Here is a problem where I can do the calculation, but I am not understanding the philosophy behind it. It is about Landau theory:

The Landau theory of phase transitions is based on the idea that the free energy of a system can be expanded as a power series of the order parameter. For a second order phase transition, the order parameter develops an expectation value, that evolves continuously from zero, so this power expansion has solid mathematics grounds. For a first order phase transition, however, since there is a jump in the order parameter, the order parameter never gets an infinitesimal value. If so, why is Landau theory still commonly used for the first order phase transition, even if the expansion seems to not be valid at the phase transition?

  • 1
    $\begingroup$ The jump that you observe can be derived by adding a cubic term (among others) to the Landau free energy expression. The continuity condition arises from forcing the Landau expression to remain an even function. I was hoping to add this as a comment but it looks like I need some fake currency for that. $\endgroup$
    – kb56
    Commented Aug 5, 2015 at 12:41
  • 1
    $\begingroup$ Yes, basically you have to assume the existence of a metastable branch of the free energy to "justify" the procedure. Note though that this assumption is simply wrong for short-range models (it can, for example, be proved rigorously that there is an essential singularity in the Ising model, in any dimension, which prevents analytic continuation of the free energy beyond the transition point). $\endgroup$ Commented Aug 5, 2015 at 18:42
  • 1
    $\begingroup$ I have no problem of the continuity of the free energy. what I am confusing is the expansion itself. Though the free energy is continuous, the "small parameter" one use is not small. $\endgroup$
    – user21090
    Commented Aug 5, 2015 at 20:37

1 Answer 1


The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has to be small! It just means that terms we throw away have to be small, which is a different thing.

Let's take an example. Suppose we have this somewhat unusual looking free energy handed to us, which is indeed analytic: $$\boxed{F(\phi,T) = \phi^2 - 2 + e^{-\frac{\phi^4}{T}} + \cosh(\phi^3)}$$ For high temperatures, the minimum of the free energy selects $\phi = 0$. Around $T \approx .3$, there is a first order transition to $\phi \neq 0$. The following two graphs give the intuitive picture (the x-axis is the order parameter, the y-axis the free energy):

example pictures

In Landau theory one usually expands these free energies. For example if we expand it to 8th order, we get $$F(\phi,T) = \phi^2 - \frac{\phi^4}{T} + \frac{\phi^6}{2} + \frac{\phi^8}{2T^2}$$ To this order, the graph for $T = .25$ looks as follows:

enter image description here

So we see that this already gives a good representation of our free energy in the region $-1 \leq \phi \leq 1$. This is because despite $\phi$ not being small, the terms we have thrown away are.

Note that if one is not interested in quantitative details but rather just wants the intuitive picture, then one can note that $F(\phi,T) = \phi^2 - \frac{\phi^4}{T} + \frac{\phi^6}{2}$ already displays the same qualitative behaviour. Moreover to this order it is easy to solve exactly and one obtains $T_c = \frac{1}{\sqrt{2}} \approx .7$ which is not a great quantitative match to the more exact $T_c = .3$, but the same physics is at play.

  • $\begingroup$ How should I be able to tell that the transition depicted in this answer is a first order phase transition? $\endgroup$
    – Jagerber48
    Commented May 15, 2018 at 1:53
  • $\begingroup$ @jgerber The value of the order parameter as a function of temperature, $\phi(T)$, is determined by it minimizing $F(\cdot,T)$. By looking at the graphs, you can hence clearly see that whilst $\phi(T) = 0$ for $T > T_c$, it discontinuously jumps to a finite value as we cross $T_c$. $\endgroup$ Commented May 17, 2018 at 22:21
  • $\begingroup$ @jgerber Note, my previous comment was an answer to your question if you are using the definition that a first order transition has a jump in the order parameter. If you are instead asking how to see that the free energy has a kink as we cross $T_c$, note that below the transition temperature, the value of $F(T) = F(phi(T),T)$ is determined by the local minima of $F(\cdot,T)$ (the one away from the origin). In particular, the rate of change of the value of this minimum as we tune T is finite at $T=T_c$. On the other hand, of course, $dF(T)/dT = 0$ for $T>T_c$ (as then $F(T) = 0$). $\endgroup$ Commented May 17, 2018 at 22:33
  • $\begingroup$ Maybe I am missing something stupid. But why do you say "despite $\phi$ not being small, the terms we have thrown away are". It seems essential that you are interested in the $\phi <1$ region. If you would be interested in a large $\phi$ region clearly your approximation would be trowing away all information. In your example there is no interesting structure at large $\phi$ but I guess there could be if we subtract an appropriately chosen term (maybe something like $\cosh(\phi^n)$ for some $n$ making two growing exponentials compete. $\endgroup$
    – Kvothe
    Commented May 19, 2020 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.