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.