Page:Thomson1881.djvu/15

 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

with similar expressions for G and H.

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,

Now, since

$$\frac{d^{n}}{dx^{n}}\frac{1}{r}=(-)^{n}\frac{n\ !}{r^{n+1}}Q_{n}$$,

where Qn is a zonal harmonic of the nth order; and since the product of two harmonics of different degrees integrated over