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

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

However, according to (16) it is:

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

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

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

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

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

This is in agreement with the generally valid theorems of the theory of H. A. Lorentz.[1]

If we set in accordance with the Stefan-Boltmann law

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

then it follows (see. (14)):

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

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

The energy density of the true radiation is:

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

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

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

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

1. See for instance, M. Abraham, Theorie der Elektrizität, II, p. 282 (1905)