Page:Thomson1881.djvu/5

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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};