The kinetic energy
$={\frac {1}{2}}\int \int \int \left(F{\frac {df}{dt}}+G{\frac {dg}{dt}}+H{\frac {dh}{dt}}\right)dx\ dy\ dz$.
Now
$F={\frac {\mu }{5}}\left[e\left(u{\frac {d^{2}}{dx^{2}}}{\frac {1}{R}}+v{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r}}+w{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r}}\right)\left({\frac {5r^{2}}{6}}{\frac {a^{2}}{2}}\right)+e{\frac {10}{3}}{\frac {u}{r}}\right.$
$\left.+e'\left(u'{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}+v'{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r'}}+w'{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r'}}\right)\left({\frac {5r'^{2}}{6}}{\frac {a'^{2}}{2}}\right)+{\frac {e'10}{3}}{\frac {u'}{r'}}\right]$,


with similar expressions for G and H.
${\frac {df}{dt}}={\frac {1}{4\pi }}\left[e\left(u{\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}+v{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r}}+w{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r}}\right)\right.$
$\left.+e'\left(u'{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}+v'{\frac {d^{2}}{dx\ dy\ }}{\frac {1}{r'}}+w'{\frac {d^{2}}{dx\ dz\ }}{\frac {1}{r'}}\right)\right]$,


with similar expressions for ${\tfrac {dg}{dt}}$ and ${\tfrac {dh}{dt}}$. Since the particles are supposed to be very small, we shall neglect those terms in F which depend on 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 ${\tfrac {df}{dt}}$ due to e'. We shall take the line joining the particle as the axis of x; and for brevity we shall denote ${\tfrac {\mu ee'}{24\pi }}$ by σ.
The coefficient of uu' in the part of the kinetic energy we are considering
:$=\sigma \int \int \int \left({\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}+{\frac {4}{r^{3}}}\right)r^{2}{\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}dx\ dy\ dz$.
Now, for values of r > R,
${\frac {1}{r'}}={\frac {1}{r}}R{\frac {d}{dx}}{\frac {1}{r}}+{\frac {R^{2}}{2\ !}}{\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}\dots$;
$\therefore {\frac {d^{2}}{dx^{2}}}{\frac {1}{r'}}={\frac {d^{2}}{dx^{2}}}{\frac {1}{r}}R{\frac {d^{3}}{dx^{3}}}{\frac {1}{r}}+\dots$.


Now, since
${\frac {d^{n}}{dx^{n}}}{\frac {1}{r}}=()^{n}{\frac {n\ !}{r^{n+1}}}Q_{n}$,
where Q_{n} is a zonal harmonic of the nth order; and since the product of two harmonics of different degrees integrated over