Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
This seems to be a misunderstanding. Autocorrelation function is derived from the correlation between $y(t)$ and $y(y+\Delta t)$ and this autocorrelation is a function of the lag $\Delta t$. It is a pair correlation function between two things, namely the value of a function and the value of a function which is obtained from the first one by a shift in $t$. If you ask about "pair correlations", what are the two things in the pair? Obviously, if something in the pair is something completely different than the original function or its shifts, the pair correlation will have nothing to do with.. – Luboš Motl Apr 13 '12 at 14:39
...with the autocorrelation of the function we started with. – Luboš Motl Apr 13 '12 at 14:39
If your pairs are those in the "radial correlation function", there won't be any relationship, either. The autocorrelation is about the delaying in time while the radial correlation function is about shifting the objects in the radial direction. Those are different variables in which objects may be moved so the correlation functions won't be related, either. – Luboš Motl Apr 13 '12 at 15:03
@Luboš Motl, There is a typo in your comment. You meant "...the correlation between $y(t)$ and $y(t+\Delta t)$..." – Vijay Murthy Apr 13 '12 at 15:29
@TopekaCanvas, What is the variable whose auto-correlation you have in mind? The radial distribution function can be related to the spatial auto-correlation function of the density. Is that what you are asking for? – Vijay Murthy Apr 13 '12 at 15:32
up vote 1 down vote accepted

You define the density auto-correlation function as $$S_{\rho\rho} = \langle \delta \rho(\mathbf{x}_1) \delta \rho(\mathbf{x}_2)\rangle$$ where $\delta \rho(\mathbf{x}) = \rho(\mathbf{x}) - \langle \rho(\mathbf{x}) \rangle$ is deviation from the local mean value.

The Fourier transform of $S_{\rho\rho}$ is related to the structure-factor $$S(\mathbf{q}) = \langle \rho \rangle^2 (2\pi)^d \delta(\mathbf{q}) + \frac{1}{V}\int d^d x_1 d^d x_2 e^{-i \mathbf{q} \cdot (\mathbf{x}_1-\mathbf{x}_2)} S_{\rho\rho}$$ where $\langle \rho \rangle$ is the average density of the whole system, i.e., $V \langle \rho \rangle = \int d^d \mathbf{x} \, \rho(\mathbf{x})$.

The structure factor $S(\mathbf{q})$ is related to the pair-correlation function $g(\mathbf{x})$ via $$S(\mathbf{q}) = \langle \rho \rangle \Big[1 + \langle \rho \rangle \int d^d \mathbf{x} \, g(\mathbf{x}) e^{-i \mathbf{q} \cdot \mathbf{x}} \Big]$$

If the system is isotropic, then $g(\mathbf{x}) = g(|\mathbf{x}|)$ is called the radial-distribution function.

Most of these relations are already in the wikipedia page linked in the question.

share|cite|improve this answer
Vijay, Qmechanic edited the question to add the Wikipedia links. – Peter Morgan Apr 13 '12 at 18:00
Thank you very much Vijay! How would you go about isolating either function? I've been trying ntegration by parts to isolate $S_{\rho\rho}$ – TopekaCanvas Apr 13 '12 at 18:43
What do you mean by isolating either function? Do you meant how $S_{\rho\rho}$ and $g$ are related in real-space? – Vijay Murthy Apr 13 '12 at 20:00
If I had an expression for $g(r)$, I was wondering if there is a way to calculate $S_{pp}$ without resorting to doing it numerically? – TopekaCanvas Apr 13 '12 at 20:35
You cannot back out $S_{\rho\rho}$ since, as I said, $g(r)$ is for an isotropic system. – Vijay Murthy Apr 14 '12 at 9:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.