Page:Zur Thermodynamik bewegter Systeme (Fortsetzung).djvu/7

 $\begin{array}{rl} H & =\frac{2\pi vi'}{c\varkappa^{2}}\int_{0}^{\pi}\sin\phi'd\phi'(1+\beta\ \cos\ \phi')\\ \\ & =\frac{4\pi v}{\varkappa^{2}}\cdot\frac{i'}{c}.\end{array}$|undefined

However, according to (16) it is:

$H=\sqrt{1-\beta^{2}}\cdot U_{0}=\varkappa\cdot\frac{4\pi v_{0}}{c}i_{0},$|undefined

where $$i_0$$ is the radiation intensity in the resting cavity. Thus it must be

$\varkappa v_{0}i_{0}=\frac{1}{\varkappa^{2}}vi'$;|undefined

or, since $$v = \varkappa v_{0}$$:

$i' = \varkappa^{2}i_{0}.$

This is in agreement with the generally valid theorems of the theory of.

If we set in accordance with the law

$i_{0} = \sigma T_{0}^4,$

then it follows (see. (14)):

$i'=\varkappa^{2}\sigma T_{0}^{4}=\frac{\sigma}{\varkappa^{2}}T^{4}.$|undefined

The constant of the law is thus to be divided by $$\varkappa^{2}$$.

The energy density of the true radiation is:

$\frac{H}{v}=\frac{4\pi}{c}\cdot\frac{i'}{\varkappa^{2}}=\frac{4\pi}{c}i_{0};$|undefined

thus it has the same value as in the resting cavity.

The total energy follows from (17): it has the value:

$U=\frac{1}{\sqrt{1-\beta^{2}}}\left(1+\frac{1}{3}\beta^{2}\right)U_{0}=\frac{4\pi v_{0}i_{0}}{c}\cdot\frac{1+\frac{1}{3}\beta^{2}}{\sqrt{1-\beta^{2}}}.$|undefined