Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/267

According to our hypothesis a, b, c are identical with μα, μβ, μγ respectively. We therefore obtain

$4\pi \mu u={\frac {d^{2}G}{dx\,dy}}-{\frac {d^{2}F}{dy^{2}}}-{\frac {d^{2}F}{d^{2}z}}+{\frac {d^{2}H}{dz\,dx}}.$

(1)

If we write

$J={\frac {dF}{dx}}+{\frac {dG}{dy}}+{\frac {dH}{dz}},$

(2)

and[1]

$\nabla ^{2}=-\left({\frac {d^{2}}{dx^{2}}}+{\frac {d^{2}}{dy^{2}}}+{\frac {d^{2}}{dz^{2}}}\right),$

(3)

we may write equation (1),

Similarly,

$\left.{\begin{aligned}4\pi \mu u&={\frac {dJ}{dx}}+\nabla ^{2}F.\\4\pi \mu v&={\frac {dJ}{dy}}+\nabla ^{2}G,\\4\pi \mu w&={\frac {dJ}{dz}}+\nabla ^{2}H.\end{aligned}}\right\}$

(4)

If we write

$\left.{\begin{aligned}F'={\frac {1}{\mu }}\iiint {{\frac {u}{r}}dx\,dy\,dz,}\\G'={\frac {1}{\mu }}\iiint {{\frac {v}{r}}dx\,dy\,dz,}\\H'={\frac {1}{\mu }}\iiint {{\frac {r}{r}}dx\,dy\,dz.}\end{aligned}}\right\}$

(5)

$\chi ={\frac {4\pi }{\mu }}\iiint {{\frac {J}{r}}dx\,dy\,dz},$

(6)

where r is the distance of the given point from the element x y z, and the integrations are to be extended over all space, then

$\left.{\begin{aligned}F&=F'+{\frac {d\chi }{dx}},\\G&=G'+{\frac {d\chi }{dy}},\\H&=H'+{\frac {d\chi }{dz}}.\end{aligned}}\right\}$

(7)

The quantity χ disappears from the equations (A), and it is not related to any physical phenomenon. If we suppose it to be zero everywhere, J will also be zero everywhere, and equations (5), omitting the accents, will give the true values of the components of ${\mathfrak {A}}$.

1. The negative sign is employed here in order to make our expressions consistent with those in which Quaternions are employed.