TITLE: Entropy of the Ising model
QUESTION [8 upvotes]: Consider the standard Ising model on $[0,N]^2$ for $N$ large. By that I mean the square-lattice Ising model without external field, inside an $N$-by-$N$ square. What is its entropy for $N$ large? It must behave asymptotically as $c(\beta)N^2$ for some constant $c(\beta)$ depending on the inverse temperature $\beta$. What is $c(\beta)$? Has it been computed?

REPLY [6 votes]: To expand on Steve Huntsman's comment, the entropy follows from Onsager's result for the free energy per site, $F=$
$$
-\beta^{-1}\left[\ln 2+ \frac{1}{2}\frac{1}{(2\pi)^2}\int_0^{2\pi}d\theta_1\int_0^{2\pi}d\theta_2 \ln(\cosh2\beta E_1\cosh2\beta E_2
-\sinh2\beta E_1\cos\theta_1-\sinh2\beta E_2\cos\theta_2)\right],
$$
and the thermodynamic relation,
$$
S=-\frac{\partial F}{\partial T},
$$
for the entropy per site.  Here $\beta=1/(k_BT)$ and $E_1$ and $E_2$ are the horizontal and vertical interaction strengths.  If you set both interaction strengths equal to 1 and use units where Boltzmann's constant equals 1, then the critical temperature is $2/\ln(\sqrt2 + 1)\approx2.269$.  If you plot $S$, you should find that it interpolates between 0 at low temperature and $\ln2$ at high temperature, as expected.  At the critical temperature, the graph has infinite slope.