# Derivation: Maximum entropy implies minimum energy

Callen claims in his book (chapter 5 in my copy at least) that the condition of minimum energy for fixed entropy is exactly equivalent to the condition of maximum entropy for fixed energy. I have seen this claim restated multiple times, and there are a bunch of answers in this site, like this one, which directly cite Callen's claim and derviation.

I have some problems with the way he derives this equivalency. Callen starts his derivation by stating that in the absence of internal constraints, $$dS = 0$$ and $$d^2S < 0$$ since entropy is maximum at this point. He then states that this implies that $$\frac{\partial^2 S}{\partial X^2} < 0$$ for all $$X$$. I understand he is roughly referring to the hessian of $$S$$, however, if this is the case, what he should probably say is that the Hessian matrix of the entropy $$\mathcal{H}(S)$$ is negative definite at that point, which means that $$v^{T}\mathcal{H}(S) v < 0~\forall v$$.

Callen finishes his derivation by stating that, since he can prove from this that

$$\frac{\partial^2 U}{\partial X^2} > 0$$ energy is a minimum at this point.

I don't think this implies the existence of a minimum, which, similarly to what was stated before requires the Hessian to be positive definite, which is stronger than all $$\partial^2_X U$$ being positive.

My question is twofold:

1) Does the partials $$\partial^2_X U$$ being positive imply that there is a minimum as Callen states?

2) If it doesn't then his derivation that the internal energy has a minimum at that point doesn't hold (as shown here, in wikipedia for instance), what would be a correct derivation of this fact?

We need to be careful about what "the partials $$\partial^2_X U$$ being positive implies a minimum" means. If we mean that we have a given set of coordinates $$X_i$$ and that $$\partial^2 U / \partial X_i^2 > 0$$ for all $$i$$, then the claim is false. For example, a Hessian matrix like

$$\begin{pmatrix} 1 & -2 \\ -2 & 1 \end{pmatrix}$$

has all $$\partial^2 U / \partial X_i^2 > 0$$ but is not positive definite.

But if we mean that if $$X$$ can be any coordinate at all, chosen from any coordinate system we want, then the claim is true: taking a unit vector $$v$$ in the direction of the coordinate $$X$$ (keeping all other coordinates in our coordinate system fixed), then

$$v^T \mathcal{H}(U) v = \frac{\partial^2 U}{\partial X^2},$$

so that if all second derivatives are positive, the Hessian is positive definite and there is a minimum.

Thankfully, Callen's proof works for any coordinate system and any coordinate, so the claimed principle is true in thermodynamics.

• Ok, I guess the question is not nicely formulated. The hessian being positive definite does imply this, but this alone doesn't guarantee a maximum, right? Aug 28, 2019 at 21:57
• I will edit the answer to try to make this more clear, sorry for the confusion Aug 28, 2019 at 21:59
• @Ignacio edited accordingly. Aug 28, 2019 at 22:30
• Hi there! In your last line you write "Callen's proof works for any coordinate system and any coordinate, so the claimed principle". I wanted to confirm why that is so. Callen uses $X$ for arbitrary independent variables, so are you saying that in equation (5.3) we should actually be thinking of the two $X$ with respect to which the derivative is taken as different: i.e. as $X_i$ and $X_j$. And then use that if $0< \frac{\partial^2 U}{\partial X_i \partial X_j}$ for all $i,j$ then we have a minimum?
– EE18
Jun 25, 2023 at 16:11
• Also, and if I am not mistaken, it seems in the derivation that $U$ or $S$ are held constant; thus is it fair to say that we are not saying that "where there is a local $S$ maximum there is a local $U$ minimum", but rather "we have a (local) maximum of the entropy at a given energy in configuration space if and only if we have a (local) minimum of the energy at that given entropy in configuration space." That is, it is a commentary about the equivalence of constrained extrema?
– EE18
Jun 25, 2023 at 16:25