Page:On Faraday's Lines of Force.pdf/46

200 and since A, B, C are always ﬁnite and vanish at an inﬁnite distance, the only solution of this equation is

$\frac{dA}{dx}+\frac{dB}{dy}+\frac{dC}{dz}=0$,

and we have ﬁnally

$\frac{d\beta}{dz}-\frac{d\gamma}{dy}=a$,

with two similar equations, shewing that $$\alpha, \beta, \gamma$$ have been rightly determined.

The function $$\psi$$ is to be determined From the condition

$\frac{d\alpha}{dx}+\frac{d\beta}{dy}+\frac{d\gamma}{dz}=\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2}\right)\psi :$

if the left-hand side of this equation be always zero, $$\psi$$ must be zero also.

THEOREM VI.

Let $$a, b, c$$ be any three functions of $$x, y, z$$, it is possible to ﬁnd three functions $$\alpha, \beta, \gamma$$ and a fourth $$V$$, so that

$\frac{d\alpha}{dx}+\frac{d\beta}{dy}+\frac{d\gamma}{dz}=0 ,$

and

$a=\frac{d\beta}{dz}-\frac{d\gamma}{dy}+\frac{dV}{dx}$,

$b=\frac{d\gamma}{dx}-\frac{d\alpha}{dz}+\frac{dV}{dy}$,

$c=\frac{d\alpha}{dy}-\frac{d\beta}{dx}+\frac{dV}{dz}$.

Let

$\frac{da}{dx}+\frac{db}{dy}+\frac{dc}{dz}=-4\pi\rho$,

and let V be found from the equation

$\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}=-4\pi\rho$,