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Obviously if you define a first order phase transition as having a discontinuity in the first derivative of the free energy then the answer is no, but I'm asking about if the following situation would still 'effectively' be a first order phase transition in all ways except for the wording of this definition.

The following diagram shows that as you go from a temperature $t$ above the critical temperature $t^*$ to a temperature $t$ below the critical temperature, a previous local minimum of the free energy 'overtakes' the initial global minimum so the system discontinuously jumps to the part of phase space where the new global minimum is - this (I think) is the 'meat' of a first order phase transition. enter image description here

In the old local minimum, the precise free energy value $\mathcal{L}_0$ at the minimum point will generically continuously vary as the temperature is changed (or in this case (trivially) continuously stay at $\eta=0$. When the system jumps into the new global minimum, this part of the graph will also have a minimum $\mathcal{L}_0$ value which continuosly varies as you vary temperature - but crucially this 'gradient of $\mathcal{L}_0$' is different in the new global minimum than in the old one - this is what causes the cusp in the $\mathcal{L}_0/T$ graph (see second graph below).

enter image description here

My question is, could we envisage a situation where accidentally/coincidentally this new global minimum has a gradient $\frac{d\mathcal{L}_0}{dt}$ which is the same as the old global minimum, therefore not causing a cusp in the $\mathcal{L}_0/T$ graph but nonetheless seeming like a first order phase transition (in that there is a discontinuous jump in the order parameter $\eta$ at the critical temperature (which is indicative of a first not second order transition according to the graph below)?

enter image description here

(I understand this would be unbelievably unlikely on a non-contrived graph)

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    $\begingroup$ I don't see why it isn't just a higher order phase transition? What makes the situation you described 1st order? $\endgroup$
    – Samuel
    Commented Oct 21, 2022 at 15:22
  • $\begingroup$ I agree that we could classify it as a higher order phase transition if we use the definition that the first derivatives of the $|matcal{L}_0(T)$ are continuous (so it's not first order). However there is a discontinuous jump in the order paramter in thie situation, whereas I thought we always got continuous order paramter jumps for continuous phase transitions? Let me add another graph to my question $\endgroup$
    – Alex Gower
    Commented Oct 21, 2022 at 15:24
  • $\begingroup$ Actually this order parameter disappearing continuously/discontinuously at the critical temperature is an alternative phase transition classification by Binder $\endgroup$
    – Alex Gower
    Commented Oct 21, 2022 at 15:36
  • $\begingroup$ But just thinking intuitively (might be wrong): if you have a discontinuously vanishing order parameter when going through ciritical temperature, doesn't that feel like having to invest energy into the system to create disorder first, before being able to raise the temperature further over the critical point, i.e. latent heat i.e. discontinuous first derivative? So I guess I am now asking the same question that you asked in the first place xD Can you have a discontinuously vanishing order parameter with a continuous first derivative of the free energy? $\endgroup$
    – Samuel
    Commented Oct 22, 2022 at 10:52
  • $\begingroup$ I agree it would have to be a pretty contrived situation but from my arguments in the OP I think it could be possible? $\endgroup$
    – Alex Gower
    Commented Oct 22, 2022 at 15:18

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I am not 100% sure that I have understood your question, but I believe that the Thouless effect occurring in the one-dimensional Ising model with interactions decaying as $1/r^2$ is something similar to what you have in mind.

Let me provide more detail. Let us consider the Ising model on $\mathbb{Z}$ with formal Hamiltonian $$ \mathcal{H} = -\sum_{i,j} \frac{\sigma_i\sigma_j}{|j-i|^2}. $$ The following scenario was first suggested by Thouless and is now known as the Thouless effect:

  1. there is a nonzero spontaneous magnetization at sufficiently low temperatures;
  2. the spontaneous magnetization exhibits a discontinuous behavior at the transition: it vanishes for all $\beta<\beta_{\rm c}$, but is strictly positive for all $\beta\geq\beta_{\rm c}$;
  3. the free energy density is infinitely differentiable in $\beta$ at $\beta_{\rm c}$ (in particular, the energy density is continuous at $\beta_{\rm c}$).

As far as I know, only 1. and 2. have been established rigorously (in this paper and this paper, respectively). That a pathological behavior must happen at $\beta_{\rm c}$ is also known rigorously (see here), but that precisely the behavior in 3. happens remains open, as far as I know.

It is also known that 1. and 2. together with the validity of correlation inequalities in this model imply the divergence of the susceptibility at $\beta_{\rm c}$ (see the discussion here, for instance). The behavior is thus dramatically different from what is seen in standard first-order transitions. Since such transitions exhibit characteristics of both first and second-order phase transitions, some people call them of mixed order. (This might help you find more on this topic, if you're interested.)

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  • $\begingroup$ Thank you, I will look into this more! One quick question - how is it that 1. and 2. imply the presence of unbounded fluctuations at $\beta_c\$ - aren't these properties present in all 'typical' first order phase transitions so why would them being present here make it different? $\endgroup$
    – Alex Gower
    Commented Oct 23, 2022 at 13:08
  • $\begingroup$ It would make more sense to me if it was 3. that implied the presence of unbounded fluctations? $\endgroup$
    – Alex Gower
    Commented Oct 23, 2022 at 13:09
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    $\begingroup$ @AlexGower You're right, this was poorly written. What one can show is the divergence of the magnetic susceptibility at $\beta_c$ as $h\downarrow 0$. This is a consequence of 1, 2 and correlation inequalities that are valid for the Ising model, but not for general systems. This is discussed in more detail on page 862 of the paper I link to (just after equation (2) therein). $\endgroup$ Commented Oct 23, 2022 at 13:19

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