In this question it is said that:

The best way to numerically work with continuous phase transitions is to study observables that have a vanishing length dimension (or mass dimension in the language of QFT)

I don't understand what that is (I don't know QFT). What is the length dimension in this context and why does $\xi/L$ has a vanishing length dimension in the Ising model?

  • 2
    $\begingroup$ Do you know what dimensionless variables are? Because I think that's what you're looking for. $\endgroup$ – Danu May 15 '15 at 18:58
  • $\begingroup$ I assume $\xi$ is the correlation length. So $\xi/L$ is dimensionless. It is generally good to work with dimensionless quantity, not just for the study of phase transitions. $\endgroup$ – Meng Cheng May 15 '15 at 18:58
  • $\begingroup$ Actually, $\xi$ diverges at the phase transition (and thus renormalization group methods work well around phase transitions, because the systems gets scale invariant). $\endgroup$ – Sebastian Riese May 15 '15 at 19:26
  • $\begingroup$ @Danu Well, I know what a dimesionless variable is. I just don't know how that applies. If a quantity is dimensionless, then in a scale invariant system, it should not depend on size. Is that it? Why? And if so, should any dimensionless quantity I could build behave that way? For example, if the susceptibility goes like $\chi\sim L^{\gamma/\nu}$, then $\chi/L^{\gamma/\nu}$ behaves that way although it has dimensions. $\endgroup$ – MyUserIsThis May 15 '15 at 23:23
  • $\begingroup$ @MengCheng I think it's the word length that confused me, sorry. For example the Binder cumulant as defined in the previous question ($\langle m^4\rangle/\langle m^2\rangle^2$) is dimensionless, why should it be better that $\xi/L$ to find the critical point? $\endgroup$ – MyUserIsThis May 15 '15 at 23:26

Perhaps in order to understand the previous answer you should follow the bellow first:

There are three different behaviours when $N\rightarrow \infty$ of $<\sigma_{0,0},\sigma,_{N,N}>$ for fixed temperature T. whether $T>T_c$, $T=T_c$ or $T<T_c$ the correlation function $<\sigma_{0,0},\sigma,_{N,N}>$ in each of this cases has a different representation those you can find them explicitly here. Now correlation length or $ξ$ is defined as $\ln t = ξ^{-1}$ where one can later use it as scale factor.

  • $\begingroup$ Hm, I had $\xi$ defined as $$\xi^2=\frac{\int_0^\infty r^2G(r)}{\int_0^\infty G(r)}$$ But anyway, I know the expression of the correlation function in the three regimes, I just fail to see why I should expect values that don't depend on the size of the system for dimensionless variables. Thank you for answering $\endgroup$ – MyUserIsThis May 15 '15 at 23:34
  • $\begingroup$ it's because your small t is defined to be $t=(T-T_{c})/T_{c}$ it measures the deviation from the critical temperature in a dimensionless way, and hence the correlation length diverges as a negative power of t namely $ξ=|t|^{-\nu}$ as $T\rightarrow T_{c}$ where $\nu$ is the critical exponent. $\endgroup$ – William May 16 '15 at 3:19

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