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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?
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Nice argument! You should write it up. –  Carl Brannen Aug 27 '11 at 15:30
    
@Ron: Do you mean Landau and Lifshitz? If so, please, cite a page. –  Physicsworks Aug 27 '11 at 17:12
    
Can we find Landau's argument in Landau & Lifshitz? If so, what volume and page? Thx. –  becko Aug 27 '11 at 18:30
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@Carl: I just did. –  Ron Maimon Aug 27 '11 at 19:59
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@Ron: Oh, yes, I've found this paper. "Theory of phase transformations. I" Zh. Eksp. Teor. Fiz. 7 19 (1937); Phys. Z. Sowjetunion 11 26 (1937) –  Physicsworks Aug 27 '11 at 20:18

2 Answers 2

You have mixed up many things. The Landau theorem only states that as soon as the transition from liquid to solid is involved with formation of a 2D or a 3D lattice, there is a cubic invariant inevitably making the transition the first order. Point. Nothing more is stated there.

The liquid-gas transition has nothing to do with these arguments, and Landau did not mention this transition in his work concerning the above theorem. By the way this is a paper in Sov. Phys. JETP Its title can be translated into English approximately like this: "To the theory of phase transitions. II" published in 1937. Have a look. Here II relates not to the order of the transition, but enumerates the papers: it is the second paper in the series.

You write "Second order transitions break symmetries, which can be discrete, like in the Ising model, or continuous, like in the x-y model.". That is not quite true. The symmetry break does not depend upon the order of the transition. The symmetry break may take place by the way of the first order transition as well. It happens by the way of the fiorst order transition much more often, than by the second order one. Ising model and x-y model are only models, not nature itself. I advice you not so much to think about models, but more about nature. Otherwise you start taking properties of models for those of nature, and this is not fruitful.

You write "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."

This is not true. Landau was somebody with an extraordinary imagination. And the thing you incriminate him is not difficult to imagine even for me, though I have much more modest possibilities than Landau had. In fact this is not the point. The point is that Landau realized that the existence of the cubic invariant "kills" the second order. By the way his paper here contains a small inexactness that has been only later understood. It shows that his result here has been at the boundary of a guess, but in general his conclusion about the first order is correct.

You write "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...". This is wrong.

First of all, Landau new about nematics quite well.

Second, the transition from liquid into nematic state is the first, rather than the second order, just for the same reason: there is a cubic invariant. The transition may be soft, and some researchers may not recognize that it is the first order, but that is how it is. You cannot avoid cubic invariant, if the order parameter is the second rank tensor, as it is the case in nematics. Have a look into the book of de Gennes, The Physics of Liquid Crystals (Oxford University Press, London, 1974).

Third, transition between certain liquid-crystalline phases may indeed be second order, but not necessarily.

Fourth, assume for a moment, that you have found some chain of transitions of possibly the second order. And what? For what purpose? What do you think to have proven by that? You did not proven that the Landau theorem is wrong, because this theorem is not what you think. It is in general the case: it is very difficult to find an error in Landau's results and statements, though it is a noble activity.

I think to have answered already to your questions. However, I make it here in a more explicit form: yes there are phases that cannot be related by the second order transition. Generally there are three classes of such phases:

1) Let us start with the example you touched: if the symmetries of the phases has a group-subgroup relation, which allows for a cubic invariant, the transition is first order except, may be, a so-called, isolated Curie point. It is sometimes formulated as follows: cube of the irreducible representation in question should contains the identity representation is the condition to have a cubic invariant. There are many transitions with such a property, not only fluid-crystal and fluid-nematic, but also between solid phases.

2) If the symmetries of the phases exhibit no group-subgroup relation the transition between them is always the first order.

3) If the symmetry before and after the transition is the same, i.e. at the isodstructural transitions the transition is the first. However, there are not too much of such transitions.

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Normally it's first order, but in a Ginzburg model it can be second order, the nematic transition is concievably second order. I am a huge fan of Landau, so you don't need to defend him to me--- he is a giant of physics. I just want to point out this thing, because I conjecture that all phase transitions may be made second order with appropriate multicritical fine tuning, and this would be interesting if true, because it would allow classifying the first order transitions based on their second order point. –  Ron Maimon Nov 7 '12 at 16:04
    
I have not mixed up anything, you are uncharitable in reading what I wrote: 1. I read the collected papers of Landau a decade ago, I know the papers from memory (although foggy, and I didn't read them all). 2. The chain of transitions is right (you said "This is wrong", it's not) They can all be second order in a model of Ginzburg/Mukhamel type (changing sign 2-derivative term and stabilizing 4-derivative term), I can write down an explicit statistical Hamiltonian if you want which starts with I and ends at IV. 3. The "cubic invariant" is the lattice structure, it doesn't preclude 2nd order. –  Ron Maimon Nov 7 '12 at 17:51

Good point, I think, but when we look at your progression, from (ii) -> (iii), you break two unrelated symmetries (lattice rotation, then lattice translation). This is fine, and you would have two distinct second order transitions. But when you try to bring the two critical points together, you correctly notice that this is going to require fine tuning of the parameters. Now, that's not to say such points don't exist. I think they tend to go under the name "multicritical points," and I seem to recall work, possibly by Dagotto, arguing that CMR transitions were governed by some sort of bicriticality (not my expertise, unfortunately.)

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.)

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Yes, I meant a multicritical point, and I agree that the intuition is that it is very difffult to tune to such a point. But Landau said it was impossible, not difficult. I was hoping that somebody could give an example where it actually is impossible, or barring that, give an argument that it is always possible to make a multicritical second-order point linking any two phases. –  Ron Maimon Aug 27 '11 at 19:54
    
I don't know the exact statement Landau made; I think he was smart enough to realize the difference between extremely unlikely and impossible. The charitable view is all I'm pointing out. There is no reason in the whole (modern!) theory of classical phase transitions, just blind chance if at all, that nature would give us two order parameter coefficients that vanish at the same thermodynamic point. –  wsc Aug 27 '11 at 20:26
    
In which case the onus on your hypothesis would be to find a transition that breaks multiple unrelated symmetries at the same thermodynamic point but fails to exhibit latent heat or any other signature of a first order transition. –  wsc Aug 27 '11 at 20:27
    
Landau said flat out impossible, and I think I know exactly what he was thinking--- the spatial symmetry is different, so there is no order parameter field which you can make in space to create a continuous transition. I thought the same thing too. So I don't think I am being uncharitable, he was just wrong. –  Ron Maimon Aug 27 '11 at 21:37
    
That second order transitions that break different symmetries can collide is not a hypothesis, it's just true. It is very easy to do this in theoretical models by fine tuning--- consider a two dimensional lattice Ising model and X-Y model and tune them to the same transition temperature. What I was wondering was whether there are phases which are always separated by first order transitions, where there is no Lagrangian which can have fields smoothly fluctuating between the two phases. –  Ron Maimon Aug 27 '11 at 21:41

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