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,

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