In the ivth part of Maxwell’s treatise he gives complete investigations of the mechanical results flowing from this theory so far as it refers to currents, but he has not given the general results in the case of Electrostatics. Nor has he shewn satisfactorily, it seems to me, that the ordinary laws of Electrostatics flow from his theory. It is these investigations we now propose to make.

35. Our notation will be as far as possible the same as Maxwell’s. Thus for the displacement at any point we use $D$, and for the e.m.f. $E$. From the connection explained in last section between $D$ and $E$ we have

where $K$ at any point is some linear vector function depending on the state of the medium at the point. If the medium change in any manner not electrical, e.g. by means of ordinary strain $K$ will in general also suffer change.

Let $w$ be the potential energy per unit volume due to the electrical configuration. Thus if a small increment $\delta D$ be given to $D$ at all points, the increment $\iiint \delta wd\mathfrak{𝔰}$ in $\iiint wd\mathfrak{𝔰}$, the potential energy of the electrical configuration in any space, = work done on the electricity in producing the change,

by the relation stated in Section IV existing between $E$ and $D$. Thus limiting the space to the element $d\mathfrak{𝔰}$

Now suppose $D=nD\prime$ so that by equation (1) $E=nE\prime$ where $E\prime$, $D\prime$ are corresponding e.m.f. and displacement respectively. Thus

Integrating from $n=0$ to $n=1$, and finally changing $D\prime$, $E\prime$ into $D$, $E$ we get

From this we get

so that by equation (2) $SE\delta D=SD\delta E$ or by equation (1)

Hence (because $E$ and $\delta E$ are quite arbitrary) $K$ is self-conjugate and therefore involves only six instead of nine coordinates²⁷.

In electrostatics the line integral of $E$ round any closed curve must be zero, for otherwise making a small conductor coincide with the curve we shall be able to maintain a current by Section IV, and so (by the same section) constantly do work on it (i.e. as a matter of fact create heat) without altering the statical configuration. Hence $E$ must have a potential, say $v$. Thus

Since in an electrostatic field there is no current in a conductor, $E=0$ throughout any such conductor and therefore $v=\text{ const.}$

36. The charge in any portion of space is defined as the amount of foreign electricity within that space. Thus the charge in any space is the surface integral of the displacement outwards. Thus if there be a charge on the element $d\Sigma$ of a surface in the dielectric this charge $=Sd{\Sigma }_a{D}_a+Sd{\Sigma }_b{D}_b$ where $a$, $b$ denote the two faces of the element (so that $d{\Sigma }_a=-d{\Sigma }_b$) and in accordance with Section 1 above $d{\Sigma }_a$ points away from the region in which the displacement is $D_a$. Thus $\sigma$ being the surface density

where the notation is used for . Similarly if there be finite volume density of foreign electricity, i.e. finite volume density of charge in any space, the charge $=-\iint SDd\Sigma =-\iiint S\nabla Dd\mathfrak{𝔰}$, so that if $D$ be the volume density

[The reason for having $+Sd{\Sigma }_a{D}_a$ before and $-Sd\Sigma D$ here is that in the former case we were considering a charge outside the region where $D_a$ is considered—between the regions $a$ and $b$ in fact—whereas in the latter case we are considering the charge inside the region where $D$ is considered. The same explanation applies to the sign of $-SDUd\Sigma$ for the surface of a conductor given below.]

In conductors, as we saw in Section IV, the displacement has virtually no meaning (except when it is changing and so the phenomenon of a current takes place) for the foreign electricity and the original electricity are not to be distinguished. Not so however with the surface of the dielectric in contact with the conductor. We may therefore regard the electricity within the body of the conductor as the original electricity so that the charge is entirely at the surface. Thus the surface density will be $-SDUd\Sigma$ where $d\Sigma$ points away from the conductor and $D$ is the displacement in the dielectric. This may be regarded as a particular case of equation (5) $D$ being in accordance with what we have just said considered as zero in the conductor.

37. All the volume integrals with which we now have to deal may be considered either to refer to the whole of space or only to the dielectrics, as the conductors (except at their surfaces) in all cases contribute nothing. The boundary of space will be considered as a surface at infinity and all surfaces where either $D$ or $E$ is discontinuous.

Putting $W=\iiint wd\mathfrak{𝔰}$ we have already found one expression for $W$, viz. 

We now give another. By equation (4)

by equation (9) Section 4 above. Thus by equations (5) and (6) Section IV,

where $ds$ is put, as it frequently will be, for an element of surface, i.e. $Td\Sigma$. The value of $W$ which we shall use²⁸ is obtained by combining these two, viz. 

So far we have merely been shewing that all the above results of Maxwell’s flow from what in Section IV has been described as his theory. We now proceed to the actual problem in hand which is proved from these results however they may be obtained. I may remark that some such investigation as the above seems to me necessary to make the logic of Maxwell’s treatise complete.

38. Suppose now that $W$ is the potential energy of some dynamical system extending throughout space. Let us give to every point of space a small displacement $\delta \eta$ vanishing at infinity and find the consequent increment $\delta W$ in $W$. If this can be put in the form

we shall have the following expression for $F$ the force per unit volume due to the system

and the following expression for $F_S$ the force per unit surface at any surface of discontinuity in $\varphi$

the notation being the same as in equation (5) Section IV.

Moreover if $\varphi$ be self-conjugate the forces both throughout the volume and at surfaces of discontinuity are producible by the stress $\varphi$ as can be seen by Section 7 above. [Compare all these statements with Section 8 above.]

For proof, we have by equation (9) Section 4

where of course the element $d\Sigma$ is taken twice, i.e. once for each face. But

where the element $ds$ is taken only once. Equating the coefficients of the arbitrary vector $\delta \eta$ for each point of space we get the required equations (10) and (11).

39. We must then put $\delta W$ where $W$ is given by equation (8) in the form given in equation (9).

We must first define $\delta$ when applied to a function of the position of a point. Suppose by means of the small displacement $\delta \eta$ any point $P$ moves to $P\prime$. Then $Q$ being the value at $P$, before the displacement $\delta \eta$, of a function of the position of a point, $Q+\delta Q$ is defined as the value of the function at $P\prime$ after the displacement. Thus even in the neighbourhood of a surface of discontinuity $\delta Q$ is a small quantity of the same order as $\delta \eta$.

Now the charge within any space, that is the quantity of foreign electricity within that space will not be altered by the strain.

To find $\delta \nabla$ we have

or, since $\delta d\rho =-Sd\rho \nabla ·\delta \eta$,

The part of $\delta W$ depending on the first two terms of equation (8) is by equation (12)

Noticing that $\delta d\mathfrak{𝔰}=-d\mathfrak{𝔰}S\nabla \delta \eta$ and that $4\pi \delta D=K\delta E+\delta KE$ we see that the last term in equation (8) contributes

Combining the last result with the first term of this we get

Thus we have

40. Now the increment $\delta K$ in $K$ is caused by two things viz. the mere rotation of the body and the change of shape of the body. Let us call these parts $\delta K_r$ and $\delta K_s$ respectively.

First consider $\delta K_r$. Suppose the rotation is $𝜖$ so that any vector which was $\omega$ becomes thereby $\omega +V𝜖\omega$. Thus the result of operating on $\omega +V𝜖\omega$ by $K+\delta K_r$ is the same as first operating on $\omega$ by $K$ and then rotating. In symbols

Substituting in equation (14)

or

It only remains to consider $\delta K_s$. $K$ is a function of the pure strain of the medium and $\delta K_s$ is the increment in $K$ due to the increment in pure strain owing to $\delta \eta$. Calling this increment of pure strain $\delta \psi$ so that by equation (3) Section 7 above

we have

Now by equations (1) and (3) Section IV,

so that $w$ is a function of the independent variables $E$, $\psi$ (because $K$ is a function of $\psi$). Therefore

This equation might have been deduced at once thus

but equation (17) is itself of importance so the above proof is preferable.

Thus finally from equation (15) we get

We therefore have for $\varphi$ in equations (10) and (11) Section IV,

This is a self-conjugate function so that as we saw in Section IV it is a stress which serves to explain forces both throughout the volume of the dielectric and over any surfaces of discontinuity in $D$ or $E$²⁹.

14. The force in particular cases

41. Let us first consider that part $-VD\Delta E∕2$ of the force $($equations (10) Section IV and (21) Section IV$)$ which does not depend on the variation of $K$ with the shape of the body.

Suppose our dielectric is homogeneous and electrically isotropic so that $K$ is a simple constant scalar. In this case

by equations (1) and (4) Section IV. Therefore by equations (5) and (6) Section IV,

From these we at once get by the theory of potential that

From this we know by the theory of potential that at the surface where the charge $\sigma$ resides $\nabla v$ is discontinuous only with regard to its normal component and at all other points is continuous. Thus

and by equation (24) $x=4\pi \sigma ∕K$ so that

Now the force $F$ per unit volume is

and the force per unit surface $F_s$ is by equation (11) Section IV,

whence by equation (26)

Thus we see that Maxwell’s theory as given in Section IV above reduces to the ordinary theory when $K$ is a single scalar. In fact two particles containing charges $ee\prime$ apparently repel one another with a force $ee\prime ∕K{r}^2$ where $r$ is the distance between them, for by equations (25), (27) and (28) the force in any charged body is that due to a field of potential $v$ given by

42. If the medium when strained remain electrically isotropic $\psi DK$ as well as $K$ must be a simple scalar. Thus with Thomson and Tait’s notation for strain, which makes the coordinates of $\psi$, $e$, $f$, $g$, $a∕2$, $b∕2$, $c∕2$ we have

Therefore $K$ is a function of $e+f+g$ only, i.e. of the density ($m$) of the medium. Thus because

we get $\psi DK=-dK∕dlogm=-k$ suppose. Hence

Thus the force $-\psi Dw\Delta$ [equations (21) Section IV and (10) Section IV] resulting from the change of $K$ with pure strain is in the case we are now considering

and is³⁰ therefore, according as $k$ is positive or negative, in the direction of or that opposite to that of the most rapid increase of the square of the electro-motive force. Thus even in the case of a fluid dielectric which has no internal charge but which forms part of a non-uniform field of (electro-motive) force the surfaces of equal pressure and therefore the free surface will if originally plane no longer remain so.

Nature of the stress

Nature of the Stress

43. We have seen that the stress which serves to explain the electrostatic forces is that given by equation (21) Section IV, viz. 

Let us first consider the part $-VD\omega E∕2$ which does not depend on the variation of $K$. Putting $\omega$ first $=UD$ and then $=UE$ we get

Therefore putting $\omega$ first $=$ any multiple of $UD+UE$ and then $=$ any multiple of $UD-UE$ we get

Lastly, since

we see that if we put $\omega =$ any multiple of $VDE$

Thus we see that the stress now considered is a tension along one of the bisectors of $D$ and $E$ (the bisector of the positive directions or the negative directions of both) $=TDTE∕2$, an equal pressure along the other bisector and a pressure $=w$ perpendicular to both these directions. When $D$ is parallel to $E$ this at once reduces to Maxwell’s case, viz. a tension in the direction of $E$ and a pressure in all directions at right angles each $=w$.

44. We have now to consider the other part of the stress, viz.

If we assume that $K$ is a function of the density ($m$) of the medium only we shall have

say, and

as in Section 14. Here however $k$ is not in general a mere scalar but a self-conjugate linear vector function. We have then in this case

which is a hydrostatic pressure or an equal tension in all directions according as $SEkE$ is positive or negative. In this case the 36 coordinates of $\psi D_1\omega ·K_1$ reduce to the 6 of $k$ for each point of space.

A more general assumption is that $\delta K_s$ (Section IV) depends only on the elongations in the directions of the principal axes of $K$. Taking $i$, $j$, $k$ as unit vectors in these directions we again have

and similarly for $j$ and $k$, so that the principal axes of the stress now considered are the principal axes of $K$.

45³¹. The most natural simple assumption for solid dielectrics seems to me to be that the medium is electrically isotropic before strain, and also isotropic with regard to the strain in the sense that if the strain be, so to speak, merely rotated, $\delta K_s$ will suffer exactly the same rotation. We may treat this problem exactly as we did (Section 11) that of stress in terms of strain for an isotropic solid. Thus splitting up $\delta \psi$ into its principal elongations, i.e. putting

we shall get, as in Section 11,

But $\delta K_s\omega =-S\delta \psi \zeta \psi D_1\zeta ·K_1\omega$ by equation (16) Section IV, so that from equation (31)

whence we see by Section 2 above that

which consists of a pressure in the directions of the lines of force $=-\alpha E^2∕8\pi$ and another pressure in all directions at right angles $=-\beta E^2∕8\pi$.

Magnetism—magnetic potential, force, induction

Magnetism
beginalign*2ex] Magnetic potential, force, induction

46. We now go on to the ordinary theory of magnetism; and here we shall merely follow Maxwell in his General Theory, so as to give an opportunity of comparing Quaternion proofs with Cartesian, as we have already done in Elasticity.

We shall not consider in detail the effect of one small magnet upon another, as this has already been done by Tait. In connection with this I am content to remark that I think the treatment of this problem can be made somewhat simpler than Tait’s by means of potential.

Suppose we have a pole $-m$ at $O$ and a pole $+m$ at $O\prime$ where $OO\prime$ is small. Calling the vector from $O$ to $O\prime \overline{OO\prime }$, let us call the vector $m\overline{OO\prime }\mu$, so that $\mu$ is the vector magnetic moment of the magnet. The potential of $-m$ at any point $P$ is $-mu$, where $u$ as usual $=P{O}^{-1}$. Similarly the potential of $+m$ is $mu\prime$, where $u\prime =PO{\prime }^{-1}$. Therefore the potential of the magnet

where of course $P$ is the variable point implied by $\nabla$. Thus the potential of a small magnet $\mu$ at any point $=S\mu \nabla u$.

Hence the potential of any magnet whose magnetic moment per unit volume at any point is $I$ is

according to the convention of Section 5 above. By equation (9) Section 4 this may be put

which shews that we may consider it due to a volume density $S\nabla I$ and a surface density $-SIUd\Sigma$³² of magnetic matter, the surface density occurring wherever there is discontinuity in $I$.

By again considering the poles $m$ and $-m$ of the small magnet $\mu$ we see that its potential energy when placed in a field of magnetic potential $\Omega$ is $-S\mu \nabla \Omega$, whence just as we obtained equation (33) we now see that the potential energy ($W$) of any magnet in such a field is

by equation (9) Section 4 above, so that the potential energy is just the same as it would be for the imaginary distribution of magnetic matter.

47. The force ($H$) on a unit magnetic pole at any point external to the magnet is given by

where

Thus we see that for all external points $H=\nabla A$, so that $A$ is called the vector magnetic potential. [It is to be observed that since $\nabla A=H+4\pi I=$ a vector, $S\nabla A=0$.] $\nabla A$ is called the magnetic induction and for it we use the single symbol $B$ so that

and also by the equation for $H$ just given

This is not the way in which Maxwell defines the magnetic force and induction, but he shews quite simply (Elect. and Mag. §§ 398–9) that his definition and the present one are identical. This can be shewn as easily without analysis at all.

48. Where $I$ is discontinuous both $H$ and $B$ are also discontinuous. From the surface density view we gave in equation (34) we see that, just as we have the expression, given in Section 14 for $\nabla v_a-\nabla v_b$, so now

so that the discontinuity in $H$ is entirely normal to the surface of discontinuity. Further from this equation we have

so that the discontinuity in $B$ is entirely tangential.

From this equation we see that for any closed surface whatever whether it include surfaces of discontinuity in $I$ or not

For adding these surfaces to the boundary of the inclosed space, in accordance with Section 4 above, we see by equation (42) that they contribute zero to the surface integral; but the total surface integral is by equation (9) Section 4 $\iiint S\nabla Bd\mathfrak{𝔰}=0$ by equation (39).

Magnetic solenoids and shells

Magnetic Solenoids and Shells

49. A magnet is said to be solenoidal if the imaginary magnetic matter of equation (34) is entirely on the surface. Thus for a solenoidal distribution

In this case the potential is by equation (34)

50. A simple magnetic shell is defined as a sheet magnetised everywhere normally to itself and such that, at any point, the magnetic moment per unit surface is a constant called the strength of the shell.

Calling the strength $\varphi$ we have for the potential energy at any point by equation (33)

Now $-Sd\Sigma \nabla u$ is the solid angle subtended by the element $d\Sigma$ at the point considered, so that

Thus if $P$ be a point on the positive side of the shell and $P\prime$ a point infinitely near $P$ but on the negative side

or what comes to the same thing

This integral may be taken along any path, e.g. along a path which nowhere cuts the shell. The same integral is true if $H$ be the magnetic force due to a whole field of which the shell is only one of several causes, for the part contributed by the rest of the field is zero on account of the infinite proximity of $P$ and $P\prime$. For future use in electro-magnetism observe that this statement cannot be made if for $H$ in the integral be substituted $B$.

51. The condition that any magnet can be divided up into such shells is at once seen to be that $I$ can be put in the form

where $\varphi$ is some scalar.

In this case the potential is by equation (33)

Remarking that the solid angle again occurs here it is needless to interpret the equation further. By equation (37) we have for the vector potential

or by equation (9) Section 4

52. The potential energy of a magnetic shell of strength $\varphi$ placed in a field of potential $\Omega$ is of importance. We see by equation (35) that it is

If then the magnets which cause $\Omega$ do not cut the shell anywhere, so that $-\nabla \Omega =\nabla A$, we shall have

by equation (8) Section 4.

Suppose now that $A$ is caused by another shell of strength $\varphi \prime$. Then by equation (49)

by equation (8) Section 4. Thus finally the potential energy $M$ of these two shells is given by

53. The general theory of induced magnetism when once the proposition (given in equation (42) Section 14) that is zero is established, is much the same whether treated by Quaternion or Cartesian notation. We shall therefore not enter into this part of the subject.

Electro-magnetism—general theory

Electro-magnetism
beginalign*2ex] General theory

54. We now propose to prove the geometrical theorems connected with Maxwell’s general theory of Electro-magnetism by means of Quaternions.

We assume the dynamical results of Chaps. V., VI. and VII., and the first six paragraphs of Chap. VIII. of the fourth part of his treatise.

These assumptions amount to the following. Connected with any closed curve in an electro-magnetic field there is a function

where $A$ is some vector function at every point of the field. The function $p$ has the following properties. If any circuit be made to coincide with the curve the generalised force acting upon the electricity in the circuit is

Again, if there be a current of electricity $\gamma$ flowing round this circuit, the generalised force $X$, corresponding to any coordinate $x$ of the position of the circuit due to the field acting upon the conductor, is

55. The first thing to be noticed is that $p$ can be transformed into a surface integral by equation (8) Section 4 above.

Next we see by the fundamental connection (Section IV above) between the e.m.f. $E$ and electricity, that $E$ must equal the line integral of $E$ round the circuit, or

We are now in a position to find $E$ in terms of $A$ and $B$, i.e. of $A$.

56. The rate of variation of $p$ is due to two causes, viz. the variation of the field ($\dot{A}$) and the motion of the circuit ($\dot{\rho }$). In the time $\delta t$ then there will be an increment $\delta A$ in $A$ and an increment $\delta d\Sigma$ in $d\Sigma$ to be considered. Thus

[This amounts to assuming that $\int \; <!--nolimits-->S\delta Ad\rho =\iint S\delta Bd\Sigma$, which of course is true by equation (8) Section 4.] Now when the circuit changes slightly we may suppose the surface over which the new integral extends to coincide with the original surface and a small strip at the boundary traced out by the motion ($\dot{\rho }\delta t$) of the boundary. Thus $\delta d\Sigma$ is zero everywhere except at the boundary and there it

57. We now come to the mechanical forces exerted on an element through which a current $C$ per unit volume flows.

We see by equation (55) that the work done by the mechanical forces on any circuit through which a current of magnitude $\gamma$ flows in any small displacement of the circuit equals $\gamma \times$ the increment in $p$ caused by the displacement. Give then to each element $d\rho$ of the circuit an arbitrary small displacement $\delta \rho$ and let $F\prime$ be the mechanical force exerted by the field upon the element. Thus as in last section

Thus the force $F\prime$ on the element $d\rho$ is $\gamma Vd\rho B$. But we may suppose this element to be an element $d\mathfrak{𝔰}$ of volume through which the current $C$ flows. Thus if for $\gamma d\rho$ we write $Cd\mathfrak{𝔰}$, and for $F\prime$, $Fd\mathfrak{𝔰}$, where $F$ is the force per unit volume exerted by the field, we get

58. So far we have been able to go by considering the electric field as a mechanical system, but to go further (as Maxwell points out) and find how $B$ or $A$ depends on the distribution of current and displacement in the field we must appeal to experiment. It has been shewn by experiment that a small circuit produces exactly the same mechanical effects on magnets as would a small magnet, at the same point as the circuit, placed with its positive pole pointing in the direction of the positive normal to the plane of the circuit when the positive direction round the circuit is taken as that of the current³³. Moreover the magnetic moment of the magnet which must be placed there is proportional to the strength of the current $\times$ the area of the circuit. Further, the effect of this circuit upon other such small circuits is the same as the mutual effects of corresponding magnets. We have now only to consider a finite circuit as split up in the usual way into a number of elementary circuits to see that a finite circuit will act upon magnets or upon other circuits exactly like a magnetic shell of strength proportional to the strength of the current and boundary coinciding with that of the circuit. The unit current in the electro-magnetic system is so taken as to make this proportionality an equality.

The one difference between the circuit and the magnetic shell is that there is no discontinuity in the magnetic potential in going round the circuit, so that by Section 14 above the line integral of $H$ round the circuit will be $4\pi \times$ the strength of the current. In symbols

for any curve, so that by equation (8) Section 4

which of course is a direct result of our original assumption that electricity moves like an incompressible fluid. Maxwell tacitly assumes this by making the assumption that only one coordinate is required to express the motion of electricity in a circuit.

59. We are now in a position to identify the $B$ we are now using with the magnetic induction for which we have already used the same symbol.

We see by equation (50) Section 14 that the mechanical force on the shell corresponding to any coordinate $x$ is

where $B\prime$ is the magnetic induction; and by equations (53) and (55) that the force on the corresponding electric circuit is

therefore $B=B\prime$ wherever there is no magnetism. And where there is magnetism $B$ is not $=H$ for $S\nabla B=0$, as we have seen. Thus $B=B\prime$ at all points. In other words the two vectors are identical and we are justified in using the same symbol for the two.

This practically ends the general theory of electro-magnetism. We content ourselves with one more application of Quaternions in this subject. We give it because it exhibits in a striking manner the advantages of Quaternion methods.

Electro-magnetic stress

Electro-magnetic phenomena explained by Stress

60. In Section 14 equation (35) we have seen that the potential energy of a magnetic element $=Id\mathfrak{𝔰}$ in a magnetic field is $SIHd\mathfrak{𝔰}$ when $H$ has a potential. Maxwell assumes that the same expression is true whether $H$ have a potential or not. Assuming this point³⁴ with him we can find the force and couple acting on the medium and a stress which will produce that force and couple. The force and couple due to the magnetism of an element is obtained by giving the element an arbitrary translation and rotation and assuming that the work done by this force and couple = the decrement in the potential energy of the element. Thus the force per unit volume is $-{\nabla }_{1}S{IH}_1$ for the decrement in the potential energy due to a small translation $\delta \rho$ is $S\delta \rho {\nabla }_{1}S{IH}_1$. Similarly the couple $M$ is given by

for the decrement $-SM\delta \omega$ in the potential energy due to a small rotation $\delta \omega$ is $-SV\delta \omega I·H=-S\delta \omega VIH$. The total force $F$ per unit volume is the sum of that just given and that given by equation (61), so that

Therefore by equation (62)

Now $S\nabla B=0$ so that $H_{1}S{\nabla }_{1}B=HS\Delta B$, and again

From these two results $($equations (65) and (67)$)$ we see by Section 7 above that the stress $\varphi$ will produce all the mechanical effects of the field.

This stress, as can be seen by giving $\omega$ the required values in equation (66), is one of pressure $-H^2∕8\pi$ in all directions at right angles to $B$ and of tension in the direction of $H$. When there is no magnetism $H=B$ so that this pressure and tension become equal and their directions at right angles to and along $B$ respectively. In fact we then have