Page:ThomsonMagnetic1889.djvu/12

So that if the sphere is placed in a magnetic field the force acting upon it is the same as that on a current ${\tfrac {1}{3}}e\omega \cos pt$.

When the sphere is moving with a uniform velocity $\omega$, equations (3) become

$-{\frac {1}{\mu \mathrm {K} }}\nabla ^{2}\mathrm {F} =-\omega ^{2}{\frac {d^{2}\mathrm {F} }{dz^{2}}}+\omega {\frac {d^{2}\psi }{dx\ dz}};$

whence

${\frac {d^{2}a}{dx^{2}}}+{\frac {d^{2}a}{dy^{2}}}+{\frac {d^{2}a}{dz^{2}}}\left(1-{\frac {\omega ^{2}}{v^{2}}}\right)=0,$

where $v$ is the velocity of propagation of electrodynamic action through the dielectric. If we put

$z=\left\{1-{\frac {\omega ^{2}}{v^{2}}}\right\}^{\frac {1}{2}}z',$

this equation becomes

${\frac {d^{2}a}{dx^{2}}}+{\frac {d^{2}a}{dy^{2}}}+{\frac {d^{2}a}{dz_{1}^{2}}}=0.$

With similar equations for $b$ and $c$, we see that a solution of these equations is

${\begin{array}{l}a=k{\dfrac {d}{dy}}{\dfrac {1}{\sqrt {x^{2}+y^{2}+z_{1}^{2}}}};\\\\b=-k{\dfrac {d}{dx}}{\dfrac {1}{\sqrt {x^{2}+y^{2}+z_{1}^{2}}}};\\\\c=0;\end{array}}$

where $k$ is a constant. Since, if $\mu =1$,

$4\pi {\frac {df}{dt}}={\frac {dc}{dy}}-{\frac {db}{dz}},$

and

${\frac {df}{dt}}=-\omega {\frac {df}{dz}},$

we have

$-4\pi \omega f=-b=k{\frac {d}{dx}}{\frac {1}{\sqrt {x^{2}+y^{2}+z_{1}^{2}}}}.$

Similarly

$-4\pi \omega g=+a=k{\frac {d}{dy}}{\frac {1}{\sqrt {x^{2}+y^{2}+z_{1}^{2}}}},$

and

$-4\pi \omega {\frac {dh}{dz}}={\frac {db}{dx}}-{\frac {da}{dy}}=k{\frac {d}{dz_{1}^{2}}}{\frac {1}{\sqrt {x^{2}+y^{2}+z_{1}^{2}}}};$