Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/308

Since $p$ is a homogeneous function of the degree $-1$ in $r$ and $a$,

	$a{\frac {dp}{da}}$	${}+r{\frac {dp}{dr}}=-p$,
or	${\frac {dp}{da}}$	${}=-{\frac {1}{a}}{\frac {d}{dr}}(pr)$,
and	$\Omega$	${}=-\iint {\frac {\phi }{a}}{\frac {d}{dr}}(pr)\,dS$.

Since $r$ and $a$ are constant during the surface-integration,

$\Omega =-{\frac {1}{a}}{\frac {d}{dr}}\left(r\iint \phi p\,ds\right)$.

But if $P$ is the potential due to a sheet of imaginary matter of surface-density $\phi$,

$P=\iint \phi p\,dS$,

and $\Omega$, the magnetic potential of the current-sheet, may he expressed in terms of $P$ in the form

$\Omega =-{\frac {1}{a}}{\frac {d}{dr}}(Pr)$.

671.] We may determine $F$, the $x$-component of the vector-potential, from the expression given in Art. 416,

$F=\iint \phi \left(m{\frac {dp}{d\zeta }}-n{\frac {dp}{d\eta }}\right)\,dS$,

where $\xi$, $\eta$, $\zeta$ are the coordinates of the element $dS$, and $l$, $m$, $n$ are the direction-cosines of the normal. Since the sheet is a sphere, the direction-cosines of the normal are

$l={\frac {\xi }{a}}$,⁠$m={\frac {\eta }{a}}$,⁠$n={\frac {\zeta }{a}}$.

so that

$m{\frac {dp}{d\zeta }}-n{\frac {dp}{d\eta }}$	${}=\left(\eta (z-\zeta )-\zeta (y-\eta )\right){\frac {p^{3}}{a}}$,
	${}=\left(z(\eta -y)-y(\zeta -z)\right){\frac {p^{3}}{a}}$,
	${}={\frac {z}{a}}{\frac {dp}{dy}}-{\frac {y}{a}}{\frac {dp}{dz}}$;

multiplying by $\phi \,dS$, and integrating over the surface of the sphere, we find

$F={\frac {z}{a}}{\frac {dP}{dy}}-{\frac {y}{a}}{\frac {dP}{dz}}$.