Page:Zur Theorie der Strahlung bewegter Koerper.djvu/5

 Thus the same quantity is absorbed by both bodies $$A$$ and $$B$$, as it is emitted; upon both surfaces (in opposite direction) the same pressure

$2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$

is exerted, so that no work is performed altogether. Thus when no external force is acting, then the system maintains its velocity.

In case one of the bodies $$A$$ or $$B$$ is replaced by a perfect mirror, the radiation condition in $$R$$ must remain exactly the same, thus besides other things, $$2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$$ is the pressure upon the mirror as well.

We ask ourselves about the energy content of space $$R$$. The radiation traveling from $$A$$ to $$B$$, has the relative velocity:

$\mathfrak{B}(-\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{-}.$

The radiation coming from $$A$$ to $$B$$, has the velocity

$\mathfrak{B}(+\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{+}.$

both have to travel the path $$\tfrac{D}{\cos\phi}$$, thus the total energy content of space $$R$$ is

The first of the two summands gives the part of the energy of space $$R$$, which was provided by the black bodies.