Page:Zur Theorie der Strahlung bewegter Koerper.djvu/4

 $2\pi\ (i_{1}+p_{1}c)\ \cos\phi\ \sin\phi\ d\phi = 2\pi\ i\ \cos\phi\ sin\phi\ d\phi$

This expression now gives us also the amount of the radiation falling upon $$B$$ in the unit of time. There, one fraction of it is absorbed, and one fraction is transformed into work. If $$B$$ would be at the absolute temperature zero, then the radiation just considered would be the only one present in $$R$$. Since in this case, no force of resistance against the motion of our system is to be expected, then the same pressure into opposite directions must be effective in $$A$$ and $$B$$, so that no work is performed in the whole. Thus in $$B$$, the energy quantity

$2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$

is absorbed, and the energy quantity

$2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\cdot c\,$

is transformed into mechanical work.

If we now imagine that $$B$$ has the same temperature as $$A$$, then also $$B$$ emanates energy

$2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$

in a certain direction. If this radiation exerts the pressure $$2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$$, then it performs work by which amount the energy provided by $$B$$ has to be diminished, so that also $$B$$ is left by the radiation quantity

$2\pi(i_{0}-p_{2}c)\cos\phi\ \sin\phi\ d\phi=2\pi i'\cos\phi\ \sin\phi\ d\phi\,$

The same energy quantity also occurs in $$A$$, and a quite similar consideration as earlier teaches, that the energy quantity

$2\pi i_{0}\cos\phi\ \sin\phi\ d\phi$

is absorbed there, because the work $$2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$$ must also be performed against the incident radiation from outside, which is also transformed into work.