I was reading CH3 of Reichl's "A Modern Course in Statistical Physics" on Ginzburg-Landau theory and don't really understand a couple of points he makes. He writes: I don't understand why the first order term in $\eta$ would imply a nonzero value of the order parameter above the transition point, and I don't really get the motivation for the $-f \eta$ term either. If you had the $\eta \ \alpha_1(Y,T)$ term then $\frac{\partial \phi}{\partial \eta} = \alpha_1 + \eta \alpha_2 + ... - f=0$ then putting $\eta=0$ doesn't work so I guess that kind of makes sense, but then you have this $-f$ which would seem to mess things up anyway.

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• Jun 6 '17 at 12:20

If the external field $f$ is nonzero, then the order parameter is always going to be nonzero. So the author is implicitly discussing the $f=0$ case, which is the nontrivial one.

Take for example the mean field theory of the Ising model: for $T\rightarrow T_c$, we have $\eta=m \to 0$, and the free energy can be approximated as follow:

$$\phi(T,m) \simeq \phi_0 (T) +NJz \left[ \left(\frac{T-T_c}{T_c} \right)\frac {m^2} 2+\frac {m^4} {12}\right]$$

where $z=2d$.

Notice the absence of a linear therm. The cubic term is also absent because of the system is symmetric under the transformation $m \rightarrow -m$.

If you impose $\partial \phi / \partial m =0$ and $\partial^2 \phi / \partial m^2 \geq 0$ (minimum conditions), you will find that for $T<T_c$ there are two solutions, while for $T \geq T_c$ there is only one, namely $m=0$.

If a linear term was present, however, you would always find a solution $m \neq 0$ also for $T\geq T_c$, even in the absence of an external field.

This is why there cannot be a linear term in the expansion of $\phi$.