When can I "change variables" in thermodynamic configuration space?

Question

Edit: This question has been asked here, but has not been answered directly as far as I can tell.

In Chapter 5 of Callen, a great deal of the motivation for introducing the Legendre transform (and thus, thermodynamic potentials) is so that intensive variables like $T,P,\mu$ can, in contexts where it's useful, be used as the independent variables. It was emphasized that the Legendre transform is a way of describing these systems in terms of intensive variables, but without losing information as would be the case if we simply inverted equations of state and replaced extensive variables by some intensive variables.

For example, it was discussed that, in writing $U = U(S,V,N)$ instead as (using the physicist's standard abuse of notation for different functions) $U = U(T,P,N)$ by elimination with $T = T(S,V,N)$ and $P = P(S,V,N)$, one loses information (turning the fundamental equation into a PDE). The Legendre transform (to $G = G(T,P,N)$) in this case lets us transform without losing information.

However, it seems that sometimes Callen nevertheless does the elimination that loses information, and I am trying to understand why this is valid. For example, in his discussion of adiabatic compression in Chapter 7.4 Callen writes (my commentary in bold)

By differentiation, we can find the two equations of state $T = T(S, V, N)$ and $P = P(S, V, N)$ [No problems here]. By knowing the initial temperature and pressure, we can thereby find the initial volume and entropy [No problems here either; we are simply solving two equations in two unknowns ($V_i$, S_i)]. Elimination of $V$ between the two equations of state gives the temperature as a function of $S$, $P$, and $N$ [Are we not losing information in this step?].

As he says in Chapter 5, does not the rewriting in terms of derivatives (intensive parameters) make our fundamental equation into a PDE? As Callen says, for $P = dY/dX$,

From the analytical point of view the relation $Y = Y(P)$ is a first-order differential equation, and its integration gives $Y = Y(X)$ only to within an undetermined integration constant.

Does the resolution to this problem perhaps have something to do with it being acceptable to (sometimes?) "locally" invert, as has been done above? What are the rules here?

Answer 1

I think the main problem in understanding what Callen did on those pages depends on transforming a possibility into an obligation.

Indeed, if we want to change thermodynamic variables keeping at the same time the whole information contained in a fundamental thermodynamic function (like entropy, Massieu functions, or thermodynamic potentials), we must use Legendre transforms. However, in some cases, we do not need to have a reversible transform. In particular, if we do not lose the memory of the original fundamental equation, it is perfectly possible to perform any change of variables (provided it is mathematically well-defined).

For example, if we want to go from a description of a thermodynamic system in terms of the entropy and its natural variables ($S(U,V,N)$) to an equivalent description in terms of $T,V,N$, we have to introduce the Massieu potential $$ \Phi (T,V,N)= S-\frac1TU. $$ Even if we forget the original $S(U,V,N)$, the new function allows us to recover it.

However, there are situations where we are not interested in a new, equivalent description. For example, we may want to express the thermodynamic state $(U,V,N)$ in terms of other variables, say $(T,V,N)$. In such a case, we need to invert the function $$ \frac1T =\left.\frac{\partial S}{\partial U}\right|_{V,N}(U,V,N) $$ and we can give the entropy as a function of $(T,V,N)$.

Without the original fundamental equation, we cannot get it back without adding the missing information under the form of some arbitrary function. But if our goal is to assign an entropy to the state with a given value of $(T,V,N)$ that's enough.