As often stated in books, near a phase transition we may express the free energy density as a power series in the order parameter $\phi(\mathbf{r})$. Up to quartic contributions, we have $$f=f_{0}-h\phi(\mathbf{r})+a_{2}\phi(\mathbf{r})^{2}+\frac{1}{2}|\mathbf{\nabla}\phi(\mathbf{r})|^{2}+a_{3}\phi(\mathbf{r})^{3}+a_{4}\phi(\mathbf{r})^{4}$$ where the second term is a coupling to an external field.

Now, I'm a bit confused about a few things:

  1. The terms with coefficients $a_{i}$ make sense to me - if we do a Taylor series we get a polynomial. However, why can't we include a term $a_{1}\phi(\mathbf{r})$?

  2. The external field term sort of makes sense, but this doesn't really stem from a power series does it? It's just something we put in by hand.

  3. Where does the gradient term come from? Is this also put in by hand - I don't see how this could come from a power series... And, why does it have to have coefficient $\frac{1}{2}$? Why can't we use $|\mathbf{\nabla}\phi(\mathbf{r})|$ and higher powers?

If anyone has any input on these points, or can suggest a good source of explanation, that would be useful :)

  • $\begingroup$ The first order term is present: it is $-h$. $\endgroup$ – valerio Jun 2 '17 at 18:40
  • $\begingroup$ @valerio92 But this represents the coupling to an external field, not the first order term in the expansion as such. For example, if there were no field ($h=0$) there would be no first order term... $\endgroup$ – klgklm Jun 2 '17 at 18:49
  1. See my answer at this question. A somehow more abstract explanation can also be found here.

  2. I took a look at K. Huang's book Statistical Mechanics (chap. 17) to be sure about this: yes, we put it by hand, and we assume that the external field $h$ is weak so that only $h$ enters and no higher powers.

  3. This is the most cumbersome point. I have always been quite surprised by that term, to be honest, and I suspect that Landau originally put it by hand using is incredible physical intuition. In Huang's book, I found no justification for it, but one can be found here:

We expect that the Gizburg Landau functional should be of the form $$f = f_0 + \alpha_2 M^2 + \alpha_4 M^4 + ... + \kappa | \nabla M|^2 + ...$$ The first line is familiar from (spatially independent) Landau theory in zero external field that we have previously discussed. The second term gives the functional some spatial dependence. Let us try to justify this form. As above, we want to expand the free energy in powers of $M$. In addition here we want to expand the free energy in increasing powers of derivatives. The justification for this is that we expect spatial changes to be rather long length scales, so increasing number of derivatives will be increasingly small. In more detail our functional needs to be a scalar, so no terms linear in $\nabla M$ can enter . We also expect the ground state should be the case where $M$ is spatially uniform — and this is true so long as $\kappa$ is positive.

One way to understand this $|\nabla M|^2$ term is to see that it is disfavoring any domain walls in the system. If $M$ is pointing in some particular direction, then it should want to point the same direction nearby. (Perhaps this is even more obviously correct in the case of Heisenberg magnets where we can imagine $M$ rotating just a little bit from place to place).

Another justification can be found here, where the author basically says that we include the gradient term to include the effect of fluctuations.

In conclusion, the gradient term takes into account the local spatial variations of the order parameter, and since we assume that fluctuations have a long length scale (we are close to the critical point) we can neglect higher order derivatives. Anyway, if we were to consider other gradient terms, they would all be even powers, because $M$ must be a scalar (see for example here).

As for the coefficient $1/2$, Huang says that it is chosen this way "to fix the scale of $m(x)$" (the order parameter), and I think he means that we arbitrarily fix it and then redefine the order parameter accordingly.

  • $\begingroup$ This is helpful, thanks. However, I'm a little unsure about point 1. We write down the free energy with the intention of calculating the configuration that minimizes it. So how can we be expanding about the minimum? That implies we already know the minimum... Also, that would seem to imply that $f=f_{0}$ is always the free energy minimum, which is not true (nor interesting). Actually I've just read that linear terms are 'total derivatives' and thus only give boundary contributions to $F=\int d^{3}\mathbf{r}f$, which can be dropped. I don't really see how that works exactly though. $\endgroup$ – klgklm Jun 2 '17 at 22:21
  • $\begingroup$ @klgklm You are right again. I was too hasty...see updated answer, even if I am not able to clarify this point completely. $\endgroup$ – valerio Jun 2 '17 at 22:43

1: Usually what we try to solve is something like $ \frac{\delta \mathcal{F}}{\delta \phi} = 0 $, where $\mathcal{F} = \int fdV $. Linear terms in $f$ will only add constant terms to the variational derivative which can be eliminated by a constant shift in the field.

3: If you start from some discrete internal energy on a lattice which have some quadratic nearest neighbor interactions, you can often rewrite the interaction term into something resembling a finite difference term and something else. In the continuum limit you will then get the gradient term in the expansion.
If the ground state is suspected to be uniform the simplest expansion is obviously only including the gradient squared. If however, you want to describe ground states with periodic order you can include higher order derivative terms and flip the sign on $\kappa$, which will take you to the realm of Swift-Hohenberg and Phase Field Crystal models.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.