Understanding the Factor 2 in the Langevin Stochastic Differential Equation

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Discussion Overview

The discussion revolves around the Langevin stochastic differential equation, specifically focusing on the factor of 2 in the expression for the correlation function of the Langevin force. Participants explore theoretical aspects and seek clarification on the underlying reasons for this factor.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the origin of the factor 2 in the correlation function = 2Dg(t-t'), noting that this is a common presentation in literature without clear justification.
  • Another participant explains that the factor 2 can be understood through the relationship between the Langevin equation and the Fokker-Planck equation, suggesting that starting with a different definition could lead to deriving the same result with g=2D.
  • A participant shares their experience with learning resources, indicating a preference for accessible materials on the topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the explanation for the factor 2, and the discussion includes multiple perspectives on how to approach the topic.

Contextual Notes

The discussion highlights the dependence on definitions and the potential for different interpretations of the Langevin equation and its relation to the Fokker-Planck equation.

Who May Find This Useful

This discussion may be useful for those studying stochastic processes, particularly in the context of physics and applied mathematics, as well as individuals seeking to understand the theoretical foundations of Brownian motion.

Niles
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Hi

I am reading about the Langevin stochastic differential equation
<br /> \frac{d}{dt}p = -\alpha p + f(t) <br />
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 receive some feedback.


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

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