Page:Motion of Electrification through a Dielectric.djvu/4

 poles. The vector-potentials of these currents are also radial, and their tensors are $$\scriptstyle{\frac{1}{2}qu}$$ and $\scriptstyle{-\frac{1}{2}qu}$. We have now merely to find their resultant when the linear element is indefinitely shortened, add on to the former qu/r, and multiply by μ0, to obtain the complete circuital vector-potential of qu, viz.:—

where r is the distance from q to the point P when A is reckoned, and the differentiation is to s, the axis of the convection-current. Both it and the space-variation are taken at P. The tensor of u is u. Though different and simpler in form (apart from the use of vectors) this vector-potential is, I believe, really the same as the one used by J. J. Thomson. From it we at once find, by the method described in § 4, the mutual energy of a pair of point-charges q1 and q2, moving at velocities u1 and u2, to be

when at distance r apart. Both axial differentiations are to be effected at one end of the line r.

As an alternative form, let ε be the angle between u1 and u2, and let the differentiation to s1 be at ds1 that to s2 at ds2, as in the German investigations relating to current-elements; then

Another form, to render its meaning plainer. Let λ1, μ1, v1 and λ2, μ2, v2 be the direction-cosines of the elements referred to rectangular axes, with the x-axis, to which λ1 and λ2 refer, chosen as the line joining the elements. Then

J. J. Thomson's estimate is

Comparing this with (8) we see that there is a notable difference.

7. The mutual energy being different, the forces on the charges, as derived by J. J. Thomson by the use of Lagrange's equations, will be different. When the speeds are constant, we shall have simply the before-described vector product (4) for the "electromagnetic force;" or

if F1 is the electromagnetic force on the first and F2 that on the second element, whilst H1 and H2 are the magnetic forces. Similar changes are needed in the other parts of the complete mechanical forces.