Editing Langevin equation (section)

==== Klein–Kramers equation ====
{{Main|Klein–Kramers equation}}
The Fokker–Planck equation for an underdamped Brownian particle is called the [[Klein–Kramers equation]].<ref name="Kramers1940">{{cite journal | last=Kramers | first=H.A. | title=Brownian motion in a field of force and the diffusion model of chemical reactions | journal=Physica | publisher=Elsevier BV | volume=7 | issue=4 | year=1940 | issn=0031-8914 | doi=10.1016/s0031-8914(40)90098-2 | pages=284–304| bibcode=1940Phy.....7..284K | s2cid=33337019 }}</ref><ref name="Risken1989">{{cite book |first=H. |last=Risken |title=The Fokker–Planck Equation: Method of Solution and Applications |publisher=Springer-Verlag |location=New York |year=1989 |isbn=978-0387504988 }}</ref> If the Langevin equations are written as
<math display="block">\begin{align}
\dot{\mathbf{r}} &= \frac{\mathbf{p}}{m}  \\
\dot{\mathbf{p}} &= -\xi \, \mathbf{p} - \nabla V(\mathbf{r}) + \sqrt{2 m \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{align}</math>
where <math>\mathbf{p}</math> is the momentum, then the corresponding Fokker–Planck equation is
<math display="block">\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</math>
Here <math>\nabla_{\mathbf{r}}</math> and <math>\nabla_{\mathbf{p}}</math> are the [[gradient operator]] with respect to {{math|'''r'''}} and {{math|'''p'''}}, and <math>\nabla_{\mathbf{p}}^2</math> is the [[Laplacian]] with respect to {{math|'''p'''}}.

In <math>d</math>-dimensional free space, corresponding to <math>V(\mathbf{r}) = \text{constant}</math> on <math>\mathbb{R}^{d}</math>, this equation can be solved using [[Fourier transform]]s. If the particle is initialized at <math>t = 0</math> with position <math>\mathbf{r}'</math> and momentum <math>\mathbf{p}'</math>, corresponding to initial condition <math>f(\mathbf{r}, \mathbf{p}, 0) = \delta(\mathbf{r}-\mathbf{r}')\delta(\mathbf{p}-\mathbf{p}')</math>, then the solution is<ref name="Risken1989"/><ref name="Chandrasekhar1943">{{cite journal|last1=Chandrasekhar|first1=S.|title=Stochastic Problems in Physics and Astronomy|journal=Reviews of Modern Physics|volume=15|issue=1|year=1943|pages=1–89|issn=0034-6861| doi=10.1103/RevModPhys.15.1|bibcode=1943RvMP...15....1C }}</ref>
<math display="block">\begin{align}
f(\mathbf{r}, \mathbf{p}, t) =& \frac{1}{\left(2 \pi \sigma_X \sigma_P \sqrt{1 - \beta^2}\right)^d} \times \\
& \quad\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{align}</math>
where
<math display="block">\begin{align}
&\sigma^2_X = \frac{k_{\mathrm{B}} T}{m \xi^2} \left[1 + 2 \xi t -  \left(2 - e^{-\xi t}\right)^2 \right]; \qquad  \sigma^2_P = m k_{\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{align}</math>
In three spatial dimensions, the mean squared displacement is
<math display="block">\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</math>