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Subsections


Langevin Equation

Problem

We have one particle surrounded by small molecules. At t=0 it had $\mathbf{r}(0)=0$. We want to calculate its position at the time t.

Assumptions:

Equation

By definition,  
 \begin{displaymath}
 \frac{d\mathbf{r}}{dt} = \mathbf{v}\end{displaymath} (1)
We already obtained equation for $d\mathbf{v}/dt$: 
 \begin{displaymath}
 \frac{d\mathbf{v}}{dt} = -\lambda\mathbf{v}+\mathbf{f}(t)\end{displaymath} (2)
with random force $\mathbf{f}(t)$: 
 \begin{displaymath}
 \left\langle \mathbf{f}(t)\right\rangle=0,\quad
 \left\lang...
 ...f{f}(0)\mathbf{f}(t)\right\rangle=\frac{6kT\lambda}{m}\delta(t)\end{displaymath} (3)

This is called Langevin equation. Langevin equation describes system with white noise acting on $\mathbf{v}$.

Solution

Let $\mathbf{v}(0)=\mathbf{v}_0$. In equilibrium  
 \begin{displaymath}
 \left\langle \mathbf{v}\right\rangle=0,\quad
 \left\langle \mathbf{v}_0^2\right\rangle=\frac{3kT}{m}\end{displaymath} (4)

Solution of Langevin equation for $\mathbf{v}$ is  
 \begin{displaymath}
 \mathbf{v}(t) = \mathbf{v}_0e^{-\lambda t} + e^{-\lambda t}\int_0^t
 \mathbf{f}(\tau)e^{\lambda\tau}\,d\tau \end{displaymath} (5)

Solution for $\mathbf{r}$ is

\begin{displaymath}
\mathbf{r}(t) = \int_0^t \mathbf{v}(\tau)\,d\tau\end{displaymath}

1.
Integration of the first term in (5):

\begin{displaymath}
\int_0^t \mathbf{v}_0e^{-\lambda\tau}\,d\tau =
 \frac{1}{\lambda}\mathbf{v}_0(1-e^{-\lambda t})
 \end{displaymath}

2.
Second term--by parts:

\begin{displaymath}
\begin{split}
 &\int_0^t\left[e^{-\lambda\tau}\int_0^\tau
 \...
 ...\lambda(\tau-t)}\right)\mathbf{f}(\tau)\,
 d\tau 
 \end{split} \end{displaymath}

We obtained:

\begin{displaymath}
\mathbf{r}(t) = \frac{1}{\lambda}\mathbf{v}_0(1-e^{-\lambda ...
 ...0^t
 \left(1-e^{\lambda(\tau-t)}\right)\mathbf{f}(\tau)\,d\tau \end{displaymath}

Averages: we know $\left\langle \mathbf{v}_0^2\right\rangle$ and $\left\langle \mathbf{f}(0)\mathbf{f}(t)\right\rangle$. The cross-average

\begin{displaymath}
\left\langle \mathbf{v}_0\mathbf{f}(t)\right\rangle=0\end{displaymath}

1.
First term gives

\begin{displaymath}
\frac{1}{\lambda^2}(1-e^{-\lambda
 t})^2\left\langle \mathbf{v}_0^2\right\rangle=\frac{3kT}{m\lambda^2}(1-e^{-\lambda t})^2 
 \end{displaymath}

2.
Second term gives

\begin{displaymath}
\begin{split}
 &\frac{1}{\lambda^2}\int_0^t d\tau \int_0^t d...
 ... \frac{1}{2\lambda}e^{-2\lambda t}-\frac32\right]
 \end{split} \end{displaymath}

Interpretation

We obtained:  
 \begin{displaymath}
 \begin{split}
 &\left\langle \mathbf{r}^2(t)\right\rangle =...
 ...
 \frac{1}{2\lambda}e^{-2\lambda t}-\frac32\right]
 \end{split}\end{displaymath} (6)
At small times

\begin{displaymath}
\left\langle \mathbf{r}^2(t)\right\rangle = \frac{3kT}{m}t^2\end{displaymath}

At large t

\begin{displaymath}
\left\langle \mathbf{r}^2(t)\right\rangle = 6Dt,\quad D = \frac{kT}{m\lambda}\end{displaymath}

The large time limit could be obtained from the Wiener equation  
 \begin{displaymath}
 \frac{d\mathbf{r}}{dt}=\mathbf{g}(t)\end{displaymath} (7)
with white noise $\mathbf{g}$:

\begin{displaymath}
\left\langle \mathbf{g}(t)\right\rangle=0,\quad
 \left\langle \mathbf{g}(0)\mathbf{g}(t)\right\rangle=12D\delta(t)\end{displaymath}

In the limit of small t Wiener equation does not work!

White noise in Wiener equation is the consequence of the averaging out a fast process--in our case, velocity relaxation!


next up previous
Next: Diffusion Approach. Focker-Planck Equation Up: Brownian Motion and Focker-Planck Previous: Brownian Motion and Focker-Planck

© 1997 Boris Veytsman and Michael Kotelyanskii
Sun Nov 2 18:50:28 EST 1997