Does $S \ge 0$ follow from Callen's thermodynamic postulates?

Question

In his textbook, Thermodynamics and an Introduction to Thermostatistics, H. B. Callen introduces entropy by the following four postulates:

Postulate I. There exist particular states (called equilibrium states) of simple systems that, macroscopically, are characterized completely by the internal energy $U$, the volume $V$, and the mole numbers $N_1, N_2, \dotsc, N_r$ of the chemical components.

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 following property: The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.

Postulate III. The entropy of a composite system is additive over the constituent subsystems. The entropy is continuous and differentiable and is a monotonically increasing function of the energy.

Postulate IV. The entropy of any system vanishes in the state for which $$(\partial U/\partial S)_{V,N_1,\dotsc,N_r} = 0 \quad \textit{(that is, at the zero of temperature)}$$

From Postulate III follows that $S$ is homogeneous first-order, that $(\partial S/\partial U)_{V,N_1,\dotsc, N_r} > 0$ (hence $T \ge 0$), and that $U$ is a single-valued, continuous, and differentiable function of $S, V, N_1, \dotsc, N_r$. Postulate IV implies that $S$ has a uniquely defined zero. (This does not mean that there is a unique zero-entropy equilibrium state, but that, unlike $U$, one cannot redefine $S$ by addition of a constant.)

Now, in the context of statistical mechanics $S$ is certainly a nonnegative function, but I wonder if $S \ge 0$ also follows directly from the above postulates?

I can see that if one also has that $S$ is a strictly concave function of $U$ then postulates III and IV would imply $S \ge 0$, but I don't see how such concavity would follow from the postulates, so that might be a red herring.

Answer 1

If I'm not mistaken, Callen pulls a fast one and redefines some terms at the end of Chapter 1, so that all equilibrium states now have $d^2 S < 0$ (bolding mine):

By straightforward differentiation, we computer the extrema of the total entropy function, and then, on the basis of the sign of the second derivative, we classify these extrema as minima, maxima, or as horizontal inflections. In an appropriate physical terminology we first find the equilibrium states and we then classify them on the basis of stability. It should be noted that in the adoption of this conventional terminology we augment our previous definition of equilibrium; that which was previously termed equilibrium is now termed stable equilibrium, whereas unstable equilibrium states are newly defined in terms of extrema other than maxima.

In other words, Callen's final definition of "equilibrium states" now includes stability, which implies that $d^2 S < 0$. In other words, the entropy function is concave, as you wanted it to be.

Answer 2

Let me start with a side remark. I think that, in analogy with the existence of a minimum for the energy, Callen's Postulate IV should be stated as

The entropy of any system is a constant in the state for which $$(\partial U/\partial S)_{V,N_1,\dotsc,N_r} = 0 \quad \textit{(that is, at the zero of temperature)}$$.

Then, exactly for the same reasons as for the energy, it is convenient to take zero such a constant as well as the energy minimum.

In any case,

• from Postulate III, the homogeneity of degree 1 of the entropy follows;
• from the homogeneity and Postulate II, the concavity of the entropy as a function of its variables follows. A discussion about this point is in the chapter on the stability of equilibrium.

Therefore, the hypersurface $(\partial U/\partial S)_{V,N_1,\dotsc,N_r} = 0$ must be the border of the convex domain of the entropy function. By the last part of Postulate II, it follows that $S \geq 0$.