Critical exponent $\nu$ of 3d Ising model

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?

• 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."? Oct 10, 2020 at 19:31
• @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
• 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. Oct 11, 2020 at 19:58