Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/184

given their relative position is as completely determined as if they formed part of the same rigid body.

512.] If we use rectangular coordinates and make $x$, $y$, $z$ the coordinates of $P$, and $x^{\prime }$, $y^{\prime }$, $z^{\prime }$ those of $P^{\prime }$, and if we denote by $l$, $m$, $n$ and by $l^{\prime }$, $m^{\prime }$, $n^{\prime }$ the direction-cosines of $PQ$, and of $P^{\prime }Q^{\prime }$ respectively, then

${\frac {dx}{ds}}=l$,		${\frac {dy}{ds}}=m$,		${\frac {dz}{ds}}=n$,
${\frac {dx^{\prime }}{ds^{\prime }}}=l^{\prime }$,		${\frac {dy^{\prime }}{ds^{\prime }}}=m^{\prime }$,		${\frac {dz^{\prime }}{ds^{\prime }}}=n^{\prime }$,

(2)

and

$l(x^{\prime }-x)+m(y^{\prime }-y)+n(z^{\prime }-z)={}$	$r\cos \theta$,
$l^{\prime }(x^{\prime }-z)+m^{\prime }(y^{\prime }-y)+n^{\prime }(z^{\prime }-z)={}$	$-r\cos \theta ^{\prime }$,
$ll^{\prime }+mm^{\prime }+nn^{\prime }=\cos \epsilon$,

(3)

where $\epsilon$ is the angle between the directions of the elements themselves, and

whence

$r{\frac {dr}{ds}}$	${}=-(x^{\prime }-x){\frac {dx}{ds}}-(y^{\prime }-y){\frac {dy}{ds}}-(z^{\prime }-z){\frac {dz}{ds}}$,
	${}=-r\cos \theta$.

(6)

Similarly

$r{\frac {dr}{ds^{\prime }}}$	${}=(x^{\prime }-x){\frac {dx^{\prime }}{ds^{\prime }}}+(y^{\prime }-y){\frac {dy^{\prime }}{ds^{\prime }}}+(z^{\prime }-z){\frac {dz^{\prime }}{ds^{\prime }}}$,
	${}=-r\cos \theta ^{\prime }$;

and differentiating $r{\frac {dr}{ds}}$ with respect to $s^{\prime }$,

$r{\frac {d^{2}r}{ds\,ds^{\prime }}}+{\frac {dr}{ds}}{\frac {dr}{ds^{\prime }}}$	${}=-{\frac {dx}{ds}}{\frac {dx^{\prime }}{ds^{\prime }}}-{\frac {dy}{ds}}{\frac {dy^{\prime }}{ds^{\prime }}}-{\frac {dz}{ds}}{\frac {dz^{\prime }}{ds^{\prime }}}$
	${}=-(ll^{\prime }+mm^{\prime }+nn^{\prime })$
	${}=-\cos \epsilon$.

(7)

We can therefore express the three angles $\theta$, $\theta ^{\prime }$, and $\eta$, and the auxiliary angle $\epsilon$ in terms of the differential coefficients of $r$ with respect to $s$ and $s^{\prime }$ as follows,

$\cos \theta$	${}=-{\frac {dr}{ds}}$,
$\cos \theta ^{\prime }$	${}=-{\frac {dr}{ds^{\prime }}}$,
$\cos \epsilon$	${}=-r{\frac {d^{2}r}{ds\,ds^{\prime }}}-{\frac {dr}{ds}}{\frac {dr}{ds^{\prime }}}$,
$\sin \theta \sin \theta ^{\prime }\cos \eta$	${}=-r{\frac {d^{2}r}{ds\,ds^{\prime }}}$.

(8)