\log C_d(L)  +  \log {\mathcal{E}}^d(\tilde f^I)-2\log 2}{  \lambda^d(L) } \ .$$

    For the one dimensional case, it is shown in Remark \[rem3.2\] that $\lambda^1(L) = O(1/L^2)$ in the limit $L\to\infty$. Moreover, the constant $C_1(L)$ approaches $1$ in the limit $L\to\infty$ by using the limiting behavior $\alpha_*(L)=O(1/L)$ in expression . For initial data $\tilde f^I$ with all of the gas molecules initially located in a small region of $\tilde{\mathbf{T}}$ with a volume fraction of order $\epsilon$, the initial entropy ${\mathcal{E}}^1(\tilde f^I)$ will satisfy ${\mathcal{E}}^1(\tilde f^I) = O(\epsilon^{-2})$. In this case, $t_{{\rm init}}$ is approximately given by $$O(-L^2 (C +\log \epsilon) )\quad \mbox{for } \epsilon\ll 1,\;L\gg1$$ and some positive constant $C$. Thus one time scale in our problems is given, or at least bounded, by $t_{{\rm init}}$. After this time, the solution satisfies $$\label{twoscale3b}
    \| \tilde f(t) - M_1\|_{L^1(\tilde{\mathbf{T}}\times{\mathbb{R}}^d,{\,\operatorname{d}\!\tilde x}{\,\operatorname{d}\!v})}  \leq 2 e^{-\lambda^d(L)\ (t- t_{{\rm init}})/2}\ ,$$ and the second time scale, is given by $2/\lambda^d(L)$, the waiting time after $t_{{\rm init}}$ for $\| \tilde f(t) - M_1\|_{L^1(\tilde{\mathbf{T}}\times{\mathbb{R}}^d,{\,\operatorname{d}\!\tilde x}{\,\operatorname{d}\!v})}$ to decrease by a factor of $1/e$; see Figure \[fig:H-decay\].

    These two times scales are quite similar to what one observes in interacting particle systems or even in card shuffling; see [@AlDi86; @Di96]. 