The part of the kinetic energy we are concerned with will evidently be
1
8
π
∫
∫
∫
2
μ
e
[
(
r
d
d
y
−
q
d
d
z
)
1
R
(
d
H
′
d
y
−
d
G
′
d
z
)
]
{\displaystyle {\frac {1}{8\pi }}\int \int \int 2\mu e\left[\left(r{\frac {d}{dy}}-q{\frac {d}{dz}}\right){\frac {1}{R}}\left({\frac {dH'}{dy}}-{\frac {dG'}{dz}}\right)\right]}
+
(
p
d
d
z
−
r
d
d
x
)
1
R
(
d
F
′
d
z
−
d
H
′
d
x
)
{\displaystyle +\left(p{\frac {d}{dz}}-r{\frac {d}{dx}}\right){\frac {1}{R}}\left({\frac {dF'}{dz}}-{\frac {dH'}{dx}}\right)}
+
(
q
d
d
x
−
p
d
d
y
)
1
R
(
d
G
′
d
x
−
d
F
′
d
y
)
d
x
d
y
d
z
{\displaystyle +\left(q{\frac {d}{dx}}-p{\frac {d}{dy}}\right){\frac {1}{R}}\left({\frac {dG'}{dx}}-{\frac {dF'}{dy}}\right)dx\ dy\ dz}
−
1
8
π
∫
∫
∫
μ
e
4
π
[
A
(
r
d
d
y
−
q
d
d
z
)
1
R
+
B
(
p
d
d
z
−
r
d
d
x
)
1
R
{\displaystyle -{\frac {1}{8\pi }}\int \int \int \mu e4\pi \left[A(r{\frac {d}{dy}}-q{\frac {d}{dz}}\right){\frac {1}{R}}+B\left(p{\frac {d}{dz}}-r{\frac {d}{dx}}\right){\frac {1}{R}}}
+
C
(
q
d
d
x
−
p
d
d
y
)
1
R
]
d
x
d
y
d
z
{\displaystyle \left.+C\left(q{\frac {d}{dx}}-p{\frac {d}{dy}}\right){\frac {1}{R}}\right]dx\ dy\ dz}
Let us take the first integral first, and take the term depending on p ; this is
μ
e
p
4
π
∫
∫
∫
d
d
z
1
R
(
d
F
′
d
z
−
d
H
′
d
x
)
−
d
d
y
1
R
(
d
G
′
d
x
−
d
F
′
d
y
)
d
x
d
y
d
z
{\displaystyle {\frac {\mu ep}{4\pi }}\int \int \int {\frac {d}{dz}}{\frac {1}{R}}\left({\frac {dF'}{dz}}-{\frac {dH'}{dx}}\right)-{\frac {d}{dy}}{\frac {1}{R}}\left({\frac {dG'}{dx}}-{\frac {dF'}{dy}}\right)dx\ dy\ dz}
.
Integrating by parts this becomes
−
μ
e
p
4
π
∫
∫
F
′
(
d
d
x
1
R
d
y
d
z
+
d
d
y
1
R
d
x
d
z
+
d
d
z
1
R
d
x
d
y
)
{\displaystyle -{\frac {\mu ep}{4\pi }}\int \int F'\left({\frac {d}{dx}}{\frac {1}{R}}dy\ dz+{\frac {d}{dy}}{\frac {1}{R}}dx\ dz+{\frac {d}{dz}}{\frac {1}{R}}dx\ dy\right)}
+
μ
e
p
4
π
∫
∫
1
R
(
d
H
′
d
x
d
y
d
z
+
d
G
′
d
x
d
x
d
z
+
d
F
′
d
x
d
z
d
y
)
{\displaystyle +{\frac {\mu ep}{4\pi }}\int \int {\frac {1}{R}}\left({\frac {dH'}{dx}}dy\ dz+{\frac {dG'}{dx}}dx\ dz+{\frac {dF'}{dx}}dz\ dy\right)}
+
μ
e
p
4
π
∫
∫
∫
1
R
d
d
x
(
d
F
′
d
x
+
d
G
′
d
y
+
d
H
′
d
z
)
{\displaystyle +{\frac {\mu ep}{4\pi }}\int \int \int {\frac {1}{R}}{\frac {d}{dx}}\left({\frac {dF'}{dx}}+{\frac {dG'}{dy}}+{\frac {dH'}{dz}}\right)}
−
F
′
(
d
2
d
x
2
+
d
2
d
y
2
+
d
2
d
z
2
)
1
R
d
x
d
y
d
{\displaystyle -F'\left({\frac {d^{2}}{dx^{2}}}+{\frac {d^{2}}{dy^{2}}}+{\frac {d^{2}}{dz^{2}}}\right){\frac {1}{R}}dx\ dy\ d}
z.
The surface-integrals are to be taken over the surface of the sphere; and the triple integral is to be taken throughout all space exterior to the sphere.
If the sphere be so small that we may substitute for the values of F',
d
F
′
d
x
{\displaystyle {\frac {dF'}{dx}}}
, &c. at the surface their values at the centre of the sphere, the first surface-integral
=
μ
e
p
F
1
′
{\displaystyle =\mu epF'_{1}}
, where
F
1
′
{\displaystyle F'_{1}}
is the value of
F
′
{\displaystyle F'}
at the centre of the sphere; the second surfaceintegral vanishes, and the triple integral also vanishes, since
d
2
d
x
2
1
R
+
d
2
d
y
2
1
R
+
d
2
d
z
2
1
R
=
0
{\displaystyle {\frac {d^{2}}{dx^{2}}}{\frac {1}{R}}+{\frac {d^{2}}{dy^{2}}}{\frac {1}{R}}+{\frac {d^{2}}{dz^{2}}}{\frac {1}{R}}=0}
and
d
F
′
d
x
+
d
G
′
d
y
+
d
H
′
d
z
=
0
{\displaystyle {\frac {dF'}{dx}}+{\frac {dG'}{dy}}+{\frac {dH'}{dz}}=0}
.