The kinetic energy
=
1
2
∫
∫
∫
(
F
d
f
d
t
+
G
d
g
d
t
+
H
d
h
d
t
)
d
x
d
y
d
z
{\displaystyle ={\frac {1}{2}}\int \int \int \left(F{\frac {df}{dt}}+G{\frac {dg}{dt}}+H{\frac {dh}{dt}}\right)dx\ dy\ dz}
.
Now
F
=
μ
5
[
e
(
u
d
2
d
x
2
1
R
+
v
d
2
d
x
d
y
1
r
+
w
d
2
d
x
d
z
1
r
)
(
5
r
2
6
−
a
2
2
)
+
e
10
3
u
r
{\displaystyle F={\frac {\mu }{5}}\left[e\left(u{\frac {d^{2}}{dx^{2}}}{\frac {1}{R}}+v{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r}}+w{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r}}\right)\left({\frac {5r^{2}}{6}}-{\frac {a^{2}}{2}}\right)+e{\frac {10}{3}}{\frac {u}{r}}\right.}
+
e
′
(
u
′
d
2
d
x
2
1
r
′
+
v
′
d
2
d
x
d
y
1
r
′
+
w
′
d
2
d
x
d
z
1
r
′
)
(
5
r
′
2
6
−
a
′
2
2
)
+
e
′
10
3
u
′
r
′
]
{\displaystyle \left.+e'\left(u'{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}+v'{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r'}}+w'{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r'}}\right)\left({\frac {5r'^{2}}{6}}-{\frac {a'^{2}}{2}}\right)+{\frac {e'10}{3}}{\frac {u'}{r'}}\right]}
with similar expressions for G and H.
d
f
d
t
=
1
4
π
[
e
(
u
d
2
d
x
2
1
r
+
v
d
2
d
x
d
y
1
r
+
w
d
2
d
x
d
z
1
r
)
{\displaystyle {\frac {df}{dt}}={\frac {1}{4\pi }}\left[e\left(u{\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}+v{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r}}+w{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r}}\right)\right.}
+
e
′
(
u
′
d
2
d
x
2
1
r
′
+
v
′
d
2
d
x
d
y
1
r
′
+
w
′
d
2
d
x
d
z
1
r
′
)
]
{\displaystyle \left.+e'\left(u'{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}+v'{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r'}}+w'{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r'}}\right)\right]}
with similar expressions for
d
g
d
t
{\displaystyle {\tfrac {dg}{dt}}}
d
h
d
t
{\displaystyle {\tfrac {dh}{dt}}}
a² and a'² .
The part of the kinetic energy we are concerned with involves the product ee': let us first calculate that part of it arising from the product of that part of F due to e with that part of
d
f
d
t
{\displaystyle {\tfrac {df}{dt}}}
x ; and for brevity we shall denote
μ
e
e
′
24
π
{\displaystyle {\tfrac {\mu ee'}{24\pi }}}
The coefficient of uu' in the part of the kinetic energy we are considering
=
σ
∫
∫
∫
(
d
2
d
x
2
1
r
+
4
r
3
)
r
2
d
2
d
x
2
1
r
′
d
x
d
y
d
z
{\displaystyle =\sigma \int \int \int \left({\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}+{\frac {4}{r^{3}}}\right)r^{2}{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}dx\ dy\ dz}
Now, for values of r > R,
1
r
′
=
1
r
−
R
d
d
x
1
r
+
R
2
2
!
d
2
d
x
2
1
r
−
…
{\displaystyle {\frac {1}{r'}}={\frac {1}{r}}-R{\frac {d}{dx}}{\frac {1}{r}}+{\frac {R^{2}}{2\ !}}{\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}-\dots }
∴
d
2
d
x
2
1
r
′
=
d
2
d
x
2
1
r
−
R
d
3
d
x
3
1
r
+
…
{\displaystyle \therefore {\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}={\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}-R{\frac {d^{3}}{dx^{3}}}{\frac {1}{r}}+\dots }
Now, since
d
n
d
x
n
1
r
=
(
−
)
n
n
!
r
n
+
1
Q
n
{\displaystyle {\frac {d^{n}}{dx^{n}}}{\frac {1}{r}}=(-)^{n}{\frac {n\ !}{r^{n+1}}}Q_{n}}
,
where Qn is a zonal harmonic of the nth order; and since the product of two harmonics of different degrees integrated over
![{\displaystyle \left.+e'\left(u'{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}+v'{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r'}}+w'{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r'}}\right)\left({\frac {5r'^{2}}{6}}-{\frac {a'^{2}}{2}}\right)+{\frac {e'10}{3}}{\frac {u'}{r'}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6d58d9a3e05265cf5d2bd7eebe54983b498c3c3)
![{\displaystyle \left.+e'\left(u'{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}+v'{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r'}}+w'{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r'}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86025ccd5c72fe2814b16101e910c58cbd61626b)
