Langevin DE

  1. Hi

    I am reading about the Langevin stochastic differential equation
    \frac{d}{dt}p = -\alpha p + f(t)
    where p is the momentum and f(t) the Langevin force. By definition <F(t)>=0 and <f(t)f(t')> = 2Dg(t-t'), where g is the second order correlation function.

    My question is, why is there a factor 2 in the expression for <f(t)f(t')>? I can't seem to find an answer in any book, but they all write the factor.

    I would be glad to recieve some feedback.

  2. jcsd
  3. atyy

    atyy 11,190
    Science Advisor

    There are two phenomenological ways of describing simple Brownian motion.

    One is with something called a Fokker-Planck equation . In the definition of the Fokker-Planck equation there is a quantity called the diffusion constant D.

    The other is with the Langevin equation. You are right that there is no need to start off with the "2" in the definition of the Langevin equation. Let's say you start off "g" instead. You will find that you can derive a Fokker-Planck equation from the Langevin equation where g=2D, with the g coming from the Langevin, and the D from the Fokker-Planck. Since they knew that, they just used 2D in the initial definition. See Eq 6.3, 6.8, 6.27 and 6.35.
  4. Thanks, that is very kind of you. These things are really interesting. I need to find a good book on this topic, uptil now I have just been using the web.

  5. atyy

    atyy 11,190
    Science Advisor

  6. Thanks, I just checked my library, and they have it. I'll pick it up Monday.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?