Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/47

are solutions of the equation

${\frac {d^{2}\psi }{dz^{2}}}+{\frac {d^{2}\psi }{d\varpi ^{2}}}-{\frac {1}{\varpi }}{\frac {d\psi }{d\varpi }}=0$.

For putting $\psi =\varpi \phi$, this equation becomes

${\frac {d^{2}\phi }{dz^{2}}}+{\frac {d^{2}\phi }{d\varpi ^{2}}}-{\frac {1}{\varpi }}{\frac {d\Phi }{d\varpi }}-{\frac {1}{\varpi ^{2}}}\Phi =0$,

that is, $\Phi \cos \phi$ is a solution of Laplace's equation.

§ 2. Expansion of the Functions.—Let the plane through the axis of the ring and a point $\mathrm {P}$, cut the ring in the two circles whose centres are $\mathrm {C}$ and $\mathrm {C} _{1}$.

Let

$\mathrm {CP} =\mathrm {R}$;$\quad \mathrm {C} _{1}\mathrm {P} =\mathrm {R}$;$\quad \mathrm {OC} =c$;

and the angle

$\mathrm {OCP} =\chi$.

Also, for convenience, put

$s={\frac {\mathrm {R} }{c}}$,

and

$l=\log {\frac {8c}{\mathrm {R} }}-2$.

When $\mathrm {R}$ is less than $c$ the above integrals may be expanded in ascending powers of $\mathrm {R} /c$ or $s$.

$\int _{0}^{\pi }{\frac {d\phi }{\sqrt {\left\{r^{2}+c^{2}-2cr\sin \theta \,\cos \phi \right\}}}}$

$={\frac {2}{\mathrm {R} _{1}}}\int _{0}^{\frac {\pi }{2}}{\frac {d\phi }{\sqrt {\left\{1-{\frac {\mathrm {R} _{1}^{2}-\mathrm {R} ^{2}}{\mathrm {R} _{1}^{2}}}\sin ^{2}\phi \right\}}}}$

$={\frac {2}{\mathrm {R} _{1}}}\left\{\log {\frac {4\mathrm {R} _{1}}{\mathrm {R} }}+{\frac {1^{2}}{2^{2}}}{\frac {\mathrm {R} ^{2}}{\mathrm {R} _{1}^{2}}}\left(\log {\frac {4\mathrm {R} _{1}}{\mathrm {R} }}-{\frac {2}{1.2}}\right)+{\frac {1^{2}3^{2}}{2^{2}4^{2}}}{\frac {\mathrm {R} ^{4}}{\mathrm {R} _{1}^{4}}}\left(\log {\frac {4\mathrm {R} _{1}}{\mathrm {R} }}-{\frac {2}{1.2}}-{\frac {2}{3.4}}\right)+\,.\,.\,.\right\}.$

[Cayley, 'Ell. Func.,' p. 54.]