Page:CunninghamExtension.djvu/21

1909.] areas, we have

$$\frac{l\ ds}{L\ dS}=\frac{r^{2}}{R^{2}}=\frac{1}{\Lambda^{2}},$$

where

in which $$\delta r,\delta R$$ are corresponding elements of a radius, $$\delta R$$ being at rest in the (R, T) system.

Hence

$$\frac{\delta r}{\delta R}=-\frac{1}{\Lambda\beta},$$

and

$$\frac{m\ ds}{M\ dS}=\frac{n\ ds}{N\ dS}=-\frac{1}{\Lambda^{2}\beta},$$

Using these and the final form of the transformation of the electric vectors in § 3, the following equations are obtained exactly as in the last section:

$$\begin{array}{l} p_{r}ds=\Lambda^{2}\left\{ P_{r}-\frac{v}{4\pi c}[EH]_{\nu}\right\} dS,\\ \\p_{\theta}ds=-\Lambda^{2}P_{\theta}dS/\beta,\\ \\p_{\phi}ds=-\Lambda^{2}P_{\phi}dS/\beta,\end{array}$$

p and P being the mechanical forces per unit area according as the element is observed in the (r, t) or (R, T) system.

If the element dS belongs to the wall of a region in which there is radiation in equilibrium $$[EH]_{\nu}$$ is zero as before if the time-average be taken, and P is normal to dS.

Thus

$$\begin{array}{l} p_{r}ds=\Lambda^{2}PL\ dS=\Lambda^{4}Pl\ ds,\\ \\p_{\theta}ds=-\frac{\Lambda^{2}}{\beta}PM\ dS=\Lambda^{4}Pm\ ds,\\ \\p_{\phi}ds=-\frac{\Lambda^{2}}{\beta}PN\ dS=\Lambda^{4}Pn\ ds.\end{array}$$

Thus p is normal to ds and is equal to $$\Lambda^{4}P.$$.

For the energy the transformation gives

$$\begin{array}{ll} \epsilon & =\frac{1}{8\pi}\left\{ e^{2}+h^{2}\right\} \\ \\ & =\frac{\Lambda^{4}}{8\pi}\left\{ E_{R}^{2}+H_{R}^{2}+\beta^{2}\left(1+\frac{v^{2}}{c^{2}}\right)\left(E_{\theta}^{2}+E_{\phi}^{2}+H_{\theta}^{2}+H_{\phi}^{2}\right)-\frac{4v\beta^{2}}{c}\left(E_{\theta}H_{\phi}-E_{\phi}H_{\theta}\right)\right\} ,\end{array}$$