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

which becomes absorbed or transformed into work.

Altogether, the energy quantity

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

is absorbed, while the energy quantity

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

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

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

and

$N={\frac {2\pi i_{0}D}{{\mathfrak {B}}^{3}}}\int _{0}^{\pi /2}{\frac {\sin \phi \ d\phi }{\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}}$ $\cdot \left[\sigma ^{2}\sin ^{2}\phi ({\mathfrak {B}}_{-}^{2}+{\mathfrak {B}}_{+}^{2})+\sigma \cos \phi {\sqrt {1-\sigma ^{2}\sin ^{2}\phi }}({\mathfrak {B}}_{-}^{2}-{\mathfrak {B}}_{+}^{2})\right]$