Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Derivation of Higher-order correlation functions from definition

+ 3 like - 0 dislike
2929 views

I'm trying to understand the definition of the n-th order correlation function. My aim is to translate the math into a numerical implementation in order to compute the correlation function $g^{(n)}$ for a distribution of spherical particles with $n=\lbrace3,4,5,\ldots\rbrace$.

I'm going to present you what I have understood for now, in order for you to get an better insight of what I'm looking for. (To reduce the verbiage of the following, I over-abbreviated sometimes $\mathbf r_1, \mathbf r_2$ into $r$, sorry if it kills the readability)

I'm used to the radial distribution (or pair distribution function) that we call $g(r)$ (which should be called rigorously $g^{(2)}(r)$). I want to derive the expression of $g^{(2)}(r)$ from the definition of $g^{(n)}$ (because I hope that if I'm able to do it for $n=2$, I will be able to do it for any $n$ !)

So let's start with the definition of $g^{(n)}$ one can find in wikipedia or any stat mech book :
$$\rho^{(n)}(\mathbf{r}_{1}\, \ldots, \, \mathbf{r}_{n}) = \rho^{n}g^{(n)}(\mathbf{r}_{1}\, \ldots, \, \mathbf{r}_{n}) \, ,$$ with $\rho$ the particle number density and $\rho^{(n)}$ the probability of (all the permutations of) elementary configuration $(\mathbf{r}_{1}\, \ldots, \, \mathbf{r}_{n})$ : $$ \rho^{(n)}(\mathbf{r}_{1}\, \ldots, \, \mathbf{r}_{n}) = \frac{N!}{(N-n)!} \frac{1}{Z_N} \int \cdots \int \mathrm{e}^{-\beta U_N} \, \mathrm{d} \mathbf{r}_{n+1} \cdots \mathrm{d} \mathbf{r}_N \, .$$ with $U_N(\mathbf{r}_{1}\, \ldots, \, \mathbf{r}_{N})$ the potential energy of the configuration and $Z_N$ the configurational integral, taken over all possible combinations of particle positions.

Rewriting the previous with $n=2$ leads to :

$$ \rho^{(2)}(\mathbf{r}_{1}\, \mathbf{r}_{2}) = N(N-1) \frac{1}{Z_N} \int \cdots \int \mathrm{e}^{-\beta U_N} \, \mathrm{d} \mathbf{r}_{3} \cdots \mathrm{d} \mathbf{r}_N \, .$$

From that starting point, I should be able to derive the expression for $g^{(2)}(\mathbf{r}_{1}\, \mathbf{r}_{2})$ but I have no clue about the origin of the Dirac Delta function that appears in the definition of $g^{(2)}(\mathbf{r}_{1}\, \mathbf{r}_{2})$ or $\rho^{(2)}(\mathbf{r}_{1}\, \mathbf{r}_{2})$

$$ \rho^{(2)}(\mathbf{r}_{1}\, \mathbf{r}_{2}) = \left\langle \sum_i\sum_j' \delta(\mathbf r_1 - \mathbf r_i) \delta(\mathbf r_2 - \mathbf r_j) \right\rangle $$ I think there is something appearing from the potential energy that was defined before, but I'm not able to understand exactly the origin.

Any help will be highly appreciated :) Thank you in advance !

(Note : related questions like "Use and understanding of higher-order correlation functions" is absolutely not helpful, and the reference on the wikipedia page leads to a paper that has nothing to deal with correlations functions IM(H)O...)

This post imported from StackExchange Physics at 2014-06-08 08:17 (UCT), posted by SE-user Pascail
asked Jun 6, 2014 in Theoretical Physics by Pascail (15 points) [ no revision ]

1 Answer

+ 4 like - 0 dislike

The delta functions mean that you are fixing the values of any two of the $N$ vectors $r_k$. By permutation symmetry of the potential, you may take these always to be $r_1$ and $r_2$, and you get $N(N-1)$ times the same integral. Hence (using the definition of the expectation in terms of the canonical ensemble) your first expression for $\rho^{(2)}$ and your second expression say exactly the same.

This doesn't help you to evaluate the remaining integral, though. As $N$ is large, this is tough and can be done only approximately, using expansions such as those discussed in the statistical mechanics book by Reichl, which I'd recommend for further reading.

answered Jun 8, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

Thank you for the clarification. As far as I understand it for now, the Dirac delta function only light up for the right value of \((r_i,r_j)\)  and the expectation value \(\langle \cdot \rangle\)hides the normalized integral on all remaining \(r\). But at some point the "selected" values of the potential should come out from this expression ? In order to numerically compute this thing from a distribution of particles, don't we have to say at some point that the value of the potential doesn't really matter and can be $U_N = 0$ ? 

By similar construction, I would guess that the expression for \(\rho^{(3)}\) is 

\(\rho^{(3)}(r_1,r_2,r_3) =\langle \sum_i \sum_j'\sum_k'' \delta(r_i-r_1)\delta(r_j-r_2) \delta(r_k-r_3) \rangle\)

(not sure about the " convention, of $k$ not being equal to $i$ nor $j$ though ...) 

Anyway I'll give a look to the Reichl. For now, the Hansen McDonald (Theory of simple liquids) was far away beyond my capacities :) 

The formulas to notice are

$$\langle f(r_1,...,r_n)\rangle= const \int dr_1...dr_n e^{-\beta U(r_1,...,r_n) } f(r_1,...,r_n)$$

and 

$$\int dr_i dr_k f(r_i,r_k)\delta(r_i-r_k)= \int dr_i f(r_i,r_i)$$ 

This turns the $n$-fold integral for the canonical expectation values into an $(n-2)$-fold integral without delta-terms.

The values of the potential do matter, of course, and determine the value of the pair correlation function.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...