Page:Motion of Electrification through a Dielectric.djvu/3

 Σμ0H0H/4&pi;, if H0 is the external field; and, by a well-known transformation, it is equivalent to ΣA0&Gamma;, if A0 is any vector whose curl is μ0H0, whilst &Gamma; is the current-density of the moving system. Further, if we choose A0 to be circuital, the polar part of &Gamma; will contribute nothing to the summation, so that we are reduced to the volume-integral of the scalar product of the circuital A0 of the one system and the density of the convection-current in the other. Or, in the present case, with a single moving charge at a point, we have simply the scalar product A0uq to represent the mutual magnetic energy; or

which is double J. J. Thomson's result.

5. When, therefore, we derive from (3) the mechanical force on the moving charge due to the external magnetic field, we obtain simply Maxwell's "electromagnetic force" on a current-element, the vector product of the moment of the current and the induction of the external field; or, if F is this mechanical force,

which is also double J. J. Thomson's result. Notice that in the application of the "electromagnetic force" formula, it is the moment of the convection-current that occurs. This is not the same as the moment of the true current, which varies according to circumstances; for instance, in the case of a small dielectric sphere uniformly electrified throughout its volume, the moment of the true current would be only $$\scriptstyle{\frac{2}{3}}$$ of that of the convection-current.

The application of Lagrange's equation of motion to (3) also gives the force on q due to the electric field so far as it can depend on M; that is, a force

$\scriptstyle{-q\mathbf{A}_{0},}$

where the time-variation due to all causes must be reckoned, except that due to the motion of q itself, which is allowed for in (4). And besides this, there may be electric force not derivable from A0, viz.

$\scriptstyle{-q\nabla\Psi_{0},}$

where &Psi;0 is the scalar potential companion to A0.

6. Now if the external field be that of another moving charge, we shall obtain the mutual magnetic energy from (3) by letting A0 be the vector-potential of the current in the second moving system, constructed so as to be circuital. Now the vector-potential of the convection-current qu. is simply qu/r; this is sufficient to obtain the magnetic force by curling; but if used to calculate the mutual energy, the space-summation would have to include every element of current in the other system. To make the vector-potential circuital, and so be able to abolish this work, we must add on to qu/r the vector-potential of the displacement current to correspond. Now the complete current may be considered to consist of a linear element qu having two poles; a radial current outward from the + pole in which the current-density is $\scriptstyle{qu/4\pi r_{1}^{2}}$;|undefined and a radial current inward to the – pole, in which the current-density is $\scriptstyle{-qu/4\pi r_{2}^{2}}$;|undefined where r1 and r2 are the distances of any point from the