On the Magnetic Effects produced by Motion in the Electric Field

On the Magnetic Effects produced by Motion in the Electric Field  (1889)  by Joseph John Thomson

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I. On the Magnetic Effects produced by Motion in the Electric Field.

By J. J. Thomson, M.A., F.R.S

In the April number of the Philosophical Magazine Mr. Heaviside discusses the question of a moving electrified sphere, and while agreeing with the results I obtained some time ago as to the magnetic force produced by such a sphere when moving slowly, differs as to the numerical magnitude of the energy possessed by the sphere and the forces acting upon it when placed in a magnetic field. The latter quantities, however, do not depend merely upon the alterations caused by the motion of the sphere in the polarization of the dielectric surrounding the sphere, but also upon the boundary conditions we adopt and upon the view we take of the motion close to the sphere of the medium in which the electric displacements occur.

This will be seen in the course of the following investigation, in which I have endeavoured to take into account the motion of the medium in which the displacements occur. I find that, in order to close the circuits in this case, it is necessary to assume effects which, as far as I know, have not been noticed.

Let us consider the case when the electric field is that due to a charged sphere moving parallel to the axis of $z$ with the velocity $w_0$, the components at $(x, y, z)$ of the velocity of the medium being $u, v, w$; then, if we supposed that the displacement -currents are due entirely to variations in the electric displacement $(f, g, h)$ caused by the motion of the sphere and the medium, the components $\xi_{1},\eta_{1},\zeta_{1}$ of these currents would be given by

$\begin{array}{c} \xi_{1}=u\frac{df}{dx}+v\frac{df}{dy}+\left(w-w_{0}\right)\frac{df}{dz},\\ \\ \eta_{1}=u\frac{dg}{dx}+v\frac{dg}{dy}+\left(w-w_{0}\right)\frac{dg}{dz},\\ \\ \zeta_{1}=u\frac{dh}{dx}+v\frac{dh}{dy}+\left(w-w_{0}\right)\frac{dh}{dz}, \end{array}$

These values, however, do not satisfy the equation

$\frac{d\xi_{1}}{dx}+\frac{d\eta_{1}}{dy}+\frac{d\zeta_{1}}{dz}=0$

unless the dielectric is moving uniformly; so that, if the circuits are to be closed, the motion of the medium must produce some other effect analogous to a current.

Since

$\begin{array}{l} \frac{d\xi_{1}}{dx}+\frac{d\eta_{1}}{dy}+\frac{d\zeta_{1}}{dz}=\frac{d}{dx}\left(f\frac{du}{dx}+g\frac{du}{dy}+h\frac{du}{dz}\right)\\ \\ \qquad+\frac{d}{dy}\left(f\frac{dv}{dx}+g\frac{dv}{dy}+h\frac{dv}{dz}\right)+\frac{d}{dz}\left(f\frac{dw}{dx}+g\frac{dw}{dy}+h\frac{dw}{dz}\right), \end{array}$

we see that the currents will be closed if we add on to the components $\xi_{1},\eta_{1},\zeta_{1}$ the components $\xi_{0},\zeta_{0},\rho_{0}$, where

$\begin{array}{c} -\xi_{0}=f\frac{du}{dx}+g\frac{du}{dy}+h\frac{du}{dz},\\ \\ -\eta_{0}=f\frac{dv}{dx}+g\frac{dv}{dy}+h\frac{dv}{dz},\\ \\ -\zeta_{0}=f\frac{dw}{dx}+g\frac{dw}{dy}+h\frac{dw}{dz}, \end{array}$

The medium is assumed to be incompressible, so that

$\frac{du}{dx}+\frac{dv}{dy}+\frac{dw}{dz}=0$

Hence the components of the total effective currents are

$\begin{array}{l} \xi+\xi_{0}=\frac{d}{dy}(vf-ug)-\frac{d}{dz}\left(uh-\left(w-w_{0}\right)f\right),\\ \\ \eta+\eta_{0}=\frac{d}{dz}\left(\left(w-w_{0}\right)g-vh\right)-\frac{d}{dx}\left(vf-ug\right),\\ \\ \xi+\xi_{0}=\frac{d}{dx}\left(uh-\left(w-w_{0}\right)f\right)-\frac{d}{dy}\left(\left(w-w_{0}\right)g-vh\right). \end{array}$

If the motion of the medium is irrotational, these conditions will be satisfied if we suppose that the motion of the dielectric gives rise to magnetic forces whose components $\alpha,\beta,\gamma$ are given by the equations

$\left.\begin{array}{l} \alpha=4\pi\left(\left(w-w_{0}\right)g-vh\right),\\ \beta=4\pi\left(uh-\left(w-w_{0}\right)f\right),\\ \gamma=4\pi(vf-ug). \end{array}\right\}$

(1)

If we suppose that the electric field is due to a number of charged spheres moving with velocities $\left(u_{1},v_{1},w_{1}\right)\left(u_{2},v_{2},w_{2}\right)$ .... respectively, and producing electric displacements whose components are $\left(f{}_{1},g_{1},h_{1}\right)\left(f_{2},g_{2},h_{2}\right)$ the component of the magnetic force parallel to $x$ will be

$4\pi\left(wg-vh-\left\{ w_{1}g_{1}+w_{2}g_{2}+\dots-v_{1}h_{1}-v_{2}h_{2}\dots\right\} \right)$

where $f, g, h$ are the resultant displacements.

Thus, since in the general case when the aether is in motion the assumption that the currents are merely due to the changes in the polarization caused by the aether moving from a place where the displacement has one value to another where it has a different one is insufficient if the circuits are closed, it is necessary to replace it by another; the assumption we shall adopt is that the motion of the polarized aether sets up magnetic forces whose components are given by equations (1).

When the aether is at rest this agrees with Maxwell's principle that the currents are equal to the rate of increase of the electric displacement. We should get these magnetic forces if, in the expression for the mean Lagrangian function of unit volume of the moving aether, there was the term

$\begin{array}{r} a\left\{ wg-vh-\sum\left(w_{1}g_{1}-v_{1}h_{1}\right)\right\} +b\left\{ uh-wf-\sum\left(u_{1}h_{1}-w_{1}f_{1}\right)\right\} \\ \\ +c\left\{ vf-ug-\sum\left(v_{1}f_{1}-u_{1}g_{1}\right)\right\} , \end{array}$

where $a, b, c$ are the components of the magnetic induction.

This term would show that there is an electromotive force parallel to $x$ equal to

$cv-bw$

and a mechanical force equal to

$c\frac{dg}{dt}-b\frac{dh}{dt}+g\frac{dc}{dt}-h\frac{db}{dt},$

if the electrified bodies are at rest.

The first of these corresponds to the well-known expression for the electromotive force on a conductor moving in a magnetic field; the second is the mechanical force on a current in a magnetic field plus the term $g\tfrac{dc}{dt}-h\tfrac{db}{dt}$.

We can deduce an important consequence of the assumption, if we consider the case of the aether moving with uniform velocity between two parallel planes charged, the one with positive, the other with negative electricity.

If $v$ is the velocity of the aether, $h$ the electric displacement at right angles to the planes, the magnetic force between the planes will be parallel to $x$, and equal to $-4\pi vh$; or if $\sigma$ is the surface-density of the electrification on the planes $-4\pi v\sigma$, the magnetic force vanishes except between the planes, so that on crossing the positively electrified surface there is an increase in the magnetic force parallel to $x$ equal to $4\pi v\sigma$. Thus the charged surface acts like a current sheet of intensity $-\sigma v$, but $-v$ is the velocity of the plane relatively to the aether; so that a charged surface moving with velocity $v$ relatively to the aether must act like a current sheet of intensity $\sigma v$.

We will now proceed to apply these results to some special cases. Let us suppose that we have a charged sphere moving along the axis of $z$ with the velocity $w_{0}$, and that it sets the aether around it in motion in the same way as an incompressible fluid is set in motion by a solid sphere of the same radius moving through it with the same velocity. If $a$ is the radius of the sphere,

$\begin{array}{l} u=\frac{1}{2}w_{0}a^{3}\frac{d^{2}}{dx\ dz}\frac{1}{r},\\ \\ v=\frac{1}{2}w_{0}a^{3}\frac{d^{2}}{dy\ dz}\frac{1}{r},\\ \\ w=\frac{1}{2}w_{0}a^{3}\frac{d^{2}}{dx^{2}}\frac{1}{r}, \end{array}$

$f=-\frac{e}{4\pi}\frac{d}{dx}\frac{1}{r},\ g=-\frac{e}{4\pi}\frac{d}{dy}\frac{1}{r},\ h=-\frac{e}{4\pi}\frac{d}{dz}\frac{1}{r};$

hence by equations (1),

$\begin{array}{l} \alpha=ew_{0}\left(1+\frac{1}{2}\frac{a^{3}}{r^{3}}\right)\frac{y}{r^{3}},\\ \\ \beta=-ew_{0}\left(1+\frac{1}{2}\frac{a^{3}}{r^{3}}\right)\frac{x}{r^{3}},\\ \\ \gamma=0. \end{array}$

Thus the lines of magnetic force are circles with their centres along and their planes at right angles to the axis of $z$.

At a distance from the centre large compared with the radius of the sphere the magnetic force is the same as that due to a current $ew_{0}$, but close to the sphere the relative motion of the sphere and aether causes it to be larger than this, and at the surface of the sphere it is the same as that due to a current $\tfrac{3}{2}ew_{0}$.

The energy due to this distribution of currents is $\tfrac{3}{7}\tfrac{e^{2}w_{0}^{2}}{a}\mu$.

Another case which can be easily solved is that of a right circular cylinder rotating with an angular velocity $\omega$, each unit length of the cylinder being charged with $\mathrm{E}$ units of electricity. If $a$ is the radius of the cylinder,

$\begin{array}{lll} u=-\omega\frac{a^{2}y}{r^{2}}, & & v=\omega\frac{a^{2}x}{r^{2}},\\ \\ f=\frac{\mathrm{E}}{2\pi}\frac{x}{r^{2}}, & & g=\frac{\mathrm{E}}{2\pi}\frac{y}{r^{2}}; \end{array}$

and by equations (1).

$\alpha=0,\ \beta=0,\ \gamma=-2\mathrm{E}\omega\frac{a^{2}}{r^{2}}$

Thus outside the rotating cylinder there is a magnetic force parallel to the axis of rotation.

If we assume that the aether outside the sphere is at rest, we can find the solution of the case of a charged metal sphere executing harmonic oscillations. Suppose the sphere to be moving parallel to the axis of $z$. the velocity at any time $t$ being represented by the real part of $\omega\epsilon^{ipt}$. Then if we take rectangular axes passing through the centre of the sphere and moving with it, the following equations are true inside the sphere if $u, v, w$ are the components of the current, $a, b, c$ those of magnetic induction, $\psi$ the electrostatic potential, $\mathrm{F,G,H}$ the components of the rector potential, and $\sigma$ the specific resistance of the metal.

$\left.\begin{array}{ll} \sigma u=-b\omega\epsilon^{ipt} & -\frac{d\mathrm{F}}{dt}-\frac{d\psi}{dx},\\ \\ \sigma v=-a\omega\epsilon^{ipt} & -\frac{d\mathrm{G}}{dt}-\frac{d\psi}{dy},\\ \\ \sigma w= & -\frac{d\mathrm{H}}{dt}-\frac{d\psi}{dz}; \end{array}\right\}$

(2)

and therefore

$\begin{array}{ll} -\frac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F}=-b\omega\epsilon^{ipt} & -\frac{d\mathrm{F}}{dt}-\frac{d\psi}{dx},\\ \\ -\frac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{G}=a\omega\epsilon^{ipt} & -\frac{d\mathrm{G}}{dt}-\frac{d\psi}{dy},\\ \\ -\frac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{H}= & -\frac{d\mathrm{H}}{dt}-\frac{d\psi}{dz}; \end{array}$

In the dielectric outside the sphere, if $f, g, h$ are the electric displacements, $\mathrm{K}$ the specific inductive capacity, and if $\partial/\partial t$ denote partial differentiation with respect to the time, the equations are

$\begin{array}{l} \frac{4\pi}{\mathrm{K}}f=-\frac{d\mathrm{F}}{dt}-\frac{d\psi}{dx}=-\left(\frac{\partial}{\partial t}-\omega\epsilon^{ipt}\frac{d}{dz}\right)\mathrm{F}-\frac{d\psi}{dx},\\ \\ \frac{4\pi}{\mathrm{K}}g=-\frac{d\mathrm{G}}{dt}-\frac{d\psi}{dy}=-\left(\frac{\partial}{\partial t}-\omega\epsilon^{ipt}\frac{d}{dz}\right)\mathrm{G}-\frac{d\psi}{dy},\\ \\ \frac{4\pi}{\mathrm{K}}h=-\frac{d\mathrm{H}}{dt}-\frac{d\psi}{dz}=-\left(\frac{\partial}{\partial t}-\omega\epsilon^{ipt}\frac{d}{dz}\right)\mathrm{H}-\frac{d\psi}{dz}; \end{array}$

and therefore

$-\frac{1}{\mu K}\nabla^{2}\mathrm{F}=-\left(\frac{\partial}{\partial t}-\omega\epsilon^{ipt}\frac{d}{dz}\right)^{2}\mathrm{F}-\left(\frac{\partial}{\partial t}-\omega\epsilon^{ipt}\frac{d}{dz}\right)\frac{d\psi}{dx},$

(3)

with a similar equation for $\mathrm{G}$.

From the form of these equations we see that the solution will take the form

$\begin{array}{l} \psi=\psi_{0}+\psi_{1}\epsilon^{ipt}+\psi'_{1}\epsilon^{-ipt}+\psi_{2}\epsilon^{2ipt}+\psi'_{2}\epsilon^{-2ipt}+\dots\\ \\ \mathrm{F}=\mathrm{F}_{1}\epsilon^{ipt}+\mathrm{F}'_{1}\epsilon^{-ipt}+\mathrm{F}_{2}\epsilon^{2ipt}+\mathrm{F}'_{2}\epsilon^{-2ipt}. \end{array}$

If we substitute these value? in the above equations, we see that we may put $\psi'_{1}\psi'_{2}\dots\mathrm{F}'_{1},\ \mathrm{F}'_{2}$ all equal to zero.

If $e$ is the quantity of electricity on the sphere,

$\psi_{0}=\frac{e}{\mathrm{K}r}$

Equating the coefficients of $\epsilon^{ipt}$ to zero in equations (2) and (3) we have

$-\frac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F}_{1}=-\frac{d\mathrm{F}_{1}}{dt}-\frac{d\psi_{1}}{dx}$ inside the sphere,

with similar equations for $\mathrm{G}_{1}$ and $\mathrm{H}_{1}$; outside the sphere we have

$-\frac{1}{\mu\mathrm{K}}\nabla^{2}\mathrm{F}_{1}=-\frac{\partial^{2}\mathrm{F}_{1}}{\partial t^{2}}-\frac{d^{2}\psi_{1}}{dt\ dx}+\frac{\omega e}{\mathrm{K}}\frac{d^{2}}{dx\ dz}\frac{1}{r}$

(4)

with similar equations for $\mathrm{G}_{1}$ and $\mathrm{H}_{1}$.

The form of equation (4) suggests that we should put

$\psi_{1}=B\frac{d}{dz}\frac{1}{r}$

A particular integral of (4) is then

$\mathrm{F}_{1}=\frac{c\mathrm{B}}{p}\frac{d^{2}}{dx\ dz}\frac{1}{r}-\frac{\omega e}{\mathrm{K}p^{2}}\frac{d^{2}}{dx\ dz}\frac{1}{r}$

The complementary function is that solution of the differential equation

$\nabla^{2}\mathrm{F}_{1}+\mu\mathrm{K}p^{2}\mathrm{F}_{1}=0$

which, when considered as a function of the angular coordinates of a point, varies as $r^{3}\tfrac{d^{2}}{dx\ dz}\tfrac{1}{r}$; this (see Proc. Math. Soc. vol. xv. p. 212) is

$\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\frac{d^{2}}{dx\ dz}\frac{1}{r}$

where

$\mathrm{E}_{2}(i\lambda r)=\frac{3\epsilon^{-i\lambda r}}{(i\lambda r)^{3}}+\frac{3\epsilon^{-i\lambda r}}{(i\lambda r)^{2}}+\frac{\epsilon^{-i\lambda r}}{i\lambda r}$

and

$\lambda^{2}=\mu\mathrm{K}p^{2}$

Thus, outside the sphere,

$\begin{array}{l} \mathrm{F}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\frac{d^{2}}{dx\ dz}\frac{1}{r}+\frac{i\mathrm{B}\epsilon^{pt}}{p}\frac{d^{2}}{dx\ dz}\frac{1}{r}-\frac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\frac{d^{2}}{dx\ dz}\frac{1}{r},\\ \\ \mathrm{G}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\frac{d^{2}}{dy\ dz}\frac{1}{r}+\frac{i\mathrm{B}\epsilon^{pt}}{p}\frac{d^{2}}{dy\ dz}\frac{1}{r}-\frac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\frac{d^{2}}{dy\ dz}\frac{1}{r},\\ \\ \mathrm{H}_{1}=\mathrm{C}\mathrm{E}_{2}(i\lambda r)r^{3}\frac{d^{2}}{dz^{2}}\frac{1}{r}-2\mathrm{C}\mathrm{E}_{2}(i\lambda r)+\frac{i\mathrm{B}}{p}\epsilon^{pt}\frac{d^{2}}{dz^{2}}\frac{1}{r}-\frac{\omega e\epsilon^{pt}}{\mathrm{K}p^{2}}\frac{d^{2}}{dz^{2}}\frac{1}{r}, \end{array}$

where $\mathrm{E}_{0}(i\lambda r)=\epsilon^{-i\lambda r}/i\lambda r$, and is introduced into the expression for $\mathrm{H}$ to make

$\frac{d\mathrm{F}}{dx}+\frac{d\mathrm{G}}{dy}+\frac{d\mathrm{H}}{dz}=0$

Inside the sphere the differential equations for $\mathrm{F}_{1}$ and $\mathrm{G}_{1}$ are of the form

$-\frac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F}_{1}=-ip\mathrm{F}$

if $\lambda_{1}^{2}=-4\pi\mu'ip/\sigma$, the solution of this equation is

$\mathrm{F}_{1}=\mathrm{D}\epsilon^{ipt}\mathrm{S}_{2}\left(\lambda_{1}r\right)r^{3}\frac{d^{2}}{dx\ dz}\frac{1}{r}$

where

$\mathrm{S}_{2}\left(\lambda_{1}r\right)=\frac{3\sin\lambda_{1}r}{\left(\lambda_{1}r\right)^{3}}-\frac{3\cos\lambda_{1}r}{\left(\lambda_{1}r\right)^{2}}-\frac{\sin\lambda_{1}r}{\lambda_{1}r}$

Similarly

$\mathrm{G}_{1}=\mathrm{D}\epsilon^{ipt}\mathrm{S}_{2}\left(\lambda r\right)r^{3}\frac{d^{2}}{dx\ dz}\frac{1}{r}$

the differential equation for $\mathrm{H}_{1}$ is

$-\frac{\sigma}{4\mu'}\nabla^{2}\mathrm{H}_{1}=-ip\mathrm{H}_{1}+\frac{\mathrm{B}}{a^{3}}$

So that

$\mathrm{H}_{1}=\mathrm{D}\epsilon^{ipt}\mathrm{S}_{2}\left(\lambda_{1}r\right)r^{3}\frac{d^{2}}{dz^{2}}\frac{1}{r}+2\mathrm{D}\epsilon^{ipt}\mathrm{S}_{0}\left(\lambda_{1}r\right)+\frac{\mathrm{B}}{ipa^{3}}\epsilon^{ipt}$

where

$\mathrm{S}_{0}\left(\lambda_{1}r\right)=\frac{\sin\lambda_{1}r}{\lambda_{1}r}$

and is introduced to make

$\frac{d\mathrm{F}}{dx}+\frac{d\mathrm{G}}{dy}+\frac{d\mathrm{H}}{dz}=0$

Since $\mathrm{F}_{1},\mathrm{G}_{1},\mathrm{H}_{1}$ are continuous when $r=a$, if $a$ is the radius of the sphere we have

$\left.\begin{array}{c} \mathrm{C}\mathrm{E}_{2}(i\lambda a)-\mathrm{DS}_{2}\left(\lambda_{1}a\right)+\frac{i\mathrm{B}}{pa^{3}}=\frac{\omega e}{p^{2}a^{3}k},\\ \\ -2\mathrm{C}\mathrm{E}_{0}(i\lambda a)-2\mathrm{DS}_{2}\left(\lambda_{1}a\right)+\frac{i\mathrm{B}}{pa^{3}}=0. \end{array}\right\}$

(5)

Since, on the assumption discussed above, the electrification on the surface of the moving sphere is equivalent to a tangential current-sheet whose intensity is $\omega\epsilon^{ipt}\sin\theta\tfrac{e}{4\pi a^{2}}$, we have as another surface-condition that the difference between the magnetic force outside and inside $=\omega\epsilon^{ipt}\sin\theta\frac{e}{a^{2}}$; hence

$-3\mathrm{C}\frac{d}{da}\mathrm{E}_{0}(i\lambda a)-3\mathrm{D}\frac{dS_{0}(\lambda_{1}a)}{da}=\frac{\omega e}{a^{2}}$

From (5) we have

$\mathrm{C}\left(\mathrm{E}_{2}(i\lambda a)+2\mathrm{E}_{0}(i\lambda a)\right)-\mathrm{D}\left(\mathrm{S}_{2}(\lambda_{1}a)+2\mathrm{S}_{0}(\lambda_{1}a)\right)=\frac{\omega e}{p^{2}a^{3}\mathrm{K}}$

From these equations we have

$\mathrm{C}=\frac{1}{\Delta}\left\{ \frac{\omega e}{\mathrm{K}p^{2}a^{3}}\frac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)-\frac{1}{3}\frac{\omega e}{a^{2}}\left(\mathrm{S}_{2}(\lambda_{1}a)-2\mathrm{S}_{0}(\lambda_{1}a)\right)\right\}$

(6)

$\mathrm{D}=\frac{1}{\Delta}\left\{ -\frac{1}{3}\frac{\omega e}{a^{2}}\left(\mathrm{E}_{2}(i\lambda a)+2E_{0}(i\lambda a)\right)-\frac{\omega e}{\mathrm{K}p^{2}a^{3}}\frac{d}{da}\mathrm{E}_{0}(i\lambda a)\right\}$

(7)

where

$\Delta=\left(\mathrm{E}_{2}(i\lambda a)+2\mathrm{E}_{0}(i\lambda a)\frac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)+\frac{d}{da}\mathrm{E}_{0}(i\lambda a)\left(\mathrm{S}_{2}(\lambda a)-2\mathrm{S}_{0}(\lambda_{1}a)\right)\right)$

Let us first consider the case where $\lambda a$ and $\lambda_{1}$ are both small. In this case $\mathrm{E}_{2}(i\lambda a)$ is large compared with $\mathrm{E}_{0}(i\lambda a)$ and $\mathrm{S}_{2}(\lambda_{1}a)$, very small compared with $\mathrm{S}_{0}(\lambda_{1}a)$; hence we see that

$\begin{array}{l} \mathrm{C}=\frac{\omega e}{\mathrm{K}p^{2}a^{3}}\mathrm{E}_{2}(i\lambda a),\\ \\ \mathrm{D}=-\frac{-\frac{\omega e}{a^{3}}\mathrm{E}_{0}(i\lambda a)}{\mathrm{E}_{2}(i\lambda a)\frac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)}, \end{array}$

Since $\mathrm{E}_{2}(i\lambda a)=\tfrac{3\epsilon^{-i\lambda a}}{(i\lambda a)^{3}}$ approximateiy, and $\mathrm{S}_{0}(\lambda_{1}a)=1$, we have

$\begin{array}{l} \mathrm{C}=-\frac{1}{3}i\omega e\lambda\epsilon^{i\lambda a},\\ \\ \mathrm{D}=\frac{\omega e}{a}\frac{ip\sigma\mathrm{K}}{4\pi}, \end{array}$

The magnetic force outside the sphere parallel to the axis of $x$ equals

$\frac{d\mathrm{H}_{1}}{dy}-\frac{d\mathrm{G}_{1}}{dz},$

or

$\begin{array}{c} -3\mathrm{C}\frac{d}{dy}\mathrm{E}_{0}(i\lambda r)\epsilon^{ipt},\\ \\ =i\omega e\lambda\epsilon^{i\lambda a}\frac{d}{dy}\frac{\epsilon^{-i\lambda a}}{i\lambda r}\cdot\epsilon^{ipt}, \end{array}$

taking the real part

$=\omega e\frac{d}{dy}\frac{\cos\left((pt-\lambda(r-a)\right)}{r}.$

Similarly the magnetic force parallel to the axis of $x$

$=-\omega e\frac{d}{dx}\frac{\cos\left((pt-\lambda(r-a)\right)}{r}.$

and the magnetic force parallel to $z$ vanishes. Thus the magnetic force is the same as that which would be produced by a current-element $\omega e\cos pt$, v being the velocity of the sphere (see Proc. Math. Soc. xv. p. 214).

The magnetic force inside the sphere parallel to $x$ equals

$\begin{array}{c} -3\mathrm{D}\frac{d}{dy}\mathrm{S}_{0}(\lambda_{1}r),\\ \\ =3\mathrm{D}\epsilon^{ipt}\frac{d}{dr}\mathrm{S}_{0}(\lambda_{1}r)\frac{y}{r}. \end{array}$

Substituting the value for $\mathrm{D}$ given by equation (7), this equals

$\frac{\omega e}{a}p^{2}\mathrm{K}y\epsilon^{ipt}$

or, taking the real part and writing $\lambda^{2}$ for $\mathrm{K}p^{2}$,

$\frac{\omega e}{a^{3}}\left(\lambda^{2}a^{2}\right)y\cos pt$

The component parallel to $y$ is

$-\frac{\omega e}{a^{3}}\left(\lambda^{2}a^{2}\right)x\cos pt$

and the $z$-component vanishes. Thus the maximum magnetic force inside the sphere is

$\frac{\omega e}{a^{2}}\cos pt(\lambda a)^{2}$

If $\lambda a$ is very small, this is very small compared with the force outside the sphere. If the velocity is uniform, $p$, and therefore $\lambda=0$, and the magnetic force inside the sphere vanishes. When there is no magnetic force inside the sphere its energy and the force acting upon it have the values assigned to them by Mr. Heaviside.

Let us next take the case where $\lambda a$ is small and $\lambda_{1}a$ large: in this case $\mathrm{C}$ and $\mathrm{D}$ have the same values as before, so that the magnetic force due to the moving sphere is the same.

We must now consider the case where $\lambda a$ and $\lambda_{1}a$ are both large; in this case we find from (6) and (7)

$\mathrm{C}=\frac{\omega e}{\mathrm{K}p^{2}a^{3}}\frac{i\lambda a}{3\epsilon^{-i\lambda a}},$

if $\lambda_{1}a/\lambda a$ and $\lambda_{1}a/\lambda^{2}a^{2}$ are both large.

Thus the magnetic force parallel to $x$ outside

$=\frac{1}{3}\frac{\omega e}{\lambda^{2}a^{2}}\frac{d}{dy}\frac{\cos\left\{ pt-\lambda(r-a)\right\} }{r},$

and that parallel to $y$

$=-\frac{1}{3}\frac{\omega e}{\lambda^{2}a^{2}}\frac{d}{dx}\frac{\cos\left\{ pt-\lambda(r-a)\right\} }{r}.$

Thus, since $\lambda^{2}a^{2}$ is large, the magnetic force, though in the same direction as that due to a current $\omega v$, is very much smaller in magnitude, and fades away to zero as $\lambda a$ increases without limit.

$\mathrm{D}=-\frac{\omega e}{3a^{2}\frac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)},$

The maximum magnetic force inside the sphere

$3\mathrm{D}\frac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)\epsilon^{ipt}=-\frac{\omega e}{a^{2}}\cos pt$

Thus in this case the magnetic force just inside the sphere is equal to $-ve$, while that outside the sphere is very much smaller. This is a striking contrast to the previous cases, where the magnetic force inside the sphere is very small compared with that outside. Thus, in this case, when the time of the oscillation is small compared with that of the electrical oscillations the distribution of magnetic force is turned inside out. The magnetic force diminishes very rapidly as we recede from the surface of the sphere. In this case the total current parallel to the axis of $z$ inside the sphere is finite, for this by equation (2) equals

$\begin{array}{l} -\frac{ip}{\sigma}2\mathrm{D}\int_{0}^{a}\mathrm{S}_{0}(\lambda_{1}r)4\pi r^{2}dr,\\ \\ =\frac{ip}{\sigma}\frac{2\mathrm{D}4\pi}{\lambda_{1}^{2}}\epsilon^{ipt}a^{2}\frac{d}{da}\mathrm{S}_{0}(\lambda_{1}a)\\ \\ =-\frac{2}{3}e\omega\cos pt. \end{array}$

So that if the sphere is placed in a magnetic field the force acting upon it is the same as that on a current $\tfrac{1}{3}e\omega\cos pt$.

When the sphere is moving with a uniform velocity $\omega$, equations (3) become

$-\frac{1}{\mu\mathrm{K}}\nabla^{2}\mathrm{F}=-\omega^{2}\frac{d^{2}\mathrm{F}}{dz^{2}}+\omega\frac{d^{2}\psi}{dx\ dz} ;$

whence

$\frac{d^{2}a}{dx^{2}}+\frac{d^{2}a}{dy^{2}}+\frac{d^{2}a}{dz^{2}}\left(1-\frac{\omega^{2}}{v^{2}}\right)=0$

where $v$ is the velocity of propagation of electrodynamic action through the dielectric. If we put

$z=\left\{ 1-\frac{\omega^{2}}{v^{2}}\right\} ^{\frac{1}{2}}z',$

this equation becomes

$\frac{d^{2}a}{dx^{2}}+\frac{d^{2}a}{dy^{2}}+\frac{d^{2}a}{dz_{1}^{2}}=0$

With similar equations for $b$ and $c$, we see that a solution of these equations is

$\begin{array}{l} a=k\frac{d}{dy}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};\\ \\ b=-k\frac{d}{dx}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};\\ \\ c=0; \end{array}$

where $k$ is a constant. Since, if $\mu=1$,

$4\pi\frac{df}{dt}=\frac{dc}{dy}-\frac{db}{dz},$

and

$\frac{df}{dt}=-\omega\frac{df}{dz},$

we have

$-4\pi\omega f=-b=k\frac{d}{dx}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}}.$

Similarly

$-4\pi\omega g=+a=k\frac{d}{dy}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}},$

and

$-4\pi\omega\frac{dh}{dz}=\frac{db}{dx}-\frac{da}{dy}=k\frac{d}{dz_{1}^{2}}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}};$

so that

$-4\pi wh=k\sqrt{1-\frac{\omega^{2}}{v^{2}}}\frac{d}{dz_{1}}\frac{1}{\sqrt{x^{2}+y^{2}+z_{1}^{2}}}.$

The displacement across any spherical surface must $=e$, so that

$\int(xf+yg+zh)d\mathrm{S}=ae$

and therefore, if $u<v$,

$\frac{k}{2\omega}\int_{0}^{\pi}\frac{\sin\theta\ d\theta}{\left(1-\frac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}=\frac{e}{\left(1-\frac{u^{2}}{v^{2}}\right)^{\frac{3}{2}}} ;$

$k=\omega e\left(1-\frac{u^{2}}{v^{2}}\right)^{-\frac{1}{2}}.$

Thus the lines of magnetic forces are circles round the axis of $z$ and the magnitude of the force equals

$\frac{\omega e\sin\theta\left(1-\frac{\omega^{2}}{v^{2}}\right)}{r^{2}\left\{ 1-\frac{u^{2}}{v^{2}}\sin^{2}\theta\right\} ^{\frac{3}{2}}}$

which is Mr. Heaviside's result. If $\omega>v$, the integral becomes infinite, the displacement will be within a cone of semi-vertical angle $\sin^{-1}\tfrac{v}{\omega}=\beta$; we must therefore only integrate within this cone, and the equation to determine $k$ is,

$\frac{k}{2\omega}\int_{\pi}^{\sin^{-1}\frac{v}{\omega}}\frac{\sin\theta\ d\theta}{\left(1-\frac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}=\frac{e}{\left(1-\frac{u^{2}}{v^{2}}\right)^{\frac{3}{2}}};$

or

$k=\frac{\omega e\left(1-\frac{u^{2}}{v^{2}}\right)^{-\frac{1}{2}}}{\cos\beta}\left(1-\frac{\omega^{2}}{v^{2}}\sin^{2}\beta\right)^{\frac{1}{2}}.$

Thus the magnetic force

$=\frac{\omega e\left(1-\frac{u^{2}}{v^{2}}\right)^{\frac{1}{2}}\left(1-\frac{\omega^{2}}{v^{2}}\sin^{2}\beta\right)^{\frac{1}{2}}}{\cos\beta r^{2}\left(1-\frac{u^{2}}{v^{2}}\sin^{2}\theta\right)^{\frac{3}{2}}}$.

Since sin $\beta=v/\omega$, this expression vanishes unless $\theta=\beta$, when it becomes infinite, so that the magnetic force and the electric displacement seem confined to the surface of a cone of semi-vertical angle $\beta$, the vertex pointing in the direction of motion.

Trinity College, Cambridge,

April 24, 1889.

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This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1940, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 70 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.