Page:Zur Theorie der Strahlung bewegter Koerper.djvu/6

My paper already cited dealt with it.

The second summand, however, represents the radiation energy gained from mechanical work. Since no work is performed at uniform motion as we have seen, this must be a work which must be performed when our system is accelerated.

One can easily determine the amount of this energy quantity in the following way. Let us imagine a system in absolute rest, and the radiation of both black surfaces $A$ and $B$ is somehow hindered: space $R$ is thus totally free of radiation. Now, in one instant the radiation from $A$ and $B$ shall be left free, and the system shall simultaneously brought to velocity $c$. From that instant on, the work $2\pi p_{1}\cos \phi \ \sin \phi \ d\phi \cdot c$ in the unit of time must be performed against the radiation emitted from $A$. Now, time ${\tfrac {D}{{\mathfrak {B}}_{-}\cos \phi }}$ is passing until this radiation arrives in $B$, and there it provides the same quantity of work itself. During this time, the work provided from outside is thus uncompensated. Also the radiation emitted from $B$ performs right from the start the work $2\pi p_{2}\cos \phi \ \sin \phi \ d\phi \cdot c$, and now the time ${\tfrac {D}{{\mathfrak {B}}_{+}\cos \phi }}$ is passing until this radiation arrives in $A$, and there the same work from outside must be performed in the opposite sense. Thus, more work was performed from the outside by

${\frac {D}{\cos \phi \cdot {\mathfrak {B}}_{-}}}2\pi p_{1}\cos \phi \sin \phi \ d\phi \cdot c{\frac {D}{\cos \phi \cdot {\mathfrak {B}}_{+}}}2\pi p_{2}\cos \phi \sin \phi \ d\phi \cdot c$

i.e., it was gained from the radiation. If we integrate this amount over all beam directions, then we obtain the expression above.

If our system is at rest in the beginning, yet the surfaces $A$ and $B$ are not hindered of emanating radiation, then radiation energy of certain amount is present at the beginning also in $A$. The previous consideration indicates,