I say that these three properties will remain, even when the velocity is not zero, and for this it is enough to show that they are not altered by the Lorentz transformation.
Indeed, let A be the intensity common to both fields, let
(
x
−
x
1
)
=
r
λ
,
(
y
−
y
1
)
=
r
μ
,
(
z
−
z
1
)
=
r
ν
,
λ
2
+
μ
2
+
ν
2
=
1.
{\displaystyle (x-x_{1})=r\lambda ,\quad (y-y_{1})=r\mu ,\quad (z-z_{1})=r\nu ,\quad \lambda ^{2}+\mu ^{2}+\nu ^{2}=1.}
These properties expressed through the equalities
{
A
2
=
∑
f
2
=
∑
α
2
,
∑
f
α
=
0
,
∑
f
(
x
−
x
1
)
=
0
,
∑
α
(
x
−
x
1
)
=
0
∑
f
λ
=
0
,
∑
α
λ
=
0
;
{\displaystyle {\begin{cases}&A^{2}=\sum f^{2}=\sum \alpha ^{2},\quad \sum f\alpha =0,\quad \sum f\left(x-x_{1}\right)=0,\quad \sum \alpha \left(x-x_{1}\right)=0\\\\&\sum f\lambda =0,\quad \sum \alpha \lambda =0;\end{cases}}}
which means again that
b
A
,
g
A
,
h
A
α
A
,
β
A
,
γ
A
λ
,
μ
,
ν
{\displaystyle {\begin{array}{ccccc}{\frac {b}{A}},&&{\frac {g}{A}},&&{\frac {h}{A}}\\\\{\frac {\alpha }{A}},&&{\frac {\beta }{A}},&&{\frac {\gamma }{A}}\\\\\lambda ,&&\mu ,&&\nu \end{array}}}
are the direction cosines of three rectangular directions, and we deduce the relations:
f
=
β
ν
−
γ
μ
,
α
=
h
μ
−
g
ν
,
{\displaystyle f=\beta \nu -\gamma \mu ,\quad \alpha =h\mu -g\nu ,}
or
(6)
f
r
=
β
(
z
−
z
1
)
−
γ
(
y
−
y
1
)
α
r
=
h
(
y
−
y
1
)
−
g
(
z
−
z
1
)
,
{\displaystyle fr=\beta \left(z-z_{1}\right)-\gamma \left(y-y_{1}\right)\quad \alpha r=h\left(y-y_{1}\right)-g\left(z-z_{1}\right),}
with the equations that we can deduce by symmetry.
If we take the equations (3) of § 1, we find:
(7)
{
x
′
−
x
1
′
=
k
l
[
(
x
−
x
1
)
+
ϵ
(
t
−
t
1
)
]
=
k
l
[
(
x
−
x
1
)
+
ϵ
r
]
,
y
′
−
y
1
′
=
l
(
y
−
y
1
)
,
z
′
−
z
1
′
=
l
(
z
−
z
1
)
.
{\displaystyle {\begin{cases}x^{\prime }-x_{1}^{\prime }=kl\left[\left(x-x_{1}\right)+\epsilon \left(t-t_{1}\right)\right]=kl\left[\left(x-x_{1}\right)+\epsilon r\right],\\\\y^{\prime }-y_{1}^{\prime }=l\left(y-y_{1}\right),\\\\z^{\prime }-z_{1}^{\prime }=l\left(z-z_{1}\right).\end{cases}}}
We found above in § 3:
l
4
(
∑
f
′
2
−
∑
α
′
2
)
=
∑
f
2
−
∑
α
2
.
{\displaystyle l^{4}\left(\sum f^{\prime 2}-\sum \alpha ^{\prime 2}\right)=\sum f^{2}-\sum \alpha ^{2}.}
So
∑
f
2
=
∑
α
2
{\displaystyle \sum f^{2}=\sum \alpha ^{2}}
entrain
∑
f
′
2
−
∑
α
′
2
.
{\displaystyle \sum f^{\prime 2}-\sum \alpha ^{\prime 2}.}
On the other hand, from equations (9) of § 1, we get:
l
4
∑
f
′
α
′
=
∑
f
α
,
{\displaystyle l^{4}\sum f^{\prime }\alpha ^{\prime }=\sum f\alpha ,}
This shows that
∑
f
α
=
0
{\displaystyle \sum f\alpha =0}
entrain
∑
f
′
α
′
=
0.
{\displaystyle \sum f^{\prime }\alpha ^{\prime }=0.}
I say now that
(8)
∑
f
′
(
x
′
−
x
1
′
)
=
0
,
∑
α
′
(
x
′
−
x
1
′
)
=
0.
{\displaystyle \sum f^{\prime }\left(x^{\prime }-x_{1}^{\prime }\right)=0,\quad \sum \alpha ^{\prime }\left(x^{\prime }-x_{1}^{\prime }\right)=0.}
Indeed, by virtue of equations (7) (and equations 9, § 1) the first parts of equations (8) are written respectively:
k
l
∑
f
(
x
−
x
1
)
+
k
ϵ
l
[
f
r
+
γ
(
y
−
y
1
)
−
β
(
z
−
z
1
)
]
,
{\displaystyle {\frac {k}{l}}\sum f\left(x-x_{1}\right)+{\frac {k\epsilon }{l}}\left[fr+\gamma \left(y-y_{1}\right)-\beta \left(z-z_{1}\right)\right],}
k
l
∑
α
(
x
−
x
1
)
+
k
ϵ
l
[
α
r
−
h
(
y
−
y
1
)
−
g
(
z
−
z
1
)
]
.
{\displaystyle {\frac {k}{l}}\sum \alpha \left(x-x_{1}\right)+{\frac {k\epsilon }{l}}\left[\alpha r-h\left(y-y_{1}\right)-g\left(z-z_{1}\right)\right].}