This property has already been given by Sommerfed [ 1] for hyperbolic motion. It generally holds for all curves of constant curvatures, as we now will demonstrate. To that end, we are using generalized coordinates: the three parameters of any family will be used as generalized spacelike coordinates, the parameter
φ
{\displaystyle \varphi }
as generalized timelike coordinate.
Such a property is geometrically predictable: because an orthogonal transformation of
S
4
{\displaystyle S_{4}}
indeed must transform a mimimal line into a minimal line. Thus
x
{\displaystyle x}
and
y
{\displaystyle y}
again mean light-point and reference-point, and
R
2
=
(
x
−
y
)
2
=
0
{\displaystyle R^{2}=(x-y)^{2}=0}
,
then also
x
(
φ
x
+
Δ
φ
)
{\displaystyle x\left(\varphi _{x}+\Delta \varphi \right)}
is effectively located with
y
(
φ
y
+
Δ
φ
)
{\displaystyle y\left(\varphi _{y}+\Delta \varphi \right)}
. In a suitable reference system which participates with the “motion” of
S
1
{\displaystyle S_{1}}
, nothing could have changed. The nature of this reference system can only be determined in § 8. Here it is sufficient, that the three family parameter of
x
{\displaystyle x}
or
y
{\displaystyle y}
remain unchanged during this motion, while
φ
x
{\displaystyle \varphi _{x}}
and
φ
y
{\displaystyle \varphi _{y}}
experience the same increase
Δ
φ
{\displaystyle \Delta \varphi }
. Thus during the transport of the potentials and fields to these four generalized coordinates, it has to be shown that these potentials and fields at the reference-point
y
{\displaystyle y}
do not depend on
φ
y
{\displaystyle \varphi _{y}}
.
In the following, we will only give the formulas for the potentials and the differentiation formulas stated in § 5, by which the (rather complicated) expressions for the field emerge from the potentials.
Light-point:
x
(
1
)
=
a
x
cos
λ
(
φ
x
−
φ
x
0
)
,
x
(
2
)
=
a
x
sin
λ
(
φ
x
−
φ
x
0
)
,
x
(
3
)
=
b
x
cos
i
φ
x
,
x
(
4
)
=
b
x
sin
i
φ
x
.
Reference-point:
y
(
1
)
=
a
y
cos
λ
(
φ
y
−
φ
y
0
)
,
y
(
2
)
=
a
y
sin
λ
(
φ
y
−
φ
y
0
)
,
y
(
3
)
=
b
y
cos
i
φ
y
,
y
(
4
)
=
b
y
sin
i
φ
y
.
{\displaystyle {\begin{aligned}{\text{Light-point:}}&x^{(1)}=a_{x}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}\right),\ x^{(2)}=a_{x}{\text{sin}}\lambda \left(\varphi _{x}-\varphi _{x}^{0}\right),\ x^{(3)}=b_{x}\cos i\varphi _{x},\ x^{(4)}=b_{x}\sin i\varphi _{x}.\\{\text{Reference-point:}}&y^{(1)}=a_{y}\cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right),\ y^{(2)}=a_{y}{\text{sin}}\lambda \left(\varphi _{y}-\varphi _{y}^{0}\right),\ y^{(3)}=b_{y}\cos i\varphi _{y},\ y^{(4)}=b_{y}\sin i\varphi _{y}.\end{aligned}}}
The transformation matrix reads:
∂
y
(
1
)
∂
a
y
⋯
∂
y
(
4
)
∂
a
y
cos
λ
(
φ
y
−
φ
y
0
)
sin
λ
(
φ
x
−
φ
x
0
)
0
0
∂
y
(
1
)
∂
b
y
⋯
∂
y
(
4
)
∂
b
y
=
0
0
cos
i
φ
y
sin
i
φ
y
∂
y
(
1
)
∂
φ
y
0
⋯
∂
y
(
4
)
∂
φ
y
0
a
y
λ
sin
λ
(
φ
y
−
φ
y
0
)
−
a
y
λ
cos
λ
(
φ
y
−
φ
y
0
)
0
0
∂
y
(
1
)
∂
φ
y
⋯
∂
y
(
4
)
∂
φ
y
−
a
y
λ
sin
(
φ
y
−
φ
y
0
)
a
y
λ
cos
λ
(
φ
y
−
φ
y
0
)
−
i
b
y
sin
i
φ
y
i
b
y
cos
i
φ
y
{\displaystyle {\begin{array}{|c|c|cccc|}{\frac {\partial y^{(1)}}{\partial a_{y}}}\cdots {\frac {\partial y^{(4)}}{\partial a_{y}}}&&\cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&{\text{sin}}\lambda \left(\varphi _{x}-\varphi _{x}^{0}\right)&0&0\\{\frac {\partial y^{(1)}}{\partial b_{y}}}\cdots {\frac {\partial y^{(4)}}{\partial b_{y}}}&=&0&0&\cos i\varphi _{y}&\sin i\varphi _{y}\\{\frac {\partial y^{(1)}}{\partial \varphi _{y}^{0}}}\cdots {\frac {\partial y^{(4)}}{\partial \varphi _{y}^{0}}}&&a_{y}{\lambda }\sin \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&-a_{y}\lambda \cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&0&0\\{\frac {\partial y^{(1)}}{\partial \varphi _{y}}}\cdots {\frac {\partial y^{(4)}}{\partial \varphi _{y}}}&&-a_{y}{\lambda }\sin \left(\varphi _{y}-\varphi _{y}^{0}\right)&a_{y}\lambda \cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&-ib_{y}\sin i\varphi _{y}&ib_{y}\cos i\varphi _{y}\end{array}}}
The reciprocal matrix reads:
∂
a
y
∂
y
(
1
)
⋯
∂
a
y
∂
y
(
4
)
cos
λ
(
φ
y
−
φ
y
0
)
sin
λ
(
φ
x
−
φ
x
0
)
0
0
∂
b
y
∂
y
(
1
)
⋯
∂
b
y
∂
y
(
4
)
=
0
0
cos
i
φ
y
sin
i
φ
y
∂
φ
y
0
∂
y
(
1
)
⋯
∂
φ
y
0
∂
y
(
4
)
1
a
y
λ
sin
λ
(
φ
y
−
φ
y
0
)
−
1
λ
a
y
cos
λ
(
φ
y
−
φ
y
0
)
−
1
i
b
y
sin
i
φ
y
1
i
b
y
cos
i
φ
y
∂
φ
y
∂
y
(
1
)
⋯
∂
φ
y
∂
y
(
4
)
0
0
−
i
b
y
sin
i
φ
y
i
b
y
cos
i
φ
y
{\displaystyle {\begin{array}{|c|c|cccc|}{\frac {\partial a_{y}}{\partial y^{(1)}}}\cdots {\frac {\partial a_{y}}{\partial y^{(4)}}}&&\cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&{\text{sin}}\lambda \left(\varphi _{x}-\varphi _{x}^{0}\right)&0&0\\{\frac {\partial b_{y}}{\partial y^{(1)}}}\cdots {\frac {\partial b_{y}}{\partial y^{(4)}}}&=&0&0&\cos i\varphi _{y}&\sin i\varphi _{y}\\{\frac {\partial \varphi _{y}^{0}}{\partial y^{(1)}}}\cdots {\frac {\partial \varphi _{y}^{0}}{\partial y^{(4)}}}&&{\frac {1}{a_{y}\lambda }}\sin \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&-{\frac {1}{\lambda a_{y}}}\cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right)&-{\frac {1}{ib_{y}}}\sin i\varphi _{y}&{\frac {1}{ib_{y}}}\cos i\varphi _{y}\\{\frac {\partial \varphi _{y}}{\partial y^{(1)}}}\cdots {\frac {\partial \varphi _{y}}{\partial y^{(4)}}}&&0&0&-ib_{y}\sin i\varphi _{y}&ib_{y}\cos i\varphi _{y}\end{array}}}
From that it follows for the arc-element of
S
4
{\displaystyle S_{4}}
(not to be confused with the arc of the spacetime lines):
d
σ
2
=
(
d
a
y
)
2
+
(
d
b
y
)
2
+
a
y
2
λ
2
(
d
φ
y
)
2
−
2
a
y
2
λ
2
d
φ
y
d
φ
y
0
+
(
λ
2
a
y
2
−
b
y
2
)
(
d
φ
y
)
2
{\displaystyle d\sigma ^{2}=\left(da_{y}\right)^{2}+\left(db_{y}\right)^{2}+a_{y}^{2}\lambda ^{2}\left(d\varphi _{y}\right)^{2}-2a_{y}^{2}\lambda ^{2}d\varphi _{y}d\varphi _{y}^{0}+\left(\lambda ^{2}a_{y}^{2}-b_{y}^{2}\right)\left(d\varphi _{y}\right)^{2}}
The applied coordinates are thus oblique-angled .
For the vectors arising in the Wiechert formulas
V
(
1
)
=
−
a
x
λ
sin
λ
(
φ
x
−
φ
x
0
)
i
b
x
2
−
a
x
2
λ
2
,
V
(
2
)
=
a
x
λ
cos
λ
(
φ
x
−
φ
x
0
)
i
b
x
2
−
a
x
2
λ
2
,
V
(
3
)
=
−
i
b
x
sin
i
φ
x
i
b
x
2
−
a
x
2
λ
2
,
V
(
4
)
=
i
b
x
cos
i
φ
x
i
b
x
2
−
a
x
2
λ
2
.
{\displaystyle {\begin{aligned}V^{(1)}&=-{\frac {a_{x}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}\right)}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}},&V^{(2)}&={\frac {a_{x}\lambda \cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}\right)}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}},\\V^{(3)}&=-{\frac {ib_{x}\sin i\varphi _{x}}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}},&V^{(4)}&={\frac {ib_{x}\cos i\varphi _{x}}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}}.\end{aligned}}}
[ 2]
R
(
1
)
=
a
x
cos
λ
(
φ
x
−
φ
x
0
)
−
a
y
cos
λ
(
φ
y
−
φ
y
0
)
,
R
(
2
)
=
a
x
sin
λ
(
φ
x
−
φ
x
0
)
−
a
y
sin
λ
(
φ
y
−
φ
y
0
)
,
R
(
3
)
=
b
x
cos
i
φ
x
−
b
x
cos
i
φ
y
,
R
(
4
)
=
b
x
sin
i
φ
x
−
b
y
sin
i
φ
y
{\displaystyle {\begin{aligned}R^{(1)}&=a_{x}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}\right)-a_{y}\cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right),\\R^{(2)}&=a_{x}\sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}\right)-a_{y}\sin \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right),\\R^{(3)}&=b_{x}\cos i\varphi _{x}-b_{x}\cos i\varphi _{y},\\R^{(4)}&=b_{x}\sin i\varphi _{x}-b_{y}\sin i\varphi _{y}\end{aligned}}}
one finds:
V
(
a
y
)
=
∑
α
=
1
4
V
(
α
)
∂
a
y
∂
y
(
α
)
=
−
a
x
λ
i
b
x
2
−
a
x
2
λ
2
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
,
V
(
b
y
)
=
∑
α
=
1
4
V
(
α
)
∂
b
y
∂
y
(
α
)
=
−
i
b
x
i
b
x
2
−
a
x
2
λ
2
sin
i
(
φ
x
−
φ
y
)
,
V
(
φ
y
0
)
=
∑
α
=
1
4
V
(
α
)
∂
φ
y
0
∂
y
(
α
)
=
−
a
x
a
y
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
b
x
b
y
cos
i
(
φ
x
−
φ
y
)
i
b
x
2
−
a
x
2
λ
2
,
V
(
φ
y
)
=
∑
α
=
1
4
V
(
α
)
∂
φ
y
0
∂
y
(
α
)
=
b
x
b
y
cos
i
(
φ
x
−
φ
y
)
i
b
x
2
−
a
x
2
λ
2
,
{\displaystyle {\begin{aligned}V^{\left(a_{y}\right)}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial a_{y}}{\partial y^{(\alpha )}}}={\frac {-a_{x}\lambda }{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}}\sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right),\\V^{\left(b_{y}\right)}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial b_{y}}{\partial y^{(\alpha )}}}={\frac {-ib_{x}}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}}\sin i\left(\varphi _{x}-\varphi _{y}\right),\\V^{\left(\varphi _{y}^{0}\right)}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial \varphi _{y}^{0}}{\partial y^{(\alpha )}}}={\frac {-{\frac {a_{x}}{a_{y}}}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+{\frac {b_{x}}{b_{y}}}\cos i\left(\varphi _{x}-\varphi _{y}\right)}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}},\\V^{\left(\varphi _{y}\right)}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial \varphi _{y}^{0}}{\partial y^{(\alpha )}}}={\frac {{\frac {b_{x}}{b_{y}}}\cos i\left(\varphi _{x}-\varphi _{y}\right)}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}},\end{aligned}}}
as well as
R
(
a
y
)
=
a
x
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
a
y
,
R
(
b
y
)
=
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
,
R
(
φ
y
0
)
=
−
a
x
λ
a
y
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
b
x
i
b
y
sin
i
(
φ
x
−
φ
x
0
)
R
(
φ
y
)
=
+
b
x
i
b
y
sin
i
(
φ
x
−
φ
x
0
)
{\displaystyle {\begin{aligned}R^{\left(a_{y}\right)}&=a_{x}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-a_{y},\\R^{\left(b_{y}\right)}&=b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y},\\R^{\left(\varphi _{y}^{0}\right)}&=-{\frac {a_{x}}{\lambda a_{y}}}\sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+{\frac {b_{x}}{ib_{y}}}\sin i\left(\varphi _{x}-\varphi _{x}^{0}\right)\\R^{\left(\varphi _{y}\right)}&=+{\frac {b_{x}}{ib_{y}}}\sin i\left(\varphi _{x}-\varphi _{x}^{0}\right)\end{aligned}}}
and eventually
V
(
a
x
)
=
V
(
b
x
)
=
V
(
φ
x
0
)
,
V
(
φ
x
)
1
i
b
x
2
−
a
x
2
λ
2
{\displaystyle V^{\left(a_{x}\right)}=V^{\left(b_{x}\right)}=V^{\left(\varphi _{x}^{0}\right)},\ V^{\left(\varphi _{x}\right)}{\frac {1}{i{\sqrt {b_{x}^{2}-a_{x}^{2}\lambda ^{2}}}}}}
(
V
{\displaystyle V}
has the parameter line
φ
x
{\displaystyle \varphi _{x}}
!)
Thus we have by § 4 (15):
I. Potentials in
y
{\displaystyle y}
:
edit
4
π
e
Φ
a
y
=
1
R
V
c
a
y
a
y
V
(
a
y
)
=
−
a
x
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
a
x
a
y
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
4
π
e
Φ
b
y
=
1
R
V
c
b
y
b
y
V
(
b
y
)
=
−
b
x
i
sin
i
(
φ
x
−
φ
y
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
4
π
e
Φ
φ
y
0
=
1
R
V
{
c
φ
y
0
φ
y
0
V
(
φ
y
0
)
+
c
φ
y
0
φ
y
V
(
φ
y
)
}
=
−
a
x
a
y
λ
2
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
4
π
e
Φ
φ
y
=
1
R
V
{
c
φ
y
φ
y
0
V
(
φ
y
0
)
+
c
φ
y
φ
y
V
(
φ
y
)
}
=
a
x
a
y
λ
2
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
b
x
b
y
cos
i
(
φ
x
−
φ
y
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
{\displaystyle {\begin{aligned}{\frac {4\pi }{e}}\Phi _{a_{y}}&={\frac {1}{RV}}c_{a_{y}a_{y}}V^{\left(a_{y}\right)}\\&=-{\frac {a_{x}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)}{a_{x}a_{y}\sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}\\{\frac {4\pi }{e}}\Phi _{b_{y}}&={\frac {1}{RV}}c_{b_{y}b_{y}}V^{\left(b_{y}\right)}\\&=-{\frac {b_{x}i\sin i\left(\varphi _{x}-\varphi _{y}\right)}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}\\{\frac {4\pi }{e}}\Phi _{\varphi _{y}^{0}}&={\frac {1}{RV}}\left\{c_{\varphi _{y}^{0}\varphi _{y}^{0}}V^{\left(\varphi _{y}^{0}\right)}+c_{\varphi _{y}^{0}\varphi _{y}}V^{\left(\varphi _{y}\right)}\right\}\\&=-{\frac {a_{x}a_{y}\lambda ^{2}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}\\{\frac {4\pi }{e}}\Phi _{\varphi _{y}}&={\frac {1}{RV}}\left\{c_{\varphi _{y}\varphi _{y}^{0}}V^{\left(\varphi _{y}^{0}\right)}+c_{\varphi _{y}\varphi _{y}}V^{\left(\varphi _{y}\right)}\right\}\\&={\frac {a_{x}a_{y}\lambda ^{2}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-b_{x}b_{y}\cos i\left(\varphi _{x}-\varphi _{y}\right)}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}\end{aligned}}}
[ 3]
Note, that
φ
y
{\displaystyle \varphi _{y}}
only appears in the relation
φ
x
−
φ
y
{\displaystyle \varphi _{x}-\varphi _{y}}
, from which one can conclude that the increase of
φ
x
{\displaystyle \varphi _{x}}
and
φ
y
{\displaystyle \varphi _{y}}
by
Δ
φ
{\displaystyle \Delta \varphi }
each, leaves the potentials unchanged. Furthermore by § 4 (18):
∂
∂
a
y
=
(
∂
∂
a
y
)
+
c
a
y
a
y
R
(
a
y
)
R
V
⋅
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
a
y
)
x
+
a
x
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
a
y
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
b
y
=
(
∂
∂
b
y
)
x
+
c
b
y
b
y
R
(
b
y
)
R
V
⋅
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
b
y
)
x
+
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
a
x
a
y
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
φ
y
0
=
(
∂
∂
φ
y
0
)
x
+
c
φ
y
0
φ
y
0
R
(
φ
y
0
)
+
c
φ
y
0
φ
y
R
(
φ
y
)
R
V
⋅
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
φ
y
0
)
x
−
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
φ
y
=
(
∂
∂
φ
y
)
x
+
c
φ
y
φ
y
0
R
(
φ
y
0
)
+
c
φ
y
φ
y
R
(
φ
y
)
R
V
⋅
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
φ
y
)
x
+
∂
∂
φ
x
.
{\displaystyle {\begin{aligned}{\frac {\partial }{\partial a_{y}}}&=\left({\frac {\partial }{\partial a_{y}}}\right)+{\frac {c_{a_{y}a_{y}}R^{\left(a_{y}\right)}}{RV}}\cdot V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}\\&=\left({\frac {\partial }{\partial a_{y}}}\right)_{x}+{\frac {a_{x}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-a_{y}}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial b_{y}}}&=\left({\frac {\partial }{\partial b_{y}}}\right)_{x}+{\frac {c_{b_{y}b_{y}}R^{\left(b_{y}\right)}}{RV}}\cdot V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}\\&=\left({\frac {\partial }{\partial b_{y}}}\right)_{x}+{\frac {b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y}}{a_{x}a_{y}\sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial \varphi _{y}^{0}}}&=\left({\frac {\partial }{\partial \varphi _{y}^{0}}}\right)_{x}+{\frac {c_{\varphi _{y}^{0}\varphi _{y}^{0}}R^{\left(\varphi _{y}^{0}\right)}+c_{\varphi _{y}^{0}\varphi _{y}}R^{\left(\varphi _{y}\right)}}{RV}}\cdot V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}\\&=\left({\frac {\partial }{\partial \varphi _{y}^{0}}}\right)_{x}-{\frac {a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial \varphi _{y}}}&=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {c_{\varphi _{y}\varphi _{y}^{0}}R^{\left(\varphi _{y}^{0}\right)}+c_{\varphi _{y}\varphi _{y}}R^{\left(\varphi _{y}\right)}}{RV}}\cdot V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}\\&=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {\partial }{\partial \varphi _{x}}}.\end{aligned}}}
By the differentiations, the relation
φ
x
−
φ
y
{\displaystyle \varphi _{x}-\varphi _{y}}
is therefore not dissolved, and
φ
y
{\displaystyle \varphi _{y}}
within that relation will therefore appear in the fields as well. But if
f
(
φ
x
−
φ
y
)
{\displaystyle f\left(\varphi _{x}-\varphi _{y}\right)}
is an arbitrary function in this relation, it follows
∂
∂
φ
y
f
(
φ
x
−
φ
y
)
=
(
∂
∂
φ
y
f
)
x
+
∂
∂
φ
x
f
=
−
f
′
+
f
′
≡
0
{\displaystyle {\frac {\partial }{\partial \varphi _{y}}}f\left(\varphi _{x}-\varphi _{y}\right)=\left({\frac {\partial }{\partial \varphi _{y}}}f\right)_{x}+{\frac {\partial }{\partial \varphi _{x}}}f=-f'+f'\equiv 0}
which was to be proven.
Light-point:
x
(
1
)
=
a
x
cos
λ
(
φ
x
−
φ
x
0
)
,
x
(
2
)
=
a
x
sin
λ
(
φ
x
−
φ
x
0
)
,
x
(
3
)
=
x
0
(
3
)
,
x
(
4
)
=
i
φ
y
.
{\displaystyle {\begin{aligned}x^{(1)}&=a_{x}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}\right),&x^{(2)}&=a_{x}{\text{sin }}\lambda \left(\varphi _{x}-\varphi _{x}^{0}\right),\\x^{(3)}&=x_{0}^{(3)},&x^{(4)}&=i\varphi _{y}.\end{aligned}}}
Reference-point:
y
(
1
)
=
a
y
cos
λ
(
φ
y
−
φ
y
0
)
,
y
(
2
)
=
a
y
sin
λ
(
φ
y
−
φ
y
0
)
,
y
(
3
)
=
y
0
(
3
)
,
y
(
4
)
=
i
φ
y
.
{\displaystyle {\begin{aligned}y^{(1)}&=a_{y}\cos \lambda \left(\varphi _{y}-\varphi _{y}^{0}\right),&y^{(2)}&=a_{y}{\text{sin }}\lambda \left(\varphi _{y}-\varphi _{y}^{0}\right),\\y^{(3)}&=y_{0}^{(3)},&y^{(4)}&=i\varphi _{y}.\end{aligned}}}
The details of the computation are analogous as under A .
4
π
e
Φ
a
y
=
−
a
x
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
a
x
a
y
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
φ
x
+
φ
y
4
π
e
Φ
3
=
0
4
π
e
Φ
φ
y
0
=
−
a
x
a
y
λ
2
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
φ
x
+
φ
y
4
π
e
Φ
φ
y
=
a
x
a
y
λ
2
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
1
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
φ
x
+
φ
y
{\displaystyle {\begin{aligned}{\frac {4\pi }{e}}\Phi _{a_{y}}&=-{\frac {a_{x}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)}{a_{x}a_{y}\sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-\varphi _{x}+\varphi _{y}}}\\{\frac {4\pi }{e}}\Phi _{3}&=0\\{\frac {4\pi }{e}}\Phi _{\varphi _{y}^{0}}&=-{\frac {a_{x}a_{y}\lambda ^{2}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-\varphi _{x}+\varphi _{y}}}\\{\frac {4\pi }{e}}\Phi _{\varphi _{y}}&={\frac {a_{x}a_{y}\lambda ^{2}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-1}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-\varphi _{x}+\varphi _{y}}}\end{aligned}}}
∂
∂
a
y
=
(
∂
∂
a
y
)
x
+
a
x
cos
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
a
y
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
φ
x
+
φ
y
∂
∂
φ
x
,
∂
∂
y
0
(
3
)
=
(
∂
∂
y
0
(
3
)
)
x
+
x
0
(
3
)
−
y
0
(
3
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
φ
x
+
φ
y
∂
∂
φ
x
,
∂
∂
φ
y
0
=
(
∂
∂
φ
y
0
)
x
+
−
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
a
x
a
y
λ
sin
λ
(
φ
x
−
φ
x
0
−
φ
y
+
φ
y
0
)
−
φ
x
+
φ
y
∂
∂
φ
x
,
∂
∂
φ
y
=
(
∂
∂
φ
y
)
x
+
∂
∂
φ
x
{\displaystyle {\begin{aligned}{\frac {\partial }{\partial a_{y}}}&=\left({\frac {\partial }{\partial a_{y}}}\right)_{x}+{\frac {a_{x}\cos \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-a_{y}}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-\varphi _{x}+\varphi _{y}}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial y_{0}^{(3)}}}&=\left({\frac {\partial }{\partial y_{0}^{(3)}}}\right)_{x}+{\frac {x_{0}^{(3)}-y_{0}^{(3)}}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-\varphi _{x}+\varphi _{y}}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial \varphi _{y}^{0}}}&=\left({\frac {\partial }{\partial \varphi _{y}^{0}}}\right)_{x}+{\frac {-a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)}{a_{x}a_{y}\lambda \sin \lambda \left(\varphi _{x}-\varphi _{x}^{0}-\varphi _{y}+\varphi _{y}^{0}\right)-\varphi _{x}+\varphi _{y}}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial \varphi _{y}}}&=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {\partial }{\partial \varphi _{x}}}\end{aligned}}}
The last line again contains the proof.
Light-point:
x
(
1
)
=
x
0
(
1
)
+
α
φ
x
,
x
(
2
)
=
x
0
(
2
)
,
x
(
3
)
=
b
x
cos
i
φ
x
,
x
(
4
)
=
b
x
sin
i
φ
x
.
{\displaystyle x^{(1)}=x_{0}^{(1)}+\alpha \varphi _{x},\ x^{(2)}=x_{0}^{(2)},\ x^{(3)}=b_{x}\cos i\varphi _{x},\ x^{(4)}=b_{x}\sin i\varphi _{x}.}
Reference-point:
y
(
1
)
=
y
0
(
1
)
+
α
φ
y
,
y
(
2
)
=
y
0
(
2
)
,
y
(
3
)
=
b
y
cos
i
φ
y
,
y
(
4
)
=
b
y
sin
i
φ
y
.
{\displaystyle y^{(1)}=y_{0}^{(1)}+\alpha \varphi _{y},\ y^{(2)}=y_{0}^{(2)},\ y^{(3)}=b_{y}\cos i\varphi _{y},\ y^{(4)}=b_{y}\sin i\varphi _{y}.}
4
π
e
Φ
y
0
1
=
α
α
(
x
0
1
−
y
0
1
)
+
α
2
(
φ
x
−
φ
y
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
,
4
π
e
Φ
2
=
0
,
4
π
e
Φ
b
y
0
=
−
i
b
x
sin
i
(
φ
x
−
φ
y
)
α
(
x
0
1
−
y
0
1
)
+
α
2
(
φ
x
−
φ
y
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
,
4
π
e
Φ
φ
y
=
α
2
−
b
x
b
y
cos
i
(
φ
x
−
φ
y
)
α
(
x
0
1
−
y
0
1
)
+
α
2
(
φ
x
−
φ
y
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
,
{\displaystyle {\begin{aligned}{\frac {4\pi }{e}}\Phi _{y_{0}^{1}}&={\frac {\alpha }{\alpha \left(x_{0}^{1}-y_{0}^{1}\right)+\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}},\\{\frac {4\pi }{e}}\Phi _{2}&=0,\\{\frac {4\pi }{e}}\Phi _{b_{y}^{0}}&={\frac {-ib_{x}\sin i\left(\varphi _{x}-\varphi _{y}\right)}{\alpha \left(x_{0}^{1}-y_{0}^{1}\right)+\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}},\\{\frac {4\pi }{e}}\Phi _{\varphi _{y}}&={\frac {\alpha ^{2}-b_{x}b_{y}\cos i\left(\varphi _{x}-\varphi _{y}\right)}{\alpha \left(x_{0}^{1}-y_{0}^{1}\right)+\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}},\end{aligned}}}
∂
∂
y
0
1
=
(
∂
∂
y
0
1
)
x
+
x
0
1
−
y
0
1
+
α
(
φ
x
−
φ
y
)
α
(
x
0
1
−
y
0
1
)
+
α
2
(
φ
x
−
φ
y
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
y
0
(
2
)
=
(
∂
∂
y
0
(
2
)
)
x
+
x
0
(
2
)
−
y
0
(
2
)
α
(
x
0
1
−
y
0
1
)
+
α
2
(
φ
x
−
φ
y
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
b
y
=
(
∂
∂
b
y
)
x
+
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
α
(
x
0
1
−
y
0
1
)
+
α
2
(
φ
x
−
φ
y
)
+
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
φ
y
=
(
∂
∂
φ
y
)
x
+
∂
∂
φ
x
.
{\displaystyle {\begin{aligned}{\frac {\partial }{\partial y_{0}^{1}}}&=\left({\frac {\partial }{\partial y_{0}^{1}}}\right)_{x}+{\frac {x_{0}^{1}-y_{0}^{1}+\alpha \left(\varphi _{x}-\varphi _{y}\right)}{\alpha \left(x_{0}^{1}-y_{0}^{1}\right)+\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial y_{0}^{(2)}}}&=\left({\frac {\partial }{\partial y_{0}^{(2)}}}\right)_{x}+{\frac {x_{0}^{(2)}-y_{0}^{(2)}}{\alpha \left(x_{0}^{1}-y_{0}^{1}\right)+\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial b_{y}}}&=\left({\frac {\partial }{\partial b_{y}}}\right)_{x}+{\frac {b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y}}{\alpha \left(x_{0}^{1}-y_{0}^{1}\right)+\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)+ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial \varphi _{y}}}&=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {\partial }{\partial \varphi _{x}}}.\end{aligned}}}
In the last formula the proof is contained again.
Light-point:
x
(
1
)
=
x
0
(
1
)
+
1
2
α
φ
x
2
,
x
(
2
)
=
x
0
(
2
)
x
(
3
)
=
x
0
(
3
)
+
x
0
(
1
)
φ
x
+
1
6
α
φ
x
3
,
x
(
4
)
=
i
(
x
0
(
3
)
+
x
0
(
1
)
φ
x
+
1
6
α
φ
x
3
)
+
i
α
φ
x
.
{\displaystyle {\begin{aligned}x^{(1)}&=x_{0}^{(1)}+{\frac {1}{2}}\alpha \varphi _{x}^{2},&x^{(2)}&=x_{0}^{(2)}\\x^{(3)}&=x_{0}^{(3)}+x_{0}^{(1)}\varphi _{x}+{\frac {1}{6}}\alpha \varphi _{x}^{3},&x^{(4)}&=i\left(x_{0}^{(3)}+x_{0}^{(1)}\varphi _{x}+{\frac {1}{6}}\alpha \varphi _{x}^{3}\right)+i\alpha \varphi _{x}.\end{aligned}}}
Reference-point:
y
(
1
)
=
y
0
(
1
)
+
1
2
α
φ
y
2
,
y
(
2
)
=
y
0
(
2
)
y
(
3
)
=
y
0
(
3
)
+
y
0
(
1
)
φ
y
+
1
6
α
φ
y
3
,
y
(
4
)
=
i
(
y
0
(
3
)
+
y
0
(
1
)
φ
y
+
1
6
α
φ
y
3
)
+
i
α
φ
y
.
{\displaystyle {\begin{aligned}y^{(1)}&=y_{0}^{(1)}+{\frac {1}{2}}\alpha \varphi _{y}^{2},&y^{(2)}&=y_{0}^{(2)}\\y^{(3)}&=y_{0}^{(3)}+y_{0}^{(1)}\varphi _{y}+{\frac {1}{6}}\alpha \varphi _{y}^{3},&y^{(4)}&=i\left(y_{0}^{(3)}+y_{0}^{(1)}\varphi _{y}+{\frac {1}{6}}\alpha \varphi _{y}^{3}\right)+i\alpha \varphi _{y}.\end{aligned}}}
∂
y
(
1
)
∂
y
0
(
1
)
⋯
∂
y
(
4
)
∂
y
0
(
1
)
1
0
φ
y
i
φ
y
∂
y
(
1
)
∂
y
0
(
2
)
⋯
∂
y
(
4
)
∂
y
0
(
2
)
=
0
1
0
0
∂
y
(
1
)
∂
y
0
(
3
)
⋯
∂
y
(
4
)
∂
y
0
(
3
)
0
0
1
i
∂
y
(
1
)
∂
φ
y
⋯
∂
y
(
4
)
∂
φ
y
α
φ
y
0
y
0
(
1
)
+
1
2
α
φ
y
2
i
(
y
0
(
1
)
+
1
2
α
φ
y
2
)
+
i
α
{\displaystyle {\begin{array}{|c|c|cccc|}{\frac {\partial y^{(1)}}{\partial y_{0}^{(1)}}}\cdots {\frac {\partial y^{(4)}}{\partial y_{0}^{(1)}}}&&1&0&\varphi _{y}&i\varphi _{y}\\{\frac {\partial y^{(1)}}{\partial y_{0}^{(2)}}}\cdots {\frac {\partial y^{(4)}}{\partial y_{0}^{(2)}}}&=&0&1&0&0\\{\frac {\partial y^{(1)}}{\partial y_{0}^{(3)}}}\cdots {\frac {\partial y^{(4)}}{\partial y_{0}^{(3)}}}&&0&0&1&i\\{\frac {\partial y^{(1)}}{\partial \varphi _{y}}}\cdots {\frac {\partial y^{(4)}}{\partial \varphi _{y}}}&&\alpha \varphi _{y}&0&y_{0}^{(1)}+{\frac {1}{2}}\alpha \varphi _{y}^{2}&i\left(y_{0}^{(1)}+{\frac {1}{2}}\alpha \varphi _{y}^{2}\right)+i\alpha \end{array}}}
V
(
1
)
=
α
φ
x
i
2
x
0
(
1
)
α
+
α
2
,
V
(
2
)
=
0
,
V
(
3
)
=
x
0
(
1
)
+
1
2
α
φ
x
2
i
2
x
0
(
1
)
α
+
α
2
,
V
(
2
)
=
i
(
x
0
(
1
)
+
1
2
α
φ
x
2
)
+
i
α
i
2
x
0
(
1
)
α
+
α
2
,
{\displaystyle {\begin{aligned}V^{(1)}&={\frac {\alpha \varphi _{x}}{i{\sqrt {2x_{0}^{(1)}\alpha +\alpha ^{2}}}}},&V^{(2)}&=0,\\V^{(3)}&={\frac {x_{0}^{(1)}+{\frac {1}{2}}\alpha \varphi _{x}^{2}}{i{\sqrt {2x_{0}^{(1)}\alpha +\alpha ^{2}}}}},&V^{(2)}&={\frac {i\left(x_{0}^{(1)}+{\frac {1}{2}}\alpha \varphi _{x}^{2}\right)+i\alpha }{i{\sqrt {2x_{0}^{(1)}\alpha +\alpha ^{2}}}}},\end{aligned}}}
V
y
0
(
1
)
=
∑
α
=
1
4
V
(
α
)
∂
y
(
α
)
∂
y
0
(
1
)
=
α
(
φ
x
−
φ
y
)
i
2
x
0
(
1
)
α
+
α
2
,
V
y
0
(
2
)
=
∑
α
=
1
4
V
(
α
)
∂
y
(
α
)
∂
y
0
(
2
)
=
0
,
V
y
0
(
3
)
=
∑
α
=
1
4
V
(
α
)
∂
y
(
α
)
∂
y
0
(
3
)
=
−
α
i
2
x
0
(
1
)
α
+
α
2
,
V
φ
y
=
∑
α
=
1
4
V
(
α
)
∂
φ
y
∂
y
(
α
)
=
−
1
2
α
2
(
φ
x
−
φ
y
)
2
+
α
(
x
0
(
1
)
−
y
0
(
1
)
)
+
α
2
i
2
x
0
(
1
)
α
+
α
2
,
}
{\displaystyle \left.{\begin{aligned}V_{y_{0}^{(1)}}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial y^{(\alpha )}}{\partial y_{0}^{(1)}}}={\frac {\alpha \left(\varphi _{x}-\varphi _{y}\right)}{i{\sqrt {2x_{0}^{(1)}\alpha +\alpha ^{2}}}}},\\V_{y_{0}^{(2)}}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial y^{(\alpha )}}{\partial y_{0}^{(2)}}}=0,\\V_{y_{0}^{(3)}}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial y^{(\alpha )}}{\partial y_{0}^{(3)}}}={\frac {-\alpha }{i{\sqrt {2x_{0}^{(1)}\alpha +\alpha ^{2}}}}},\\V_{\varphi _{y}}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial \varphi _{y}}{\partial y^{(\alpha )}}}=-{\frac {{\frac {1}{2}}\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)^{2}+\alpha \left(x_{0}^{(1)}-y_{0}^{(1)}\right)+\alpha ^{2}}{i{\sqrt {2x_{0}^{(1)}\alpha +\alpha ^{2}}}}},\end{aligned}}\right\}}
[ 4]
R
(
1
)
=
x
0
(
1
)
−
y
0
(
1
)
+
1
2
α
2
(
φ
x
−
φ
y
)
2
,
R
(
2
)
=
x
0
(
2
)
−
y
0
(
2
)
,
R
(
3
)
=
x
0
(
3
)
−
y
0
(
3
)
+
x
0
(
1
)
φ
x
−
y
0
(
1
)
φ
y
+
1
6
α
(
φ
x
3
−
φ
y
3
)
,
R
(
4
)
=
i
(
x
0
(
3
)
−
y
0
(
3
)
+
x
0
(
1
)
φ
x
−
y
0
(
1
)
φ
y
+
1
6
α
[
φ
x
3
−
φ
y
3
]
)
+
i
α
(
φ
x
−
φ
y
)
{\displaystyle {\begin{aligned}R^{(1)}&=x_{0}^{(1)}-y_{0}^{(1)}+{\frac {1}{2}}\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)^{2},\\R^{(2)}&=x_{0}^{(2)}-y_{0}^{(2)},\\R^{(3)}&=x_{0}^{(3)}-y_{0}^{(3)}+x_{0}^{(1)}\varphi _{x}-y_{0}^{(1)}\varphi _{y}+{\frac {1}{6}}\alpha \left(\varphi _{x}^{3}-\varphi _{y}^{3}\right),\\R^{(4)}&=i\left(x_{0}^{(3)}-y_{0}^{(3)}+x_{0}^{(1)}\varphi _{x}-y_{0}^{(1)}\varphi _{y}+{\frac {1}{6}}\alpha \left[\varphi _{x}^{3}-\varphi _{y}^{3}\right]\right)+i\alpha \left(\varphi _{x}-\varphi _{y}\right)\end{aligned}}}
R
y
0
(
1
)
=
x
0
(
1
)
−
y
0
(
1
)
+
α
2
(
φ
x
−
φ
y
)
2
,
R
y
0
(
2
)
=
x
0
(
2
)
−
y
0
(
2
)
,
R
y
0
(
3
)
=
−
α
(
φ
x
−
φ
y
)
,
R
φ
y
=
−
α
(
x
0
(
1
)
−
y
0
(
1
)
)
(
φ
x
−
φ
y
)
−
1
6
α
2
(
φ
x
−
φ
y
)
3
,
−
α
(
x
0
(
3
)
−
y
0
(
3
)
)
−
α
2
(
φ
x
−
φ
y
)
{\displaystyle {\begin{aligned}R_{y_{0}^{(1)}}&=x_{0}^{(1)}-y_{0}^{(1)}+{\frac {\alpha }{2}}\left(\varphi _{x}-\varphi _{y}\right)^{2},\\R_{y_{0}^{(2)}}&=x_{0}^{(2)}-y_{0}^{(2)},\\R_{y_{0}^{(3)}}&=-\alpha \left(\varphi _{x}-\varphi _{y}\right),\\R_{\varphi _{y}}&=-\alpha \left(x_{0}^{(1)}-y_{0}^{(1)}\right)\left(\varphi _{x}-\varphi _{y}\right)-{\frac {1}{6}}\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)^{3},\\&\quad -\alpha \left(x_{0}^{(3)}-y_{0}^{(3)}\right)-\alpha ^{2}\left(\varphi _{x}-\varphi _{y}\right)\end{aligned}}}
In relation to that it has to be remarked, that a decomposition with respect to those parameter lines is physically invalid , because the parameter
y
0
(
3
)
{\displaystyle y_{0}^{(3)}}
is always related to minimal directions instead of spacelike ones. Therefore, we can only conclude the constancy of the potentials and the fields from the things stated above, which result is not influenced by the previous circumstance. We have again:
∂
∂
φ
y
=
(
∂
∂
φ
y
)
x
+
∂
∂
φ
x
{\displaystyle {\frac {\partial }{\partial \varphi _{y}}}=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {\partial }{\partial \varphi _{x}}}}
and because of the appearance of
φ
y
{\displaystyle \varphi _{y}}
only in the relation
φ
x
−
φ
y
{\displaystyle \varphi _{x}-\varphi _{y}}
, the proof has been given.
(C) 1. Hyperbolic motion.
edit
Light-point:
x
(
1
)
=
x
0
(
1
)
,
x
(
2
)
=
x
0
(
2
)
,
x
(
3
)
=
b
x
cos
i
φ
x
,
x
(
4
)
=
b
x
sin
i
φ
x
.
{\displaystyle x^{(1)}=x_{0}^{(1)},\ x^{(2)}=x_{0}^{(2)},\ x^{(3)}=b_{x}\cos i\varphi _{x},\ x^{(4)}=b_{x}\sin i\varphi _{x}.}
Reference-point:
y
(
1
)
=
y
0
(
1
)
,
y
(
2
)
=
y
0
(
2
)
,
y
(
3
)
=
b
y
cos
i
φ
y
,
y
(
4
)
=
b
y
sin
i
φ
y
.
{\displaystyle y^{(1)}=y_{0}^{(1)},\ y^{(2)}=y_{0}^{(2)},\ y^{(3)}=b_{y}\cos i\varphi _{y},\ y^{(4)}=b_{y}\sin i\varphi _{y}.}
∂
y
(
1
)
∂
y
0
(
1
)
⋯
∂
y
(
4
)
∂
y
0
(
1
)
1
0
0
0
∂
y
(
1
)
∂
y
0
(
2
)
⋯
∂
y
(
4
)
∂
y
0
(
2
)
=
0
1
0
0
∂
y
(
1
)
∂
b
y
⋯
∂
y
(
4
)
∂
b
y
0
0
cos
i
φ
y
sin
i
φ
y
∂
y
(
1
)
∂
φ
y
⋯
∂
y
(
4
)
∂
φ
y
0
0
−
b
y
i
sin
i
φ
y
b
y
i
cos
i
φ
y
{\displaystyle {\begin{array}{|c|c|cccc|}{\frac {\partial y^{(1)}}{\partial y_{0}^{(1)}}}\cdots {\frac {\partial y^{(4)}}{\partial y_{0}^{(1)}}}&&1&0&0&0\\{\frac {\partial y^{(1)}}{\partial y_{0}^{(2)}}}\cdots {\frac {\partial y^{(4)}}{\partial y_{0}^{(2)}}}&=&0&1&0&0\\{\frac {\partial y^{(1)}}{\partial b_{y}}}\cdots {\frac {\partial y^{(4)}}{\partial b_{y}}}&&0&0&\cos i\varphi _{y}&\sin i\varphi _{y}\\{\frac {\partial y^{(1)}}{\partial \varphi _{y}}}\cdots {\frac {\partial y^{(4)}}{\partial \varphi _{y}}}&&0&0&-b_{y}i\sin i\varphi _{y}&b_{y}i\cos i\varphi _{y}\end{array}}}
V
(
1
)
=
0
,
V
(
2
)
=
0
,
V
(
3
)
=
−
sin
i
φ
x
,
V
(
4
)
=
cos
i
φ
x
.
{\displaystyle V^{(1)}=0,\ V^{(2)}=0,\ V^{(3)}=-\sin i\varphi _{x},\ V^{(4)}=\cos i\varphi _{x}.}
d
V
(
1
)
d
s
=
0
,
d
V
(
2
)
d
s
=
0
,
d
V
(
3
)
d
s
=
−
cos
i
φ
x
b
x
,
d
V
(
4
)
d
s
=
−
sin
i
φ
x
b
x
.
{\displaystyle {\frac {dV^{(1)}}{ds}}=0,\ {\frac {dV^{(2)}}{ds}}=0,\ {\frac {dV^{(3)}}{ds}}=-{\frac {\cos i\varphi _{x}}{b_{x}}},\ {\frac {dV^{(4)}}{ds}}=-{\frac {\sin i\varphi _{x}}{b_{x}}}.}
R
(
1
)
=
x
0
(
1
)
−
y
0
(
1
)
,
R
(
2
)
=
x
0
(
2
)
−
y
0
(
2
)
,
R
(
3
)
=
b
x
cos
i
φ
x
−
b
y
cos
i
φ
y
,
R
(
4
)
=
b
x
sin
i
φ
x
−
b
y
sin
i
φ
y
.
{\displaystyle {\begin{aligned}R^{(1)}&=x_{0}^{(1)}-y_{0}^{(1)},&R^{(2)}&=x_{0}^{(2)}-y_{0}^{(2)},\\R^{(3)}&=b_{x}\cos i\varphi _{x}-b_{y}\cos i\varphi _{y},&R^{(4)}&=b_{x}\sin i\varphi _{x}-b_{y}\sin i\varphi _{y}.\end{aligned}}}
(
R
V
)
=
b
y
sin
i
(
φ
x
−
φ
y
)
,
1
+
R
d
V
d
s
=
b
y
b
x
cos
i
(
φ
x
−
φ
y
)
{\displaystyle (RV)=b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right),\ 1+R{\frac {dV}{ds}}={\frac {b_{y}}{b_{x}}}\cos i\left(\varphi _{x}-\varphi _{y}\right)}
V
y
0
(
1
)
=
V
y
0
(
2
)
=
0
,
V
b
y
=
∑
α
=
1
4
V
(
α
)
∂
y
(
α
)
∂
b
y
=
−
sin
i
(
φ
x
−
φ
y
)
,
V
φ
y
=
∑
α
=
1
4
V
(
α
)
∂
y
(
α
)
∂
φ
y
=
−
i
b
y
cos
i
(
φ
x
−
φ
y
)
.
{\displaystyle {\begin{aligned}V_{y_{0}^{(1)}}&=V_{y_{0}^{(2)}}=0,\\V_{b_{y}}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial y^{(\alpha )}}{\partial b_{y}}}=-\sin i\left(\varphi _{x}-\varphi _{y}\right),\\V_{\varphi _{y}}&=\sum _{\alpha =1}^{4}V^{(\alpha )}{\frac {\partial y^{(\alpha )}}{\partial \varphi _{y}}}=-ib_{y}\cos i\left(\varphi _{x}-\varphi _{y}\right).\end{aligned}}}
(
d
V
d
s
)
y
0
(
1
)
=
(
d
V
d
s
)
y
0
(
2
)
=
0
,
(
d
V
d
s
)
b
y
=
−
cos
i
(
φ
x
−
φ
y
)
b
x
,
(
d
V
d
s
)
φ
y
=
−
i
b
y
b
x
sin
i
(
φ
x
−
φ
y
)
,
{\displaystyle {\begin{aligned}\left({\frac {dV}{ds}}\right)_{y_{0}^{(1)}}&=\left({\frac {dV}{ds}}\right)_{y_{0}^{(2)}}=0,\\\left({\frac {dV}{ds}}\right)_{b_{y}}&=-{\frac {\cos i\left(\varphi _{x}-\varphi _{y}\right)}{b_{x}}},\\\left({\frac {dV}{ds}}\right)_{\varphi _{y}}&=-i{\frac {b_{y}}{b_{x}}}\sin i\left(\varphi _{x}-\varphi _{y}\right),\end{aligned}}}
R
y
0
(
1
)
=
x
0
(
1
)
−
y
0
(
1
)
,
R
y
0
(
2
)
=
x
0
(
2
)
−
y
0
(
2
)
,
R
b
y
=
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
,
R
φ
y
=
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
.
{\displaystyle {\begin{aligned}R_{y_{0}^{(1)}}&=x_{0}^{(1)}-y_{0}^{(1)},&R_{y_{0}^{(2)}}&=x_{0}^{(2)}-y_{0}^{(2)},\\R_{b_{y}}&=b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y},&R_{\varphi _{y}}&=ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right).\end{aligned}}}
Furthermore
V
(
x
0
(
1
)
)
=
V
(
x
0
(
2
)
)
=
V
(
b
x
)
=
0
,
V
(
φ
x
)
=
1
i
b
x
{\displaystyle V^{\left(x_{0}^{(1)}\right)}=V^{\left(x_{0}^{(2)}\right)}=V^{\left(b_{x}\right)}=0,\ V^{\left(\varphi _{x}\right)}={\frac {1}{ib_{x}}}}
Thus:
4
π
e
Φ
y
0
(
1
)
=
V
y
0
(
1
)
R
V
=
0
,
4
π
e
Φ
y
0
(
2
)
=
V
y
0
(
2
)
R
V
=
0
,
{\displaystyle {\frac {4\pi }{e}}\Phi _{y_{0}^{(1)}}={\frac {V_{y_{0}^{(1)}}}{RV}}=0,\ {\frac {4\pi }{e}}\Phi _{y_{0}^{(2)}}={\frac {V_{y_{0}^{(2)}}}{RV}}=0,}
4
π
e
Φ
b
y
=
V
b
y
R
V
=
−
1
b
y
sin
i
(
φ
x
−
φ
y
)
sin
i
(
φ
x
−
φ
y
)
=
−
1
b
y
,
4
π
e
Φ
φ
y
=
i
cos
i
(
φ
x
−
φ
y
)
sin
i
(
φ
x
−
φ
y
)
.
{\displaystyle {\begin{aligned}{\frac {4\pi }{e}}\Phi _{b_{y}}&={\frac {V_{b_{y}}}{RV}}=-{\frac {1}{b_{y}}}{\frac {\sin i\left(\varphi _{x}-\varphi _{y}\right)}{\sin i\left(\varphi _{x}-\varphi _{y}\right)}}=-{\frac {1}{b_{y}}},\\{\frac {4\pi }{e}}\Phi _{\varphi _{y}}&=i{\frac {\cos i\left(\varphi _{x}-\varphi _{y}\right)}{\sin i\left(\varphi _{x}-\varphi _{y}\right)}}.\end{aligned}}}
∂
∂
y
0
(
1
)
=
(
∂
∂
y
0
1
)
x
+
R
y
0
(
1
)
R
V
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
y
0
1
)
x
+
x
0
(
1
)
−
y
0
(
1
)
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
y
0
(
2
)
=
(
∂
∂
y
0
(
2
)
)
x
+
R
y
0
(
2
)
R
V
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
y
0
2
)
x
+
x
0
(
2
)
−
y
0
(
2
)
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
b
y
=
(
∂
∂
b
y
)
x
+
R
b
y
R
V
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
b
y
)
x
+
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
i
b
x
b
y
sin
i
(
φ
x
−
φ
y
)
∂
∂
φ
x
,
∂
∂
φ
y
=
(
∂
∂
φ
y
)
x
+
R
φ
y
R
V
V
(
φ
x
)
∂
∂
φ
x
=
(
∂
∂
φ
y
)
x
+
∂
∂
φ
x
.
{\displaystyle {\begin{aligned}{\frac {\partial }{\partial y_{0}^{(1)}}}&=\left({\frac {\partial }{\partial y_{0}^{1}}}\right)_{x}+{\frac {R_{y_{0}^{(1)}}}{RV}}V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}=\left({\frac {\partial }{\partial y_{0}^{1}}}\right)_{x}+{\frac {x_{0}^{(1)}-y_{0}^{(1)}}{ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial y_{0}^{(2)}}}&=\left({\frac {\partial }{\partial y_{0}^{(2)}}}\right)_{x}+{\frac {R_{y_{0}^{(2)}}}{RV}}V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}=\left({\frac {\partial }{\partial y_{0}^{2}}}\right)_{x}+{\frac {x_{0}^{(2)}-y_{0}^{(2)}}{ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial b_{y}}}&=\left({\frac {\partial }{\partial b_{y}}}\right)_{x}+{\frac {R_{b_{y}}}{RV}}V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}=\left({\frac {\partial }{\partial b_{y}}}\right)_{x}+{\frac {b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y}}{ib_{x}b_{y}\sin i\left(\varphi _{x}-\varphi _{y}\right)}}{\frac {\partial }{\partial \varphi _{x}}},\\{\frac {\partial }{\partial \varphi _{y}}}&=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {R_{\varphi _{y}}}{RV}}V^{\left(\varphi _{x}\right)}{\frac {\partial }{\partial \varphi _{x}}}=\left({\frac {\partial }{\partial \varphi _{y}}}\right)_{x}+{\frac {\partial }{\partial \varphi _{x}}}.\end{aligned}}}
In the last line, the proof for the constancy of the field is contained, since
φ
y
{\displaystyle \varphi _{y}}
only appears in the relation
φ
x
−
φ
y
{\displaystyle \varphi _{x}-\varphi _{y}}
.
By § 5 (16)
4
π
e
F
¯
α
β
=
1
(
R
V
)
2
[
R
¯
d
V
¯
d
s
]
α
β
−
1
+
R
d
V
d
s
(
R
V
)
3
[
R
V
¯
]
α
β
{\displaystyle {\frac {4\pi }{e}}{\bar {F}}_{\alpha \beta }={\frac {1}{(RV)^{2}}}\left[{\bar {R}}{\frac {d{\bar {V}}}{ds}}\right]_{\alpha \beta }-{\frac {1+R{\frac {dV}{ds}}}{(RV)^{3}}}[{\overline {RV}}]_{\alpha \beta }}
in which
[
R
¯
d
V
¯
d
s
]
α
β
=
R
¯
α
(
d
V
¯
d
s
)
β
−
R
¯
β
(
d
V
¯
d
s
)
α
{\displaystyle \left[{\bar {R}}{\frac {d{\bar {V}}}{ds}}\right]_{\alpha \beta }={\bar {R}}_{\alpha }\left({\frac {d{\bar {V}}}{ds}}\right)_{\beta }-{\bar {R}}_{\beta }\left({\frac {d{\bar {V}}}{ds}}\right)_{\alpha }}
etc.
or, if we set
T
¯
α
=
(
R
V
)
(
d
V
¯
d
s
)
α
−
(
1
+
R
d
V
d
s
)
V
¯
α
{\displaystyle {\bar {T}}_{\alpha }=(RV)\left({\frac {d{\bar {V}}}{ds}}\right)_{\alpha }-\left(1+R{\frac {dV}{ds}}\right){\bar {V}}_{\alpha }}
it follows
4
π
e
F
¯
α
β
=
1
(
R
V
)
3
[
R
T
¯
]
α
β
{\displaystyle {\frac {4\pi }{e}}{\bar {F}}_{\alpha \beta }={\frac {1}{(RV)^{3}}}\left[{\overline {RT}}\right]_{\alpha \beta }}
For the computation of
T
¯
α
{\displaystyle {\bar {T}}_{\alpha }}
we have in the present case
T
y
0
(
1
)
=
0
,
T
y
0
(
2
)
=
0
,
T
b
y
=
(
R
V
)
(
d
V
d
s
)
b
y
−
(
1
+
R
d
V
d
s
)
b
V
b
y
=
0
,
T
φ
y
=
(
R
V
)
(
d
V
d
s
)
φ
y
−
(
1
+
R
d
V
d
s
)
V
φ
y
=
−
i
b
y
2
b
x
{\displaystyle {\begin{aligned}T_{y_{0}^{(1)}}&=0,\quad T_{y_{0}^{(2)}}=0,\\T_{b_{y}}&=(RV)\left({\frac {dV}{ds}}\right)_{b_{y}}-\left(1+R{\frac {dV}{ds}}\right)_{b}V_{b_{y}}=0,\\T_{\varphi _{y}}&=(RV)\left({\frac {dV}{ds}}\right)_{\varphi _{y}}-\left(1+R{\frac {dV}{ds}}\right)V_{\varphi _{y}}=-i{\frac {b_{y}^{2}}{b_{x}}}\end{aligned}}}
Thus the remarkable effect arises in hyperbolic motion, that the vector
T
{\displaystyle T}
(of the tangent of the world line of reference point
W
{\displaystyle W}
) is directed in the parallel direction. We will see, what follows from that. Namely, we have
4
π
e
F
y
0
(
1
)
φ
y
2
=
4
π
e
F
y
0
(
1
)
b
y
=
4
π
e
F
y
0
(
2
)
b
y
≡
0
{
(magnetic
components),
{\displaystyle {\frac {4\pi }{e}}F_{y_{0}^{(1)}\varphi _{y}^{2}}={\frac {4\pi }{e}}F_{y_{0}^{(1)}b_{y}}={\frac {4\pi }{e}}F_{y_{0}^{(2)}b_{y}}\equiv 0\ \left\{{\begin{matrix}{\text{(magnetic}}\\{\text{components),}}\end{matrix}}\right.}
4
π
e
F
y
0
(
1
)
φ
y
=
1
(
R
V
)
3
R
y
0
(
1
)
T
φ
y
=
x
0
(
1
)
−
y
0
(
1
)
i
b
x
b
y
sin
3
i
(
φ
x
−
φ
y
)
4
π
e
F
y
0
(
2
)
φ
y
=
1
(
R
V
)
3
R
y
0
(
2
)
T
φ
y
=
x
0
(
2
)
−
y
0
(
2
)
i
b
x
b
y
sin
3
i
(
φ
x
−
φ
y
)
4
π
e
F
b
y
φ
y
=
1
(
R
V
)
3
R
b
y
T
φ
y
=
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
i
b
x
b
y
sin
3
i
(
φ
x
−
φ
y
)
}
(electric
components)
{\displaystyle \left.{\begin{aligned}{\frac {4\pi }{e}}F_{y_{0}^{(1)}\varphi _{y}}&={\frac {1}{(RV)^{3}}}R_{y_{0}^{(1)}}T_{\varphi _{y}}={\frac {x_{0}^{(1)}-y_{0}^{(1)}}{ib_{x}b_{y}\sin ^{3}i\left(\varphi _{x}-\varphi _{y}\right)}}\\{\frac {4\pi }{e}}F_{y_{0}^{(2)}\varphi _{y}}&={\frac {1}{(RV)^{3}}}R_{y_{0}^{(2)}}T_{\varphi _{y}}={\frac {x_{0}^{(2)}-y_{0}^{(2)}}{ib_{x}b_{y}\sin ^{3}i\left(\varphi _{x}-\varphi _{y}\right)}}\\{\frac {4\pi }{e}}F_{b_{y}\varphi _{y}}&={\frac {1}{(RV)^{3}}}R_{b_{y}}T_{\varphi _{y}}={\frac {b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y}}{ib_{x}b_{y}\sin ^{3}i\left(\varphi _{x}-\varphi _{y}\right)}}\end{aligned}}\right\}\ {\begin{matrix}{\text{(electric}}\\{\text{components)}}\end{matrix}}}
Thus the field (
F
α
β
{\displaystyle F_{\alpha \beta }}
indeed contains, as predicted,
φ
y
{\displaystyle \varphi _{y}}
only in the relation
φ
x
−
φ
y
{\displaystyle \varphi _{x}-\varphi _{y}}
) is not only constant in the varying reference system (this time, it is a system comoving with the reference-point, as can be easily seen), but it is also purely electrical.
Computation by the method of vectorial splitting of § 1.
edit
4
π
e
A
x
=
4
π
e
Φ
y
0
(
1
)
=
0
,
4
π
e
A
y
=
4
π
e
Φ
y
0
(
2
)
=
0
,
{\displaystyle {\frac {4\pi }{e}}{\mathfrak {A}}_{x}={\frac {4\pi }{e}}\Phi _{y_{0}^{(1)}}=0,\ {\frac {4\pi }{e}}{\mathfrak {A}}_{y}={\frac {4\pi }{e}}\Phi _{y_{0}^{(2)}}=0,}
4
π
e
A
b
=
−
1
b
y
,
4
π
e
i
φ
=
4
π
e
Φ
φ
y
c
φ
y
φ
y
=
1
−
b
y
2
i
cos
i
(
φ
x
−
φ
y
)
sin
i
(
φ
x
−
φ
y
)
=
1
b
y
cos
i
(
φ
x
−
φ
y
)
sin
i
(
φ
x
−
φ
y
)
{\displaystyle {\begin{aligned}{\frac {4\pi }{e}}{\mathfrak {A}}_{b}&=-{\frac {1}{b_{y}}},\\{\frac {4\pi }{e}}i\varphi &={\frac {4\pi }{e}}{\frac {\Phi _{\varphi _{y}}}{\sqrt {c\varphi _{y}\varphi _{y}}}}={\frac {1}{\sqrt {-b_{y}^{2}}}}{\frac {i\cos i\left(\varphi _{x}-\varphi _{y}\right)}{\sin i\left(\varphi _{x}-\varphi _{y}\right)}}={\frac {1}{b_{y}}}{\frac {\cos i\left(\varphi _{x}-\varphi _{y}\right)}{\sin i\left(\varphi _{x}-\varphi _{y}\right)}}\end{aligned}}}
furthermore
4
π
e
H
x
=
4
π
e
H
y
=
4
π
e
H
b
=
0
,
{\displaystyle {\frac {4\pi }{e}}{\mathfrak {H}}_{x}={\frac {4\pi }{e}}{\mathfrak {H}}_{y}={\frac {4\pi }{e}}{\mathfrak {H}}_{b}=0,}
−
4
π
e
i
E
x
=
4
π
e
F
y
0
(
1
)
φ
y
c
φ
y
φ
y
=
−
x
0
(
1
)
−
y
0
(
1
)
b
x
b
y
2
sin
3
i
(
φ
x
−
φ
y
)
,
−
4
π
e
i
E
y
=
4
π
e
F
y
0
(
2
)
φ
y
c
φ
y
φ
y
=
−
x
0
(
2
)
−
y
0
(
2
)
b
x
b
y
2
sin
3
i
(
φ
x
−
φ
y
)
,
−
4
π
e
i
E
b
=
4
π
e
F
b
y
φ
y
c
φ
y
φ
y
=
−
b
x
cos
i
(
φ
x
−
φ
y
)
−
b
y
b
x
b
y
2
sin
3
i
(
φ
x
−
φ
y
)
,
{\displaystyle {\begin{aligned}-{\frac {4\pi }{e}}i{\mathfrak {E}}_{x}&={\frac {4\pi }{e}}{\frac {F_{y_{0}^{(1)}\varphi _{y}}}{\sqrt {c\varphi _{y}\varphi _{y}}}}=-{\frac {x_{0}^{(1)}-y_{0}^{(1)}}{b_{x}b_{y}^{2}\sin ^{3}i\left(\varphi _{x}-\varphi _{y}\right)}},\\-{\frac {4\pi }{e}}i{\mathfrak {E}}_{y}&={\frac {4\pi }{e}}{\frac {F_{y_{0}^{(2)}\varphi _{y}}}{\sqrt {c\varphi _{y}\varphi _{y}}}}=-{\frac {x_{0}^{(2)}-y_{0}^{(2)}}{b_{x}b_{y}^{2}\sin ^{3}i\left(\varphi _{x}-\varphi _{y}\right)}},\\-{\frac {4\pi }{e}}i{\mathfrak {E}}_{b}&={\frac {4\pi }{e}}{\frac {F_{b_{y}\varphi _{y}}}{\sqrt {c\varphi _{y}\varphi _{y}}}}=-{\frac {b_{x}\cos i\left(\varphi _{x}-\varphi _{y}\right)-b_{y}}{b_{x}b_{y}^{2}\sin ^{3}i\left(\varphi _{x}-\varphi _{y}\right)}},\end{aligned}}}
which formulas have already been given by Sommerfeld .[ 5]
↑ Sommerfeld , l. c. (33), p. 673 for the potentials and p. 677 for the fields.
↑
b
x
2
>
a
x
2
λ
2
{\displaystyle b_{x}^{2}>a_{x}^{2}\lambda ^{2}}
is the condition for the ratio
b
x
a
x
{\displaystyle {\frac {b_{x}}{a_{x}}}}
; because
V
{\displaystyle V}
must be timelike.
↑ The
c
a
y
a
y
{\displaystyle c_{a_{y}a_{y}}}
etc. mean the coefficients of
d
σ
2
{\displaystyle d\sigma ^{2}}
.
↑ Since the Wiechert-Schwarzschild formulas only contain
V
α
=
∑
β
=
1
4
c
α
β
V
(
β
)
,
R
α
=
∑
β
=
1
4
c
α
β
R
(
β
)
{\displaystyle V_{\alpha }=\sum _{\beta =1}^{4}c_{\alpha \beta }V^{(\beta )},\ R_{\alpha }=\sum _{\beta =1}^{4}c_{\alpha \beta }R^{(\beta )}}
etc.
↑ L. c. (33), p. 675 and 677.