# Page:Thomson1881.djvu/5

for points outside the sphere,

 $\left.\begin{array}{ll} F & =\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx^{2}}\frac{1}{R}+\frac{2\mu ep}{3R}\\ \\G & =\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dy}\frac{1}{R}\\ \\H & =\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dz}\frac{1}{R}\end{array}\right\}$ (2)

Now, by 'Electricity and Magnetism,' § 634, T the kinetic energy

$=\frac{1}{2}\int\int\int(Fu+Gv+Hw)dx\ dy\ dz$,

in our case,

$=\frac{1}{2}\int\int\int(F\frac{df}{dt}+G\frac{dg}{dt}+H\frac{dh}{dt})dx\ dy\ dz$.

Now

$\frac{1}{2}\int\int\int F\frac{df}{dt}dx\ dy\ dz$.

substituting for F and $\frac{df}{dt}$

$F=\frac{1}{2}\frac{\mu e^{2}p^{2}}{20\pi}\int\int\int\left(\frac{5r^{2}}{6}-\frac{a^{2}}{2}\right)\left(\frac{d^{2}}{dx^{2}}\frac{1}{r}\right)^{2}dx\ dy\ dz$,

since the term

$\frac{\mu e^{2}p^{2}}{12\cdot\pi}\int\int\int\frac{1}{r}\frac{d^{2}}{dx^{2}}\frac{1}{r}dx\ dy\ dz$,

evidently vanishes.

Transforming to polars and taking the axis of x as the initial line, the above integral

 $\begin{array}{l} =\frac{\mu e^{2}p^{2}}{40\pi}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{a}^{\infty}\left(\frac{5r^{2}}{6}-\frac{a^{2}}{2}\right)\frac{\left(3\cos^{2}\theta-1\right)^{2}}{r^{4}}\sin\theta\ dr\ d\theta\ d\phi\\ \\=\frac{4\mu e^{2}p^{2}}{75a}\end{array}$ $\frac{1}{2}\int\int\int G\frac{dg}{dt}dx\ dy\ dz=\frac{\mu e^{2}p^{2}}{40\pi}\int\int\int\left(\frac{5}{6}r^{2}-\frac{a^{2}}{2}\right)\left(\frac{d^{2}}{dx\ dy}\frac{1}{r}\right)^{2}dx\ dy\ dz$.

By transforming to polars, as before, we may show that this

$=\frac{\mu e^{2}p^{2}}{25a}$

Similarly,

$\frac{1}{2}\int\int\int H\frac{dh}{dt}dx\ dy\ dz=\frac{\mu e^{2}p^{2}}{25a}$;