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,^[1] 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:^[2]

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

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

${\overset {\pi /2}{\int _{0}}}{\frac {D}{\cos \phi }}\left[{\frac {2\pi (i_{0}+p_{1}c)\cos \phi \sin \phi \ d\phi }{{\mathfrak {B}}_{-}}}+{\frac {2\pi (i_{0}-p_{2}c)\cos \phi \sin \phi \ d\phi }{{\mathfrak {B}}_{+}}}\right]$ $=2\pi i_{0}D\int _{0}^{\pi /2}\sin \phi \ d\phi \left({\frac {1}{{\mathfrak {B}}_{-}}}+{\frac {1}{{\mathfrak {B}}_{+}}}\right)+$ $+2\pi cD\int _{0}^{\pi /2}\sin \phi \ d\phi \left({\frac {p_{1}}{{\mathfrak {B}}_{-}}}-{\frac {p_{2}}{{\mathfrak {B}}_{+}}}\right).$

(1)

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

1. See J. Larmor, Boltzmann-Festschrift, p. 595, 1904.
2. See F. Hasenöhrl, l. c.