest on n points of support, the equation of the
three moments supplies n — 2 equations connecting the n stress
18—2
276 BENDING OF RODS. [ART. 21
couples Llt L2,...Ln at the points of support. Two more equations
are therefore necessary to find the n couples, and these may be
deduced from the conditions at the extremities.
If one end of the rod is free, and at a distance c from the
nearest point of support, the stress couple Ln at that point of
support is found, by taking moments about it, to be Ln — — \wcz.
If an extremity rest on a point of support the stress couple at
that point is zero.
If an extremity be built into a wall so that the tangent to the
rod at that point is fixed in a horizontal position we may imagine
that the fixture is effected by attaching that end of the rod to two
points of support indefinitely close together. The required condi-
tion at that end then follows at once from the equation of the three
moments. Let Xn+1 be the couple at the wall, Ln that at the
nearest point of support and let c be the distance, then writing
a = c, b = 0 in the equation of the three moments we have
The pressures on the points of support may be obtained by
combining equations (10), (11) and (3). If JR? be the pressure on
the rod measured upwards at B, we find by eliminating ET2, Ut'
The case in which W= 0 has also been attained in Vol. L Art. 145.
The weight W has been included in the equation of the three
moments to facilitate the calculations. It is however evident that
we may regard the point of the rod to which the weight W is
attached as a point of support at which the pressure is known.
Such a point may be included in the equation of the three
moments as one of the three consecutive points A, B, C. The
deflection at each of these points being unknown, the extended
equation of the three moments fails to determine the stress couple.
But the pressure being known, the equation (12) gives an additional
equation connecting the stress couples, and the extended equation
of the three moments then gives the deflection.
Yielding of the supports. In some cases the points of support
are the tops of vertical columns which are themselves elastic. Let
the bases of the columns be on the same level, h^, h^ &c., z-^Zi &c.
their original altitudes and their altitudes under pressure; crx,
ART. 22] EQUATION OF THE THREE MOMENTS. 277
o-o &c. their sectional areas, then for any column h — z = hRIE<r.
We have therefore the additional equations
-#2 -Ri R2 R,
yi = Zi-zz = -?--±, t/8 = _2_f^ &a
Pa Pi p2 ps
where pl = El o-j/^ &c. and E is Young's modulus.
22. Ex. 1. A uniform rod, weight W, is supported at its extremities; the de-
flection at its middle point is observed and found to be h. Show that the value of the
constant K for the rod is given by 48ft . K= 5o?W, where 2a is the length of the rod.
If the inclination to the horizon of the tangent at either end of the rod be observed
by a level and found to be 6, show that the value of K is also given by K=o?W/Qe.
This example shows how the value of K may be found by experiment for any
given rod.
Ex. 2. A uniform heavy rod is supported at its extremities A, C and at its
middle point B ; A and C are at the same level and B such that the pressures on
the three supports are equal. Prove that the depth of B below AC is 7/15ths of the
whole central deflection of the beam AC when supported only at its ends.
This example shows that when a long heavy bridge is supported on three columns
of equal strength, their summits ought not to be on the same level.
Ex. 3. A heavy rod rests on a series of points of support which are very nearly
in a horizontal line. Let A, B be any consecutive two of these points, a their
distance apart, yl , j/2 their altitudes above a horizontal plane. Let L^ , L2 be the
stress couples, 6lt 02 the inclinations of the rod to the horizon at A, B. Prove that
K (tan 02 - tan 0j) = \ (Lj + L2) a + ^ wa3,
The stress couples having been found, the first of these equations enables us to
find the inclination of the rod at any point of discontinuity when the inclination at
some point is known. The second determines the inclination at any one point.
Ex. 4. A uniform slightly elastic rod rests on five supports in the same
horizontal line, two at the ends and one at each of the points found by dividing the
rod into four equal parts. The second and fourth supports from either end are now
removed. Prove that the ratio of the new to the old pressure on an end support is
as 21 : 11. [Coll. Ex. 1893.]
Ex. 5. A uniform bridge of weight W formed of a single uniform plank is
supported at its ends : a man of weight W stands on the bridge at a point whose
distances from the ends are a and 6. Prove that the deflection just under the man
is ab { W (a2 + Sab + 62) + 8 JKa&}/24 (a + b) E,
where E is the bending modulus. [Coll. Ex. 1893.]
Ex. 6. A naturally straight weightless wire of flexural rigidity C has its ends A
and B built in horizontally at the same level, and is slightly bent by a weight W
attached to it at a point Q. Prove that the deflection y at a point P in AQ is given
by the equation Cy = | ^ BQ* . AP* (3 A Q . BP - BQ . AP).
Find the points of inflection of the axis of the wire and show that the point at which
the axis is horizontal is in the longer segment, and that its distance from the
corresponding support is bisected by one of the inflections. [St John's Coll. 1893.]
Ex. 7. A uniform heavy beam rests on three points of support, A and C at its
ends and B at the middle. The middle support is at first so placed at a depth yl
278 BENDING OF HODS. [ART. 24
below AC that the beam is entirely supported by A and C. The support B is then
gradually raised to a height y^ above AC such that the beam is wholly supported by
B. Prove that as B is being raised, the pressure at B is proportional to the height
raised. Prove also that the ratio t/x : y2 is equal to 5:3. [Fidler's Treatise on
Bridge Construction, 1893.]
33. Britannia Bridge. Ex. 1. A uniform heavy beam ABC is supported at
its extremities A, C and at its middle point B, and the three points are in one
horizontal line. Prove that 3/16ths of the weight is supported at either end and
5/8ths at the middle point. We notice that the pressure at the middle support is
more than three times that at either end.
Prove also that the stress couple is a maximum at a point which divides either
span in the ratio of 3 : 5, but the stress couple at either of these points is 9/16ths of
the stress couple at the central point of support. Prove that the latter is equal to
the stress couple at the middle point of a beam supported at each end whose length
is equal to that of either span.
Prove that there is a point of contrary flexure in each span dividing it in th«
ratio 1 : 3.
Ex. 2. A uniform beam is supported at its extremities and at two other points
dividing the beam into three equal spans, all the four points being on the same
level. Prove that the pressures on the supports are in the ratios 4:11:11:4.
Ex. 3. A uniform beam ABCDE is supported at its extremities A, E and at
three points B, C, D, all five being on the same horizontal line. To assimilate this
problem in some measure to the case of the Britannia Bridge the two middle spans
are supposed to be twice the lengths of the outside ones, i.e. BC=CD=z2AB=s2DE.
Prove that the pressures on A, B, C are in the ratios 4 : 27 : 34.
The examples in this article are taken from a treatise on The Britannia and
Conway Tubular Bridges by Edwin Clark, resident engineer, 1850.
The tubes AB, BC, CD, DE, which form the four spans of the Britannia Bridge,
were raised separately into their proper places and then rigidly connected into one
long tube. The connections at B and D were such that the tubes adjacent to each
had a common tangent. The junction at C was however so arranged that the
tangents to BC and CD should make a small angle with each other. The object of
this was to diminish the inequality between the pressure on C and that on either B
or D. It was found convenient to make the angle between the tangents equal to
2 tan"1 -002. In Example 3, given above, the analytical condition to be satisfied at
C is that the tangents to BC and CD should be continuous, but in the bridge the
condition is that these tangents should make a known small angle with each other.
24. Ex. 1. A rod without weight is supported at its extremities A, C and at
some other point B, all three being in the same horizontal line. Given weights P,
Q are suspended at the points D, E, bisecting AB and BC. Show that the inclination
to the horizon of the tangent at A and the deflection y at the weight P are given by
32 (a + b) K tan a= -Pa3 (a + 26) + QaP,
16S(a+b)Ky= -P(7a + 166) a»+9Qa2&3,
where AB=a, BC=b.
It appears from this result that when the point of support B bisects A C and
Q=3P the tangent at A should be horizontal. Moseley describes three experiments
with different rods supported on knife edges by which this curious result has been
verified. See his Mechanical Principles of Engineering and Architecture, 1855.
ART. 25] EXAMPLES.
279
Ex. 2. A uniform thin rod of length 2 (a + b) rests on two points of support in
a horizontal line whose distance apart is 2a. Show that, if the middle point and
the two free ends are on the same horizontal line, fe/a must be the positive root of
the cubic Sr3 + 9r2 - 3r - 5 = 0.
Ex. 3. A uniform heavy rod rests on any number of points of support in the
same horizontal line. Let A, B, C, D, E be any consecutive five of these, and let
their distances apart be a, b, c, d. Prove that the pressures R2, Ea, B4, at B, G, D
are connected by the linear relation aR3+/3R3 + yR4=: Jtc5, where
a=a2 (b + c) (c + d) (b + c + d),
S = (a + b) (b + c) (c + d) (b + c + d) (a + b+c) (a + b+c + d).
Ex. 4. Prove that the deflection y at any point Q between B and C, in Ex. 3, is
given by
- &Kby=BQ . GQ {£„ (b + CQ) +L8 (b+BQ) + ±wb (W + BQ . CQ)}.
Ex. 5. A wire is bent into the form of a circle of radius c, and the tendency at
every point to become straight varies as the curvature. Show that, if it be made to
rotate about any diameter with a small angular velocity u, it will assume the form
of an ellipse whose axes are 2c ( 1 ± -^ — ) , m being the mass of a unit of length,
V *3P /
and /t/c the couple necessary to bend the straight line into the circle. [Math. T. 1868.]
Ex. 6. A heavy elastic flexible wire originally straight is soldered perpen-
dicularly into a vertical wall. If the deflection is not small prove that the
difference between the tension at any point P and the weight of a portion of the
wire whose length is the height of P above the free end is proportional to the square
of the curvature at P. [May Exam.]
Ex. 7. A flexible wire is pushed into a smooth tube forming an arc of a circle,
and lies in a principal plane of the tube ; prove that it will only touch it in a series
of isolated points, and that if it only touch the inner circumference at one point,
the pressure there will be 4E cos a (sin a — sin 7) /a2 sin2 a, where a is the inner radius
of the tube, 2a the angle subtended at the centre by the wire, y the angle at which
either end of it meets the wire, and E the coefficient of flexibility. [Math. T. 1871.]
Ex. 8. Three very slightly flexible rods are hinged at the extremities so as to
form a triangle, and are attracted by a centre of force attracting according to the
law of nature situated in the centre of the inscribed circle. Show that the curvature
of any side, as AB, at the point of contact of the inscribed circle varies as
cos \A + cos %B - cos ^ G
cos \G
Ex. 9. Equal distances AB, BC, CD are measured along a light rod which is
supported horizontally by pegs at B, D below the rod and C above. A weight is
now hung on at A, producing at that point a deflection. Find how much B must be
moved horizontally towards A that the deflection may be unaltered when the peg D
is removed. [Coll. Exam. 1888.]
25. Ex. 1. A uniform heavy rod rests symmetrically on 2m + 1 supports placed
at equal distances apart, and the altitudes are such that the weight of the rod is
280 BENDING OF RODS. [ART. 26
equally distributed over the supports. Show that the altitude yp of the support
distant pa from the middle point, is given by
where a is the distance between two consecutive points of support and /3a is the
length of the rod beyond either of the terminal supports.
We first see by taking moments about the pth support that the stress couple Lp
at that point is a quadratic function of p. The extended equation of the three
moments is Lp + ±Lp+l + LM + $ wa? = (yp^ - 2yp+l + yp) 6-fiT/a2.
By an easy finite integration, or by the rules of algebra, it follows that yp is a
biquadratic function of p. Since there can be no odd powers of p, we have
The values of A and B are then found by applying the equation of the three
moments to any two convenient spans.
Ex. 2. A uniform heavy rod rests on m supports placed at the same level at
equal distances a from each other, one being at each end. Prove that the stress
couple at the nth point of support is
(l-J^U^-fl -&*-»)*»-»
*- —
where h and k are the roots of ft2 + 4ft + 1=0. Prove also that the pressure on the
nth support is aBn=3wa?- 6Ln except when n=l or m.
The equation of the three moments is an equation of differences and may be
solved in the usual manner by assuming Ln=A + Bhn + Ck*. The constants B, C
are determined by the conditions that Ln=0 when n=l and n=m. It is also
evident that h= -tan-j^w, fc= -
26. Flexural rigidity not constant. If the rod is not uniform the equation
of the three moments takes a more complicated form. We shall first suppose the
flexural rigidity K to vary from point to point of the rod, but the weight per unit
of length to remain constant. We start as before with the equation (2), Art. 18.
Let us multiply this equation by (b - x)/K and integrate over the length BQ. Since
where ya and /3 have the same meaning as before, we find
y3-b^jl(L,-U^-^)^^-wf^^/^dx ...... (I.).
Substituting for 17, from (10), this becomes
+ W. G,
The left-hand side of (I.) is the elevation of the point C of support above the
tangent at B. The equation obtained by integrating over the length AB is similarly
?/! + a/3 = I^A + LyA' + wA",
where A, A', A" are obtained from B, B', B" by writing a for b.
ART. 28] A BENT BOW. 281
Eliminate /3 and we have
which is the equation of the three moments when the flexural rigidity is not uniform.
When the weight w per unit of length is not constant, we may include the weight
in the term W. We put W=wdx and integrate that term throughout each of the
spans.
27. A bent bow. A uniform inextensible rod, used as a bow,
is slightly bent by a string tied to its extremities. It is required to
find its form.
Taking the string as the axis of so, the statical equation is
•
where T is the tension of the string. Let A, B be the extremities
of the rod, C a point on the rod at which the tangent is parallel to
the string. Let OC be the axis of y. Then since dyjdx vanishes
when x = Q and decreases algebraically as x increases, d2y/da? is
negative. In forming (1), p has been taken as positive, we must
therefore give the second term the negative sign. Putting T = Kn-
for brevity, the equation gives y = h cos nx ............... (2),
C
where h is the versine of the arc formed by the bow. It is obvious
that unless the conditions of the problem make h small, we cannot
reject the terms containing (dyjdxf in the expression for p in
equation (1).
The form of the curve given by the equation (2) is sketched in
the diagram. It appears therefore that the bow may take the form
ACB, the string being attached at A and B. It may also take the
form ACB' with the string attached at A and B', and so on.
28. We may easily find a second approximation to the
solution of the differential equation. This is perhaps necessary,
for, owing to the smallness of the inclination of the rod to the
string, if the ordinates near B were slightly decreased, a considerable
change might be made in the distance OB.
If we substitute for p its full value, the differential equation
--rfyi+' (3).
282 BENDING OF EODS. [ART. 29
Expanding the right-hand side we have
We see that the terms on the right-hand side are of the third
order of small quantities. We therefore assume as a trial solution
y =. k cos cnx + Bk3 cos Sena ............... (4)
where A; is a small quantity analogous to h, and c, B are as yet
undetermined constants. Substitute in the differential equation
and neglect all powers of k above the third, we then have
(1 — c8) n2k cos cnx + (1 — 9c2) Bk*n*c* cos 3
= — |n2 (k cos cnx) (&Vna sin2 cnx)
= — f n*k*c2 (cos cnx — cos Scnx}.
The equation is therefore satisfied if we put
The solution to the third order of small quantities is therefore
y = k cos cnx — -fr$n*k9 cos Scnx ............... (5)
where c exceeds unity by the small quantity ^-n2F. Let, as before,
h represent the distance 0(7; we have y = h when x = 0, hence
h = k-&nW ........................ (6).
Let the lengths of the string and the rod be 2a and 21, then
when x = a, y = 0, and the least value of a is given by cna = %7r..
We also have
when terms above the order k3 are neglected. Eliminating a, we
have I = J (1 - -ftn*) (1 + ^cW) = ^ (1 + ^n*> . . .(8)
when the fourth powers of fc are neglected.
Smce »• = ?/* we have
when the fourth powers of h/l are rejected. This equation
determines the tension necessary to produce a given deflection 00 = h.
29. Let us regard the half OB of the bow as a uniform rod
having one end C and the tangent at 0 fixed while the other end
B is acted on by a force T whose direction is parallel to the
tangent at C.
ART. 31] A BENT BOW. 283
Let the length I be given, then the equation (8) shows that k
is imaginary unless I exceeds 7r/2n. Let n0 = irj^l and let T0 = Kn<?
be the corresponding value of the force T. It follows that the rod
cannot begin to bend unless the force exceeds T0, where T9 = K7^l^l-.
Let T = T0 (1 + £) where | is a small quantity, then
The equation (8) gives
Since dk/dt; is infinite when k = 0, we see that k (and therefore
also A) increases much more rapidly than the force does. A slight
increase in the force makes a considerable change in the value ofk.
30. When the terms containing dyjdx are included in equa-
tion (1), we have -K—2L-— = Ty (10),
where accents denote differential coefficients with regard to x.
Multiplying by y' and integrating, we find
where i/r is the acute angle made by the tangent at any point P
with the string of the bow.
Let y = h cos <£, then sin ^i|r = ^nh sin <j>. The equation may be
written in the form d-^r/ds = ri*y. Put e = ^nh, substitute for y
and i/r and integrate between the limits s = 0 to s = I, we then have
_ fr d<j>
~J0 (l-^sin2^
nl
If the length I and the force T are given, ri* = TjK is also known.
This equation then determines e and therefore h.
The integral (12) is lessened by writing unity for the denomi-
nator. We then have nl > %TT. Since ri*=T/K it immediately
follows that the tension or force must exceed the value of 7rl&Y4 P.
This is the result already arrived at in Art. 29, and it has now
been proved without the use of series. The equations (8) and (9)
of Art. 28 may be obtained by expanding the integral (12) in
powers of e2 and neglecting all powers of e above the second.
31. The importance of the case considered in Art. 29 lies in
its application to the theory of thin vertical columns. The rod
may be regarded as a vertical column having the tangent at its
lower end C fixed in a vertical position, while a weight, much
greater than that of the column, is supported on the upper ex-
tremity. It appears from what precedes that if the weight on
284 BENDING OF EODS. [ART. 32
the summit is gradually increased, the column will remain erect,
without bending, until the weight becomes nearly equal to a
certain quantity depending on the flexibility and dimensions of
the column.
Since the constant K is equal to Eal? (Art. 13) it follows that
the bending weight, for columns of the same kind, varies as the
fourth power of the diameter directly, and as the square of the
length inversely. This result is usually called Euler's* law.
Columns yield under pressure in two ways, first the materials
may be crushed, and secondly the column may bend and then
break across. In some cases both effects may occur at once. If
the column is short it follows from Euler's law that the bending
weight is large, so that short columns yield by crushing. Long
columns on the other hand break by bending and are not crushed.
Many experiments have been made to test the truth of Euler's
law. The results have not been altogether confirmatory, possibly
because Euler's law applies only to uniform thin columns, in which
the central line in the unstrained state is a vertical straight line.
For the details of these experiments we must refer the reader to
works on engineering. See also Mr Hodgkinson's Experimental
researches on the strength of pillars, Phil. Trans. 1840.
In this investigation we have supposed that the weight has been placed centri-
cally over the axis of the column. The weight of the column itself has also been
neglected and no allowance has been made for the shortening of the column due to
the weight it has to support.
32. Heavy columns. Ex. 1. A vertical column in the form of a paraboloid
of latus rectum 4m with its vertex upwards is fixed in the ground. Show that it
will bend under its own weight when slightly displaced if the length be greater than
IT (2Emlwfi, where w is the weight of a unit of volume, E the weight which would
stretch a bar of the same material and unit area to twice its natural length.
Ex. 2. A vertical cylindrical column of radius r is fixed in the ground. Show
2 /9£r2\ 1
that it will bend under its own weight if its length be greater than <? ( — — I ,
^ low /
where c is the least root of «7_i (c) = 0.
Let A be the area, r the radius of a section of the column (supposed to be thin
and straight) at a distance x from the base G, then (Art. 13), K=EAk*. When the
* Euler, Berlin Memoirs, 1757. Petersburg Commentaries, 1778. Lagrange,
Acad. de Berlin, 1769. Poisson, Trait€ de M&canique, 1833. See also Thomson
and Tait, vol. i. Art. 611, where some figures are given. Also the Proceedings of the
Roy. Irish Acad. 1873, where Sir B. Ball notes an error in Poisson's analysis. In
the Proc. London Math. Soc. 1893, vol. xxiv., Prof. Love discusses the stability of
columns. A discussion of Euler's theory is contributed to the Canadian Society of
Civil Engineers, 1890, by C. F. Findlay, C.E.
ART. 33]
THEORY OF A BENT ROD.
285
column is a paraboloid Ak*=%irr* and r*=4m(l-x), when the column is a cylinder
Ak* is constant. In the figure of Art. 27, let x', y' be the coordinates of any point
P' between P and B. Taking moments about P the differential equation is seen to
be
where A' is the area of the section at f. Differentiating this equation with regard
to x, we find after some reduction
where $=l-x, pP=w!2mE, and the column is supposed to be a paraboloid.
At the free end where |=0, we have l~dyjdZ = Q and, since the stress couple is
there zero, d2yld^=0. At the base where |=Z we have dyjd^=0 and this leads to
the condition that the column cannot begin to bend unless lfi>ir.
When the column is a cylinder, the differential equation becomes
which may be reduced to Bessel's form. To effect this put dyjd^ = ^z, p=|'t, we
then see that X = £, /* = §••
Both these examples are due to Prof. Greenhill, Proceedings of the Camb. Phil.
Soc. 1881, vol. iv.
33. Theory of a bent circular rod. A uniform thin straight rod without
weight is bent without tension into the form of a circular arc of great radius; it is
required to find the stress couple at any point P. See Art. 15.
We shall obtain a particular solution of this problem by making an hypothesis
which simplifies the process, and which we afterwards verify by showing that all the
equations of equilibrium are satisfied.
We assume (1) that all filaments of matter parallel to the length of the rod are
bent into circles with their centres on a straight line perpendicular to the plane of
bending. This straight line will be referred to as the axis of bending. We assume
(2) that the particles of matter which in the unstrained rod lie in a normal section
continue to lie in a plane when bent, (3) that this plane is normal to the system of
circles above described.
B'
Let ABCD be a short length of the straight rod bounded by two normal planes
AOC, HMD. To examine the small changes which this length undergoes we take
the plane AOC as that of yz and let some perpendicular straight line OH be the axis
of *. To avoid confusing the figure only the lines on the positive octant have been
£86 BENDING OF RODS. [ART. 33
drawn. Let the plane of xz be the plane of bending, so that the axis of y is parallel
to the axis of bending. Thus OA is the axis of z, 00 that of y. Let QR be any
elementary filament parallel to the axis of x, let (0, y, z), (a;, y, z) be the coordinates
of Q and JR. Let the positions of these points and lines in the bent rod be denoted
by corresponding letters with accents. According to the hypothesis A'O'C', B'M'D'
are normal to all the filaments of the bent rod, and (when produced) these planes
intersect in the axis of bending. Any filament, such as Q'R', is a circular arc whose
unstretched length is OM.
The rod being bent without tension, the filaments near A'B' are compressed
while those on the opposite side of the rod are extended. There is therefore some
surface such that the filaments which lie on it have their natural length. This
surface is usually called the neutral surface, and the lines on it parallel to the length
of the rod are called neutral lines. Since the filaments on this surface are circular
arcs of the same length with their centres on the axis of bending, the neutral surface
is a cylinder which cuts the plane of yz in a straight line parallel to the axis of
bending. Let the origin Of be taken on the neutral surface, the axis of x is there-
fore a tangent to a neutral line, and the unstretched length of every filament, such
as Q'R', is equal to OM or O'M '. Let p be the radius of curvature of this neutral
line. Since the rod is thin, all the linear dimensions of the mass ABGD are small
compared with p.
When the unstretched length QR has been compressed or stretched into the
length Q'R', it remains sensibly parallel to the axis of x, but its distances from the
planes xz, xy may have been altered. Let these distances be y'=y + v, z'=z + w, and
let the stretched length Q'R' be x'=x + u. Since R' lies in a plane normal to the
neutral line at M', we have x'=(p-z-w) sin -=x-- - '— .
The difference x'-x represents the stretch of the fibre QR whose unstretched
length is x. The tension per unit of sectional area is therefore equal to - E - .
P
When the rod is only slightly deformed by the bending (as in Art. 17) the displace-
ment w must be small compared with z. We may then, as a first approximation,
equate the tension to - Ezjp.
Since the rod has been bent without altering its length, the resultant tension
across the section AOC is zero, and we have
tf(Ezlp)dydz=0.
It immediately follows that the centre of gravity of the section lies in the plane of
xy. The neutral surface therefore passes through the centre of gravity of every
normal section. In a cylindrical rod therefore, bent without tension, the central line
is also a neutral line.
Since the elementary tensions have no components parallel to the axes of y or *,
it follows that the shear is zero.
If L be the moment about the axis of y of the tensions which act across the
section AOC, measured positively from z to x, we have
P P
where wfc2 is the moment of inertia of the sectional area about the axis of y, Le.
about a straight line drawn through the centre of gravity of the section perpendi-
cular to the plane of bending, see Art. 13. Since the rod is a uniform cylinder bent
into a circular arc, the corresponding couples about O'C', M'D' balance each other.
ART. 34] THEORY OF A BENT ROD. 287
In the same way the moment about the axis of z of the tensions which act across
the section AOC is \\yzdydz.E\p. This couple cannot be balanced by the equal
couple about M'B' because their axes are not parallel. It is therefore necessary that
this moment should vanish. It follows that the rod will not remain in the plane of
bending unless the product of inertia of the area of the normal section about the
axis of y and any perpendicular straight line in its plane is zero. In other words,
the plane of bending must be perpendicular to a principal axis of the section at its
centre of gravity.
34. If we suppose, as already explained in Art. 8, that each fibre or filament of
the rod is contracted or extended in the same manner as if it were separated from
the rest of the rod, the mutual pressures of these filaments transverse to the length
of the rod and also the tangential actions are zero. Each element of the rod is
therefore in equilibrium, and the surface conditions are also satisfied. Each
filament is slightly displaced, like those discussed in Art. 8, and slightly turned
round. These displacements are those represented by v, w, and are such that,
when the fibres are stretched independently of each other, the body remains
continuous.
The expressions for the coordinates y'=y+v, z'=z + w, of Q' in terms of the
coordinates y, z of Q may be deduced from the theorems given in Art. 8. It
follows from that article that when the filament QR is stretched into the filament
Q'R by a tension Nx, the rectangular base QLMN remains rectangular and similar
to its original form, and is of such size that corresponding sides are connected by
the relation (Q'L1 - QL)/QL = - NJE'.
Let <f> be the angle which the side Q'L' makes with the axis of y, measured posi-
tively from z to y ; then
Rejecting the squares of the small quantities v, w and remembering that QL=dy,
dv Nr dw
we have -=- = — =f> , - tan d> = -3— .
ay E' ay
Treating the side Q'N' in the same way, we have — -= - -£ , tan0 = -r- .
az Hi az
Substituting for Nx its value -E(z + w)lp, and neglecting wjp as before, we find by
E yz+f(z) E z*+F(y)
integration v=_JL_m, w=_ __M.
Equating the two values of tan <f> and substituting for v and w, we find that
It follows fh&tf(z) = az + b, and therefore
_E (y + a)z + b _ E
'~~~ ~
p ~2E' p
The terms containing 6 and a2 — c represent a translation of the section as a whole,
those containing the first powers of y, z represent a rotation through an angle Ea/E'p.
If neither of these displacements exist, we may omit these terms.
The expressions thus found for u, v, w, give the displacements of Q referred to
the axes O'M ', O'A', QfC'. They also give those of R referred to corresponding
axes with M ' for origin. The displacements of R referred to the axes with & as
origin are therefore given by
xz E yz a;2 E z*-y*
™» "=¥>' w=*p + W-vT'
where x, j/, z are the coordinates of R.
288
BENDING OF RODS.
[ART. 36
35. If the section of the beam is a rectangle having the sides EF, GH perpen-
dicular to the plane of bending, we see by examining the expression for v and w that
these sides become curved when the rod is bent, and that they have their convexities
E F
O
H
turned towards the centre of curvature of the rod. The sides EG, FH which before
bending were parallel to the plane of bending remain straight lines but are inclined
to the plane of bending and tend outwards on the concave side of the rod.
The expressions found in Art. 34 for the displacements u, v, w agree with those
given by Saint- Venant for one case of bending. But what has been said in that
article is not to be taken for a complete discussion of his problem ; for that the reader
should consult a treatise on the theory of Elasticity.
The second and third assumptions of Art. 33 are included in the first, either if
the circle is complete, or if proper forces are applied at the extremities of the arc.
The first assumption may be regarded as following from the statement in the
enunciation that the rod is uniform, without weight and bent into the form of a
circular arc.
In the theory of Bernoulli and Euler these assumptions are applied to the case
of any thin rod *. The theory thus extended leads to the result that the bending
moment is proportional to the curvature and this result agrees with experiment.
But other results of the theory are not so nearly in agreement with facts. To
obtain a correct theory it is necessary to have recourse to the general equations of
equilibrium of an elastic solid. In this treatise the expression for the bending
moment is intended to rest on experiment (Art. 13), and the bending of a circular
arc has been considered merely as the simplest example of the theory of elasticity.
36. Airy's Problem. In using standards of length two considerations have
attracted attention, (1) the application of supports in such a manner as to produce
no irregularities of flexure and (2) the application of such supports as will permit
the expansive or contractive effects of temperature. The importance of the former
was made known by Eater, that of the latter by Baily. Freedom of expansion is
usually secured by supporting the body on rollers. Excessive flexure is avoided by
making the rollers rest on levers which are so arranged that the weight of the body
is either equally distributed over the points of support or distributed in such ratios
as may be thought proper.
The flexure is so small that the mere curvature of the central line does not
produce a sensible alteration of its length. If however the measured length is
marked on the upper surface of the measuring rod, this length may be either stretched
or shortened by the curvature of the central line. There may therefore be a small
error in each length measured by the rod, which would be multiplied indefinitely
when the whole distance measured is great. The problem is to determine how this
* Prof. Pearson shows in The Quarterly Journal for 1889 that the results of the
Bernoulli-Eulerian theory give fairly approximate formulae for the stress and strain
of beams whose diameter is one-tenth, or less, of their length.
ART. 37] CIRCULAR RODS. 289
error may be avoided. Airy's principle is that the extension of each element of the
upper surface of the measuring rod is proportional to the bending moment L. He
therefore infers that the supports of the rod should be so arranged that Jl/dr = 0, the
limits of integration being from one end of the measured length to the other.
We may deduce the correctness of this principle from the theory given in Art. 33.
The extension of the filament QR has been shown to be approximately QR (zip),
•where p is the radius of curvature of the central line and z the distance from the
central line of the projection of QR on the plane of bending. If then z be the half
thickness of the rod, the extension of an element dx on the surface is zdx/p. Since
L=K/p, it immediately follows that the extension of any element on the surface of
a uniform rod is proportional to the bending moment.
Ex. 1. A bar, of length a, is supported at two points symmetrically placed, and
the marks defining the extremities of the measured length are close to its ends ;
prove that the distance between the points of support should be a/^/3.
Ex. 2. A standard of length a is supported on m rollers placed at equal distances,
and the weight is equally distributed over the rollers. The measuring marks are
placed at distances e from the ends. If D be the distance between two consecutive
rollers, prove that D ,J(m>-\)=aJ(l- SeS/a3).
Memoirs of the Royal Astronomical Society, Vol. rv., 1846, and Monthly Notices,
Vol. vi., 1845.
37. Bending of Circular rods. The natural form of a thin
inextensible rod is a circular arc; supposing it to be slightly flexible,
it is required to find the deviation from the circular form produced
by any forces*.
Let AB be the arc of the circle when undeformed, 0 its centre,
a its radius. Let P be any point on
the circle, P' the corresponding point
on the rod when bent. Let a, 6 be the
polar coordinates of P; a (1 + u), 6 + $
those of P', referred to 0 as origin.
If /a be the radius of curvature at P',
we have by a theorem in the differential
11 1 / , dto\ /1N
--- = -- \u + ^a> •••(!)>
pa a\ dfrj
calculus
p
where the squares of u are neglected. Let us represent either side
of this equation by qfa.
If the central line be extensible, let d^ and ds be the un-
stretched and stretched lengths of an element of arc, then
d*! = add, (ds? = (adu)* + a? (1 + u? (d9 + d(f>y.
* The case of a circular arc is important because the periods of its vibrations,
both when inextensible and extensible, can be found. See the second volume of the
Author's Rigid Dynamics, where also the expression for the work of the stresses is
found in a different manner.
B. 8. II.
19
290 BENDING OF RODS. [ART. 38
Neglecting the squares of small quantities, this gives
ds = a (1 + w) dO + ad<f>.
If p be the proportional elongation of the elementary arc
ds-dSl_d(f>
P' ds, ~d6*
If the rod is inextensible, we have p = 0.
The equations of equilibrium of an inextensible rod may be
formed by either of the methods described in Arts. 10, 11. Taking,
for example, the three equations marked (4) in Art. 11, and joining
them to L = K^, p = Q ..................... (3),
(Z>
we have five equations to find T, U, L, u, <f> in terms of 6.
38. If the rod is slightly extensible as well as flexible, the equations become
somewhat changed. The arc ds in the equations of equilibrium in Art. 11 means
now the stretched length of the element, while F and G represent the impressed
forces referred to a unit of length of the stretched rod. The equation p = 0 must
also be replaced by another connecting p with the tension.
The relations which connect L and T with p and q are perhaps most easily
deduced from the expression for the work done by the stresses when the rod is
deformed. If Wds^ be the work done by the stresses when the element is stretched
and bent, we have Wds1= - ^ ( Hp* +^£) ........................... (4),
\ a /
where H and K are the constants of tension and flexural rigidity. This result
follows at once from those given in Art. 16 of this volume and in Art. 493 of Vol. i.,
when we assume that the work due to a deformation of bending is independent of
that of stretching.
From this expression for W we may deduce the values of T and L. Keeping
one end P' of an element P'Q' fixed, let the element be further stretched, without
altering the curvature, so that its length ds becomes ds', then dp= — - - . The
uS-t
work done by the tension T at the end Q' is - T (ds' - ds), and that done by the
(j[gf _ (%9
couple at Q' is - L - . The sum of these is dW . dsl . We therefore have
......................... (5).
p
Next let the element, without altering its length, receive an increase of curvature
so that the radius of curvature is changed from p to p' ; then — = — — . The
a p p
tension at Q' does no work, while the work of the couple L at Q' is -L { — -- ) ds.
\P P/
A\sods = (l+p)dslt .• -£«J!L_JL .............................. (6).
a dq 1 + p
These expressions give for a slightly extensible and flexible rod
(7).
ART. 40] CIRCULAR RODS. 291
The equations of equilibrium found in Arts. 10 and 11 when joined to (7)
supply five equations from which L, T, U, «, <f> may be found
39. Ex. One end of a heavy, slightly flexible inextensible wire, in the form
of a circular quadrant, is fixed into a vertical wall, so that the plane of the wire is
vertical and the tangent at the fixed end horizontal. Assuming that the change of
curvature at any point is proportional to the moment of the bending couple there,
prove that the horizontal deflection at the free end is irwatl8E, where E is the
flexural rigidity, w the weight of a unit of length, and a the radius of the circle.
[Trin. Coll. 1892.]
Let A be the free end of the rod, B the end fixed into the wall, 0 the centre.
Taking moments about any point P for the side PA, Art. 10, we arrive at
E d*u
•where AOP=&, and OP=a(l+u). The constants of integration are determined
from the conditions that u and dujdO vanish at B, and the deflection required is
the value of au when 0=0.
4O. To find the work when a thin rod, whose central line in the natural state is
a circle of radius a, is stretched and bent so that the central line becomes a circle of
radius p, by a method analogous to that used in Art. 33 for a straight rod.
The figure of Art. 33 may be used in what follows, except that the lines OM,
AB, CD must be supposed to be small arcs of circles.
Let OM be an element of the central line of the unstrained solid, O'M' the same
element when the rod is deformed. Let the tangents to OM, O'M' be the axes of x
and x', and let the planes of xz, x'z' be the planes of the circles. Let QR be any
filament parallel to OM, Q'R' its position in the strained rod. Let y, z ; y', z' be
the coordinates of Q, R ; Q', R', each referred to its own set of axes.
If dslt ds be the lengths of OM, O'M' and 1+p stand for dsjdsl as before, the
tension of O'M' per unit of area is Ep. If dcr^, d<r be the lengths of QR, Q'R',
we have dffl=d»l ( 1 — 1, da=ds ( I -- ) .................. (1),
and the resultant tension of all the fibres which cross the area dydz is therefore
The work done by this tension when the filament is pulled from its unstretched
length dffi to the length d<r, is
-$Edydz (^ - lYd*! ........................... (3).
The difference z'-z is a small fraction of z; for a straight rod it has been
shown to be of the order 22/p, Art. 34. As a first approximation we take z'=,z.
Substituting for dsjds1 and for 1/p their values 1 + p and (1 4- g)/a, and neglecting
all powers of z/a above the second, we find that the work is
1 ............ (4).
Integrating this over the area u of the section, and remembering that 0 is the
centre of gravity of the area, we have for the whole work
........................ (5);
when the higher powers of p, q are rejected this becomes
........................... (6)-
19—2
292 BENDING OF RODS. [ART. 43
41. In the same way we find that the tension of the fibres which cross the
areacfydzis Edydz Yp-q (1+J>) •£ + f -J j- J ........................ (?)•
Remembering that 0 is the centre of gravity of the section, we find by an obvious
integration that the resultant tension T and the resultant couple L are given by
(8).
These reduce to the forms given in Art. 38 when the product pq is neglected.
42. If we examine the expressions for the work, tension and couple given by
equations (6) and (8) of Arts. 40, 41 we see that they contain two constants of
elasticity, viz. Eta and EwW. These were represented by the letters H, K in the
corresponding expressions in Art. 38.
When the rod is such that the constant of elasticity Eta is infinite or very great,
a small change in the proportional extension p alters the product Eup very con-
siderably. If, therefore, the tension is finite or not very great, p must be very
nearly equal to zero. It follows that in all the geometrical relations of the figure
we may regard p as equal to zero. At the same time the product Etap which occurs
in the tension is not to be regarded as zero, but as a quantity analogous to the
singular form oo . 0. If the tension is finite, the term Ep2 which occurs in the
work is zero.
Since the other constant of elasticity, viz. EwW/a?, is not necessarily large in
thin rods, it does not follow that q must be small, because Eta is large.
Rods in which Eu is very great are said to be inextensible. Such rods may be
bent, and the bending couple is proportional to the change of curvature.
43. Very flexible rod. When the flexibility of the rod is such that it may be
made to pass through several small rings not nearly in one straight line the
integrations of the differential equation become more intricate. To simplify the
problem we suppose that though weights may be attached to any points, the rod
itself is without weight.
Let A, B, C &c. be a series of small smooth rings through which the rod is
passed. Let the stress couple at A be Llt and let Tlt U^ be the tension and shear
at the same point. Let Lit T3, Ut be the corresponding stresses at £ and so on.
A
The stress Llt Tlt Uj acting at A may be reduced to a single resultant ^ acting
along some straight line A'B', whose position is found in Vol. i. Art. 118. If J" be
any point between the rings A and B, the stress at f must be equivalent to the
same force, for otherwise the portion AP1 of the light rod would not be in
ART. 44] VERY FLEXIBLE ROD. 293
equilibrium. In the same way the stress at every point of the rod between the
rings B and C is equivalent to a single resultant Fz acting along some other
straight line B'G', and so on for every portion of the rod lying between contiguous
rings. These straight lines may be called the lines of pressure. We shall suppose
the forces JFlf Fa, &c. when positive to be pulling forces, so that for instance
the action of AP on PB is equivalent to the force F1 acting in the direction B'A'.
The stress forces at points one on each side of any ring, as B, being F1 and Fz, it
follows that the pressure on the ring B is the resultant of ^ acting along B'A' and
F2 reversed and thus made to act along B'G'. The pressure at B therefore acts
along B'B, and this line is a normal to the rod at B.
Let us consider the equilibrium of any portion BP of the rod, where Pis a point
between B and G. Let if/ be the angle the tangent PH makes with B'C', and let
B'G' be the axis of £. Let 17 be the perpendicular distance of P from that axis.
Let L, T, U be the stress couple, tension and shear at P. Then
U=-Fasin^, L=F^j ..................... (1).
Taking moments about P for the portion BP we have —=K~= F2t] ...... (2) .
Multiplying both sides by sin ^ = drj/ds and integrating, we find - 2K cos ^ = Fpf* + G.
This result may be written in the form
2KF2cosf + FJr)2=I ... .............................. (3),
where I is a constant for the portion BP of the rod. We notice that in this
equation F3 cos \f/ is the tension and F2t) the stress couple at the point P.
A similar equation holds for each portion of the rod which lies between
contiguous rings. If P move along the rod and pass through the ring G, the
tension and stress couple undergo no sudden changes of value, though the shear is
altered discontinuously. It follows that jF2cos^ and F2rj are the same on both
sides of C and that therefore I is the same for both portions of the rod. The
constant I has therefore the same value throughout the whole length of the rod.
If one extremity of the rod is free, let A be the ring nearest the free end. The
tension and the stress couple at A are therefore zero ; hence, by equation (3) the value
of I is zero. In this case, since the stress at A is reduced to the shear only, the
line of pressure between the rings A and B is the normal at A.
Since pcos^=(dsjd\f/) (d£/da) = d£/d^, we have by (2)
d£_jBTcos^_ JTcos \f/ ..
d$ F27) (I - 2KF2 cosffi'"
where % is measured positively opposite to the direction of F2. Putting \f/=v-20,
we reduce this to the difference of two elliptic integrals,
F2|=i I (1 - c2 sin2 «)* dO - - I - — - r ,
7V ij (l-c2sin-'0)*
where P=I + 2KFa and c2i2=42TF2.
44. To show that these results supply a sufficient number of equations, let us
suppose, as an example, that both ends of the rod are free and that it has been
made to pass through five small rings at A, B, C, D, E.
Beginning at the ring A, the line of pressure A'B' is the normal at A; let 8 be
the angle it makes with any fixed straight line in the plane of the rings. Taking
AB' as the axis of £ and A as origin, the coordinates of B, viz. £, 17, are known
functions of 6. The equations (3) and (4) give £, 17, in terms of f and Flt the
constant in (4) being determined from the condition that when £=0 the value of ^
is known, viz. in this case ^ is a right angle. Equating these two values of £ and 77
294 BENDING OF RODS. [ART. 45
we have two equations to determine F1 and the value of \(/ at B. The tangent at
B having been found, the normal BB' can be drawn and the position of B
determined.
In the figure of Art. 43 we have Fl sin A'B'B = F2 sin BB'C'. When therefore we
repeat the process just described and take B'C' as a second axis of | and the foot of
the perpendicular from B on B'C' as the origin, with the object of finding F2 and the
value of ^ at C, we really have sufficient equations to find the angle BB'C' also.
In the same way we next take C'D' as the axis of £ and finally D'E'. But since
this last line of pressure must be the normal at E, the value of ^ at E must be a
right angle. This supplies a final equation from which 0 may be found.
Ex. A light rod DE is made to pass through two small rings A, C in the same
horizontal line at a distance apart equal to 26, and has a weight W applied at a
point B so that the vertical through B bisects AC at right angles. If 20 be the
angle between the normals at A and C prove that
On rods in three dimensions.
45. Measures of Twist. Let PK be a normal to the
central line of an elastic rod at any point P, and let K lie on
the outer boundary of the rod, the portion PK is called a
transverse of the rod. This name is due to Thomson and Tait.
Let P, P\ P" &c. be a series of adjacent points on the central
line of the unstrained rod, and let each of the arcs PP', P'P" &c.
be infinitesimal. Any transverse PK having been drawn at the
first of these points, let the plane KPPf intersect the normal
plane at P' in a second transverse P'K'. Let the plane K'P'P"
intersect the normal plane at P" in a third transverse, and so on.
We thus obtain a series of transverses, any consecutive two of
which lie in a tangent plane to the central line.
If the rod when unstrained is straight and cylindrical it is
obvious that all the transverses thus drawn lie in a plane passing
through the central line. It is also clear that the extremities
Kt K' &c. of the transverses then trace out a straight line on the
surface of the rod parallel to the central line.
Let these transverses be fixed in the material of the rod and
move with it when the rod is strained. The normal section at P
of the rod being fixed, let the elements lying between the normal
planes at P, P', P" &c. be twisted round the tangents PP',
P'P" &c. respectively, so that the points K, K', K" &c. trace
out a spiral line on the outer boundary of the rod. The twist
of the elementary portion of the rod which lies between the
ART. 47] RESOLVED CURVATURE. 295
normal planes at P, P' is measured by the infinitesimal angle
which the transverse P'K' makes with the plane KPP't or, what
is ultimately the same, by the angle which the planes KPP',
PP'K' make with each other. If the arc PP' of the central line
be ds, and if the angle which the planes KPP', PP'K' make with
each other be d-%, the ratio d-^/ds represents the twist of the portion
ds of the rod referred to a unit of length, and is usually called the
twist at P.
It is sometimes useful to so choose the transverses PK, PK' &c. in the unstrained
rod that the angle which the planes KPP', PP'K' make with each other has any
convenient value. Let dxi be this angle and let dx^r-yds, then TJ is an arbitrary
function of the arc «. If dx or rdt be the corresponding angle in the strained rod,
the twist is measured by r - r^ .
46. Resolved Curvature. Let a straight rod be strained
by bending, so that the central line takes the form of a curve of
double curvature. If de be the angle between the normal planes
to the central axis at P, P', the curvature at P is measured by
the ratio dejds, and the central line is said to be curved in the
osculating plane.
It is sometimes more convenient to resolve the curvature in
two directions at right angles. Let the normal planes at P, P'
intersect each other in a straight line CO, then GO intersects the
osculating plane at right angles in some
point C. Since PC, P'G are two con-
secutive normals lying in the osculating
plane, the point G is the centre of the
circle of curvature ; let GP = p. Let us
now draw a plane through the tangent
PP' to the central line making an arbi-
trary angle <j> with the osculating plane, and let this plane cut CO
in Q. Then since PQ, P'Q are two consecutive normals to the
central line, the point Q is the centre of a circle of curvature
drawn in the plane QPP'. If the radius PQ of this circle be R,
we have from the right-angled triangle QGP, -p = - cos </>.
It follows that the curvature in a plane drawn through the
tangent may be deduced from the curvature in the osculating plane
by the same rule that we use in statics to resolve a force.
47. Let us draw two planes through the tangent at P to the
central line, and let these be at right angles to each other. Let
296 BENDING OF RODS. [ART. 48
the resolved curvatures of the central line in these planes be
called K and X. Then the curvature in the osculating plane is
VX^ + X2), and the tangent of the angle the osculating plane
makes with the first of these planes is X/«.
These two planes intersect the normal plane at P to the
central line in two straight lines at right angles. Let these be
PK, PL, the straight line PK being perpendicular to the plane
in which the resolved curvature is K.
The three straight lines PK, PL, PP' thus form a convenient
system of orthogonal axes to which we may refer that part of the
rod which lies in the immediate neighbour-
hood of P. The resolved curvatures of the
central line in the planes perpendicular to
PK, PL, being K, \ and the twist about
PP' being r, it follows that in passing from
the point PtoP' the three axes are screwed
into positions P'K', P'L', P'P" by a combination (1) of the rota-
tions icds, \ds, rds about the axes PK, PL, PP', and (2) of a
translation of the origin P along the tangent to P'. It should
be noticed that each of the three quantities K, X, r is of — 1
dimension as regards space.
The quantities K, \ are the resolved curvatures of the strained
rod and are the same as the resolved bendings produced by the
forces, only when the unstrained rod is straight. To find the
bending produced by the external forces when the unstrained rod
is itself curved we must subtract from K, X the resolved curvatures
of the unstrained rod.
48. Since icds, \ds, rds are rotations about the axes of
reference, we know by the parallelogram of angular velocities
that they may be resolved about other axes by the parallelogram
law. If then we wish to refer the strains to a different set of
axes, say PKlt PLlt PT1} we change K, X, r into Klt \, TX by the
usual formula? for the transformation of coordinates or for the
resolution of forces. In this way we may refer the bending and
twist in the neighbourhood of^P to any arbitrary system of axes
having the origin at P. These generalized axes may be screwed
from their positions at the origin P to those at P' by the three
rotations x^ds, Xjcfo, Tids and the translation ds along the
tangent.
ART. 51] RELATIONS OF STRESS TO STRAIN. 297
49. In many of the applications of analytical geometry to physical problems
it is found advantageous to make the coordinate axes moveable, so that they may
always be in the most convenient position. Thus if a point travel along the strained
rod and successively occupy the positions P, P', P", &c., the axes change their
directions in space. To specify the motion of these axes we may either use a
second system of axes fixed in space or we may refer the motion to the moving
axes themselves in the manner above described. The first method requirnfi
the use of the formulae of transformation of axes which are often complicated, in
the second we avoid the introduction of a second system of axes. Moving axes are
of great importance in Dynamics and are also of much use in discussing the
geometrical properties of curves and surfaces. For these applications the reader is
referred to the second volume of the Author's Treatise on Rigid Dynamics,
50. Ex. 1. A straight line is marked on the surface of a thin unstrained
cylindrical rod, parallel to the central line. If the rod is bent along any curve on
a spherical surface so that the marked line is laid in contact with the spherical
surface, show that the twist is zero.
If the rod is laid on a cylindrical surface so that the marked line is in contact
with the cylinder, show that the twist is sin a cos a/a, where a is the radius of the
cylinder and a is the angle the rod makes with the axis of the cylinder. Both these
results are given by Thomson and Tait, Art. 126.
If P, P' be two consecutive points on the central line, the transverses PK, P'K'
are normals to the surface. The first result follows, because the transverses pass
through the centre of the sphere, so that the angle between the planes KPP,
PP'K' is zero. Since the radius of curvature at any point of a helix lies on the
normal to the cylinder on which the helix is drawn, the second result follows from
the ordinary expression for the radius of geometrical torsion.
Ex. 2. A straight thin rod has a straight line marked along one side. If the
rod is bent and laid on a surface so that this line lies in contact with a geodesic,
show that the twist at any point P is A sin 6 cos 0, where A is the difference of the
curvatures of the principal sections of the surface at P and 0 is the angle the rod
makes with either line of curvature.
51. Relations of stress to strain. Let P be any point on
the central line ; the mutual action of the parts of the rod on each
side of the normal section at P can be reduced to a force and a
couple with any convenient point of that section as base.
Let three rectangular axes be taken at the point P to which
we may refer the strains and stresses in the neighbouring portion
of the rod. Let K, L, T be the components of the stress couple
about these axes. If the unstrained rod is straight, let K, X, T be
the resolved parts of the curvature and twist about the axes ; if
the unstrained rod is itself curved, then K, X, r represent the
changes in the curvature and twist produced by the external forces.
We shall now assume the two following principles*:—
(1) that the changes in the twist and curvature of the rod in
* See Thomson and Tait, 1883, Art. 591.
298 BENDING OF RODS. [ART. 53
the neighbourhood of P are independent of the force and are
functions only of the couple ;
(2) that the couples K, L, T are linear functions of the strains
K, \, T.
These assumptions are necessary because we do not in this
place enter into the theory of elasticity.
If we suppose, as usual, that the strains are so small that we
may neglect all powers but the lowest which enter into the
equations, the second principle is equivalent to the assumption
that when K, L, T are expanded in powers of K, \ T the lowest
powers in the series are the first.
52. Since the three couples K, L, T are each expressed in
terms of K, \, T by a different linear equation, it might be supposed
that we shall have to deal with nine constants. But if the elastic
forces form a conservative system we may reduce these to six by
using the work function.
Let Wds be the work function of an element of the rod
bounded by the normal sections at P, P'. Supposing the end P
fixed, let one strain, say X, become \ + d\, the other two remaining
unaltered. Since the element of the rod has been rotated about
the axis of the couple L through an angle equal to d\ . ds, the work
done by the couple L is Ld\ds, while that done by each of th.e
couples K and T is zero. We therefore have dsd W = LdXds.
Similar expressions hold when K and T are increased by die and
dr, so that in general
K = d W/dic, L = d W/d\ T = d W/dr.
Since K, L, T are linear functions of K, \, T it follows that W is
a quadratic function of K, \, T, Le.
W = % (A K* + £X2 + Or2 + 2a\r + 26™ +
53. We have already seen that if we refer the strains to
another set of axes the quantities K, \, r are changed by the
ordinary formulae for transformation of coordinates, Art. 48.
Since a homogeneous quadratic expression can always be cleared
of the terms containing the products of the variables, it follows
that by a proper choice of the axes of reference the expressions for
W, and therefore those for K, L, T may be reduced to the simplified
forms W = £ (A&? + AV + (fo8),
ART. 54]
HELICAL TWISTED RODS.
299
These axes are called the principal axes of stress and the constants
Alt Bit C1} are the principal flexure and torsion rigidities.
In what follows it will generally be assumed that the tangent
to the central line at P is one principal axis of stress at P ; this is
of course the axis of torsion. If also the constants of rigidity for
the other two principal axes are equal, we have
where the suffixes have been dropped as being no longer required.
The expression for the work is not complete if the rod is
extensible, for we have not yet taken account of the extension or
stretching, of the element PP' of the rod. This additional term is
given in Vol. I. on the supposition that the tension obeys Hooke's
law. It will not be required in the problems considered in this
chapter.
54. Helical twisted rods. A uniform thin rod, naturally
straight, whose principal stress axes at any point are the central line
and any two perpendicular axes, is bent into the form of a helix of
given angle and receives at the same time a given uniform twist.
It is required to find the force and couple which must be applied at
one extremity, the other being fixed, that the rod may retain the
given strains.
Let APQ be an arc of the helix, A the fixed extremity, Q the
terminal at which the forces are applied. Let AMB be a circular
section of the cylinder on which the helix lies, OZ the axis of the
cylinder.
The mutual action of the portion AP of the helix and the
portion PQ consists of a force and a couple. From the uniformity
of the figure it is clear that
the force and couple must
be the same in magnitude
wherever the point P is
taken on the helix, and
that their direction and
axis respectively must
make the same angles
with the principal axes of
the curve at P.
The stress force at P
may be resolved into two
300 BENDING OF RODS. [ART. 54
components, one acting along the generating line MPF of the
cylinder and the other acting parallel to the plane XT. The
latter, if it is not zero, must be in equilibrium with the compo-
nent at Q parallel to the same plane. This however is impossible,
because as P is moved along the helix the direction of the com-
ponent at P makes always the same angles with the principal
axes at P and is therefore changed, while that of the component
at Q remains unaltered. Both these components must therefore
be zero. It follows that the resultant stress force at any point
P must act along the generator through that point
Let R be the stress force at any point of the rod. The force
R may be transferred to the axis OZ of the cylinder by introducing
the couple Ra acting in the plane OZFM. The force R thus
becomes independent of the position of P.
Let us now turn our attention to the stress couples at P. Let
Px be drawn perpendicular to the axis of the cylinder and let
TPz be a tangent at P to the helix. Then by the known
properties of the curve, the plane TPx is the osculating plane
at P. Let Py be the binormal. If p = I/K be the radius of
curvature of the helix, the strains round Py and Pz are re-
spectively K and r, each being measured in the positive direction
round the axes, i.e. from z to x and x to y. There is no strain
round Px because the rod is naturally straight. If A and G are
the constants of flexure and torsion, the corresponding stress
couples are AK and Cr. These couples may be resolved into two
components, one having the generator PF for axis and the other
having its axis parallel to the plane of XY.
Let the resultant of the latter couple and of the couple Ra be
called H. The couple H at P together with the force R acting
along OZ must be in equilibrium with the corresponding reversed
couple H' and the reversed force R at Q. The forces are equal
and opposite, hence the couples H, H' must be in equilibrium.
Since the axis of the couple H always makes the same angle with
OM, its direction is altered when the point P is moved along the
helix while that of the couple Q is fixed. Equilibrium cannot
exist for all positions of P unless both H and H ' are zero. The
stress at P is therefore equivalent to a wrench whose force acts
along the axis OZ of the cylinder and whose couple acts in a plane
perpendicular to that axis.
Consider the equilibrium of the portion AP of the helix. The
ART. 55] SPIRAL SPRINGS. 301
fibres of the rod which are nearest to OZ are compressed and those
more remote are stretched. Hence the arc QP tends to turn AP
round the binormal Py in the direction z to x. Also, as P travels
along the wire in the direction APQ the positive direction of the
torsion is from a; to y, hence the twist couple exerted by PQ on
AP is in the same direction, viz. x to y.
The stress couples which act at P on the portion AP of the
rod are therefore L = 0, M = AK, N = Cr round Px, Py, Pz respec-
tively. These together with the force at P are equivalent to a
wrench, let G be the couple measured clockwise round OZ and let
R be the force acting along OZ. By equating the moments of
these about HP and also about a parallel to MT drawn through
P, we find that
G = AK cos 0, + Gr sin a,
Ra = — AK sin a + Gr cos a.
Here R tends to pull out the spiral AP, and G to twist it round
OZ from A to B.
These equations determine R and G when the angle a of the
spiral, the curvature K and the twist r of the material are known.
By giving the proper twist, we can make G = 0 and then the
spring can be maintained in the spiral form by a force R only.
55. Spiral Springs. A thin rod or wire, whose natural form
is a given helix and whose principal axes of stress at any point are
the tangent to the central line and any two perpendicular axes, is
bent into the form of another given helix. It is required to find the
forces and couples which must be applied at one end, the other being
fixed, that the rod may retain the given form.
Let Oj, a be the radii of the cylinders on which the unstrained
and strained helices lie; a1} a. the angles of the helices. Let the
axes of the two cylinders be coincident and let it be taken as the
axis of Z, the plane of XY being perpendicular to it.
Let P, P' «&c. be a series of consecutive points on the central
line of the unstrained rod and let Pf, P'%' &c. be the principal
normals at these points. The angle between the consecutive
planes £PP', PP'f is efesin ^coso,/^ where ds is the arc PP'.
Let PW, P'v' be the binormals at the same points, then the
curvature of the unstrained rod, measured, as in Art. 47, round
the binormal, is ds cos2 ^/Oj. Let P£ P'? be the tangents to the
helix taken positively in the direction in which s is measured.
302 BENDING OF RODS. [ART. 55
Let the axes of f , 77 be fixed in the material of the rod and be the
transverses of reference. When the rod is strained, let Px, Py, Pz
be the principal normal, binormal and tangent at P, then P£
coincides with Pz, and P£ makes some angle <j> with Px. The
figure of Art. 54 may be used to represent the strained position
of the rod, the axes Pff, Pi), P£ are not drawn but may easily be
supplied by the description just given.
The stress at the point P of the strained rod consists of (1) a
force which we may suppose to be resolved into two components,
one along the generating line of the cylinder and the other
parallel to the plane of XY. (2) A couple C (r — Ta), whose axis
is the tangent Pz, and two couples AK and — AK^ whose axes are
Py and Prj respectively ; where
sin «! cos at _ cos2 o^ _ cos2 a
Ti = > KI = > K = ,
a, Oi a
and rds is the elementary angle between the planes |PP', PP'£'
in the strained rod.
Examining first the stress force, we find, as in Art. 54, that
the component parallel to the plane of XT is zero. The stress
force at every point P therefore acts along the generating line of the
cylinder ; let this force be R, and let it be transferred to the axis of
the cylinder by introducing a couple Ra.
Taking next the stress couples, we find by the same reasoning
that the component about any axis parallel to the plane of X Y is
zero. Let us first equate to zero the moment about Px; since
Px is perpendicular to Py, Pz and to the axis of the couple Ra,
and makes with Prj an angle \TT + <j>, we have /Ci sin <£ = 0. Since
KI is not zero (as it was in Art. 54), it follows that <f> = 0. The axes
P£ and Px therefore coincide and the couples AK and — AK^ have
a common axis Py, viz. the binormal of the strained helix. The
angle rds is also equal to the angle between the consecutive
osculating planes to the strained helix, i.e. r = sin a cos a/a.
Equating to zero the moment about a perpendicular to the
plane passing through Px and the generator of the cylinder, we
have Ra = — A sin a (K — K^) + (7 cos a (r — Tt) (1).
Equating the moment about a generator to the corresponding
moment at the terminal we have
G = A cos a (/c — KJ + (7 sin a (T — TJ) (2).
ART. 57] EQUATIONS OF EQUILIBRIUM. 303
The curvatures and torsions are
_ cos2 C*! _ cos8 a _sina1cosa1 sin a cos a
KI — - , K — - , T! , T = .
Oi a Oi a
56. If the spiral spring have a great many turns so that
«j and a are both small, we have when the squares of alt a. are
neglected Ra = — Aa { ) + G (
\a 0^1 \a c^y
If there be no couple G but only a force at each end pulling the
spiral out, we deduce from these equations that a = alt so that the
spring occupies a cylinder of the same radius as before the strain.
We also have Ra = C 1 ,
a
which is independent of the constant of flexure. It appears there-
fore that the spring resists the pulling out chiefly by its torsion.
It is stated by both Saint- Venant and Thomson and Tait, that
this result was first obtained by Binet in 1815.
Let I be the length of the spiral spring, h the elongation of its
axis produced by the force R ; then
I sin a — I sin «j = h.
Rejecting as before the squares of ax and a we find that R=C.j-:,
lei
This expression determines the force required to produce a given
elongation in a given spring of small angle.
57. Equations of Equilibrium. To form the general equa-
tions of equilibrium* in three dimensions of a strained rod.
Let P, P' &c. be a series of consecutive points on the central line
of the unstrained rod. Let a series of transverses PK, P'K' &c.
be drawn such that the angle of twist TX is either zero or some
arbitrary function of the arc s. Taking the transverse PL per-
pendicular to PK, let the resolved curvatures about these lines
be \i and Kt. If these transverses are the principal flexure and
torsion axes at each point of the rod they form a convenient
* The general equations of a rod in Cartesian coordinates may be found in the
Treatise on Natural Philosophy by Thomson and Tait, 1879. The intrinsic equa-
tions, or those referred to moving axes, are given in the Treatise on Statics by
Minchin, 1889.
304 BENDING OF RODS. [ART. 57
system of coordinate axes. If not let some other system of axes,
Pf, Pij, P£, be chosen which are connected with the trans verses
PK, P'K' &c. in a known manner. Let 01, #2> #3 be the resolved
parts of KI} \!, T! about the axes P£, Prj, P£. Then, as explained
in Art. 48, the axes at P may be brought into positions parallel to
those at P' by rotations O^ds, 02ds, 03ds about themselves.
When the rod is strained the axes P£, Pij, P£ will move with
the material of the rod and assume new positions in space. Left
these be Pa, Py, Pz. Let w^ds, o)2ds, a)3ds be the rotations by
which the axes at P in the strained rod are brought into positions
parallel to those at P'. The differences (o^ — 0j) ds, (o>.2 — #2) ds,
(0,3 — 08) ds may be used to measure the strains produced by the
external forces.
Let jKj, .R2, R*', L!, L3, L3 be the stress forces and couples
which act at P on the element PP' in front of P and let them be
estimated as positive when they act in the negative directions of
the axes at P. Then R^ + dR^ &c., Ll + dL1 &c., are the corre-
sponding forces and couples at P' and act on the element PP',
behind P', in the positive directions of the axes at P'. Besides
these the element is acted on by the impressed forces F:ds, F2ds,
Fsds and the impressed couples (if any) G-ids, Gzds, G3ds.
Since MI, R2, R3 are the components of a vector, viz. the stress
force at P, the differences of the resolved parts at P and P' along the
same set of axes are given by the rule for resolving vectors* ; we
therefore have -v-1 — Q)8-Z^ + o^R3 + Fl = 0 (1),
~2 -<»>£, + 0,3^ + ^ = 0 (2),
ds
dR3
~ds
(3).
* The following proof of the rule is the same as that given in the second volume
of the Author's Rigid Dynamics.
Describe a sphere of unit radius whose centre is at P and let the axes Px, Py, Pz
cut its surface in x, y, z. Let parallels to the corresponding axes at P' drawn
through P cut the surface in a/, y', z'. Thus we have two spherical triangles xyz,
x'y'z', all whose sides are quadrants. Also x, y, z are brought into coincidence with
x", y', t? by the combined effect of the rotations Wjds, u.2ds, u3ds about Px, Py, Pz
respectively.
Let U, V, W be the components of the vector at P in the directions of the axes
x, y, z; U+dU, &c. the components of the vector at P7 along the axes x', y', z'.
The difference of the resolved parts along the axis of a; is then
'- U.
ART. 58] EQUATIONS OF EQUILIBRIUM.
305
In the same way since Ll} L2, L3 are the components of a vector,
we have ~ - &>3£a + &>2L3 + Gl + fj,R3 - vR, = 0 ......... (4)
-~ - wilt + w3A + G3 + vE^ -\R3 = Q ......... (5)
3 (6),
where \, /*, v are the direction cosines of the arc PP' referred to
the axes at P.
The relations between the couples Llt &c., and the strains
MI — 0i, &c., may be deduced from the expression for the work W
given in Art. 52, by writing a>1-01> &c. for K, r, X. Supposing for
the sake of brevity that the axes are the principal flexure and
torsion axes, we have
L1 = A(co1-01), La = B(a>2-e,\ i, = 0(w,-08)...(7).
If the axes are the tangent at P to the central line and two per-
pendicular axes, we have X = 0, p, = 0 and v = 1 ; but in all cases
\, fjb, v are known from the given conditions of the rod.
We thus have nine equations to determine the quantities
It1} R2, R3; L1} L2, L8; G>I} &>2, »3. If the rod is extensible there
will be another equation supplied by Hooke's law.
58. The meaning of these equations will be made clear if we
apply them to the simpler case in which the rod is uniform and
when unstrained is straight and without twist. In this case
#! = (), 0a = 0, #8==0, and MJ, &>2, &>3 are the components of the
curvature and twist. Let us also take the tangent PT as the
axis of x and the principal flexure axes PK, PL as axes of y and z
The rotations about Px, Py cannot alter the arc xy, but the rotation about Pz will
move y' away from x by the arc u3ds. In the same
way the rotations about Px and Pz cannot alter the 3
arc xz, but the rotation about Py will move z" to-
wards x by the arc u^ds. Therefore
xy' =xy + u3ds, xz' =xz- u^ds.
Also the cosine of the arc xx' differs from unity by
the square of a small quantity. Substituting we
find that the difference of the resolved parts along
the axis of x is dU'- Vtaads + Ww^ds.
If U, V, W stand for Bl, Ez, E3 we join to this the
force F^ds ; equating the result to zero and dividing
by ds, we obtain the first of the six equations. If
U, V, W stand for Z,1 , L2 , L3 we add the couple G^ds
and the moments of the forces J?j + dJR, &c. acting at P. We thus obtain in the
same way the fourth of the six equations.
E. s. ii. 20
306 BENDING OF RODS. [ART. 59
so that \ = 1, /i = 0, y = 0. The equations (1) to (6) then take the
forms
0, -w.A + ca.lH-tf, =0,
0,3^ + ^ = 0, -0,^3 + 0,3^+^-^3 = 0,
as
and the equations (7) become L1 = Aco1, L2 = Bo)2, L3 = Ca)z.
From these modified equations we can immediately deduce the
equations of the bending of a rod in two dimensions as given in
Art. 11. Let the plane of the rod be the plane of xy and as
before let the axis of x be the tangent, then &»i = 0, o,2=0,
o»8 = l//j. The stress forces Ri, R2 are respectively T, U, while
.Rs = 0; the stress couples A = 0, £a = 0, Ls = K(oi. Also (^ = 0,
£2 = 0, #, = 0 and in the notation of Art. 11, F1 = F, F2= G,
Fs = 0. The two first and the sixth equations immediately reduce
to those given in Art. 11, while the third, fourth and fifth are
identities.
59. Supposing the rod to be uniform and when unstrained to
be straight and without twist, we find, by eliminating Ll} Za, L3,
three equations of the form
These are the same as the three equations of the motion of a rigid
body about a fixed point (with s instead of t). This analogy is
due to Kirchhoff, but we cannot properly discuss it here.
ASIATICS.
On Astatic Couples.
1. THE conditions of astatic equilibrium in two dimensions
have already been investigated in the first volume of this treatise.
We have now to consider what other conditions are necessary
when the body is displaced in any manner in three dimensions*.
* The subject of Asiatics appears to have been first studied by Moebius, who
published his results in his Lehrbuch der Statik, 1837. Moigno also, in his Statique,
has discussed the subject at great length. Minding in the fifteenth volume of
Crelle's Journal gave the theorem that, whenever the body is so placed that the
forces admit of a single resultant, that resultant intersects two conies fixed in the
body. Many proofs have been given of this curious theorem ; we may mention that
by Darboux, Tait's proof by quaternions modified by Minchin ; Lannor's proof with
the use of the six coordinates of a line.
Darboux published in the Memoires de la Society des Sciences physiques et
naturelles de Bourdeaux, t. n. [2e Serie], 2* Cahier, a very long paper on this subject.
In contradiction to Moebius, he showed that when one point of a body is fixed there
are in general four positions, and only four, in which the body can be placed so that
the forces are in equilibrium. These he called the initial positions of the body.
His investigation is rather long and a different proof is given here. He also
introduced the idea of a central ellipsoid analogous to the momenta! ellipsoid
used in discussing moments of inertia. This result is given in Art. 14 of the text,
and the general lines of his argument have been followed in that article. By the
use of this ellipsoid he gave a geometrical turn to the proof of Minding's theorem,
but it remained rather complicated. Extending the theory by considering all
positions of the body, he showed that Poinsot's central axis formed a complex
of the second order, such that each straight line is the intersection of two perpen-
dicular tangent planes to the conies used by Minding. The first part of this result
was subsequently arrived at by Somoff in 1879.
The theorems on Asiatics given by Moigno may be found in his Lecons de
Mgcanique Analytique, 1868, which he tells us are chiefly founded on the methods of
Cauchy. As his demonstrations are different from those given in this treatise, it
may be useful to indicate the plan of his work. First, by a transformation of axes,
he obtains the twelve equations of equilibrium given in Art. 11. Thence he deduces
the conditions that a system of forces can be astatically reduced to a single force by
considering what single force can be in equilibrium with the system. Supposing
these conditions not to be satisfied, he shows that the system can be reduced to two
forces provided two conditions are satisfied. These conditions agree with the two
last determinantal equations given in Art. 73. He next shows that the system can
always be reduced to a force and two couples and that the point of application of
the force may be arbitrarily chosen on a plane fixed in the body. This plane is
defined to be the central plane. He then shows that if the arbitrary point is
properly chosen the directions of the forces and of the arms may be simplified in the
manner described in Art. 27. This point is defined to be the central point. Pro-
ceeding next to consider the case in which the body is so placed that the forces
admit of a single resultant, he shows that that single resultant must intersect two
conies fixed in the body. He next discusses the case in which the equilibrium is
astatic only for displacements of the body round a given axis ; following the same
plan as before, he enquires into the conditions that the system can be reduced to
one, two or three forces. He concludes with an application to magnetic forces and
investigates the positions of the central plane and central point.
20—2
308 ASTATICS. [ART. 4
2. We shall suppose, as before, that each force acting on the
body retains the same direction in space, the same magnitude and
continues to act at the same point of the body, for all displace-
ments.
The forces of a couple remain parallel, equal, and unaltered in
magnitude as the body is moved, but the length of the arm is
not necessarily the same. Let A, B be the points of application
of the forces, then the distance AB is unaltered, and is called the
astatic arm of the couple. If in any position of the body the
inclination of the astatic arm to the forces is 0, the arm of the
couple is AB sin 6.
The product of either force into the astatic arm is called the
astatic moment of the couple. The astatic moment is of course un-
altered by any change in position of the body. Representing the
astatic moment by K, the actual moment in any position of the
body is K.sinO.
The angle 6 which the astatic arm makes with the force is
called the astatic angle of the couple.
Two couples are said to have the same astatic effect when they
are equivalent in all positions of the body.
For the sake of brevity the couple whose force is P and astatic
arm is AB is represented by the symbol (P, AB).
3. The astatic effect of a couple is not altered if we replace it
by another having the same astatic moment, the astatic arms being
parallel, and the forces acting in the same direction in space as
before.
Let the astatic arm AB be moved to a new position A'B' in
the body. The extremities of the astatic arm of a couple are
fixed in the body and move with it ; thus as the body is displaced,
AB and A'B' continue to be parallel to each other. The astatic
angles of the two couples continue therefore to be equal to each
other. Since the astatic moments are equal, it follows that the
actual moments of the couples are equal. The two couples are
therefore equivalent.
It may be noticed that we cannot in general turn the astatic arm of a couple
through any angle in the manner explained in Vol. i. Art. 92; for the planes of the
couples may not remain parallel to each other, unless the displacements of the body
are restricted to be parallel to the original plane of the couples.
4. To find the astatic resultant of two couples whose forces are
parallel but whose astatic arms are inclined at any angle.
ART. 6] ASTATIC COUPLES. 309
Let AB, A'B' be the astatic arms of the couples, the forces at
A, A' being supposed to act in the same direction in space.
Through any point 0 draw OL, OM to represent the directions
of AB, A'B' and let the lengths of OL, OM be proportional to the
astatic moments of the couples. We shall now prove that the
diagonal ON of the parallelogram described on OL, OM will
represent in direction the astatic arm of the resultant couple and
in length the magnitude of the astatic moment of that couple.
Let the straight lines OL, LN be fixed in the body. By Art. 3
the two couples may be replaced by two others having OL, LN
for their astatic arms and having the four forces all equal. The
two forces acting at L being equal and opposite may be removed,
so that the two given couples are equivalent to two equal and
opposite forces acting respectively at 0 and N. These two forces
constitute a single couple having ON for its astatic arm and
having its astatic moment proportional to the length of ON. The
proposition is therefore proved.
From this proposition we infer that the theorems used to
compound forces apply also to compound the astatic arms of
couples having their forces parallel. It is hardly necessary to
add that the forces of the resultant couple are parallel to those of
the two constituents.
5. To find the astatic resultant of two couples whose astatic
arms are parallel but whose forces are inclined at any angle.
Let AB, A'B' be the parallel astatic arms of the couples, both
AB, A'B' pointing in the same direction in the body. Through
any point 0 draw 00 parallel to AB and also two straight lines
OL, OM parallel to the forces at A and A' and proportional to
the astatic moments of the couples. We shall prove that the
diagonal ON of the parallelogram OLM represents the moment
of the resultant couple, the plane of the couple is parallel to the
plane NOG, and the astatic arm is in the direction of 00.
Let the couples be referred to a common astatic arm along 00,
the forces at 0 are then represented by OL and OM. Proceeding
as in Art. 4 the results stated are easily seen to be true.
6. Working rule. Uniting these two propositions we may
construct a rule to resolve or compound couples.
When the forces are parallel we resolve or compound lengths,
measured along the astatic arms and proportional to the astatic
310 ASTATICS. [ART. 8
moments, by the parallelogram law, the new forces "being supposed
to act parallel to their former directions.
When the arms are parallel we resolve or compound lengths,
measured along the directions of the forces and proportional to
the astatic moments, by the parallelogram law, the new arms
being parallel to their former directions.
7. There is one resolution of a couple which will be found
useful afterwards.
Let Ox, Oy, Oz be any set of Cartesian axes, not necessarily
rectangular. Let (x, y, z) be the coordinates of any point D, and
let OD = r. Then a couple whose astatic arm is r and forces ± P
may be resolved into three other couples whose astatic arms are
situated in the axes of coordinates and whose lengths are equal to
ar, y, z. The forces of these couples are parallel to that of the
original couple and their astatic moments are Px, Py, Pz.
Let us now take any three points A, B, G on the axes and let
OA = a, OB = b, OC=c. These three couples may be replaced by
three others having OA, OB, OC for their astatic arms. It follows
that any force P acting at any point D may be replaced by four
parallel forces acting at any four points A, B, G and 0 whose
magnitudes are respectively equal to Pxfa, Py/b, Pzfc and
Conversely, since these four parallel forces may be compounded
into a single force equal to their sum and acting at the centre of
gravity of A , B, C, 0, it is evident that they are equivalent to the
force P acting at the point (x, y, z). See Vol. L, Art. 80.
8. Two couples cannot be astatically compounded together into a single resultant
couple unless either the four forces are parallel or the two astatic arms are parallel.
If possible let three couples be in astatic equilibrium. Transfer these parallel to
themselves so that one force of each couple acts at the point 0. Let OA, OB, OC
be the astatic arms, let OP, OQ, OR be the directions of the forces. Then as the
body is displaced, OA, OB, OC are fixed in the body, OP, OQ, OR are fixed
in space.
If the four forces of any two of the three couples are parallel, the forces of their
resultant couple are also parallel to them, by Art. 4. Thus equilibrium could not
exist unless all the six forces were parallel to each other. In what follows, we may
therefore suppose that no two of the three lines OP, OQ, OR are coincident. In the
same way no two of the three arms OA, OB, OC are coincident.
Place the body so that OC, OR are in one straight line. Since in this position
the couples (P, OA), (Q, OB) are in equilibrium, the planes POA, QOB coincide.
Thus OA, OB lie in the plane POQ and continue to lie in that plane as the body is
turned round OC. It follows that the axis OC must be perpendicular to this plane
and therefore to both OA and OB. Similarly OA is perpendicular to both OB and OC.
ART. 9] THE TWELVE ELEMENTS. 311
Supposing as before that 00, OE are in one straight line, it is clear that the
body may be turned round OC until OA coincides with OP. The axis OB must
then coincide with OQ, for otherwise equilibrium could not exist. Summing up, the
axes OA, OB, OC are at right angles and the body can be so placed that the forces
of the respective couples act along their astatic axes.
Keferring to the figure of Art. 76, Vol. i., we see that if the couple (P, OA) is a
stable couple, the couple (Q, OB) must be unstable, for otherwise they would not act
in opposite directions when the body is rotated about OC. Similarly by rotating
the body about OB we see that (B, OC) is an unstable couple. Therefore (B, OC)
cannot balance (Q, OB) when the body is rotated about OA. The three couples
cannot therefore be in equilibrium in all positions of the body.
The Central Ellipsoid.
9. To reduce any number of forces astatically to a single force
and three couples.
Let the forces be P1} P2, &c. and let their points of application
be Mlt M2, &c. respectively. Let Ox, Oy, Oz be any axes, not
necessarily rectangular, which are fixed in the body and move
with it. Let (#, y, z) be the coordinates of the point of applica-
tion M of any one force P, and let OM=r.
Take three arbitrary points A, B, C on the axes of coordinates ;
let OA = a, OB = b, OC = c. By Art. 7 the force P acting at x, y, z,
is equivalent to an equal and parallel force acting at 0, together
with three astatic couples whose arms are OA, OB, OC respec-
tively, whose astatic moments are Px, Py, Pz and whose forces
are parallel to P.
In this way all the forces may be brought to act at the origin
parallel to their original directions. These may be compounded
together into a single force, whose magnitude and direction in
space are the same for all positions of the body. Let us represent
this force by R.
Each force P will also give a couple having OA for its astatic
arm. , Compounding the forces at the extremities of this common
arm, all these couples reduce to a single couple. The arm 0 A of
this couple is fixed in the body while the magnitude and direction
in space of the forces are the same for all positions of the body.
Let us represent the magnitude of either of its forces by F.
The couples having OB, OC, for their astatic arms may be
treated in the same way. Their astatic arms also are fixed in the
body, while the magnitude and direction in space of the forces are
always the same. Let these forces be Q and H.
312 ASTATICS. [ART. 10
Summing up, we see that a system of forces can be reduced to
a principal force R acting at any assumed base point 0, together
with three couples (F, OA\ (G, OB) and (H, 00), having their
astatic arms arranged along any three assumed straight lines
OA, OB, OC fixed in the body and not all in one plane.
It may be seen that this reasoning, as far as we have gone, is
the same as that used in the corresponding proposition when the
body is fixed in space (Vol. I., Art. 257). The difference is, that
when the body has only one position in space these three couples
may be compounded into a single couple. But no single couple
can be found which is equivalent to these, when the body may
assume any position in space (Art. 8).
10. Consider any one position of the forces and of the body.
In this position let X, T, Z, be the components along the axes of
any force P. To find the resultant force M, we bring all these P's
to act at the base 0. The force R is therefore the resultant of
2X, 2F, 2.Z acting at 0 along the axes. To avoid the continual
recurrence of the symbol 2 it will be convenient to represent these
components by X0, F0, Z0.
To find the force F we seek the resultant of all the forces
similar to Pac/a acting at A. The force F is therefore the resultant
of the three forces ^Xx/a, 2F»/a, ^Zas/a acting at A parallel to
the axes. In the same way the forces G and H are the resultants
of "ZXyjl, 2Yy/b, ZZy/b and of 2Xz/c, 2Yz/c, ^Zzjc. It will be
found convenient to represent the summations S2T&, ^Xy &c. by
the symbols Xx, Xy, &c.
In this way the three couples (F, a), (G, b), (H, c) are resolved
into nine elementary couples whose astatic moments are repre-
sented by the constituents of either of the following determinantal
figures
couple (F, a) = ZXx, 2Yx, 2Zx = Xx, Yxy Zx
couple (G, b) = 2Xy, 2Fy, 2Zy = Xv, Yy, Zy
couple (H,c) = 2Xz, ZYz, 2Zz = Xz> Yz, Z>
where the common arms of the three couples in the first, second
and third rows are OA, OB, 00 respectively. Thus the small
letter or suffix indicates the axis on which the astatic arm is
situated, while the large letter indicates the direction of the force.
This convenient notation is the same as that used by Darboux.
These will be referred to afterwards as the nine elementary
ART. 11]
CONDITIONS OF EQUILIBRIUM.
313
couples. Together with X0, Y0, Z0, the three force components, we
thus have twelve elementary quantities for each base point.
For the sake of brevity we shall represent the couples (F, a),
(G, b\ (H, c) by the symbols Kx, Ky> Kz.
As we are chiefly concerned with the astatic moments of the
couples, the forces and arms are separately of only slight import-
ance. It is often convenient to choose the arms of all the couples
to be unity and positive. The signs of the forces alone then
determine the signs of the moments. In other cases it is found
advantageous to make the forces of all the couples equal to the force
R. The forces then divide out of the equations, leaving relations
between lengths only.
It will be found useful to remember that the direction ratios of
any one of the forces F, G, H are proportional to the constituents
of the corresponding row of the determinantal figure. An inter-
pretation of the symbols when taken in columns will be found
later on.
The figure represents the relation of the elementary couples
to the axes. To avoid complication the forces at 0 are omitted.
The directions of the forces at the extremities A, B, G of the
astatic arms are shown by the arrow-head, while each arrow-head
is marked by the astatic moment of the corresponding couple.
11. Conditions of equilibrium. // a system of forces be in
astatic equilibrium each of the twelve elements is zero.
Resolving parallel to the axes we have X0 = 0, F0 = 0, Z0 = 0.
Taking moments about the axes of coordinates we have
But the body must be in equilibrium in all positions. Instead
of turning the body round any axis, let us turn every force in
314 ASIATICS. [ART. 13
the opposite direction round a parallel axis through its point of
application. First let the rotation be about an axis parallel to x
through a right Bangle. The X forces are all unchanged, but the
Y forces now act parallel to z in the positive direction while the
Z forces act parallel to y in the negative direction. Hence, writing
T for Z and — Z for Y in the equations of moments already found,
we have
Yv + Zt = 0,. XZ-YX = 0, -Zx-Xy = Q.
Joining these to the preceding equations we find Zx — 0,
.3^ = 0, Xz = 0, Tx = 0, i.e. every constituent with an as in it
(except Xx) is zero.
In the same way by turning the system round y we find that
all the constituents are zero except Xx> Yv, Zz. But we also find
that YV + ZZ = 0, ZZ + XX = 0, Xx+Yy = 0. Hence each of the
three XX) Yy, Zz is also zero. Thus all the twelve elements are
zero.
That these conditions of equilibrium are sufficient as well as necessary follows
at once from the previous article. Thus, since the force F is the resultant of Xx/a,
YJa, Zx\a, it is clear that F is zero. Similarly G and II are zero. Since X0 , ra ,
Z0 , are zero the principal force E is zero, so that the body is in equilibrium in all
positions.
We may however also arrive at the same result independently. The body and
forces in any one position being referred to axes a:, y, z, let the twelve elements be
zero. The axes x, y, z remaining fixed in space, let the body be moved about the
origin into any other position, and let the coordinates of the point (x, y, z) become
(x', y', z'). Since x, y, z are linear functions of x', y', z' whose coefficients are
independent of the coordinates, it is evident that the twelve elements 2Xx' &c. are
also zero. The six statical equations of equilibrium referred to in Art. 11 are
therefore satisfied in this new position of the body.
12. If two systems of forces be referred to the same origin and
axes they cannot be astatically equivalent unless the twelve elements
are equal each to each.
Let the twelve elements of the two systems be Xx &c., Xx &c.
If we reverse the forces of the second system, the two systems
together would be in equilibrium. Hence Xx — Xx = 0, &c. = 0.
Thus all the elements are equal each to each.
13. Ex. 1. If the same system of forces can be astatically represented in either
of two ways, viz. (1) by three forces (F, G, H) acting at (A, B, <J) or (2) by three other
forces (F', G', H') acting at (A', B', C'), prove that (unless the system can be reduced
to two astatic forces instead of three) the planes ABC, A'B'C' must coincide.
Let us first suppose that the three forces F, G, H, are not all parallel to one
plane. Take the plane A'B'C' as the plane of xy. We have Zz, Ye, Zt, the same
for both systems. But since the ordinates of the points of application of F', G', H',
ART. 14] THE CENTRAL ELLIPSOID. 315
are zero, each of the three Xt, Yt, Z, must be zero. Consider the equation Zt=Q.
Place the body in such a position that the forces F, G act parallel to the plane
A'B'C'. This is possible since a plane can be drawn parallel to any two straight
lines. Then by hypothesis the direction of H will not be parallel to the plane
A'B'C'. The components of the forces F, G, parallel to the axis of z are now zero.
Hence Zt must be zero for the single force H. Thus either H=Q or the ordinate
of its point of application is zero. Supposing .F, 0, H to be all finite, it follows
that C lies in the plane A'B'C'. By similar arguments we prove that the other points
A, B also lie in the same plane A'B'C'.
Next, let us suppose all the three forces F, G, H are parallel to one plane. In
this case one of the forces as H can be resolved into two components/ and g parallel
to F and G respectively. Each of the two sets of parallel forces (/, F) and (g, G)
can be replaced by a single force at its centre of parallel forces. The system
F, G, H can therefore be reduced to two astatic forces.
Ex 2. If a system of forces F, G, H, acting at the corners of a triangle ABC,
can be reduced to two astatic forces F", G' acting at two points A', B', then either the
forces F, G, H are all parallel to one plane or the triangle ABC is evanescent.
We need only to examine the case in which F, G, H are all finite, for, if one be
zero, the other two are necessarily parallel to one plane.
The system P, G' can be regarded as the limiting case of a triangle of forces
F', G', H' acting at the corners of a triangle A'B'C' where H' is zero and the position
of C' is arbitrary. If then the forces F, G, H are not all parallel to the same plane
it would follow from Ex. 1 that all the corners A, B, C lie in the plane A'B'C'. But
this is impossible since C' is an arbitrary point, unless the triangle ABC is evanes-
cent and lies in the straight line A'B'.
14. The central ellipsoid. A base point 0 having been
chosen, the rectangular axes Ox, Oy, Oz are arbitrary. We shall
now show that there is one system of axes which will enable us
to analyse the system of forces more simply than any other.
Let Ox', Oy', Ozf be a second system of axes also fixed in the
body. Let A', R, G' be points taken arbitrarily on these axes,
let their distances from 0 be a', V, c'. Let F', 0', H' be the
forces which act at A', B', C'. We shall suppose both systems of
axes to be rectangular.
As the body is moved about, the forces F', 0', H' keep their
directions in space unaltered, so that as regards the body the
points of application and the magnitude of each force are the
only elements fixed. Let us then find the magnitude of the
force F' which acts at A', the forces at 0, A, B, G being regarded
as given. To effect this we shall resolve the arms of each of the
nine elementary couples along OA', OB', OC', keeping the forces
unaltered. We shall reserve for examination only those com-
ponents whose arms are along OA'.
Let (I, m, n) be the direction cosines of the axis Ox. Then
316 ASIATICS. [ART. 16
the groups of couples (Xx, Xy, Xz); (Yx, Yy, Yz); (Zx, Zy, Zz)
yield three component couples having their forces parallel to
X, T, Z respectively. Their astatic moments are (Art. 6),
XJ, + Xym + Xzn = Llt
YJ, + Tym + Yzn = L2,
ZJ, + Zym + Zzn = L3.
These couples have a common arm OA' and their forces are at
right angles. Compounding them we have
(F'a')*=(Xxl+Xym+ 3»2+( 7J.+ Yym+Yzn)* + (ZJ, + Zym+ Zzri)\
The direction cosines of the force F' are proportional to the three
moments Llt La, L3.
We notice that this expression for F'a' contains only the
direction cosines of OA', and does not depend on the position
of OB' or 0(7', except only that these must be at right angles to
OA'. We are thus able to consider the couple whose arm is OA'
apart from those whose arms are OB' and 00'.
Let us measure along OA' a length OP', such that OP1 is
inversely proportional to the astatic moment of the couple whose
arm is OA'. For convenience we shall suppose the product of
OP' and this astatic moment to be unity. Thus OP' . F'a = 1.
Let OP' = p, and let £, rj, £, be the coordinates of P' referred to
the original axes Ox, Oy, Oz. Then £=lp, rj = mp, £=np. We
therefore find for the locus of P' the quadric
1 = (X£ + Xyr, + X£f + ( Yx£ + Yyr, + FZ02 + (Z^ + Z& + Zjff.
15. This quadric may be regarded as defined by a statical
property, viz. if any radius vector be taken as the axis Ox', the
astatic moment of the corresponding couple (Ff, a') is measured
by the reciprocal of that radius vector. It follows that whatever
coordinate axes Ox, Oy, Oz are chosen we must have the same
quadric. The equations of the quadric when referred to different
sets of axes may be different, but the quadric itself is always the
same. The quadric is therefore to be regarded as fixed in the
body. Any point of the body may be chosen as the base 0, and
every such base has a corresponding quadric whose centre is at
the base. This quadric is called the central ellipsoid of that point.
It is also called Darboux's ellipsoid.
16. Let us represent the astatic moment of the couple whose
astatic arm is directed from a given base along the radius vector
p by the symbol Kp. In the same way the astatic moments, Fa,
ART. 17 J THE CENTRAL ELLIPSOID. 317
Gb and He, of the couples whose astatic arms are directed along
the axes will be represented by Kx, Ky, Kz. With this notation
we have
XJ + IV + Zx- = F*a? = KJ,
Xy* + Fj,2 + ZJ = G*b* = Ky\
X? + F/ + Zf = H V = #/ ;
XyX, + YyY, + ZyZ, = KyK, COS «,
XZXX + YZYX + Z2ZX = KZKX cos &
X^ + YxYy +ZxZy = KxKy cos 7;
where a, & 7, are the angles between the directions of the forces
(G, H), (H, F), (F, G) of the couples KX) Ky, K2.
Expanding the squares in the equation of the central ellipsoid
at the origin, it may be written in the form
K*& + Ky*rf + K?? + 2KyKe cos cwtf + 2K,KX cos /9# + 2ZaJT1, cos 7^7 = 1.
Also if K' be the moment of the couple corresponding to the
arm OA', whose direction cosines are I, m, n, we have
K!*=K2V + Zy2m8 + Z,2n2 + IKyKjnn cos a + IKJLjiL cos 0 + IKyKylm cos 7.
It may be useful to state the rule by which the signs of any of the astatic
moments Kx, Ky, Kz are determined. The directions of the forces being fixed in
space, there is for each line of action a positive and a negative direction determined
by reference to some axes fixed in space. The astatic arms are measured in
the body, and for each of these also there is a positive and a negative direction. Now
imagine the couple moved parallel to itself until either extremity of its astatic arm
is placed at the origin, so that one force acts at the origin. The moment is then
the product of the astatic arm into the other force, when each is taken with its
proper sign.
17. Ex. 1. Show that the discriminant of the central ellipsoid at the origin is
equal to (6VFGH)2, where F is the volume of the tetrahedron OABC.
Prove also that the minors of the coefficients of £2, if-, f 2 in the discriminant are
(.Kj,.^ sin a)2, (K,KX sin /3)2 and (JBT.^,, sin 7)*, respectively.
If parallels to the directions of the forces F, G, H are drawn from the centre of a
sphere to cut the surface, the arcs joining the points of intersection form a spherical
triangle whose sides are a, /3, 7. If 6, ij>, \f/ be the opposite angles, the minors of
the coefficients of Tjf, f£, £ij in the discriminant are respectively
- KyKgK^ sin £ sin 7 cos 8, - KJLyKJ sin 7 sin a cos <j> and - K-^KyK^ sin a sin /9 cos ^.
Ex. 2. An astatic arm OP moves about any given base point 0 so that its
corresponding astatic moment is constant. Show that OP traces out a cone in the
body coaxial with the central ellipsoid at 0.
Ex. 3. If Ox, Oy, Oz be any rectangular axes meeting at a fixed origin 0,
Kx, Ky, Kg the corresponding astatic moments, prove that K£ + Ky*+K? is invari-
able for all such axes.
Since this expression is the first invariant of the central ellipsoid at 0 the pro-
perty follows at once. It also follows from the geometrical property of an ellipsoid,
that the sum of the squares of the reciprocals of three diameters at right angles is
constant.
318 ASIATICS. [ART. 19
18. If we refer the central ellipsoid to its principal diameters
as axes of reference, the equation loses the terms containing the
products of the coordinates. If F, G, H represent the forces of
the three couples for this position of the axes, the equation is
The quadric is therefore in general an ellipsoid. If one of the
three forces is zero, i.e. if one of the couples is absent, the quadric
reduces to a cylinder.
Since the terms containing the products £77, ^ ££ are absent,
it follows that if the three forces F, G, H are all finite, their
directions are at right angles to each other. If one force is zero,
the other two must be at right angles.
Summing up, we see that whatever point of the body we choose
as base, there are always three straight lines at right angles, faced
in the body, such that, when these are taken as the astatic arms of the
couples, the forces of the couples act in directions at right angles
to each other and are faced in space.
In this way we have for each base point two convenient
systems of rectangular axes, one fixed in the body, viz. the astatic
arms of the couples, the other fixed in space, viz. the directions of
the forces.
The axes fixed in the body are called the principal axes of the
base. The couples are then called the principal couples.
19. The initial position. The base point 0 being regarded
as fixed, and the body referred to principal axes, it is evident that
we may turn the body about 0 until the system of axes fixed in
the body coincides in position with the system fixed in space.
The peculiarity of this position of the body is that the forces
of each of the three couples act along the astatic arm of that
couple. The moments of the couples are therefore zero. The
forces Plf P2, &c. of the given system reduce to the single resul-
tant R whose line of action passes through the given base.
This is called an initial position of the body and the couples
are then said to be in their zero positions.
The body being placed in an initial position, it is clear that if
we turn it round any one of the astatic arms through two right
angles, the same property will recur again, i.e. the force of each
couple will act along its astatic arm. Thus any base being given
there are at least four corresponding initial positions.
ART. 21] THE INITIAL POSITIONS. 319
Though in all these four positions of the body the two systems
of axes coincide in position, yet the positive direction of an axis
of one system may be the same as either the positive or the
negative direction of an axis of the other system. It is usual to
choose the positive directions of one system so that in one of these
four positions of the body the two systems of axes may have the
same positive directions as well as coincide in position. This
initial position is called the positive initial position.
20. When the body is placed in a positive initial position the
nine elementary couples described in Art. 10 are reduced to
Xx Yx = 0 ZX = Q,
Xy = 0 Yy Fz = 0,
Xz = 0 Yy = Q Zz.
The equation to the central ellipsoid then takes the simple
form Xx*? + Yjrf + Z*? = 1.
If (I, m, n) be the direction cosines of any other arm OA' the
direction cosines of the force F' acting at its extremity are
proportional to Xxl, Yym, Zzn,
and (F'aJ = XXH2 + Yjm* + Z/n*.
Thus the direction and magnitude of F' have been found. If
the body is now moved into any other position, F' continues to act
in the same direction in space and therefore continues to make
the same angles with F, G, H that it made in the initial position.
21. There are no other positions besides the four initial positions in which a body
can be placed so that the system of forces may reduce to a single resultant which passes
through the given base, except when the central ellipsoid at the given base point is a
surface of revolution.
Let OA, OB, OC be the principal axes at the given base O. Let OF, OG, OH be
three straight lines at right angles drawn parallel to the forces of the corresponding
couples. In order to use conveniently the formulae of spherical trigonometry we
suppose these axes to cut the surface of a sphere whose centre is at 0 in the six
points A , B, C, F, G, H. The planes of the couples are the planes which contain
the astatic arms and the forces, and are therefore the planes of the spherical arcs
AF, BG, CH. If their astatic moments are Kx=Fa, Kv=Gb, Kt=Hc their
moments in any position of the body are -Kj-sin AF, Kysin BG and Kt sin CH.
When the body is in an initial position the spherical triangles coincide. Starting
from this position, the body may be brought into any other by turning it round
some axis 01. If this axis intersect the sphere in I, the spherical arcs IA, IB, 1C
are respectively equal to IF, IG, IH, and if 2w is the angle of rotation, the angles
AIF, BIG, CIH are each equal to 2«. Join A F, BG, CH by arcs of great circles
and draw the perpendicular arcs IL, IM, IN.
If this position of the body can be one of equilibrium when the base is fixed, the
320 ASTATICS. [ART. 22
three couples must balance each other. Resolving the axis of each of these along
and perpendicular to 01, the moments of the three latter components are respect-
ively JTgSin .4.FCOS IL, Ey&inBGcosIM, and .KjSmC'.H'cos IN. Since the three
components are in equilibrium, these moments must be proportional to sin MIN,
sin NIL, sin LIM, that is to sin BIG, sin CIA and sin AIB.
C
F
For brevity let a, j8, 7 represent the arcs IA, IB, 1C. Since BC is a right angle
we have cos /3 cos 7 + sin /3 sin 7 cos BIG = cos BG=0,
.-. sin2 /3 sin2 7 sin2 Bl C = sin2 /3 sin2 7 - cos2 /3 cos2 7
= 1 - cos2 p - cos2 7
= cos2 a.
Again, sin ^F cos JL = 2sin^.Fcosa = 2 sin a cos a sin w.
Similar expressions hold for the other angles.
Substituting these values in the condition of equilibrium, and dividing out the
common factors, we have K.f=Kyi=K*. Thus the proposed position of the body
cannot be one of equilibrium when the base is fixed unless the ellipsoid is a sphere.
This argument assumes that none of the factors divided out are zero. We must
therefore examine separately the case in which I lies on one of the principal planes.
If I lies on BG, the first component is zero, and the other two are Ky sin BG cos IM
and Kt sin CHcosIN. The condition of equilibrium is that these moments should
be equal ; hence IT,,2 sin2 £ cos2 /3 = K/ sin2 7 cos2 7.
Since /3 and 7 are complementary, this requires that Kvz=Kg2, i.e. the ellipsoid is
one of revolution.
Lastly, if I is at the point C, each of the three component couples is zero. The
component having 01 for its axis is then the sum or difference of the couples
Kxsin2u, Kvsin2u. Since this component also must vanish we again have
KX2=KV*, i.e. the ellipsoid is one of revolution.
22. Ex. 1. The body being placed in a positive initial position, prove that the
direction of F' is parallel to the normal to the ellipsoid Xij? + Yv-f + Z£*='L drawn
at the point where OA' cuts the ellipsoid. This ellipsoid is called the second central
ellipsoid of Darboux.
Ex. 2. The body being placed in a positive initial position, a straight line OQ is
drawn from the base parallel and proportional to the force F1 for all positions of
OA' in the body. Prove that the locus of Q is the ellipsoid
This is called the third central ellipsoid of Darboux.
ART. 25] THE CENTRAL PLANE. 321
Prove also that, if the arms OA', OB', OC' be at right angles, the corresponding
forces 1", G', H' are parallel to a system of conjugate diameters in the third ellipsoid.
This and the last example are due to Darboux.
Ex. 3. When the body is in a positive initial position for any base, prove that
the direction of the force corresponding to any astatic arm OA' is parallel to the
eccentric line of OA' in the central ellipsoid of the given base.
The Central Plane and the Central Point.
23. To compare the central ellipsoids at different points of the
body.
Suppose the forces to be referred to any base 0 and any axes
Ox, Oy, Oz, and that the nine elementary couples and the three
force-components are known for these axes. We shall now find
the corresponding quantities when some point 0', whose coordi-
nates are (p, q, r), is taken as the base.
Through 0' we draw axes O'x', O'y, O'zf parallel to (x, y, z).
The nine elementary couples may be transferred to these new
axes without any change (Art. 3). But the three force-components
will introduce new couples. By Art. 9 the component X0 acting
at 0 may be transferred to the origin 0' if we introduce the new
couples (X0, — p), (X0, — g\ (X0, — r), the coordinates of 0 referred to
0' being (—p, —q, —r). Similar reasoning applies to the components
F0, Z0. Hence we have for the nine elementary couples at 0'
Xx' = Xx-X0p, Yx'=Yx-Y0p, Zx' = Zx-Z0p;
Xy = Xy — X(,q, Yy = Yy ~ Ytf, Zy = Zy ~ Ztf,
Xz'=Xz-X0r, Yz' = Yz-Y0r, Zz'=Zz-Z0r.
The equation of the central ellipsoid at 0' is therefore, by
Art. 14,
{(Xx - X0p) % + (Xy - X0q) r)' + (Xz - Z.r)
+ i(F. - Y0p) f + (Yy - Y0q) j+(Yg- Y0r)
+ {(Zx - Z0p) £' + (Zy - Z0q) V' + (Zz - Z0r) ?}* = 1 5
the origin of the running coordinates £', »/, % being 0'.
24. If the principal force R is zero, we have X0 = 0, F0 = 0,
Z0 = 0. Int his case the central ellipsoid at 0' is the same as that
at 0. Thus the central ellipsoids at all base points are similar
and similarly situated.
25. The Central Plane. If the principal force R be not zero
R. s. n. 21
322
ASTATICS.
[ART. 26
+ (F. - F0r) ? = 0,
+ (Z, - ZQr) ? = 0
the form of the central ellipsoid will depend on the position of the
base point. We notice that the three planes
(X. - X0p) £ + (Xy - X0q) rj' + (X. - X0r) ? = 0,
(F* - Y0p) ?+(Yy- F.g)
(Z. - Z0p) ? + (Zv - Z0q)
are conjugate planes.
If the central ellipsoid is a cylinder the conjugate planes pass
through the axis of the cylinder, and the equations to the three
conjugate planes are then not independent. We thus have the
determinantal equation
X.-X,p, Xy-X0q,
Yx-Y0p, Yv-Y9q,
Zx-Z0p, Zy-Z0q,
This equation may be written in the form
Xz-X0r,
Yz-Y0r,
X
= 0
.(2).
XT' Y
x Ay ^z
Y Y Y Y
^ o -*# J y z
f7 f7 17 ?7
&Q " 'T ^"U ^ Z
1 p q r
When p, q, r are regarded as the running coordinates, this is evi-
dently the equation to a plane. The peculiarity of this plane is
that, if any point on it is chosen as base, the central ellipsoid is a
cylinder. This plane is called the central plane.
26. Since the central ellipsoid at every point is fixed in the
body the locus of base points at which the ellipsoid is a cylinder is
also fixed. The central plane is therefore fixed in the body. In
discussing its properties we may put the body into any position
we please.
Take any point 0 on the central plane as base, and let the
body be placed in an initial position. By Art. 20 all the nine
elementary couples, except Xx, Yy, Zz, are zero. Since the
ellipsoid is a cylinder one of the three Xx, Yy, Zz is also zero,
say Xx = 0. Substituting in the second form of the equation to
the central plane given in Art. 25, we see that it becomes
pX0YyZz = Q. If any one of the three X0, Yy, Zz is zero, the
equation to the plane is indeterminate, but if all these are finite,
the equation to the central plane is p = 0. It follows therefore
that the infinite axis of the central ellipsoid at any point of the
central plane is perpendicular to that plane.
ART. 28] THE CENTRAL POINT. 323
27. This leads to a simplified reduction of the forces
Pit P2, &c. Let us take the base of reference 0 at any point
of the central plane, and the principal diameters of the central
cylinder as axes of coordinates. The moment of that principal
couple whose astatic axis is along the infinite axis of the cylinder
is measured by the reciprocal of that axis, and is therefore zero.
Thus all the forces have been reduced to two couples (instead of
three) and a force R. The astatic arms of the couples lie in the
central plane and the forces of one couple are perpendicular to
those of the other.
«
28. The Central Point. It has been proved in Art. 9 that a
system of forces may be reduced to a principal force R at the base
of reference and three couples having their arms directed along
any three straight lines at right angles. Let us now enquire if a
base 0' can be found such that each of the forces of the couples
is perpendicular to the principal force.
If one system of axes O'A, O'B, O'G at any base 0' possess this
property, then every system of axes at that base will also possess
the same property. To prove this, let O'A', O'B', O'G' be any
other such system of axes. To deduce the forces at A', B', G'
from those at A, B, G, we resolve the arms OA, OB, OG in the
directions OA', OB', OG' and transfer the forces parallel to them-
selves, see Art. 6. Since each of the forces at A, B, G is
perpendicular to the force R, it follows that the forces at A',
B', G', which are compounded of these, are also perpendicular
to R
Let Ox, Oy, Oz be any given rectangular axes, and let p, q, r
be the coordinates of 0'. Through 0' draw a system of axes
O'x', O'y', O'z' parallel to Ox, Oy, Oz. Then, by what has just
been proved, the couples corresponding to these axes must have
their forces perpendicular to R. If the nine corresponding ele-
mentary couples are Xx &c., the conditions of perpendicularity are
X0XX + Y9Yxf + ZfiZ,? = o,
and two similar equations obtained by writing y and z for x in the
suffixes. Substituting for Xx', &c. their values given in Art. 23,
= X0XX + Y0YX
= X0Xy + Y07y
R-r = X0X, + Y9Y,
21—2
324 ASTATICS. [ART. 30
Since these give only one set of values for p, q, r there is but one
point which possesses the given property. This point is called
the central point.
29. The central point lies on the central plane. To prove this
let us consider the principal axes at the central point. Since the
forces of the three couples are at right angles to each other, they
cannot all, if finite, be perpendicular to the principal force. One
of these must therefore vanish. The central ellipsoid is therefore
a cylinder, i.e. the central point lies on the central plane.
That the central point lies in the central plane may also be
proved by substituting its coordinates in the equation (2) of the
central plane found in Art. 25. These coordinates p, q, r are
given in Art. 28, and a simple inspection shows that the equation
is satisfied.
Thus it appears that there is a certain point, lying on the central
plane, such that the forces of the two principal couples at that point
are at right angles to each other and to the principal force. This
point is called the central point.
The central point in the three-dimensional theory has not the same signification
as the central point denned in Vol. i., Art. 160, with reference to two dimensions.
In the latter the displacements of the body are confined to one plane, and for such
displacements the single resultant always passes through a central point fixed in
the body. In the former the displacements are unrestricted so that the lines of
action of the forces do not necessarily remain in one plane.
The preceding theorems on the central plane and central point are generally
given in treatises on Astatics, though the demonstrations in each may be different.
30. We may express the formulae for the coordinates of the
central point in the form of a working rule.
As already explained in Art. 9 the forces are represented by
P1} P2, &c. Their points of application are Mlt Mz, &c. and their
coordinates are (x1} yl} ^), (#2, y2, z2), &c. Also let the direction
cosines of Plt P2, &c. be respectively (c^, b1} ca), («2, 62, Ca), &c.
Then Xx = P^x^ + P^a^ + ... X0 = P& + P^ + ...
Yx = PA*i + P2M2 + . . . F0 = PA + P262 + . . .
Zx = P^OBj, + P^Xv + . . . Z, = Pad + P2C2 + . . .
Let 012, #13, &c., be the inclinations of the forces (Plt P2), (Plt P8>
&c. Then cos #12 = a&a + &A + dc.2, &c.
ART. 32] SUMMARY. 325
Substituting in the expression for p, Art. 28, we have
= PlQ1Xl + P»Q&+...
P1Q1 + A& + ...
where Q1 = P, + P2 cos 012 + P3 cos 013 + . . .
Q2 = Pj cos flu + P2 + P8 cos #23 + ...
&c. &c.
It is evident that Qa is the sum of the resolved parts of all the
forces in the direction P1} Q2 is the sum of the resolved parts in
the direction P2, and so on.
The equation just arrived at is the common formula for the
centre of gravity of weights P^, P2Q.2 &c. Similar equations
hold for q and r. Hence we have this rule. To find the central
point of any number of forces, we first multiply each force by the
sum of the resolved parts of all the forces along the direction of that
force. We then place weights proportional to these products at the
points of application of the forces. The centre of gravity of these
weights is the central point required.
31. Ex. Show that the equation to the central plane, referred to any axes,
when expressed in terms of the forces and their mutual inclinations takes the form
where
and F=
(7,, a.,, (I3
blt 62, &3
, C2, C3
The coefficient Lx is derived from M by writing unity for each of the x's in the
determinant, Lv is derived from M by writing unity for each y, and so on.
To prove this, we start with the equation (2) of the central plane given in Art.
25 and make the same substitutions as in Art. 30. On writing down the determi-
nant it will be seen that the determinants Lx, Lv, L, may be obtained from the
determinant M by the rule just stated. The determinantal sum M when expanded
takes the form of a series of products of triplets of the forces. To find the
coefficient of P^Pg we put all the other forces equal to zero ; the determinant then
assumes the known form of the product of the two determinants just written down.
32. Summary. It will be convenient if we now sum up shortly the gradual
steps made in reducing a system of forces to its simplest equivalents.
1. In Art. 9 the forces were reduced to a force R at an arbitrary base point 0
together with three couples whose arms Ox, Oy, Oz are arbitrary.
2. In Art. 18 it was shown that at the arbitrary base the arms Ox, Oy, Ot
could be chosen at right angles to each other so that the forces of each couple are
at right angles to the forces of the other two couples. These arms are called the
principal axes at O and are fixed in the body.
3. In Art. 25 it was shown that, if the base point 0 is placed anywhere on a
certain plane fixed in the body, the forces can be reduced to the single force R
together with two couples. The arms of these couples are at right angles and lie in
326 ASTATICS. [ART. 33
the plane. The forces also of each couple are perpendicular to those of the other.
This plane is called the central plane.
4. In Art. 28 it was shown that if the base point is placed at a certain point on
the central plane the forces of the couples are perpendicular to the force E. Thus
the forces of the original system can finally be reduced to a force B together with
two couples whose arms are at right angles arid such that the forces of each couple
are not only perpendicular to those of the other bnt are also perpendicular to the
force jR. This base point is called the central point.
The principal axes at the central point are two straight lines lying in the central
plane and a third, perpendicular to that plane. The two former are called the central
lines of the central plane. The latter is sometimes called the central axis. But it
must not be confused with Poinsot's central axis with which it coincides only when
the body is properly placed. It bears indeed a certain resemblance to Poinsot's
central axis, for the system is reduced to a force and two couples (instead of one)
such that the forces of the couples are perpendicular to the force.
33. Analogy to Moments of Inertia. Ex. 1. If K be the astatic moment of
the couple corresponding to any astatic arm OP drawn from the central point 0,
prove that the astatic moment K' of the couple corresponding to any parallel arm
O'P' drawn from any point 0' is given by K'2=K2 + E'2p" where p is the projection
of 00' on either astatic arm.
Thus, a motion of the base 0 in a direction perpendicular to the astatic arm does
not alter the magnitude of the astatic moment, but a motion along the arm from the
central point increases the moment.
Ex. 2. If -ETj, K2, K3 be the astatic moments corresponding to the principal
astatic axes Ox, Oy, Oz drawn from any point 0, prove that the astatic moment K
corresponding to any arm OP making angles a, /3, y with the axes is given by
J&=K? cos2 o +Z,2 cos?p+Ksz cos87.
It appears from these two propositions that the theory of astatic moments
of couples has an analogy to the theory of moments of inertia. The square of the
astatic moment about an arm drawn from 0 in any direction OP corresponds
to the moment of inertia of a rigid body with regard to a plane drawn through
0 perpendicular to OP. By noticing this correspondence we may deduce the
analogous propositions in the two theories one from the other.
It is clear from these two propositions that the mass of the rigid body is
analogous to the square of the principal force R, and that the centre of gravity
must be at the central point. For any base in the central plane the moment
of the couple whose astatic arm is perpendicular to that plane is zero, hence the
rigid body must be a lamina whose plane is the central plane of the forces.
The analogy may be made more distinct by adding another proposition. Let
0 be the central point, Oy, Oz the principal astatic axes in the central plane,
Ox that perpendicular. The astatic moment K about any axis OP, whose
direction cosines are I, m, n, is given by
Z2=Z22m2 + Z32«2 (1).
Let a lamina be placed in the plane of yz with its centre of gravity at 0, having
the axes of x, y, z for its principal axes of inertia; and let K£, K£ be its moments
of inertia at the origin with regard to the planes respectively perpendicular to the
axes of y and z. The equation (1) then shows that K2 is the moment of inertia of
the lamina with regard to a plane drawn through 0 perpendicular to OP.
ART. 34] THE CONFOCALS. 327
Let 0' be any other point whose coordinates are £, 77, f, and let O'P' be parallel
to OP. The astatic moment K' at 0' corresponding to the arm O'P' is given by
where p is the projection of 00' on OP. This is also the formula which gives the
moment of inertia of the lamina with regard to a plane drawn through 0'
perpendicular to O'P', provided R2 is the mass of the body.
It follows that the moment of inertia of the lamina with regard to a plane
drawn through any point 0' perpendicular to any straight line O'P' represents the
square of the astatic moment at the base 0' for the arm O'P'.
Since the moments of inertia for all arms through 0' represent the squares of
the astatic moments for the same arms, it follows that they have the same maxima
and minima and are connected together by the same rules. The principal axes of
inertia at 0' are therefore the same in direction as the principal astatic axes at 0'.
That the principal astatic moments at 0' are the normals to the confocals (4) of
Art. 34, and that the astatic moments are the three values of M given by the cubic,
follow at once from the properties of the principal axes of inertia, see Rigid
Dynamics, Vol. i. Art. 56.
Since the moments of inertia of the lamina about the axes of y and z are
respectively K^ and K^, it follows that the lamina might take the form of a
homogeneous elliptic disc, whose semi-axes of y and z are respectively 2K2/R and
2K3jRt and whose mass is R*. The boundary is therefore similar to the imaginary
focal conic.
The Confocals.
34. To investigate the mode in which the central ellipsoids at
different bases are arranged about the central point.
Let the central point be chosen as the origin and the principal
diameters of the central ellipsoid as axes of coordinates. Let the
infinite axis be the axis of x, then the plane of yz is the central
plane.
As we are enquiring into the positions of the neighbouring
central ellipsoids, and as these are fixtures in the body, we may
put the body itself into any position we may find convenient
Let it be placed in its positive initial position with the central
point as the base.
In this position all the nine elementary couples are zero,
except Yy and Z,. Also XQ = R, 70 = 0, Z, = 0. The central
ellipsoid at the origin is FyV + Z/%2 = I (1).
The central ellipsoid at any point 0' whose coordinates are p, q, r,
is FyV 2 + Zft* + & (P% + W + r O2 =l (2)>
where (f , i) ', f ) are referred to axes meeting at 0' parallel to the
axes as, y, z, Art. 23.
Let an astatic arm O'A' move about 0' so that the correspond-
328 ASTATICS. [ART. 35
ing couple (F't OA'} has a constant astatic moment equal to M,
and in any position let (I, m, n) be its direction cosines. Then,
since the moment M (Art. 14) is the reciprocal of the correspond-
ing radius vector of the central ellipsoid, we see that I, m, n are
connected together by -the relation
Fj,2ma + Zfn* + -R2 (pi + qm + rn)8 = M '• ;
Now, after division by .R2, the left-hand side of equation (3)
expresses the square of the perpendicular drawn from the central
point on a tangent plane to the ellipsoid
and the right-hand side of (3) expresses the square of the perpen-
dicular from the central point on a plane through 0' parallel to
that tangent plane. The equation (3) therefore shows that this
tangent plane passes through 0'. Hence we infer that if O'A'
move about 0', so that the corresponding astatic moment is constant
and equal to M, then O'A' is always perpendicular to a tangent
plane drawn from 0' to touch the confocal (4).
These tangent planes all touch the enveloping cone of the
confocal (4), and the axis O'A' traces out the reciprocal cone of
this enveloping cone. These two cones are known to be co-axial
and their axes (Art. 17, Ex. 2) are in the same directions as those
of the central ellipsoid at 0'.
If M is so chosen that the confocal (4) passes through the
point 0', the enveloping cone becomes the tangent plane and
therefore the cone traced out by O'A' reduces to the normal at 0'.
Hence the principal diameters of the central ellipsoid at any
point 0' are the three normals to the three quadrics which pass
through 0' confocal to the quadric (4). Also the astatic moments
of the three corresponding couples are the values of M given by the
cubic (4) when we write for %, rj, £ the coordinates of 0'.
35. Instead of using the three confocals we may use any one of
them, say the ellipsoid. By known properties of solid geometry the
three normals at any point 0' are (1) the normal to the ellipsoid,
(2) parallels to the principal diameters of the section of the
ellipsoid diametral to 00'.
Let Mlt Mz, Af3 be the three values of M given by the cubic
(4), J/j being the greatest. Let D2, Ds be the lengths of the
ART. 36] THE CONFOCALS. 329
principal semidiameters of the section of the ellipsoid, D2 being
parallel to the normal at 0' to the confocal M2, and Ds parallel to
the normal to M3. Then it is known by solid geometry that
Thus Mz, M3 are known in terms of M^ and quantities connected
with the ellipsoid.
36. As these confocals play an important part in the theory
of astatic forces, it is necessary to state distinctly their position.
Let the body be referred to the central point as origin, and the
principal diameters of the central cylinder as axes, the plane of yz
being the central plane. Let K%, K3 be the astatic moments of
the couples whose astatic arms are along y and z. These astatic
moments are the same for all positions of the body and are
represented by Yy and Zz when the body is in its initial position.
The equation to the confocals is therefore
£2 if £2 1
The focal conies of these are obtained in the usual manner by
putting M =
thus have •
K,
?
= KZt £=0; and M =
1
H*>
I
= 0. We
_
£?
If we take as the standard case K8 > K^ the first is a hyper-
bola, the second an ellipse, and the third is imaginary. The two
first are represented in the diagram by the dotted lines. These
330 ASTATICS. [ART. 37
conies will be referred to as the focal conies, and a straight line
intersecting both conies may be called a focal line.
The figure represents the positive octant of a set of confocal
quadrics intersecting in 0'. The semi #-axes are represented by
OAlf OA?, OA8 and are respectively equal to MJR, MzjR, M3/R.
As is well knowo the vertices Fs, F2 of the two focal conies lie
between Alt A3 and A,. We have OF, = KJR, OFS = K9/R
If K2=Q, the ellipsoid and the hyperboloid of one sheet are surfaces of revolution.
The hyperboloid of two sheets reduces to any two planes through Oz, and the hyper-
bolic conic becomes the axis of z. The central plane is now indeterminate and
is any plane through the astatic arm of K3.
If both JT2=0 and K3=Q, the ellipsoid becomes a sphere, one hyperboloid is a
right cone, and the other any two planes through the axis of the cone.
37. Theorem on focal lines. A straight line is drawn from any point P on
one focal conic to any point Q on the other, it is required to prove that
where al) a2, Oj are the direction cosines of PQ, and p is the perpendicular distance
from the origin.
We know that the tangent planes drawn through any right line to the two
confocals which that line touches are at right angles to each other, see Salmon's
Solid Geometry, Art. 172. Since the focal conies are evanescent confocals, the
planes through PQ and the tangents at P and Q to the conies are at right angles.
If p, p' are the perpendiculars on these planes, I, m, n ; V, m', n' their direction
cosines, we have
E *p* = KfP - (Kf - K£) ra2, JPp'» = Kfl'* + (K3*- K*) m'2.
Since the straight lines p, p' and PQ are mutually at right angles, this becomes
Kz* (1 - m2 - m'2) + Kf (1 - n2 - n'2) = Kfaf + Kfaf.
The theorem may be more easily proved by taking as the coordinates of P and Q
(x, y, z) and (x', y', z') where
Rx'=K3coatj>, Ry' = (K^-K^sm<p, Rz'=0.
The direction cosines a2, a3 and the length p may then be found by elementary
formulae, and it will be seen that the relation to be proved is satisfied.
It follows from this theorem that every focal line is a generator of the right
circular cylinder whose radius is p and whose axis passes through the common
centre of the conies and is parallel to the focal line.
Ex. 1. Show that four real focal lines can be drawn parallel to a given
straight line.
Let a generator parallel to the given straight line travel round the hyper-
bolic conic and trace out a cylinder. This will cut the plane of the other conic
in a hyperbola. Each branch of this hyperbola passes inside the elliptic
conic, because it goes through the focus; it therefore cuts the ellipse in two
points.
Ex. 2. If a straight line PQ intersect one focal conic and if its distance from
the central point be p, where p is given in the theorem above, show that that
straight line will intersect the other conic also.
ART. 38] ARRANGEMENT OF POINSOT'S AXES. 331
If possible let PQ intersect one focal conic in P and not intersect the other.
Describe two cylinders whose bases are the focal conies and whose generators are
parallel to PQ. By Ex. 1 these intersect in four lines, and each of these four is also
a generator of the right circular cylinder whose radius is p. Now by supposition
PQ lies on one of the elliptic cylinders and also on the circular cylinder, hence
these two quadric cylinders intersect each other in five lines, which is impossible.
Ex. 3. The locus of all the straight lines drawn from any given point P on the
hyperbolic conic to intersect the elliptic conic is a right cone, the tangent of whose
semi-angle is (Ka2 - JK^)/K3Rz where z is the ordinate of P.
Ex. 4. Show that four real focal lines can be drawn through a given point P,
and that they are the intersections of the two quadric cones
- . _
where (p, q, r) are the coordinates of P and £, 77, £ are referred to parallel axes
meeting at P. "\
Ex. 5. Prove that the circular sections of the central ellipsoid whose centre is
at 0' are perpendicular to the generating lines at 0' of the hyperboloid of one
sheet. [Darboux.]
Ex. 6. If the base is situated on one of the principal planes at the central
point, show that one principal axis at that base is perpendicular to that plane
and the astatic moment of the corresponding couple is the same for all base
points in that plane.
Ex. 7. If the b^ase is situated on one of the principal axes at the central point,
prove that the three principal axes at the base are parallel to those at the central
point.
Ex. 8. If a straight line is a principal axis at every point of its length, prove
that it is one of the principal axes at the central point.
Ex. 9. Find the locus of the base point 0' at which the central ellipsoid is a
surface of revolution.
In order that two of the three quantities M1,M.2,MS, in Art. 35 may be equal we
must have either D2=0 or D2=D8. In the first case 0' lies on the elliptic focal
conic. In the second case 0' is at an umbilicus U and the locus is therefore the
hyperbolic focal conic. In both cases the unequal axis is a tangent to the focal
conic.
The same results follow from the equation to the central ellipsoid in the form
see Art. 34. By applying the usual analytical conditions that this is a surface of
revolution we obtain the required relation between p, q, r.
Arrangement of Poinsot's central axes.
38. In whatever position the body is placed relatively to the
forces it has been shown in Vol. I. that the forces acting on the
body can be simplified into a single force, acting along a straight
line called by Poinsot the central axis, and a couple round that
axis. As the body takes different positions relative to the forces
332
ASTATICS.
[ART. 38
Poinsot's axis also moves relatively to both. In order to determine
the arrangement of Poinsot's axes for all possible positions of the
body and forces it will be convenient to have two systems of axes,
one fixed in the body and the other fixed relatively to the forces.
Let the axes fixed in the body be the principal axes at the
central point. These we shall represent by Ox, Oy, Oz. Following
the same notation as before, the forces are represented by the
astatic couples (G, b), (H, c), whose astatic arms are placed along
y and z, together with a force R acting at 0. The astatic
moments of these couples are represented by K2, K3 respectively.
Let the axes fixed in space be parallel to the forces R, G, H
These are represented by Ox', Oy', Oz'. We shall sometimes
speak of them as the axes of the forces.
Let the direction cosines of either set of axes relatively to the
other be given by the diagram. The positive
directions of these axes are so chosen that by
turning one set round the common origin the
positive directions of x, y, z may be made to
coincide with those of x, y', z'. The advantage
of this choice is, that in the determinant of direction cosines every
constituent is equal to its minor with the proper sign as given by
\H
R
G
H
x y
Cl °2
the ordinary rules of determinants. Without losing the simplicity
of the other relations of these constituents, we thus avoid any
ambiguity of sign in the minors.
ART. 39]
ARRANGEMENT OF POINSOT'S AXES.
333
In the figure the axes are represented in the manner usually
adopted in spherical trigonometry. The axes Ox, Oy, Oz and
Ox, Oy', Oz1 cut the sphere in x, y, z and R, G, H respectively ;
the angles being represented by arcs of great circles. The
Eulerian angular coordinates of R referred to x are 0 = xR,
ilr = yxR, <f> = MRG. Since the angle between any two planes
is equal to the arc joining their poles, it is easy to see that
z!G = 6, Iz = yl
39. To find the position of Poinsot's axis referred to the axes
of the forces, and also the moment of the forces about it.
Let Px" be the required Poinsot's axis, F the moment of the
couple round it. The axis Px" is parallel to Ox, let its coordi-
nates referred to x, y', z , be 77 ', £'.
The couples K%, Kz have their astatic arms on the axes y, z, and
their forces parallel to y', z'. To refer these couples to the axes
x', y', z1 we resolve the arms and move the forces parallel to
themselves (Art. 6). Thus we replace the two couples by six
others whose arms are arranged along the axes of x', y, z'. In
the figure the forces at 0 are omitted to avoid complication, the
arrows indicate the directions of the other forces of each of the six
couples ; and each arrow-head (as in Art. 10) is marked by the
astatic moment of the corresponding couple.
By hypothesis all these couples together with a force R acting
at 0 are equivalent to the couple F round Px" and a force equal
to R acting along Px". Taking moments about the axes
Ox, Oy', Oz! we have T = Ktba-K& : (1),
(2),
.....(3).
Another proof. We may also obtain these results very simply without resolving
the couples. Let the arms OB, 00 of the couples be taken as unity so that the
334 ASTATICS. [ART. 40
forces G, H acting at B and C are measured by the astatic moments Z"2, K3,
Art. 10. The axes Ox't Oy', Oz' being the axes of reference, the coordinates of B
and C are respectively aa, ba, c2; aa, 63, cs. Since K3 acts parallel to Oy', its
moment about Oz' is K2a2, and since K3 acts parallel to Oz' its moment about
Oy' is -JTjaj. In the same way their moments about Ox' are K3b3 and -K2c3.
Equating these to the moment of B acting along Px" and of T we have the same
results as before.
40. When the body is rotated about Ox', the direction cosines
Oz, as are invariable. It follows that the straight line whose
position is determined by the equations (2) and (3) is fixed
relatively to the forces. Hence we infer, that, when the body is
rotated about an axis passing through the central point and parallel
to the principal force, Poinsot's aocis always coincides with a
straight line fixed in space.
This straight line traces out a right circular cylinder in the
body whose radius p is given by the equation
&p = Kjaf + K,*af ..................... (4).
This cylinder is fixed in the body and moves with it. In one
complete revolution of the body each generator in turn passes
through the straight line fixed in space and becomes the Poinsot's
axis for that position of the body.
Referring to the figure of Art. 38, the axis of this cylinder cuts
the sphere of reference in R. We may also imagine the sphere of
such size that the cylinder envelopes it along the circular boundary
of the figure. In the figure the direction of the force R and the
generators of the cylinder are supposed to be perpendicular to the
plane of the paper.
As the body turns round OR as its axis, the dotted part of the
figure remains fixed in space while the part indicated by the
continuous lines moves round R.
Let a plane through the axis of the cylinder and the straight
line fixed in space cut the sphere in the arc RP. Let RP
produced backwards cut the circle GH in P'. Then the position
of P or P' may be found from the equations
(5).
In every position of the body Poinsot's central axis is a
straight line drawn through P perpendicular to the plane of the
circle OH. Here P is distinguished from P' by the sign of either
t] or £" as given by the equations (2) and (3).
ART. 41] ARRANGEMENT OF POINSOT'S AXES. 335
It follows from these results, that all the straight lines, each of
which would be a Poinsot's axis if the body were properly placed,
may be classified as the generators of a system of right circular
cylinders. The axes of these cylinders pass through the central
point and are always parallel to the direction of the principal
force.
Conversely, a straight line being given in the body, it may be
required (when possible) to place the body in such a position that
the straight line may be a Poinsot's axis. To effect this, we turn
the body about the central point until the given straight line
is parallel to the principal force. If alt az, a3 are the direction
cosines of the given straight line referred to the principal axes of
the body at the central point, then, in this position of the body,
«!, a2, a3 are also the direction cosines of the principal force. If
the distance of the given straight line from the central point does
not satisfy equation (4) the straight line cannot be a Poinsot's
axis. If however the equation is satisfied, we turn the body
round the principal force as an axis of rotation through the angle
GP determined by equation (5), or, which is the same thing, we
turn the body until the given straight line passes through the
point 97', £" in the plane y'z determined by the equations (2), (3).
The body has then been placed in the required position. When
the straight line fixed in the body has been made parallel to the
principal force the body may be inverted, so that the given straight
line is again parallel to the force but points in the opposite
direction. If the condition (4) is satisfied in one case, it is
satisfied in the other. Thus if the construction yield one position
in which the given straight line is a Poinsot's axis, it will yield
another.
41. In every position of the body the couple-moment of
Poinsot's axis is given by
F = K3 cos Gz — KZ cos Hy
= Kz (cos i/r sin <f> + sin ty cos <f> cos &)
+ Kz (sin i/r cos <£ + cos i/r sin $ cos 6),
by using the spherical formulae for the triangles GIz and Ely.
This may be written in the form
where ro is the maximum value of T, and </> = <£„ determines the
336 ASTATICS. [ART. 43
position of the body when the couple-moment is zero. We easily
S S cos 6
find tan 0o = — ^ -- z — =- tan ilr ......... (7),
K 2 cos 6 + K3
cos 02 sin2 + Kz cos 0 + Ktf cos2
o," ...(8).
Make the arc MN0 = <£„> then the arc NQG = $ — 4>o
F = F0 sin ^(r. As the body rotates about the axis OR, both
M and j^0 move with it. When <f> — <£0 = 0 or IT, the point ^
coincides with either P7 or P; the couple-moment vanishes and
the system is equivalent to a single resultant. As the body is
turned from either of these opposite positions through any angle
the couple F increases and its magnitude varies as the sine of the
angle of rotation. The couple reaches a maximum in either of the
positions given by <£ — <f>0 = + |TT and then decreases again. Thus
there are in general two positions of the body in which the couple-
moment F has a given value, and two more in which it has the
same value with an opposite sign.
42. We may interpret this result in a slightly different
manner. We may ascribe to each generator a certain couple-
moment F peculiar to itself, which becomes the couple-moment"
when the body is so placed that that generator is a Poinsot's axis.
Make MN-i = MN0 + GP, then for any generator of the cylinder,
say the one which passes through P, we have F = F0 sin N±P.
It will be useful to state this result in words. Through the
line of action of R draw two planes, one passing through the two
generators whose couple-moments are each zero, and the other
arbitrary and cutting the cylinders in two other generators. If F
be the couple-moment for these last two generators and % the angle
between the planes, then F = F0 sin ^ where F0 is given by either of
the forms in equation (8).
43. In what precedes it has been supposed that both the direction and the line
of action of the principal force E are given in the body. In this case the body can
only be rotated about Ox' as an axis. If the direction of R is not given, but only its
line of action, the body can also be inverted by rotating it through two right angles
about an axis perpendicular to Oxf. To avoid complicating the figure it will be more
convenient to effect this last change by rotating the forces in the opposite direction,
each about its point of application, so that the angles between their directions
remain unaltered.
The effect of this inversion is easily seen to be, that the positive directions of x'
and of one of the two y', z' are reversed. As it will be convenient that they should
AET. 44] ARRANGEMENT OF POINSOT's AXES. 337
have the same positive directions in space as before, we shall represent the effect of
the inversion by changing the signs of the force R and of that of one of the astatic
moments K^Ka. The sign of the couple-moment T about Poinsot's axis also must
be changed (even if its magnitude remains unaltered) when the positive direction of
x in space is to be the same after inversion as before.
One result of these changes is that the arc FP (Art. 40) takes up another
position (say Q'Q, not drawn in the figure of Art. 38) making the same angle with
GR as before, but on the other side. The angle 00 and the couple T0 are also
changed. Thus the positions in which Poinsot's couple vanishes are changed by
the inversion of the body.
44. To find the equation of Poinsot's axis referred to the prin-
cipal axes at the central point.
Following the notation already described in Art. 39, the
equations of Poinsot's axis referred to the axes of the forces
are Rrf = -K*az, R? = -K3a3 ............... (1),
and the couple-moment F is given by r = K3b3 — K2c2 ........ (2).
Transforming these to the axes fixed in the body, we obviously have
R (Cjf + C217 + CgO = - K3a3.
Eliminating f , 77, £ in turn, and remembering that each constituent
of the determinant of transformation in Art. 38 is equal to its
minor, we have
R ( — v)as + £aa) = — K^a^ + K3ajbl \
R ( ! - fa + £«») = - K&h + Kaasbt I ............ (3).
R ( — |oa + ^aj) = - K.£tzC3 + Kaatba)
These may also be written in the form
)3 + K&\
KscA ..... ....(4).
Any two of these are the equations to Poinsot's axis when the
relative positions of the body and the forces are given by the
direction cosines a^, &c. They are also the equations of the fixed
generator of the circular cylinder, Art. 40.
Adding together the squares of the equations (3), we obtain
the equation of the cylinder traced out by Poinsot's axis as the
body is turned round Ox. This cylinder is easily seen to be a
right circular cylinder and its radius p is given by
KjaJ + K?a? ..................... (5),
as already proved in Art. 40.
When the body is so placed that the forces reduce to a single
K. s. ii. 22
338 ASTATICS. [ART. 45
resultant, the equations (4) may be put into a more convenient
form. Since F = 0, the first of those equations reduces to
R ( - 1703 + £a2) = - KJ>3 +
also, by (2) 0 = - JiT2c2 +
Subtracting the squares, we have
Let us seek the intersection of the single resultant with the plane
of xy\ putting therefore £= 0, the two first of equations (4) become
_
„ 2 _ „ 2 — 2 "S > . UJ £r 2 — 1 ...... \/'
A straight line drawn through the point thus determined parallel
to the force R is the single resultant.
Adding these equations together and remembering that
b/ + a./ = 1 - c32 = cx2 + c22,
772 £2 1
we have, after division by c^*, K'2-K^ + 7T2 = !& ......... ^
This is the equation of a focal conic, Art. 36. The single resultant
therefore intersects the focal conic in the plane of xy. In the
same way, it intersects that in the plane of xz. We thus arrive at
a theorem due to Minding, viz. that when the body is so placed that
the forces are equivalent to a, single resultant, the line of action of
that resultant is a focal line. A fuller consideration of this mode
of proof and of Minding's theorem will be found a little further on.
An apparent exception arises when either a3=0 or a2=0. Supposing that a3=0
the equations (3) become -Ba2f= - K2a2clt JJa1f=^r2a2c2-
Since - c2 = a^b3 - ajb^ , we have T = K3b3 - Kzcz = (KS + K^) bs = 0.
Thus either &3 = 0 or K9 + K2a1 = Q. Joining the former to T=0, we have c2=0.
The latter is impossible if K3 is greater than Ka ; if K3 is less than Zf2 the focal
conic (7) is a hyperbola and the single resultant is parallel to an asymptote. Thus
in both cases the single resultant intersects the focal conic.
Ex. 1. Show that the single resultant intersects the plane of the imaginary focal
conic in the conic £2 + £-2 = JL (_L _ i) .
This conic is fixed in the body when aa is given.
Ex. 2. Show that the circular cylinder (5) intersects the plane of xy in the
conic whose equation is
45. The direction of the principal force E, and a point £, rj, f on a generator
of the circular cylinder being given referred to the principal axes of the body, it is
required to find the couple-moment about that generator when the body is so placed
that the generator is a Poinsot's axis.
ART. 46] ARRANGEMENT OF POINSOT'S AXES, 339
For the sake of brevity let us write
Multiplying the second and third of equations (4) Art. 44 by Kfa and K
respectively we have
- Taj + K3*a3 (Rr - Fa3) =
The couple-moment r is therefore given by
(Kfaf + Xfafl T=B (Kfaq + E*a3r-KzK3p) .................. (l).
If the line of action of B only is given and the force may act either way along
it, we obtain another value of F by inverting either the body or the forces. If F be
the couple-moment after inversion we have by Art. 43
Ittttf+XfafiFmBiEjaa+Kfaf+Kfa) ............... (2).
The force B then acts along the negative direction of its line of action.
We may write (1) in the form
We therefore see that the plane through the line of action of B and the two genera-
tors whose couple-moments are zero (Art. 41) is
- (KJ-K*) a2a£+K3 (K^+KJ a#, - Kz (K^ + K,) a2f=0 ......... (4).
Conversely, when the magnitude of the couple F is given, either of the equations
(1) or (3) enables us to find the generators which have the given moment F when the
body is so placed that one of them is a Poinsot's axis. When F is given, either of
these equations represents a plane intersecting the circular cylinder (5) in two
straight lines which are parallel to the principal force. These are the generators
required ; see also Art. 41. If we change the sign of F we obtain another plane,
parallel to the former, giving two other generators, each of whose couple-moments
has the given magnitude but an opposite sign. These four are obviously sym-
metrically arranged round the principal force.
Another construction for Poinsot's axis and moment is indicated in the follow-
ing examples.
Ex. 1. A straight line OQ is drawn through the central point 0 perpendicular
to the plane containing the force B and its corresponding fixed generator. Prove
that p, q, r are the coordinates of the point Q in which this straight line cuts the
circular cylinder. Prove also that Q is one of the poles of the great circle repre-
sented by PF in the figure of Art. 38.
Ex. 2. Let OS be the straight line whose direction cosines are proportional to
-K2K3, K£az, K3*a3, when referred to the principal axes of the body at the central
point 0; thus OS is fixed in the body when the position of OB is given. If
^ be the angle contained by the lines OQ, OS, prove that
r _ jg,ag,a+g,y+jr,V) *
cos0~| Kjaf + Kfaf \ '
Show also that the straight line OS lies in the plane containing the force B and the
two generators whose couple-moments are zero.
46. If the magnitude of the couple-moment F is given as well as the line
of action of B, we may obtain other cylinders which will intersect the right cylinder
already found in the corresponding Poinsot's axes.
The first of equations (4) Art. 44 is
B(-r,at+£aa)-ral=- KJ)3 + Ktct,
and r=--fiT2c2 +#„&,.
22—2
340 ASTATICS. [ART. 48
Hence subtracting the squares, as in Art. 44,
Now by Art. 38, V-cj2=ci2~a32> hence, substituting for Cj2 from the second of
equations (4), -we have
Again 6^*-Cjs=a22-61a, substituting for b^ from the third of equations (4), we have
^(-^a + W-IXp-r2 {flt-fr^ + qaj-ra,}' .
K?-K* K*
Lastly, the last two of equations (4) give
{-gf-fra + ^J-ra,}8,
-°i ......... (3)'
The three surfaces (1) (2) and (3) are cylinders, for the equation to any one of them
shows that an expression of the first degree in |, 17, f is some function of another
expression of the first degree. Also the axis of each cylinder is parallel to the
straight line £/«! = 17/03 =f/a3> i-e. the axis of each is parallel to the line of action
of the force B.
It may be noticed that the direction cosines blt 62, 63; elt c2, cs have been
eliminated so that the equations to these cylinders contain only the principal force
R, the direction cosines of R and Poinsot's couple F.
47. Supposing that the coordinates (|, 17, f) of some point on the cylindrical
locus (5) are given, and that the line of action of the force R is also known, any one
of the equations (1), (2), (3), of Art. 46 may be regarded as a quadratic to find the
couple-moment when the body is so placed that the corresponding generator is a
Poinsot's axis.
If we seek the corresponding equations when the forces are inverted we change
the signs of R, F and one of the K'B (Art. 43). But these changes leave the
quadratics unaltered. Thus the two values of T given by any one of these
quadratics correspond to the two directions in which R can act along the same
given line of action.
Ex. The given point (f , 17, f) being supposed to be on the circular cylinder, prove
that the three quadratics (1) (2) (3) of Art. 46 reduce to the same, viz.
Prove also that the roots of this quadratic are given by
T (JT, V + Kfafl = R (Kfaq + JT,
where p, q, r have the meanings specified in Art. 45.
48. Winding's Theorem. By joining any one of the three cylinders (1), (2),
(3) to the circular cylinder we have sumcient equations to find the generators which
can have a given couple-moment and are also parallel to any given straight line.
It will often be more convenient to use the intersections of the cylinders with one
of the coordinate planes. Thus putting f =0, the cylinder (1) cuts the plane of xy
(J?7?a> + ra1)2-ra (Bfaj - Ta2)2
in the conic -1 — ^.8_^2 - + — fT~y-=saf ........................ W-
When the forces are equivalent to a single resultant we have F = 0 and in that
772 PI
case equation (1) reduces to the focal conic ' y2 + ^a = -55 .......... (2)-
The single resultant therefore intersects the focal conic in the plane of xy.
Similarly it intersects that in the plane of xz. See Art. 44.
ART. 52] ARRANGEMENT OF POINSOT'S AXES. 341
40. Conversely, let a straight line intersect both focal conies, then by Art. 37
it is a generator of the circular cylinder. If the direction cosines of this straight
line are «!, a2, as, the corresponding couple-moment T is given by the quadratic (1)
of Art. 48.
This quadratic gives two values of T. Multiplying (2) by B2a32 and subtracting
the result from (1) we find that one root is F=0 and that the other is given by
(ijV + JTsV) r=2Ba8{jrifta2 + Z82(Vi1-£a!1)} ............... (8).
The result is that the couple-moment for the generator is zero for one of the
two directions in which the force E can act along that generator.
These two values of T follow also from equations (1) and (2) Art. 45, for when
the value of F given by (1) is zero, the value given by (2) agrees with that shown in
equation (3) of this article.
Finally, we see that if any straight line can be the line of action of a single
resultant force that line must intersect both the focal conies, and if a straight line
intersect both the focal conies it can be the line of action of a single resultant if the
body be properly placed.
6O. Ex. 1. The direction of the principal force E being given by the direction
cosines alt a*, a3 referred to the principal axes at the central point show that each
oftheplanes ( I- * \Kf+ (i- £ ) ^-
Vh a?; \at ojj a,
passes through the line of action of E and intersects the focal conies in four points,
which are the corners of a parallelogram formed by the focal lines, two of which are
parallel to the direction of E. Prove also that the focal lines parallel to the given
direction of E are the corresponding single resultants.
This follows easily from Art. 45.
Ex. 2. If the body IB so placed that the force E acts along an asymptote of the
hyperbolic focal conic, prove (1) that the circular cylinder contains the elliptic focal
conic on its surface ; (2) that as the body is turned round OE Poinsot's axis lies in
the plane containing E and parallel to the force H which corresponds to the
greater astatic moment K9 ; (3) that Poinsot's couple T is always zero as the body
is turned round OB provided the force E acts in the proper direction, but is zero
only when the plane of the hyperbolic conic contains the force H if B act in the
other direction.
51. Relations of Poinsot's axis to the confocals. The manner in which the
single resultant is connected with the confocals is given by Minding's theorem.
We may also find the relations of Poinsot's axis with the same confocals in the
general case in which the couple is not zero. To effect this we require the following
lemma in solid geometry.
62. Lemma. Let the squares of the semi-axes of two confocals be a2+X,
/32 + X,7*+X and aa + X', /32+X', -ys-f X'. Let the direction cosines of any straight line
be (I, m, n) and its distance from the origin be p. If two planes at right angles can
be drawn through the straight line to touch the two confocals, then
It follows that when the confocals are given the left-hand side is constant for all
straight lines.
Let (l't m', n'), (I", m", n") be the direction cosines of the tangent planes, and
p, p' the lengths of the perpendiculars on them. Then
P2=(a2 + X) V* +(/9*+ X) m* +(-ya + *) »",
p'2= (a2 + X') t'» + (j8« + X') m"2 +(72 + X') n"».
342 ASTATICS. [ART. 55
Noticing that (?=p*+pf* we find by addition
p2 = 08 (J'2 + J"2) + fil (m'2 + m"2) + -f (n* + n"7) + X + X'.
Hence since P+l'* + l"*=l &o., we have
p2 + a2P + pfrn? + -y^i2 = a8 + jS2 + 7s + X + X'.
63. Let ns now apply this Lemma to any generator of the cylinder. Let
a, /3, 7 be the semi-axes of the imaginary focal conic, then, by Art. 36,
a2=0, p=-Kfl&t -f=-Kfl&.
The values of X, X' are the squares of the semi-major axes of the two confocala ; let
these be represented by MJR? and M^'/R2 as in Art. 35. The direction cosines of
any generator are (alt aa, as) and its distance p from the central point is given by
Ripi=K£a£+K£aJ. Hence, substituting, the left-hand side of the equation in the
Lemma reduces to zero. We therefore have M^ + M/2 = Kz* + K3*.
If therefore any two planes at right angles are drawn through a possible Poinsot's
axis and two confocals are drawn to touch these planes, the sum of the squares of the
semi-major axes of these confocals is constant. This constant when multiplied by R* is
the sum of the squares of the astatic moments of the principal couples at the central
point.
From this we may deduce as a corollary a theorem discovered by Darboux.
Let a plane be drawn through any possible Poinsot's axis to touch one of the focal
conies, then a perpendicular plane through the same axis will touch another focal conic.
For in the limit these conies may be regarded as the bounding rims of two flat
confocals whose semi-major axes are respectively KJR and K^R.
64. Ex. 1. If a possible Poinsot'a axis touch two confocals prove that the sum
of the squares of their semi-major axes is equal to K22 + K3* after division by .R2.
If a straight line touch two confocals, and tangent planes are drawn at the points
of contact, these planes are known to be at right angles. If we apply the general
theorem in Art. 53 to these two tangent planes, the result follows at once.
Ex. 2. If a possible Poinsot's axis intersect one of the focal conies prove that it
must intersect the other also.
For suppose it intersects the plane of xy in the elliptic focal conic, it may be
regarded as touching the confocal surface whose semi-major axis is K3/R. Hence
it also touches the confocal surface whose semi-major axis is KZ/R (by the last
example), i.e. it intersects the plane of xz in the hyperbolic focal conic.
Reduction to Three and to Four Forces.
55. We have seen that the forces of any astatic system may
be reduced to two couples and a single force. This representation
of the forces, though very simple in its character, may not always
be convenient. These couples and the force have an intimate
relation to the central point and central plane, and the positions
of this point and plane may not suit the circumstances of the
problem we wish to consider.
We shall now examine some other representations of an astatic
system. We shall show that the forces may be reduced to three
forces which act at three arbitrary points in the central plane.
ART. 56] REDUCTION TO THREE FORCES. 343
These points however must not in general lie in one straight line.
We shall show that the forces of the system may also be reduced
to four forces which act at any four points fixed in the body at
which we may find it convenient to apply them. The four points
must not in general lie in one plane.
We can see another advantage of these representations of the
forces. For the points of application may be regarded as the
corners of a triangle or tetrahedron of reference. We are thus
enabled to use the systems of coordinates called trilinear and
tetrahedral with considerable effect.
56. To show that all the forces of any system may be reduced
to three forces which act at three points lying in the central plane.
Following the same notation as in Art. 9, let the forces of the
system be P,, P2, &c. and let MI, M2, ... be their points of
application. Let these be referred to any axes Ox, Oy, Oz, either
rectangular or oblique, which are fixed relatively to the body.
Let the coordinates of Mlt M2, &c. be (a^, y1} zj, (x2, y^, z^), &c.
Let Oaf, Oy', Oz, be another system of axes, not necessarily
rectangular, to which we may refer the forces. These are fixed
relatively to the forces. Let the components of the forces along
these be (X\, Y\, Z\\ (Z'2, Fa, £'2), &c.
Consider the system of parallel forces X\, X'2, &c. All these
are astatically equivalent to a single force 2X' acting at their
centre of parallel forces. In the same way the two other systems
of parallel forces, viz. Y\, Y72 &c. and Z\, Z\ &c., are equivalent to
2F' and ^Z' each acting at its own centre of parallel forces in
directions parallel to y' and / respectively. These forces we may
represent by F, G, H, and their points of application by A, B, C.
The centre of parallel forces is known to possess the astatic
quality If then we move the arbitrary axes Ox, Oy', Oz in any
manner about the origin, keeping their inclination to each other
unaltered, the system will yet be equivalent to the same three forces
F, 0, H acting at the same three points A, B, C in directions always
parallel to the axes Ox', Oy', Oz',
To find the coordinates of these points we may therefore
consider any one position of the forces and the body. In this
position let X, Y, Z be the components of any force P resolved
along the axes Ox, Oy, Oz. Then
m'Y+m"Z, Z' = &c.
344 ASTATICS. [ART. 57
where (I, m, ri), (V, m', n7), (I", m", n")> are the direction ratios of
the axes (x, y, z) referred to (of, y', a').
Let (x, y, z) be the coordinates of A, then
1*2, Tx + l"
with similar values for yl and z^. Taking the same notation
as in Art. 10 we write $Xx = Xx &c., $X = X0 &c. We thus
have Fx, = IXX + l'Yx + l"Zx
Hence it appears that the point A lies on the plane
Yx Zx
Yy Zy
= 0
.(2).
*z "z
*0
In the same way the points B and C also lie on this plane.
57. We notice that the directions of the axes Ox', Oy', Oz,
are perfectly arbitrary except that they cannot all lie in one plane.
We may therefore obtain an infinite variety of triangles ABC
with corresponding forces at the corners. Any one of these may
be called an astatic triangle, and the points A, B, C, may be
called astatic points.
We may obviously make the inclinations of the forces F, 0, H
to each other whatever we please, though of course the position of
the triangle ABC is dependent on our choice of these inclinations.
It is generally most convenient to make the forces F, G, H act in
directions at right angles to each other.
We have seen that when we want to find the positions of
A, B, C we may consider the body to have some fixed position
relative to the forces. For this position Xx &c. are all constant
whatever the positions of the axes x', y', z' may be. The equation
(2) therefore gives, as the locus of the points A, B, C, a plane fixed
in the body. We also see that the locus is a unique plane
except when all the coefficients are zero. An independent and
elementary proof that the plane ABC is unique has been given
in Art. 13.
Comparing the equation (2) with that found in Art. 25 we
notice that this plane is the same as that already called the
ART. 59] REDUCTION TO THREE FORCES. 345
central plane. It follows that all the astatic triangles lie on the
central plane.
58. To find the central plane and one astatic triangle with
rectangular forces.
The theorem proved in Art. 56 supplies us with a useful
method of finding the position of the central plane. To effect
this we resolve all the forces of the system into any three direc-
tions we may find convenient. Taking the forces in these three
directions separately we have three sets of parallel forces. We
then find the centre of parallel forces of each set by any method
we may find convenient. We thus arrive at three points which
we call A, B, G The plane through A, B, C is the central plane.
We have also found one astatic triangle.
Suppose the system referred to rectangular axes Ox, Oy, Oz
and consider any position of the body relative to the forces.
All the ^-components form a system of parallel forces which may
be collected into a single astatic force %X = F acting at a point A
whose coordinates are
In the same way the y-components may be collected into a force
= G acting at a point B whose coordinates are
The ^--components may be similarly treated.
These three points lie on the central plane. The forces
F, G, H act in directions at right angles to each other and their
magnitudes have been found.
If the principal force is finite, the axes may always be so
chosen that 2X, 27, 2Z are not zero. If the principal force is
zero, the coordinates of the three points are either infinite or take
an indeterminate form; and in this case the central plane is either
at an infinite distance or is indeterminate in position. Thus
whenever there is a central plane this construction may be used
to find it.
59. Referring to the table of elementary couples given in
Art. 10 these expressions for (x, y, z) &c. give a new interpreta-
tion to those symbols. It has been shown in Art. 10 that the
346 ASTATICS. [ART. Gl
constituents in any row of that table are the components of the
corresponding couples. It has now been proved that the consti-
tuents in any column are proportional to the coordinates of an
astatic point with rectangular forces, Art. 57.
60. To reduce all the forces of any system to four forces
which act at four given points not all in one plane.
Let A, B, C, D be any four points fixed in the body. These
we shall regard as the corners of the tetrahedron of reference.
Let Pi, P2, &c., be any forces acting on the body and let
Mlt M2, &c. be their points of application. We propose to replace
each of these by four forces acting at the corners A, B, G, D
parallel to the original direction of the force. Consider DA, DB,
DC to be a system of oblique axes, let £, 77, f, be the coordinates
of any point M and let DA = a, DB = b, D<7= c. Then by Art. 7
the forces acting at A, B, C, D are respectively
P£/a, Pifb, PQc, P-PZ/a-Pv/b-Pyc.
Now £/c is equal to the ratio of the perpendiculars drawn from
M and C on the face ABC, and this ratio is the tetrahedral
coordinate of M, Representing the four tetrahedral coordinates
of M by a, @, 7, 8, and remembering that their sum is unity we see
that the four forces at the corners A, B, C, D, are respectively Pa,
Py9, P7, PS.
We therefore have the following working rule. Any force P
acting at the point whose tetrahedral coordinates are a, /3, 7, 8 may
be replaced by four parallel forces acting at the corners of the
tetrahedron of reference whose magnitudes are respectively Pa, P/3,
P% PS.
The several forces acting at each corner may now be com-
pounded together. The result is that any system of forces can be
replaced by four forces, one at each corner of the tetrahedron.
61. We may prove in the same way that a force P acting at
any point M in the plane ABC may be replaced by three parallel
forces respectively equal to Pa, P/3, Py, and acting at A, B, C,
where a, /8, «y, are the areal coordinates of M referred to the
triangle ABC.
We may also deduce this result from the general theorem for a
tetrahedron. We notice that tetrahedral coordinates become areal
when the point considered lies in a coordinate plane. We may
ART. 63] TRANSFORMATION OF TRIANGLES. 34-7
therefore disregard the coordinate 8 and treat the tetrahedral
coordinates a, /8, 7, as if they were areal.
62. To show that the system can be reduced to three forces
acting at any three points in the central plane which form a
triangle.
Let the system be reduced to three forces acting at the corners
A, B, G of some astatic triangle ; then this triangle lies in the
central plane. Let A', H, (7, be any three points in the same
plane, but not in a straight line, and let D' be a fourth point not
in that plane. Regarding A'B'G'D' as the tetrahedron of refer-
ence we shall transfer the forces from A, B, G to the corners of
this tetrahedron.
To find the force at U, we multiply each force by its 8 coordi-
nate. Since this coordinate is zero for each of the points A, B, (7,
the resultant force at D' is zero.
63. Transformation of Triangles. One astatic triangle
ABC and the rectangular forces F, G, H at its corners being given,
it is required to transfer this representation to any other triangle
A'B'C' and to find the rectangular forces F', G', H' at its corners.
Let axes drawn through any point 0 parallel to either of these
sets of forces be called the axes of those forces. We thus have
two sets of rectangular axes. Let their mutual direction cosines
be given in the usual way by the diagram.
Then any force F may be resolved into the components
Fl, Fm, Fn, acting respectively parallel to the p, ™ TT,
axes of F', G', H'. Treating the forces G, H in
the same way we have F' = Fl + Gl' + HI", * l m n
" I' m' n'
We also have H ' l" m" n"
G = F'l' + G'm
The point of application of the force F' is the centre of the
parallel forces Fl, GV, HI" which act at A, B, C. Thus the point
A' at which F' acts is the centre of gravity of three weights
(positive or negative) proportional to Fl, Gl', HI" placed at the
corners A, B, G of the given triangle. By properly choosing these
ratios we can place the corner A' at any point we please.
The areal coordinates of the corners of either triangle referred
to the other can also be found very simply by using the theorem
of Art. 61. Let (au &, 7^, (03, &, 7«). («s, &, 7s) be the areal
348 ASTATICS. [ART. 65
coordinates of the points A', B', C' referred to the given triangle
ABC. If we transfer the forces F', G', H' back again to the
triangle ABC, the three forces at A will be F'a1} G'a2, H'a3. But
these are the components of F. The forces at B, G, may be
similarly found.
Hence F'a^Fl F'fr^Gl' F'^ = Hl"t
G'OI = Fm G'& = Gm' G'y2 = Hm",
H'as = Fn H'& = Gn' H'js = Hn".
By choosing the nine direction cosines in any way which their
mutual relations permit we can use these formulae to transform
from one triangle to another.
If the forces of the two triangles are oblique we regard (I, m, n), (I', m', n'),
(I", m", n"), as the direction ratios of F, G, H referred to the axes I", G', H'. The
direction ratios of F', G', H' referred to the axes of F, G, H, are proportional to the
minors of (I, I', I") &c. If these direction ratios be (\, \', X") (/it, ft', ft,") (v, v', v") we
have F=Fr\+G'i*. + H'v, G=&G., H=&c.,
instead of the expressions given above. With this exception all the other equations
in this article apply to oblique forces.
64. The imaginary focal Conic. Let us suppose that the forces of the two
triangles ABC, A'B'C' are rectangular. The nine direction cosines are connected
by relations such as lm + l'm' + l"m"=Q &c. Hence the coordinates of A', B', C' are
connected by the three equations
<V4 , ft^2 , 7i72_0 «2Q3 , ftA , !?73_0 asai , &A . 78>i_A m
]?* + QZ + H* ' F2 + ~G* H* ' 'J2+'G2+ ^2-u—(L)-
If therefore A' be taken at any point (a, /3, 7), both B' and C' must lie on the
straight line ^ + |f + g|=0 (2),
where a, /3, y are current coordinates. Taking B' anywhere on this line, then C' is
found as the intersection of two straight lines.
This straight line (2) is evidently the polar line of (alt fa, -ft), with regard to the
a2 fi2 •y2
imaginary come ^+ ^_ + X_ = 0 (3).
Thus the three astatic points are always at the corners of a self-conjugate triangle
with regard to this conic.
The statical property of this conic is that each side of every astatic triangle with
rectangular forces is the polar line of the opposite corner. But as two different
conies cannot have the polar lines of every point the same in each conic, it follows
that this conic is unique. Whatever astatic triangle ABC we take as the triangle
of reference, the conic given by this equation is the same.
65. Ex. 1. Show that, whatever astatic triangle with rectangular forces is
taken as the triangle of reference, the quantities
(1) •F2+G2 + IFi, (2) FGH&, (3) a*G*H* + WEPF2 + c^G*.
are invariable, where a, b, c are the sides, A the area of the triangle, and /'', G, H
the forces.
ART. 68] THE IMAGINARY FOCAL CONIC. 349
We have also the invariant property that the centre of gravity of three weights,
proportional to F*, Ga, H2, placed at the corners is the same for all triangles.
Ex. 2. Show that, whatever astatic triangle with oblique forces is taken as the
triangle of reference, the quantities
(1)
(2)
(3) a'*G'H' {F* (cos a - cos /3 cos 7) - F'H' (cos /3 - cos y cos a)
- F'G' (cos 7 - cos o cos /3) - G'H' sin2 a}+&c. + &c.
are invariable, where a, /3, y are the mutual inclinations of the forces and
li.= 1 - cos* a - cos2 j8 - cos2 7+2 cos a cos /3 cos 7.
We notice that ft is six times the volume of the tetrahedron formed by unit lines
drawn from any point parallel to the forces. It follows that /* cannot vanish unless
the astatic forces are parallel to one plane.
Ex. 3. A system of forces is equivalent to a force R, acting at a point 0, and two
couples, whose astatic moments are K2, K3, and whose astatic arms are placed along
the rectangular axes OY, OZ, the forces of the couples being perpendicular to each
other and to the force R, see Art. 32. If these are transferred to an astatic triangle
A'B'C' situated in the plane yz, the coordinates of the corners being (ijl, fj), (773, &),
(773 , fs) and the rectangular forces F', G', H', prove that
F'=Rl F'Tii=KzV F'^=Ksl"
G'=Rm
where I, m, n &o. are the nine direction cosines of F', G', H', as in Art. 63.
If the forces F', G', H' are all equal, prove that the sum of the distances of the
three corners from each of the axes of y and z is zero.
66. To find the Central Point. The astatic triangle ABC with rectangular
forces F, G, H being given, show that the central point is the centre of gravity of
three weights proportional to F'2, G'2, H2 placed at the corners.
This follows easily from the theorem proved in Art. 30. We multiply each
force, such as F, by the resolved part of all the forces along it, i.e. by F', the
product is F3. The rule asserts that the central point is the centre of gravity of
the three products F*, G2, H2, placed at the points of application of F, G, H.
Ex. If the forces F, G, H of an astatic triangle are not rectangular prove that
the central point is the centre of gravity of three weights proportional to
F(F+GcoBy + HcoB{i), G(Fcoay+G + Heoaa), H(Fooaft+G coaa + H)
placed at the corners, where a, ft, y are the angles between the forces (G, H), (H,F),
(F, G).
This result follows at once from the general theorem given in Art. 30.
67. The central point coincides with the centre of the imaginary conic. To find
the centre of the conic we follow the rule given in treatises on Conies. Differentiating
the equation of the conic (Art. 64) with regard to the areal coordinates a, /3, 7
separately, and equating the results, we find that a, /3, 7 are proportional to
F2, G2, H2. The result follows at once.
68. The imaginary conic being given, it is required to find the central lines and
the principal moments of the system.
Let the system of forces be reduced to its simplest form (Art. 32J, i.e. let the
350 ASTATICS. [ART. 71
forces be represented by a force R acting at the central point 0 together with two
astatic couples whose arms are placed along the central lines Oy, Oz. Let the
astatic moments be Ka, Kt.
Consider the origin O as one corner of an astatic triangle and produce the arms
of the couples to very distant points B and C, replacing the forces by two others,
viz. G and H, both very small. Then OBG is an infinitely large astatic triangle
with rectangular forces. Let OB = b, OG=c, then bG=K% and cH=Ka, also F=R.
We shall now use this triangle to find the equation to the imaginary conic by the
formula given in Art. 64.
Let 77, f be the Cartesian coordinates referred to the rectangular axes Oy, Oz of
any point. Let o, /3, 7 be the areal coordinates of the same point referred to the
infinitely large triangle OBG. Then o = l, /3 =??/&, 7 ={"/«. The conic
«a . P , v_0
£•2 + (J3 + 2J2-'
therefore reduces to —^ + ~ + -^ = 0.
We therefore infer (1) that the centre of the imaginary conic is the central point,
(2) the principal diameters are the central lines of the system, (3) that the lengths
of the principal semidiameters are K% \/ - 1/R and Ks >J — 1/JJ.
Referring to Art. 36, we see that the imaginary conic is the same as the
imaginary focal conic.
69. Ex. 1. If ABC be an astatic triangle with rectangular forces show that
either central line makes an angle 0 with the side BG where
4AF3 (H26 cos C - G2c cos B)
~ cos 2C + ^GV cos 2B '
and A is the area of the triangle.
Ex. 2. If a triangle having its orthocentre at the central point be projected
orthogonally on the central plane, prove that the projection is a possible astatic
triangle with rectangular forces, provided the self-conjugate circle projects into the
real conic -^ + ^ = ™ •
70. Transformation of tetrahedra. The forces being referred to one
tetrahedron as ABCD, it is required to refer them to any other tetrahedron as
A'B'C'D'.
If the coordinates of the corners of the first tetrahedron with regard to the
second are known, the transference may be effected at once by using the rule given
in Art. 60. But if the coordinates of the second tetrahedron with regard to the first
are given, we may proceed in the following manner.
Let the tetrahedral coordinates ot A'B'C'D' referred to the first tetrahedron be given
by the diagram, and let the whole determinant be A. Then
the coordinates of A referred to the second tetrahedron are '
the minors of the several terms in the row opposite A after
division by A. The coordinates of B are the minors of the
terms in the row opposite B after division by A, and so on.
The coordinates of the corners of the first tetrahedron
are now known and the transference may be effected as before.
71. Ex. 1. If one corner as D be changed to D' without altering the opposite
face show that the direction of the force at D' is parallel to the force at D, and that
ART. 72] THE IMAGINARY FOCAL CONIC. 351
their magnitudes are inversely proportional to the distances of D and D' from the
unchanged face. See the rule in Art. 60.
If D' lie in the plane BDC show that the force at A is unaltered.
Ex. 2. The forces at the corners of a tetrahedron ABCD are F, G, H, L
respectively ; it is required to find the central plane, the angles between the forces
being given.
Let the cosine of the angle between two forces F, G be represented by cosFG
and so on. Let /, g, h, I be the minors of the four constituents in the leading
diagonal of the determinant.
1 , cosFG, coaFH, coaFL
coaFG, 1 , ooaGH, cosGL
cosFff, cosGH, 1 , coaHL
cosFL, cosGL, cosHL, 1
Then the central plane divides any side as AB in a point P such that
F.AP _ /I
G.BP- V g'
Resolve the force F into three others Flt F%, F3, acting parallel to G, H, L.
Consider the three sets of parallel forces, viz. (G, FJ, (H, F2), (L, Fa). We may
collect each into its own centre of parallel forces and thus obtain three points on the
central plane, Art. 58. The central plane therefore cuts AB in a point P where
F1 . AP=G . BP. But since J^, F2, F3 are in equilibrium with - F, we have by
Art. 48 of Vol. i., F1*]Fa=glf. The result follows at once.
72. If the forces F, G, H, of an astatic triangle ABC are rectangular and of
finite magnitude, and if the area ABC is not zero, prove that the system cannot be
reduced to fewer than three forces.
If possible let the forces be reduced to two, P and Q, and let these act at D and
E in the plane of the triangle. Let p, q, r be the perpendiculars from A, B, C on DE.
Turn the forces about their points of application until the force at A is perpen-
dicular to the plane ABC, then the forces at B and G act in that plane. Taking
moments about DE we have .F-p=0. Similarly Gg=0, Hr=Q. But this is impos-
sible if the area of the triangle is not zero.
That the points of application D, E must lie in the plane ABC follows from
Art. 57, for DE may be regarded as one side of an astatic triangle, the third force
being zero. We may also prove this in an elementary manner. Place the body so
that the direction of the force P is parallel to the plane of ABC, while the other Q
is not parallel ; this is possible provided P and Q are not parallel to each other.
Then, as in Art. 13, taking the plane of ABC as that of xy, we have Zt the same for
the three forces F, G, H and the two P, Q. The ordinate of E is therefore zero.
In the same way the ordinate of D is zero.
If the forces P and Q are parallel to each other, they cannot form a couple because
their components parallel to F, G and H are not zero. They can therefore be re-
duced to a single force. Proceeding as above we easily show that its point of appli-
cation lies in the triangle ; thence we deduce as before that the area of ABCia zero.
That the three forces F, G, H cannot be reduced to two, P, Q, also follows from
the invariants of an astatic triangle. Regarding DE as one side of the triangle,
the third force being zero, we see that the second invariant of Art. 65 is zero. It
follows that FGH&. is also zero, which is impossible unless either the area A or one
of the forces F, G, H is zero.
352
ASTATICS.
[ART. 73
73. To investigate the condition that the forces of an astatic
system can be reduced to two forces.
We have seen in Art. 57 that the forces of the system can be
reduced to three forces, viz. X0, F0, Z0, acting at three points
A, B, G whose coordinates (a^, y1} z^) (x2, yt, z^) (xs, ys, z3) are given
1 ~\jr "^ T7" TT" "Y" ^^ T7"
Dy j\ (,Xi = -A<B -A(>2/i ~ •A-y -^0*1 = -A-Z)
Y x = Yx F07/2 = F F0z2 = Fz,
p n? 17 _ f7 17 ^ 7"
We shall suppose in the first instance that the principal force
is not zero, and that the axes are so chosen that X0, F0, Z<> are all
finite.
If the three points A, B, C lie in a straight line we may make
a further reduction. We can replace each of these forces by two
other forces parallel to it and of proper magnitude, acting at any
two points M1} M2, which lie in the straight line. By compound-
ing the three forces at Ml} and also those at M2, the whole system
can be reduced to two forces. In order therefore that the system
of forces may be reducible to two forces it is sufficient that the
three points A, B, C should lie in a straight line.
It is also necessary, for otherwise the system is equivalent to
an astatic triangle with rectangular forces. Now by Art. 72 such
a system cannot be reduced to two forces unless either the triangle
is evanescent or one at least of the forces X0> F0, Z0, is zero.
If the three points A, B, G lie in a straight line a plane can be
drawn through that straight line and the origin. Hence
= 0.
Yv, F."
r<y,Zz
The projection of these points on any coordinate plane must also
lie in a straight line. We therefore have
X0, Xy, XZ
Fo, Yy, Y,
Z0> Z,
y>
= 0,
Xx, X0,X2
\ V V Y
-* X) •* C> -* Z
= 0,
Y Y
-*• y> •*• o
= 0.
The second of these four equations expresses the fact that
A, B, C lie in a plane perpendicular to that of yz, the third that
they lie in a plane perpendicular to that of xz, and so on.
Since no two of these four planes coincide, except when the
points A, B, G lie in a coordinate plane, any two of the last
three equations are sufficient to express the fact that the three
ART. 76] REDUCTION OF TWO FORCES. 353
points A, B, C lie in a straight line except when the three force
components are zero.
These determinants are the coefficients of the several terms in
the equation to the central plane. That plane is therefore inde-
terminate.
Expressions for these determinants in terms of the forces,
without the intervention of coordinate axes, have been given in
Art. 31.
74. To find the equivalent forces. We have seen that they may be made to act
at any two points Mlt M2 which lie on the straight line ABC. The equation of
this straight line is evidently — = - — — = * L. This straight line is called
*2-*i 2/2-2/1 *a-*i
the central line of the two forces.
If two forces, not parallel to each other, are together astatically equivalent to
two other forces, we may prove in an elementary manner that the four points of
application lie in one straight line.
Let Plt P2 acting at Mlt Mz be equivalent to Qlt Qz acting at N1} N3. Make P1
act parallel to N^a and take moments about N^a . It immediately follows that
M2 lies on Nfl^. Similarly M2 lies on NjNs. Thus the central line is fixed in the
body.
Take any two distinct points Mlt M2 on the central line. Let the coordinates
of the points thus chosen be (/, g, h) and (/, g', h'). Let (F, G, H), (F, G', H') be
the components of the forces at these two points. The forces will then be known
when we have found (F, G, H) and (F', G', H').
Since this system of two forces is equivalent to the given system, the twelve
elements must be the same for each system (Art. 12).
We therefore have
Xx = Ff+F'f, Xy=Fg + F'g', X
Yx=Gf+G'f, Yv=Gg + G'g', Y
Zx=Hf+H'f, Zv=Hg + H'g', Z,=Hh + H'h',
Any six of these equations determine F, G, H; F', G', H' when/, g, h and/', g', h
are given.
75. To show that whatever points are chosen on the central line, the forces at
those points are always parallel to the same plane.
Supposing the system to be already reduced to two forces Pl , P2 acting at some
two points M j , M2 , let us replace these by two other forces Qt , Q8 acting at any other
points #1 , NZ on the central line. The force Ql is the resultant of two forces which
act parallel to P1 and Pa ; it is therefore parallel to any plane to which Pl and Pa
are both parallel. In the same way the force Q3 is parallel to the same plane.
It should also be noticed that the resultant of the two forces Plf Pa, when
transferred parallel to themselves to act at the same point, is a force fixed in
direction and magnitude.
76. Referring to the determinantal conditions given in Art. 73, we see that if
we substitute £, T>, {for the terms in any row in the first
determinant (repeated here in the margin) we have the
equation of the plane containing the origin and the central
line of the two resultant forces.
Xt
Y.
= 0
R. 8. II.
354 ASTATICS. [ART. 78
If however we substitute £, rj, f for the terms in any column of the same deter-
minantal equation, we have the equation of the plane to which the two resultant
forces are parallel whatever be their points of application.
The first of these theorems follows at once from the values of x1 , &c. given in
Art. 73. The second is easily proved by substituting in _, _
J4 Jf f ~ \)
the terms of the first and second columns the values of Xx r r,
&c. given in Art. 74, and in the third column £, 77, f. After ' '
an obvious reduction and division byfg'-f'g, the equation
reduces to the form shown in the margin, which is the plane required. There is no
exceptional case when the divisor vanishes, for the equation to the plane then takes
the form 0=0.
77. We have hitherto assumed that X0 , Y0, Z0 are all finite. The case in which
any one or any two are zero may be treated as a limiting case and the corresponding
conditions may be derived from those obtained when X0 , F0 , Z0 have finite but general
values. As long as the conditions thus obtained are not nugatory they will be the
conditions required. If however the principal force R is zero, the three compo-
nents X0, Y0, Z0 vanish for all axes and the reasoning in Art. 73 fails from the
beginning.
The equations of Art. 74 supply a method of arriving at the conditions that the
given forces can be reduced to two forces without making any assumption about the
principal force. The body being in any position, let the components of the two
forces be, as before (F, G, H), (F, G', H'), and let their points of application be
(/. 9' ft)» (/'» ff'i ^')- ^ne required conditions may then be deduced from the twelve
equations given in Art. 74. It is evident by simple inspection that the four
determinants! equations given in Art. 73 are satisfied.
If the principal force is zjero and the system can be reduced to two forces, those
two forces must be equal and opposite, i.e. they must form a couple. Let ±F, ±G,
iff be the resolved parts of the forces of this couple, (/, g, h) (/', g', h') the coor-
dinates of the extremities of its astatic arm. Then equating the nine finite elements
of the system to those of the couple we have
Xx = F(f'-f), Xv = F(g'-g), X,=F(h'-h)
¥x=G(f-f), Yv=G(g'-g), Yt=G(h'-h)
Zx=H(f-f), Zv = H(g'-g), Z,=H(h'-h).
The necessary and sufficient conditions that the system should be equivalent to two
forces are therefore that (Xx, Yx, Zx), (Xv, Yy, Zv), (Zz, Yt, Zf), should be each
proportional to the direction cosines of one straight line. This straight line is
parallel to the forces of the couple.
78. Ex. 1. Show that any force F acting at a point A may be replaced by forces
Px, P2 acting parallel to F at any two points Mlt M3 such that AM^M^ is a straight
line. Show also that these forces are
AM<, AM,
P^ = FAM^7M1 and P>=FAM^M,'
Ex. 2. Two given forces Plt P2, acting at the points Mlt My, are changed into
two forces Qlt Q2 which are at right angles to each other, and act at two other
points Nlt Na in the straight line M^M^. If yl , t/2 are the distances of Nj , N2 from
the central point of the forces Plf P2, prove that E4y1y3 = -(P1P2Dsiae)2 where
.R2=pi2 + p22+2p1p2cos0, D is the distance M^M^ and 0 is the inclination of the
forces Pj, P2 to each other. It follows that the product y^ is the same for all
equivalent rectangular forces.
ART. 79] REDUCTION OF TWO FORCES. 355
Ex. 3. In all transformations of two forces Plf P2 into two others in which the
points of application remain on the same straight line, the quantities
(1)
(2)
(3)
are invariable, where x1 , x% are the distances of the points of application M a , Mt
from any fixed point on the central line, D is the distance M^ and 0 is the
angle made by the forces with each other.
Ex. 4. A system consists of two forces Plt P8 acting at Jflf Mz and the inclina-
tion of the forces to each other is 0. Show that (1) the central point 0 is the centre
of gravity of weights proportional to Pl (Pl + P2 cos 6) and P2 (Pz cos B + P2) placed at
Mlt Ma. (2) The central ellipsoid at 0 is two parallel planes perpendicular to
Ifjifj. (3) The principal axes at 0 are M^M^ and any two perpendicular straight
lines.
79. To determine the conditions that the forces of an astatic
system reduce to a single force.
Let the single force be Pl} let it act at the point (x^ ylt zj,
and let its components be Xlt Ylt Z±. Comparing the elements
at any base we have
Xai = Xia;1, Xv = X1yl, Xz = X1z1, &c.
Hence we see that the constituents in any column of any of the
four determinants of Art. 73 bear to each other the ratios
(Xlt Yly Z^) of the components of the single force and that these
ratios must be the same for every column.
We also notice that the constituents in any row of any of the
four determinants bear to each other the ratios (xlt ylt z^) or
(1, ylt ZT) &c. of the coordinates of the point of application.
We have twelve elementary equations and six arbitrary
quantities (X1} Y1} Z^)t (xly ylt z^) leaving six conditions to be
satisfied by the elements of the system.
Since X0 = X1} &c., it is clear that the single equivalent force
is equal and parallel to the principal force, Art. 11. Also, since
the coordinates of the central point depend on the twelve ele-
ments, it is evident that the central points of the two equivalent
systems coincide, Art. 28. Thus it follows that the point of
application of the equivalent single force is the central point of
the system.
23—2
NOTES.
NOTE A, Art. 149. Green's theorem. We may deduce from equation (5) of
Art. 149 an extension of Gauss' theorem, Art. 106. Let P, Q, R be the components
of a vector I and let V=l, so that, by (1), 17=0. We then have
If therefore the components P, Q, R of & vector satisfy the condition
£ + £ + ? = 0 (2)
dx dy dz
the surface integral or flux of the vector taken through any closed surface is zero.
It is of course obvious that when 7 as in Gauss' theorem represents the force due
to an attracting body, P=dVjdx &c. , where V is now the potential of the body, and
(2) becomes Laplace's equation.
I. Let two surfaces S, S' be bounded by the same rim. Let that side of either
be called the positive side towards which the normals are drawn.
Since these surfaces enclose a space the surface integral of the vector taken over
both surfaces is zero, provided the normals are drawn all outwards or all inwards,
i.e. provided their positive sides are opposed to each other. Reversing the directions
of the normals for one surface, it follows that the surface integrals for two surfaces
with the same rim or boundary are equal provided their positive sides are the same.
II. Let a curve, such that the direction of the vector I at any point of the curve
is a tangent, be called a vectorial curve (Art. 47). Let a tube or filament be formed
by drawing vectorial curves through any small closed curve, as in Art. 126. Let
a, a-' be the areas of the normal sections at any two points P, P'.
By the extension of Gauss' theorem just proved, the surface integral of the
vector over the boundary of the tube PP1 is zero. The surface integral taken over
the whole space PP', as in Art. 127, is Ta' -la where I, T are the magnitudes of the
vector at the bounding sections. Hence when the vector is such that its components
satisfy the equation (2), the flux across every section of a vectorial filament is the same.
III. It is shown in Art. 149 that in some cases a volume integral can be
replaced by a surface integral. We may also show that in some cases a surface
integral can be replaced by a line integral taken round the rim of the surface.
Let X, Y, Z be the components of a vector whose line integral is to be taken
round a closed curve. Let S be a continuous surface bounded by this curve as its
rim. Let P, Q, R be the components of another vector related to X, Y, Z by the
. dZ dY _ dX dZ . dY dX
equations "=~j T~> V= j 3~~ • **—-* 3— (*>)•
dy dz dz dx dx dy
The theorem to be proved is that the surface integral of the vector (P, Q, R) taken
over the surface <S is equal to the line integral of the vector (X, Y, Z) taken round
NOTES. 357
the rim. Let (I, m, ri) be the direction cosines of the normal to dS, the theorem
then asserts that $(Pl+Qm + Rn)dS = $(Xdx+Ydy + Zdz) ...................... (4).
That side of S is called the positive side towards which the normals (I, m, n) are
drawn. The line integral is to be taken clock-wise when viewed from the positive
side.
If we construct an infinitely small sphere whose centre C is at (xyz), the
components of the vector (X, Y, Z) at the point x+%, y + rj, z + f&re by Taylor's
theorem X' = X+^ + 1, + t, F=F+ + &c., Z'=&C.
OX dig at ax
The sum of the moments of the vector round a parallel to the axis of z drawn
through C, taken for every element of volume dv of the sphere, is
where %vk* has been written for the equal integrals \%?dv, ffidv. It is obvious that
in a sphere $frdv = Q, $Kdv=0, &c. = 0.
It follows that if (X, Y, Z) are the components of one vector, (P, Q, R) are
the components of another vector connected with the former at every point by a
geometrical relation which is independent of all coordinates.
We shall now prove that the theorem (4) is true for any area which is so small
that it may be regarded as plane. Taking the plane of xy to contain the area, we
have
where the third expression follows from the second by an integration between
limits in the manner described in Art. 149. Thus, if AB, drawn parallel to x, cut
the rim in A, B, I I — dxdy=f(YB- YA)dy. But at B, dy is positive and at A
dy is negative, hence taking the integral round the rim and the/efore giving dy its
proper sign, this becomes \Ydy. Since 1=0, m=0, n=l and dz=Q, this equation
asserts that the flux of the vector (P, Q, R) parallel to the positive direction of the
axis of z is equal to the line integral round the rim taken clock- wise.
To prove the theorem for a surface of finite size we add the results obtained for
each element of area. Let two adjacent elements meet along the arc AB. When
integrating round each element we pass over AB in opposite directions so that the
signs of dx, dy, dz in one integration are opposite to those in the other. The sum
of the integrals may therefore be found by integrating round both elements as if
they were one, omitting the arc AB. The same reasoning applies to all the elements
and the sum of the line integrals may be found by integrating round the rim.
The surface integrals of the vector (P, Q, R) taken over two surfaces bounded
by the same rim are each equal to the same line integral. Hence the surface integral
of the vector (P, Q, R) for any closed surface is zero. This also follows at once from
the extension of Gauss' theorem, for the vector (P, Q, R) as denned by (3) evidently
satisfies the condition (2).
The following results show how some volume integrals can be replaced by
surface integrals.
(1) The volume of a solid enclosed by a surface S is $JV cos <f>dy where d<r is an
element of the surface, and <j> is the angle the outward normal at d<r makes with
the radius vector produced. [Gauss.]
(2) The potential at the origin, of the solid (if of unit density) is
[Smith's Prize, 1871.]
358 NOTES.
(3) The integral Jcos^&r/r3 is 4w or 0 according as the origin is inside or
outside 8.
(4) The x component of attraction is $cos<t>'dff/r where <f>' is the angle the
normal at d<r makes with x. [Gauss.]
In Arts 358, 360, 362 and Note M there are some examples of surface integrals
replaced by line integrals.
NOTE B, Art. 190. Potential of a thin circular ring. When the law of force
is the inverse square of the distance, Dickson puts the potential at any point E into
the form F2=, , A — ,
where £(/> + p') is the mean of the greatest and least distances of R from the ring,
M is the mass, and K is the complete elliptic integral of the first kind to modulus
OP/a. See the figure of Art. 185.
NOTE C, Art. 211. Attraction of a solid ellipsoid. In the text the potential
at an internal point P is found first and the axial components of force are deduced
by differentiation. The following method of finding the components of force is so
simple as to deserve attention.
Through P we pass an ellipsoid concentric with and similar to the boundary ot
the solid. The attraction at P of the portion of the solid external to this ellipsoid
has been proved to be zero in Art. 68. It is therefore necessary only to find the
attraction at P of the portion of the solid bounded by this ellipsoid. The problem
is thus reduced to that of finding the attraction of an ellipsoid at a point on its
surface. Let the semi-axes of this ellipsoid be ma, nib, me.
We now construct an elementary cone whose vertex is P and whose base is an
element Q of the surface. If dw be the solid angle of the cone, its attraction at P
is JpT^dwdr/r2 taken between the limits r=0 and r=r. The attraction is therefore
prdw.
The axial components of the attraction of the whole ellipsoid at P are therefore
X= — pJrXdw, F= — pjryudw, Z — — pjrvdto (1),
where (X, ft, t>) are the direction cosines of QP and the integrations are to be taken
so as to include all the elementary cones which lie on one side of the tangent plane
at P.
Let (£, 17, f) be the coordinates of P when referred to the centre. Since Q lies on
the ellipsoid we have - — =-, ~ H ^r- + - — ^-5— = 1 (2).
m2a m262 mjcj
Since the point (£, ij, f) lies on the surface this gives
This value of r has to be substituted in the expressions (1) and the integrations
effected. As the radius vector turns round P, it is evident by (3) that no values of
X, /*, F make r imaginary. Since the value of r determined by X, n, v differs only
in sign from that determined by - X, - /*, — F, the equation (3) represents the surface
twice over. Since the signs of X, Y, Z depend on the signs of the products r\, r/t,
rv, it is clear that if we integrate the equations (1) taking all positions of the radius
vector and not merely those on one side of the tangent plane, we shall obtain in
each case twice the required attraction. We therefore have
NOTES. - 359
where (X, /j., v) have all possible values. It is obvious that the term containing the
product X/i disappears on integration, for the elements corresponding to (X, /j.) and
(X, - fi) destroy each other. In the same way the term containing the product X?
disappears. We therefore have
X2
if c*
These may be written in the form
X=-Ap£, Y=-Bprj, Z=-Cp£. (4).
We notice that the constants A, B, C are functions of the ratios of the axes and
are therefore the same for all similar ellipsoids.
The integrals given above for A, B, C may also be written in the form
•(5),
where the integration extends over the whole surface of the ellipsoid. It easily
follows that A+B+C = 4ir Aa* + Bb2 + Cc* =]r2du ...................... (6),
where r is the radius vector of the bounding ellipsoid drawn from the centre as origin.
The potential is seen by an easy integration to be V=%p {D- A^-Bff- Cf2},
where D=jV2dw, since £/>D must evidently be the potential at the centre.
NOTE D, Art. 218. Other laws of force. The potential of a thin homogeneous
homoeoid at an internal point (f^f ) when the force varies as the inverse *cth power
of the distance can be found, free from all signs of integration, when K is an even
integer > 2. Let up be the surface density at any point Q, where p is the perpen-
dicular from the centre on the tangent plane at Q. The potential is
2>At
-(K-l)(K-
where
and
The general term is ~ ff-EfVf and i (n) = l . 2.3....n.
—
The series has ^ (K — 2) terms. Thus for the law of the inverse fourth power it
reduces to the first term ; for the law of the inverse sixth power, there are two
terms and so on.
At an external point P' whose coordinates are £', 17', f , we have
o * c
Here a', b',c' are the semi-axes of the confocal drawn through P', and e*= a'8 - a*=<fec.
It should be noticed that the differentiations implied in the operator V are to be
performed on (£', if, f ) on the supposition that a', 6', c' are constant. The potential
at an external point may be deduced from that at an internal point by a method
which is practically one of inversion. See Art. 203. [Phil. Trans. 1895.]
360 NOTES.
NOTE E, Art. 250. Heterogeneous ellipsoid. When the attracted point P
lies within the substance of the ellipsoid, a little more explanation may be added.
Through P we describe an ellipsoid similar to the external surface of the given body.
Let it be defined by m=n as in Art. 241. The potential of the inner portion at P is
Fa = 2 f"8 dm3 I du (1 - m2)" ~ l (m2 - 1 + E)n F (u).
JO J A,
£2
Now X satisfies -ol— +&c. = l (Art. 204) and since P lies on the ellipsoid (n),
71 & T~ A.
4hb+*c. = l. It follows that X=0 and therefore, since X^m2, X1=0 (Art. 249).
*
Next consider the shell outside the ellipsoid (n). As explained in Art. 240, we
put Xx=0 and integrate from m2=n2 to m2=l. We have therefore
F2= 2 I \ dm" | du (1 - m2)*"1 (m2 - 1 + B)n F (u).
A f°°
Addhig Fj and F2 we have F=S I dm2 I du[e&c.].
The order of the integrations may evidently be reversed, and the argument may be
continued as in Art. 250, and in the result we have X=0.
NOTE F, Art. 264. Other laws of force. When the law of force is the inverse
*th power of the distance we require the expansion of l]RK~l. There are two ways
of extending Legendre's series.
First we may continue to make the expansion in powers of h and put
If K - 1 is an odd integer, say equal to 2m + 1, we have
0 - 1 ^P
v*~ 1. 3. 5... (2m- 1) dp" «•+"•
If x-l is an even integer, say equal to 2m +2, we have
1 d"1 sin (n + m+l)0
™~ 2. 4. 6. ..2m dp™ sin0 '
where p= cos 9. The four most important theorems relating to the function Qn are
given in Art. 282, Ex. 3.
Secondly, we may retain Legendre's functions of p as the coefficients, but cease
to expand in powers of h. We then have when K is even and greater thau 2
(1 - 2ph + hrf t*-1) (1 - A2)" ~s '
There is a similar expansion when K is odd and >1, except that Pn is replaced by
sin (n + l)0/sin 6 and that the coefficients of the function \f/ (h) are different.
The function $(h) is an integral rational function of h containing only even
powers, the highest being hK~4. Thus the function does not increase in complexity
as n increases, but has always the same number of terms.
When the body considered is a thin spherical surface or a circular ring, h is the
ratio of the radius to the distance of the attracted particle. Thus \{/ (h) is constant
for an integration over the surface of any portion of a sphere or along the circum-
ference of the ring.
When the law of force is the inverse fourth power /t=4 and if/ (h) — (2n + 1) ;
when the law is the inverse sixth, /t=6, and
1 . 3^ (h) = (2n + 1){ - (In - 1) k*+ 2n+3).
NOTES. 361
The general value of \ft (h) is given in the Proceedings of the Mathematical Society,
vol. xxvi., 1895, page 481.
NOTE G, Art. 281. Legendre's theorem. There is another proof of the
theorem J Pnydp — 2/(2n + 1) which is in general use. We have
We multiply both sides of this equation by dp and integrate between the limits
-1 and +1. We then have, by Art. 278,
dP _r/i
/
Integrating the left-hand side, we have
{log (1 + h) - log (1 -
Both series being convergent, we find the value of JPn2dp by equating the
coefficients of Ti2™ on each side.
We may deduce some other interesting results from the equation of differences
(n + 1) Pn+1 - (2n + 1) pPn + nPn_x = 0.
Multiplying both sides by PK and integrating between the limits - 1 and + 1, we
have (2n + 1) fpPnPK dp = (n + 1) J Pn+1 PKdp + n$ Pn_t PK dp.
It follows from Art. 278 that $pPnPKdp is zero except when K and n differ by
unity. In that case we have Jy PK-Pn+i dp = .^ + (" * '+ as in page 219.
In the same way we may show that lp2PnPKdp is zero except when K and n
are equal or differ by 2. In these cases
~ (2n- 1) (2n+ 1) (2n+3) ' «««» ~ (2n + l) (2» + 3) (2n + 5) '
where the limits of the integrals are - 1 to + 1.
We may, by successive induction, deduce from the equation of differences, that
_^A(m-r)A(r) A(n-r) 2n + 2m-4r+l
m n~' A(m + n-r) 2n + 2m - 2r + 1 Jn+n-2"
where S expresses summation from r=0 to the lesser of the two quantities m, n.
.. . 1.3.5... (2m -1) .. . m + 1 ,.
A= ' •'• m> =
We may interpret A (m), when m is zero or a negative integer, by supposing this
relation to hold generally, so that putting m=0 we have .4(0) = 1. Similarly
.4(-l) = 0, and hence, when m is any negative integer, ^(m) = 0.
In the series r is supposed to vary from r—Q to either m or n. If however r is
taken beyond these limits, for instance if r= -1 or m+1, then (in consequence of
the property of the function A just stated) the coefficient of the corresponding term
is zero. Hence practically we may consider r to be unrestricted in value.
We notice that in this expansion the suffixes of P are all even or all odd according
as m + n is even or odd. If then we multiply by Pl and integrate the product
between the limits -1 and +1, we have JP,PmPndj>— 0 if l + m + n is odd
(Art. 278).
Supposing l + m + n to be even, it follows (by subtracting the even number 21)
that m + n- 1 is also even and that there may be a term on the right-hand side in
which the suffix is given by m + n-2r=l. This term, after multiplication by Pj,
supplies the integral JP{2dp and is not zero. We then find
P
P
362 NOTES.
where s = $(l + m + n). In order that this integral may not be zero, no one of the
quantities I, m, n must be greater than the sum of the other two, and l + m + n must
be an even integer.
The reader may consult a paper by the late Prof. J. C. Adams in the Proceedings
of the Royal Society 1878, No. 185. The value of the integral is also given by
Ferrers as an example on page 156 of his Treatise on Spherical Harmonics, 1877.
By using the results referred to in Art. -292 we also find
1.3 .5 ... (2m + 2n-2s + l)
/" dmPm+n , _m(nt + l) ... (m + s-l) 1.3.5 ... (2m + 2n-2s +
*J l dpm P~ 1.2...S ' 1.3.5...(2n-2« + l)
when l=n — 2s. When n — I is odd or l>n, the integral is zero.
CP ^ 1.2.3...K
cPmdp = — _ 3 5 — +m+n •
when K>m. The integral is zero if /c -m is odd, or if K<m. In both integrals the
limits are - 1 to + 1.
When the law of force is the inverse /cth power of the distance, the equation of
differences takes the form
as explained in Art. 282, Ex. 3. We may use this equation in a similar manner to
find J*(p)Q»*4p and \<t>(p)pzQJdp where ~
NOTE H, Art. 288. Laplace's theorem. Laplace deduces the equation
lYmYndw = Q from the equation (7) of Art. 284. What follows is an extension
of his method, M6canique Celeste, livre troisi&me 12. Let us write (7) in the form
dfdY\ d dY
where 6 = l-/t2, c=l/(l-/*2), p=m(m + l), p' = n(n + l).
Multiplying these equations by Yn , Ym respectively and subtracting, we find
Integrating by parts, we find that the unintegrated parts cancel, we therefore have
where the quantities in square brackets are to be taken between limits, the first
between /&=±1, the second from 0 = 0 to 2ir.
Now b = l-fjf, if therefore Ym, Yn and their differential coefficients with regard
to fjL are finite all over the sphere, the first integral is zero.
The range of <j> from 0 to 2w, carries a point P round the sphere on a small circle
to the point from which P started. If then the quantities c, Ym, Yn, and their
differential coefficients with regard to <f> are "one rallied" on the sphere, the second
quantity in square brackets is the same at both limits, and the second integral is zero.
It follows that if p and p' are unequal (that is, if neither m=n nor m + n= -1)
the integral |FmFndw = 0.
If we generalise (7) and write it in the form
NOTES. 363
where a, 5, c, e and A are given finite functions of p, <f>, but not of p, while p is a
given function of m, the function Ym is not now a Laplace's function, but the
equation (p~p')}\YmYnAd/jid<t> = 0 (5)
will in certain cases be true. This may be proved by the same reasoning as before.
The unintegrated parts cancel and the integrated parts vanish provided (1) b and e
are zero when ju= ±1, (2) Ym, Yn and their differential coefficients with regard to ft
and <f> are finite one-valued functions of /*, <j>. Other cases in which the integrated
parts are zero will suggest themselves to the reader and need not be particularised
here.
We may also extend the theorem to the case in which the integration is effected
only over the area within some closed curve drawn on the sphere, provided Ym, Yn
. 1 dYm 1 dYn 1 dYm I dYn .
are such that => — -^ = ==- -j-* , =r- -3-= = •=- — r-f at all points of the boundary.
Ym d/j. Yn d/j. Ym d<j> Yn d<f>
For example, the equation (5) is true if Ym and Yn vanish at all points of the
boundary.
The equation (5) is also true if both Ym and Yn satisfy the condition
dY I e A dY ,
at all points of the boundary, where ^ is the angle the arc & makes with the
elementary arc of the boundary and X is an arbitrary function of 6, <f> but not of m
or n. When 6 = l/c = l-M2 and «=0, X=0, this implies that the space variation
of Y perpendicular to the boundary is zero.
NOTE I, Art. 329. Magnetic sphere. The expression in the text for the
potential applies obviously to an external point. At an internal point, the potential,
by the same rule, is equal to %vlr cos 0. This also follows at once from the result
given in the next article for an ellipsoid. The force due to a uniformly magnetised
solid sphere at an internal point P is therefore -$irl. The direction, when taken
positively, is opposite to the direction of magnetisation, and tends to demagnetise
the body.
NOTE K, Art. 342. Magnetic forces. Kelvin, when speaking of the two
definitions of resultant force in a crevasse (1) tangential and (2) perpendicular to
the lines of magnetisation, sometimes calls the former " the polar definition " and
the latter "the electromagnetic definition" (Reprint &c. Art. 517). This latter force
is called " the magnetic induction " by Maxwell and this phrase has been generally
adopted in the text. A slight modification has however been made in Art. 342 and on
a few other occasions when the change seemed to make the meaning of the context
clearer. Maxwell's phrase is not entirely unobjectionable and it is much to be
desired that some short term could be generally agreed to.
NOTE L, Art. 345. The magnetic induction. At a point outside a magnetic
body the magnetic force and the magnetic induction are the same. It follows that
their components satisfy Laplace's equation, and we have
, -
ax dy az ax ay dz
At a point inside a magnetic body, we have by substitution (Art. 345)
dX, dY, dZ, /dX ,\ fdA , \
-1 + -r-1 + -r-1 = ( T- + &°- ) + 4jr(— + &c. )
dx dy dz \dx ] \dx J
364 NOTES.
Since the magnetic force is by definition that due to Poisson's two distributions,
the sum of the terms in the first bracket on the right-hand side is equal to 4?rp
(Arts. 105, 41). The sum of the terms in the second bracket is - 4irp (Art. 339).
We therefore have
dX dY dZ^_ dXj dYl d_Zi_Q ^
dx dy dz dx dy dz
It follows that the components (X, Y, Z) of the magnetic force satisfy different
differential equations according as the point under consideration is external or
internal. The components of the magnetic induction satisfy the same equation (viz.
Laplace's equation) whether the point is inside or outside.
Since the equation satisfied by the components of the magnetic induction is the
same as the condition (2) given in Note A, page 356, it follows immediately that
the surface integral of the magnetic induction taken through any closed surface is
zero. This surface may be wholly within or wholly without or partly within and
partly without the magnetic body. See also Art. 488.
It also follows that the surface integrals of the magnetic induction taken through
any two surfaces having the same rim are equal. See Note A.
NOTE M, Art. 358. Vector potential. Since the surface integral of the
magnetic induction depends on the closed rim and not on the form of the surface
(Note L, page 363), it should be possible to find the induction through a closed
curve, without constructing a surface to act as a diaphragm.
This is effected by finding a vector A whose components F, G, H satisfy the
equations
_ dH dG v dF dH dG dF
AI = -J— -j— , •*!— "T> j?~ » •"! = ~j>T "~ ~j~ (3)i
dij af of af a£ drj
where (Xlt Ylt ZJ are the components of the induction at a point P whose
coordinates are (£, 17, £). Then, as proved in Note A, page 356, the induction
through any closed surface is equal to the line integral of the vector (FGH) round
the rim. This new vector is called by Maxwell the vector potential of magnetic
induction. [See his Electricity, Art. 405.]
The relations (3) are satisfied at an external point for a simple lamellar shell of
unit strength by taking
F=[-t <?=/X H=[^ (4)
J R J R J R
where the integration extends round the rim of the shell, and R is the distance of
an element of the rim (xyz) from a point (£ijf) in space. This follows at once from
the values of X, Y, Z given in Art. 358.
Example. Prove that for a simple magnetic shell of strength m, in the form of
a small circle of radius a and centre 0, the vector potential at a point P is
approximately
mira*p ( 3 a2 15 a V
rs | 2 r2 ~8~
where r = OP and p is the distance of P from the axis of the shell. [Coll. Ex. 1896.]
To prove this we take the plane of the circle as the plane of xy, the centre as
origin and the plane of xz to contain P. We then have x — a cos <f>, y = asm <j>, z=Q
and .R2=r2-2apcos0 + a2. Substituting in (4) and expanding the denominator in
powers of ajr, we see that F=0. Rejecting all odd powers of cos<f> in the
expansion for G we find at once that G has the value given in the enunciation.
NOTES.
365
We must refer this to axes of x, y which are independent of the position of P if
we wish to use equations (3). We then have
F=-Ar,lp, G = Ayp, H=0,
where (£, if, 0) are the coordinates of P and^2=^2 + ijs.
For an elementary lamellar shell, the vector potential is A — Msindli3, where
r=OP, 0 is the angle r makes with the axis Oz and M. = va?m. The direction of the
vector is perpendicular to the plane POz and its positive direction is clockwise
round Oz.
For an elementary magnet whose moment is M, centre 0, and axis the axis of z,
we assume the magnitude of the vector to be M sin 0/r2 and its direction to be as
just described. The components are then evidently F=——£, G = —j\, H=0.
Since the potential of an elementary magnet is M cos 0/r2, it is not difficult to see
that the equations (3) are satisfied.
To find the components of the vector potential of a small magnet when the
direction cosines of the axis are X, /*, v, we resolve the magnet into M\, Mfj., Mv.
The F component of MX is zero, those of Mfj,, Mv are Hpf-jR3 and -Mvij/R3
respectively. The F component for a magnetic body at P is therefore
where R is the distance of any point (xyz) of the body from the point P in space
whose coordinates are (£, ij, f) and M=Idv, Art. 32(6.
NOTE N, Art. 397. Electrified sphere. The figure has been drawn by Dickson
to show the distribution of electrical density on the surface of a sphere under the
influence of a point-charge at S (where OS = 10, 0.4 = 6). Let a radius vector from
the centre 0 cut the curve drawn inside the circle in P, the circle itself in Q, and
60°
the dotted circle outside in R. The length PQ then represents the density of the
(negative) charge at any point Q of the sphere, when uninsulated; while the length
QR would represent the uniform density of an equal (positive) charge freely
distributed on the sphere, when the point-charge at S is absent and the sphere
insulated. Consequently, if the sphere be initially uncharged and at zero potential,
366 NOTES.
and if the point-charge be then brought to S, QE - PQ will represent the (positive)
density at the point Q. This density will be negative from A to F, at which latter
point the total density is zero. If the whole figure be rotated about OS, F will
trace out the line of no force. For the data given, the angle FOS is about 56J°,
and if the tangent from S touch the circle at T, the angle SOT will be about 53£°.
NOTE P, Art. 486. Discontinuity. The result in Ex. 8 is interesting as it
exhibits a discontinuity. The difficulty thus introduced would disappear if we
supposed the value of K to be continuous but to change rapidly from K to K'. See
some brief remarks on this subject in chap. xm. of the second volume of the
Author's treatise on Rigid Dynamics (Art. 620 of the fifth edition).
INDEX TO ATTEACTIONS.
The numbers refer to the articles.
AIBT. Clairaut's theorem to a second approximation, 304, note.
BEKTBAND. Eelation of force to the curvatures of level surfaces, 128.
BIANCO. Eemarks on the history of "potential," 39, note.
BIOT. Terrestrial magnetism explained by a central magnet, 335.
CAPACITY. Electrical, defined, 371. Condensers, 417. Several cases, 418, 419, &c.
Capacity found by inversion, 432. Specific inductive capacity, 371, 473.
Effect of a change of dielectric, 474, 483. Plane, cylindrical and other con-
densers, 417, 419, 478, 479, 484, Ac. Spheres, &c., 486.
CENTBOBARIC BODIES. Denned, 137. The fixed point is the centre of gravity and
every axis is a principal axis, 137. The law of force is the inverse square or
the direct distance, 137. The boundary of the body is a single closed
surface and the centre of gravity is inside, 140, 141.
CLAIBATJT'S THEOKEM. Expression for gravity, 304. Potential at any external
point, 307. Second approximations, 309.
CONDENSERS. Green's solution, first and second approximation, 417. Examples,
392, 418. Cylindrical condenser, 419, 479. Energy of condensers, 447. With
dielectrics, 484.
CONDUCTOR. Denned, 366. Conductor with a cavity, 386. Two conducting
spheres, 374. Ellipsoid, 376. Disc, 382. Eod, 385. Concentric spheres, 392.
Sphere acted on by a point-charge, 397; diagram, page 365. Cylinders, 407,
&c. Nearly spherical conductor, 420. Enclosed in a nearly spherical shell,
421. A nearly spherical solid of revolution in a uniform field of force, 421,
Ex. 4. Spheres intersecting orthogonally, 423, 436, and at an angle irjn, 433.
Theory of a system of conductors, 438, &c. Mutual potential energy, 446.
Junction of conductors, 448. Introduction of a conductor, 449.
CONES. Attraction of sections at the vertex, 25.
CYLINDER. Various problems, 24. Infinite circular cylinder, attraction at any
internal or external point, 55, 56. Heterogeneous cylindrical shell, 58.
Elliptic shell, 72. Solid elliptic cylinders, 232, &c. Potential of an elliptic
cylinder, 237. Potential of a heterogeneous cylinder, 333. A magnetic
cylinder and the magnetism induced in any field of force, 333.
CYLINDER OF HODS. Limiting case of a cylinder and consideration of the resulting
discontinuity, 52.
DARWIN. Clairaut's theorem to a second approximation with references, 304, note.
DICKSON. Potential of a circular ring, Note B, page 358. Diagram of the distribu-
tion of electricity on a sphere, Note N, page 365.
DIELECTRIC. See INDUCTION. Defined, 473. Substitution of a solid dielectric for
air, 474. Plane and cylindrical dielectrics, 478, 479, 482. Poisson's con-
ditions, 481. Kelvin's theorem, 483. Various problems and results, 484,
485, 486. Extension of Poisson's theorem for dielectrics, 492.
368 INDEX TO ATTRACTIONS.
The numbers refer to the articles.
DIFFERENTIATION. Method applied to find attractions, 92.
DIP. Tangent is twice that of the magnetic latitude, 324.
DIRECT DISTANCE. Magnitude and direction of attraction, 7, 8.
Disc. See STBATUM, PLANE CONDUCTORS. Attraction of a circular disc at a point
on the axis, 21. Attraction of an infinite disc, 22. Table-land, 23. Elliptic
disc at focus, 30. Elliptic disc at any point, for the law of the inverse cube,
29. Bectangular disc, 30. Disc bounded by two parallel lines, 30. Confocal
level surfaces, 51. Infinite heterogeneous disc, 93. Elliptic discs, special
laws of density, 228. Any law of density, 251, 252. Condition that the
level surfaces are confocals, 263. Electrified elliptic disc, 382. See
ELLIPSOIDAL CONDUCTOR.
DYSON. Anchor ring, 193. Ellipsoids, 247, note.
EAKNSHAW. Points of equilibrium are unstable, 119.
ELECTRICAL PROBLEM. Enunciation, 372. Green's method of solution, 154, &c. ;
another proof, 393, &c. Method of inversion, 168, <fec.
ELLIPSOID. Potential of a solid ellipsoid at an internal point, 211, &c. and Note C,
page 358. At an external point, 222, 225. Other laws of force, 218, and
Note D, page 359. Level surfaces, 216. Spheroids, 219, 220. Heterogeneous,
with similar strata, 239. Any law of density, 245, 247, Note E, page 360.
References, 248, note. Nearly spherical ellipsoids, first and second approxi-
mations, 220, 221. Potential of a magnetic ellipsoid, 330. Induced magnetism
in an ellipsoid, 331.
ELLIPSOIDAL CONDUCTOR. Surface density Mpjiirabc, 376. Quantity on a portion
of the ellipsoid, 377. Potential, 378. Only one arrangement, 381. Elliptic
disc, 382. Quantity on the portion bounded by parallel chords, 385.
Insulated rod, 385.
ELLIPSOIDAL SHELL. Internal attraction of a homoeoid is zero, 68. The converse
theorem, 73. Attraction at an external point close to the surface, 71, 209.
Theorem on the polar plane of an external point, 69. Potential of a
homoeoid at an internal point, 196. The fundamental integrals I and J, 200.
Potentials of confocal homoeoids at corresponding points, 203. Level
surfaces, 205. Lines of force, 207. Thin homoeoid, external point, 208,
209. Linear and quadratic law of density, 231. Any law, 247.
ELLIPTIC COORDINATES. Poisson's theorem, 110. Potential of an ellipsoid, 230 ;
of an elliptic disc, 251.
EQUILIBRIUM. Points of equilibrium are unstable, 119. The separating cone, 120.
Level surfaces near a point of equilibrium are quadrics, 120. Eepelling
particles lie on the surface of the containing vessel, 121.
EQUALLY ATTRACTIVE BODIES. Bodies attract equally if their potentials are equal
over an including surface, 129, &c. Also, if of equal mass and have the same
level surfaces, 131. Their centres of gravity and principal axes coincide,
136. A prolate spheroid and a straight line, 228. Discs and homoeoids, 228.
EVERETT. Units of attraction, 6. Referred to, 363, note. Numerical values of
magnetic force, intensity and permeability in soft iron, 473.
FARADAY. Magnetic induction, 342, note. Dielectrics, 465, note.
FERRERS. Attraction of a stratum, 94, note. Heterogeneous ellipsoids and
ellipsoidal shells, 246. Spherical Harmonics, 248, note. Electricity on
spherical bowl, 451, note. Expression for JPzPmPndp, limits ±1. Note G,
page 361.
INDEX TO ATTRACTIONS. 369
The numbers refer to the articles.
FILAMENT. See LINES OF FORCE. Definition, 126.
FLUX. The product Fdtr, 107.
FOCALOED. Defined, 194. Potentials, 228.
FUNDY. Attraction of the tide in the Bay of Fundy, 38.
GALLOP. Electricity on a circular disc and spherical bowl, 461, note.
GAUSS. Mean potential over a sphere, 84. The theorem \Fd<r= ±4;r3f, 106. Also
deduced from Green's theorem, 153. Terrestrial magnetism, 336, &c.
GREEN. Attraction of a stratum, 142, 147. A volume integral replaced by a surface
integral, 149, and a surface integral by a line integral, Note A, page 356.
Equivalent layer, 154. Infinite and multiple-valued functions, 158, 161.
Kelvin's extensions, 163. Green's mathematical papers, 248, note. Green's
method of solving electrical problems, 156, 393. Green's theoren on a
condenser, 417. Origin of the name "potential," 39, note.
HANSTEEN. Terrestrial magnetism explained by two magnets, 335.
HAUGHTON. Problem on a fluid earth with a spheroidal crust, 313.
HOBSON. Two memoirs on attractions, 248, note.
HOMOEOID. Defined, 194. See ELLIPSOIDAL SHELL.
HOMOTHETIC SHELL. Defined by Chasles, 194.
IMAGE. Defined, 396.
INDUCTION. Magnetic force and magnetic induction defined, 342. How related,
345, see Note L, page 363. Other names, Note K, page 363. Coefficients of in-
duction, 438. Magnetic induction in a solid, 465. Boundary condition, 468,493.
Induction problems, 486. Surface integral of magnetic induction, 488, Note
M, page 364.
INVERSE PROBLEMS. Find the curved rod such that the attraction of the arc PQ at
0 (1) bisects the angle POQ, 20, (2) passes through a fixed point, 20,
(3) passes through the intersection of the tangents at P, Q, 20. Given the
potential, find the body, 164.
INVERSION. Kelvin's point inversion, 168. Geometrical properties, 172, &c. In-
version from a line, i.e. in two dimensions, 181, &c. The cylindrical trans-
formation r'=Arn, O'=n9, 184, <&c.
IVORY. Geometrical property of confocal ellipsoids, 202. Theorem on attractions,
222. Application to infinite cylinders, 235. Finite cylinders, 236.
JELLET. Potentials for different laws of force, 96.
KELVIN, LORD. Theorems on attraction, 111, note, 121. Attraction of a film,
142, note. Centrobaric bodies, 135, note. Extension of Green's theorems,
163. Method of Inversion, 168. Magnetism, 314, note. Magnetic force and
magnetic induction, 342 and Note K, page 363. Solenoids, 347. Lamellar
shells, 350. Electricity on a sphere, 403, note. Electricity on a circular
disc and spherical bowl, 451, note. On two spheres by successive images,
460, note. Induced magnetism, 465, note. Theorem on dielectrics, 483.
LACHLAN. Theorem on inversion, 179.
LAMB. Potential of an elliptic cylinder, note to 238.
LAME. Poisson's equation in orthogonal and elliptic coordinates, 109, 110.
LAMELLAR SHELLS. Defined, 350. Theory, 350 — 354. Elementary rule to find
the magnetic force at a given point, 356. Cartesian components of force,
358, see VECTOR POTENTIAL. Potential of a lamellar body, 359. Mutual
potential energy of two thin shells expressed by integrations round the
rims, 360. Force due to a thin circular lamellar shell, 362. Mutual potential
R. 8. II. 24
370 INDEX TO ATTRACTIONS.
The numbers refer to the articles.
of a thin circular shell and a small magnet on the axis, 362, Ex. 3. Other
laws of force, 362. Eelation of lamellar shells to electric currents, 361.
LAPLACE. The equation V2F=0, 95. Corresponding equation for other laws of
force, 96. Properties of functions which satisfy his equation, 133, &c.
Laplace's functions and his second equation, 284. Three fundamental
theorems on Laplace's functions, 288 — 290. Extension of Theorem I., Note H,
page 362. Expansion of the potential in a series of Laplace's functions, 283.
General expression for Yn, 286. Properties of the surface r=a(l+j3SYn),
293. Various expansions, 292. See SOLID OF BEVOLUTION. Laplace's rule to
find the potential of certain heterogeneous bodies, 297. Clairaut's theorem,
304.
LEGENDBE'S FUNCTIONS. Theorems on these functions, 264, &o. Expressions for
Pn, 269—271. Four equations, 273. Boots of Pn=0, 274, 276. The integral
lf(p)Pndp, 278—280. The integral $Pn2dp, and others, 281, 282, Note G,
page 361. Expression for Pn (p) with any axis of reference, 287. Expansion
ofp", dPJdp, d'2Pn/dp2 &o., 292. See SOLID OF BEVOLUTION. Expansion of
the potential for other laws of force, Note F, page 360. Expressions for
$PtPmPndp, limits ±1, \pKPmdp, and \Pm(d«PnldpK)dp, Note G, page 361.
LEVEL SURFACES. Definition and theorems, 45, &c. Of a rod, 49, 51. Of a
homoeoid, 205, 206. Cut at right angles, 123. Eankine's theorem, 125.
To trace level surfaces, 134.
LINES OF FOKCE. Definition, 47. Direction in which a particle tends to move, 48,
114. Of a rod, 49. Of a homoeoid, 207. Attraction varies inversely as the
area of a tube of force, 127, extension, Note A, page 356. To trace lines of
force, 134. Lines of force due to a rectilinear row of particles, 323. Also
due to a series of parallel infinite rods in one plane, 323.
MAcCuLLAGH. Potential at a distant point, 135.
MACLAUBIN. Attraction of confocal ellipsoids at an external point, 224.
MAGNET. Potential, 316, 322. Eesolution, 317. Mutual action of two small
magnets, couples, 318, forces, 320. Potential energy, 322. Lines of force
and level curves, 323. Examples of magnets acting on each other, 324.
MAGNETIC BODY. Elementary rule, 327. Applied to a rod, a sphere, an ellipsoid, a
cylinder, a lamina, 328—332. Earth's magnetism, 325, 335. Mutual potential
energy of two bodies, 334. Bodies not uniformly magnetised, 346. Induced
magnetism in spherical shells, uniform field, also a magnet inside, 487.
MAXWELL. Treatise referred to, 314, note; 363, note; 370; a nearly spherical
conductor enclosed in a nearly spherical shell, 421, Ex. 2. Electricity on a
sphere, 403, note ; orthogonal spheres, <fec., 423, note. Two spheres, 460, note.
Stress in dielectrics, 465, note.
MOUNTAINS. Attraction of, 33 — 38. Density of earth, 35. Pyramid of Egypt, 37.
MUBPHY. Electricity on two spheres by successive influence, 460, note. Two
spheres in contact, 464.
PERMEABILITY, MAGNETIC. Defined, 469. Boundary condition in induced magnetism,
469. Eelation to specific inductive capacity, 473.
PLANA. Attraction of a circular ring, 191.
PLANE CONDUCTOBS. Acted on by a point-charge, 412. Quantity of electricity on a
portion of the plane, 413. Two planes, 414. Acted on by a perpendicular
and also a parallel rod, 416. A circular disc acted on by a point-charge (1) in
its plane, 451, &c., (2) on the axis, 454.
INDEX TO ATTRACTIONS. 371
The numbers refer to the articles.
PLATTFAIR. Cylinder of greatest attraction, 24. Attraction of a lamina, 27. Solid
of greatest attraction, 31. Bectangular disc, 30. Density of Schehallien, 35.
POISSON. The theorem V2F=-47rp, 105. Deduced from Gauss' theorem, 108.
Polar, cylindrical and oblique Cartesian coordinates, 108. Elliptic coor-
dinates, 110. Mean potential through the volume of a sphere, 84. Potential
at a distant point, 135, note. Attraction of a film, 142, note. Level surfaces
of a homoeoid, 205. Ivory's theorem, 222. Heterogeneous ellipsoids, 248.
Attraction of ellipsoids, 248, note. Magnetism, 314, note. Eepresents
magnetism by an equivalent solid and superficial distribution, 339. These
expressed in various kinds of coordinates, 340. Electricity on a sphere acted
on by a point-charge, 403, note. Electricity on two separate spheres, 459,
note; 464, Ex. 4. Induced magnetism, 465, note. Poisson's conditions
for dielectrics, 481.
POTENTIAL. Origin of the name, 39, note. Geometrical definition (1) at an external
point, 39, (2) at an internal point, 101, &c. Definition derived from work, 44.
Kesolved force, 41. Other laws of force, 43. Potential of a rod, 49, 50.
Discs and cylinders at points on their axes, 53. Infinite cylinders at any
point, 55, 56. Heterogeneous infinite cylinders, 58. Mutual potential of two
systems, 59. Gauss and Poisson on the mean potential of a body, 84.
Belation between the potential of the same body for different laws of force,
96 — 98. Potential cannot be an absolute minimum, &c., 111. Consideration
of an internal point, 113. Various theorems, 115, 116. At a distant point,
135. MacCullagh's theorem, 135. Centrobaric bodies, 137. Potential
constant in a cavity, 99, 139. Continuity at the surface, 103, 146. Given
the potential, find the body, 164, &c. Potential given over two concentric
spheres, find it generally, 299. Poisson's general expression for the potential
of a magnetic body, 339. See SOLENOIDS, LAMELLAE SHELLS, CONDUCTORS.
Potential energy of an electric system, 495.
BANKINE. The angles of intersection of the sheets of a level surface, 125.
KECTILINEAB FIGURES. See Discs. Potential of a lamina found in terms of
potentials of the sides, 257. Potential of a solid in terms of those of the
faces, 258. Potentials of all rectilinear figures can be found in finite terms,
259. Solid angle subtended by a triangle at any point, 262.
RINGS, CIRCULAR AND ANCHOR. Polar line of P divides a uniform ring into parts
equipotential at P, 73. Potential at any point for the law of the inverse
distance, 55, 66. General method of inversion for any law of force by using
an ellipse, 186. Several formula for the potential of a ring, 190, also Note
B, page 358. Theorems of Plana and Poincar^, 191. Anchor rings, 192.
EGBERTS' THEOREM. Potential of a lamina for different laws of force, 98.
ROD. Components of attraction, 10, &c. Infinite rod and the attraction of
cylinders, 14. Singular form, 15. Other laws of attraction, 16. Various
problems, 16. Condition that two curvilinear rods equally attract the origin,
17, &c. Inverse rods, 20. Potential, 49, 50. A cylinder of rods, 52.
Magnetic rod, 328. Electrified rod, 385.
RODRIGUES. Potential of a homogeneous ellipsoid, 225, note. Legendre's
functions, 271.
SATURN. Figure of Saturn, acted on by the ring, 310. Measurements by Herschel
and Bessel, 310.
SCREENS. Electrical, 390.
372 INDEX TO ATTRACTIONS.
The numbers refer to the articles.
SIEMENS. Instrument to find the depth of the sea under a ship, 23.
SIMILAB BODIES. Their attractions and potentials compared, 94.
SOLENOIDS. Defined, 347. Potential, 348. Condition that magnetism is sole-
noidal, 349.
SOLID ANGLES. How measured, 26. Solid angle subtended by a triangle at any
point, 262. Of a cone, 264. Normal attraction of a disc, 27.
SOLID OF GREATEST ATTRACTION. Playfair's theorem, 31.
SOLID OF EEVOLUTION, Expression for the potential (1) in Legendre's functions,
300, (2) in a definite integral, 302. Potentials of a thin ring, solid anchor
ring, oblate spheroid, &c., in Legendre's functions, 303.
SPHERES. Potential of uniform shell, 64. Annulus, 74. Theorem of Cavendish,
73. Discussion -of a discontinuity, 75. Attraction of a segment at the centre
of the base, 76. Solid sphere, 78, 79. Attraction of a shell on an element of
itself, 79. Mutual pressure of two parts of a shell, 79. Potential of a shell
near the rim, 79. Attraction for other laws of force, 80. Eccentric shells,
81. Heterogeneous shells, V = Va\r', 86. Stokes' theorem, X+X'— - V/a,
87. Find the law of force that the attraction may be the same as that of a
single particle, 89. Laplace's expressions for the potential of a thin shell
with any law of density, 294. Also solid sphere, 296. Nearly spherical
bodies, 297, 420. Potential of a magnetic sphere, 329.
SPHERICAL CONDUCTORS. Two spheres joined by a wire, 374. Concentric spheres,
392. Single sphere acted on by a point-charge, 397. Diagram, page 365.
Lines of force and level surfaces, 405. Sphere in uniform field, 406. Sphere
surrounded by a ring, 406, 422. Quantity on a segment and potential, 406.
Nearly spherical bodies, 420, 421. Two orthogonal spheres, 423. Acted on
by a point-charge, 426. Geometrical properties, 431. Spheres intersecting
at an angle ir/n, 433. Three orthogonal spheres, 436. Spherical bowl, 456,
&c. Two separate spheres, 459 — 463. Examples, 464.
SPHERICAL HARMONICS. Defined, 267. Zonal harmonics, 267. Tesseral surface
harmonics, 286. Sectorial, <fec., 286.
SPHEROIDS. Potential at an internal point, 219. Nearly spherical, first and second
approximations, 220, 221. Potential of an oblate spheroid found in Legendre's
functions, 303.
STOKES. Attraction of a spherical shell, 87. Theorems on potentials, 116, 117,
111, note. Potential of a body in general, 283, note. Generalisation of
Clairaut's theorem, 304, note.
STRATUM. See Discs. Green's theorem Z'-Z=47rm, 142, 147. Attraction of a
stratum on an element of itself X' + X=2F, 142. Green's equivalent stratum,
154. Linear and quadratic layers on an ellipsoid, 231.
SURFACE. Condition that two surfaces equally attract the origin, 28. Surface of
equilibrium defined, 46. Surface integral, Note A, page 356. See INDUCTION.
SUSCEPTIBILITY, MAGNETIC. Defined, 465. Relation to permeability, 469.
TABLE-LAND. Bouguer's rule to find the attraction, 23. Other authors, 23.
TERRESTRIAL MAGNETISM. Gauss' investigation, 335, &c. Dip, 324, Ex. 2. Horizontal
force, 325. Biot and Hansteen, 335.
TETRAHEDRON. Potential in quadriplanar coordinates found in terms of the
potentials of the faces, 262. Potentials of the triangular faces, 257,
262, Ex. 2.
THOMSON, J. J. Magnetism, 314, note. Law of magnetic attraction, 322. Referred
INDEX TO ATTRACTIONS. 373
The numbers refer to the articles.
to, 363, note. Discussion of some results of Kirchhoff, 460, note. Diagram of
permeability referred to, 473.
TRIANGLE. Attraction of the sides, 16. Position of equilibrium when attracted by
the sides produced indefinitely, 16. A particle at the centre of inscribed
circle is in unstable equilibrium, 121, Ex. 5.
TUBES OF FORCE. Defined, 126. See LINES OF FOECE.
UNITS. Theoretical and astronomical, 2. o.o.s. system, 3. Foot, pound and
second system, 4. Dimensions of K and m, 5.
VECTOR POTENTIAL. Induction through a surface equal to an integral round the
rim, 358, Note M, page 364.
WORK. Relation to potential, 44. Mutual work, 59, &c., 439, &c. Potential
energy of conductors, with examples, 450. Potential energy of an electric
system, 495.
INDEX TO THE BENDING OF RODS.
/EOLOTROPIC. Defined, 6.
AIRY'S PROBLEM. How a standard of length should be supported, 36.
BALL, SIR R. Notice of an error made by Poisson, 31, note.
BENT BOW. Its equation and the tension, 27 — 30.
BENT ROD. Two methods of forming the equations of equilibrium, 10, 11. The
experimental law, 13. A heavy rod rests on n supports with weights, find
the stresses, 21, 22. Inequality of pressures, 22, &c. Altitudes of the supports
to equalize the pressures, 25. Problems on heavy rods, 24, &c.
BRITANNIA BRIDGE. Problems on the bridge, 23. How the inequality of pressure
was diminished, 23.
CENTRAL AXIS. Defined, 1.
CIRCULAR RODS. Equations to find the deformation, 37. Extensible circular rods,
38. Expressions for the tension, bending moment and work, 40 — 42.
Limiting case when the rod is inextensible, 42.
CLAPEYHON. The equation of the three moments, 19, note.
COLUMNS. Theory of their flexure and Euler's laws, 31. Hodgkinson's experi-
mental researches, 31. Greenhill's problems, 32.
CONTRACTION. Found by theory (1) for a stretched rod, 7, (2) for a circular rod, 34.
DEFLECTION. A heavy rod rests on n supports, find the deflection, 17.
EQUATIONS OF EQUILIBRIUM. In two dimensions, 10, 11. In three dimensions, 57.
EULER. His laws on columns &c., 31.
EXPERIMENT. Hooke's law, 5. Bending of a rod, 13.
FINDLAT. Euler's laws, 31, note.
FLEXURAL RIGIDITY. Defined, 13. Its magnitude found by experiment, 22 ; variable,
26. Principal flexural rigidity, 53.
HELICAL TWISTED RODS. A straight rod is bent into a helix, 54.
HEPPEL. History of the equation of the three moments, 19, note.
HOOKE'S LAW. Enunciation, 5. Corresponding contraction, 8, 34.
ISOTROPIC. Defined, 6.
KELVIN. See THOMSON AND TAIT. Resilience, 16.
KIHCHHOFF. His analogy, 59.
374 INDEX TO THE BENDING OF RODS.
The numbers refer to the articles.
LAGRANGE. His memoir referred to, 31, note.
LAME:. His constants X, /*, of elasticity, 7.
LIMITS OF ELASTICITY. Defined, 14.
LOVE. His treatise on elasticity, 8. Stability of columns, 31, note.
MINCHIN. Equations in three dimensions, 57.
MOSELEY. Problem on a rod compared with experiment, 24.
NEUTRAL LINE. Defined, 33. If a straight rod is bent without tension the central
line is neutral, 33.
PEABSON. Yielding of the supports, 19, note. Correctness of the Bernoulli-Eulerian
theory, 36, note.
POISSOK. His treatise referred to, 31, note.
RESILIENCE. Defined, 16.
RESOLVED CUBVATUBE. Theory, 46. Represented by two constants, K, X, 47.
ROTATING BOD. Problem, 24, Ex. 5.
SAINT- VENANT. His results on bending, 35. Spiral spring, 56.
SPIBAL SPBTNGS. Theory, 55.
STANDABDS OF LENGTH. How supported, 36.
STBESS AND STBAIN. Their relations to each other in three dimensions, 51, 52.
Principal axes of stress, 63.
THEOBY. Stretching of rods, 7, 8. Bending of circular rods, 33.
THOMSON AND TAIT. Equations in three dimensions, 57. Stress and strain, 51.
Spiral spring, 56. Figures of bending, 31, note.
THBEE MOMENTS. Equation or theorem, 19. Extension of the theorem, 20.
Method of use, 21. Corresponding theorem for three pressures, 24.
TWIST. In three dimensions defined and measured, 45. Twists of a rod placed
on a sphere, a cylinder, any surface, 50. Principal torsion rigidity, 53.
VECTOBS. Rule for resolving a variable vector at a moving point, 57, note.
VEBY FLEXIBLE BOD. Theory of a light rod passing through several small rings
not in a straight line, 24, Ex. 6, 7, 43.
WEBB. Equation of three moments with variable flexural rigidity, 19, note ; 26.
WOBK. The work of bending (1) a straight rod, 16, (2) a circular rod, 40. Re-
lations of stress to strain in three dimensions deduced from work, 52.
YIELDING OF THE SUPPORTS. Rod supported by columns, 21.
INDEX TO ASTATICS.
CENTBAL ELLIPSOID. Definition and equation, 14; its discriminant, 17. Com-
parison of ellipsoids at different bases, 23. Locus of bases at which it is a
surface of revolution, 37, Ex. 9. Other central ellipsoids, 22.
^CENTRAL PLANE. Definition and equation, 25. It is fixed in the body, 26. Equa-
tion expressed in terms of the points of applications and mutual inclinations
of the forces, 31.
Working rule to find the central plane, 58.
CENTRAL POINT. Definition and coordinates, 28. It lies in the central plane, 29.
Not the same point as in two dimensions, 29.
Working rule to find the central point, 30, 66.
INDEX TO ASTATICS. 375
The numbers refer to the articles.
CONE or EQUAL ASTATIC MOMENT. Its axes coincide with those of the central
ellipsoid, 17.
CONFOCALS. The principal astatic axes are the normals, 34. The confocal conies,
36. Relation to Poinsot's axis, 51.
COUPLES. Conditions of equilibrium, 3, 8. Astatic angle, 2. Working rule to
find resultant couple, 6.
DABBOUX. His memoir, 1. His ellipsoids, 15, 22. Theorems, 22, 37.
ELEMENTS. The twelve elements, 10. Interpretation (1) in rows, 10, (2) in
columns, 59.
EQUILIBRIUM." Equilibrium of three couples, 3, 8. General conditions of astatic
equilibrium, 11, 12. If three forces balance three others, the six points of
application lie in a plane, 13.
FOCAL LINES. Definition, 36. Distance from the centre, 37. Four focal lines can
be drawn (1) through every point, 37, (2) parallel to a given line, 37. Locus
of focal lines which pass through a point on a focal conic, 37. Minding's
theorem, 44.
FOUE FORCES. Reduction of a system to four forces, 60. Working rule, 60. Trans-
formation of tetrahedra, 70. Intersections of the central plane with the
edges, 71.
IMAGINARY CONIC. Defined, 36, 64. Its centre, 67, &c.
INITIAL POSITION. Definition, 19, 20. These are the only positions of equilibrium
with the base fixed, 21.
INVARIANTS. Of astatic moments, 17; of astatic triangles, 65; of two forces, 79.
LARMOB. Proof of Minding's theorem referred to, 1, note.
MINCHIN. Quaternions, 1, note.
MINDING'S THEOREM. Proof, 44. Further consideration, 48. Minding's memoir, 1.
MOEBIUS. First studied Asiatics, 1, note.
MOIGNO. His treatise, 1, note.
MOMENTS OF INERTIA. The analogy to astatic moments for all arms, 33.
POINSOT'S AXIS. Its position in space as the body turns round R and locus in the
body, 38, 40. Its equation referred (1) to the axes of the forces, 39 ; (2) to
axes in the body, 44, 45. Three elliptic cylinders, 46. Case in which the
principal force acts along an asymptote of a focal conic, 44, 50. Its relation
(1) to confocal surfaces, 51; (2) to the focal conies, 54. To place the body so
that a given straight line may be (when possible) a Poinsot's axis, 44.
POINSOT'S COUPLE MOMENT. Its magnitude referred (1) to the axes of the forces,
39 ; (2) to axes in the body, 41, 45. Found by a quadratic, 47. Axis of no
moment, 41, 44, 50.
POINTS ASTATIC. Definition, 67.
PRINCIPAL ASTATIC AXES. Principal couples, 18. Principal axes at various points, 37.
PRINCIPAL FORCE. Definition, 9.
REDUCTION OF A SYSTEM OF FORCES. To three couples and a principal force, 9.
To three rectangular couples and a force, 18. To two rectangular couples
and a force, 27. To two rectangular couples, with forces perpendicular to
the principal force, 29. Summary, 32. Reduction to four forces. 60;
three forces, 56 ; two forces, 73 ; one force, 79.
SINGLE RESULTANT. To place a body so that the forces are equivalent to a single
resultant at a given point, 19, 37, 40, 44, &o.
SOMOFF. His treatise referred to, 1, note.
376 INDEX TO ASTATICS.
The numbers refer to the articles.
THREE FORCES. Kednction to three possible, 66. Working rule, 60. Astatic points
lie on the central plane, 13, 57. Position of the central plane of three
forces, 68. Transformation of astatic triangles, 63. The imaginary conic,
64. The invariants, 65. The central point, 66; the central point is the
centre of the conic, 67. Determination of the central lines and principal
moments, 68. A reduction to fewer forces than three not generally possible,
72.
TRIANGLE ASTATIC. Definition, 67.
Two FORCES. Conditions that a reduction to two forces is possible, 73. The forces
are parallel to a fixed plane, 75. Invariants of two forces, 78.
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