aus Wikisource, der freien Quellensammlung
∂
r
∂
x
0
ω
r
d
(
r
j
1
)
=
d
(
∂
r
∂
x
0
ω
j
1
)
−
r
j
1
⋅
δ
(
∂
r
∂
x
0
ω
r
)
,
=
d
(
∂
r
∂
x
0
ω
j
1
)
−
ω
j
1
⋅
δ
∂
r
∂
x
0
−
j
1
∂
r
∂
x
0
δ
ω
+
j
1
∂
r
∂
x
0
ω
δ
r
r
,
{\displaystyle {\begin{array}{ll}{\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}{\frac {\omega }{r}}d\left(rj_{1}\right)&=d\left({\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\omega j_{1}\right)-rj_{1}\cdot \delta \left({\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}{\frac {\omega }{r}}\right),\\\\&=d\left({\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\omega j_{1}\right)-\omega j_{1}\cdot \delta {\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}-j_{1}{\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\delta \omega +j_{1}{\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}{\frac {\omega \delta r}{r}},\end{array}}}
so erhält man sofort:
(
ϵ
{\displaystyle \epsilon }
)
X
d
t
=
D
v
1
[
d
(
∂
r
∂
x
0
ω
j
1
)
−
ω
j
1
⋅
δ
∂
r
∂
x
0
−
j
1
∂
r
∂
x
0
δ
ω
−
∂
j
1
∂
x
0
ω
δ
r
]
.
{\displaystyle {\mathfrak {X}}\ dt={\mathsf {Dv}}_{1}\left[d\left({\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\omega j_{1}\right)-\omega j_{1}\cdot \delta {\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}-j_{1}{\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\delta \omega -{\frac {\partial j_{1}}{\partial {\mathfrak {x}}_{0}}}\omega \delta r\right].}
Hiefür aber kann, weil einerseits
δ
∂
r
∂
x
0
=
∂
δ
r
∂
x
0
,
{\displaystyle \delta {\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}={\frac {\partial \delta r}{\partial {\mathfrak {x}}_{0}}},}
und andererseits
∂
r
∂
x
0
δ
ω
=
∂
r
∂
x
0
d
ω
d
r
δ
r
=
∂
ω
∂
x
0
δ
r
{\displaystyle {\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\delta \omega ={\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}{\frac {d\omega }{dr}}\delta r={\frac {\partial \omega }{\partial {\mathfrak {x}}_{0}}}\delta r}
ist, auch geschrieben werden:
(
ζ
{\displaystyle \zeta }
)
X
d
t
=
D
v
1
[
d
(
∂
r
∂
x
0
ω
j
1
)
−
ω
j
1
∂
r
∂
x
0
−
j
1
∂
ω
∂
x
0
δ
r
−
∂
j
1
∂
x
0
ω
δ
r
]
,
{\displaystyle {\mathfrak {X}}\ dt={\mathsf {Dv}}_{1}\left[d\left({\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\omega j_{1}\right)-\omega j_{1}{\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}-j_{1}{\frac {\partial \omega }{\partial {\mathfrak {x}}_{0}}}\delta r-{\frac {\partial j_{1}}{\partial {\mathfrak {x}}_{0}}}\omega \delta r\right],}
oder (was dasselbe ist):
(
η
{\displaystyle \eta }
)
X
d
t
=
D
v
1
[
d
(
∂
r
∂
x
0
ω
j
1
)
−
∂
∂
x
0
(
ω
j
1
δ
r
)
]
,
{\displaystyle {\mathfrak {X}}\ dt={\mathsf {Dv}}_{1}\left[d\left({\frac {\partial r}{\partial {\mathfrak {x}}_{0}}}\omega j_{1}\right)-{\frac {\partial }{\partial {\mathfrak {x}}_{0}}}\left(\omega j_{1}\delta r\right)\right],}
oder, falls man für
ω
{\displaystyle \omega }
seinen Werth (
β
{\displaystyle \beta }
) substituirt:
(
ϑ
{\displaystyle \vartheta }
)
X
d
t
=
4
A
2
D
v
1
[
−
d
(
∂
ψ
∂
x
0
d
ψ
d
r
j
1
)
+
∂
∂
x
0
(
d
ψ
d
r
j
1
⋅
δ
ψ
)
]
.
{\displaystyle {\mathfrak {X}}\ dt=4A^{2}{\mathsf {Dv}}_{1}\left[-d\left({\frac {\partial \psi }{\partial {\mathfrak {x}}_{0}}}{\frac {d\psi }{dr}}j_{1}\right)+{\frac {\partial }{\partial {\mathfrak {x}}_{0}}}\left({\frac {d\psi }{dr}}j_{1}\cdot \delta \psi \right)\right].}
Nun ist aber:
d
ψ
d
r
j
1
=
−
∂
1
ψ
{\displaystyle {\frac {d\psi }{dr}}j_{1}=-\partial _{1}\psi }
. Somit folgt:
(
ι
.
)
X
d
t
=
4
A
2
D
v
1
[
d
(
∂
ψ
∂
x
0
⋅
∂
1
ψ
)
−
∂
∂
x
0
(
∂
1
ψ
⋅
δ
ψ
)
]
.
E
b
e
n
s
o
w
i
r
d
:
(
ϰ
.
)
Y
d
t
=
4
A
2
D
v
1
[
d
(
∂
ψ
∂
y
0
⋅
∂
1
ψ
)
−
∂
∂
y
0
(
∂
1
ψ
⋅
δ
ψ
)
]
,
(
λ
.
)
Z
d
t
=
4
A
2
D
v
1
[
d
(
∂
ψ
∂
z
0
⋅
∂
1
ψ
)
−
∂
∂
z
0
(
∂
1
ψ
⋅
δ
ψ
)
]
.
{\displaystyle {\begin{array}{lllll}(\iota .)&&{\mathfrak {X}}\ dt=4A^{2}{\mathsf {Dv}}_{1}\left[d\left({\frac {\partial \psi }{\partial {\mathfrak {x}}_{0}}}\cdot \partial _{1}\psi \right)-{\frac {\partial }{\partial {\mathfrak {x}}_{0}}}\left(\partial _{1}\psi \cdot \delta \psi \right)\right].&&{\mathsf {Ebenso\ wird}}:\\\\(\varkappa .)&&{\mathfrak {Y}}\ dt=4A^{2}{\mathsf {Dv}}_{1}\left[d\left({\frac {\partial \psi }{\partial {\mathfrak {y}}_{0}}}\cdot \partial _{1}\psi \right)-{\frac {\partial }{\partial {\mathfrak {y}}_{0}}}\left(\partial _{1}\psi \cdot \delta \psi \right)\right],\\\\(\lambda .)&&{\mathfrak {Z}}\ dt=4A^{2}{\mathsf {Dv}}_{1}\left[d\left({\frac {\partial \psi }{\partial {\mathfrak {z}}_{0}}}\cdot \partial _{1}\psi \right)-{\frac {\partial }{\partial {\mathfrak {z}}_{0}}}\left(\partial _{1}\psi \cdot \delta \psi \right)\right].\end{array}}}
Durch
(
ι
.
)
{\displaystyle (\iota .)}
ist der Beweis geliefert für die Formel (11.):
Um den Beweis der folgenden Formel (12.) zu erhalten, bedarf es ebenfalls ziemlich mühsamer Betrachtungen. Es waren [vergl. den Anfang des vorhergehenden Abschnitts (pag. 158)] mit
(
μ
1
{\displaystyle \mu _{1}}
.)
x
0
=
C
1
+
C
11
x
0
+
C
12
y
0
+
C
13
z
0
,
y
0
=
C
2
+
C
21
x
0
+
C
22
y
0
+
C
23
z
0
,
z
0
=
C
3
+
C
31
x
0
+
C
32
y
0
+
C
33
z
0
{\displaystyle {\begin{array}{l}x_{0}=C^{1}+C^{11}{\mathfrak {x}}_{0}+C^{12}{\mathfrak {y}}_{0}+C^{13}{\mathfrak {z}}_{0},\\y_{0}=C^{2}+C^{21}{\mathfrak {x}}_{0}+C^{22}{\mathfrak {y}}_{0}+C^{23}{\mathfrak {z}}_{0},\\z_{0}=C^{3}+C^{31}{\mathfrak {x}}_{0}+C^{32}{\mathfrak {y}}_{0}+C^{33}{\mathfrak {z}}_{0}\end{array}}}