Page:Thomson1881.djvu/16

the surface of a sphere vanishes, we may substitute in the integral ${\tfrac {d^{2}}{dx^{2}}}{\tfrac {1}{r}}$ for ${\tfrac {d^{2}}{dx^{2}}}{\tfrac {1}{r'}}$; then, transforming to polars, the integral

:$=\sigma \int _{0}^{2\pi }\int _{0}^{\pi }\int _{R}^{\infty }4Q_{2}^{2}{\frac {1}{r^{2}}}\sin \theta d\phi \ dv\ dr$ $={\frac {16\pi \sigma }{5R}}$;

for values of r < R,

${\frac {1}{r'}}={\frac {1}{R}}+{\frac {rQ_{1}}{R^{2}}}+{\frac {rQ_{2}}{R^{3}}}+\dots$.

Now $r^{n}Q_{n}$ is a solid harmonic of the nth order; hence ${\tfrac {d^{2}}{dx^{2}}}\left(r^{n}Q_{n}\right)$ is a solid harmonic of the (n-2)th order; and in particular ${\tfrac {d^{2}}{dx^{2}}}\left(r^{4}Q_{4}\right)$ is a solid harmonic of the second order; and, by the same reasoning as before, we may substitute in the integral ${\tfrac {1}{R^{5}}}{\tfrac {d^{2}}{dx^{2}}}\left(r^{4}Q_{4}\right)$ for ${\tfrac {d^{2}}{dx^{2}}}{\tfrac {1}{r'}}$. Now

$r^{4}Q_{4}={\frac {35x^{4}-30x^{2}\left(x^{2}+y^{2}+z^{2}\right)+3\left(x^{2}+y^{2}+z^{2}\right)^{2}}{8}}$; $\therefore {\frac {d^{2}}{dx^{2}}}\left(r^{4}Q_{4}\right)=12x^{2}-6\left(y^{2}+z^{2}\right)=12r^{2}Q_{2}$.

So for values of r < R the integral becomes

${\frac {\sigma }{R^{5}}}\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{R}24Q_{2}^{2}r^{3}\sin \theta \ d\phi \ d\theta \ dr$ $={\frac {24\sigma \pi }{5R}}$.

Adding this to the part of the integral for r > R, we get for the coefficient of uu' ,${\tfrac {8\sigma \pi }{R}}$. The coefficients of uv' and uw' vanish by inspection.

The coefficient of vv'

$=\sigma \int \int \int r^{2}{\frac {d^{2}}{dx\ dy}}{\frac {1}{r}}{\frac {d^{2}}{dx\ dy}}{\frac {1}{r'}}dx\ dy\ dz$.

Now when r > R we may, by the same reasoning as before, substitute ${\tfrac {d^{2}}{dx\ dy}}{\tfrac {1}{r}}$ for ${\tfrac {d^{2}}{dx\ dy}}{\tfrac {1}{r'}}$, in the integral, and it becomes

$\sigma \int \int \int {\frac {9r^{2}x^{2}y^{2}}{r^{10}}}dx\ dy\ dz$,