that only the vectors denoted by accented symbols shall remain, by substituting from these latter formulae: thus
g
=
g
′
+
v
4
π
c
2
(
c
′
+
4
π
v
g
)
,
{\displaystyle g=g'+{\frac {v}{4\pi c^{2}}}(c'+4\pi vg),}
so that
ϵ
−
1
g
=
g
′
+
v
4
π
c
2
c
′
,
{\displaystyle \epsilon ^{-1}g=g'+{\frac {v}{4\pi c^{2}}}c',}
where ε is equal to
(
1
−
v
2
/
c
2
)
−
1
{\displaystyle \left(1-v^{2}/c^{2}\right)^{-1}}
, and exceeds unity;
and
b
=
b
′
−
4
π
v
(
h
′
−
v
4
π
c
2
b
)
{\displaystyle b=b'-4\pi v\left(h'-{\frac {v}{4\pi c^{2}}}b\right)}
so that
ϵ
−
1
b
=
b
′
−
4
π
v
h
′
;
{\displaystyle \epsilon ^{-1}b=b'-4\pi vh';}
giving the general relations
ϵ
−
1
(
a
,
b
,
c
)
=
(
ϵ
−
1
a
′
,
b
′
−
4
π
v
h
′
,
c
′
+
4
π
v
g
′
)
{\displaystyle \epsilon ^{-1}(a,\ b,\ c)=\left(\epsilon ^{-1}a',\ b'-4\pi vh',\ c'+4\pi vg'\right)}
ϵ
−
1
(
f
,
g
,
h
)
=
(
ϵ
−
1
f
′
,
g
′
+
v
4
π
c
2
c
′
,
h
′
−
v
4
π
c
2
b
′
)
.
{\displaystyle \epsilon ^{-1}(f,\ g,\ h)=\left(\epsilon ^{-1}f',\ g'+{\frac {v}{4\pi c^{2}}}c',\ h'-{\frac {v}{4\pi c^{2}}}b'\right).}
Hence
4
π
d
f
′
d
t
′
=
d
c
′
d
y
′
−
d
b
′
d
z
′
{\displaystyle 4\pi {\frac {df'}{dt'}}={\frac {dc'}{dy'}}-{\frac {db'}{dz'}}}
4
π
ϵ
d
g
′
d
t
′
=
d
a
′
d
z
′
−
(
d
d
x
′
+
v
c
2
ϵ
d
d
t
′
)
c
′
{\displaystyle 4\pi \epsilon {\frac {dg'}{dt'}}={\frac {da'}{dz'}}-\left({\frac {d}{dx'}}+{\frac {v}{c^{2}}}\epsilon {\frac {d}{dt'}}\right)c'}
4
π
ϵ
d
h
′
d
t
′
=
(
d
d
x
′
+
v
c
2
ϵ
d
d
t
′
)
b
′
−
d
a
′
d
y
′
{\displaystyle 4\pi \epsilon {\frac {dh'}{dt'}}=\left({\frac {d}{dx'}}+{\frac {v}{c^{2}}}\epsilon {\frac {d}{dt'}}\right)b'-{\frac {da'}{dy'}}}
−
(
4
π
c
2
)
−
1
d
a
′
d
t
′
=
d
h
′
d
y
′
−
d
g
′
d
z
′
{\displaystyle -\left(4\pi c^{2}\right)^{-1}{\frac {da'}{dt'}}={\frac {dh'}{dy'}}-{\frac {dg'}{dz'}}}
−
(
4
π
c
2
)
−
1
ϵ
d
b
′
d
t
′
=
d
f
′
d
z
′
−
(
d
d
x
′
+
v
c
2
ϵ
d
d
t
′
)
h
′
{\displaystyle -\left(4\pi c^{2}\right)^{-1}\epsilon {\frac {db'}{dt'}}={\frac {df'}{dz'}}-\left({\frac {d}{dx'}}+{\frac {v}{c^{2}}}\epsilon {\frac {d}{dt'}}\right)h'}
−
(
4
π
c
2
)
−
1
ϵ
d
c
′
d
t
′
=
(
d
d
x
′
+
v
c
2
ϵ
d
d
t
′
)
g
′
−
d
f
′
d
y
′
.
{\displaystyle -\left(4\pi c^{2}\right)^{-1}\epsilon {\frac {dc'}{dt'}}=\left({\frac {d}{dx'}}+{\frac {v}{c^{2}}}\epsilon {\frac {d}{dt'}}\right)g'-{\frac {df'}{dy'}}.}
Now change the time-variable from t' to t" , equal to
t
′
−
v
c
2
ϵ
x
′
{\displaystyle t'-{\frac {v}{c^{2}}}\epsilon x'}
; this will involve that
d
d
x
′
+
v
c
2
ϵ
d
d
t
′
{\displaystyle {\frac {d}{dx'}}+{\frac {v}{c^{2}}}\epsilon {\frac {d}{dt'}}}
is replaced by
d
d
x
′
{\displaystyle {\frac {d}{dx'}}}
, while the other differential operators remain unmodified; thus the scheme of equations reverts to the same type as when it was referred to axes at rest, except as regards the factors ε on the left-hand sides.