# Notes on the Electro-magnetic Theory of Moving Charges

XVI. Notes on the Electro-magnetic Theory of Moving Charges.

By W. B. Morton, B.A.[1]

1. This subject has been brought into prominence recently by the use which Mr. Larmor has made of moving electrons in his dynamical theory of the aether. The matter was investigated in 1881 by Prof. J. J. Thomson,[2] who showed that a point charge moving so slowly that the electric displacement it carries is not sensibly disturbed generates magnetic force like a current element according to Ampere's rule; and by Mr. Heaviside,[3] who investigated the matter more generally in 1889, and showed that in steady rectilinear motion at any speed less than that of light, the lines of displacement continue to be radial but are concentrated towards the plane perpendicular to the direction of motion. The displacement at distance r, in a direction making an angle Θ with the line of motion is proportional to

${\displaystyle {\frac {1}{r^{2}\left(1-{\frac {u^{2}}{V^{2}}}\sin ^{2}\theta \right)^{\frac {3}{2}}}}}$,

where u is the velocity of the moving charge and V the velocity of light. The lines of magnetic force are circles round the line of motion.

2. This solution of course represents the state of affairs at a great distance from a small charged conductor of any shape. It would also give us the distribution of charge on a moving sphere if it were correct to assume that the lines of displacement meet the charged surface at right angles. This assumption was made by Prof. Thomson and, at first, by Mr. Heaviside, but the latter, quoting a suggestion of Mr. G. F. C. Searle, subsequently pointed out that when there is motion the electric force is no longer derived from a potential function, and as a consequence does not meet the equilibrium surface at right angles. Substituting, the correct surface condition, he showed that the charged conductor, whose motion would give at all points the radial distribution found for a point charge, was not a sphere but a spheroid of certain ellipticity.

3. It seemed of some interest to inquire what the distribution of charge on a moving sphere would be. The surface-density at a point of the surface is now the normal component of the displacement at that point. By carrying the investigation a step further I have found that, if the conductor be a sphere or any ellipsoid, the ordinary static arrangement of charge is unaltered by the motion; t. e. the number of tubes of displacement leaving each element of the surface is unchanged, but the tubes no longer leave the surface at right angles. We may imagine that the motion has the effect of deforming the tubes, keeping their ends on the conductor fixed. The proof of this, involving a consideration of the general case, is here given and is followed by a note on the energy of a moving charge in a magnetic field.

4. Suppose we have any distribution of charge moving with uniform velocity u parallel to the axis of z, and that the field has assumed its steady configuration. We shall denote ${\displaystyle 1-{\frac {u^{2}}{V^{2}}}}$ by k², V being the velocity of light. Then since we have a steady state,

${\displaystyle {\frac {d}{dt}}=-u{\frac {d}{dz}}}$.

Also, since each element of charge produces a magnetic field with no z-component, we have ${\displaystyle \gamma =0}$ in the general case also. Using these two data, the equations connecting the displacement (f, g, h) and the magnetic force (α, β, γ) become

 ${\displaystyle -{\frac {d\beta }{dz}}=-4\pi u{\frac {df}{dz}}}$, ${\displaystyle {\frac {d\alpha }{dz}}=-4\pi u{\frac {dg}{dz}}}$, ${\displaystyle {\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}=-4\pi u{\frac {dh}{dz}}}$, ${\displaystyle {\frac {dg}{dz}}-{\frac {dh}{dy}}=-{\frac {u}{4\pi V^{2}}}{\frac {d\alpha }{dz}}}$, ${\displaystyle {\frac {dh}{dx}}-{\frac {df}{dz}}=-{\frac {u}{4\pi V^{2}}}{\frac {d\beta }{dz}}}$, ${\displaystyle {\frac {df}{dy}}-{\frac {dg}{dx}}=0}$.

These equations together with

${\displaystyle {\frac {df}{dx}}+{\frac {dg}{dy}}+{\frac {dh}{dz}}=0}$

are satisfied by

 ${\displaystyle f=-{\frac {d\phi }{dx}},\ g=-{\frac {d\phi }{dy}},\ h=-k^{2}{\frac {d\phi }{dz}}}$, ${\displaystyle \alpha =4\pi u{\frac {d\phi }{dy}},\ \beta =-4\pi {\frac {d\phi }{dx}}}$,

where φ is any function satisfying

${\displaystyle {\frac {d^{2}\phi }{dx^{2}}}+{\frac {d^{2}\phi }{dy^{2}}}+k^{2}{\frac {d^{2}\phi }{dz^{2}}}=0}$.
These results have been obtained by Prof. Thomson and Mr. Heaviside. The particular case of a point charge, e, is got by putting
${\displaystyle \phi ={\frac {e}{4\pi {\sqrt {k^{2}(x^{2}+y^{2})+z^{2}}}}}}$.

Evidently in the general case φ must vanish at infinity.

5. Mr. Heaviside points out that φ=constant is the condition holding at a surface of equilibrium. The matter may be stated thus : — If we suppose the field to terminate at the surface of a conductor, inside which the vectors vanish, we must see that the "curl" relations of the field are not violated for circuits which lie partly inside the empty space enclosed by the conductor. In particular, if there is a vector whose line integral round every circuit in the field vanishes, the lines of this vector must meet the surface at right angles. Otherwise we should have a finite value for the integral round a circuit drawn close to the surface outside and completed inside. In other words, if a vector is derived from a potential function, this function must be constant over the surface. In the ordinary static case it is the electric force (X, Y, Z) which is so derived; but in the case of a steadily moving field it is the vector (X, Y, ${\displaystyle {\frac {Z}{k^{2}}}}$) which meets the surface at right angles.

6. Let F(x y z)=C be the equation of the charged surface. Then φ(x y z) has to be constant over this surface and satisfy

${\displaystyle {\frac {d^{2}\phi }{dx^{2}}}+{\frac {d^{2}\phi }{dy^{2}}}+k^{2}{\frac {d^{2}\phi }{dz^{2}}}=0}$.

Put z=kζ, then φ is a function of x, y, ζ, which is constant when F(x, y, kζ)=C, and which satisfies

${\displaystyle {\frac {d^{2}\phi }{dx^{2}}}+{\frac {d^{2}\phi }{dy^{2}}}+k^{2}{\frac {d^{2}\phi }{d\zeta ^{2}}}=0}$.

Therefore if we regard (x y ζ) as Cartesian coordinates of a point, φ is the potential at external points of an electrostatic free distribution on the surface F(x, y, kζ)=C. The components of electric force due to this distribution, at a point (x y ζ) on the surface, are

${\displaystyle -{\frac {d\phi }{dx}},\ -{\frac {d\phi }{dy}},\ -{\frac {d\phi }{d\zeta }}}$.

This force acts in the normal to the surface, and is proportional to the surface-density at (x y ζ), which we shall call σ'. Therefore

${\displaystyle \left({\frac {d\phi }{dx}},\ {\frac {d\phi }{dy}},\ {\frac {d\phi }{d\zeta }}\right)=-A\sigma '\left({\frac {dF}{dx}},\ {\frac {dF}{dy}},\ {\frac {dF}{d\zeta }}\right)/{\sqrt {\left({\frac {dF}{dx}}\right)^{2}+\left({\frac {dF}{dy}}\right)^{2}+\left({\frac {dF}{d\zeta }}\right)^{2}}}}$

But

${\displaystyle {\frac {d}{d\zeta }}=k{\frac {d}{dz}}}$;

therefore, denoting differentiation with respect to x y z by subscripts 1 2 3,

${\displaystyle (\phi _{1},\ \phi _{2},\ \phi _{3})=-A\sigma '(F_{1},\ F_{2},\ F_{3})/{\sqrt {F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}}$.

Now let σ be the surface-density at (x y z) on the moving conductor F(x y z)=C, then equating σ to the normal component of (f g h),

${\displaystyle \sigma ={\frac {fF_{1}+gF_{2}+hF_{3}}{\sqrt {F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}}}$,

or putting in the values we have found for (f g h) in terms of φ

 ${\displaystyle \sigma =-{\frac {\phi _{1}F_{1}+\phi _{2}F_{2}+k^{2}\phi _{3}F_{3}}{\sqrt {F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}},}$ ${\displaystyle =A\sigma '{\sqrt {\frac {F_{1}^{2}+F_{2}^{2}+k^{2}F_{3}^{2}}{F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}}}$.

Now the perpendicular from the origin on the tangent plane to F(x y z)=C at the point (x y z) is

${\displaystyle p={\frac {xF_{1}+yF_{2}+zF_{3}}{\sqrt {F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}}}$,

and the perpendicular from the origin on the tangent plane to F(x, y, kζ)=C at (x y ζ) is

${\displaystyle p'={\frac {x{\frac {dF}{dx}}+y{\frac {dF}{dy}}+\zeta {\frac {dF}{d\zeta }}}{\sqrt {\left({\frac {dF}{dx}}\right)^{2}+\left({\frac {dF}{dy}}\right)^{2}+\left({\frac {dF}{d\zeta }}\right)^{2}}}}}$
 ${\displaystyle ={\frac {xF_{1}+yF_{2}+k\zeta F_{3}}{\sqrt {F_{1}^{2}+F_{2}^{2}+k^{2}F_{3}^{2}}}}}$ ${\displaystyle ={\frac {xF_{1}+yF_{2}+zF_{3}}{\sqrt {F_{1}^{2}+F_{2}^{2}+k^{2}F_{3}^{2}}}}}$; ${\displaystyle \therefore {\frac {\sigma }{\sigma '}}={\frac {Ap}{p'}}}$.

If now F(x y z) = C is an ellipsoid, then we know that ${\displaystyle \sigma \propto p}$, therefore also ${\displaystyle \sigma '\propto p'}$, that is the arrangement of charge on the moving ellipsoid is the same as if it were at rest.

7. Applying the above to the ellipsoid (a b c), we find that φ as a function of (x y ζ) is the potential of a free distribution on the ellipsoid

 ${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}+{\frac {k^{2}\zeta ^{2}}{c^{2}}}=1}$; ${\displaystyle \therefore \ \phi =C\int _{\mu }^{\infty }{\frac {d\lambda }{\sqrt {(a^{2}+\lambda )(b^{2}+\lambda )\left({\frac {c^{2}}{k^{2}}}+\lambda \right)}}}}$ ${\displaystyle =Ck\int _{\mu }^{\infty }{\frac {d\lambda }{\sqrt {(a^{2}+\lambda )(b^{2}+\lambda )\left(c^{2}+k^{2}\lambda \right)}}}}$,

where μ is given by

${\displaystyle {\frac {x^{2}}{a^{2}+\mu }}+{\frac {y^{2}}{b^{2}+\mu }}+{\frac {\zeta ^{2}}{{\frac {c^{2}}{k^{2}}}+\mu }}=1}$,

or

${\displaystyle {\frac {x^{2}}{a^{2}+\mu }}+{\frac {y^{2}}{b^{2}+\mu }}+{\frac {z^{2}}{c^{2}+k^{2}\mu }}=1}$.

Deteimining the value of the constant C so that the density at a point shall be ${\displaystyle {\frac {ep}{4\pi abc}}}$, we get

${\displaystyle \phi ={\frac {e}{8\pi }}\int _{\mu }^{\infty }{\frac {d\lambda }{\sqrt {(a^{2}+\lambda )(b^{2}+\lambda )\left(c^{2}+k^{2}\lambda \right)}}}}$.

Putting b=a, c=ka, we get

${\displaystyle \phi ={\frac {e}{4\pi {\sqrt {k^{2}(x^{2}+y^{2})+z^{2}}}}}}$.
Showing that, as Mr. Heaviside pointed out, the field of a point charge is given when the conductor is an oblate spheroid whose axes have the ratio 1:k.

For a sphere the integral becomes

${\displaystyle \phi ={\frac {e}{8\pi k'a}}\log {\frac {\theta +k'a}{\theta -k'a}}}$

where

${\displaystyle k'={\sqrt {1-k^{2}}}={\frac {u}{V}}}$,

and θ is given by

${\displaystyle {\frac {k^{2}(x^{2}+y^{2})}{\theta ^{2}-k'^{2}a^{2}}}+{\frac {z^{2}}{\theta ^{2}}}=1}$.

To test the value of φ let us make k' approach zero, i. e. the motion becomes infinitely slow. θ is then =r.

Then

 ${\displaystyle \phi ={\frac {e}{8\pi a}}{\underset {k'=0}{Lt}}\log {\frac {(r+k'a)-\log(r-k'a)}{k'}}}$ ${\displaystyle ={\frac {e}{8\pi a}}\cdot {\frac {2a}{r}}={\frac {e}{4\pi r}}}$.

8. The mutual energy of a moving charge and external magnetic field has been given by Mr. Heaviside for the case of motion which is very slow compared with the velocity of radiation. It is eu A · cos (uA), where A is the circuital vector potential of the external field. Mr. Larmor, in the second part of his "Dynamical Theory" (Phil. Trans. 1895, p. 717), concludes that the same expression holds good for motion at any speed. He seems, however, to overlook the fact that in the general case the displacement-currents in the medium — being no longer derivable from a potential function — will make their appearance in the result as well as the convection-current eu.

If (F G H) is the vector potential, the part of the energy corresponding to the displacement currents will be

${\displaystyle \int (F{\dot {f}}+G{\dot {g}}+H{\dot {h}})d\tau }$,

which in the case we have been considering becomes

 ${\displaystyle -u\int (F{\frac {df}{dz}}+G{\frac {dg}{dz}}+H{\frac {dh}{dz}})d\tau }$ ${\displaystyle =u\int (F{\frac {d^{2}\phi }{dx\ dz}}+G{\frac {d^{2}\phi }{dy\ dz}}+k^{2}H{\frac {d^{2}\phi }{dz^{2}}})d\tau }$.
But by a well-known transformation, when we take the integral through all space, we have
${\displaystyle \int (F{\frac {d^{2}\phi }{dx\ dz}}+G{\frac {d^{2}\phi }{dy\ dz}}+H{\frac {d^{2}\phi }{dz^{2}}})d\tau }$
${\displaystyle =\int {\frac {d\phi }{dz}}({\frac {dF}{dx}}+{\frac {dG}{dy}}+{\frac {dH}{dz}})d\tau }$
${\displaystyle =0}$ since (F G H) is circuital.
${\displaystyle \therefore }$ the expression for this part of the energy reduces to
${\displaystyle -u(1-k^{2})\int H{\frac {d^{2}\phi }{dz^{2}}}d\tau =-{\frac {u^{2}}{V^{2}}}\int H{\frac {d^{2}\phi }{dz^{2}}}d\tau }$.

Therefore if the velocity u ceases to be negligible in comparison with V, we have a correction of the second order in the ratio ${\displaystyle {\frac {u}{V}}}$ in addition to the expression involving the convection-current simply. It also appears from the above that the force on the moving charge cannot, unless this term be neglected, be expressed in terms of the magnetic intensity at the charge, but will depend on the entire field.