# Reason for ignoring the collisional integral in the BBGKY hierarchy $f_2$?


EDIT

As suggested in the comments - here is a definition of the terms in this equation:

• $\vec q_i$ is the position of the $i$th particle.
• $\vec p_i$ is the momentum of the $i$th particle.
• $U(|\vec q_i-\vec q_j|)$ is the interaction potential between the $i$th and the $j$th particles.
• $f_n(\vec q_1,...,\vec q_n,\vec p_1,...,\vec p_n)$ (argument dropped in above) is the n-particle distribution function (i.e. the probability that you would find any arbitrary $n$ particles with position $\vec q_1,...,\vec q_n$ and momenta $\vec p_1,...,\vec p_n$).
• $dV_3$ is a phase space volume, $dV_3=d\vec q_3 d\vec p_3$.

In the Boltzmann picture of collisions, the interaction $U$ is short-ranged, meaning that if $|\vec q_i - \vec q_j| > d$, then $U(|\vec q_i-\vec q_j|)\approx 0$. A consequence of this is that the three-particle collision integral will only be nonzero when a third particle happens to be within a distance $d$ of particles 1 or 2. Now we must keep in mind that the ultimate goal here is to derive an expression for the two-particle collision integral in the first BBGKY equation, $$\left( \frac{\partial }{\partial t} + \frac{\vec p_1}{m}\cdot\frac{\partial }{\partial \vec r_1}\right) f_1 = \int dV_2 \frac{\partial U(|\vec q_1-\vec q_2|)}{\partial q_1}\cdot\frac{\partial }{\partial \vec p_1} f_2$$ Whatever $f_2$ emerges from solving the two-particle BBGKY equation will end up in the RHS integral above. This integral itself is only nonzero when the colliding particles 1 and 2 are within $d$ of one another. So the three-particle term in the second BBGKY equation only affects the one-particle distribution when all three particles are within $d$ of one another. Neglecting the integral over $f_3$ is an assumption that such interactions with a third particle are extremely rare, which is reasonable for dilute gases.