I have now come across two different definitions for the spinodal curve which, together with the coexistence line, encloses two metastable phase regions. The first definition is from Tong on page 139. In considering the van der Waal's liquid, one throws away the unphysical phase, which has negative compressibility. However, for every phase transition except the critical one, there is always a part of the gaseous phase and the liquid phase which is jumped over. This is the inner red curve in the figure below. Tong identifies this with the metastable superheated liquid and super cooled gas.Inner red curve is the spinodal curve according to Tong

This definition makes sense to me, and it is also rather general. Another definition I have come across is that the spinodal region is where the second derivative of the thermodynamic potential one is considering has a negative second derivative. If one uses the Helmholtz free energy, then this overlaps with the first definition for the van der Waal's gas. However, the correct thermodynamic potential for the van der Waal's gas is the Gibbs free energy, which selects out the state. What I meant by "correct" thermodynamic potential is, given the standard assumptions on the system (for van der Waal's conducting and flexible walls), the thermodynamic potential which selects out the thermodynamically favorable state.

In that case, the ``metastable region'' would now encompass part of the unphysical phase as well.

My question is, which of these two definitions are correct from a mathematical and experimental standpoint. I am guessing it is Tong's definition experimentally since we never see states with negative compressibility. Mathematically, I have read the argument that one should think of stable and unstable fixed points on a thermodynamic potential (as a function of order parameter). Then close to the unstable fixed point, the second derivative will be negative, and vice versa for the stable fixed point. So to be explicit, take the Gibbs free energy around a local minimum or maximum and expand it:

$$G(V+\delta V) = G(V) + \delta V G'(V) + \frac{(\delta V)^2}{2} G^{(2)}(V) + \mathcal{O}((\delta V)^3)$$ Since we are at an extremum, the first derivative is zero. Hence the Gibbs free energy will be lowered for at a maximum $G^{(2)}(V)<0$, hence unstable, and increased at a minimum $G^{(2)}(V)>0$, hence stable. Again note that this definition perfectly overlaps with Tong's definition if we use rather the Helmholtz free energy $A$. However, $A$, does not select out the favorable state.

  • $\begingroup$ It should probably be mentioned that one can rigorously prove that systems with short-range interactions typically have an essential singularity at the phase transition. In particular, this implies that the description of metastable states via an analytic extension of the stable branches of the pressure is wrong in general for non-mean-field models (the van der Waals model being itself mean-field). $\endgroup$ Apr 7, 2023 at 19:34
  • $\begingroup$ @YvanVelenik I do not think the argument is very compelling. The presence of an essential singularity does not forbid the possibility of metastable states avoiding the phase transition. Superheated and supercooled states do exist in the laboratory, and they look like analytic extensions of the stable phases. Therefore, it is not obvious what should be wrong with experimental facts. $\endgroup$ Apr 7, 2023 at 22:14
  • $\begingroup$ @YvanVelenik What do you mean by "essential singularity at the phase transition"? I am assuming you are not talking about the singular behavior of the thermodynamic potential, since that would be there by definition for every phase transition. $\endgroup$ Apr 7, 2023 at 22:14
  • $\begingroup$ I mean that the pressure has an essential singularity at the phase transition. An essential singularity is a particular type of singularity that precludes any possibility of analytic continuation through this point. In particular, there does not exist an analytic function coinciding with the pressure along the gas branch and extending beyond the phase transition point (in contrast to what happens in the van der Waals model). $\endgroup$ Apr 8, 2023 at 8:15
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 Sure, metastable states exist. What this proves is that they should not be described in this way: metastability is really a dynamical phenomenon. In fact, in simple models (say, the planar Ising model), the dynamics of metastability has been studied in detail and yields an infinite family of $C^\infty$ continuations of the pressure that all have physical significance. See, for instance, this paper or this account for physicists. $\endgroup$ Apr 8, 2023 at 8:23

1 Answer 1


The two definitions mentioned in this question are not different. Isothermal compressibility is defined as $$ \chi_T=-\frac{1}{V}\left. \frac{\partial V}{\partial P} \right|_T=\frac{1}{V} \left( \left. \frac{\partial^2 F}{\partial V^2}\right|_T \right) ^{-1}. $$ Therefore, negative compressibility would imply a Helmholtz free energy concave with respect to the volume.

The observation about the correct free energy signals a misunderstanding about thermodynamic potentials. There is no right or wrong thermodynamic potential, but only thermodynamic potentials that are more useful in some conditions than others. Usefulness is not a synonym for correctness.

So, it is true that the most simple experimental conditions are in terms of pressure and temperature. But this fact does not forbid describing a system in terms of temperature and volume as independent variables. Indeed, this is the most helpful representation if we want to represent the coexistence of phases with different densities. Thermodynamics and the theory of Legendre transform to ensure the equivalence with alternative descriptions in terms of pressure and temperature.

A final word of caution about van der Waals' metastable region. It is true that the real systems exhibit the two metastable regions around the spinodal decomposition boundary. However, there is no reason that the extent of such metastable regions would be quantitatively described by van der Waals theory.

  • $\begingroup$ The example you mention is exactly the case where I said that the two definitions overlap. As for what I mean by "correct" thermodynamic potential, I have clarified this, as well as the second definition. The two definitions I mention do not overlap in general. The link I gave explicitly talks about the second derivative of the Gibbs free energy being negative. However, what I can perhaps assume from your answer, is that you feel like it should always be the Helmholtz free energy that should determine the metastable region? $\endgroup$ Apr 8, 2023 at 10:09

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