Consider the standard Ising model on $[0,N]^2$ for $N$ large. By that I mean the squarelattice 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?

3$\begingroup$ Start with the partition function, see e.g. p. 4801 (search for the page numbers) of amazon.com/ModernCourseStatisticalPhysics/dp/0471595209. Then do some thermodynamics. $\endgroup$ – Steve Huntsman Jul 24 '11 at 14:11
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.

$\begingroup$ Thanks for expanded explanation. I need to do some reading before I can make sense of this answer. $\endgroup$ – Boris Bukh Jul 25 '11 at 17:36

$\begingroup$ The double integral expression for F looks a bit unpleasant, but it's really just some sort of hypergeometric fuction, i.e. a solution to a linear second order ODE. It just happens not to be expressible in terms of anything more elementary. Feel free to email me if you have questions. $\endgroup$ – Will Orrick Jul 25 '11 at 20:23