Editing Langevin equation (section)

=== Trajectories of free Brownian particles ===
Consider a free particle of mass <math>m</math> with equation of motion described by
<math display="block"> m \frac{d \mathbf{v}}{dt} = -\frac{\mathbf{v}}{\mu} + \boldsymbol{\eta}(t), </math>
where <math>\mathbf{v} = d\mathbf{r}/dt</math> is the particle velocity, <math>\mu</math> is the particle mobility, and <math>\boldsymbol{\eta}(t) = m \mathbf{a}(t)</math> is a rapidly fluctuating force whose time-average vanishes over a characteristic timescale <math>t_c</math> of particle collisions, i.e. <math>\overline{\boldsymbol{\eta}(t)} = 0</math>. The general solution to the equation of motion is
<math display="block"> \mathbf{v}(t) = \mathbf{v}(0) e^{-t/\tau} + \int_0^t \mathbf{a}(t') e^{-(t-t')/\tau} dt', </math>
where <math>\tau = m\mu</math> is the correlation time of the noise term. It can also be shown that the [[Autocorrelation|autocorrelation function]] of the particle velocity <math>\mathbf{v}</math> is given by<ref>{{cite book | author = Pathria RK | author-link = Raj Pathria|year = 1972 | title = Statistical Mechanics | publisher = Pergamon Press | location = Oxford | isbn = 0-08-018994-6 | pages = 443, 474&ndash;477}}</ref>
<math display="block"> \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} + \left[\frac{3k_\text{B}T}{m} - v^2(0)\right] \Big[e^{-|t_2-t_1|/\tau} - e^{-(t_1+t_2)/\tau}\Big],
\end{align} </math>
where we have used the property that the variables <math>\mathbf{a}(t_1')</math> and <math>\mathbf{a}(t_2')</math> become uncorrelated for time separations <math>t_2'-t_1' \gg t_c</math>. Besides, the value of <math display="inline">\lim_{t \to \infty} \langle v^2 (t) \rangle = \lim_{t \to \infty} R_{vv}(t,t)</math> is set to be equal to <math>3k_\text{B}T/m</math> such that it obeys the [[equipartition theorem]]. If the system is initially at thermal equilibrium already with <math>v^2(0) = 3 k_\text{B} T/m</math>, then <math> \langle v^2(t) \rangle = 3 k_\text{B} T/m</math> for all <math>t</math>, meaning that the system remains at equilibrium at all times.

The velocity <math>\mathbf{v}(t)</math> of the Brownian particle can be integrated to yield its trajectory <math>\mathbf{r}(t)</math>. If it is initially located at the origin with probability 1, then the result is
<math display="block"> \mathbf{r}(t) = \mathbf{v}(0) \tau \left(1-e^{-t/\tau}\right) + \tau \int_0^t \mathbf{a}(t') \left[1 - e^{-(t-t') / \tau}\right] dt'.</math>

Hence, the average displacement <math display="inline">\langle \mathbf{r}(t) \rangle = \mathbf{v}(0) \tau \left(1-e^{-t/\tau}\right)</math> asymptotes to <math>\mathbf{v}(0) \tau</math> as the system relaxes. The [[mean squared displacement]] can be determined similarly:
<math display="block"> \langle r^2(t) \rangle = v^2(0) \tau^2 \left(1 - e^{-t/\tau}\right)^2 - \frac{3k_\text{B}T}{m} \tau^2 \left(1 - e^{-t/\tau}\right) \left(3 - e^{-t/\tau}\right) + \frac{6k_\text{B}T}{m} \tau t. </math>

This expression implies that <math>\langle r^2(t \ll \tau) \rangle \simeq v^2(0) t^2</math>, indicating that the motion of Brownian particles at timescales much shorter than the relaxation time <math>\tau</math> of the system is (approximately) [[T-symmetry|time-reversal]] invariant. On the other hand, <math>\langle r^2(t \gg \tau) \rangle \simeq 6 k_\text{B} T \tau t/m = 6 \mu k_\text{B} T t = 6Dt</math>, which indicates an [[Irreversible process|irreversible]], [[Dissipation|dissipative process]].

[[File:Phase Potrait and 2D Hist.jpg|thumb|upright=1.5|This plot corresponds to solutions of the complete Langevin equation for a lightly damped harmonic oscillator, obtained using the [[Euler–Maruyama method]]. The left panel shows the time evolution of the phase portrait at different temperatures. The right panel captures the corresponding equilibrium probability distributions. At zero temperature, the velocity slowly decays from its initial value (the red dot) to zero, over the course of a handful of oscillations, due to damping. For nonzero temperatures, the velocity can be kicked to values higher than the initial value due to thermal fluctuations. At long times, the velocity remains nonzero, and the position and velocity distributions correspond to that of thermal equilibrium.]]