# Page:SearleEllipsoid.djvu/10

The value of $\mathbf{\Psi}$ in terms of $h$ thus becomes

 $\mathbf{\Psi}=\frac{q\alpha}{\mathrm{K}}\int_{h}^{\infty}\frac{dh}{h^{2}-l^{2}}$. (17)

Equation (11) now becomes

 $\frac{x^{2}}{h^{2}}+\frac{\rho^{2}\alpha}{h^{2}-l^{2}}=1$, (18)

so that instead of the cylindrical coordinates $x$ and $\rho(=\sqrt{y^{2}+z^{2}})$ we, can take $h$ and $\phi$ where

 $x=h\cos\phi,\ \rho=\frac{\sqrt{h^{2}-l^{2}}}{\sqrt{\alpha}}\sin\phi$. (19)

From (18) we have in terms of $h$ and $\phi$

$\frac{dh}{dx}=\frac{(h^{2}-l^{2})\cos\phi}{h^{2}-l^{2}\cos^{2}\phi},\ \frac{dh}{d\rho}=\frac{h\sqrt{h^{2}-l^{2}}\sin\phi\sqrt{\alpha}}{h^{2}-l^{2}\cos^{2}\phi}$.

Hence

 $\mathrm{E}_{1}=-\frac{d\mathbf{\Psi}}{dh}\cdot\frac{dh}{dx}=\frac{\alpha q\ \cos\phi}{K\sqrt{h^{2}-l^{2}\cos^{2}\phi)}}$, (20)
 $\mathrm{E}_{\rho}=-\frac{1}{\alpha}\frac{d\mathbf{\Psi}}{dh}\cdot\frac{dh}{d\rho}=\frac{qh\ \sin\phi\sqrt{\alpha}} {\mathrm{K}\sqrt{h^{2}-l^{2}}(h^{2}-l^{2}\cos^{2}\phi)}$, (21)
 $\mathrm{H}=\mathrm{K}u\mathrm{E}_{\rho}=\frac{quh\ \sin\phi\sqrt{\alpha}}{\sqrt{h^{2}-l^{2}(h^{2}-l^{2}\cos^{2}\phi)}}$, (22)

I now pass on to calculate the total energy possessed by the ellipsoid when in motion along its axis of figure. In making the calculation I shall suppose that $a^2>ab^2$, i.e., that $l^2$ is positive. The case in which $a^2 can be deduced by the appropriate mathematical transformation.

I have shown {§ 22} that the total energy, viz. the volume integral of $\frac{\mathrm{K}\mathbf{E}^{2}+\mu\mathbf{H}^{2}}{8\pi}$, due to the motion of a charge on any surface, is

$\mathrm{W}=\frac{1}{2}q\mathbf{\Psi}_{0}+2\mathrm{T}$,

where $\Psi_{0}$ is the value of the convection-potential at the surface of the body, and T is the magnetic part of the energy, viz., the volume integral of $\mu H^{2}/8\pi$.