Page:Zur Theorie der Strahlung in bewegten Körpern.djvu/19

 or

We can determine the first integral by the aid of equation (21), the second one has the value:

thus it becomes

$Q=h\epsilon_{0}\varkappa_{1}-h\epsilon_{0}\frac{\delta w}{c}\left(\frac{1+\beta_{1}^{2}}{2\beta_{1}(1-\beta_{1}^{2})^{2}}-\frac{1}{4\beta_{1}^{2}}\log\frac{1+\beta_{1}}{1-\beta_{1}}\right),$|undefined

where $$\varkappa_1$$ is the value, which emerges from $$\varkappa$$ when $$\beta_1$$ is replaced by $$\beta$$ (see equation (21)). One easily convince oneself, that the bracket expression in the last equation has the value $$d\varkappa_{1}/d\beta_1 $$; thus it becomes:

$Q=h\epsilon_{0}\left(\varkappa_{1}-\delta\beta\frac{dx_{1}}{d\beta_{1}}\right)=h\epsilon_{0}\varkappa=h\epsilon.$|undefined

Our earlier assertion is proven by that. At the beginning, when the velocity of our system was $$w$$, the energy quantity contained in $$B$$ had the amount $$h(\epsilon+\epsilon') = h\epsilon_{0}(\varkappa+\tau)$$; when the velocity is changed, and the equilibrium of radiation in $$R$$ is reestablished, then this energy quantity has the amount $$h\epsilon_{0}(\varkappa_{1}+\tau_{1})$$ (where $$\tau_{1}$$ is formed from $$\beta_{1}$$ in the same way again, as $$\tau$$ from $$\beta$$). As we know, the fraction $$h\epsilon_{0}\varkappa_{1}$$ of the heat reservoir of bodies $$A$$ and $$B$$ stems from that; since they (as we have just proven) absorb the fraction $$h\epsilon_{0}\varkappa$$ from the initially existing radiation, we see that when the velocity is changed from $$w$$ to $$w_1$$, the boundary of the cavity gives off the heat

$h\epsilon_{0}(\varkappa_{1}-\varkappa)$