# How to deduce the formula of the Correlation Length on a periodic lattice?

Sometimes in Monte Carlo simulations we need to compute the correlation length, but this is a hard task without a formula. However, for instance, in an periodic cubic lattice of $$L^3$$ spins, some papers present a formula of the correlation length as follows:

$$\xi = \frac{1}{2 \sin(\pi/L)} {\left (\frac{\chi}{F}-1 \right)}^{1/2}$$

where $$\chi$$ is the susceptibility, which is $$\hat{G}(0)$$ (i.e. the Fourier transformed two-point correlation function at $$\vec{k} = (0,0,0)$$), and $$F$$ is $$\hat{G}(\vec{k}_{min})$$, where $$\vec{k}_{min} = (2\pi/L,0,0)$$. My guess is that it comes from the following definition $${\xi}^2 = \frac{\sum_{\vec{r}} r^2 G(\vec{r})}{\sum_{\vec{r}}G(\vec{r})},$$

but I can't get to the formula.

To me this looks like a rearranged version of the Ornstein-Zernike form of the Fourier-transformed correlation function, which reads $$\hat{G}(k) = \frac{\hat{G}(0)}{1+(k\xi)^2}$$ for any sufficiently small $$k$$. Obviously, this form applies to an isotropic system rather than a cubic lattice.
The assumptions behind that equation are discussed in this question and my answer to it; both the question and the answer give further pointers to the literature. Your equation reduces to this one, if we make the assumption that $$k=k_\text{min}$$ is small, so that $$\sin\frac{1}{2}k_\text{min}\approx \frac{1}{2}k_\text{min}$$. (I note that the lattice spacing is taken to be unity, making both $$L$$ and $$k$$ dimensionless).
I guess that the lattice structure of the problem is the reason your equation involves a $$\sin$$ function, while mine does not, but I don't know enough about the context to say for sure.
• ....the appearance of that $\sin(\pi/L)$ is still a mystery for me. – O. Daniel Jun 6 '19 at 1:52