I am studying Brownian motion, specifically Langevin equation. This equation includes a force expressed by a white noise, say $\xi(t)$. One of the hypothesis is that it is $\delta$-correlated (since it is a white noise). This is usually written as: \begin{equation} \langle\xi(t)\xi(t+\tau)\rangle=\sigma^2\delta(\tau) \end{equation} Where $\delta$ is the Dirac delta. But I have problems with this "Dirac-delta".

Two deductions are made from the equation above. On one hand, we can compute the spectral density of $\xi(t)$: \begin{equation} S_\xi(\omega)=\int_{\infty}^{\infty} \langle\xi(t)\xi(t+\tau)\rangle e^{i\pi\omega\tau}= \int_{\infty}^{\infty} \sigma^2\delta(\tau)e^{i\pi\omega\tau} =\sigma^2. \end{equation} This result is clear, from the definition of $\delta$.

But, on the other hand, let evaluate the autocorrelation in $t=0$. This is, setting $\tau=0$, we have \begin{equation} \langle \xi(t)\xi(t+\tau) \rangle|_{\tau=0}=\sigma^2\delta(0) =\sigma^2. \end{equation}

Both results seem suitable , however, I don't understand these two definitions of $\delta$. Specifically, I don't understand how explain that first $\delta$ works as a true Dirac-$\delta$ while then it works like a indicator function. Strictly speaking, $\delta(0)=\infty$, isn't it?

Summarizing, I need a way to explain why both calculus are correct. Or, if they are not, an explanation.

  • $\begingroup$ Wiener-Khinchin theorem $\endgroup$ – DanielSank Feb 21 '15 at 17:41
  • $\begingroup$ I know this theorem. In fact, today I've been working on it. I know the integral equality stems from Wiener-Khinchin theorem. The problem is the third equality, when $\delta$ works as a indicator function. How can $\delta$ have two meanings at the same time? Or what is wrong in my explanation above? $\endgroup$ – Carles Pérez Feb 21 '15 at 17:51
  • $\begingroup$ I don't understand what you're asking. You have three lines of equations in your post. The first one more or less defines "white noise" as you sort of say yourself in the post. The second line uses the first one to do an integral. The third line has no $\delta$ in it at all, so I don't know what you mean when you say that in the third like $\delta$ acts as an "indicator function". All of that said, where did you get that third line? $\endgroup$ – DanielSank Feb 21 '15 at 18:00
  • $\begingroup$ Sorry, I should explained it better. I edited the third equation in order to make it more understandable. The question is: if $\xi(t)$ is $\delta$-correlated, then $C_\xi(\tau)=\sigma^2\delta(\tau)$. What happens with $C_\xi(0)$? $\endgroup$ – Carles Pérez Feb 21 '15 at 18:25
  • 1
    $\begingroup$ In your third line you have essentially assumed $\delta(0)=1$, which is false. $\endgroup$ – DanielSank Feb 22 '15 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.