2
$\begingroup$

I'm trying to understand the physics of phase transitions, specially at the critical point, but I find myself stuck.

For an hydrostatic system, I studied the stability conditions, that lead to (in the case of Helmholtz free energy) $$ \left(\frac{\partial^{2} F}{\partial T^{2}}\right)_{V} \leq 0, \qquad \left(\frac{\partial^{2} F}{\partial V^{2}}\right)_{T} \geq 0 $$ where we are assuming that the second derivatives are not null. If they are, we have to expand the analysis to the first derivative that is different from zero.

In class, we have used the second condition to proof (not in an actually formal way) that in a P-V diagram, the critical point has to be an inflection point. I leave here the detailed "proof":

Outside the critical point, we have: \begin{equation} dF = -SdT - p dV \quad\rightarrow\quad \left( \frac{\partial F }{\partial V} \right)_{T} = - p \end{equation} At the critical point, we know that the second derivative of any thermodynamic potential has to be zero, so: $$ \left( \frac{\partial^2 F }{\partial V^2} \right)_{T} = 0 \quad\rightarrow\quad \left( \frac{\partial p }{\partial V} \right)_{T} = 0 $$ The third derivative will be as well 0, as the free energy has to be minimum (and not an inflection point). Hence, $$ \left( \frac{\partial^4 F }{\partial V^4} \right)_{T} \geq 0 \quad\rightarrow\quad - \left( \frac{\partial^3 p }{\partial V^3} \right)_{T} \geq 0 $$

and we conclude that, indeed, in the P-V plane the critical point is an inflection point.

Well, everything would be fine except that the teacher didn't explained why the second derivative of any thermodynamic potential has to be zero at the critical point. I have searched all over the internet and have not found any satisfactory answer. I have checked the Callen, but it deduces this idea differently.

I would appreciate if someone could help me.

EDIT: I define the critical point as the end point of a phase equilibrium curve

P.D: So sorry for my bad English :)

$\endgroup$
8
  • $\begingroup$ @Aplateofmomos The end point of a phase equilibrium curve $\endgroup$ May 25, 2022 at 14:14
  • $\begingroup$ physics.stackexchange.com/questions/512544/…, seems so the answer lies in stat mech $\endgroup$
    – Babu
    May 25, 2022 at 14:17
  • $\begingroup$ @Aplateofmomos I've read that question and the asnwers to it before posting my question... Yeah, it seems that the answer is in stat mech, but I would like a more accurate answer as the given in that question doesn't satisfy me... $\endgroup$ May 25, 2022 at 14:27
  • $\begingroup$ Tough. Idk enough stat mech to answer. But if you want to understand what this partial derivative conditions mean in terms of the surfaces of F, then you can check out this answer I wrote $\endgroup$
    – Babu
    May 25, 2022 at 14:28
  • $\begingroup$ @Aplateofmomos Thank you! :) $\endgroup$ May 25, 2022 at 14:34

1 Answer 1

1
$\begingroup$

Let me start by saying that the statement

at the critical point, we know that the second derivative of any thermodynamic potential has to be zero

is not unconditionally true. It is enough to think about the behavior of the second derivative of the Helmholtz free energy with respect to temperature. It is proportional to the constant-volume specific heat, which does not vanish at the critical point.

Providing an answer to this question has been an interesting challenge. I have reached my conclusion only after trying different solution strategies and searching the literature.

The short answer is that we cannot prove the vanishing of the second derivative with respect to the volume of the Helmholtz free energy at the critical point if the definition is just the endpoint of the coexistence region.

Indeed, the relevant thermodynamic potential function of the pressure should be an increasing concave function. For all the temperatures $T$ along the coexistence curve, the corresponding $g(T,p)$ will have a point with different left and right derivatives. There is no discontinuity of the derivative at and above the critical temperature.

Correspondingly to this picture in the pressure domain, the Legendre transformation of $g(T,p)$, say $f(T, V)$, is a convex function of the volume $V$. Above and at the critical temperature, such a function has a continuous second derivative. Below the critical temperature, there is an interval of volumes where the graph is a straight line. The extension of such a straight line (the affine part of the convex function) vanishes at the critical point.

The key point is that such a general picture of the behavior of the function $f(T,V)$, does not imply the vanishing of the second derivative $-\frac{\partial^2{f}}{\partial{V}^2}$ automatically.

Therefore, the vanishing of such a second derivative (i.e., the horizontal inflection point of the critical isotherm) can be obtained only by adding another condition beyond the constraints originating from the principles of Thermodynamics. Support for such a position comes from a recent paper by J.C.Obeso-Jureidini, D.Olascoaga, and V.Romero-Rochín. The authors show an additional hypothesis is needed to prove the divergence of the isothermal compressibility at the critical point.

A better approach to the problem would be to start with a more physically-based definition of the critical point. Such a choice does not imply renouncing mathematical rigor. I would suggest starting from a definition of the critical point as the point in the thermodynamical space where the thermodynamic potential $f(T, V)$ would cease to be a strictly convex function of the volume due to the appearance of a concave intruder which eventually is suppressed by the phase separation restoring the convexity in the form of a finite interval of rectilinear behavior. In a way, such a choice is equivalent to a definition of the critical point in terms of the horizontal inflection point. This way, the emphasis is put on the physical mechanism (the incipient instability of the one-phase region) at the basis of both the critical behavior and the coexistence region.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.