How is the free energy of Kosterlitz-Thouless transition smooth yet non-analytic?

Question

Here is an answer by @tparker which makes the following remark

"... a Kosterlitz-Thouless transition, at which the free energy density is smooth but non-analytic..."

The expression for the Helmholtz free energy for the KT transition is $$F=E-TS=(\pi J-2k_BT)\ln \Big(\frac{R}{a}\Big)\tag{1}$$ where $J$ is a parameter that depends upon the system in which the vortex is located, $R$ is the system size, and $a$ is the radius of the vortex core. But $F(T)$ seems to be both smooth and analytic as a function of $T$. What's the caveat that I failed to catch?

Note I must admit that I am not well-familiar with KT transition. Got interested in 'crossovers' and that led me to the discussions in the post linked above.

Answer 1

First, the expression given in the OP is not the expression for the actual free energy, only what comes out of the naive heuristic energy/entropy argument.

In reality, renormalization group computations lead to the following predictions: first, the correlation length should blow up at the transition as $$ \xi \simeq A\exp\bigl( B/\sqrt{t} \bigr) $$ for $t>0$ ($\xi$ is infinite for $t\leq 0$), where $t=(T-T_{\rm BKT})/T_{\rm BKT}$ is the reduced temperature. Observe how this is dramatically faster than the more common power-law divergence of the correlation length at a usual critical point.

Second, the singular part of the free energy should satisfy $f_{\rm sing} \sim \xi^{-2}$, that is, $$ f_{\rm sing} \simeq C \exp\bigl( -2B/\sqrt{t} \bigr) $$ for $t>0$ small.

Note that the function $$ t\mapsto \begin{cases} \exp\bigl( -2B/\sqrt{t} \bigr) & \text{for }t>0\\ 0 & \text{for }t\leq 0 \end{cases} $$ is infinitely differentiable but not analytic at $t=0$, since one does not recover the original function by summing its Taylor series. This is what is meant by "smooth yet not analytic" in this context.

I am not a specialist, so I won't go into more detail here. There are no mathematically rigorous proofs of the above claims in the XY model (even the proof of the existence of the Kosterlitz-Thouless phase transition requires rather sophisticated mathematical arguments). There are, however, other simpler examples of phase transitions in which this type of "smooth but non analytic behavior" are found and for which rigorous results are available.

If you want to read more about these issues in the XY model, you can look at Kosterlitz's original paper (see also his recent review). You can also read about that in several textbooks, for instance this one (Itzykson and Drouffe) and this one (Kardar).

Answer 2

The typical heuristic argument here is to look at the case for an infinite system, i.e. at the limit $R\rightarrow \infty$.

For $T<\pi J/(2 k_{\mathrm{B}})$, the first term ($E$) dominates and the free energy will diverge $F\rightarrow \color{red}{+} \infty$. It can only lower $F$ by having the lowest $E$ and hence no vortices.

For $T>\pi J/(2 k_{\mathrm{B}})$, the entropy wins and the free energy will diverge $F\rightarrow \color{red}{-}\infty$, i.e. you want to be as “messed up as possible” so you create vortices.

To find the critical temperatures between these two régimes, set $F=0$ and get an expression for $T_{\mathrm{c}}$.

It should be noted that this is only possible because both the vortex energy and the entropy have the same logarithmic scaling. With a different energy scaling for the vortex, you could have vortices $\forall T$ or $\not \exists T$. So no phase transition.