These expressions satisfy Maxwell 's equations, when
d
d
t
=
−
v
∂
∂
x
{\displaystyle {\frac {d}{dt}}=-v{\frac {\partial }{\partial x}}}
and lead to equation
(
1
−
A
2
v
2
)
∂
2
U
∂
x
2
+
∂
2
U
∂
y
2
+
∂
2
U
∂
z
2
=
0.
{\displaystyle \left(1-A^{2}v^{2}\right){\frac {\partial ^{2}U}{\partial x^{2}}}+{\frac {\partial ^{2}U}{\partial y^{2}}}+{\frac {\partial ^{2}U}{\partial z^{2}}}=0.}
But if
v
{\displaystyle v}
t
{\displaystyle t}
d
d
t
=
∂
∂
t
−
v
∂
∂
x
.
{\displaystyle {\frac {d}{dt}}={\frac {\partial }{\partial t}}-v{\frac {\partial }{\partial x}}.}
If our value shall generally hold for
x
{\displaystyle x}
∂
X
∂
t
{\displaystyle {\frac {\partial X}{\partial t}}}
v
∂
X
∂
x
{\displaystyle v{\frac {\partial X}{\partial x}}}
Now it is
∂
X
∂
t
=
∂
2
∂
x
∂
t
U
(
1
−
A
2
v
2
)
,
{\displaystyle {\frac {\partial X}{\partial t}}={\frac {\partial ^{2}}{\partial x\partial t}}U\left(1-A^{2}v^{2}\right),}
thus it must be
∂
∂
t
[
∂
U
∂
x
(
1
−
A
2
v
2
)
]
{\displaystyle {\frac {\partial }{\partial t}}\left[{\frac {\partial U}{\partial x}}\left(1-A^{2}v^{2}\right)\right]}
v
∂
2
U
∂
x
2
(
1
−
A
2
v
2
)
{\displaystyle v{\frac {\partial ^{2}U}{\partial x^{2}}}\left(1-A^{2}v^{2}\right)}
or
A
2
x
∂
v
∂
t
{\displaystyle A^{2}x{\frac {\partial v}{\partial t}}}
1
−
A
2
v
2
{\displaystyle 1-A^{2}v^{2}}
Also the values of
Y
,
Z
{\displaystyle Y,Z}
M
,
N
{\displaystyle M,N}
[
2
x
2
−
(
1
−
A
2
v
2
)
ϱ
2
]
A
2
∂
v
∂
t
{\displaystyle \left[2x^{2}-\left(1-A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}{\frac {\partial v}{\partial t}}}
3
x
(
1
−
A
2
v
2
)
{\displaystyle 3x\left(1-A^{2}v^{2}\right)}
and
(
1
−
A
2
v
2
)
[
(
x
2
+
(
1
−
A
2
v
2
)
ϱ
2
)
]
∂
v
∂
t
{\displaystyle \left(1-A^{2}v^{2}\right)\left[\left(x^{2}+\left(1-A^{2}v^{2}\right)\varrho ^{2}\right)\right]{\frac {\partial v}{\partial t}}}
−
[
2
x
2
−
(
1
−
A
2
v
2
)
ϱ
2
]
A
2
v
2
∂
v
∂
t
{\displaystyle -\left[2x^{2}-\left(1-A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}v^{2}{\frac {\partial v}{\partial t}}}
must be small against
3
x
(
1
−
A
2
v
2
)
v
2
{\displaystyle 3x\left(1-A^{2}v^{2}\right)v^{2}}
This condition is fulfilled, when the dimensions of the space, in which the energy comes essentially into consideration, are sufficiently small. Because the terms to be neglected all contain the linear dimensions in a higher power. Though
d
v
/
d
t
{\displaystyle dv/dt}
v
{\displaystyle v}
When this neglect is allowed, then we can put for the change of kinetic energy
d
d
t
(
m
2
v
2
)
=
m
v
d
v
d
t
=
K
d
r
d
t
d
t
=
m
d
r
d
t
d
2
r
d
t
2
,
{\displaystyle {\frac {d}{dt}}\left({\frac {m}{2}}v^{2}\right)=mv{\frac {dv}{dt}}=K{\frac {dr}{dt}}dt=m{\frac {dr}{dt}}{\frac {d^{2}r}{dt^{2}}},}
![{\displaystyle {\frac {\partial }{\partial t}}\left[{\frac {\partial U}{\partial x}}\left(1-A^{2}v^{2}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/261dc89eef32cfdb66fd4da993113a4be78d2e11)
![{\displaystyle \left[2x^{2}-\left(1-A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}{\frac {\partial v}{\partial t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6225731fdd15f4f3e443f03cfe51eaf7a2a0dcc)
![{\displaystyle \left(1-A^{2}v^{2}\right)\left[\left(x^{2}+\left(1-A^{2}v^{2}\right)\varrho ^{2}\right)\right]{\frac {\partial v}{\partial t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cae8ae2e144643d5e0e8e38b93bb6d1dd95ecd75)
![{\displaystyle -\left[2x^{2}-\left(1-A^{2}v^{2}\right)\varrho ^{2}\right]A^{2}v^{2}{\frac {\partial v}{\partial t}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3df7c54151e6612a20826a245c274ec8755258f7)
