Emergence of symmetry in phase transition Goldenfeld's book on phase transitions contains the following claim (in the context if discussing the Landau symmetry principle): "Although a symmetry is either present or absent, it can emerge either continuously or discontinuously as the coupling constants are changed.
Could somebody clarify this statement? It seems contradictory, and there is no discussion in the book.
 A: He is simply saying that the order parameter can be either continuous or discontinuous at the transition (continuous vs first-order phase transition).
As a simple example, consider the two-dimensional nearest-neighbor Potts model with $q$ colors: for each $i\in\mathbb{Z}^2$, $\sigma_i\in\{1,\dots,q\}$ and the Hamiltonian is $H=-\sum_{\langle i,j\rangle} \delta_{\sigma_i,\sigma_j}$.
This model undergoes a phase transition at a finite temperature $T_c(q)$:


*

*When $T>T_c(q)$, there is a unique Gibbs state, invariant under permutation of the colors. In particular, the fraction of vertices of each of the $q$ colors are equal.


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*

*When $T<T_c(q)$, there are $q$ distinct (extremal) Gibbs states $\mu^1,\dots,\mu^q$. In the state $\mu^k$, the fraction of vertices of color $k$ is strictly larger than the fraction of vertices of any of the other $q-1$ colors: the symmetry under permutation of the $q$ colors is broken.


$\hskip2in$
So, when $T>T_c(q)$, the symmetry under permutation of the $q$ colors is present. When $T<T_c(q)$, it is absent. However, the way things occur at $T_c(q)$ depends dramatically on the number of colors. To quantify that, it is useful to introduce a quantity to detect the presence/absence of the symmetry, that is, an order parameter. An natural choice is
$$
m = \sum_{k=1}^q e^{2ik\pi/q} \rho_k
$$
where $\rho_k$ is the density of vertices with color $k$. In terms of $m$, the behavior discussed above can be rephrased as:


*

*$m(T,q) = 0$ for all $T>T_c(q)$

*$m(T,q) \neq 0$ for all $T<T_c(q)$
The behavior at $T_c(q)$ can then be shown to be the following:


*

*When $q\geq 5$, at $T_c(q)$ there are $q+1$ (extremal) Gibbs states: the $q$ ordered low-temperature states coexist with the disordered high-temperature state. The order parameter $m$ is discontinuous at $T_c(q)$: the transition is of first order.

*When $2\leq q\leq 4$, there is a unique, disordered, Gibbs state at $T_c(q)$. This state is still symmetric under permutation of the colors. The order parameter $m$ is still equal to $0$ at $T_c(q)$ and depends continuously on $T$: the transition is continuous. As $T$ increases toward $T_c(q)$, the difference between the $q$ phases becomes smaller and smaller and the $q$ phases become indistinguishable at $T_c(q)$.


So, in the first case, the symmetry appears discontinuously, while in the second case it "appears continuously" (in the sense explained above).
A: He probably means that order parameter may or may not present a discontinuity.
For example, in the Ising model for the paramagnetic to ferromagnetic phase transition, the order parameter is the magnetisation $M$. I will consider spins that can either be up or down, no 3D orientation. When $M=\pm1$, all the spins are aligned in the same together, so there is a definite symmetry with respect to the fully random case $M=0$.
As you decrease the temperature $T$ to and below the Curie temperature $T_c$, the curve depends on the value of the external magnetic field $H$. 
In the complete absence of a field $H = 0$, then the magnetisation curve picks up from $0$ to a finite value discontinuously only at $T=T_c$.  But if there is a background $H\neq 0$, this can "help" the symmetry manifestation start sooner, so you have a continuous function.  At $T=0$ and at $T\rightarrow \infty$, however, the two curves agree: so they agree on the symmetry being present or absent.
Depending on how rigourous your definition of "phase transition" is, only the $H=0$ case would qualify as such, because it displays a discontinuity. 

