# Entropy without thermodynamic limit

I am reading K. Huang Statistical Mechanics (2Ed , Wiley, 1987) Chapter 14 Ising Model. The paragraph of title Absence of Spontaneous Magnetization in One Dimension bothers me.

Problem discussed is as follows. One dimensional chain of classical spins each denoted by $s_i$. Each spin can only take values 1 and -1. Energy of configuration is given by $$E(s) = -\epsilon\sum_{i=1}^{N-1} s_i s_{i+1}$$
where $\epsilon$ is a positive constant and $N$ is number of spins in the system.

The argument presented in the book considers the following transition:

• initially all spins are aligned in the same direction $\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow$

the energy is at the absolute minimum, but the entropy is zero.

• final state have a wall $\uparrow\uparrow\uparrow\uparrow\downarrow\downarrow\downarrow\downarrow$

the energy increases by $2\epsilon$, but the entropy increases by $k \log (N-1)$, since we have $N-1$ choices to place the wall.

Clearly, the author uses the microcanonical ensemble, where $S = k \log \Gamma$ with $\Gamma$ denoting the number of configurations which are characterized by the same value of energy ($-(N-1)\epsilon$ for the initial state and $-(N-3)\epsilon$ for the final state).

The thing that bothers me is why the author does not take the thermodynamic limit in the expression for entropy, a procedure which was introduced and justified in the earlier chapters.

In the introductory chapters the entropy was defined as a function which value can be obtained by the following procedure $$s = k\lim_{N \rightarrow \infty} \frac{1}{N} \log(N-1),$$ where $s$ denotes the entropy concentration $s=\frac{S}{N}$ and its equal to zero in the considered case.

Can you please explain to how to obtain the author's value of entropy and preserve the thermodynamic limit?

The thermodynamic limit in this case is sort of trivial. Kerson is assuming that the reader notices that as $N\rightarrow\infty$, the entropy of a single domain wall grows unboudedly, while the the energy of the domain wall remains a (small) constant. Therefore, the free energy $A=E-TS$ is going to be dominated by the $-TS$ entropic term at thermodynamically large $N$, unless $T=0$. Since the free energy change associated with creating a domain wall is negative, the domain wall will form spontaneously; the lowest-energy state (with all the spins aligned) is not the equilibrium state.