Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/225

The surface-integral for the corresponding element of the surface $\alpha +d\alpha$ will be

${\frac {dV}{d\alpha }}{\frac {D_{1}^{2}}{c}}d\beta \ d\gamma +{\frac {d^{2}V}{d\alpha ^{2}}}{\frac {D_{1}^{2}}{c}}d\alpha \ d\beta \ d\gamma$

since $D_{1}$ is independent of $\alpha$. The surface-integral for the two opposite faces of the element of volume, taken with respect to the interior of that volume, will be the difference of these quantities, or

${\frac {d^{2}V}{d\alpha ^{2}}}{\frac {D_{1}^{2}}{c}}d\alpha \ d\beta \ d\gamma$

Similarly the surface-integrals for the other two pairs of forces will be

${\frac {d^{2}V}{d\beta ^{2}}}{\frac {D_{2}^{2}}{c}}d\alpha \ d\beta \ d\gamma$ and ${\frac {d^{2}V}{d\gamma ^{2}}}{\frac {D_{3}^{2}}{c}}d\alpha \ d\beta \ d\gamma$

These six faces enclose an element whose volume is

$ds_{1}ds_{2}ds_{3}={\frac {D_{1}^{2}D_{2}^{2}D_{3}^{2}}{c^{3}}}d\alpha \ d\beta \ d\gamma$

and if $\rho$ is the volume-density within that element, we find by Art. 77 that the total surface-integral of the element, together with the quantity of electricity within it, multiplied by 4$\pi$ is zero, or, dividing by $d\alpha \ d\beta \ d\gamma$,

${\frac {d^{2}V}{d\alpha ^{2}}}D_{1}^{2}+{\frac {d^{2}V}{d\beta ^{2}}}D_{2}^{2}+{\frac {d^{2}V}{d\gamma ^{2}}}D_{3}^{2}+4\pi \rho {\frac {D_{1}^{2}D_{2}^{2}D_{3}^{2}}{c^{2}}}=0,$

(11)

which is the form of Poisson’s extension of Laplace’s equation re erred to ellipsoidal coordinates.

If $\rho =0$ the fourth term vanishes, and the equation is equivalent to that of Laplace.

For the general discussion of this equation the reader is referred to the work of Lamé already mentioned.

149.] To determine the quantities $\alpha ,\ \beta ,\ \gamma$ we may put them in the form of ordinary elliptic functions by introducing the auxiliary angles $\theta ,\ \phi$ and $\psi$, where

${\begin{array}{lcc}\lambda _{1}=b\sin \theta ,&&(12)\\\\\lambda _{2}={\sqrt {c^{2}\sin ^{2}\phi +b^{2}\cos ^{2}\phi }},&&(13)\\\\\lambda _{3}={\frac {c}{\sin \psi }}.&&(14)\end{array}}$

If we put $b=kc$, and $k^{2}+k'^{2}=1$, we may call $k$ and $k'$ the two complementary moduli of the confocal system, and we find

$\alpha =\int _{0}^{\theta }{\frac {d\theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}},$

(15)