The Mathematical Theory of Electricity and Magnetism (5th ed, 1927) — part 10 of 39

155. Although no sufficient reason has been found compelling us to ascribe electric action to the presence of an intervening medium, we are still free to assume, as a hypothesis, that such a medium exists and that electric action is transmitted through this medium. As various electric and electro- magnetic phenomena are discussed we shall examine what properties would have to be attributed to the medium to account for these properties. If it is found that contradictory properties would have to be ascribed to the medium, then the hypothesis of action through an intervening medium will have to be abandoned. If the properties are found to be consistent, then the hypotheses of action at a distance and action through a medium are still both in the field, but the latter becomes more or less probable just in proportion as the properties of the hypothetical medium seem probable or improbable. We shall return to the general question of the existence of a medium in Chapter XX.

156. Since electric action takes place even across the most complete vacuum obtainable, we conclude that if this action is transmitted by a medium, this medium must be the ether. Assuming that the action is transmitted by the ether, we must suppose that at any point in the electro- static field there will be an action and reaction between the two parts of the ether at opposite sides of the point. The ether, in other words, is in a state of stress at every point in the electrostatic field. Before discussing the particular system of stresses appropriate to an electrostatic field, we shall investigate the general theory of stresses in a medium at rest.

General Theory of Stresses in a medium at rest.

157. Let us take a small area dS in the medium perpendicular to the axis of x. Let us speak of that part of the medium near to dS for which x is greater than its value over dS as x+, and that for which x is less than this value as a?, so that the area dS separates the two regions x+ and x. Those parts of the medium by which these two regions are occupied exert forces upon one another across dS, and this system of forces is spoken of as the stress across dS. Obviously this stress will consist of an action and reaction, the two being equal and opposite. Also it is clear that the amount of this stress will be proportional to dS.

Let us assume that the force exerted by x+ on #_ has components

Ixxdk, JxydS, lzZdS.

154-157]

General Theory of Stress

143

then the force exerted by x_ on x+ will have components

— PxxdS, —T^ydS, — PxzdS.

The quantities Pxx, Pxy, Pxz are spoken of as the components of stress perpendicular to Ox. Similarly there will be components of stress Pyx> Pyy, PyZ perpendicular to Oy, and components of stress Pzx, Pzy, Pzz perpendicular to Oz.

Let us next take a small parallelepiped in the medium, bounded by planes

x = £, x = £ + d x ;

y = v, y = v + dy;

z=%, z = £ + dz.

The stress acting upon the parallelepiped across the face of area dydz in the plane x = g will have components

• (Pxx)z-t dydz, - (Pxy)x^dydz, - (Pxz)x^ dydz,

while the stress acting upon the parallelepiped across the opposite face will have components

(Pxz)x-z+dzdydz, (Pxy)x~s+dzdydz, (Pxz)x^+dXdydz.

Compounding these two stresses, we find that the resultant of the stresses acting upon the parallelepiped across the pair of faces parallel to the plane of yz, has components

dx

Fig. 47.

dP

—^ dxdydz,

000

dx

dxdydz.

Similarly from the other pairs of faces, we get resultant forces of com- ponents

IX dxdydz,

dP

yy

and

dy dPzx

dz

dxdydz,

dy dPzy

dxdydz,

dPv

yz

dy dPzz

dxdydz,

dxdydz, ■— dxdydz.

For generality, let us suppose that in addition to the action of these stresses the medium is acted upon by forces acting from a distance, of amount H, H, Z per unit volume. The components of the forces acting on the parallelepiped of volume dxdydz will be

S dxdydz, H dxdydz, Z dxdydz.

Compounding all the forces which have been obtained, we obtain as equations of equilibrium

and two similar equations.

dx

V-tyX , V-*ZX n

By dz

.(79)

144 The State of the Medium in the Electrostatic Field [ch. vt

158. These three equations ensure that the medium shall have no motion of translation, but for equilibrium it is also necessary that there should be no rotation. To a first approximation, the stress across any face may be supposed to act at the centre of the face, and the force B, H, Z at the centre of the parallelepiped. Taking moments about a line through the centre parallel to the axis of Ox, we obtain as the equation of equilibrium

Pyz-Hy = 0 (80).

This and the two similar equations obtained by taking moments about lines parallel to Oy, Oz ensure that there shall be no rotation of the medium. Thus the necessary and sufficient condition for the equilibrium of the medium is expressed by three equations of the form of (79), and three equations of the form of (80).

159. Suppose next that we take a small area dS anywhere in the medium. Let the direction cosines of the normal

to dS be ± I, ± m, ± n. Let the parts of the medium close to dS and on the two sides of it be spoken of as S+ and $_, these being named so that a line drawn from dS with direction cosines

• 1, + m, + n will be drawn into S+, and

one

with direction cosines — I, — m, — n will be drawn into &. Let the force exerted by S+ on & across the area dS have components

FdS, GdS, HdS, Fig. 48.

then the force exerted by #_ on S+ will have components

-FdS, -GdS, -HdS.

The quantities F, G, H are spoken of as the components of stress across a plane of direction cosines I, m, n.

To find the values of F, G, H, let us draw a small tetrahedron having three faces parallel to the coordinate planes and a fourth having direction cosines I, m, n. If dS is the area of the last face, the areas of the other faces are IdS, mdS, ndS and the volume of the parallelepiped is % \f2lmn (dS)2 . Resolving parallel to Ox, we have, since the medium inside this tetrahedron is in equilibrium,

£ V2K (dS)% H - ldSPxx - mdSPyx - ndSP2X + FdS = 0, giving, since dS is supposed vanishingly small,

F=lPxx + mPyx + nPzx (81)

and there are two similar equations to determine G and H.

158-160] General Theory of Stress 145

160. Assuming that equation (80) and the two similar equations are satisfied, the normal component of stress across the plane of which the direction cosines are I, m, n is

IF + mQ + nH=PPxx + m2Pyy + n2Pzz + 2mnPyz + 2nlPzx + 2lmPxy.

The quadric

x2Pxx + y2Pyy + z2Pzz + 2yzPyz + 2zxPzx + 2xyPxy = l (82)

is called the stress-quadric. If r is the length of its radius vector drawn in the direction I, m, n, we have

r2 (lPxx + m\y + nPzz + 2mnPyz + 2nlPzx + 2lmPxy) = 1.

It is now clear that the normal stress across any plane I, m, n is measured by the reciprocal of the square of the radius vector of which the direction cosines are I, m, n. Moreover the direction of the stress across any plane I, m, n is that of the normal to the stress-quadric at the extremity of this radius vector. For r being the length of this radius vector, the coordinates of its extremity will be rl, rm, rn. The direction cosines of the normal at this point are in the ratio

rlPxx + rmPxy + mPzx : rlPxy + rmPyy + rnPyz: rlPzx + rmPyZ + rnPzz

or F : G : H, which proves the result.

The stress-quadric has three principal axes, and the directions of these are spoken of as the axes of the stress. Thus the stress at any point has three axes, and these are always at right angles to one another. If a small area be taken perpendicular to a stress axis at any point, the stress across this area will be normal to the area. If the amounts of these stresses are P1} Pi, P3) then the equation of the stress-quadric referred to its principal ixes will be

Clearly a positive principal stress is a simple tension, and a negative principal stress is a simple pressure.

As simple illustrations of this theory, it may be noticed that

(i) For a simple hydrostatic pressure P, the stress-quadric becomes an imaginary sphere

p(e+ri2+{2)=-i:

The pressure is the same in all directions, and the pressure across any plane is at right angles to the plane (for the tangent plane to a sphere is at right angles to the radius vector).

(ii) For a simple pull, as in a rope, the stress-quadric degenerates into two parallel planes

P£2 = l.

J. 10

146 The State of the Medium in the Electrostatic Field [ch. vi

The Stresses in an Electrostatic Field.

161. If an infinitesimal charged particle is introduced into the electric field at any point, the phenomena exhibited by it must, on the present view of electric action, depend solely on the state of stress at the point. The phenomena must therefore be deducible from a knowledge of the stress- quadric at the point. The only phenomenon observed is a mechanical force tending to drag the particle in a certain direction — namely, in the direction of the line of force through the point. Thus from inspection of the stress- quadric, it must be possible to single out this one direction. We conclude that the stress-quadric must be a surface of revolution, having this direction for its axis. The equation of the stress-quadric at any point, referred to its principal axes, must accordingly be

Z?r + W+£2) = 1 (83),

where the axis of f coincides with the line of force through the point. Thus the system of stresses must consist of a tension Pt along the lines of force, and a tension ^ perpendicular to the lines of force — and if either of the quantities i? or P2 is found to be negative, the tension must be interpreted as a pressure.

Since the electrical phenomena at any point depend only on the stress- quadric, it follows that R must be deducible from a knowledge of i? and I. Moreover, the only phenomena known are those which depend on the magnitude of R, so that it is reasonable to suppose that the only quantity which can be deduced from a knowledge of Px and P^ is the quantity R — in other words, that 7? and P% are functions of R only. We shall for the present assume this as a provisional hypothesis, to be rejected if it is found to be incapable of explaining the facts.

162. The expression of i? as a function of R can be obtained at once by considering the forces acting on a charged conductor. Any element dS

R2

of surface experiences a force — dS urging it normally away from the con- ductor. On the present view of the origin of the forces in the electric field, we must interpret this force as the resultant of the ether-stresses on its two sides. Thus, resolving normally to the conductor, we must have

^rdS=(P_)BdS-(P1)0dS,

where (P)e, (7?)0 denote the values of i? when the intensity is R and 0 respectively. Inside the conductor there is no intensity, so that the stress-quadrics become spheres, for there is nothing to differentiate one direction from another. Any value which (/?)„ may have accordingly arises

1(51-164] Stresses in Electrostatic Field 147

simply from a hydrostatic pressure or tension throughout the medium, and this cannot influence the forces on conductors. Leaving any such hydrostatic pressure out of account, we may take (i?)0 = 0, and so obtain (P1)R in the form

R2

«=8^ •. <84>-

163. We can most easily arrive at the function of R which must be taken to express the value of R2, by considering a special case.

Consider a spherical condenser formed of spheres of radii a, b. If this condenser is cut into two equal halves by a plane through its centre, the two halves will repel one another. This action must now be ascribed to the stresses in the medium across the plane of section. Since the lines of force are radial these stresses are perpendicular to the lines of force, and we see at once that the stress perpendicular to the lines of force is a pressure. To calculate the function of R which expresses this pressure, we may suppose b — a equal to some very small quantity c, so that R may be regarded as constant along the length of a line of force. The area over which this pressure acts is it (b2 — a2), and since the pressure per unit area in the medium perpendicular to a line of force is — R, the total repulsion between the two halves of the condenser will be — R2Tr(b2 — a2).

The whole force on either half of the condenser is however a force 2tt<t2

per unit area over each hemisphere, normal to its surface. The resultant of

all the forces acting on the inner hemisphere is ira2 x 2ira2, or putting

2ira2a = E, so that E is the charge on either hemisphere, this force is E2/2a2.

Similarly, the force on the hemisphere of radius b is E2/2b2. Thus the re-

'1 1 \ sultant repulsion on the complete half of the condenser is \E2 ( - — yA . Since

this has been seen to be also equal to — R.7T (b2 — a2), we have

W , R2

^=--^7-„ = -27T(7

on taking a = b in the limit.

Thus in order that the observed actions may be accounted for, it is necessary that we have

7?a R3

P — zL i P — - —

^"Stt' *~ 8tt'

Moreover, if these stresses exist, they will account for all the observed mechanical action on conductors, for the stresses result in a mechanical force 27ro-2 per unit area on the surface of every conductor.

164. It remains to examine whether these stresses are such as can be transmitted by an ether at rest.

10—2

148 The State of the Medium in the Electrostatic Field [oh. vi

As a preliminary we must find the values of the stress-components Pxx, Pxy,... referred to fixed axes Ox, Oy, Oz.

The stress-quadric at any point in the ether, referred to its principal axes, is seen on comparison with equation (83) to be

£(p-^2-n=i (85).

07T

Here the axis of f is in the direction of the line of force at the point. Let the direction-cosines of this direction be llt mu i. Then on transforming to axes Ox, Oy, Oz we may replace f by lxx + mxy + nxz.

Equation (85) may be replaced by

07T

and on transforming axes £2 + rf + £2 transforms into a? + y2 + z2. Thus the transformed equation of the stress-quadric is

R2

^ {2 (l,x + my + n&f - (x2 + y2 + z2)} = 1.

Comparing with equation (82), we obtain

£.«=?- (2V-1) (86),

Pxy = ^(2limi) (87),

07T

and similar values for the remaining components of stress.

Or again, since X = l^R, Y = rn^R, Z = ^R,

these equations may be expressed in the form

Pxx = ^ {X2 - Y2 - Z2),

07T

_XY

xy 4,

In this system of stress-components, the relations Pxy = Pyx are satisfied, as of course they must be since the system of stresses has been derived by assuming the existence of a stress-quadric. Thus the stresses do not set up rotations in the ether (cf. equation (80)).

In order that there may be also no tendency to translation, the stress- components must satisfy equations of the type

Optx | OPcy , 0*xz a /OQ\

-dx~ + ~dy- + ~dF = ° (88)'

expressing that no forces beyond these stresses are required to keep the ether at rest (cf. equation (79)).

164-166] Stresses in Electrostatic Field 149

On substituting the values of the stress-components, we have

dx dy dz

' - ss Is {X' - T' - ^ + 1 <2Xr> + &<***>}

= J. \2X (— + - + ~\ + 2F (^ - d-I) + 2Z (^ - d-l)\

8tt\ \dx dy dz) \dy dx J ' \dz dx))'

On putting

Y -d-Z V- JZ 7-JJ.

*~ dx' ' dy1 dz

we find at once that

dX_dY= &V_ 92F_ dy dx dxdy dxdy '

dX_dZ_ &V_ d*V

dz dx dxdz dxdz '

aX dY d_Z = _/cPV d^v &Z\ = 0

dx dy dz \ dx2 dy2 dz'2 ) ' shewing that equation (88) is satisfied.

165. Thus,  to   recapitulate,  we   have    found  that  a  system  of  stresses 
     

consisting of

(i) a tension -6— per unit area in the direction of the lines of force,

(ii) a pressure — per unit area perpendicular to the lines of force,

is one which can be transmitted by the medium, in that it does not tend to set up motions in the ether, and is one which will explain the observed forces in the electrostatic field. Moreover it is the only system of stresses capable of doing this, which is such that the stress at a point depends only on the electric intensity at that point.

Examples of Stress.

166. Assuming this system of stresses to exist, it is of value to try to picture the actual stresses in the field in a few simple cases.

Consider first the field surrounding a point charge. The tubes of force are cones. Let us consider the equilibrium of the ether enclosed by a frustum of one of these cones which is bounded by two ends p, q. If G)p, o)q are the areas of these ends, we find that there are tensions of

150 The State of the Medium in the Electrostatic Field [ch. vi

amounts

B

pWp

Bg'COg

8tt ' 8tt

so that the forces on the two ends have as resultant a force tending to move the ether inwards towards the charge. This tendency- is of course balanced by the pressures acting on the curved surface, each of which has a component tending to press the ether inside the frustum away from the charge.

Since Rpa>p = Rqcoq, the former is the greater.

Fig. 49.

167. A  more  complex  example  is  afforded 
     

by two equal point charges, of which the lines of force are shewn in fig. 50.

Fia. 50.

The lines of force on either charge fall thickest on the side furthest removed from the other charge, so that their resultant action on the charges amounts to a traction on the surface of each tending to drag it away from the other, and this traction appears to us as a repulsion between the bodies.

We can examine the matter in a different way by considering the action

and reaction across the two sides of the plane which bisects the line joining

the two charges. No lines of force cross this plane, which is accordingly

made up entirely of the side walls of tubes of force. Thus there is a pressure

R?

— per unit area acting across this plane at every point. The resultant of

o7T

all these pressures, after transmission by the ether from the plane to the charges immersed in the ether, appears as a force of repulsion exerted by the charges on one another.

166-169] Energy in the Electrostatic Field 151

Energy in the Medium.

168. In setting up the system of stresses in a medium originally un- stressed, work must be done, analogous to the work done in compressing a gas. This work must represent the energy of the stressed medium, and this in turn must represent the energy of the electrostatic field. Clearly, from the form of the stresses, the energy per unit volume of the medium at any point must be a function of R only. To determine the form of this function, we may examine the simple case of a parallel plate condenser,

R2

and we find at once that the function must be 5—.

07T

We have now to examine whether the energy of any electrostatic field

R2 can be regarded as made up of a contribution of amount 5— per unit volume

o7T

from every part of the field.

In fig. 51, let PQ be a tube of force of strength e, passing from P at potential VP to Q at potential VQ. The ether inside this tube of force

R2

being supposed to possess energy — per unit volume,

the total energy enclosed by the tube will be

co ds,

L

p 87T

where w is the cross section at any point, and the integration is along the tube. Since Rco = 4nre, this expression

= \ e I R ds

= -*e]PTsds

= ie(Vf-VQ).

This, however, is exactly the contribution made by the charges + e at P, Q to the expression ^ 2eF. Thus on summing over all tubes of force, we find that the total energy of the field \SeV may be obtained exactly, by

R2

assigning energy to the ether at the rate of 3— per unit volume.

Energy in a Dielectric.

169. By  imagining  the  parallel  plate   condenser  of  §  168  filled  with 
     

dielectric of inductive capacity K, and calculating the energy when charged,

KR2 we find that the energy, if spread through the dielectric, must be — —

per unit volume.

152 The State of the Medium in the Electrostatic Field [ch. vi

Let us now examine whether the total energy of any field can be regarded as arising from a contribution of this amount per unit volume. The energy contained in a single tube of force, with the notation already used, will be

QKR2 , o) as,

I

P 8?r

KR

or, since -r — = P, where P is the polarisation, this energy

47T

= | \RPcods

rQ

= I e\ Rds

J p

= 2 e ( Kp — ' Ka)j

so that the total energy is \XeV, as before. Thus a distribution of energy of

7T7?2

amount -=- — per unit volume will account for the energy of any field.

Crystalline dielectrics.

170. We have seen (§ 152) that in a crystalline dielectric, the com- ponents of polarisation and of electric intensity will be connected by equations of the form

4tt/ = KUX + K2l Y + K31Z \

4tt# = K12X + K22Y + K32Z \ (89).

4ttA = KIZX + KsiY+ K3SZ J

The energy of any distribution of electricity, no matter what the dielectric

may be, will be ^ %EV. If V1} V2 are the potentials at the two ends of

a unit tube, the part of this sum which is contributed by the charges at the

ends of this tube will be £ (Fx — TQ. If 9/ds denote differentiation along the

fdV . fdV

tube, this may be written — \ \ — ds, or again — \ I — Pco ds, where P is the

polarisation, and a> the cross section of the tube. Thus the energy may be

dV supposed to be distributed at the rate of — \ -^- P per unit volume. If e is the

angle between the direction of the polarisation and that of the electric

dV intensity, we have — -~- = R cos e, so that the energy per unit volume

= \RP cos e = $(/X + gY+hZ) (90).

In a slight increase to the electric charges, the change in the energy of the system is, by § 109, equal to 1,V8E, so that the change in the energy per unit volume of the medium is

BW=XSf+ Y8g + Z8h. TllUS ~df=X> Tg=Y> lh=Z <91>-

169-171] Maxwell's Displacement Theory 153

From formulae (89) and (90), we must have W = %(fX + gY+hZ)

= i {KUX* + (K12 + K21) XY+ ...}, from which

\x = ~L [KnX + * ( /fi2 + z 2i) Y + * (^13 + *■> z)-

We must also have

8JF 8TT §£ dW dg dW dh dX df dX+ dg dX + dh dX

= ^{KuX + K21Y + K3lZ}. Comparing these expressions, we see that we must have

-**-12 = -ft-nt -^-13 = -"L31> -^-23 == -^32«

The energy per unit volume is now

W = ±r(KnX* + 2K12XY+...) (92).

Maxwell's Displacement Theory.

171. Maxwell attempted to construct a picture of the phenomena occurring in the electric field by means of his conception of "electric dis- placement." Electric intensity, according to Maxwell, acting in any medium — whether this medium be a conductor, an insulator, or free ether — produces a motion of electricity through the medium. It is clear that Maxwell's conception of electricity, as here used, must be wider than that which we have up to the present been using, for electricity, as we have so far under- stood it, is incapable of moving through insulators or free ether. Maxwell's motion of electricity in conductors is that with which we are already familiar. As we have seen, the motion will continue so long as the electric intensity continues to exist. According to Maxwell, there is also a motion in an insulator or in free ether, but with the difference that the electricity cannot travel indefinitely through these media, but is simply displaced a small distance within the medium in the direction of the electric intensity, the extent of the displacement in isotropic media being exactly proportional to the intensity, and in the same direction.

The conception will pei-haps be understood more clearly on comparing a conductor to a liquid and an insulator to an elastic solid. A small particle immersed in a liquid will continue to move through the liquid so long as there is a force acting on it, but a particle immersed in an elastic solid will be merely "displaced" by a force acting on it. The amount of this displacement will be proportional to the force acting, and when the force is removed, the particle will return to its original position.

154 The State of the Medium in the Electrostatic Field [ch. vi

Thus at any point in any medium the displacement has magnitude and direction. The displacement, then, is a vector, and its component in any direction may be measured by the total quantity of electricity per unit area which has crossed a small area perpendicular to this direction, the quantity being measured from a time at which no electric intensity was acting.

172. Suppose, now, that an electric field is gradually brought into existence, the field at any instant being exactly similar to the final field except that the intensity at each point is less than the final intensity in some definite ratio k. Let the displacement be c times the intensity, so that when the intensity at any point is kR, the displacement is c/cR. The direction of this displacement is along the lines of force, so that the electricity may be regarded as moving through the tubes of force : the lines of force become identical now with the current-lines of a stream, to which they have already been compared.

Let us consider a small element of volume cut off by two adjacent equipotentials and a tube of force. Let the cross section of the tube of force be co, and the normal distance between the equipotentials where they meet the tube of force be ds, so that the element under consideration is of volume cods. On increasing the intensity from kR to (k + d/c) R, there is an increase of displacement from ckR to c (k + c?«) R, and therefore an additional dis- placement of electricity of amount cRdic per unit area.

Thus of the electricity originally inside the small element of volume, a quantity cRcodic flows out across one of the bounding equipotentials, whilst an equal quantity flows in across the other. Let Vi; V.A be the potentials of these surfaces, then the whole work done in displacing the electricity originally inside the element of volume cods, is exactly the work of transferring a quantity cRd/c of electricity from potential Vx to potential V%. It is therefore cRco(Vl — V^dic and, since V2 — V1 = /cRds, this may be written as cR2cods/cdK. Thus as the intensity is increased from 0 to R, the total work spent in displacing the electricity in the element of volume cods

= I cR2 (cods) Kdic = %cR2 . cods. Jo

This work, on Maxwell's theory, is simply the energy stored up in the

nnrl t.he» r^lST>ln.PAmpn+•. n.+. nnv

47T

-ds-

R2 element of volume cods of the medium, and is therefore equal to 3— cods.

07T

Thus c must be taken equal to j— , and the displacement at any point is

measured by

R_

47T*

171-174] MaxwelVs Displacement Theory 155

If the element of volume is taken in a dielectric of inductive capacity K,

77D2 77"

the energy is -= — , so that c = j— , and the displacement is

KR

4nr '

173. It is now evident that Maxwell's "displacement" is identical in magnitude and direction with Faraday's "polarisation" introduced in Chap. v.

Denoting either quantity by P, we had the relation

//

P cose dS = E (93),

expressing that the normal component of P integrated over any closed surface is equal to the total charge inside. On Maxwell's interpretation of

the quantity P, the surface integral 1 1 P cos e dS simply measures the total

quantity of electricity which has crossed the surface from inside to outside. Thus equation (93) expresses that the total outward displacement across any closed surface is equal to the total charge inside.

If we now follow Maxwell in supposing that electricity is of two kinds,

(i) the kind which appears as a charge on an electrified body, (ii) the kind which Maxwell imagines to occupy the whole of space, and to undergo displacement when electric action takes place,

then it appears that any increase of electricity of kind (i) inside any closed

surface is accompanied by an exactly equal decrease of electricity of kind (ii).

In other words the sum total of the two kinds of electricity inside any closed

surface remains constant.

174. It will be understood that Maxwell's theory of electrical displace- ment attempts to give a physical picture of the processes of the electric field, ., but that the truth of the picture is by no means essential to the mathematical theory of electricity. The displacement theory is historically important because 0 it led Maxwell to the hypothesis of displacement currents which form the ~ foundation of his electromagnetic theory of light (Chap. xvn). But we shall see later that the general electromagnetic theory can be developed without v the preliminary displacement theory. The displacement theory has served as part of the scaffolding by which the electromagnetic theory was constructed; whether the scaffolding ought now to be discarded remains an open question.

CHAPTER VII

GENERAL ANALYTICAL THEOREMS

Green's Theorem.

175. A theorem, first given by Green, and commonly called after him, enables us to express an integral taken over the surfaces of a number of bodies as an integral taken through the space between them. This theorem naturally has many applications to Electrostatic Theory. It supplies a means of handling analytically the problems which Faraday treated geometrically with the help of his conception of tubes of force.

176. Theorem. If u, v, w are continuous functions of the Cartesian coordinates x, y, z, then

1

l l(lu + mv + nw) dS = — l\[^- + ^~ + ^-j dxdydz (94).

Here 2 denotes that the surface integrals are summed over any number of closed surfaces, which may include as special cases either

(i) one of finite size which encloses all the others, or (ii) an imaginary sphere of infinite radius,

and I, m, n are the direction-cosines of the normal drawn in every case from the element dS into the space between the surfaces. The volume integral is taken throughout the space between the surfaces.

r r ^

Consider first the value of I hr- dxdydz. Take any small prism with its

axis parallel to that of x, and of cross section dydz. Let it meet the surfaces at P, Q, R, S, T, U, ... (fig. 53), cutting off areas dSP, dSQ, dSR, ....

The contribution of this prism to I M ^- dxdydz is dydz I »- dx, where the integral is taken over those parts of the prism which are between the surfaces.

du , fQdu

Thus ^dx=\ ^dx+ ~dx+...

J OX J p ox } R ox

p I/O/ J R

= — Up + UQ—UR + US'

175-177]

Green's Theorem

157

where uP, uQ, uR>... are the values of u at P, Q, R,.... Also, since the pro- jection of each of the areas dSP, dSQ)... on the plane of yz is dydz, we have

dydz = lPdSP = - lQdSQ = lRdSR =..., where lP, lQ, lR,... are the values of I at P, Q, R,.... The signs in front of lp, Iq, Ir>-" are alternately positive and negative, because, as we proceed along PQR..., the normal drawn into the space between the surfaces makes angles which are alternately acute and obtuse with the positive axis of x.

Fig. 53.

Thus

dydz\—dx = dydz(—Up + uQ — uR+...)

= — IpUpdSp — lQUQdSQ — lRuRdSR — (95),

and on adding the similar equations obtained for all the prisms we obtain

fffd£dxdydz = -zfjludS (96),

the terms on the right-hand sides of equations of the type (95) combining so as exactly to give the term on the right-hand side of (96).

We can treat the functions v and w similarly, and so obtain altogether

///(is + S + i) dxdydz = ~ 2/i (lu +mv+ nw) dS-

proving the theorem.

177. If   u,  v,  w  are  the  three  components  of  any  vector  F,  then  the 
     

expression

du dv dw dx dy dz

is denoted, for reasons which will become clear later, by div F. If N is the component of the vector in the direction of the normal (I, m, n) to dS, theD

N = he + mv + ww.

158 General Analytical Theorems [ch. vh

Thus Green's Theorem assumes the form

jffdiyFdxdydz = -^ff]STdS (97).

A vector F which is such that div F = 0 at every point within a certain region is said to be " solenoidal " within that region. If F is solenoidal within any region, Green's Theorem shews that

jJNdS = 0,

where the integral is taken over any closed surface inside the region within which F is solenoidal. Two instances of a solenoidal vector have so far occurred in this book — the electric intensity in free space, and the polarisa- tion in an uncharged dielectric.

178. Integration through space external to closed surfaces. Let the outer surface be a sphere at infinity, say a sphere of radius r, where r is to be made infinite in the limit. The value of

jj(lu + mv + nw) dS

taken over this sphere will vanish if u, v, and w vanish more rapidly at infinity than — . Thus, if this condition is satisfied, we have that

///

— + — + ^— j dxdydz = — % jj (lu + rnv + nw) dS,

where the volume integration is taken through all space external to certain closed surfaces, and the surface integration is taken over these surfaces, I, m, n being the direction-cosines of the outward normal.

179. Integration through the interior of a closed surface. Let the inner surfaces in fig. 53 all disappear, then we have

///(^ + 1 + Tz) dxdydz = "/I {lu + mv + nw) dS'

where the volume integration is throughout the space inside a closed surface, and the surface integration is over this area, I, m, n being the direction- cosines of the inward normal to the surface.

180. Integration through a region in which u, v, w are discontinuous. The only case of discontinuity of u, v, w which possesses any physical import- ance is that in which u, v, w change discontinuously in value in crossing certain surfaces, these being finite in number. To treat this case, we enclose each surface of discontinuity inside a surface drawn so as to fit it closely on

177-180]

Green's Theorem

159