
#1
Jul2112, 02:59 PM

P: 1,863

Hi
I am reading about the Langevin stochastic differential equation [tex] \frac{d}{dt}p = \alpha p + f(t) [/tex] where p is the momentum and f(t) the Langevin force. By definition <F(t)>=0 and <f(t)f(t')> = 2Dg(tt'), 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
Jul2112, 04:51 PM

Sci Advisor
P: 8,005

There are two phenomenological ways of describing simple Brownian motion.
One is with something called a FokkerPlanck equation http://www.pma.caltech.edu/~mcc/Ph127/b/Lecture17.pdf . In the definition of the FokkerPlanck 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 FokkerPlanck equation from the Langevin equation where g=2D, with the g coming from the Langevin, and the D from the FokkerPlanck. Since they knew that, they just used 2D in the initial definition. See http://web.phys.ntnu.no/~ingves/Teac...loads/kap6.pdf Eq 6.3, 6.8, 6.27 and 6.35. 



#3
Jul2112, 04:57 PM

P: 1,863

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
Jul2112, 04:59 PM

Sci Advisor
P: 8,005

Langevin DE
I've been learning stuff from Kurt Jacob's book http://books.google.com/books?id=bx8...gbs_navlinks_s . It's was quite accessible for me (I'm a biologist).




#5
Jul2112, 05:03 PM

P: 1,863

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



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