Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/335

698.] 697.] Let us next suppose that the axis of one of the shells is turned about $$C$$ as a centre, so that it now makes an angle $$\theta$$ with the axis of the other shell (Fig. 48). We have only to introduce the zonal harmonics of $$\theta$$ into this expression for $$M$$, and we find for the more general value of $$M$$,

This is the value of the potential energy due to the mutual action of two circular currents of unit strength, placed so that the normals through the centres of the circles meet in a point $$C$$ in an angle $$\theta$$, the distances of the circumferences of the circles from the point $$C$$ being $$c_1$$ and $$c_2$$, of which $$c_1$$ is the greater.

If any displacement $$dx$$ alters the value of $$M$$, then the force acting in the direction of the displacement is $$X = \frac{dM}{dx}$$.

For instance, if the axis of one of the shells is free to turn about the point $$C$$, so as to cause $$\theta$$ to vary, then the moment of the force tending to increase $$\theta$$ is $$\Theta$$, where

Performing the differentiation, and remembering that where $$Q_1^\prime$$ has the same signification as in the former equations,

698.] As the values of $$Q_i^\prime$$ occur frequently in these calculations the following table of values of the first six degrees may be useful. In this table $$\mu$$ stands for $$\cos \theta$$, and $$\nu$$ for $$\sin \theta$$.




 * $$Q_1^\prime = 1$$,
 * $$Q_2^\prime = 2 \mu $$,
 * $$Q_3^\prime = \frac{3}{2} (5 \mu^2 -1 ) = 6(\mu^2 - \frac{1}{4} \nu^2)$$,
 * $$Q_4^\prime = \frac{5}{2} \mu (7\mu^2 - 3) = 10 \mu (\mu^2 - \frac{3}{4} \nu^2)$$,
 * $$Q_5^\prime = \frac{15}{8} (21\mu^4 - 14\mu^2 +1 ) = 15 (\mu^4 -\frac{3}{2} \mu^2 \nu^2 + \frac{1}{8}\nu^4 )$$,
 * $$Q_6^\prime = \frac{21}{8} \mu (33 \mu^4 - 30\mu^2 +5) = 21 \mu (\mu^4 - \frac{5}{2} \mu^2\nu^2 + \frac{5}{8} \nu^4) $$.
 * }
 * $$Q_5^\prime = \frac{15}{8} (21\mu^4 - 14\mu^2 +1 ) = 15 (\mu^4 -\frac{3}{2} \mu^2 \nu^2 + \frac{1}{8}\nu^4 )$$,
 * $$Q_6^\prime = \frac{21}{8} \mu (33 \mu^4 - 30\mu^2 +5) = 21 \mu (\mu^4 - \frac{5}{2} \mu^2\nu^2 + \frac{5}{8} \nu^4) $$.
 * }
 * $$Q_6^\prime = \frac{21}{8} \mu (33 \mu^4 - 30\mu^2 +5) = 21 \mu (\mu^4 - \frac{5}{2} \mu^2\nu^2 + \frac{5}{8} \nu^4) $$.
 * }