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this question comes from an exercise of Sethna's book "Statistical Mechanics: Entropy, Order Parameters and Complexity". it is in page 282 question 12.2

In 3d Ising model, the spin correlation function is measured to be in the form $$C(r,T) = {r^{ - 1.026}}C\left( {r{{\left( {T - {T_c}} \right)}^{0.65}}} \right)$$ where $C(r,T)$ is found to be roughly $exp(-x)$

The question is what is the critical exponent $\nu$ in this case.

I found this question pretty easy as we consider $$C(\vec r,T) \sim {r^{ - \tau }}{e^{ - r/\xi }}$$ and $$\xi(T)\sim |T-T_c|^{-\nu}$$ we will get $$\xi(T)\sim |T-T_c|^{-0.65}$$ and thus $\nu=0.65$.

However, I found a solution manual of Sethna's book, it gives $\nu=0.59$. It really puzzles me where is this 0.59 comes from. Did I miss something?

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    $\begingroup$ Does your copy of the solution manual also say "This solution is to an older version of the exercise. It may not correspond to the current version of the text."? $\endgroup$
    – Seth Whitsitt
    Oct 10, 2020 at 19:31
  • $\begingroup$ @SethWhitsitt Thanks for your kind reply! Yes there is. I can only find the 2006 version of the book. in the webpage of Sethna, i vaguely remember he mentioned there is a 2004 version. But i can't find that one. I think there might be some differences between the old and new one. After googling some references, i found that my word claims this critical exponent to be about 0.63 which is very clear to 0.65 $\endgroup$
    – FaDA
    Oct 11, 2020 at 12:40
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    $\begingroup$ Yes, I think your reasoning that one should have $\nu = 0.65$ from the given data is correct, and Sethna's solution manual is working with a previous edition. The exact value of $\nu$ is not known, but it has been determined to be $\nu \approx 0.629971(4)$ numerically. $\endgroup$
    – Seth Whitsitt
    Oct 11, 2020 at 19:58

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