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

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

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