In thermodynamics one says that in equilibrium the corresponding thermodynamic potential is minimized. Why?

For example take the case of a canonical ensemble. Based on the assumption that the macroscopic state is given by the state which has the most microscopical states (i.e. entropy is maximized) and that the energy of the system is fixed one finds that $\rho = e^{-\beta H}/Z$ for some $\beta$. Then the Helmholtz free energy is given by $F = -\frac{1}{\beta} \ln Z$.

Now why does the free energy $F$ become minimal (if I keep $T = const.$)?

  • $\begingroup$ I know the Wikipedia article. Ok one could replace the question by "why does the principle of minimum energy holds". The principle of minimum energy follows from the principle of maximum entropy. But the maximal entropy is already used to derive the canonical ensemble and hence to define $F$. So I don't see why $F$ should become minimal due to the maximization of entropy. $\endgroup$ – toaster Jan 11 at 13:49
  • $\begingroup$ One common definition of the free energy $F=E-TS$. More explicitly, if we start with the entropy $S(E)$, not yet assuming that it's maximized, we can define $1/T=dS/dE$ as usual and then define $F=E-TS$. Given $S(E)$, the definition is unambiguous, and no equilibrium or canonical-ensemble assumption has been made. $\endgroup$ – Dan Yand Jan 11 at 14:21
  • $\begingroup$ If you do not assume the canonical ensemble you don't have a function S(E). You only have a sensible function S(E) if you restrict the possible states. This is done by saying that the only allowed states are $e^{-\beta H}/Z$ (or some other ensemble). Now you have a definite entropy for a given energy and therefore obtain a function S(E). Furthermore if you do not restrict to the canonical ensemble you cannot derive the first law. So when doing thermodynamics the canonical ensemble is always implicit. $\endgroup$ – toaster Jan 11 at 15:32
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    $\begingroup$ Entropy is an overloaded word, with different closely-related meanings. After thinking about your latest comment, I realize that my comment actually mixed two different meanings. I started with the standard definition of $S(E)$ as a property of the model (the log of the number of microstates with energy $\leq E$, for example), but my statement "not yet assuming that it's maximized" used an inequivalent-but-also-standard definition in which $S$ is a property of the state. I suspect that the answer to your original question involves a careful disentangling of related-but-different meanings. $\endgroup$ – Dan Yand Jan 11 at 16:33

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