Page:Zur Thermodynamik bewegter Systeme (Fortsetzung).djvu/2

then (15) becomes

$-\beta \varkappa {\frac {\partial H}{\partial \varkappa }}+\beta H-\beta v\left({\frac {\partial H}{\partial v}}\right)_{S_{0}}=0,$

or when $\beta$ is different from zero:

$\varkappa {\frac {\partial }{\partial \varkappa }}\left({\frac {H}{\varkappa }}\right)+v\left({\frac {\partial }{\partial v}}\right)_{S_{0}}\left({\frac {H}{\varkappa }}\right)=0.$

It follows from this equation, that $H/\varkappa$ must be a function of $v/\varkappa$, which of course must also depend on $S_{0}$. Furthermore, $H$ must be identical with $U_{0}$ for $\beta =0,\ \varkappa =1$. We satisfy these requirements when we put

${\frac {H}{\varkappa }}=F\left(S_{0},{\frac {v}{\varkappa }}\right)$.

$F\left(S_{0},{\tfrac {v}{\varkappa }}\right)$ is evidently the energy amount of the resting system, when it is adiabatically expanded from $v$ to $v/\varkappa$; if we denote this energy value with $U'_{0}$, then

$H={\sqrt {1-\beta ^{2}}}\cdot U'_{0}.$

Now, if we assume in accordance with Lorentz's hypothesis, that the velocity change is accompanied with a volume change proportional to ${\sqrt {1-\beta ^{2}}}$, then $U'_{0}$ is the energy of the resting body; then we remove the prime and thus put:

$H={\sqrt {1-\beta ^{2}}}\cdot U{}_{0}.$

(16)

8. Summary of results.

With the aid of equations (2) and (10), momentum and total energy ($U$) can be expressed by the state variables of the resting system. (We have to consider here, that equations (1), (3), (4) and (5) may not be applied now; they only hold for velocity changes at constant volume.)