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Brownian motion as a prototype

The original Langevin equation describes Brownian motion, the apparently random movement of a particle in a fluid due to collisions with the molecules of the fluid,

$m{\frac {d^{2}\mathbf {r} }{dt^{2}}}=-\lambda {\frac {d\mathbf {r} }{dt}}+{\boldsymbol {\eta }}\left(t\right).$

The degree of freedom of interest here is the position $\mathbf {r}$ of the particle, $m$ denotes the particle's mass. The force acting on the particle is written as a sum of a viscous force proportional to the particle's velocity (Stokes' law), and a noise term ${\boldsymbol {\eta }}\left(t\right)$ (the name given in physical contexts to terms in stochastic differential equations which are stochastic processes) representing the effect of the collisions with the molecules of the fluid. The force ${\boldsymbol {\eta }}\left(t\right)$ has a Gaussian probability distribution with correlation function

$\left\langle \eta _{i}\left(t\right)\eta _{j}\left(t^{\prime }\right)\right\rangle =2\lambda k_{B}T\delta _{i,j}\delta \left(t-t^{\prime }\right),$

where $k_{B}$ is Boltzmann's constant, $T$ is the temperature and $\eta _{i}\left(t\right)$ is the i-th component of the vector ${\boldsymbol {\eta }}\left(t\right)$. The δ-function form of the correlations in time means that the force at a time $t$ is assumed to be completely uncorrelated with it at any other time. This is an approximation; the actual random force has a nonzero correlation time corresponding to the collision time of the molecules. However, the Langevin equation is used to describe the motion of a "macroscopic" particle at a much longer time scale, and in this limit the $\delta$-correlation and the Langevin equation become exact.

Another prototypical feature of the Langevin equation is the occurrence of the damping coefficient $\lambda$ in the correlation function of the random force, a fact also known as Einstein relation.

Trajectories of free Brownian particles

Consider a free particle of mass $m$ with equation of motion described by

$m{\frac {d\mathbf {v} }{dt}}=-{\frac {\mathbf {v} }{\mu }}+\mathbf {F} (t),$

where $\mathbf {v} =d\mathbf {r} /dt$ is the particle velocity, $\mu$ is the particle mobility, and $\mathbf {F} (t)=m\mathbf {a} (t)$ is a rapidly fluctuating force whose time-average vanishes over a characteristic timescale $t_{c}$ of particle collisions, i.e. ${\overline {\mathbf {F} (t)}}=0$. The general solution to the equation of motion is

$\mathbf {v} (t)=\mathbf {v} (0)e^{-t/\tau }+\int _{0}^{t}\mathbf {a} (t')e^{-(t-t')/\tau }dt',$

where $\tau =m\mu$ is the relaxation time of the Brownian motion. As expected from the random nature of Brownian motion, the average drift velocity $\langle \mathbf {v} (t)\rangle =\mathbf {v} (0)e^{-t/\tau }$ quickly decays to zero at $t\gg \tau$. It can also be shown that the autocorrelation function of the particle velocity $\mathbf {v}$ is given by

Simulated squared displacements of free Brownian particles (semi-transparent wiggly lines) as a function of time, for three selected choices of initial squared velocity which are 0, 3kT/m, and 6kT/m respectively, with 3kT/m being the equipartition value in thermal equilibrium. The colored solid curves denote the mean squared displacements for the corresponding parameter choices.

${\begin{aligned}R_{vv}(t_{1},t_{2})&\equiv \langle \mathbf {v} (t_{1})\cdot \mathbf {v} (t_{2})\rangle \\&=v^{2}(0)e^{-(t_{1}+t_{2})/\tau }+\int _{0}^{t_{1}}\int _{0}^{t_{2}}R_{aa}(t_{1}',t_{2}')e^{-(t_{1}+t_{2}-t_{1}'-t_{2}')/\tau }dt_{1}'dt_{2}'\\&\simeq v^{2}(0)e^{-|t_{2}-t_{1}|/\tau }+{\bigg [}{\frac {3k_{B}T}{m}}-v^{2}(0){\bigg ]}{\Big [}e^{-|t_{2}-t_{1}|/\tau }-e^{-(t_{1}+t_{2})/\tau }{\Big ]},\end{aligned}}$

where we have used the property that the variables $\mathbf {a} (t_{1}')$ and $\mathbf {a} (t_{2}')$ become uncorrelated for time separations $t_{2}'-t_{1}'\gg t_{c}$. Besides, the value of $\lim _{t\to \infty }\langle v^{2}(t)\rangle =\lim _{t\to \infty }R_{vv}(t,t)$ is set to be equal to $3k_{B}T/m$ such that it obeys the equipartition theorem. Note that if the system is initially at thermal equilibrium already with $v^{2}(0)=3k_{B}T/m$, then $\langle v^{2}(t)\rangle =3k_{B}T/m$ for all $t$, meaning that the system remains at equilibrium at all times.

The velocity $\mathbf {v} (t)$ of the Brownian particle can be integrated to yield its trajectory (assuming it is initially at the origin)

$\mathbf {r} (t)=\mathbf {v} (0)\tau {\big (}1-e^{-t/\tau }{\big )}+\tau \int _{0}^{t}\mathbf {a} (t'){\Big [}1-e^{-(t-t')/\tau }{\Big ]}dt'.$

Hence, the resultant average displacement $\langle \mathbf {r} (t)\rangle =\mathbf {v} (0)\tau {\big (}1-e^{-t/\tau }{\big )}$ asymptotes to $\mathbf {v} (0)\tau$ as the system relaxes and randomness takes over. In addition, the mean squared displacement can be determined similarly to the preceding calculation to be

$\langle r^{2}(t)\rangle =v^{2}(0)\tau ^{2}{\big (}1-e^{-t/\tau }{\big )}^{2}-{\frac {3k_{B}T}{m}}\tau ^{2}{\big (}1-e^{-t/\tau }{\big )}{\big (}3-e^{-t/\tau }{\big )}+{\frac {6k_{B}T}{m}}\tau t.$

It can be seen that $\langle r^{2}(t\ll \tau )\rangle \simeq v^{2}(0)t^{2}$, indicating that the motion of Brownian particles at timescales much shorter than the relaxation time $\tau$ of the system is (approximately) time-reversal invariant. On the other hand, $\langle r^{2}(t\gg \tau )\rangle \simeq 6k_{B}T\tau t/m=6\mu k_{B}Tt=6Dt$, which suggests that the long-term random motion of Brownian particles is an irreversible dissipative process. Here we have made use of the Einstein–Smoluchowski relation $D={\mu \,k_{B}T}$, where $D$ is the diffusion coefficient of the fluid.