# Page:Thomson1881.djvu/4

$F=\frac{\mu ep}{4\pi}\int\int\int\frac{1}{PQ}\frac{d^{2}}{dx^{2}}\frac{1}{\rho}dx\ dy\ dz$;

Now $\frac{d^{2}}{dx^{2}}\frac{1}{\rho}=\frac{Y_{2}}{\rho^{3}}$, where Y2 is a surface harmonic of the second order. And when ρ > R,

$\frac{1}{PQ}=\frac{1}{\rho}+\frac{R}{p^{2}}Q_{1}+\frac{R^{2}}{\rho^{3}}Q_{2}+\dots$;

and when ρ < R,

$\frac{1}{PQ}=\frac{1}{R}+\frac{\rho}{R^{2}}Q_{1}+\frac{\rho^{2}}{R^{3}}Q_{2}+\dots$;

where Q1, Q2, &c. are zonal harmonics of the first and second orders respectively referred to OP as axis.

Let Y'2 denote the value of Y2 along OP. Then, since $\int Y_{n}Q_{m}ds$, integrated over a sphere of unit radius, is zero when n and m are different, and $\frac{4\pi}{2n+1}Y_{n}^{'}$ when n=m, Y'n being the value of Yn at the pole of Qn, and since there is no electric displacement within the sphere,

 $F=\frac{\mu ep}{4\pi}\times\frac{2\pi Y_{2}^{'}}{5}\left\{ \int_{R}^{\infty}\frac{R^{2}}{\rho^{4}}d\rho+\int_{a}^{R}\frac{\rho\ d\rho}{R^{3}}\right\}$ $=\frac{\mu ep}{5}Y_{2}^{'}\left(\frac{5}{6R}-\frac{a^{2}}{2R^{3}}\right)$,

or, as it is more convenient to write it,

$=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx^{2}}\frac{1}{R}$.

By symmetry, the corresponding values of G and H are

 $G=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dy}\frac{1}{R}$, $H=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dz}\frac{1}{R}$.

These values, however, do not satisfy the condition

$\frac{dF}{dx}+\frac{dG}{dy}+\frac{dH}{dz}=0$.

If, however, we add to F the term $\frac{2\mu ep}{3R}$, this condition will be satisfied; while, since the term satisfies Laplace's equation, the other conditions will not be affected: thus we have finally