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In Kinetic Theory, one studies the evolution of a system of $N$ particles interacting with each other. We use the notation $\boldsymbol{w}_{i}$ to describe the coordinates in phase-space of each particle. Liouville's Equation gives the incompressible evolution of the $N-$Body distribution function

$$f^{(N)} (\boldsymbol{w}_{1},...,\boldsymbol{w}_{N})$$

One can then define the reduced distribution function $f_{i}$ , for $1 \leq i \leq N$ as

$$f_{i} (\boldsymbol{w}_{1},...,\boldsymbol{w}_{i}) = \int d \boldsymbol{w}_{i+1} ... \boldsymbol{w}_{N} \; f^{(N)} (\boldsymbol{w}_{1},...,\boldsymbol{w}_{N}) $$

A system is said to be separable, if we have $f^{(N)} (\boldsymbol{w}_{1},...,\boldsymbol{w}_{N}) = \prod\limits_{i = 1}^{N} f_{1} (\boldsymbol{w}_{i})$.

In order to measure the $\textit{distance}$ to separability, one first introduces the two-body correlation function $g_{2} (\boldsymbol{w}_{1} , \boldsymbol{w}_{2})$ defined as

$$f_{2} (\boldsymbol{w}_{1},\boldsymbol{w}_{2}) = f_{1} (\boldsymbol{w}_{1}) \, f_{1} (\boldsymbol{w}_{2}) + g_{2} (\boldsymbol{w}_{1},\boldsymbol{w}_{2}) $$

This two-body correlation function plays a crucial role in the collision term appearing in the first equation of the BBGKY hierarchy.

My question is : What is the definition of the three-body correlation function $g_{3} (\boldsymbol{w}_{1},\boldsymbol{w}_{2},\boldsymbol{w}_{3})$ ?

My feeling would be that it is defined as

$$ \begin{aligned} f_{3} (\boldsymbol{w}_{1},\boldsymbol{w}_{2},\boldsymbol{w}_{3}) = &f_{1} (\boldsymbol{w}_{1}) \, f_{1} (\boldsymbol{w}_{2}) \, f_{1} (\boldsymbol{w}_{3}) \\& + \; f_{1}(\boldsymbol{w}_{1}) \, g_{2} (\boldsymbol{w}_{2},\boldsymbol{w}_{3}) + f_{1} (\boldsymbol{w}_{2}) \, g_{2} (\boldsymbol{w}_{1},\boldsymbol{w}_{3}) + f_{1} (\boldsymbol{w}_{3}) \, g_{2} (\boldsymbol{w}_{1},\boldsymbol{w}_{2}) \\& + \; g_{3} (\boldsymbol{w}_{1},\boldsymbol{w}_{2},\boldsymbol{w}_{3}) \end{aligned}$$

Is this definition correct ?

Then, would you have a nice expression for the general $n-$body correlation function $g_{n} (\boldsymbol{w}_{1},...,\boldsymbol{w}_{n})$ ?

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Yes, your three-body expression is consistent with the BBGKY hierarchy. Regarding $n$-body correlations, there may be a way to express the collision integrand in terms of $g$ and $f$ functions, but I do not know it and can not remember ever seeing one. That said, I'll heartily upvote any answer that manages to write it out.

Most treatments of the problem opt for a diagrammatic representation, which tends to communicate the ideas behind many-body collision integrals in a more compact (if initially difficult-seeming) notation. I found this note to be a digestible introduction to this approach. I can provide additional references in comments if this turns out to be what you are looking for.

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  • $\begingroup$ Thanks for this reference, that I will read with great attention. As I come from a background with very few quantum notions, I tend to be scared by diagrammatic extensions, but it definitely seems to be worth the effort. Regarding the general expression of the n-body correlation function $g_{n}$, I came up with my own writing, which I will write in an answer to my question. However, the expressions tend to be rather complicated. $\endgroup$ – jibe Nov 5 '14 at 9:37
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In order to generalize the nice writing of the $3-$body correlation function, we may define recursively the $n-$body correlation function $g_{n}$ as

$$ f_{n} (1,..,n) = \sum_{k=0}^{n} \frac{1}{k!(n\!-\!k)!} \sum_{\sigma \in \mathcal{S}_{n}} f_{1} (\sigma [1]) \, ... \, f_{1} (\sigma [k]) \, g_{n-k} (\sigma [k\!+\!1],...,\sigma [n]) \, , $$ where $\mathcal{S}_{n}$ is the set of all permutations of ${1,...,n}$. The prefactor in ${ 1/ (k! (n\!-\!k)!) }$ ensures the correct normalization of the function. Indeed, we rely on the fact the correlation functions $g_{k}$ are invariant under the permutations of its arguments. In the sum $\sum_{k=0}^{n}$, the case ${k=n}$ corresponds to the case of separability for which ${f_n = f_{1} ...f_{1}}$. As $k$ decreases, we are adding higher order correlation terms.

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