# Page:Zur Thermodynamik bewegter Systeme (Fortsetzung).djvu/6

The energy content of the cavity is:

$U=2\pi v\int_{0}^{\pi}\frac{J\sin\phi\ d\phi}{c'},$

where $c'$ means the relative velocity:

$c'=c\left(-\beta\cos\phi+\sqrt{1-\beta^{2}\sin^{2}\phi}\right).$

If we set in the previous integral according to (19):

$J = i + J\beta\ \cos\varphi,$

then it becomes:

$U=2\pi v\int_{0}^{\pi}\frac{i\ \sin\phi\ d\phi}{c'}+2\pi v\beta\int_{0}^{\pi}\frac{J\cos\phi\ \sin\phi\ d\phi}{c'}.$

The second summand is equal to:

$q\cdot\frac{2\pi v}{c^{2}}\int_{0}^{\pi}\frac{J}{c'}\cdot c\ \cos\varphi\cdot\sin\phi\ d\phi=q\mathfrak{G},$

as one can most simply recognize by comparison with the penultimate equation of p. 11 of my first report. In consequence of (2) it is therefore:

$H=2\pi v\int_{0}^{\pi}\frac{i\ \sin\phi\ d\phi}{c'}.$

The quantity $H$ is thus identical with the energy of the true radiation, which was indeed to be expected. If we introduce $i'$ and $\phi'$ by means of (20) and (21), then

$J_{abs}=i'\frac{\varkappa^{2}}{(1-\beta^{2}\sin^{2}\phi)^{2}\left(\frac{c}{c'}\right)^{4}}=i'\frac{\varkappa^{2}}{(1-\beta\cos\varphi)^{4}},$

because $c'\sqrt{1-\beta^{2}\sin^{2}\phi}=c(1-\beta\cos\varphi)$ (see F. Hasenöhrl, Ann. d. Phys., 15, p. 347, Gl. 7 [1904]). That the absolute radiation intensity is changing with direction proportional to $(1-\beta \cos \varphi)^4$, was already demonstrated by v. Mosengeil in another way (Ann. d. Phys., 22, p. 875, eq. 11 [1907]).