 Ballistic-to-diffusive Brownian displacement

For a free one-dimensional particle following Langevin dynamics of mass $m$ and drag coefficient $\zeta$, write $\tau =m/\zeta$ and $D=k_{B}T/\zeta$, where $T$ is the temperature and $k_B$ is the Boltzmann constant. Prepare the initial velocity in thermal equilibrium, independently of future noise. The stationary velocity autocorrelation function is $\mathbb{E}[V_s{V}_u]=(k_{B}T/m)e^{-|s-u|/\tau }$. Twice integrating this covariance gives the mean squared displacementFor $t\ll \tau$ this is $(k_{B}T/m)t^2+O(t^3)$: the particle initially keeps its thermal velocity. For $t\gg \tau$ the leading term is $2Dt$, recovering ordinary diffusion. A particle prepared with zero initial velocity instead has the transient in inertial Brownian displacement with zero initial velocity. The distinction is initial preparation, rather than a change in the long-time diffusion coefficient.^[ref]