aus Wikisource, der freien Quellensammlung
Potentialfunction einer nicht homogenen Kugel.
Für
V
′
{\displaystyle V'\!}
nehmen wir am besten den Ausdruck (7) des vorigen Paragraphen. Dann findet sich
(2)
2
∂
V
′
∂
r
′
=
∓
a
r
′
2
F
(
π
)
−
a
r
′
2
F
(
0
)
±
(
1
+
a
2
r
′
2
)
∫
0
π
F
(
θ
)
d
θ
(
a
2
+
r
′
2
−
2
a
r
′
cos
θ
)
1
2
±
r
′
2
−
a
2
r
′
∫
0
π
(
−
r
′
+
a
cos
θ
)
a
2
+
r
′
2
−
2
a
r
′
cos
θ
)
3
2
F
′
(
θ
)
d
θ
,
{\displaystyle {\begin{aligned}2{\frac {\partial V'}{\partial r'}}=&\mp {\frac {a}{r'^{2}}}F(\pi )-{\frac {a}{r'^{2}}}F(0)\\&\pm \left(1+{\frac {a^{2}}{r'^{2}}}\right)\int \limits _{0}^{\pi }{\frac {F(\theta )d\theta }{(a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {1}{2}}}}\\&\pm {\frac {r'^{2}-a^{2}}{r'}}\int \limits _{0}^{\pi }{\frac {(-r'+a\cos \theta )}{a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}F'(\theta )d\theta ,\\\end{aligned}}}
und daraus wird für
r
′
=
a
±
0
{\displaystyle r'=a\pm 0\!}
(3)
2
(
∂
V
′
∂
r
′
)
a
±
0
=
∓
1
a
F
(
π
)
−
1
a
F
(
0
)
±
2
∫
0
π
F
′
(
θ
)
d
θ
2
a
sin
1
2
θ
±
lim
r
′
2
−
a
2
r
′
∫
0
π
(
−
r
′
+
a
cos
θ
)
F
′
(
θ
)
d
θ
(
a
2
+
r
′
2
−
2
a
r
′
cos
θ
)
3
2
.
{\displaystyle {\begin{aligned}2\left({\frac {\partial V'}{\partial r'}}\right)_{a\pm 0}=&\mp {\frac {1}{a}}F(\pi )-{\frac {1}{a}}F(0)\pm 2\int \limits _{0}^{\pi }{\frac {F'(\theta )d\theta }{2a\sin {\frac {1}{2}}\theta }}\\&\pm \lim {\frac {r'^{2}-a^{2}}{r'}}\int \limits _{0}^{\pi }{\frac {(-r'+a\cos \theta )F'(\theta )d\theta }{(a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}.\\\end{aligned}}}
Das letzte Integral ist noch zu transformiren. Wir schreiben
a
cos
θ
−
r
′
=
(
a
−
r
′
)
−
a
(
1
−
cos
θ
)
=
(
a
−
r
′
)
−
2
a
sin
1
2
θ
2
.
{\displaystyle {\begin{aligned}a\cos \theta -r'&=(a-r')-a(1-\cos \theta )\\&=(a-r')-2\,a\sin {\frac {1}{2}}\theta ^{2}.\\\end{aligned}}}
Demnach ist
lim
r
′
2
−
a
2
r
′
∫
0
π
(
−
r
′
+
a
cos
θ
)
F
′
(
θ
)
d
θ
a
2
+
r
′
2
−
2
a
r
′
cos
θ
)
3
2
=
lim
(
r
′
+
a
)
r
′
∫
0
π
−
(
r
′
−
a
)
2
F
′
(
θ
)
d
θ
(
a
2
+
r
′
2
−
2
a
r
′
cos
θ
)
3
2
−
lim
(
r
′
2
−
a
2
)
r
′
∫
0
π
2
a
sin
1
2
θ
2
(
2
a
sin
1
2
θ
)
3
F
′
(
θ
)
d
θ
.
{\displaystyle {\begin{aligned}&\lim {\frac {r'^{2}-a^{2}}{r'}}\int \limits _{0}^{\pi }{\frac {(-r'+a\cos \theta )F'(\theta )d\theta }{a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}\\=&\lim {\frac {(r'+a)}{r'}}\int \limits _{0}^{\pi }{\frac {-(r'-a)^{2}F'(\theta )d\theta }{(a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}\\-&\lim {\frac {(r'^{2}-a^{2})}{r'}}\int \limits _{0}^{\pi }{\frac {2\,a\sin {\frac {1}{2}}\theta ^{2}}{\left(2\,a\sin {\frac {1}{2}}\theta \right)^{3}}}F'(\theta )d\theta .\\\end{aligned}}}
Der letzte Bestandtheil der rechten Seite verschwindet, wenn das Integral einen endlichen Werth hat, d. h. wenn
F
′
(
0
)
=
0
{\displaystyle F'(0)=0\!}
ist. Den ersten Bestandtheil zerlegen wir weiter. Es ist nemlich