Estimating the Free Energy of a Kink In statistical mechanics, people often estimate whether or not a certain feature will occur by estimating the feature's free energy. For example, in the 1D Ising model, they want to estimate the probability of a kink occurring, or in the 2D XY model, the probability of a vortex forming. They generally say something like:

The energy of a kink is $E$, and it can occur in $N$ different places. Therefore the free energy of a kink is $F=E-TS=E-T\ln(N)$. If this is negative then a kink will occur, but if it's positive then a kink will not occur.

How can we formally see that this is a valid way of estimating the probability of a kink? If I were presented with this problem and asked to find the probability of a kink, I would probably do something like:

The energy of a kink is E, and it can occur in N different places. Therefore, the probability of a kink is $\frac{Ne^{-\beta E}}{Z}$ according to the Boltzmann law. If this is much greater than the probability of no kink then a kink will occur, but if it is much less than the probability of no kink then a kink will not occur.

But I don't see the connection between this method and the method given in the books. I would like to start out with very basic statistical mechanics facts, such as (here, $p_i$ is the probability of the $i$th microstate)


*

*$p_i=\frac{e^{\beta E_i}}{Z}$

*$F = -T\ln(Z)$

*$F=\sum_i p_i E_i+T\sum_i p_i \ln(p_i)$ is minimized in equilibrium.
or similar, and then derive the heuristic rule using free energy estimates.
 A: Imagine that we instead focus on each kink site individually; it lives in thermal equilibrium with a bath at temperature $T$ and therefore has two states; one state without a kink with probability $\propto 1$, the other state with a kink of energy $E$ with probability $\propto e^{-E/\tau}.$ Therefore kinks are essentially fermions; the probability for this site to have no kink is $$p_0=\frac{1}{1+e^{-E/\tau}}.$$
For all $N$ of the sites to have no kinks we find a probability $P_0 = p_0^N,$ a special case of what would be, for non-interacting kinks, an overall binomial distribution. Anyway therefore if we wanted a $P_0$ of, say, $1/e$, we would have to find the temperature such that $$\left(\frac{1}{1+e^{-E/\tau}}\right)^N = \frac1e,$$or,$$1+e^{-E/\tau} = e^{1/N}\approx 1 + \frac1N + \frac12\frac1{N^2}+ \dots.$$ Keeping only the first-order term for large $N$ gives $E = \tau\ln N;$ the effect of choosing a different probability goes like $-\ln(\ln(1/P_0)),$ so we can treat it as relatively unimportant.
A: It's a bit gauche to answer twice but it'd be nice to answer a different part of your question: why can the free energy be used here at all?
Suppose some macroscopic variable $x$ changes. In this ensemble we have a small system $\text s$ which can share energy with a much larger "environment" system $\text e$. Whether it does so as a result of this change, depends on whether it has to: we have $\delta E_\text e = -\delta E_\text s = -\frac{dE_\text s}{dx}\delta x$ by energy conservation, where the choice of a total derivative is entirely intentional (that is what it needs to be). 
The total change in entropy due to the change $\delta x$ is given by$$
\delta S = \delta S_\text e + \delta S_\text s \approx \frac1T~\delta E_\text e + \frac{dS_\text s}{dx} \delta x.$$ Assuming the environment temperature does not change as a result of $\delta x$ we can then find $$\delta S = \delta x~\frac{d}{dx}\left(S_\text s - \frac{E_\text s}{T}\right)=-\frac{\delta x}{T}~\frac{dF}{dx}.$$Assuming the environment is not at a negative temperature, the overall system therefore increases entropy whenever it follows any sort of change which decreases the smaller system's corresponding free energy.
What your argument is doing is it is establishing the change in free energy $\Delta F$ due to establishing a kink, and then looking at the sign of this change to determine if the overall entropy increases. This also has an interpretation in terms of an entropic force $T~dS_\text s/dx$ balancing out the applied force $-dE_\text s/dx.$ (These both should probably be interpreted as generalized force a la Lagrangian mechanics.) When these sum to zero the system does not favor or disfavor kinks; when the entropic force is larger than the "anti-kink force" which costs energy to produce kinks, then the system prefers to have them despite their energy cost; when the entropic force is smaller than the anti-kink force, the system prefers to be kink-free.
