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

$4\pi \mu u=\frac{d^{2}G}{dxdy}-\frac{d^{2}F}{d{y}^2}-\frac{d^{2}F}{d^{2}z}+\frac{d^{2}H}{dzdx}.$

(1)

If we write

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

(2)

and^[1]

${\mathrm{\nabla }}^2=-(\frac{d^2}{d{x}^2}+\frac{d^2}{d{y}^2}+\frac{d^2}{d{z}^2}),$

(3)

we may write equation (1),

Similarly,

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

(4)

If we write

$\begin{array}{c}{F}^{\prime }=\frac{1}{\mu }\iiint \frac{u}{r}dxdydz,\\ G^{\prime }=\frac{1}{\mu }\iiint \frac{v}{r}dxdydz,\\ H^{\prime }=\frac{1}{\mu }\iiint \frac{r}{r}dxdydz.\end{array}\}$

(5)

$\chi =\frac{4\pi }{\mu }\iiint \frac{J}{r}dxdydz,$

(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

$\begin{array}{cc}F& =F^{\prime }+\frac{d\chi }{dx},\\ G& =G^{\prime }+\frac{d\chi }{dy},\\ H& =H^{\prime }+\frac{d\chi }{dz}.\end{array}\}$

(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 $𝔄$.

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