aus Wikisource, der freien Quellensammlung
Potentialfunction einer nicht homogenen Kugel.
{
r
2
∂
V
∂
r
−
(
r
2
∂
V
∂
r
)
r
+
d
r
}
sin
θ
d
θ
d
φ
{\displaystyle \left\lbrace r^{2}{\frac {\partial V}{\partial r}}-\left(r^{2}{\frac {\partial V}{\partial r}}\right)_{r+dr}\right\rbrace \sin \theta \,d\theta \,d\varphi }
=
−
∂
(
r
2
∂
V
∂
r
)
∂
r
sin
θ
d
r
d
θ
d
φ
.
{\displaystyle =-{\frac {\partial \left(r^{2}{\frac {\partial V}{\partial r}}\right)}{\partial r}}\sin \theta \,dr\,d\theta \,d\varphi .}
r
sin
θ
d
r
d
φ
{\displaystyle r\,\sin \theta \,dr\,d\varphi }
r
sin
(
θ
+
d
θ
)
d
r
d
φ
{\displaystyle r\,\sin(\theta +d\theta )\,dr\,d\varphi }
N
=
∂
V
r
∂
θ
{\displaystyle N={\frac {\partial V}{r\,\partial \theta }}}
N
=
−
(
∂
V
r
∂
θ
)
θ
+
d
θ
{\displaystyle N=-\left({\frac {\partial V}{r\,\partial \theta }}\right)_{\theta +d\theta }}
{
∂
V
r
∂
θ
sin
θ
−
(
∂
V
r
∂
θ
sin
θ
)
θ
+
d
θ
}
r
d
r
d
φ
{\displaystyle \left\lbrace {\frac {\partial V}{r\,\partial \theta }}\sin \theta -\left({\frac {\partial V}{r\,\partial \theta }}\sin \theta \right)_{\theta +d\theta }\right\rbrace r\,dr\,d\varphi }
=
−
∂
(
∂
V
∂
θ
sin
θ
)
∂
θ
d
r
d
θ
d
φ
.
{\displaystyle =-{\frac {\partial \left({\frac {\partial V}{\partial \theta }}\sin \theta \right)}{\partial \theta }}dr\,d\theta \,d\varphi .}
Endlich handelt es sich noch um zwei Seitenflächen, rechtwinklig gegen den Parallelkreis. Jede von ihnen hat den Flächeninhalt
r
d
r
d
θ
{\displaystyle r\,dr\,d\theta }
N
=
∂
V
r
sin
θ
d
φ
{\displaystyle N={\frac {\partial V}{r\,\sin \theta \,d\varphi }}}
N
=
(
∂
V
r
sin
θ
d
φ
)
φ
+
d
φ
{\displaystyle N=\left({\frac {\partial V}{r\,\sin \theta \,d\varphi }}\right)_{\varphi +d\varphi }}
{
∂
V
r
sin
θ
∂
φ
−
(
∂
V
r
sin
θ
∂
φ
)
φ
+
d
φ
}
r
d
r
d
θ
{\displaystyle \left\lbrace {\frac {\partial V}{r\,\sin \theta \,\partial \varphi }}-\left({\frac {\partial V}{r\,\sin \theta \,\partial \varphi }}\right)_{\varphi +d\varphi }\right\rbrace r\,dr\,d\theta }
=
−
1
sin
θ
∂
2
V
∂
φ
2
d
r
d
θ
d
φ
.
{\displaystyle =-{\frac {1}{\sin \theta }}{\frac {\partial ^{2}V}{\partial \varphi ^{2}}}dr\,d\theta \,d\varphi .}
Fassen wir diese Beiträge zusammen, so wird aus der linken Seite der Gleichung (2):
−
d
r
d
θ
d
φ
{
∂
(
r
2
∂
V
∂
r
)
∂
r
sin
θ
+
∂
(
sin
θ
∂
V
∂
θ
)
∂
θ
+
1
sin
θ
∂
2
V
∂
φ
2
}
.
{\displaystyle -dr\,d\theta \,d\varphi \left\lbrace {\frac {\partial \left(r^{2}{\frac {\partial V}{\partial r}}\right)}{\partial r}}\sin \theta +{\frac {\partial \left(\sin \theta \,{\frac {\partial V}{\partial \theta }}\right)}{\partial \theta }}+{\frac {1}{\sin \theta }}{\frac {\partial ^{2}V}{\partial \varphi ^{2}}}\right\rbrace .}
M
=
ρ
r
2
sin
θ
d
r
d
θ
d
φ
.
{\displaystyle M=\rho \,r^{2}\,\sin \theta \,dr\,d\theta \,d\varphi .}
−
r
2
sin
θ
d
r
d
θ
d
φ
{\displaystyle -r^{2}\,\sin \theta \,dr\,d\theta \,d\varphi }
(3)
1
r
2
{
∂
(
r
2
∂
V
∂
r
)
∂
r
+
1
sin
θ
∂
(
sin
θ
∂
V
∂
θ
)
∂
θ
+
1
sin
θ
2
∂
2
V
∂
φ
2
}
=
−
4
π
ρ
.
{\displaystyle {\frac {1}{r^{2}}}\left\lbrace {\frac {\partial \left(r^{2}{\frac {\partial V}{\partial r}}\right)}{\partial r}}+{\frac {1}{\sin \theta }}{\frac {\partial \left(\sin \theta \,{\frac {\partial V}{\partial \theta }}\right)}{\partial \theta }}+{\frac {1}{\sin \theta ^{2}}}{\frac {\partial ^{2}V}{\partial \varphi ^{2}}}\right\rbrace =-4\,\pi \,\rho .}
