Page:Thomson1881.djvu/19

by the preceding work, that the coefficient of uu' is zero; the coefficient of vv', 3σπ; and the coefficient of ww', 5σπ. Adding, we get the whole kinetic energy due to the vector-potential arising from e and the electric displacement arising from e'

$={\frac {\pi \sigma }{2R}}(8uu'+(5+3)vv'+(5+3)ww')$ $={\frac {4\pi \sigma }{R}}(uu'+vv'+ww')$.

We can get that part of the kinetic energy due to the vector-potential arising from e' and the electric displacement from e by writing e' for e, and u', v', w' for u, v, w respectively. Hence, that part of the kinetic energy which is multiplied by ee'

$={\frac {8\pi \sigma }{R}}(uu'+vv'+ww')$;

or, substituting for σ its value,

$={\frac {\pi ee'}{3R}}(uu'+vv'+ww')$.

Or if q and q' be the velocities of the spheres, and ε the angle between their directions of motion, this part of the kinetic energy

$={\frac {\mu ee'}{3R}}qq'\cos \epsilon$,

and the whole kinetic energy due to the electrification

$=\mu \left({\frac {2}{15}}{\frac {e^{2}q^{2}}{a}}+{\frac {2}{15}}{\frac {e'^{2}q'^{2}}{a_{1}}}+{\frac {ee'}{3R}}qq'\cos \epsilon \right)$

(6)

If x, y, z be the coordinates of the centre of one sphere, x', y', z' those of the other, we may write the last part of the kinetic energy in the form

${\frac {\mu ee'}{3R}}\left({\frac {dx}{dt}}{\frac {dx'}{dt}}+{\frac {dy}{dt}}{\frac {dy'}{dt}}+{\frac {dz}{dt}}{\frac {dz'}{dt}}\right)$.

By Lagrange's equations, the force parallel to the axis of x acting on the first sphere

$={\frac {dT}{dx}}-{\frac {d}{dt}}\left({\frac {dT}{d\cdot {\frac {dx}{dt}}}}\right)$ $={\frac {\mu ee'}{3}}\left\{\left({\frac {dx}{dt}}{\frac {dx'}{dt}}+{\frac {dy}{dt}}{\frac {dy'}{dt}}+{\frac {dz}{dt}}{\frac {dz'}{dt}}\right){\frac {d}{dx}}{\frac {1}{R}}-{\frac {d}{dt}}{\frac {\frac {dx'}{dt}}{R}}\right\}$,