Page:Zur Theorie der Strahlung in bewegten Körpern.djvu/18

$c/c_{-}\cos \alpha \,$ or $c/c_{+}\cos \alpha \,$. When the velocity has now the value $w_{1}$, then this relation becomes equal to $c/c_{-,1}\cos \alpha _{1}\,$ or $c/c_{+,1}\cos \alpha _{1}\,$. Thus, from the radiation (31) present in $R$ at the beginning, the fraction

$2\pi \sin \psi _{1}d\psi _{1}ih{\frac {c_{-,1}^{3}\cos \alpha }{c\cdot c_{-}^{3}}}$

or

$2\pi \sin \psi _{1}d\psi _{1}i'h{\frac {c_{+,1}^{3}\cos \alpha }{c\cdot c_{+}^{3}}}$

is now absorbed. If we want to obtain the fraction $Q$ of the whole energy contained in $R$ at the beginning, which is absorbed by $A$ and $B$ after a velocity change of $\delta w$, then we have to integrate these expressions with respect to $\psi _{1}$ from 0 to $\pi /2$. We preliminarily insert the values from (25) for $i$ and $i'$, and thus we obtain

$Q={\overset {\pi /2}{\underset {0}{\int }}}2\pi \sin \psi _{1}d\psi _{1}i_{0}h\left({\frac {c_{-,1}^{3}}{c_{-}^{4}}}+{\frac {c_{+,1}^{3}}{c_{+}^{4}}}\right).$

Now it is according to (1):

$c_{-,1}^{2}=c^{2}+(w+\delta w)^{2}-2c(w+\delta w)\cos \varphi =c_{-}^{2}-2\delta w(c\ \cos \varphi -w),$

or by use of (5)

$c_{-,1}^{2}=c_{-}^{2}-2\delta wc_{-}\cos \psi .$

Since we can also set $c_{-,1}\cos \psi _{1}\,$ instead of $c_{-}\cos \psi \,$ (within the expression multiplied with the infinitely small magnitude $\delta w$) it eventually becomes:

$c_{-}=c_{-,1}\left(1+\delta w{\frac {\cos \psi _{1}}{c_{-,1}}}\right).$

Quite analogously it is given:

$c_{+}=c_{+,1}\left(1-\delta w{\frac {\cos \psi _{1}}{c_{+,1}}}\right).$

If we use these equations, it becomes.

$Q=2\pi i_{0}h{\overset {\pi /2}{\underset {0}{\int }}}\sin \psi _{1}d\psi _{1}\left({\frac {1-4\delta w{\frac {\cos \psi _{1}}{c_{-,1}}}}{c_{-,1}}}+{\frac {1+4\delta w{\frac {\cos \psi _{1}}{c_{+,1}}}}{c_{+,1}}}\right).$