= Ballistic-to-diffusive Brownian displacement
{title2=$\mathbb E[(X_t-X_0)^2]=2D[t-\tau(1-e^{-t/\tau})]$}
For a free one-dimensional particle following <Langevin dynamics> of <mass> $m$ and drag coefficient $\zeta$, write $\tau=m/\zeta$ and $D=k_BT/\zeta$, 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 $\mathbb E[V_sV_u]=(k_BT/m)e^{-|s-u|/\tau}$. Twice integrating this <covariance> gives the <mean squared displacement>
$$
\mathbb E[(X_t-X_0)^2]=2D\left[t-\tau(1-e^{-t/\tau})\right].
$$
For $t\ll\tau$ this is $(k_BT/m)t^2+O(t^3)$: 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>.<On the Theory of the Brownian Motion by G. E. Uhlenbeck and L. S. Ornstein>[1930, velocity relaxation model; the displayed displacement formula follows by integrating its equilibrium velocity covariance]{ref}
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