= Solution
For one Cartesian <velocity> component of a free Brownian particle, the inertial <Langevin equation> is
$$
m\dot v=-\zeta v+\xi(t),\qquad
\langle\xi(t)\rangle=0,\qquad
\langle\xi(t)\xi(t')\rangle=2\zeta k_BT\delta(t-t').
$$
The random <force> and viscous drag obey the fluctuation-dissipation relation. With a <thermal equilibrium> <velocity> distribution, the <relaxation time> is $\tau=m/\zeta$ and
$$
\langle v(t)v(0)\rangle=\frac{k_BT}{m}e^{-|t|/\tau}.
$$
Integrating <velocity> twice gives
$$
\boxed{\langle[x(t)-x(0)]^2\rangle
=2\frac{k_BT}{\zeta}\left[t-\tau(1-e^{-t/\tau})\right].}
$$
For $t\gg\tau$, this is Einstein's <diffusion> law, $\langle\Delta x^2\rangle=2Dt$, with \b[$D=k_BT/\zeta$]. In three dimensions $\langle|\Delta\mathbf r|^2\rangle=6Dt$. At $t\ll\tau$ the motion is ballistic, $\langle\Delta x^2\rangle\simeq(k_BT/m)t^2$. A sphere in the Stokes-drag regime has $\zeta=6\pi\eta a$, giving $D=k_BT/(6\pi\eta a)$.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2014/paper-1-tsp2014-brownian.png]
{title=Original equilibrium Langevin mean-square displacement, showing the ballistic-to-diffusive crossover and the factor-of-two one-coordinate diffusion convention}
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