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!


2 Answers 2


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.

  • $\begingroup$ Thanks a lot! Your answer makes a lot of sense to me. I am just wondering whether there is a reason for all textbooks to use $E$ (instead of $S$) as the thermodynamic potential? The arguments usually go like this: use $S(E, V, N)=k_B*\ln(\Omega)$ to solve for $E(S, V, N)$ and then take partial derivatives to obtain other quantities (using $dE=TdS-pdV$ ). Then by using the Legendre transformation, we can get the thermodynamic potentials for canonical $F(T, V, N)=E-TS$ and grand canonical ensemble $J(T, V, \mu)=E-TS-\mu N$ correctly. What is wrong with this argument? $\endgroup$
    – NuKuYul
    Feb 5, 2022 at 2:04
  • $\begingroup$ They are all equivalent. In terms of mathematical derivation, there is no real difference using $S(U, V, N)$ or $U(S, V, N)$, because $\partial S/\partial U>0$. You can use either of them as a fundamental relation for a thermodynamic system. The only reason to prefer $U(S, V, N)$ over $S(U,V,N)$ is that the differential $dU=TdS-pdV+\mu dN$ can be easily interpreted as the first law in a quasi-static process, and following a traditional presentation of thermodynamics this seems very natural. But that's more about pedagogy. $\endgroup$
    – Meng Cheng
    Feb 5, 2022 at 3:04
  • $\begingroup$ Callen's classic book on thermodynamics introduces $S$ first, and uses the entropy maximization as the starting point for logical development of thermodynamics. $\endgroup$
    – Meng Cheng
    Feb 5, 2022 at 3:06
  • $\begingroup$ I see. Thank you again for the detailed explanation! $\endgroup$
    – NuKuYul
    Feb 5, 2022 at 3:19

Strictly speaking, thermodynamic potential should be a name reserved for the internal energy as a function of the extensive variables ($S,V,N$, for simple systems) and its Legendre transforms (Helmholtz free energy, enthalpy, Gibbs free energy, to stick on the most well known). Thermodynamics allows for other fundamental equations (i.e. those functions of the state which encode the whole thermodynamic information about a system), originating from the entropy as a function of its extensive variables. Legendre transforms are collectively known as Massieu's functions (see Callen's textbook on thermodynamics).

Even if it is not widely appreciated, the natural set of fundamental equations for statistical mechanics is not the set of thermodynamic potentials, but the set of Massieu functions. In therm of them, each fundamental equation of the different ensembles displays the same relation with the appropriate partition function.

Indeed, starting with the entropy $S(U,V,N)$, we have the following Massieu functions relevant for the canonical and grand-canonical ensemble: $$ \begin{align} S_1\left(\frac{1}{T},V,N\right) &= S-\frac{U}{T}=-\frac{F}{T}\\ S_2\left(\frac{1}{T},V,\frac{\mu}{T}\right) &= S-\frac{U}{T}+\frac{\mu}{T}N=\frac{PV}{T}=-\frac{J}{T}. \end{align} $$ It is immediate to see that in terms of these functions, the relation between partition function and thermodynamics is uniform: $$ \begin{align} S&= k_B \ln \Omega\\ S_1&= k_B \ln Z\\ S_2&= k_B \ln \Xi \end{align} $$ where $\Xi$ is the grand-canonical partition function.


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