# Langevin DE

1. Jul 21, 2012

### Niles

Hi

$$\frac{d}{dt}p = -\alpha p + f(t)$$
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.

Best,
Niles.

2. Jul 21, 2012

### atyy

There are two phenomenological ways of describing simple Brownian motion.

One is with something called a Fokker-Planck equation http://www.pma.caltech.edu/~mcc/Ph127/b/Lecture17.pdf [Broken] . 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 http://web.phys.ntnu.no/~ingves/Teaching/TFY4275/Downloads/kap6.pdf Eq 6.3, 6.8, 6.27 and 6.35.

Last edited by a moderator: May 6, 2017
3. Jul 21, 2012

### Niles

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.

Best.
Niles.

4. Jul 21, 2012

### atyy

5. Jul 21, 2012

### Niles

Thanks, I just checked my library, and they have it. I'll pick it up Monday.