does not depend at first order on the choice of the threshold ${\varepsilon}$ - a phenomenon known as cutoff and conjectured to hold for a large class of Markov chains as soon as the mixing time is of a larger order than the relaxation time (which is defined as the inverse of the spectral gap of the generator).

However the patterns of convergence to equilibrium in the symmetric and asymmetric cases are very different. Let us for simplicity focus on the case with a density of particles $k=\alpha N$, $\alpha\in (0,1)$. In the symmetric case, the time scale associated with the hydrodynamic profile is $N^2$ and the limit is given by the heat equation [@KOV] (which takes an infinite time to relax to its equilibrium profile which is flat) and microscopic mixing occurs on a larger time scale $N^2\log N$.

In the asymmetric setup the hydrodynamic limit is given by the inviscid Burgers’ equation with a shorter time scale $N$ [@Reza] (see also [@LabbeKPZ; @LabLac16] for adaptations of this result to the segment). The equilibrium profile for this equation is reached after a finite time and in this case, the mixing time is of order $N$ and corresponds exactly to 