[Physics] Is this Landau’s other critical phenomena mistake

statistical mechanicssymmetrythermodynamics

There was an old argument by Landau that while the liquid gas transition can have a critical point, the solid-liquid transition cannot. This argument says that the solid breaks translational symmetry, and it is impossible to do this in a second order transition.

But this argument is subtly false. Second order transitions break symmetries, which can be discrete, like in the Ising model, or continuous, like in the x-y model. The reason Landau said it is because it is hard to imagine breaking all the translational and rotational symmetries all at once to make a second order liquid-solid point.

But nowadays we know about nematics, and we can imagine the following chain of second-order transitions:

fluid (I)-> fluid with broken x-y-z rotational symmetry with a z-directional order (II)
-> fluid broken translational symmetry in the same direction
-> broken x-y direction rotational symmetry in the y-direction
-> broken y- direction translational symmetry
-> broken x-direction translational symmetry

Each of these transitions can be second order, and together, they can make a solid from a fluid. the question is, how badly does Landau's argument fail.

• Are there any two phases which cannot be linked by a second order phase transition?
• Are there always parameters (perhaps impossible to vary in a physical system) which will allow the second order points to be reached?
• Is it possible to make the second order transitions collide by varying other parameters, to bring them to one critical point (in the example, a critical point between fluid and solid).
• Do these critical points exist in any system?

The point though, and I think the point Landau is making, is that that sort of fine tuning is unlikely to be relevant for real phase transitions, with only a handful of knobs (That is, you write down the action density and it's something like $a \phi^2+b \psi^2+$interactions etc. but for reasons beyond our control, $a(g)$ and $b(g)$ must both go to zero at the same value of $g$... On the other hand, some folks believe there are "deconfined" quantum critical points, where no fine-tuning is necessary to drive two unrelated order parameters to zero at the same "tuning" parameter point.)