Page:Thomson1881.djvu/19

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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\},