The water-gas phase transition is said to be similar to the ferromagnetic-paramagnetic phase transition (same set of critical exponents = same universality class). In the former case the order parameter is the difference in the densities, while in the latter it is the magnetization density.

In the magnetic case the $O(N)$ symmetry is broken spontaneously - hence we talk about two distinct phases as suggested by Landau, i.e. the ferromagnetic phase (which has $O(N-1)$ symmetry) and the paramagnetic phase (which has the full $O(N)$ symmetry). Now for the water-gas case, what symmetry is spontaneously broken? Both liquid water and gaseous water have translational and rotational symmetry..

Also, sometimes it is suggested that water and liquid should not be considered as distinct phases, since one can join the two phases by an excursion in the parameter space which does not encounter any singularities in the free energy. If so, then shouldn't we consider the paramagnetic and ferromagnetic phase to be just one phase as well?

How do we reconcile the concept of phases being characterized by symmetry (and their breakings), or being characterized by excursions in the parameter space, since they seem to give contradictory results? Thanks.


2 Answers 2


I'll make an attempt of partial answer here, and perhaps extend the question a bit : I think liquid water-gas water are already phases that spontaneously break symmetry of say your "water Hamiltonian". Since you can go continuously (without any phase transition) from one phase to the other phase (by going around the critical point at high temperature and pressure), these must have the same symmetry.

I think that a very related picture is to consider the ferromagnetic Ising model (which shares the same universality class as the water-gas critical point) but with a longitudinal field h that couples uniformly to z-spins. The Hamiltonian of this spin system is invariant under time-reversal symmetry.

If one has zero field, then we know that at a certain critical temperature ($T_c$), the spin-system (assume it's an infinite system) exhibits a spontaneous symmetry breaking - i.e. the system chooses an order parameter (local z-moment) that is not time reversal symmetrical - from a paramagnetic phase to a ferromagnetic phase (or vice-versa).

Now, if one turns on the longitudinal field ($h$), the Hamiltonian is no longer invariant under time-reversal symmetry, which means that even if you were to start from say the infinite temperature and finite h-field, as you lower your temperature, your system will never exhibit spontaneous symmetry breaking, since the h-field that you are applying already breaks time reversal symmetry. In other words, the positive and negative magnetization phases break time reversal symmetry, but do not break any symmetry of your Hamiltonian with finite h.

Hence, the phase with positive magnetization can be related to the phase with negative magnetization by going continuously around the critical point (as for water-gas !). Moreover, like water-gas transition, there is a line of first order transition that terminates at the critical point : this line for the Ising model is the h=0 line with $T<T_c$. If your are not going along that line, then you can never have a spontaneous symmetry breaking.

Therefore (and this is my point), the phase of your system that you want to look at and that has a different symmetry state must be along that line of first order transition (and is not the gas or water phases). In the case of the Ising model, it's just a ferromagnet breaking time-reversal symmetry of your Hamiltonian (h=0 line). However, for water I do not know how you characterize this phase. I guess a logical question to ask now would be : what does the Hamiltonian for the water looks like... And what are it's degrees of freedom. Then, one can answer your question about what symmetry is in fact broken.

Hope this helps a bit.

  • $\begingroup$ Nice answer, but the last part seems a bit misguided: there will not be any symmetry of the Hamiltonian even along the first order region, since it will be broken by RG-irrelevant perturbations. The main point is that in the presence of an Ising symmetry, the Ising critical point only needs the tuning of a single parameter; in the absence of an Ising symmetry, one requires tuning two parameters (like in the water phase diagram). In the latter case, even when one tunes to criticality, there will still be RG-irrelevant symmetry-breaking terms present. $\endgroup$ Commented Aug 6, 2021 at 16:57

There is no spontaneous symmetry breaking at the water-gas phase transition, because it's a first-order transition and symmetry breaking typically happens at second-order phase transitions. Physicists usually think of a phase as a region of parameter space that's connected by paths that don't cross any phase transitions, so a physicist would indeed say that water and gas are in the same phase.

But you have to specify which paths are allowed. Typically, your Hamiltonian has some symmetry and you only consider paths that respect that symmetry throughout. (So instead of considering the full parameter space, you're only considering the lower-dimensional submanifold of parameter space that respects the Hamiltonian's symmetry.) If you confine your attention to that submanifold, you indeed get distinct phase. But if you allow completely arbitrary paths which don't have to respect the symmetry, than almost every single system that's ever been studied is in the same phase - the topologically trivial phase! (Topological systems are the only exception.) This is because all phase transition surfaces are submanifolds of parameter space with large codimension, so it's easy to find a path that avoids touching them.

So just as the distinction between SPT phases and trivial phases only makes sense if you impose some symmetry constraint on your allowed adiabatic evolution, indeed the distinction between any two (non-topological) phases only makes sense if you impose some symmetry constraint.


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.