Page:Philosophical magazine 21 series 4.djvu/311

applied to Electric Currents. 287

in unit of volume is

$E=C\mu \left(\alpha ^{2}+\beta ^{2}+\gamma ^{2}\right),$

where C is a constant to be determined.

Let us take the case in which

$\alpha ={\frac {d\phi }{dx}},\ \beta ={\frac {d\phi }{dy}},\ \gamma ={\frac {d\phi }{dz}}$

(35)

Let

$\phi =\phi _{1}+\phi _{2}\,$

(36)

and let

${\frac {\mu }{4\pi }}\left({\frac {d^{2}\phi _{1}}{dx^{2}}}+{\frac {d^{2}\phi _{1}}{dy^{2}}}+{\frac {d^{2}\phi _{1}}{dz^{2}}}\right)=m_{1}\ \mathrm {and} \ {\frac {\mu }{4\pi }}\left({\frac {d^{2}\phi _{2}}{dx^{2}}}+{\frac {d^{2}\phi _{2}}{dy^{2}}}+{\frac {d^{2}\phi _{2}}{dz^{2}}}\right)=m_{2};$

(37)

then $\phi _{1}$ is the potential at any point due to the magnetic system $m_{1}$, and $\phi _{2}$ that due to the distribution of magnetism represented by $m_{2}$. The actual energy of all the vortices is

$E=\Sigma C\mu \left(\alpha ^{2}+\beta ^{2}+\gamma ^{2}\right)dV,$

(38)

the integration being performed over all space.

This may be shown by integration by parts (see Green's 'Essay on Electricity,' p. 10) to be equal to

$E=-4\pi C\Sigma \left(\phi _{1}m_{1}+\phi _{2}m_{2}+\phi _{1}m_{2}+\phi _{2}m_{1}\right)dV.$

(39)

Or since it has been proved (Green's 'Essay,' p. 10) that

${\begin{array}{c}\Sigma \phi _{1}m_{2}dV=\Sigma \phi _{2}m_{1}dV,\\E=-4\pi C\left(\phi _{1}m_{1}+\phi _{2}m_{2}+2\phi _{1}m_{2}\right)dV.\end{array}}$

(40)

Now let the magnetic system $m_{1}$ remain at rest, and let $m_{2}$ be moved parallel to itself in the direction of $x$ through a space $\delta x$; then, since $\phi _{1}$ depends on $m_{1}$ only, it will remain as before, so that $\phi _{1}m_{1}$ will be constant; and since $\phi _{2}$ depends on $m_{2}$ only, the distribution of $\phi _{2}$ about $m_{2}$ will remain the same, so that $\phi _{2}m_{2}$ will be the same as before the change. The only part of E that will be altered is that depending on $2\phi _{1}m_{2}$, because $\phi _{1}$ becomes $\phi _{1}+{\tfrac {d\phi _{1}}{dx}}\delta x$ on account of the displacement. The variation of actual energy due to the displacement is therefore

$\delta E=-4\pi C\Sigma \left(2{\frac {d\phi _{1}}{dx}}m_{2}\right)dV\ \delta x.$

(41)

But by equation (12), the work done by the mechanical forces on $m_{2}$ during the motion is

$\delta W=\Sigma \left({\frac {d\phi _{1}}{dx}}m_{2}dV\right)\delta x;$

(42)

and since our hypothesis is a purely mechanical one, we must