When deriving the Boltzmann equation from the BBGKY hierarchy one of the starting points is the equation: $$ \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} \newcommand{\braoket}[3]{\left<#1\right|#2\left|#3\right>} \l \p{}{t}+\f{\vec{p}_1}{m}\cdot \p{}{\vec{q}_1}+\f{\vec{p}_2}{m}\cdot \p{}{\vec{q}_2}-\p{U(|\vec{q}_1-\vec{q}_2|)}{\vec{q}_1}\cdot \l \p{}{\vec{p}_1}-\p{}{\vec{p}_2}\r\r f_2$$ $$=\int dV_3\l \p{U(|\vec q_1-\vec q_3|)}{\vec q_1}\cdot \p{}{\vec p_1}+\p{U(|\vec{q}_2-\vec{q}_3|)}{\vec{q}_2}\cdot \p{}{\vec p_2}\r f_3$$ The term on the RHS is called the collisional integral and scales as $nd^3/\tau_c$ compared to the last term on the LHS which scales as $1/\tau_c$. Due to this scaling in the derivation of the Boltzmann relation the RHS is often set to zero. I understand completely the mathematics behind this but not the physics. So please can someone explain the physics of why the RHS (concerning collisions with other particles in the system) can be neglected compared to the last term on the LHS (concerning collisions between the two particles under consideration).


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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