I am having some trouble understanding and interpreting the noise term in the Langevin equation for a colloidal particle in a fluid. By the Langevin model, I mean the following model as the equation of motion of a colloidal particle in a fluid:

\begin{equation} m \frac{d^2 x}{dt^2}= - \gamma v + \eta (t) \end{equation}

where $x$, $v$ are the position and velocity of the particle respectively. The constant $\gamma$ depends on the size of the object and the viscosity of the medium, and $\eta (t)$ is 'Gaussian White noise' and is a stochastic process.

The autocorrelation function of $\eta(t)$ is written as: \begin{equation} <\eta(t) \eta(t+\tau)>=k\delta(\tau) \end{equation}

What I understand from the term 'Gaussian noise' is that the function $\eta(t)$ will take random values at any time $t$, corresponding to a Gaussian distribution with zero mean and some variance, and the value of $\eta(t)$ at any other time is independent and identically distributed (identical to the distribution at t).

Is my understanding of the noise term correct? Or is the value of the noise term impulsive at every instant of time, unlike what I have described in the above paragraph? (The autocorrelation function of the noise term seems to indicate so).

If the noise term is impulsive in nature, then why is it called Gaussian noise? On the other hand, if the noise term is finite valued at every instant of time, then how can the autocorrelation function, which is the expectation value of the function taken at two instants of time be impulsive in nature?

  • $\begingroup$ "...and the value of $\nu(t)$ at any other time is independent and identically distributed (identical to the distribution at t)." Can you explain where this statement comes from? I don't think $\nu$ is defined anywhere in the post. Is it supposed to be $v$? If so, I'm still not sure I understand where that statement is coming from. $\endgroup$
    – DanielSank
    Commented Apr 28, 2016 at 18:30
  • $\begingroup$ @DanielSank Sorry, I meant $\eta(t)$ instead of $\nu(t)$. And this is what I understand the force to be; not what I have read from some source. Thanks for correcting me. $\endgroup$
    – Harsha
    Commented Apr 28, 2016 at 18:32

1 Answer 1


The name "Gaussian noise" actually has to do with the higher order correlations in the noise, such as: $$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \rangle, $$ $$\langle \eta(t) \eta(t+\tau_1) \eta(t+\tau_2) \eta(t+\tau_3) \rangle, $$ and so on.

If the noise is Gaussian then all of these higher order correlations can be rewritten in terms of the two-term correlation function, as expressed in Isserlis' theorem a.k.a. Wick's theorem. There is some subtelty here due to your stationary process being continuous-time but there should be some similar expression in terms of products of delta functions, I suppose.

As to whether the $\eta(t)$ is finite, no, it is not finite. However if you smooth it with any nonzero smoothing timescale, then the result will be finite. From a frequency-domain point of view, if you look at the fourier transform of a segment of $\eta(t)$ over a small time interval, it will have finite fourier components but they will extend up to infinite frequency.

  • $\begingroup$ Thank you for answering. So is the value taken by $\eta(t)$ finite valued at every instant of time, or is it impulsive? $\endgroup$
    – Harsha
    Commented Apr 29, 2016 at 8:39
  • $\begingroup$ Ah, I missed that part of your question; I'll edit my answer. $\endgroup$
    – Nanite
    Commented Apr 29, 2016 at 12:05

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