Page:Thomson1881.djvu/5

 for points outside the sphere,

{{MathForm2|(2)|$$\left.\begin{array}{ll} F & =\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx^{2}}\frac{1}{R}+\frac{2\mu ep}{3R}\\ \\G & =\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dy}\frac{1}{R}\\ \\H & =\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dz}\frac{1}{R}\end{array}\right\} $$}}

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

in our case,

Now

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

since the term

evidently vanishes.

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

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

Similarly,