Page:On Faraday's Lines of Force.pdf/45

Rh In the same way it may be shewn that the values of $$\alpha, \beta, \gamma$$ satisfy the other given equations. The function $$\psi$$ may be considered at present as perfectly indeterminate.

The method here given is taken from Prof. W. Thomson’s memoir on Magnetism (Phil. Trans. 1851, p. 283).

As we cannot perform the required integrations when $$a, b, c$$ are discontinuous functions of $$x, y, z,$$ the following method, which is perfectly general though more complicated, may indicate more clearly the truth of the proposition.

Let $$A, B, C$$ be determined from the equations

$\frac{d^2A}{dx^2}+\frac{d^2A}{dy^2}+\frac{d^2A}{dz^2}+a=0$,

$\frac{d^2B}{dx^2}+\frac{d^2B}{dy^2}+\frac{d^2B}{dz^2}+b=0$,

$\frac{d^2C}{dx^2}+\frac{d^2C}{dy^2}+\frac{d^2C}{dz^2}+a=0$,

by the methods of Theorems I. and II., so that $$A, B, C$$ are never inﬁnite, and vanish when $$x, y$$, or $$z$$ is infinite.

Also let

$\alpha=\frac{dB}{dz}-\frac{dC}{dy}+\frac{d\psi}{dx}$,

$\beta=\frac{dC}{dx}-\frac{dA}{dz}+\frac{d\psi}{dy}$,

$\gamma=\frac{dA}{dy}-\frac{dB}{dx}+\frac{d\psi}{dz}$,

then

$\frac{d\beta}{dz}-\frac{d\gamma}{dy}=\frac{d}{dx}\left(\frac{dA}{dx}+\frac{dB}{dy}_\frac{dC}{dz}\right)-\left(\frac{d^2A}{dx^2}+\frac{d^2A}{dy^2}+\frac{d^2A}{dz^2}\right)$

$=\frac{d}{dx}\left(\frac{dA}{dx}+\frac{dB}{dy}+\frac{dC}{dz}\right)+a$

If we ﬁnd similar equations in $$y$$ and $$z$$, and differentiate the ﬁrst by $$x$$, the second by $$y$$, and the third by $$z$$, remembering the equation between $$a, b, c,$$ we shall have

$\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2}\right)\left(\frac{dA}{dx}+\frac{dB}{dy}+\frac{dC}{dz}\right)=0;$