= Velocity relaxation time
{title2=$\tau_v=m/\zeta$}
For a particle of <mass> $m$ with <linear drag> $-\zeta v$, the velocity relaxation time is the <exponential relaxation time> $\tau_v=m/\zeta$. Under a constant applied <force> $F$ and zero-mean random forcing, the averaged <Langevin equation> is $m\,d\langle v\rangle/dt=-\zeta\langle v\rangle+F$. Solving gives
$$
\langle v(t)\rangle-\frac{F}{\zeta}=\left(\langle v(0)\rangle-\frac{F}{\zeta}\right)e^{-t/\tau_v}.
$$
Thus the initial mean-velocity deviation decays by a factor $e^{-1}$ after time $\tau_v$. Since the force-per-velocity coefficient $\zeta$ has units $\mathrm{kg\,s^{-1}}$, $m/\zeta$ has units of time. If the damping parameter is instead defined as a rate $\gamma=\zeta/m$, the same time is $1/\gamma$. For a sphere obeying <Stokes's law>, $\tau_v=m/(6\pi\eta R)$. The <fluctuation-dissipation relation for a Langevin particle> gives the same decay time for its equilibrium velocity <covariance>.
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