We have the Langevin equation, that describes the motion of a particle in a viscous medium, given by

\begin{equation}\label{Langevin} \frac{dv}{dt} = -\gamma v + \zeta(t) \end{equation}

With the conditions that

\begin{equation} \langle \zeta(t) \rangle = 0 \end{equation}

\begin{equation} \langle \zeta(t)\zeta(t') \rangle = \Gamma \delta(t-t') \end{equation}

And, if we make the time discrete, by putting $t = n\tau$ we can obtain the relation

\begin{equation} v_{n+1} = av_n + \sqrt{\tau \Gamma}\xi_n \end{equation}

where $a = (1 - \tau \gamma)$ with the conditions

\begin{equation} \langle \xi_i \rangle = 0 \end{equation}

\begin{equation} \langle \xi_i \xi_j \rangle = \delta_{ij} \end{equation}

My question is that I didn't know how I obtain the discrete equation from the continuous equation. I understand the $a$ but why the square root appears? What transformation between $\xi$ and $\zeta$ I should do?


1 Answer 1


The short answer to your question is $\zeta(t) \rightarrow \sqrt{\frac{\Gamma}\tau}\xi_n$. The easiest reason to give for the square root is dimensional analysis. $\zeta$ is dimensionful, but $\xi$ is dimensionless, so using dimensional analysis in the variance equations will give you the square root.

In order to deduce this, you should think about how one discretizes differential equations. Working out this procedure, one gets $v(t) \rightarrow v_n$, $\zeta(t) \rightarrow \zeta_n$, $\frac{dv}{dt} \rightarrow \frac{v_{n+1} - v_n}{\tau}$, and $\delta(t-t') \rightarrow \frac{1}{\tau}\delta_{ij}$. The $\frac{1}{\tau}$ can be remembered based on dimensional considerations, or you can integrate/sum both sides in order to properly derive it. Now, simply plug these in and find the relationship between $\zeta_n$ and $\xi_n$.

  • $\begingroup$ I understand now what you just said. I wasn't make any correction on the Dirac delta turning to a Kronecker delta. In this way the correct transformation is $\zeta(t) \rightarrow \sqrt{\frac{\Gamma}{\tau}}\xi_n$. This is just because of the $\tau$ on the passage from de derivative to a rational number. It helped a lot! Thank's! $\endgroup$
    – user78217
    Commented Nov 1, 2016 at 13:06
  • $\begingroup$ Sorry about my typo, you are correct about the $\zeta$ transformation. $\endgroup$
    – Aaron
    Commented Nov 1, 2016 at 15:24

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