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In the 2D Ising model the value for $\alpha$ is $0$, but I fail to see how we can have this if the specific heat of the system actually has a divergence in the critical temperature. I've seen this leads to a logarithmic divergence of it, but I can't understand that. Wouldn't that mean the near the critical point $c$ behaves like $c\sim (t_c-t)^{-\alpha}=1$, while $c$ is obviously non constant?

What exactly does the logarithmic divergence mean? I see that if $\alpha$ was small, then $(t_c-t)^{-\alpha}=e^{-\alpha\log(t_c-t)}\sim-\log(t_c-t)+o(\alpha^2)$. Is that what the logarithmic divergence means?

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    $\begingroup$ Your interpretation is correct: $\alpha=0$ means log dependence here, because $x^{-\alpha}$ diverges "faster" than $\log x$ for any positive $\alpha$ (i.e. $\lim_{x\rightarrow 0} x^\alpha \log x=0$ for $\alpha>0$. $\endgroup$
    – Meng Cheng
    Commented May 15, 2015 at 17:31
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    $\begingroup$ And is there any deeper reason for it instead of being just because the $\log$ function diverges slower than any power? $\endgroup$
    – MyUserIsThis
    Commented May 15, 2015 at 17:56
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    $\begingroup$ I don't think so. You can just say there is a log divergence, but just to unify with other $\alpha>0$ cases it is convenient to say this is $\alpha=0$. $\endgroup$
    – Meng Cheng
    Commented May 15, 2015 at 17:59
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    $\begingroup$ @MengCheng Another reason why people say $\alpha = 0$ for the 2D Ising model, even though strictly speaking $\alpha$ isn't defined because the heat capacity divergence is only logarithmic, is that $\nu = 1$ for the 2D Ising model, so the Josephson identity $\alpha = 2 - \nu d$ formally gives $\alpha = 0$. $\endgroup$
    – tparker
    Commented Nov 28, 2017 at 8:40

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