Page:ThomsonMagnetic1889.djvu/6

6 the components of the rector potential, and $$\sigma$$ the specific resistance of the metal.

{{MathForm2|(2)|$$\left.\begin{alignat}{2} \sigma u & = -b\omega\epsilon^{ipt} && -\dfrac{d\mathrm{F}}{dt}-\dfrac{d\psi}{dx},\\ \\ \sigma v & = -a\omega\epsilon^{ipt} && -\dfrac{d\mathrm{G}}{dt}-\dfrac{d\psi}{dy},\\ \\ \sigma w & = && -\dfrac{d\mathrm{H}}{dt}-\dfrac{d\psi}{dz}; \end{alignat}\right\} $$}}

and therefore

$\begin{alignat}{2} -\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{F} & =-b\omega\epsilon^{ipt} && -\dfrac{d\mathrm{F}}{dt}-\dfrac{d\psi}{dx},\\ \\ -\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{G} & =a\omega\epsilon^{ipt} && -\dfrac{d\mathrm{G}}{dt}-\dfrac{d\psi}{dy},\\ \\ -\dfrac{\sigma}{4\pi\mu'}\nabla^{2}\mathrm{H} & =-\dfrac{d\mathrm{H}}{dt} && -\dfrac{d\psi}{dz}. \end{alignat}$|undefined

In the dielectric outside the sphere, if $$f, g, h$$ are the electric displacements, $$\mathrm{K}$$ the specific inductive capacity, and if $$\partial/\partial t$$ denote partial differentiation with respect to the time, the equations are

$\begin{array}{l} \dfrac{4\pi}{\mathrm{K}}f=-\dfrac{d\mathrm{F}}{dt}-\dfrac{d\psi}{dx}=-\left(\dfrac{\partial}{\partial t}-\omega\epsilon^{ipt}\dfrac{d}{dz}\right)\mathrm{F}-\dfrac{d\psi}{dx},\\ \\ \dfrac{4\pi}{\mathrm{K}}g=-\dfrac{d\mathrm{G}}{dt}-\dfrac{d\psi}{dy}=-\left(\dfrac{\partial}{\partial t}-\omega\epsilon^{ipt}\dfrac{d}{dz}\right)\mathrm{G}-\dfrac{d\psi}{dy},\\ \\ \dfrac{4\pi}{\mathrm{K}}h=-\dfrac{d\mathrm{H}}{dt}-\dfrac{d\psi}{dz}=-\left(\dfrac{\partial}{\partial t}-\omega\epsilon^{ipt}\dfrac{d}{dz}\right)\mathrm{H}-\dfrac{d\psi}{dz}; \end{array}$|undefined

and therefore

with a similar equation for $$\mathrm{G}$$.

From the form of these equations we see that the solution will take the form

$\begin{array}{l} \psi=\psi_{0}+\psi_{1}\epsilon^{ipt}+\psi'_{1}\epsilon^{-ipt}+\psi_{2}\epsilon^{2ipt}+\psi'_{2}\epsilon^{-2ipt}+\dots\\ \\ \mathrm{F}=\mathrm{F}_{1}\epsilon^{ipt}+\mathrm{F}'_{1}\epsilon^{-ipt}+\mathrm{F}_{2}\epsilon^{2ipt}+\mathrm{F}'_{2}\epsilon^{-2ipt}. \end{array}$

If we substitute these values in the above equations, we see that we may put $$\psi'_{1}\ \psi'_{2}\dots\mathrm{F}'_{1},\ \mathrm{F}'_{2}$$ all equal to zero.

If $$e$$ is the quantity of electricity on the sphere,

$\psi_{0}=\dfrac{e}{\mathrm{K}r}.$