Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/224

The denominator of this fraction is the product of the squares of the semi-axes of the surface $\lambda _{1}$.

If we put

$D_{1}^{2}=\lambda _{3}^{2}-\lambda _{2}^{2},\ D_{2}^{2}=\lambda _{3}^{2}-\lambda _{1}^{2},\ \mathrm {and} \ D_{3}^{2}=\lambda _{2}^{2}-\lambda _{1}^{2},$

(5)

and if we make $a=0$, then

${\frac {ds_{1}}{d\lambda _{1}}}={\frac {D_{2}D_{3}}{{\sqrt {b^{2}-\lambda _{1}^{2}}}{\sqrt {c^{2}-\lambda _{1}^{2}}}}}$

(6)

It is easy to see that $D_{2}$ and $D_{3}$ are the semi-axes of the central section of $\lambda _{1}$ which is conjugate to the diameter passing through the given point, and that $D_{2}$ is parallel to $ds_{2}$, and $D_{3}$ to $Ds_{3}$.

If we also substitute for the three parameters $\lambda _{1},\lambda _{1},\lambda _{1}$ their values in terms of three functions $\alpha ,\ \beta ,\ \gamma$ defined by the equations

${\begin{array}{lll}{\frac {da}{d\lambda _{1}}}={\frac {c}{{\sqrt {b^{2}-\lambda _{1}^{2}}}{\sqrt {c^{2}-\lambda _{1}^{2}}}}}&&\lambda _{1}=0\ \mathrm {when} \ \alpha =0,\\\\{\frac {d\beta }{d\lambda _{2}}}={\frac {c}{{\sqrt {\lambda _{2}^{2}-b^{2}}}{\sqrt {c^{2}-\lambda _{2}^{2}}}}}&&\lambda _{2}=b\ \mathrm {when} \ \beta =0,\\\\{\frac {d\gamma }{d\lambda _{3}}}={\frac {c}{{\sqrt {\lambda _{3}^{2}-b^{2}}}{\sqrt {\lambda _{3}^{2}-c^{2}}}}}&&\lambda _{3}=c\ \mathrm {when} \ \gamma =0;\end{array}}$

(7)

then

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

(8)

148.] Now let $V$ be the potential at any point $\alpha ,\ \beta ,\ \gamma$, then the resultant force in the direction of $ds_{1}$ is

$R_{1}=-{\frac {dV}{ds_{1}}}=-{\frac {dV}{d\alpha }}{\frac {d\alpha }{ds_{1}}}=-{\frac {dV}{d\alpha }}{\frac {c}{D_{2}D_{3}}}.$

(9)

Since $ds_{1},\ ds_{2}$, and $ds_{3}$ are at right angles to each other, the surface-integral over the element of area $ds_{2}\ ds_{3}$ is

${\begin{array}{ll}R_{1}ds_{2}ds_{3}&={\frac {dV}{d\alpha }}{\frac {c}{D_{2}D_{3}}}\cdot {\frac {D_{3}D_{1}}{c}}\cdot {\frac {D_{1}D_{2}}{c}}\cdot d\beta \ d\gamma \\\\&={\frac {dV}{d\alpha }}{\frac {D_{1}^{2}}{c}}d\beta \ d\gamma .\end{array}}$

(10)

Now consider the element of volume intercepted between the surfaces $\alpha ,\ \beta ,\ \gamma$, and $\alpha +d\alpha ,\ \beta +d\beta ,\ \gamma +d\gamma .$. There will be eight such elements, one in each octant of space.

We have found the surface-integral for the element of surface intercepted from the surface $\alpha$ by the surfaces $\beta$ and $\beta +d\beta$, $\gamma$ and $\gamma +d\gamma$.