any $z> 1-(2{\beta})^{-1}$ we have $$\label{liquib}
\lim_{{\varepsilon}\to 0}\lim_{N\to \infty}\frac{\log \left( \sum_{i= (z-{\varepsilon})N}^{(z+{\varepsilon})N} \xi(i) \right) }{\log k_N} \Rightarrow 1-2{\beta}(1-z),$$ where the convergence holds in probability under the equilibrium measure $\pi_{N,k}$ when $N$ tends to infinity and ${\varepsilon}$ tends to zero in that order. The typical distance to zero of the left-most particle at equilibrium is also given by this profile in the sense that it is typically $o(N)$ when ${\beta}\le 1/2$ and of order $N(1-\frac 1 {2{\beta}})$ when ${\beta}\ge 1/2$. While we only give heuristic justification for these statements concerning equilibrium, it is worth mentioning that they can be made rigorous by using the techniques exposed in Section \[Sec:Prelim\].

To estimate the mixing time, we assume that the system gets close to equilibrium once the number of particles on any “mesoscopic” interval of the form $[ (z-{\varepsilon})N, (z+{\varepsilon})N ]$ is close to its equilibrium value. While the mean number of particle is of order $k^{1-2{\beta}(1-z)}$ (cf. ), the typical equilibrium fluctuation around this number should be the given by the square root due to near-independence of different particles and thus be equal to $k^{\frac{1}{2}-{\beta}(1-z)}$.

To estimate the surplus of particles in this interval at time $t N^2 (\log k)^{-1}$ for $t> 