Page:Zur Theorie der Strahlung in bewegten Körpern.djvu/26

 Since the decrease of energy is equal to the performed work, we have

$d(\epsilon v)=-\frac{1}{3}\epsilon dv.$

or

$vd\epsilon+\frac{4}{3}\epsilon dv=0.$

Our contradiction is solved, when the density of the true radiation doesn't remain constant, but (as at the isothermic change) is always equal to $$\epsilon_{0}\varkappa$$, where $$\epsilon_{0}$$ remains constant and $$\varkappa$$ is the already mentioned function of the momentary velocity. Then it is the temperature which remains constant. When we insert this into the previous differential equation $$\epsilon-\epsilon_{0}\varkappa$$, and dividing $$\epsilon_{0}$$ away by the constant, then it remains

$vd\varkappa+\frac{4}{3}\varkappa dv=0,$

from which it follows:

$v=v_{0}\varkappa^{-3/4}.$

Herein, $$v_0$$ is the volume, when the velocity is equal to zero, and $$\varkappa= 1$$. Namely, this result is true including magnitudes of order $$\beta^2$$. If we insert for $$\varkappa$$ its value from (23), then it becomes:

$v=v_{0}\left(1+\frac{2}{3}\beta^{2}\right)^{-3/4}=v_{0}\left(1-\frac{1}{2}\beta^{2}\right).$

The simplest assumption is now, that for example the dimensions of matter are invariable perpendicular to their direction of motion, while the dimension coinciding with the direction of motion depends on the translation velocity by the factor $$1-\tfrac{1}{2}\beta^{2}$$. The agreement with the assumption of and  is thus a complete one.

I would like to allow myself, to remark that I derived this result also by another way, where the knowledge of the value of the radiation pressure was not