1/{\beta}$ (note that $t=1/{\beta}$ is the time of macroscopic equilibrium where most particles are packed on the right), we consider the number of particles that end up there after keeping a constant drift of order $z (\log k) /(N t)$, which is smaller than $b_N$. Neglecting interaction between particles and making a Brownian approximation for the random walk with drift, we obtain that the expected number of particles following this strategy is given by $$k \exp\left(- \log k \frac{({\beta}t-z)^2}{2 t} \right)=k^{1-\frac{({\beta}t-z)^2}{2 t}}.$$

Hence equilibrium should be attained when this becomes negligible with respect to the typical fluctuation $k^{\frac{1}{2}-{\beta}(1-z)}$ for all values of $z$ where we find particles at equilibrium. That is, when the inequality $$\label{fluqueton}
 1-\frac{({\beta}t-z)^2}{2 t}< \frac{1}{2}-{\beta}(1-z),$$ is valid for all $z\in[0,1]$ if ${\beta}\le 1/2$ or for all $z\ge 1-\frac{1}{2{\beta}}$ if $\beta\ge 1/2$. A rapid computation show that one only needs to satisfy the condition for the smallest value of $z$ (either $1$ or $1-\frac{1}{2{\beta}}$), which boils down to finding the roots of a degree two polynomial. This yields that we must have $t>t_0$ where $$t_0:=\begin{cases}
       \frac{2}{{\beta}}+\frac{1}{{\beta}^2} &\text{ if } {\beta}\le 1/2,\\
       \left( \frac{\sqrt{2}+2\sqrt{{\beta}}}{2{\beta}} \right)^2 &\text{ if } {\beta}\ge 1/2.
      \end{cases}$$

\[pluscomplique\] Describing the equilibrium “logarithmic profile” of particles 