Page:CunninghamExtension.djvu/22

so that for the radiation in equilibrium

$\epsilon ={\frac {\Lambda ^{4}}{3}}E\left\{1+2{\frac {c^{2}+v^{2}}{c^{2}-v^{2}}}\right\}=\Lambda ^{4}{\frac {3c^{2}+v^{2}}{3\left(c^{2}-v^{2}\right)}}E.$

Thus

$p=\Lambda ^{4}P={\frac {1}{3}}\Lambda ^{4}E={\frac {c^{2}-v^{2}}{3c^{2}+v^{2}}}\epsilon .$

This is the same equation as that obtained in § 3. Exactly in the same way for the momentum is found the result

$g={\frac {4}{3}}{\frac {\Lambda ^{4}v}{c^{2}-v^{2}}}E={\frac {4v}{3c^{2}+v^{2}}}\epsilon .$