225. The pull on the dielectric is that due to the tensions of the lines of force which cross its boundary. In air these lines of force are the same as if we had charges e, e' at P, P' entirely in air, so that the whole tension in the direction PJP of the lines of force in air is

ee'

pp'2'

(K-l)

4a2(/iT + iy

This system of tensions shews itself as an attraction between the dielectric and the point charge. If the dielectric is free to move and the point charge fixed, the dielectric will be drawn towards the point charge by this force, and conversely if the dielectric is fixed the point charge will be attracted towards the dielectric by this force.

202 Methods for the Solution of Special Problems [oh. vm

Inversion.

226. The geometrical method of inversion may sometimes be used to deduce the solution of one problem from that of another problem of which the solution is already known.

Geometrical Theory.

227. Let 0 be any point which we shall call the centre of inversion, and

Fig. 67.

let AB be a sphere drawn about 0 with a radius K which we shall call the radius of inversion.

Corresponding to any point P we can find a second point P', the inverse to P in the sphere. These two points are on the same radius at distances from 0 such that OP . OP' = K\

As P describes any surface PQ ..., P' will describe some other surface PQ'..., each point Q' on the second surface being the inverse of some point Q on the original surface. This second surface is said to be the inverse of the original surface, and the process of deducing the second surface from the first is described as inverting the first surface.

It is clear that if P'Q'... is the inverse of PQ..., then the inverse of P'Q'..- will bePQ....

If the polar equation of a surface referred to the centre of inversion as origin be / (r, 8, <p) = 0, then the equation of its inverse will be

f[ — , 0, <£J=0. For the polar equation of the inverse surface is by

definition / (r, 0, </>) = 0, where rr' = K- for all values of 6 and </>.

226, 227]

Inversion

203

Inverse of a sphere. Let chords PP', QQ', ... of a sphere meet in 0 (fig. 68). Then

0P.0P' = 0Q.0Q'=... = t\

where t is the length of the tangent from 0 to the sphere. Thus, if t is the radius of inversion, the surface PQ... is the inverse of P'Q'..., i.e. the sphere

Fig. 68.

is its own inverse. With some other radius of inversion K, let P"Q".. the inverse of PQ .... then

0P.0P"=0Q.0Q'=... = K\

OP" OQ" IP

• ••• - ^

be

so that

Thus the inverse of a

OP' OQ' "

and the locus of P", Q", ... is seen to be a sphere sphere is always another sphere.

A special investigation is needed when the sphere passes through 0. Let OS be the diameter through 0, and let 8' be the point inverse to S. Then, if P' is the inverse of any point P on the circle,

0P.0P' = 0S.0S\ OP _ OS' or 0S~0P"

so that POS, S'OP' are similar triangles. Since OPS is a right angle, it follows that OS'P' is a right angle, so that the locus of P' is a plane through S' perpen- dicular to OS'. Thus the inverse of a sphere which passes through the centre of inversion is a plane, and, conversely, the inverse of any plane is a sphere which passes through the centre of inversion.

Fig. 69.

204 Methods for the Solution of Special Problems [ch. viii

228. If P, Q are adjacent points on a surface, and P', Q' are the corre- sponding points on its inverse, then OPQ, OQ'P' are similar triangles, so that PQ, P'Q' make equal angles with OPP'. By making PQ coincide, we find that the tangent plane at P to the surface PQ and the tangent plane at P' to the sur- face P'Q' make equal angles with OPP'. Hence, if we invert two surfaces which intersect in P, we find that the angle

between the two inverse ' surfaces at P' is equal to the angle between the original surfaces at P, i.e. an angle of intersection is not altered by inversion.

Also, if a small cone through 0 cuts off areas dS, dS' from the surface PQ... and its inverse P'Q'..., it follows that

d# OP' dS'- OP'*'

Fig. 70.

Electrical Applications.

229. Let PP', QQ' be two pairs of inverse points (fig. 70). Let a charge e at Q produce potential Vp at P, and let a charge e at Q' produce potential Vp at P', so that

VP' =

then

Take

then

P~PQ* ~ P'Q'

Il-i ?Q -i op

VP e-P'Q'~ e' OQ"

eOQ K ' Vp OP K VP

K ~ OF'

Now let Q be a point of a conducting surface, and replace e by crdS, the charge on the element of surface dS at Q. Let Vp denote the potential of the whole surface at P, and let Vp denote the potential at P' due to a charge e' on each element dS ' of the inverse surface, such that

e' OQ'

adS K

K

Then, since Vp = Vp -^p, for each element of charge, we have by addition

VP'= Vv

K

Thus charges e' on dS', etc. produce a potential

VPK

OP'

at P'.

228-230] Inversion 205

Now suppose that P is a point on the conducting surface Q, so that VP becomes simply the potential of this surface, say V. The charges e on dS', etc. now produce a potential

Qpi at jt ,

so that if with these charges we combine a charge — VK at 0, the potential produced at P' is zero. Thus the given system of charges spread over the surface P'Q' ..., together with a charge — VK at the origin, make the surface P'Q' ... an equipotential of potential zero. In other words, from a knowledge of the distribution which raises PQ... to potential V, we can find the distribution on the inverse surface P'Q' . . . when it is put to earth under the influence of a charge — VK at the centre of inversion.

If e, e' are the charges on corresponding elements dS, dS' at Q, Q', we have seen that

e' a'dS' K OQ' /OQ'

"~ 0Q~ K "V i

e adS 0Q~ K ~V OQ' dS' OQ'" whlle dS = W

„ a (0Q'-% K3 „ocl,

Hence 7-(w) =W> (132)'

giving the ratio of the surface densities on the two conductors.

Conversely, if we know the distribution induced on a conductor PQ ... at potential zero by a unit charge at a point 0, then by inversion about 0 we obtain the distribution on the inverse conductor P'Q'... when raised to

potential -^.. As before, the ratio of the densities is given by equation (132).

Examples of Inversion.

230. Sphere. The simplest electrical problem of which we know the solution is that of a sphere raised to a given potential. Let us examine what this solution becomes on inversion.

If we invert with respect to a point P outside the sphere, we obtain the distribution on another sphere when put to earth under the influence of a point charge P. This distribution has already been obtained in § 214 by the method of images. The result there obtained, that the surface-density varies inversely as the cube of the distance from P, can now be seen at once from equation (132).

So also, if P is inside the sphere, we obtain the distribution on an uninsulated sphere produced by a point charge inside it, a result which can again be obtained by the method of images.

When P is on the sphere, we obtain the distribution on an uninsulated plane, already obtained in § 208.

206 Methods for the Solution of Special Problems [ch. vm

231. Intersecting  Planes.     As  a  more  complicated  example  of  inversion, 
     

let us invert the results obtained in § 212. We there shewed how to find

Fig. 71.

7T

the distribution on two planes cutting at an angle — , when put to earth

lb

under the influence of a point charge anywhere in the acute angle between them. If we invert the solution we obtain the distribution on two spheres, cutting at an angle nr\n, raised to a given potential. By a suitable choice of the radius and origin of inversion, we can give any radii we like to the two spheres.

If we take the radius of one to be infinite, we get the distribution on a plane with an excrescence in the form of a piece of a sphere : in the par- ticular case of n = 2, this excrescence is hemispherical, and we obtain the distribution of electricity on a plane face with a hemispherical boss. This can, however, be obtained more directly by the method of § 219.

Spherical Harmonics.

232. The problem of finding the solution of any electrostatic problem is equivalent to that of finding a solution of Laplace's equation

throughout the space not occupied by conductors, such as shall satisfy certain conditions at the boundaries of this space — i.e. at infinity and on the surfaces of conductors. The theory of spherical harmonics attempts to provide a general solution of the equation V2F = 0.

This is no convenient general solution in finite terms : we therefore examine solutions expressed as an infinite series. If each term of such a series is a solution of the equation, the sum of the series is necessarily a solution.

231-233] Spherical Harmonics 207

233. Let  us  take  spherical  polar  coordinates  r,  6,  <f>,  and  search  for 
     

solutions of the form

V = RS,

where R is a function of r only, and S is a function of 6 and $ only.

Laplace's equation, expressed in spherical polars, can be obtained analyti- cally from the equation

d2V d-v a2r dx* + df + d? ~ °

by changing variables from x, y, z to r, 6, <f>, but is most easily obtained by applying Gauss' Theorem to the small element of volume bounded by the spheres r and r + dr, the cones 6 and 6 + d6, and the diametral planes (j> and <f> + dj>. The equation is found to be

r» dr \ dr) + r* sin 6 dd V™ d0J+ r2 sin2 0 d(f>* " ' Substituting the value F = RS, we obtain

^/2^\ R d_( . „dS\ E c^S r2 dr V 3r J r2 sin 6 dd &m dd)+r* sin2 6 d<fr " '

or, simplifying,

i 9 /,as\ l a / . .as\ , ji a^_

JR 3r V dr) + 8 sin 6 d0 [8m dd)+S sin2 6 d<f>*

The first term is a function of r only, while the last two terms are inde- pendent of r. Thus the equation can only be satisfied by taking

1 d ( 3>R>

R dr X

*Tr)=K <133>'

1 3 / . adS\ , 1 82# „ /1Q.,

where K is a constant. Equation (133), regarded as a differential equation for R, can be solved, the solution being

&-*** + £» (135),

where A, B are arbitrary constants, and n (n + 1) = K. After simplification equation (134) becomes

^Hl)+S5n>5+"<"+1>s=° <186>

Any solution of this equation will be denoted by Sn, the solution being a function of n as well as of 6 and <f>. The solution of Laplace's equation we have obtained is now

V = RS = (Ar» + J^Sn,

and by the addition of such solutions, the most general solution of Laplace's equation may be reached.

208 Methods for the Solution of Special Problems [oh. Tin

234. Definitions. Any solution of Laplace's equation is said to be a spherical harmonic.

A solution which is homogeneous in x, y, z of dimensions n is said to be a spherical harmonic of degree n.

A spherical harmonic of degree n must be of the form rn multiplied by a function of 6 and $>, it must therefore be of the form ArnSn, where Sn is a solution of equation (136).

Any solution 8n of equation (136) is said to be a surface-harmonic of degree n.

235. Theorem. If V is any spherical harmonic of degree n, then yjrm+i fa a spherical harmonic of degree — (n + 1).

For V must be of the form ArnSn, so that

V ASn

rzn+i rn+i

which is known to be a solution of Laplace's equation, and is of dimensions — (n + 1) in r. Conversely if V is a spherical harmonic of degree — (n + 1), then r2n+1 V is a spherical harmonic of degree n.

236. Theorem. If V is any spherical harmonic of degree n, then

fis+t+uy

dafdyW where s, t, and u are any integers, is a spherical harmonic of degree n — s — t — u.

dv dn-v dv A For a?+5E + ^"a'

so that on differentiation s times with respect to x, t times with respect to y, and u times with respect to z,

gs+t+u+2"|7 fls+t+u+2y fis+t+u+^y

daf+2dytdzu + dx*dyt+*dzu + dxsdytdzu+2 = '

°r V' [dtfdyw) = °'

which proves the theorem.

237. Theorem. If Sm,Sn are two surface harmonics of different degrees m, n, then

\ \SnSmda) = 0,

where the integration is over the surface of a unit sphere. In Green's Theorem (§ 181),

(<£V2¥- - ¥V«<D) dxdydz = - \(<P ^ - ¥ ^) dS,

dn dn J

put <£ = rnSn> ^ = rmSm, and take the surface to be the unit sphere.

234-239] Spherical Harmonics 209

Then V23> = 0, Va¥ = 0, ^-=-5- = - nrn-1#n, and ^- = - mrm-1/Sm.

on dr 9n

Thus the volume integral vanishes, and the equation becomes

[j(nrm+n-ignSm _ mrm+n-i^n/Sfm) da) = 0,

or, since n is,, by hypothesis, not equal to m,

onomd(o = 0.

Harmonics of Integral Degree.

238. The  following  table  of  examples  of  harmonics  of  integral  degrees  7i=0,  —1,  -2, 
     

• 1, is taken from Thomson and Tait's Natural Philosophy.

_ , . ?/ , r+3 .», r + 2 rz(x2-y2) 2rxyz

n-0. 1, tan-1^, log , tan "^ log , , \ , *./, . ., , %,,.

#' 0r-2 a; ° r— z (x2+y2)2 (V+y2)2

Also if V0 is any one of these harmonics, --^, -^-^, --^ are harmonics of degree — 1, so

that r -tt-^ , r-yr-^, r-^~ are harmonics of degree zero. As examples of harmonics derived

ox dy oz

in this way may be given

rx ry zx zy x x

x2~+y2> X2 + 1J2' X2 + 1J2' X2+y2' T + l' T^z'

By differentiating any harmonic V0 any number * of times, multiplying by r2,_1 and differentiating again s - 1 times, we obtain more harmonics of degree zero.

n= — 1. Any harmonic of degree zero divided by r or differentiated with respect to

x, y or z, e.g.

1 1 , .y 1 . r+z x x

■ , - tan 1 - , - log

r' r x' r ° r — z' x2+y2' r(r+z)'

n= - 2. By differentiating harmonics of degree — 1 with respect to x, y or z we obtain harmonics of degree — 2, e.g.

x y z z , . y z , r+z -q> H-> ~%, -, tan-1'2-, -log .

11 = 1. Multiplying harmonics of degree —2 by r3, we obtain harmonics of degree 1, e.g

. y , r--z n

x, y, z, a tan-1-, slog- -— 2?\

' ^ ' a? r — z

Rational Integral Harmonics.

239. An important class of harmonic consists of rational integral algebraic functions of x, y, z. In the most general homogeneous function of x, y, z of degree n there are \ (n + 1) (n + 2) coefficients. If we operate with V2 we are left with a homogeneous function of x, y, z of degree n — 2, and therefore possessing \n (n — 1) coefficients. For the original function to be a spherical harmonic, these %n(n— 1) coefficients must all vanish, so that we must have ^n(n — 1) relations between the original ^(w + l)(?i + 2) coefficients. j 14

210 Methods jo r the Solution of Special Problems [ch. viii

Thus the number of coefficients which may be regarded as independent in the original function, subject to the condition of its being a harmonic, is

±(n + I)(n + 2)-4in(n-l), or 2n + 1. This, then, is the number of independent rational harmonics of degree n.

For instance, when n = 1 the most general harmonic is

Ax + By + Cz,

possessing three independent arbitrary constants, and so representing three independent harmonics which may conveniently be taken to be x, y and z.

When n = 2, the most general harmonic is

ax2 + by2 + cz2 + dyz + ezx -Yfxy,

where a, b, c are subject to a + b +c = 0. The five independent harmonics may conveniently be taken to be

yz, zx, xy, x2 — y2, x2 — z2.

When n = 0, 2n + 1 = 1. Thus there is only one harmonic of degree zero, and this may be taken to be V— 1.

Corresponding to a rational integral harmonic Vn of positive degree n,

y there is the harmonic -—^ of degree — (n + 1). These harmonics of degree

— (n + 1) are accordingly 2/i + 1 in number. Thus the only harmonic of

this kind and of degree — 1 is

Consider now the various expressions of the type

gs+t+u /J>

.(137),

da? dyf dzu \r where s + t + u = n.

These, as we know, are harmonics of degree — (n + 1), and from § 235

y it is obvious that they must be of the form ~^i , where Vn is a rational

integral harmonic of degree n. Since - is harmonic, V2 ( - J = 0, so that

d2 [l\ fd2 d2\fl\ /loox

The most general harmonic obtained by combining the harmonics of type (137) is

2^u3^a^(r) (139)'

but by equation (138) this can be reduced at once to the form

dz pq da? By* \rj p q dxdy \r) '

239, 240] Spherical Harmonics 211

where p + q = n — 1 and p + q = n. This again may be replaced by

dz „=0 pdxPdyf^1-P\rJ PZ0 p dxP oyn~v \r J '

so that there are 2w + 1 arbitrary constants in all, and it is obvious

on examination that the harmonics, multiplied by all the coefficients

Bp, ... Bp', ... are independent. Thus, by differentiating - n times, we have

arrived at 2n + 1 independent rational integral harmonics, and it is known that this is as many as there are.

Expansion in Rational Integral Harmonics.

240. Theorem*. The value of any finite single-valued function of position on a spherical surface can he expressed, at every point of the surface at which the function is continuous, as a series of rational integral harmonics, provided the function has only a finite number of lines and points of discontinuity and of maxima and minima on the surface.

Let F be the arbitrary function of position on the sphere, and let the sphere be supposed of radius a. Let P be any point outside the sphere at a distance / from its centre 0, and let Q be any point on the surface of the sphere.

p

iiG. 72.

Let PQ be equal to R, so that

R* =f* + a2 - 2a/ cos POQ.

We have the identity

f2-a* [fdS _a

.(140),

4>wa JJ R3 f ""

where the integration is taken over the surface of the sphere, a result which it is easy to prove by integration.

A point charge e placed at P induces surface density -- — „3 on the surface of

the sphere (§ 214), and the total induced charge is -~i- The identity is therefore

obvious from electrostatic principles.

• The proof of this theorem is stated in the form which seems best suited to the requirements of the student of electricity and makes no pretence at absolute mathematical rigour.

14—2

212 Methods for the Solution of Special Problems [ch. vm

Now introduce a quantity u denned by

f*-a? CfFdS

u='

a- 4<ira

[[FdS

.(141),

so that u is a function of the position of P. If P is very close to the sphere, /2 — a2 is small, and the important contributions to the integral arise from those terms for which R is very small : i.e. from elements near to P.

If the value of F does not change abruptly near to the point P, or oscillate with infinite frequency, we can suppose that as P approaches the sphere, all elements on the sphere from which the contribution to the integral (141) are of importance, have the same F. This value of F will of course be the value at the point at which P ultimately touches the sphere, say Fp. Thus in the limit we have

(/2 - a2) FP reds

4>ira J R*

u =

.(142),

a

= Fp-f , by equation (140),

= FP,

when in the limit / becomes equal to a.

If the value of F oscillates with infinite frequency near to the point P, we obviously may not take F outside the sign of integration in passing from equation (141) to equation (142).

If the value of F is discontinuous at the point P of the sphere with which P ultimately coincides, we again cannot take F outside the sign of integration. Suppose, however, that we take coordinates p, 3 to express the position of a point P' on the surface of the sphere very near to P, the coordinate p being the distance PP", and 3 being the angle which PP' makes with any line through P in the tangent plane at P. Then F may be regarded as a function of p, 3, and the fact that F is discontinuous at P is expressed by saying that as we approach the limit p = 0, the limiting value of F (assuming such a limit to exist) is a function of 3 — i.e. depends on the path by which P is approached. Let F (3) denote this limit. Then

u--

_/2-aa f F(3)Pdpd3

Ana

i

M

4-rra

Aira

Ztt]

F(3)

r/2-

/

rli

dS~

■13

d3

1^

2tt

F(3) ( -.) d3, by equation (140).

On passing to the limit and putting a—f, we find that

u=±fF(S)43 ....

•(143),

240]

Spherical Harmonics

213

i.e. u is the average value of F taken on a small circle of infinitesimal radius surrounding

O

P. In particular, if F changes abruptly on crossing a certain line through Py having a value Fi on one side, and a value F2 on the other, then the limiting value of u is

u = $(Fl + F2).

If we take 0 to denote the angle POQ, -^■=(/2-2a/cos0+a2)-^ 1/ a2-2afcos6-h

1

7L

, a2 - 2a/ cos 6 /a2 - 2a/ cos 0\3

•1- o »z r -g I ~ I —

f"

P

or, arranging in descending powers of/

.(144),

in which i?, P±, R, ... are functions of 6, being obviously rational integral functions of cos 6. When 6 = 0,

and when 0 = ir,

so that when 6 = 0, and when # = ir,

••• J i

1 /_ a a

p = p— — l

— P— P= — P=- — 1

It is clear, therefore, that the series (144) is convergent for 0 = 0 and 6 = 7r, and a consideration of the geometrical interpretation of this series will shew that it must be convergent for all intermediate values*.

Differentiating equation (144) with respect to / we get

1

d

R

a cos 6 —f

~ R3 df

a

a2

Wys-Z%Ti-

(145).

If we multiply this equation by 2/ and add corresponding sides to equation (144), we obtain

F Multiplying this equation by — -r — , and integrating over the surface of the

sphere, we obtain

p-a? [[FdS _™2n + l

47ra

R3 o 4tt

FR

a1

f

n+l

dS,

• Being a power series in cos 6 it can only have a single radius of convergence, and this cannot be between cos 0 = 1 and cos0=-l.

214 Methods for the Solution of Special Problems [ch. vin

or, by equation (141),

*=i^!<2"+i>/M7Hrf&

If the function F is continuous and non-oscillatory at the point P, then on passing to the limit and putting f =a, we obtain

^i{2n + l)ffFPndS (146).

0 *' J

4vra2

If Fis discontinuous and non-oscillatory, then the value of the series on the right is not F, but is the function defined in equation (143).

Now it is known that 1/r is a spherical harmonic, so that we have

where the differentiation is with respect to the coordinates of Q. Hence. 1/R must be of the form (cf. § 233)

1 «,/ , . B

h = x{At"+?&)8* <147^

where Sn is a surface harmonic of order n. Comparing with equation (144), and remembering that a in this equation is the same as the r of equation (147), we see that PlX, regarded as a function of the position of Q, is a surface harmonic of order n, and we have already seen that it is a series of powers

CO

of cos 9, or of - , the highest power being the nth, so that rnPn is a rational integral harmonic of order n. It follows that

FrnPndS,

being the sum of a number of terms each of the form rnPn, is also a rational integral harmonic of order n, say Vn. On the surface of the sphere

Vn = anfJFPndS, so that equation (146) becomes

'-as!5^7- (148>'

which establishes the result in question.

241. Theorem. The expansion of an arbitrary function of position on the surface of a sphere as a series of rational integral harmonics is unique.

For if possible let the same function F be expanded in two ways, say

F=tWn (149),

F=XWn' (150),

where Wn, Wn' are rational integral harmonics of order n. Then the function

u = 2(W;i-Wn)

240-243]

Spherical Harmonics

215

is a spherical harmonic, which vanishes at every point of the sphere. Since V-u = 0 at every point inside the sphere it is impossible for u to have either a maximum or a minimum value inside the sphere (cf. § 52), so that u = 0 at every point inside the sphere. Since Wtl — Wn' is a harmonic of order n, it must be of the form rn8n, where Sn is a surface harmonic, so that

u=lrnSn=0.

Thus u is a power series in r which vanishes for all values of r from r = 0 to r = a. Thus Sn = 0 for all values of n. Hence Wn = Wn', and the two expansions (149) and (150) are seen to be identical.

242. It is clear that in electrostatics we shall in general only be concerned with functions which are finite and single-valued at every point, and of which the discontinuities are finite in number. Thus the only classes of harmonics which are of importance are rational integral harmonics, and in future we confine our attention to these. We have found that

(i) The rational integral harmonics of degree n are (2w + 1) in number,

and may all be derived from the harmonic - by differentiation.

(ii) Any function of position on a spherical surface, which satisfies the conditions which obtain in a physical problem, can be

expanded as a series of rational integral harmonics, p"p P'

and this can be done only in one way.

243. Before considering these harmonics in detail, we may try to form some idea of the physical concep- tions which lead to them most directly.

The function - is the potential of a unit charge

at the origin. If, as in § 64, we consider two charges

• e at points 0', 0" at equal small distances a, — a from the origin along the axis of x, we obtain as the potential at P,

e e e e

O'O O' Fia. 73.

V=

OP 0"P ~~ OP" OP' = -e.PP

1(1 dx\r

axis

If we take - e . PP" = 1, we have a doublet of strength - 1 parallel to the

r) /I \

of x, and the potential at P is ^- f - J . In fact this potential is exactly

x

the same as — 3 already found in § 64.

216 Methods for the Solution of Special Problems [ch. vm

Thus the three harmonics of order — 1 obtained by dividing the rational

integral harmonics of order 1 by r3, namely ^-(-J. k~ (-), k~ (-) , are

simply the potentials of three doublets each of unit strength, parallel to the negative axes of x, y, z respectively.

If in fig. 73 we replace the charge e at 0' by a doublet of strength e parallel to the negative axis of x, and the charge — e at 0" by a doublet of strength — e parallel to the negative axis of x, we obtain a potential

-<-).

dx2 \r/

If instead of the doublets being parallel to the axis of x, we take them parallel to the axis of y, we obtain a potential

a2 /i>

dxdy \r,

So we can go on indefinitely, for on differentiating the potential of a system with respect to x we get the potential of a system obtained by replacing each unit charge of the original system by a doublet of unit strength parallel to the axis of x. Thus all harmonics of type

fiS+t+U

f

I)

dxsdytdzw V?

(cf. § 236) can be regarded as potentials of systems of doublets at the origin, and, as we have seen (§ 239), it is these potentials which give rise to the rational integral harmonics.

244. For  instance  in  finding  a  system  to  give  potential  ~-^  (  -  j ,  we  may  replace  the 
     

charge 0 in fig. 73 by a charge — at distance 2a from 0 and - — at 0. The charge at 0'

may be similarly treated, so that the whole system is seen to consist of charges

E, -2E, E,

at the points x= — b, 0, b where 6 = 2a, and E2 = j-2.

A system of this kind placed along each axis gives a charge -6E at the origin and a charge E at each corner of a regular octahedron having the origin as centre. The potential

~ dx2 \rj 8y2 \rj dz2 \r ) = 0, so that such a system sends out no lines of force.

245. The most important class of rational integral harmonics is formed by harmonics which are symmetrical about an axis, say that of x. There is one harmonic of each degree n, namely that derived from the function

— (-)

dxn\r)'

These harmonics we proceed to investigate.

243-247] Spherical Harmonics 217

Legendre's Coefficients. 246. The function

,- n =— (151)

V a2 - 2ar cos ^ + r2

can, as we have already seen (cf. equation (144)), be expanded in a convergent series in the form

1 1 r r2 rn

v a2 - 2ar cos <9 + r2 a x a2 a3 an+1

if a is greater than r. Here the coefficients %,!%,... are functions of cos 0, and are known as Legendre's coefficients. When we wish to specify the particular value of cos 0, we write F^, as Pn (cos 0).

Interchanging r and a in equation (152) we find that, if r > a,

= - + pA + P^+ (1-53).

Va2 - 2ar cos 0 + r2 r r2 r3

We have already seen that the functions I{, P^, ... are surface harmonics, each term of the equations (152) and (153) separately satisfying Laplace's equation. The equation satisfied by the general surface harmonic Sn of degree n. namely equation (136), is

dSn\ d2Sn

Hw)+;Jb+"<"+i>-a

sm6d0\ d0 J sm20d<f>2

In the present case i^ is independent of <p, so that the differential equation satisfied by Pn is

or, if we write fi for cos 0,

rA(1-^?£}+Hn+1)^° (15i)-

This equation is known as Legendre's equation. 247. By actual expansion of expression (151)

so that on picking out the coefficient of rn, we obtain

D_1.3...2w-1 1.3... 2n~ 3 n_2 1.3...2n-5 n_4_

^~ JH ^ ~~27(n-2)\ * +2.4.(ti-4)!/" ""

(155).

Thus Pn is an even or odd function of fx according as n is even or odd. It will readily be verified that expression (155) is a solution in series of equation (154).

218 Methods for the Solution of Special Problems [ch. viii

Let us take axes Ox, Oy, Oz, the axis Ox to coincide with the line 0 = 0, then \xr = r cos 6 = x. Then it appears that Pnrn is a rational integral function of x, y, and z of degree n, and, being a solution of Laplace's equation, it must be a rational integral harmonic of degree n. We have seen that there can only be one harmonic of this type which is also symmetrical about an axis ; this, then, must be Pnrn.

248. If  we  write 
     

(a2 - 2ar/M + r2)-^ =/(a)

we have, by Maclaurin's Theorem,

/(a)=/(0) + a

dm

da

a2 p/(a)l

a = 0

2

da2

• ...

a=0

.(156).

If P is the point whose polar coordinates are a, 0 and Q is the point r, 6, then f{a) = -p-~ . The Cartesian co- ordinates of P may be taken to be a, 0, 0 ; let those of Q be #, y, z. Then /(a) = , ■■ ■ , so that as regards

V(a? - a)2 + 2/2 + .z2

differentiation of /(a),

3_

8a

9_

dx'

Fig. 74.

Thus

3»

( aart Ja=o V ' 1 dx- Uo 9*n

(0)

3^n V«2 + y2 + £2

jp /r

9#n vr,

so that equation (156) becomes

1

d_(l\ a2 ^ /l a Bx \r) + 2 ! &Z2 Vr

and on comparison with expansion (153), we see that

n n! dxn\r)'

giving the form for Pn which we have already found to exist in § 245.

249. A  more  convenient  form  for  Pn  can  be  obtained  as  follows. 
     

Let 1 - hy = (1 - 2h/j, + A2)* (157),

so that

y = fM + h

y2-\

.(158).

247-251] Spherical Harmonics 219

From this relation we can expand y by Lagrange's Theorem (cf. Edwards, Differential Calculus, § 517) in the form

y=/1+Aa_ + ...+_y (V)+--

Differentiating with respect to fx,

From equation (157), however, we find

jp = (1 - 2hfj, + h*)-*=l + hP1 + ... + hnPn + .... Equating the coefficients of hn in the two expansions, we find

. fi-Rn©V-i>- ass).

250. This last formula supplies the easiest way of calculating actual values of /£. The values of i?, P2,...P7 are found to be

■^ (A4) = A*t

i^(^) = 1(3^-1),

/£(/) = M°>8-3/)» J30») = £(35/^-3(y + 3),

/>(/.) = £(63^ -70^ + 15//),

P6 O) = JB (23V - B15fi* + 105/a" - 5),

P7 (fi) = TV (42 V - 693/i6 + 315/a8 - 35/*).

251. The equation (/i2 — l)n = 0 has 2n real roots, of which n may be regarded as coinciding at /* = 1, and w at /* = - 1. By a well-known theorem, the first derived equation,

!>'-i)"=o,

will have 2n — 1 real roots separating those of the original equation. Passing to the nth derived equation, we find that the equation

|>-iy=o

has n real roots, and that these must all lie between /u. = — 1 and p. == + 1. The roots are all separate, for two roots could only be coincident if the original equation (/*2 — l)n = 0 had n + 1 coincident roots.

Thus the n roots of the equation Pn (fx,) = 0 are all real and separate and lie between /x = — 1 and /m = + 1.

220 Methods for the Solution of Special Problems [en. vm

252. Putting  jx  =  1,  we  obtain 
     

l+P1h + PJi2+...=

\fl-2h + h2 = l+h + h2+ ...,

so that i?=i?= ... = 1. Similarly, when /a = -1, we find (cf. § 240) that

_£= + £ = -£=... =-1.

We can now shew that throughout the range from /* = — 1 to /t = + l, the numerical value of i^ is never greater than unity. We have

(1 - 2h cos 6 + h?)' i = (1 - heie)-$ (1 - Ae"ie) "^

x ( 1 + \zhe~id + i^| /i2e-2ifi + ...) ,

so that on picking out coefficients of hn,

D 1.3...2»-10 a 1 1.3...2n-30 , ON . ,

P-= 2.4...2n 2coBwg + 2-2.4...2n-22cOB(n-2)g + ""

Every coefficient is positive, so that Pn is numerically greatest when each cosine is equal to unity, i.e. when 6 = 0. Thus Pn is never greater than unity.

Fig. 75 shews the graphs of P1} P^, Pz, P*, from p. = — 1 to /* = + 1, the value of 6 being taken as abscissa.

» = o

6=1

6 =

3n

0=7,

y- + i

0

\ \ ^v1

\ \ 'V^ /\

P7 /

'Jz

fi = 0 Fig. 75.

^"72-

i*.= -1

252, 253 J Spherical Harmonics 221

Relations between coefficients of different orders. 253. We have

(l-2h(j. + h*)-i = l + 2hnPn (160).

Differentiating with regard to h,

(fx-h)(l-2hfM + hi)-* = Cknhn-'Pn (161),

i

00

so that (ji - h) (1 + ^hnPn) = (1 - 2hfi + h?) tnhn^Pn.

i i

Equating coefficients of hn, we obtain

(n+l)Pn+1 + nPn_1 = (2n+l)fiPn (162).

This is the difference equation satisfied by three successive coefficients.

Again, if we differentiate equation (160) with respect to p.

op

so that, by combining with (161),

1 Ofl

Equating coefficients of hn,

r> 0-tn Orn—\ /I nn\

nPn = f*d^--d]T (16o>-

Differentiating (162), we obtain

op

Eliminating /x -^ from this and (163),

(fc.+ l)J»-?gi-^ (164).

By integration of this we obtain

/■sco**-w£;t*0t) <165>-