Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/189

In the language of quaternions, the resultant force on $ds$ is the vector part of the product of the directrix multiplied by $ds$.

Since we already know that the directrix is the same thing as the magnetic force due to a unit current in the circuit $s^{\prime }$, we shall henceforth speak of the directrix as the magnetic force due to the circuit.

518.] We shall now complete the calculation of the components of the force acting between two finite currents, whether closed or open.

Let $\rho$ be a new function of $r$, such that

then by (17) and (20)

and equations (11) become

$R=-{\frac {d\rho }{dr}}\cos \epsilon +r{\frac {d^{2}}{ds\,ds^{\prime }}}(Q+\rho )$, $S=-{\frac {dQ}{ds^{\prime }}}$, $S^{\prime }={\frac {dQ}{ds}}$.

(26)

With these values of the component forces, equation (13) becomes

${\frac {d^{2}X}{ds\,ds^{\prime }}}$	${}=-{}$	$\cos \epsilon {\frac {d\rho }{dr}}{\frac {\xi }{r}}+\xi {\frac {d^{2}}{ds\,ds^{\prime }}}(Q+\rho )-l{\frac {dQ}{ds^{\prime }}}+l^{\prime }{\frac {dQ}{ds}}$,
	${}={}$	$\cos \epsilon {\frac {d\rho }{dx}}+{\frac {d^{2}(Q+\rho )\xi }{ds\,ds^{\prime }}}+l{\frac {d\rho }{ds^{\prime }}}-l^{\prime }{\frac {d\rho }{ds}}$.

(27)

519.] Let

	$F=\int _{0}^{s}l\rho \,ds$,	$G=\int _{0}^{s}m\rho \,ds$,	$H=\int _{0}^{s}n\rho \,ds$,	(28)
	$F^{\prime }=\int _{0}^{s^{\prime }}l^{\prime }\rho \,ds^{\prime }$,	$G^{\prime }=\int _{0}^{s^{\prime }}m^{\prime }\rho \,ds^{\prime }$,	$H^{\prime }=\int _{0}^{s^{\prime }}n^{\prime }\rho \,ds^{\prime }$.	(29)

These quantities have definite values for any given point of space. When the circuits are closed, they correspond to the components of the vector-potentials of the circuits.

Let $L$ be a new function of $r$, such that

and let $M$ be the double integral