I was reading Theory of Simple Liquids, when I came across BBGKY hierarchy. In deriving the expression for the hierarchy, they integrate an integration of N variables over N-n variables to make the corresponding function of N variables the same function but with reduced number of variables n. I am quite unable to follow the argument, and after going through many different references with different notations and all, I have completely messed up my understanding. It will be great if someone can tell me what exactly are we doing in the derivation of BBGKY?


1 Answer 1


There's really two questions here, one about the definition of distribution functions and one about the derivation of the BBGKY hierarchy. I will address them in turn.


Let's define, for convenience, $\mathbf{r}^n = \mathbf{r}_1,\dots,\mathbf{r}_n$ and $\mathbf{r}^{(N-n)} = \mathbf{r}_{n+1},\dots,\mathbf{r}_N$.

Next, let's denote the phase-space probability density $f^{[N]}(\mathbf{r}^N, \mathbf{p}^N; t)$. This means that $f^{[N]}\mathrm{d}\mathbf{r}^N\mathrm{d}\mathbf{p}^N$ is the probability of having the system within a small volume $\mathrm{d}\mathbf{r}^N\mathrm{d}\mathbf{p}^N$ centered around the phase-space point $(\mathbf{r}^N, \mathbf{p}^N)$. In other words, it is the probability of having each particle $i$ within the volume $\mathrm{d}\mathbf{r}_i\mathrm{d}\mathbf{p}_i$ centered at $(\mathbf{r}_i, \mathbf{p}_i)$.

If we integrated over $\mathbf{r}_1$ and $\mathbf{p}_1$, say, we would have a very similar quantity $f^{[N-1]}\mathrm{d}\mathbf{r}^{(N-1)}\mathrm{d}\mathbf{p}^{(N-1)}$ for the remaining particles. After all, all we did was to stop tracking particle $1$.

Right, this was elementary, so let's move to the reduced phase-space distribution function $f^{(n)}(\mathbf{r}^n, \mathbf{p}^n; t)$. Now, $f^{(n)}$ does not label the particles the same way as $f^{[N]}$. The motivation is instead that if we see $n$ particles, we don't really care if they are the particles $1, \dots, n$ or those with labels $N-n+1, \dots, N$ for we are assuming that they are all identical. So we define $f^{(n)}\mathrm{d}\mathbf{r}^{n}\mathrm{d}\mathbf{p}^{n}$ to give the probability of having $n$ particles (of a total of $N$) in the phase-space volume $\mathrm{d}\mathbf{r}^{n}\mathrm{d}\mathbf{p}^{n}$. In other words, the probability of having a molecule (any molecule) in each of the $n$ elements of the phase-space $\mathrm{d}\mathbf{r}_i\mathrm{d}\mathbf{p}_i$ centered at $(\mathbf{r}_i, \mathbf{p}_i)$.

So, there are $N$ choices for the molecule at $\mathbf{r}_1$, $N-1$ choices for $\mathbf{r}_2$ and so on up to $n$. This is to say that $f^{(n)}(\mathbf{r}^n, \mathbf{p}^n; t) = N(N-1)\cdots(N-n+1)f^{[n]}(\mathbf{r}^n, \mathbf{p}^n; t)$, which we can write in the more conventional $$f^{(n)}(\mathbf{r}^n, \mathbf{p}^n; t) = \frac{N!}{(N-n)!}\int\!\!\!\!\int f^{[N]}(\mathbf{r}^N, \mathbf{p}^N; t) \mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}$$

BBGKY Hierarchy

BBGKY hierarchy then follows by inspecting a system of $N$ identical particles in a potential with only 1- and 2-body forces (i.e. an external force and pair potentials: No angulars, dihedrals or any of the like here, these are typically intramolecular in nature anyway). The equilibrium version of the hierarchy is slightly easier to derive (starting from differentiating over one position $f^{[N]} = \frac{e^{-\beta\mathcal{H}}}{Z}$ and then integrating over a part of the phase-space), but here I'll write out the more general version starting with the Liouville equation:

$$\left(\frac{\partial}{\partial t} + \sum_{i=1}^N \frac{\mathbf{p}_i}{m} \cdot\frac{\partial}{\partial \mathbf{r}_i} + \sum_{i=1}^N \mathbf{X}_i \cdot\frac{\partial}{\partial \mathbf{p}_i}\right) f^{[N]} = -\sum_{i=1}^N \sum_{j=1}^N \mathbf{F}_{ij} \cdot \frac{\partial f^{[N]}}{\partial \mathbf{p}_i}$$

where $\mathbf{F}_{ij}$ is the force between particles $i$ and $j$ and $\mathbf{X}_{i}$ the external force on particle $i$. Integrating over the particles $n+1, \dots N$, $$\int\!\!\!\!\int\left(\frac{\partial}{\partial t} + \sum_{i=1}^N \frac{\mathbf{p}_i}{m} \cdot\frac{\partial}{\partial \mathbf{r}_i} + \sum_{i=1}^N \mathbf{X}_i \cdot\frac{\partial}{\partial \mathbf{p}_i}\right) f^{[N]}\mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)} = -\sum_{i=1}^N \sum_{j=1}^N \int\!\!\!\!\int \mathbf{F}_{ij} \cdot \frac{\partial f^{[N]}}{\partial \mathbf{p}_i} \mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}$$

All the terms where $i>n$ vanish as they are over differentiated quantities (the probability of finding a particle outside the box or at infinite velocity has to be zero; think of the fundamental theorem of calculus), so

$$\left(\frac{\partial}{\partial t} + \sum_{i=1}^n \frac{\mathbf{p}_i}{m} \cdot\frac{\partial}{\partial \mathbf{r}_i} + \sum_{i=1}^n \mathbf{X}_i \cdot\frac{\partial}{\partial \mathbf{p}_i}\right) f^{[n]} = -\sum_{i=1}^n\sum_{j=1}^N \int\!\!\!\!\int \mathbf{F}_{ij} \cdot \frac{\partial f^{[N]}}{\partial \mathbf{p}_i} \mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}$$ Note the inner sum on the right hand side still goes over all $N$ particles, so we separate it into its own term and write

$$\left(\frac{\partial}{\partial t} + \sum_{i=1}^n \frac{\mathbf{p}_i}{m} \cdot\frac{\partial}{\partial \mathbf{r}_i} + \sum_{i=1}^n \mathbf{X}_i \cdot\frac{\partial}{\partial \mathbf{p}_i}\right) f^{[n]} = -\sum_{i=1}^n\sum_{j=1}^n \int\!\!\!\!\int \mathbf{F}_{ij} \cdot \frac{\partial f^{[N]}}{\partial \mathbf{p}_i} \mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)} - \sum_{i=1}^n\sum_{j=n+1}^N \int\!\!\!\!\int \mathbf{F}_{ij} \cdot \frac{\partial f^{[N]}}{\partial \mathbf{p}_i} \mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}$$

The pair force term can be pulled out of the left integral (as $i, j$ do not coincide with the integration variables) and the rest of the term may be replaced with a $f^{[n]}$. The sum over $j$ in the right integral, on the other hand, is really just $(N-n)$ terms repeating because $f^{[N]}$ is symmetric with respect to interchange of variables. So we have

$$\left(\frac{\partial}{\partial t} + \sum_{i=1}^n \frac{\mathbf{p}_i}{m} \cdot\frac{\partial}{\partial \mathbf{r}_i} + \sum_{i=1}^n \left(\mathbf{X}_i+\sum_{j=1}^n\mathbf{F}_{ij}\right) \cdot\frac{\partial}{\partial \mathbf{p}_i}\right) f^{[n]} = -(N-n)\sum_{i=1}^n \int\!\!\!\!\int \mathbf{F}_{i, n+1} \cdot \frac{\partial f^{[N]}}{\partial \mathbf{p}_i} \mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}$$

which finally can be written as (sorry for the long derivation, but I'm not sure if it can be found in this much detail elsewhere, I hope this helps),

$$\left(\frac{\partial}{\partial t} + \sum_{i=1}^n \frac{\mathbf{p}_i}{m} \cdot\frac{\partial}{\partial \mathbf{r}_i} + \sum_{i=1}^n \left(\mathbf{X}_i+\sum_{j=1}^n\mathbf{F}_{ij}\right) \cdot\frac{\partial}{\partial \mathbf{p}_i}\right) f^{[n]} = -(N-n)\sum_{i=1}^n \int\!\!\!\!\int \mathbf{F}_{i, n+1} \cdot \frac{\partial f^{[n+1]}}{\partial \mathbf{p}_i} \mathrm{d}\mathbf{r}_{n+1}\mathrm{d}\mathbf{p}_{n+1}$$

Now we can write $f^{(n)}$ in place of $f^{[n]}$ (and get rid of the ugly factor $(N-n)$), as is often done, but this is not necessary.


You mentioned the Theory of Simple Liquids, which is a modern classic and a good book, particularly for reference. I personally, however, often find that Hill's work is more approachable (be it his research articles, or in this case the book Statistical Mechanics; available through Dover i.e. dirt cheap), although it must be said that the notation he uses is somewhat old and so somewhat clumsy.

  • $\begingroup$ Just one point: In the third step for BBGKY, how did you use the fact that the probability of finding a particle outside the box and that of having infinite velocity is zero to change the limit of the sum? Also, how does the Fundamental Theorem of Calculus help us in this? Thanks a lot, @alarge! $\endgroup$
    – Shuppar
    Dec 2, 2014 at 8:14
  • 1
    $\begingroup$ @Shuppar $\int\!\!\int\sum_{i=1}^N\frac{\mathbf{p}_i}{m}\cdot\frac{\partial}{\partial \mathbf{r}_i} f^{[N]}\mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}=\int\sum_{i=1}^N \frac{\mathbf{p}_i}{m}\cdot\int\frac{\partial}{\partial\mathbf{r}_i} f^{[N]}\mathrm{d}\mathbf{r}^{(N-n)}\mathrm{d}\mathbf{p}^{(N-n)}$. Now $\int\frac{\partial}{\partial\mathbf{r}_i} f^{[N]}\mathrm{d}\mathbf{r}_i=(f^{[N]})_{\mathbf{r}_i=-\infty}^{\infty}=0$ (for at infinity, i.e. outside the box, the probability distribution must vanish). Similarly for the other terms where you integrate over $\mathbf{p}_i$. $\endgroup$
    – alarge
    Dec 2, 2014 at 9:27

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