Langevin equation

An electric circuit consisting of a resistor and a capacitor.

There is a close analogy between the paradigmatic Brownian particle discussed above and Johnson noise, the electric voltage generated by thermal fluctuations in a resistor.[9] The diagram at the right shows an electric circuit consisting of a resistance R and a capacitance C. The slow variable is the voltage U between the ends of the resistor. The Hamiltonian reads ${\displaystyle {\mathcal {H}}=E/k_{\text{B}}T=CU^{2}/(2k_{\text{B}}T)}$, and the Langevin equation becomes

$${\displaystyle {\frac {dU}{dt}}=-{\frac {U}{RC}}+\eta \left(t\right),\;\;\left\langle \eta \left(t\right)\eta \left(t'\right)\right\rangle ={\frac {2k_{\text{B}}T}{RC^{2}}}\delta \left(t-t'\right).}$$

This equation may be used to determine the correlation function

$${\displaystyle \left\langle U\left(t\right)U\left(t'\right)\right\rangle ={\frac {k_{\text{B}}T}{C}}\exp \left(-{\frac {\left|t-t'\right|}{RC}}\right)\approx 2Rk_{\text{B}}T\delta \left(t-t'\right),}$$

which becomes white noise (Johnson noise) when the capacitance C becomes negligibly small.

The dynamics of the order parameter ${\displaystyle \varphi }$ of a second order phase transition slows down near the critical point and can be described with a Langevin equation.[7] The simplest case is the universality class "model A" with a non-conserved scalar order parameter, realized for instance in axial ferromagnets,

$${\displaystyle {\begin{aligned}{\frac {\partial \varphi \left(\mathbf {x} ,t\right)}{\partial t}}&=-\lambda {\frac {\delta {\mathcal {H}}}{\delta \varphi }}+\eta \left(\mathbf {x} ,t\right),\\{\mathcal {H}}&=\int d^{d}x\left[{\frac {1}{2}}r_{0}\varphi ^{2}+u\varphi ^{4}+{\frac {1}{2}}(\nabla \varphi )^{2}\right],\\[2pt]\left\langle \eta \left(\mathbf {x} ,t\right)\eta \left(\mathbf {x} ',t'\right)\right\rangle &=2\lambda \delta \left(\mathbf {x} -\mathbf {x} '\right)\delta \left(t-t'\right).\end{aligned}}}$$

Other universality classes (the nomenclature is "model A",..., "model J") contain a diffusing order parameter, order parameters with several components, other critical variables and/or contributions from Poisson brackets.[7]

$${\displaystyle m{\frac {dv}{dt}}=-\lambda v+\eta (t)-kx}$$

A particle in a fluid is described by a Langevin equation with a potential energy function, a damping force, and thermal fluctuations given by the fluctuation dissipation theorem. If the potential is quadratic then the constant energy curves are ellipses, as shown in the figure. If there is dissipation but no thermal noise, a particle continually loses energy to the environment, and its time-dependent phase portrait (velocity vs position) corresponds to an inward spiral toward 0 velocity. By contrast, thermal fluctuations continually add energy to the particle and prevent it from reaching exactly 0 velocity. Rather, the initial ensemble of stochastic oscillators approaches a steady state in which the velocity and position are distributed according to the Maxwell–Boltzmann distribution. In the plot below (figure 2), the long time velocity distribution (orange) and position distributions (blue) in a harmonic potential (${\textstyle U={\frac {1}{2}}kx^{2}}$) is plotted with the Boltzmann probabilities for velocity (red) and position (green). In particular, the late time behavior depicts thermal equilibrium.

Consider a free particle of mass ${\displaystyle m}$ with equation of motion described by

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

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

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

where ${\displaystyle \tau =m\mu }$ is the correlation time of the noise term. It can also be shown that the autocorrelation function of the particle velocity ${\displaystyle \mathbf {v} }$ is given by[10]

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, 3k_BT/m, and 6k_BT/m respectively, with 3k_BT/m being the equipartition value in thermal equilibrium. The colored solid curves denote the mean squared displacements for the corresponding parameter choices.

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

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

The velocity ${\displaystyle \mathbf {v} (t)}$ of the Brownian particle can be integrated to yield its trajectory ${\displaystyle \mathbf {r} (t)}$. If it is initially located at the origin with probability 1, then the result is

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

Hence, the average displacement ${\textstyle \langle \mathbf {r} (t)\rangle =\mathbf {v} (0)\tau \left(1-e^{-t/\tau }\right)}$ asymptotes to ${\displaystyle \mathbf {v} (0)\tau }$ as the system relaxes. The mean squared displacement can be determined similarly:

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

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

This plot corresponds to solutions of the complete Langevin equation obtained using the Euler–Maruyama method. The left panel shows the time evolution of the phase portrait of a harmonic oscillator at different temperatures. The right panel captures the corresponding equilibrium probability distributions. At zero temperature, the velocity rapidly decays from its initial value (the red dot) to zero 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.

If the external potential is conservative and the noise term derives from a reservoir in thermal equilibrium, then the long-time solution to the Langevin equation must reduce to the Boltzmann distribution, which is the probability distribution function for particles in thermal equilibrium. In the special case of overdamped dynamics, the inertia of the particle is negligible in comparison to the damping force, and the trajectory ${\displaystyle x(t)}$ is described by the overdamped Langevin equation

$${\displaystyle \lambda {\frac {dx}{dt}}=-{\frac {\partial V(x)}{\partial x}}+\eta (t)\equiv -{\frac {\partial V(x)}{\partial x}}+{\sqrt {2\lambda k_{\text{B}}T}}{\frac {dB_{t}}{dt}},}$$

where ${\displaystyle \lambda }$ is the damping constant. The term ${\displaystyle \eta (t)}$ is white noise, characterized by ${\displaystyle \left\langle \eta (t)\eta (t')\right\rangle =2k_{\text{B}}T\lambda \delta (t-t')}$ (formally, the Wiener process). One way to solve this equation is to introduce a test function ${\displaystyle f}$ and calculate its average. The average of ${\displaystyle f(x(t))}$ should be time-independent for finite ${\displaystyle x(t)}$, leading to

$${\displaystyle {\frac {d}{dt}}\left\langle f(x(t))\right\rangle =0,}$$

Itô's lemma for the Itô drift-diffusion process ${\displaystyle dX_{t}=\mu _{t}\,dt+\sigma _{t}\,dB_{t}}$ says that the differential of a twice-differentiable function f(t, x) is given by

$${\displaystyle df=\left({\frac {\partial f}{\partial t}}+\mu _{t}{\frac {\partial f}{\partial x}}+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\right)dt+\sigma _{t}{\frac {\partial f}{\partial x}}\,dB_{t}.}$$

Applying this to the calculation of ${\displaystyle \langle f(x(t))\rangle }$ gives

$${\displaystyle \left\langle -f'(x){\frac {\partial V}{\partial x}}+k_{\text{B}}Tf''(x)\right\rangle =0.}$$

This average can be written using the probability density function ${\displaystyle p(x)}$;

$${\displaystyle \int \left(-f'(x){\frac {\partial V}{\partial x}}p(x)+{k_{\text{B}}T}f''(x)p(x)\right)dx=\int \left(-f'(x){\frac {\partial V}{\partial x}}p(x)-{k_{\text{B}}T}f'(x)p'(x)\right)dx=0,}$$

where the second term was integrated by parts (hence the negative sign). Since this is true for arbitrary functions ${\displaystyle f}$, it follows that

$${\displaystyle {\frac {\partial V}{\partial x}}p(x)+{k_{\text{B}}T}p'(x)=0,}$$

thus recovering the Boltzmann distribution

$${\displaystyle p(x)\propto \exp \left({-{\frac {V(x)}{k_{\text{B}}T}}}\right).}$$

The probability density function p(k) of the turbulent kinetic energy k is investigated for diabatic atmospheric surface layer (ASL) flows.[11] In the latter reference, authors show when the velocity components are near-Gaussian and their squared amplitudes are nearly independent, the resulting p(k) is shown to be γ-distributed. therefore, a nonlinear Langevin equation that preserves a γ-distributed p(k), but allows linear relaxation of k to its mean state, is proposed and tested using multiple ASL data sets.

A Fokker–Planck equation is a deterministic equation for the time dependent probability density ${\displaystyle P\left(A,t\right)}$ of stochastic variables ${\displaystyle A}$. The Fokker–Planck equation corresponding to the generic Langevin equation described in this article is the following:[12]

$${\displaystyle {\frac {\partial P\left(A,t\right)}{\partial t}}=\sum _{i,j}{\frac {\partial }{\partial A_{i}}}\left(-k_{\text{B}}T\left[A_{i},A_{j}\right]{\frac {\partial {\mathcal {H}}}{\partial A_{j}}}+\lambda _{i,j}{\frac {\partial {\mathcal {H}}}{\partial A_{j}}}+\lambda _{i,j}{\frac {\partial }{\partial A_{j}}}\right)P\left(A,t\right).}$$

The equilibrium distribution ${\displaystyle P(A)=p_{0}(A)={\text{const}}\times \exp(-{\mathcal {H}})}$ is a stationary solution.

The Fokker–Planck equation for an underdamped Brownian particle is called the Klein–Kramers equation.[13][14] If the Langevin equations are written as

$${\displaystyle {\begin{aligned}{\dot {\mathbf {r} }}&={\frac {\mathbf {p} }{m}}\\{\dot {\mathbf {p} }}&=-\xi \,\mathbf {p} -\nabla V(\mathbf {r} )+{\sqrt {2m\xi k_{\mathrm {B} }T}}{\boldsymbol {\eta }}(t),\qquad \langle {\boldsymbol {\eta }}^{\mathrm {T} }(t){\boldsymbol {\eta }}(t')\rangle =\mathbf {I} \delta (t-t')\end{aligned}}}$$

where ${\displaystyle \mathbf {p} }$ is the momentum, then the corresponding Fokker–Planck equation is

$${\displaystyle {\frac {\partial f}{\partial t}}+{\frac {1}{m}}\mathbf {p} \cdot \nabla _{\mathbf {r} }f=\xi \nabla _{\mathbf {p} }\cdot \left(\mathbf {p} \,f\right)+\nabla _{\mathbf {p} }\cdot \left(\nabla V(\mathbf {r} )\,f\right)+m\xi k_{\mathrm {B} }T\,\nabla _{\mathbf {p} }^{2}f}$$

Here ${\displaystyle \nabla _{\mathbf {r} }}$ and ${\displaystyle \nabla _{\mathbf {p} }}$ are the gradient operator with respect to r and p, and ${\displaystyle \nabla _{\mathbf {p} }^{2}}$ is the Laplacian with respect to p.

In ${\displaystyle d}$-dimensional free space, corresponding to ${\displaystyle V(\mathbf {r} )={\text{constant}}}$ on ${\displaystyle \mathbb {R} ^{d}}$, this equation can be solved using Fourier transforms. If the particle is initialized at ${\displaystyle t=0}$ with position ${\displaystyle \mathbf {r} '}$ and momentum ${\displaystyle \mathbf {p} '}$, corresponding to initial condition ${\displaystyle f(\mathbf {r} ,\mathbf {p} ,0)=\delta (\mathbf {r} -\mathbf {r} ')\delta (\mathbf {p} -\mathbf {p} ')}$, then the solution is[14][15]

$${\displaystyle {\begin{aligned}f(\mathbf {r} ,\mathbf {p} ,t)={\frac {1}{\left(2\pi \sigma _{X}\sigma _{P}{\sqrt {1-\beta ^{2}}}\right)^{d}}}\exp \left[-{\frac {1}{2(1-\beta ^{2})}}\left({\frac {|\mathbf {r} -{\boldsymbol {\mu }}_{X}|^{2}}{\sigma _{X}^{2}}}+{\frac {|\mathbf {p} -{\boldsymbol {\mu }}_{P}|^{2}}{\sigma _{P}^{2}}}-{\frac {2\beta (\mathbf {r} -{\boldsymbol {\mu }}_{X})\cdot (\mathbf {p} -{\boldsymbol {\mu }}_{P})}{\sigma _{X}\sigma _{P}}}\right)\right]\end{aligned}}}$$

where

$${\displaystyle {\begin{aligned}&\sigma _{X}^{2}={\frac {k_{\mathrm {B} }T}{m\xi ^{2}}}\left[1+2\xi t-\left(2-e^{-\xi t}\right)^{2}\right];\qquad \sigma _{P}^{2}=mk_{\mathrm {B} }T\left(1-e^{-2\xi t}\right)\\&\beta ={\frac {k_{\text{B}}T}{\xi \sigma _{X}\sigma _{P}}}\left(1-e^{-\xi t}\right)^{2}\\&{\boldsymbol {\mu }}_{X}=\mathbf {r} '+(m\xi )^{-1}\left(1-e^{-\xi t}\right)\mathbf {p} ';\qquad {\boldsymbol {\mu }}_{P}=\mathbf {p} 'e^{-\xi t}.\end{aligned}}}$$

In three spatial dimensions, the mean squared displacement is

$${\displaystyle \langle \mathbf {r} (t)^{2}\rangle =\int f(\mathbf {r} ,\mathbf {p} ,t)\mathbf {r} ^{2}\,d\mathbf {r} d\mathbf {p} ={\boldsymbol {\mu }}_{X}^{2}+3\sigma _{X}^{2}}$$

A path integral equivalent to a Langevin equation can be obtained from the corresponding Fokker–Planck equation or by transforming the Gaussian probability distribution ${\displaystyle P^{(\eta )}(\eta )d\eta }$ of the fluctuating force ${\displaystyle \eta }$ to a probability distribution of the slow variables, schematically ${\displaystyle P(A)dA=P^{(\eta )}(\eta (A))\det(d\eta /dA)dA}$. The functional determinant and associated mathematical subtleties drop out if the Langevin equation is discretized in the natural (causal) way, where ${\displaystyle A(t+\Delta t)-A(t)}$ depends on ${\displaystyle A(t)}$ but not on ${\displaystyle A(t+\Delta t)}$. It turns out to be convenient to introduce auxiliary response variables ${\displaystyle {\tilde {A}}}$. The path integral equivalent to the generic Langevin equation then reads[16]

$${\displaystyle \int P(A,{\tilde {A}})\,dA\,d{\tilde {A}}=N\int \exp \left(L(A,{\tilde {A}})\right)dA\,d{\tilde {A}},}$$

where ${\displaystyle N}$ is a normalization factor and

$${\displaystyle L(A,{\tilde {A}})=\int \sum _{i,j}\left\{{\tilde {A}}_{i}\lambda _{i,j}{\tilde {A}}_{j}-{\widetilde {A}}_{i}\left\{\delta _{i,j}{\frac {dA_{j}}{dt}}-k_{\text{B}}T\left[A_{i},A_{j}\right]{\frac {d{\mathcal {H}}}{dA_{j}}}+\lambda _{i,j}{\frac {d{\mathcal {H}}}{dA_{j}}}-{\frac {d\lambda _{i,j}}{dA_{j}}}\right\}\right\}dt.}$$

The path integral formulation allows for the use of tools from quantum field theory, such as perturbation and renormalization group methods.