Page:Zur Thermodynamik bewegter Systeme.djvu/13

when

$dv=-v{\frac {\beta d\beta }{1-\beta ^{2}}}$ and $dU_{0}=-p_{0}dv\,$

(12)

(the latter relation generally holds for adiabatic state changes, according to (7)).

We notice that according to (10) and (2)

${\begin{array}{rl}H=&U-\beta \phi =U-q{\mathfrak {G}}=U-\beta ^{2}(pv+U)\\=&U(1-\beta ^{2})-\beta ^{2}pv\end{array}}$

If we also insert for $U$ its value from (5), then it is given

$H=(1-\beta ^{2})H-(1-\beta ^{2})\beta {\frac {\partial H}{\partial \beta }}-\beta ^{2}pv$

or

$(1-\beta ^{2}){\frac {\partial H}{\partial \beta }}=-\beta (H+pv).$

(13)

Then it is

$(1-\beta ^{2}){\frac {\partial p}{\partial \beta }}=(1-\beta ^{2}){\frac {\partial }{\partial \beta }}\left(p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}\right)=$ ${\begin{array}{l}=-\beta p_{0}{\frac {\partial }{\partial U_{0}}}(H+pv)+\beta {\frac {\partial }{\partial v}}(H+pv)=\\\\=-\beta \left(p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}\right)-\beta v\left(p_{0}{\frac {\partial p}{\partial U_{0}}}-{\frac {\partial p}{\partial v}}\right)+\beta p=\\\\=\beta v\left({\frac {\partial p}{\partial v}}-p_{0}{\frac {\partial p}{\partial U_{0}}}\right).\end{array}}$

If we insert this value into (11) and consider (12), then we see that indeed $dp=0$.

The simultaneous change of $T$ is:

$dT={\frac {\partial T}{\partial v}}dv+{\frac {\partial T}{\partial U_{0}}}dU_{0}+{\frac {\partial T}{\partial \beta }}d\beta =$ $=-v{\frac {\beta d\beta }{1-\beta ^{2}}}\left({\frac {\partial T}{\partial v}}-p_{0}{\frac {\partial T}{\partial U_{0}}}\right)+{\frac {\partial T}{\partial \beta }}d\beta .$