From Wikisource
Jump to: navigation, search
This page has been validated.

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}.


\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<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


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.