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Consider the Ising model with nearest neighbours interactions on a rectangular lattice $L\times M$.

If $L=M$ (2-dimensional square lattice), it is known (e.g. by Peierls argument or Onsager explicit solution) that the model exhibits a phase transition when $L=M\to\infty$.

If instead we fix $L=1$ (1-dimensional line) and let $M\to\infty$, the model does not exhibit a phase transition.

My question is: which relation among the side lenghts $L,M$ guarantes the presence/absence of a phase transition? For example what about the case $L=\log M$ ?

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  • $\begingroup$ Due to an argument by Landau you can not have a phase transition in 1D systems. users-phys.au.dk/fogedby/statphysII/no-PT-in-1D.pdf I think that the limiting procedure for the size of both dimensions to go to infinity will not change anything in Peierls argument. $\endgroup$
    – Bubble
    Sep 29 '14 at 21:47
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    $\begingroup$ Thank you. I looked again the Peierls argument and I think you are right: it suffices that $L,M\to\infty$, no matter their relation $\endgroup$
    – tituf
    Sep 30 '14 at 3:21
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Any increasing sequence $(\Lambda_n)_{n\geq 1}$ of finite subsets of $\mathbb{Z}^d$, $d\geq 2$, such that $\bigcup_{n\geq 1} \Lambda_n =\mathbb{Z}^d$ will do. All sequences $(\mu_{\Lambda_n}^+)_{n\geq 1}$ of finite-volume Gibbs measures in $\Lambda_n$ with $+$-boundary condition converge to the same infinite-volume Gibbs measure $\mu^+$, under which there is spontaneous magnetization as soon as the inverse temperature $\beta$ is large enough.

This can be proved easily using the FKG inequality (see, for example, the chapter on the Ising model here).

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