Page:Zur Theorie der Strahlung bewegter Koerper.djvu/16

 which becomes absorbed or transformed into work.

Altogether, the energy quantity

$\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{i_{0}}{i}+\mathfrak{B}_{+}^{2}\frac{i_{0}}{i'}\right)=M$|undefined

is absorbed, while the energy quantity

$\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{p_{1}c}{i}-\mathfrak{B}_{+}^{2}\frac{p_{2}c}{i'}\right)=N$|undefined

is transformed into work. One can convince oneself, that exactly the same expressions are valid, when for example the black surface $$B$$ is replaced by a mirror. Namely, the radiation coming from $$A$$ to $$B$$ is not absorbed in $$B$$, but is partly reflected, i.e., from the totality of the energy incident in $$B$$, the fraction $$\tfrac{i'}{i}$$ is reflected; it comes back to $$A$$, and the fraction $$\frac{i_{0}}{i'}$$ is absorbed there, so that eventually the fraction $$\tfrac{i'}{i}\cdot\tfrac{i_{0}}{i'}=\tfrac{i_{0}}{i}$$ of the radiation propagating at the beginning in the direction from $$A$$ to $$B$$, is absorbed again. Thus $$M$$ and therefore also $$N$$ remain unchanged.

Now, if one uses equations (4), (5), (7) and (8), then

$M=\frac{2\pi i_{0}D}{\mathfrak{B}^{4}}\int_{0}^{\pi/2}\sin\phi\ d\phi(\mathfrak{B}_{-}^{3}+\mathfrak{B}_{+}^{3})$|undefined

and