aus Wikisource, der freien Quellensammlung
Wirkung sämmtlicher Theilchen
ε
′
{\displaystyle \varepsilon '\,}
auf ein Theilchen
ε
{\displaystyle \varepsilon \,}
.
Zur Abkürzung wollen wir setzen
∑
ε
′
r
v
′
2
=
W
,
{\displaystyle \sum {\frac {\varepsilon '}{r}}v'^{2}=W,}
∑
ε
′
r
d
x
1
d
t
=
u
1
,
{\displaystyle \sum {\frac {\varepsilon '}{r}}{\frac {dx_{1}}{dt}}=u_{1},}
∑
ε
′
r
d
y
1
d
t
=
u
2
,
{\displaystyle \sum {\frac {\varepsilon '}{r}}{\frac {dy_{1}}{dt}}=u_{2},}
∑
ε
′
r
d
z
1
d
t
=
u
3
.
{\displaystyle \sum {\frac {\varepsilon '}{r}}{\frac {dz_{1}}{dt}}=u_{3}.}
Dann haben wir
(3)
D
=
−
ε
c
2
v
2
V
+
ε
c
2
W
−
2
ε
c
2
(
u
1
d
x
d
t
+
u
2
d
y
d
t
+
u
3
d
z
d
t
)
.
{\displaystyle D=-{\frac {\varepsilon }{c^{2}}}v^{2}V+{\frac {\varepsilon }{c^{2}}}W-2{\frac {\varepsilon }{c^{2}}}\left(u_{1}{\frac {dx}{dt}}+u_{2}{\frac {dy}{dt}}+u_{3}{\frac {dz}{dt}}\right).}
Die Functionen
V
,
W
,
u
1
,
u
2
,
u
3
,
{\displaystyle V,\,W,\,u1,\,u2,\,u3,\,}
genügen der Gleichung von Laplace , folglich auch
D
{\displaystyle D\,}
, insofern es von
x
,
y
,
z
{\displaystyle x,\,y,\,z}
abhängig ist:
(4)
∂
2
D
∂
x
2
+
∂
2
D
∂
y
2
+
∂
2
D
∂
z
2
=
0.
{\displaystyle {\frac {\partial ^{2}D}{\partial x^{2}}}+{\frac {\partial ^{2}D}{\partial y^{2}}}+{\frac {\partial ^{2}D}{\partial z^{2}}}=0.}
Wir wollen noch die Aenderung von
V
{\displaystyle V\,}
herstellen, die in dem Zeitelement
d
t
{\displaystyle dt\,}
dadurch zu Stande kommt, dass die Theilchen
ε
′
{\displaystyle \varepsilon '\,}
sich bewegen, und
x
,
y
,
z
{\displaystyle x,\,y,\,z}
constant genommen werden. Es findet sich
∂
V
∂
t
=
−
∑
ε
′
∂
(
1
r
)
∂
x
1
d
x
1
d
t
−
∑
ε
′
∂
(
1
r
)
∂
y
1
d
y
1
d
t
−
∑
ε
′
∂
(
1
r
)
∂
z
1
d
z
1
d
t
.
{\displaystyle {\frac {\partial V}{\partial t}}=-\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial x_{1}}}{\frac {dx_{1}}{dt}}-\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial y_{1}}}{\frac {dy_{1}}{dt}}-\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial z_{1}}}{\frac {dz_{1}}{dt}}.}
Nun haben wir aber
∂
(
1
r
)
∂
x
1
=
−
∂
(
1
r
)
∂
x
,
{\displaystyle {\frac {\partial \left({\frac {1}{r}}\right)}{\partial x_{1}}}=-{\frac {\partial \left({\frac {1}{r}}\right)}{\partial x}},}
folglich
−
∑
ε
′
∂
(
1
r
)
∂
x
1
d
x
1
d
t
=
∑
ε
′
∂
(
1
r
)
∂
x
d
x
1
d
t
=
∂
u
1
∂
x
,
{\displaystyle -\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial x_{1}}}{\frac {dx_{1}}{dt}}=\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial x}}{\frac {dx_{1}}{dt}}={\frac {\partial u_{1}}{\partial x}},}
und ebenso
−
∑
ε
′
∂
(
1
r
)
∂
y
1
d
y
1
d
t
=
∑
ε
′
∂
(
1
r
)
∂
y
d
y
1
d
t
=
∂
u
2
∂
y
,
{\displaystyle -\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial y_{1}}}{\frac {dy_{1}}{dt}}=\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial y}}{\frac {dy_{1}}{dt}}={\frac {\partial u_{2}}{\partial y}},}
−
∑
ε
′
∂
(
1
r
)
∂
z
1
d
z
1
d
t
=
∑
ε
′
∂
(
1
r
)
∂
z
d
z
1
d
t
=
∂
u
3
∂
z
.
{\displaystyle -\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial z_{1}}}{\frac {dz_{1}}{dt}}=\sum \varepsilon '{\frac {\partial \left({\frac {1}{r}}\right)}{\partial z}}{\frac {dz_{1}}{dt}}={\frac {\partial u_{3}}{\partial z}}.}