# Ising model observables

Is there a formula or equation relating $\langle E\rangle$ and $\langle M\rangle$ (average spin per site) and $\langle E^2\rangle$ to temperature $T$ for the square lattice Ising model at zero magnetic field? I have done some simulations and know what the graphs are supposed to look like approximately, but is there some neat expression for them?

And is there known what the exact functions are for finite grid sizes?

• At the end of Square-lattice_Ising_model, you have the value of free energy, internal energy, and magnetization, in the limit $N \rightarrow +\infty$ Commented Jun 14, 2013 at 8:26
• The definitions of internal energy, and free energy, are general thermodynamic definitions, from the partition function. I suppose, in your case, that your $<E>$ corresponds to the internal energy. Internal energy is the total energy contained by a thermodynamic system.(Gibbs) free energy measure the work obtainable from a thermodynamic system. Commented Jun 14, 2013 at 8:37
• What is the definition of free energy, and internal energy for the square lattice? My Energy is the product of its spin and the sum of its four neighboring spins. Commented Jun 14, 2013 at 8:44
• The magnetization at $T>Tc$ should be zero. You have to go to the beginning of the Wikipedia article - Definition of the model - to see the used variables $K,L,J,J*$. Commented Jun 14, 2013 at 8:48

Onsager computed the partition function of the 2D periodic square lattice (toroidal boundaries) Ising model. It is arguably one of the most elegant proof of modern statistical mechanics.

The original paper is available on the APS website below: (you will need institutional access)

L. Onsager, "Crystal Statistics. I. A Two-Dimensional Model with an Order-Disorder Transition", Phys. Rev. (65), 1944. Link

Although I found it laying around on some university server: http://www.colorado.edu/physics/phys7230/phys7230_sp08/Onsager1944.pdf

In essence, he obtains the partition function $$Z(\beta,N,H=0)=(2\cosh(2\beta J) e^I)^N$$ with $$I=\frac{1}{2\pi}\int_0^\pi d\phi\ln\left(\frac{1}{2}\left[1+(1-\kappa^2\sin^2\phi)^{1/2}\right]\right)$$ where $$\kappa=\frac{2\sinh(2\beta J)}{\cosh^2(2\beta J)}$$

Traditionnal canonical ensemble techniques can be applied from there. Note that the free energy associated to $$Z(\beta,N,0)$$ is non analytic and that a phase transition arises when $$\kappa=1$$ . This correctly predicts that $$T_c=\frac{2J}{k\ln(1+\sqrt{2})}$$

Yes, these equations exist and can be derived from the partition function in JGab's answer.

The internal energy per spin is: $$u(\beta) = - \frac{\partial}{\partial \beta} \left( \ln(2) + \frac{1}{8 \pi^2} \int_{0}^{2\pi} \mathrm{d} q_1 \int_{0}^{2\pi} \mathrm{d} q_2 \\\ln \left[ \big( 1 - \sinh(2 \beta J) \big)^2 + \sinh(2 \beta J) \left( 2 - \cos q_1 - \cos q_2 \right) \right] \right)$$

The spontaneous magnetization per spin is: $$m_s(T) = \begin{cases} \left( 1 - \sinh^{-4} (2 \beta J) \right)^{\frac{1}{8}} & \text{for } T < T_c \\ \hskip 1.5cm 0 & \text{for } T \geq T_c \end{cases}$$ Note that this is strictly speaking not the magnetization $\langle m \rangle$ you asked for, but its limit $$m_s = \lim_{B \rightarrow 0^+} \lim_{N \rightarrow \infty} \langle m \rangle$$ where the two limits do not commute. See this question for a discussion of this important point.

• To complete this excellent answer, note that $\langle m\rangle = 0$ for $B=0$, due to sign symmetry. So one needs to introduce an external field $B$ to get a non-zero answer, which explains why the limit $B\rightarrow 0^+$ is taken here,
– a06e
Commented May 29, 2022 at 17:36