Page:Zur Theorie der Strahlung in bewegten Körpern. Berichtigung.djvu/2

 electromagnetic momentum which coincides with the direction of the system and which is contained in the cavity, we have to calculate expression (1) with $$cos \varphi$$, then to integrate with respect to $$\varphi$$ from O to $$\pi$$, and then to multiply the result with the volume of cavity $$h$$. If we additionally substitute for $$c'$$ its value, then the momentum becomes

$\begin{array}{rl} G & =\frac{2\pi i_{0}}{c^{2}}h\overset{\pi}{\underset{0}{\int}}\frac{\sin\varphi\cos\varphi\ d\varphi}{(1+\beta^{2}-2\beta\cos\varphi)^{2}}\\ \\ & =\frac{\epsilon_{0}h}{c}\left(\frac{1}{2\beta}\frac{1+\beta^{2}}{(1-\beta^{2})^{2}}-\frac{1}{4\beta^{2}}\log\frac{1+\beta}{1-\beta}\right).\end{array}.$|undefined

Now, the longitudinal electromagnetic mass is given by $$\tfrac{1}{c}\tfrac{dG}{d\beta}$$; thus it becomes equal to (when higher terms are neglected):

$\frac{4}{3}\frac{h\epsilon_{0}}{c^{2}}.$|undefined

This is half of the value given by me.

After it was sought in vain after a principal difference, I found that this difference stems from a calculation error, unfortunately committed by me in my paper. At p. 362, line 6 from above,

not $\frac{2\beta_{1}}{c^{2}(1-\beta_{1}^{2})^{2}}\overset{\pi/2}{\underset{0}{\int}}\dots$ shall be stated, but $\frac{4\beta_{1}}{c^{2}(1-\beta_{1}^{2})^{2}}\overset{\pi/2}{\underset{0}{\int}}\dots$,|undefined

therefore the heat absorbed by the walls of the cavity when the system is accelerated, is:

$Q=h\epsilon_{0}\left(\varkappa_{1}-2\delta\beta\frac{\partial\varkappa_{1}}{\partial\beta}\right);$|undefined

however, since furthermore the walls have given off the heat $$h\epsilon_{0}\varkappa_1$$, we can say that the walls of the cavity (when accelerated by $$\delta_{\tau}$$) have altogether given off the heat

$2h\epsilon_{0}\delta\beta\frac{\partial\varkappa}{\partial\beta}=2h\epsilon_{0}\delta\varkappa$