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I am reading a paper that deals with the Boltzmann equation. They add a collision which is supposed to account for collisions which happen when particles are within a radius of $d$ from each other. The authors then state that the collision term is something along the lines of (2D): $$\left( \frac{\partial f}{\partial t} \right)_{\text{coll}} = 2d \, f(x,v',t) \int dv' |\textbf{v} - \textbf{v}'| \, f(x,v',t)$$

What I don't understand is why the two function $f$ are evaluated at the same $x$. Shouldn't we integrate over a circle of radiud $d$? Why are we integrating over the velocity instead?

The authors state something along the lines of: In the reference system of particle 1, particle 2 must lie within a rectangle of width $2d$ and length $|\textbf{v}_1 - \textbf{v}_2| \, dt$ for a collision to happen...

Any help would be much appreciated.

By the way, I have changed the notation a bit, but the paper is this: http://arxiv.org/pdf/0907.4688.pdf

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  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0907.4688 $\endgroup$ – Qmechanic Mar 12 '14 at 17:20
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(Full disclosure, I didn't RTFA, and I don't have time to.)

Just to review, $f\left(x,v,t\right)dxdv$ is the number of particles with positions between $x$ and $x+dx$ and velocities between $v$ and $v+dv$.

First, why is there no integral over position? In principle, there should be. However, assuming that $d$ is small, $f\left(x,v,t\right)\approx f\left(x+d,v,t\right)$; I.e. $f$ shouldn't change much over small distances. If that's not the case, you probably shouldn't use the Boltzmann Transport Equation.

My guess is that the collision term has the form of "number of collisions" = "number of particles at x, v" * "number of particles those particles can collide with". The former is just $f\left(x,v,t\right)$. The latter is that integral.

Say two particles have the same x and v; then they're moving in parallel and won't collide. (Even if they did collide, it wouldn't change anything.) Suppose their velocities are just a little different. Collisions will happen, but they'll be infrequent. Think about a car chase where one car is moving just a little faster than the other. The faster car will catch the slower one, but it'll take longer, so fewer chases will end in a given time period. If the difference in velocities was large, the catch will happen much more quickly; particles with very different velocities collide much more often. (Just to be clear, I really do mean velocity and not speed; direction is important.)

That's what the $\left|v-v^\prime\right|dt$ term encapsulates. If the velocities are the same, it's zero. If the velocities are very different, it's large. You can get that term by rigorous means if you'd like. Combine that with the number of particles with velocity $v^\prime$, integrate over all possible $v^\prime$, and you have the "number of particles those particles can collide with".

I think that's the gist of it anyway.

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