# Is an entropy maximum being unique a tacit assumption in thermodynamics?

In Chapter 1 of his famous textbook on thermodynamics, Callen gives (among various other posulates) the following postulate:

Postulate II There exists a function ( called the entropy S) of the extensive parameters of any composite system, defined for all equilibrium states and having the foil owing property: The values assumed by the extensive parame- ters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.

My question is: is it tacitly assumed that this entropy maximum is unique? That is, that there are not two different entropy-maximizing equilibrium states on the manifold? Perhaps this is related to toy models of the ferromagnet and what not that I've worked out in statistical mechanics courses without some deep understanding unfortunately.

Edit: I guess this question can be recapiutaled as, "can Callen's maximum ($$S_i\leq S_{max}$$) be interpreted as the existence of a strict global maximum ($$S_i< S_{max}$$)"?

• IIRC, in the information theoretic formulation (i.e. MaxEnt principle) one can show that (in the simple cases of equilibrium ensembles such as canonical, grand canonical etc.) the equilibrium density matrix which maximizes the entropy functional is unique. But I have to look that up again. Is that what you are asking? Commented May 12, 2023 at 9:31

This is more about reasonable approximation rather than precise uniqueness.

The entropy function will in general have several local maxima. Of these, one is the highest so that is the 'true' equilibrium. Could two of these maxima have exactly the same $$S$$? In the messy real, non-idealized universe, that is extremely unlikely. But could the values of $$S$$ be very close for two or more maxima? That can happen.

Landau theory of phase transitions considers the situations with multiple equivalent extrema. More specifically, the first order transitions correspond to the existence of a single global minimum of Landau function, which changes its position, whereas the second order phase transitions correspond to simultaneous existence of several minima, with the system being stuck in one of them - the minima then merge at the phase transition. Magnetization of a ferromagnet is a typical example, as all its directions are equivalent and all can be treated as thermodynamic equilibria (image source).

Landau function is often associated with the free energy, and if we stick to phenomenological formulation of thermodynamics (which what Callen presents here), then there is probably no way of distinguishing them. From the statistical mechanics viewpoint they are not the same, since the susceptibilities determined from a partition function always display analytical behavior (which probably could be linked to existence of a unique extremum/ground state.) One then says that states with different extrema correspond to partial sums over the phase space, where the system spends very long time - hence mentioning of ergodicity breaking in this context (the system does not visit all the microstates, and time averaging is not equivalent to the ensemble average.)

However, I think that the quoted statement by Callen is made too early on in the discussion, to go into such cases.

Remark
As pointed above, Callen's books is an axiomatic presentation of phenomenological thermodynamics, using the entropy definition due to Clausius, Gibbs and Lewis (Gibbs I): see this answer for different definitions of entropy.

In that chapter Callen assumes that the equilibrium state given the externally controlled parameters is unique. In that case the true maximum of the entropy or true minimum of the potential(s) is unique but there are systems in which the entropy maximum (and the local minimum) is not unique given those parameters. Later on Callen mentions metastable states. For those there can be more than one local maxima (minima) a famous example being pure gaseous $$H_2$$ and $$O_2$$ whose mixture can exist together essentially "forever" but the slightest spark will violently turn the mixture into $$H_2O$$. Of course, it is possible to say that the relaxation time is not really infinite just very very very slow, hence the mixture is not really in equilibrium but that does not help much while you are observing it and die before anything would happen. There are other cases of metastable cases, for example supercooled liquid, or nitroglycerin...

Callen writes

In actuality, few systems are in absolute and true equilibrium. In absolute equilibrium all radioactive materials would have decayed completely and nuclear reactions would have transmuted all nuclei to the most stable of isotopes. Such processes, which would take cosmic times to complete, generally can be ignored. A system that has completed the relevant processes of spontaneous evolution, and that can be described by a reasonably small number of parameters, can be considered to be in metastable equilibrium. Such a limited equilibrium is sufficient for the application of thermodynamics.

Others handle issue claiming that there is always a true set of macroscopic parameters that would define a unique equilibrium we just have to find them. Maybe, but even then they are not always controllable externally, and then are called internal variables.

To complete the points already indicated, there may be the "stability condition" aspect which limits the form of the function $$S=S(U,V)$$.

Chapter 8 of Callen's book clearly states that the surface $$S(U,V)$$ must be concave:

To recapitulate, stability requires that the entropy surface lies everywhere below its family of tangent planes.

As a consequence, one is led to the local conditions of stability. For example $$({\frac{\partial^2 S}{\partial U^2}})_{V,N}<0$$......

These conditions must be verified over the entire surface and not only at the maximums. (Otherwise, there would be a phase change).

Under these conditions, it seems difficult for the function $$S$$ to have two maximum because it would necessarily have a convex part?

Hope it can help and sorry for my poor english.