Let
| \begin{equation*} \left( \frac{\partial S}{\partial U}\right) := \frac{1}{T} \quad \mbox{and} \quad \left( \frac{\partial S}{\partial N}\right) := - \frac{\mu}{T} \quad . \end{equation*} | (2.36) |
So
| \begin{equation*} \beta = \frac{1}{kT} \quad \mbox{, and} \quad \gamma = -\frac{\mu}{kT} \quad . \end{equation*} | (2.37) |
Following again the procedure for the calculation of the free energy we find
| \begin{equation*} \Omega = U - \mu N- T S \end{equation*} | (2.38) |
and
| \begin{equation*} \Omega(T,V,\mu) = -k T \ln(Z(T, V, \mu)) \qquad . \end{equation*} | (2.39) |
© J. Carstensen (Stat. Meth.)