I always thought that first-order transitions cannot have symmetry breaking. But the water-to-ice transition seems to break this idea. We do know that it has a latent heat of freezing, but we also know that the rotational symmetry of water is broken as it crystallizes into ice. This would make it a second-order transition, which should have no latent heat. How is this resolved? A possible explanation I thought of is that the freezing of ice is a two-step process: first, the temperature is brought to the critical temperature below, and then some symmetry-breaking field is turned on that causes the first-order transition, and the latent heat observed comes from this. Does this explanation make sense?
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2$\begingroup$ No, freezing water is a first order phase transition, regardless what you think. Any crystallization is going from disordered liquid to ordered crystal. $\endgroup$– Jon CusterCommented Nov 10 at 2:03
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Your first assumption is wrong, you can have a first order transition with symmetry breaking. Perhaps your misleading intuition was guided by the Ising model and the usual quartic Landau theory, where the only way of getting new solutions in the ordered phase is by a pitchfork bifurcation.
You can check that this is not necessarily the case. You could have unstable solutions at a finite distance already existing before the transition and gaining stability suddenly at the transition. A simple example in the context of Landau theory is with a real order parameter $x$ with reflection symmetry $x\to-x$ with the free energy and the transition when $r=0$: $$ f(x) = x^2(r+(x^2-1)^2) $$
Alternatively, you could imagine a saddle-node bifurcation where in the broken phase, new solutions suddenly spawn. A toy model illustrating this would be the mean field $q$-state Potts model with $q>2$ ($q=2$ is the usual Ising model).
Thus for your solid-liquid transition, you do not need to artificially add a continuous transition.
