# What information is lost in the symmetrization necessary to derive the BBGKY hierarchy?

The book on Kinetic theory I'm reading derives the BBGKY hierarchy after introducing the reduced distribution functions

$f_s(q^1,p_1,q^2,p_2,\dots,q^s,p_s):=\int\ \rho\ \ \mathrm d q^{s+1} \mathrm d p_{s+1}\cdots \mathrm d q^N \mathrm d p_N,$

where $\rho$ is the phase space probability function. In fact, in the derivation, to get rid of most terms, he reasonably states that we must make the assumption that the $f_s$ are symmetric, i.e.

$\forall k.\ f_s(\dots,q^k,p_k,q^{k+1},p_{k+1},\dots)=f_s(\dots,q^{k+1},p_{k+1},q^k,p_k,\dots).$

The step from going from the particle trajectory $\pi$ (which solves the Hamiltonian equations of motions) to the distribution $\rho$ (which solves the continouty equation with the vector field, generated by multiple $\pi$-streamlines, as flow) can be a bit confusing. The fact that the author concerns himself two times with this and doesn't just say "we can clearly assume that.." suggests to me that it might not be so simple to put asside. He offers a symmetrization method $f_s\mapsto \frac{1}{s!}\sum_{\pi}f_s$, where the sum permutes the arguments.

Now firstly, I wonder how this works with multi-particle distribution functions. What do the equations look like for an $f_{s,t}$ which describe $s$ particle of one sort and $t$ particles of another sort?

Also, as this implies even the Boltzmann equation rests on this assumptions which therefore apparently isn't initially part of the Hamiltonian description - what information is lost in the process of symmetrization?

Step back a bit and look at the Hamiltonians you're considering. Let's assume the Hamiltonian is given in the form $$\mathcal{H}(\overline{p}, \overline{q}) = \mathcal{H}(p_1, q_1, \dots, p_N, q_N)$$ All this assumption of permutability is saying is that if I exchange any two $p_i, q_i$ with another $p_j, q_j$ I get the same Hamiltonian. Because $\rho$ satisfies the equation $$\frac{d}{d t} \rho = \frac{\partial}{\partial t} \rho + \{\mathcal{H}, \rho\}_{(\overline{p}, \overline{q})}$$ if the Hamiltonian has that symmetry, then $\rho$ by necessity has that same symmetry to satisfy the equation. So the statement is, in a sense, saying that there are no "special" particles in the system.

Now, if you have multiple species that are interacting, you can break the Hamiltonian into a sum of the individual species Hamiltonians, plus the interactions between the two species. The interchange relationship then will only happen on a species-by-species basis, so I can only permute $p_i, q_i$ for one species, but not with $P_j, Q_j$ of another. In practice, you handle this by factoring the distribution functions into the product of one species distribution with another, and having coupling terms proportional to the product of the two, viz. $$\rho = \rho_s(\overline{p}, \overline{q})\rho_t(\overline{P}, \overline{Q})$$ The virtue here is that if you have $$\mathcal{H} = \mathcal{H}(\overline{p}, \overline{q}) + \mathcal{H}(\overline{P}, \overline{Q}) + \mathcal{V}_{int}$$ if there is no interaction you have the densities evolving separately, and if there is an interaction you get the product and can integrate over, for example, the Coulomb interaction of ions with electrons.

I've not worked out the implications myself for this, but I should imagine you have to integrate over the $(p,q)$ and $(P,Q)$ spaces separately just fine, but must be careful if you ever try to permute a $p$ with a $P$, for instance. The result would be a coupled distribution function

The short answer then is that if the Hamiltonian treats the particles indistinguishable, so that they all interact the same way, there's no information loss.

If the particles are all the same (we don't need to invoke indistinguishability at that stage I think), then there is no loss of information since the permutation of two particles in a reduced distribution function will ask exactly the same question as before the permutation probability-wise.

Note that usually, reduced distribution functions are defined with combinatorial factors that take into account the fact that when you decide to trace over the degrees of freedom of $m$ particles you have $C^N_m$ ways to choose them.

I think that this is the only thing the author of your book is saying.

• Thanks for the answer. Well, the question is exactly about the question what scenarios there are in which $f$ isn't symmetric. I could imagine that all particles are the same, but the first and second slot of $f$ still describe them under different circumstances. And regarding the symmetrization factors, this should be just the $\frac{1}{s!}$ I wrote above, no? – Nikolaj-K Dec 10 '13 at 9:21
• If the particles are not the same then f is not symmetric but otherwise it should be. Think of it more as a constraint than an actual property it acquires at some point. – gatsu Dec 10 '13 at 12:06