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:

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

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