The Liouville equation of motion is written in terms of an $N$ particle distribution $f_N$. \begin{equation} \frac{\partial f_N}{\partial t}=\{H,f_N\} \end{equation} Where $\{\cdot ,\cdot \}$ is the Poisson bracket and $f_N=f_N(q_1,\dots ,q_N,p_1,\dots ,p_N)$. Let us now define an $n$ particle probability distribution function $f_n$ with $n< N$. \begin{equation} f _n(q_1,\dots,q_n,p_1,\dots ,p_n,t)=\frac{N!}{(N-n)!}\int \prod ^N_{i=n+1}d q^id p_if_{N+1}( q_1,\dots ,q_{N+1},p_1 ,\dots,p_{N+1},t) \end{equation} Now $f_n$ satisfies, \begin{equation} \frac{\partial f _n}{\partial t}=\{H_n,f_n\}+\sum ^n_{i=1}\int dq_{n+1}dp_{n+1}\frac{\partial U(q_i-q_{n+1})}{\partial q_i}\frac{\partial f _{n+1}}{\partial p_i}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (*) \end{equation} With the $n$-body Hamiltonian $H_n$, \begin{equation} H_n=\sum ^n_{i=1}\bigg(\frac{p_i^2}{2m}+U(q_i)\bigg)+\sum _{i<j\leq n}U(q_{ij}) \end{equation} And $q_{ij}=q_i-q_j$. Here $(*)$ is the BBGKY hierarchy. I am reading out of the following notes.

Despite reading the linked notes and wikipedia pages etc I am struggling to understand how the BBGKY hierarchy is related to the Liouville equation. In particular taking $n=N$ does not (to my naive understanding) regenerate the Liouville equation. Why do we not require $f_{N+1}$ in the Liouville equation by the logic of the $n$ particle distribution function? Lastly is the Boltzmann equation defined for $f_N$ or $f_1$ (or is it irrelevant, the equation holding in any case?).

Any help on the BBGKY formalism is appreciated so much!


1 Answer 1


The Liouville equation for the $N$ particle system, describes the time evolution of the phase space N-particle probability density, which you can also neatly rewrite with the Liouville operator: $f^{N}(t)= e^{-iLt}f^{N}(0).$ Now almost always we're interested in a smaller subset of only $n$ particles, for which then we have to define a reduced phase space distribution function, obtained by integrating out the remaining degrees of freedom ($6(N-n)).$ So the reduced $f^n$ has the form: $$ f^n (\mathbf{r}^n,\mathbf{p}^n,t) = \frac{N!}{(N-n)!}\int f^N(\mathbf{r}^N,\mathbf{p}^N,t)d\mathbf{r}^{N-n}d\mathbf{p}^{N-n} $$ For our purposes here, to derive an equation of motion for $f^n,$ let's consider a simpler general Hamiltonian: $$ \frac{\partial H}{\partial \mathbf{r}_i}=-\mathbf{X}_i-\sum_{j=1}^{N}\mathbf{F}_{ij} $$ With $\mathbf{X}$ denotes the external forces due to an external field and $\mathbf{F}_{ij}$ a pairwise interparticle interaction. Inserting into the Liouville equation, for $f^N,$ with a slight rearrangement: $$ \left(\frac{\partial}{\partial t}+\sum_{i=1}^{N}\frac{\mathbf{p}_i}{m}\frac{\partial}{\partial\mathbf{r}_i}+\sum_{i=1}^N \mathbf{X}_i\frac{\partial}{\partial \mathbf{p}_i}\right)f^N=-\sum_{i=1}^{N}\sum_{j=1}^{N}\mathbf{F}_{ij}\frac{\partial f^N}{\partial \mathbf{p}_i} $$ Now by integrating out the unwanted $6(N-n)$ degrees of freedome, we obtain: $$ \left(\frac{\partial}{\partial t}+\sum_{i=1}^{n}\frac{\mathbf{p}_i}{m}\frac{\partial}{\partial\mathbf{r}_i}+\sum_{i=1}^n \mathbf{X}_i\frac{\partial}{\partial \mathbf{p}_i}\right)f^n=-\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbf{F}_{ij}\frac{\partial f^n}{\partial \mathbf{p}_i}- \underbrace{\frac{N!}{(N-n)!}\sum_{i=1}^n \sum_{j=n+1}^N \iint \mathbf{F}_{ij} \frac{\partial f^N}{\partial \mathbf{p}_i} d\mathbf{r}^{N-n}}_{(*)} d\mathbf{p}^{N-n} $$ Notice that all the terms with $i>n$ have vanished, to ensure a valid density function $f^n$ for the subspace. Working with identical particles here, we know $f^N$ is symmetric w.r.t. interchange of individual particle labels and the second sum in $(*)$ from $n+1$ to $N$ can be replaced by $N-n$ times the value of any one term (identical particles). With this simplification we can re-write the above equation as: $$ \left(\frac{\partial}{\partial t}+\sum_{i=1}^{n}\frac{\mathbf{p}_i}{m}\frac{\partial}{\partial\mathbf{r}_i}+\sum_{i=1}^n \left(\mathbf{X}_i +\sum_{j=1}^n \mathbf{F}_{ij}\right)\frac{\partial}{\partial \mathbf{p}_i}\right)f^n=-\sum_{i=1}^n \iint \mathbf{F}_{i,n+1} \frac{\partial f^{n+1}}{\partial \mathbf{p}_i} d\mathbf{r}_{n+1} d\mathbf{p}_{n+1} $$ We have derived the BBGKY hierarchy, for a relatively less general Hamiltonian. But note that it is an exact expression as it stands, linking the one particle phase space density to the two particle density which itself linked the three particle...all the way up to the N particle phase space density. Although such expression doesn't lend itself to much use in practice (for one thing, we don't know any of the $f^n$'s), it can become useful by providing some form of a closure relation, in other words without a closure relation, all BBGKY does is: expresses one unknown density function $f^n$ in terms of another unknown $f^{n+1}.$ It does not however represent an equivalent formulation of the Liouville equation for $f^N$ when you set $n=N$, it only gives you a link with the density function of subspaces of N, which you still don't know. If no closure is provided, BBGKY is not really useful! Have a look at the book of JP Hansen and IR McDonald for a more in depth coverage.

  • $\begingroup$ Brilliant answer! One quick follow up question, what is a closure relation? $\endgroup$
    – RedPen
    Commented Aug 26, 2015 at 10:40
  • 2
    $\begingroup$ @RedPen A closure relation would be one that removes the recursion in BBGKY, such that $f^n$ can be defined in terms of another function $g^n$ of the same subspace. To get yourself acquainted with such relations, have a look at the Hypernetted Chain Approximation which is a closure to the Ornstein-Zernicke relation, which relates the total correlation function to the direct corr. function such that we can eliminate one of the functions and define the other. $\endgroup$
    – Ellie
    Commented Aug 26, 2015 at 14:55
  • 2
    $\begingroup$ @RedPen I answered a question about a simple closure relation some time ago, which you may or may not find useful: physics.stackexchange.com/questions/149561 $\endgroup$
    – alarge
    Commented Aug 28, 2015 at 21:18
  • $\begingroup$ @alarge Thank you very much alarge, both your answers were extremely helpful indeed! I have asked a general question about how to use such formula here chemistry.stackexchange.com/questions/35633/… $\endgroup$
    – RedPen
    Commented Aug 29, 2015 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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