Page:CunninghamExtension.djvu/21

areas, we have

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

where

$\Lambda ={\frac {r^{2}-c^{2}t^{2}}{k^{2}}},$ ${\frac {n\ ds}{N\ dS}}={\frac {m\ ds}{M\ dS}}={\frac {r\ \delta r}{R\ \delta R}},$

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

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