Page:ThomsonMagnetic1889.djvu/6

$\mathrm {G,H}$ the components of the rector potential, and $\sigma$ the specific resistance of the metal.

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

(2)

and therefore

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

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

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

and therefore

$-{\dfrac {1}{\mu K}}\nabla ^{2}\mathrm {F} =-\left({\dfrac {\partial }{\partial t}}-\omega \epsilon ^{ipt}{\dfrac {d}{dz}}\right)^{2}\mathrm {F} -\left({\dfrac {\partial }{\partial t}}-\omega \epsilon ^{ipt}{\dfrac {d}{dz}}\right){\dfrac {d\psi }{dx}},$

(3)

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,

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