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

Rh 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 '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

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

and

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

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

$$\int_0^\pi\frac{d\phi}{\sqrt{\left\{r^2 + c^2 - 2cr\sin\theta\,\cos\phi\right\}}}$$