I mean that, is there anything more fundamental to yields the result that the random force in Langevin Equation is delta-correlated?
As is shown in the picture of a textbook below, its formula (3.4) is given by the assumption that "impacts are independent".However, it is still daunted for me to derive delta-correlated function from it.
Maybe there are fundamental concepts or derivation steps I should work on, which I will be appreciated of if you could point out generously. enter image description here

  • $\begingroup$ I am trying to learn the stochastic dynamics on my own, which is short of physical scheme.Thus I hope someone can show more detail in mathemaical approach to get me out of that…… $\endgroup$
    – Yan
    Aug 3, 2016 at 16:32
  • $\begingroup$ Welcome to Physics.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. 4) If you get a satisfactory answer, remember to accept it by clicking on the green checkmark. $\endgroup$
    – auden
    Aug 3, 2016 at 16:56

1 Answer 1


Delta-correlation is just an approximation. The actual forces that they represent are not truly delta-correlated. However, typical atomic-scale force autocorrelations last ~ 1 picosecond, so it's a pretty good approximation.

EDIT. To clarify, imagine dividing time up into tiny slices (~ 1 ps). The stochastic force experienced by the particle at time $t_i$ will be random with a mean of zero (i.e. Eq. 3.3). At the next time slice, $t_{i+1}$, a different stochastic force will be acting also with a mean zero. If the collisions that cause these two forces are independent then their product must have zero mean. This follows from the fact that when two random variables $X$ and $Y$ are independent of each other, the expectation of their product is the product of their means, $$ \langle XY\rangle=\langle X\rangle\langle Y\rangle $$ or, in this instance, $$ \langle F_a(t_i)F_a(t_{j\neq i}) \rangle = \langle F_a(t_i)\rangle\langle F_a(t_{j\neq i})\rangle=0 $$ See the Expectation of product of random variables.

  • $\begingroup$ Sorry maybe I hadn't express myself very clearly.Delta-function is not what I am confused with, but how to derive it from more fundamental condition, such as "impacts are independent" in which the mathematics mainly catch my thought. Could you offer some hints over that?Thank you so much. $\endgroup$
    – Yan
    Aug 4, 2016 at 0:39
  • $\begingroup$ @Yan I've updated my answer, let me know if it's still unclear. $\endgroup$
    – lemon
    Aug 4, 2016 at 6:02
  • $\begingroup$ Indeed, your answer after edit has improved so much ,yet still hadn't shown where delta-function could come from……right? $\endgroup$
    – Yan
    Aug 4, 2016 at 6:32
  • 1
    $\begingroup$ @Yan The delta function comes from the fact that autocorrelation of the noise is so very short-lived. Naturally this needs not always be so. In general if you take a fully described system without a stochastic component, using the Mori-Zwanzig procedure you can drop the fast degrees of freedom into a stochastic variable, often with autocorrelation (i.e. generalized Langevin equation), and when this correlation time is short, you get Langevin equation with the delta function. Look up the derivation of the classical Caldeira-Leggett model for a concrete example (e.g. Tuckerman's book/notes). $\endgroup$
    – alarge
    Aug 5, 2016 at 8:46
  • $\begingroup$ Via your comment, I get a general framework that I should work on .But still some details I am pretty not sure:did "the fast degrees of freedom" is similar to the decription "the fast variable"?(Newcomer……you know) $\endgroup$
    – Yan
    Aug 5, 2016 at 9:54

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