Page:SearleEllipsoid.djvu/10

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

Equation (11) now becomes

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

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

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

Hence

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 $$a^2<ab^2$$ 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 $$\mu H^{2}/8\pi$$.