Page:MortonCharge.djvu/3

with uniform velocity u parallel to the axis of z, and that the field has assumed its steady configuration. We shall denote $1-{\frac {u^{2}}{V^{2}}}$ by k², V being the velocity of light. Then since we have a steady state,

${\frac {d}{dt}}=-u{\frac {d}{dz}}$.

Also, since each element of charge produces a magnetic field with no z-component, we have $\gamma =0$ in the general case also. Using these two data, the equations connecting the displacement (f, g, h) and the magnetic force (α, β, γ) become

$-{\frac {d\beta }{dz}}=-4\pi u{\frac {df}{dz}}$,

${\frac {d\alpha }{dz}}=-4\pi u{\frac {dg}{dz}}$,

${\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}=-4\pi u{\frac {dh}{dz}}$,

${\frac {dg}{dz}}-{\frac {dh}{dy}}=-{\frac {u}{4\pi V^{2}}}{\frac {d\alpha }{dz}}$,

${\frac {dh}{dx}}-{\frac {df}{dz}}=-{\frac {u}{4\pi V^{2}}}{\frac {d\beta }{dz}}$,

${\frac {df}{dy}}-{\frac {dg}{dx}}=0$.

These equations together with

${\frac {df}{dx}}+{\frac {dg}{dy}}+{\frac {dh}{dz}}=0$

are satisfied by

$f=-{\frac {d\phi }{dx}},\ g=-{\frac {d\phi }{dy}},\ h=-k^{2}{\frac {d\phi }{dz}}$, $\alpha =4\pi u{\frac {d\phi }{dy}},\ \beta =-4\pi {\frac {d\phi }{dx}}$,

where φ is any function satisfying

${\frac {d^{2}\phi }{dx^{2}}}+{\frac {d^{2}\phi }{dy^{2}}}+k^{2}{\frac {d^{2}\phi }{dz^{2}}}=0$.