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]).