# Page:Thomson1881.djvu/12

The part of the kinetic energy we are concerned with will evidently be

 $\frac{1}{8\pi}\int\int\int2\mu e\left[\left(r\frac{d}{dy}-q\frac{d}{dz}\right)\frac{1}{R}\left(\frac{dH'}{dy}-\frac{dG'}{dz}\right)\right]$ $+\left(p\frac{d}{dz}-r\frac{d}{dx}\right)\frac{1}{R}\left(\frac{dF'}{dz}-\frac{dH'}{dx}\right)$ $+\left(q\frac{d}{dx}-p\frac{d}{dy}\right)\frac{1}{R}\left(\frac{dG'}{dx}-\frac{dF'}{dy}\right)dx\ dy\ dz$ $-\frac{1}{8\pi}\int\int\int\mu e4\pi\left[A(r\frac{d}{dy}-q\frac{d}{dz}\right)\frac{1}{R}+B\left(p\frac{d}{dz}-r\frac{d}{dx}\right)\frac{1}{R}$ $\left.+C\left(q\frac{d}{dx}-p\frac{d}{dy}\right)\frac{1}{R}\right]dx\ dy\ dz$

Let us take the first integral first, and take the term depending on p; this is

$\frac{\mu ep}{4\pi}\int\int\int\frac{d}{dz}\frac{1}{R}\left(\frac{dF'}{dz}-\frac{dH'}{dx}\right)-\frac{d}{dy}\frac{1}{R}\left(\frac{dG'}{dx}-\frac{dF'}{dy}\right)dx\ dy\ dz$.

Integrating by parts this becomes

 $-\frac{\mu ep}{4\pi}\int\int F'\left(\frac{d}{dx}\frac{1}{R}dy\ dz+\frac{d}{dy}\frac{1}{R}dx\ dz+\frac{d}{dz}\frac{1}{R}dx\ dy\right)$ $+\frac{\mu ep}{4\pi}\int\int\frac{1}{R}\left(\frac{dH'}{dx}dy\ dz+\frac{dG'}{dx}dx\ dz+\frac{dF'}{dx}dz\ dy\right)$ $+\frac{\mu ep}{4\pi}\int\int\int\frac{1}{R}\frac{d}{dx}\left(\frac{dF'}{dx}+\frac{dG'}{dy}+\frac{dH'}{dz}\right)$ $-F'\left(\frac{d^{2}}{dx^{2}}+\frac{d^{2}}{dy^{2}}+\frac{d^{2}}{dz^{2}}\right)\frac{1}{R}dx\ dy\ d$z.

The surface-integrals are to be taken over the surface of the sphere; and the triple integral is to be taken throughout all space exterior to the sphere.

If the sphere be so small that we may substitute for the values of F', $\frac{dF'}{dx}$ , &c. at the surface their values at the centre of the sphere, the first surface-integral $=\mu epF'_{1}$, where $F'_1$ is the value of $F'$ at the centre of the sphere; the second surfaceintegral vanishes, and the triple integral also vanishes, since

$\frac{d^{2}}{dx^{2}}\frac{1}{R}+\frac{d^{2}}{dy^{2}}\frac{1}{R}+\frac{d^{2}}{dz^{2}}\frac{1}{R}=0$

and

$\frac{dF'}{dx}+\frac{dG'}{dy}+\frac{dH'}{dz}=0$.