182
MR. W. B. MORTON ON THE
with uniform velocity u parallel to the axis of z , and that the field has assumed its steady configuration. We shall denote
1
−
u
2
V
2
{\displaystyle 1-{\frac {u^{2}}{V^{2}}}}
by k ², V being the velocity of light. Then since we have a steady state,
d
d
t
=
−
u
d
d
z
{\displaystyle {\frac {d}{dt}}=-u{\frac {d}{dz}}}
.
Also, since each element of charge produces a magnetic field with no z-component, we have
γ
=
0
{\displaystyle \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
−
d
β
d
z
=
−
4
π
u
d
f
d
z
{\displaystyle -{\frac {d\beta }{dz}}=-4\pi u{\frac {df}{dz}}}
,
d
α
d
z
=
−
4
π
u
d
g
d
z
{\displaystyle {\frac {d\alpha }{dz}}=-4\pi u{\frac {dg}{dz}}}
,
d
β
d
x
−
d
α
d
y
=
−
4
π
u
d
h
d
z
{\displaystyle {\frac {d\beta }{dx}}-{\frac {d\alpha }{dy}}=-4\pi u{\frac {dh}{dz}}}
,
d
g
d
z
−
d
h
d
y
=
−
u
4
π
V
2
d
α
d
z
{\displaystyle {\frac {dg}{dz}}-{\frac {dh}{dy}}=-{\frac {u}{4\pi V^{2}}}{\frac {d\alpha }{dz}}}
,
d
h
d
x
−
d
f
d
z
=
−
u
4
π
V
2
d
β
d
z
{\displaystyle {\frac {dh}{dx}}-{\frac {df}{dz}}=-{\frac {u}{4\pi V^{2}}}{\frac {d\beta }{dz}}}
,
d
f
d
y
−
d
g
d
x
=
0
{\displaystyle {\frac {df}{dy}}-{\frac {dg}{dx}}=0}
.
These equations together with
d
f
d
x
+
d
g
d
y
+
d
h
d
z
=
0
{\displaystyle {\frac {df}{dx}}+{\frac {dg}{dy}}+{\frac {dh}{dz}}=0}
are satisfied by
f
=
−
d
ϕ
d
x
,
g
=
−
d
ϕ
d
y
,
h
=
−
k
2
d
ϕ
d
z
{\displaystyle f=-{\frac {d\phi }{dx}},\ g=-{\frac {d\phi }{dy}},\ h=-k^{2}{\frac {d\phi }{dz}}}
,
α
=
4
π
u
d
ϕ
d
y
,
β
=
−
4
π
d
ϕ
d
x
{\displaystyle \alpha =4\pi u{\frac {d\phi }{dy}},\ \beta =-4\pi {\frac {d\phi }{dx}}}
,
where φ is any function satisfying
d
2
ϕ
d
x
2
+
d
2
ϕ
d
y
2
+
k
2
d
2
ϕ
d
z
2
=
0
{\displaystyle {\frac {d^{2}\phi }{dx^{2}}}+{\frac {d^{2}\phi }{dy^{2}}}+k^{2}{\frac {d^{2}\phi }{dz^{2}}}=0}
.