I am a bit confused by the relation between thermodynamic potential and partition functions.
From my understanding, we can generate all thermodynamical quantities by taking partial derivatives to the thermodynamic potential (or $\ln(Z)$ where $Z$ is the partition function). Thus, I was expecting them to be proportional to each other.
For canonical ensembles, the partition function is $Z=\sum e^{-\beta E}$ and the corresponding thermodynamic potential is the Helmholtz free energy $F=-k_BT\ln(Z)$.
Similarly, for the grand canonical ensembles, the partition function is $Z=\sum e^{-\beta E-\alpha N}$ and the corresponding thermodynamic potential is the grand potential $J=-k_BT\ln(Z)$.
However, for microcanonical ensemble, it does not seem to work anymore. The partition function $Z=\sum1=\Omega$ so it will be the entropy which is proportional to $\ln(Z)$. However, the thermodynamic potential is the internal energy $U$ in this case. It seems a bit wired to me.
I am wondering whether there is some deep reason behind this? Thanks!
Best Answer
The thermodynamic potential for microcanonical ensemble should be $S$ instead of $U$, for two reasons: first, by definition microcanonical ensemble describes an isolated system whose energy is conserved. So in this sense, $U$ is treated like an external condition (like the temperature of a canonical ensemble). Second, thermodynamic potentials satisfy extremal principles: that is, the thermodynamic equilibrium state is the minimum/maximum of the potential. The entropy $S$ does exactly that for isolated systems.