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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




 * $$ \begin{align}

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_BT}{m} - v^2(0)\bigg] \Big[e^{-|t_2-t_1|/\tau} - e^{-(t_1+t_2)/\tau}\Big], \end{align} $$

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 $$3k_BT/m$$ such that it obeys the equipartition theorem. Note that if the system is initially at thermal equilibrium already with $$v^2(0) = 3 k_B T/m$$, then $$ \langle v^2(t) \rangle = 3 k_B T/m$$ 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


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

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, $$\langle r^2(t \gg \tau) \rangle \simeq 6 k_B T \tau t/m = 6 \mu k_B T t = 6Dt$$, 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.