Thus
d
2
w
1
d
s
2
−
w
1
{
(
d
w
1
d
s
)
2
+
(
d
w
2
d
s
)
2
+
(
d
w
3
d
s
)
2
−
(
d
w
4
d
s
)
2
}
=
w
¨
x
(
1
−
w
2
)
1
2
+
3
w
˙
x
(
w
w
˙
)
(
1
−
w
2
)
1
2
+
w
x
(
1
−
w
2
)
1
2
{
(
w
w
¨
)
+
3
(
w
w
˙
)
1
−
w
2
}
{\displaystyle {\begin{array}{l}{\frac {d^{2}w_{1}}{ds^{2}}}-w_{1}\left\{\left({\frac {dw_{1}}{ds}}\right)^{2}+\left({\frac {dw_{2}}{ds}}\right)^{2}+\left({\frac {dw_{3}}{ds}}\right)^{2}-\left({\frac {dw_{4}}{ds}}\right)^{2}\right\}\\\\\qquad ={\frac {{\ddot {w}}_{x}}{\left(1-w^{2}\right)^{\frac {1}{2}}}}+{\frac {3{\dot {w}}_{x}(w{\dot {w}})}{\left(1-w^{2}\right)^{\frac {1}{2}}}}+{\frac {w_{x}}{\left(1-w^{2}\right)^{\frac {1}{2}}}}\left\{(w{\ddot {w}})+{\frac {3(w{\dot {w}})}{1-w^{2}}}\right\}\end{array}}}
Now Abraham[1] x component of the reaction being
K
x
=
2
e
2
3
(
1
−
w
2
)
[
w
¨
x
+
w
˙
x
3
(
w
w
˙
)
1
−
w
2
+
w
x
1
−
w
2
{
(
w
w
¨
)
+
3
(
w
w
˙
)
2
1
−
w
2
}
]
;
{\displaystyle K_{x}={\frac {2e^{2}}{3\left(1-w^{2}\right)}}\left[{\ddot {w}}_{x}+{\dot {w}}_{x}{\frac {3(w{\dot {w}})}{1-w^{2}}}+{\frac {w_{x}}{1-w^{2}}}\left\{(w{\ddot {w}})+{\frac {3(w{\dot {w}})^{2}}{1-w^{2}}}\right\}\right];}
accordingly,
K
x
d
t
=
2
e
2
3
c
2
[
d
2
w
1
d
s
2
−
w
1
{
(
d
w
1
d
s
)
2
+
(
d
w
2
d
s
)
2
+
(
d
w
3
d
s
)
2
−
(
d
w
4
d
s
)
2
}
]
d
s
=
K
1
d
s
.
{\displaystyle {\begin{array}{ll}K_{x}dt&={\frac {2e^{2}}{3c^{2}}}\left[{\frac {d^{2}w_{1}}{ds^{2}}}-w_{1}\left\{\left({\frac {dw_{1}}{ds}}\right)^{2}+\left({\frac {dw_{2}}{ds}}\right)^{2}+\left({\frac {dw_{3}}{ds}}\right)^{2}-\left({\frac {dw_{4}}{ds}}\right)^{2}\right\}\right]ds\\\\&=K_{1}ds.\end{array}}}
To complete the symmetry of the result, we define quantities
K
t
{\displaystyle K_{t}}
K
4
{\displaystyle K_{4}}
K
t
d
t
=
2
e
2
3
c
2
[
d
2
w
4
d
s
2
−
w
4
{
(
d
w
1
d
s
)
2
+
(
d
w
2
d
s
)
2
+
(
d
w
3
d
s
)
2
−
(
d
w
4
d
s
)
2
}
]
d
s
=
K
4
d
s
.
{\displaystyle {\begin{array}{ll}K_{t}dt&={\frac {2e^{2}}{3c^{2}}}\left[{\frac {d^{2}w_{4}}{ds^{2}}}-w_{4}\left\{\left({\frac {dw_{1}}{ds}}\right)^{2}+\left({\frac {dw_{2}}{ds}}\right)^{2}+\left({\frac {dw_{3}}{ds}}\right)^{2}-\left({\frac {dw_{4}}{ds}}\right)^{2}\right\}\right]ds\\\\&=K_{4}ds.\end{array}}}
we then have
w
x
K
x
+
w
y
K
y
+
w
z
K
z
=
K
t
,
{\displaystyle w_{x}K_{x}+w_{y}K_{y}+w_{z}K_{z}=K_{t},}
w
1
K
1
+
w
2
K
2
+
w
3
K
3
+
w
4
K
4
=
0.
{\displaystyle w_{1}K_{1}+w_{2}K_{2}+w_{3}K_{3}+w_{4}K_{4}=0.}
Laue[2]
(
K
x
δ
x
+
K
y
δ
y
+
K
z
δ
z
−
K
δ
x
)
d
t
{\displaystyle \left(K_{x}\delta x+K_{y}\delta y+K_{z}\delta z-K\ \delta x\right)dt}
is an integral invariant for the whole group of spherical wave transformations. It will be sufficient to prove this for the case of the transformation
x
′
=
x
r
2
−
t
2
,
y
′
=
y
r
2
−
t
2
,
z
′
=
z
r
2
−
t
2
,
t
′
=
t
r
2
−
t
2
.
{\displaystyle x'={\frac {x}{r^{2}-t^{2}}},\ y'={\frac {y}{r^{2}-t^{2}}},\ z'={\frac {z}{r^{2}-t^{2}}},\ t'={\frac {t}{r^{2}-t^{2}}}.}
↑ Theorie der Elektricität , Vol. II, p. 123.↑ Ann. d. Phys. , Bd. 28, p. 436 (1908). 
![{\displaystyle K_{x}={\frac {2e^{2}}{3\left(1-w^{2}\right)}}\left[{\ddot {w}}_{x}+{\dot {w}}_{x}{\frac {3(w{\dot {w}})}{1-w^{2}}}+{\frac {w_{x}}{1-w^{2}}}\left\{(w{\ddot {w}})+{\frac {3(w{\dot {w}})^{2}}{1-w^{2}}}\right\}\right];}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebfee8513f31da574eb8f1a689d33c414da2b09f)
![{\displaystyle {\begin{array}{ll}K_{x}dt&={\frac {2e^{2}}{3c^{2}}}\left[{\frac {d^{2}w_{1}}{ds^{2}}}-w_{1}\left\{\left({\frac {dw_{1}}{ds}}\right)^{2}+\left({\frac {dw_{2}}{ds}}\right)^{2}+\left({\frac {dw_{3}}{ds}}\right)^{2}-\left({\frac {dw_{4}}{ds}}\right)^{2}\right\}\right]ds\\\\&=K_{1}ds.\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42c33d3d7c899663c20417f1e1f06c8785b40f7d)
![{\displaystyle {\begin{array}{ll}K_{t}dt&={\frac {2e^{2}}{3c^{2}}}\left[{\frac {d^{2}w_{4}}{ds^{2}}}-w_{4}\left\{\left({\frac {dw_{1}}{ds}}\right)^{2}+\left({\frac {dw_{2}}{ds}}\right)^{2}+\left({\frac {dw_{3}}{ds}}\right)^{2}-\left({\frac {dw_{4}}{ds}}\right)^{2}\right\}\right]ds\\\\&=K_{4}ds.\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3032b311161199cc496ba452580aff6d6ecf4481)
