Is every first order phase transition related to a second order phase transition by the tuning of some coupling? In practice is the expectation that the this tuning can usually be done in the lab. (By playing with the temperature, density, magnetic and electric fields and/or particle numbers.)
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2 Answers
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I suppose the answer is no. Consider a transition between the solid-state phase and the gas or liquid phase. Gas and liquid phases are translationally invariant. In a solid crystal, translational invariance is broken due to the presence of lattice. According to the acepted point of view, a continuous transition between phases with different symmetries is impossible. This in particular is a reason for the absence of critical points in solid-liquid phase transitions.
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$\begingroup$ Good point. Any point that is locked in by first order phase transitions cannot be tuned to the second order phase transitions without passing through a first order phase transition. Of course you could just do that, I did not add the criteria that you could not. Thinking of it in these terms I guess my question was whether every phase diagram has a second order phase transition. $\endgroup$– KvotheOct 8, 2020 at 16:51
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3$\begingroup$ I think your statement is confusing. Almost the opposite is in fact true: Second order phase transitions are (always, I think, perhaps not for QCP and/or topological phase transitions) related to spontaneous symmetry breaking. In the Ising model this is clear. You go from a phase with Z_2 symmetry to a phase where the lattice breaks this symmetry. However, if you turn on a magnetic field then this field already explicitly breaks the symmetry and so you can move around the critical point and in fact both phases will be considered to have the same symmetry, namely no Z_2 symmetry. ... $\endgroup$– KvotheOct 10, 2020 at 11:43
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$\begingroup$ For the liquid-vapor transition again you could theoretically write an hamiltonian, I think, where there would be a phase transition between liquid and vapor breaking a Z_2 symmetry. This hamiltonian would have to be constrained to move on the line of the first order phase transition. In practice in the lab you always have access to both the pressure and the temperature (and not just to some weird combination of them restricting you to the first order phase transition line) and so basically you always have an explicit breaking of the Z_2 symmetry. ... $\endgroup$– KvotheOct 10, 2020 at 11:48
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$\begingroup$ Just like in the Ising case using a magnetic field. So the two regions are considered to be in the same phase. A good answer on this question is given in physics.stackexchange.com/a/61421/130040. In conclusion I don't think this is a very good answer. $\endgroup$– KvotheOct 10, 2020 at 11:48
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There is an Ising model I've modified that displays both first and second-order phase transitions, and I do employ tuning to magnetize the cellular automaton.
Tuning in this example is the frustration of the system between its Moore and von Neumann neighborhods. When the neighborhoods are tuned from 4 and 5 cells up to 6, the system enters a ferromagnetic transition.
