# Why is number of collisions equal to number of scattered particles times number of scatterers?

In Landau, Lifshitz "The Classical Theory of Fields" $\S12$ "Invariant cross-section" the following is said1:

... according to the usual definition of the cross-section $\sigma$, the number of collisions occuring in volume $dV$ in time $dt$ is $$d\nu=\sigma v_{\text{rel}}n_1n_2dVdt,$$ where $v_{\text{rel}}$ is the velocity of particle 1 in the rest system of particle 2 ...

I'm trying to make sense of this formula, and here's what I got so far: $\sigma v_{\text{rel}}n_1dt$ is total number of scattered particles in time $dt$, and $n_2dV$ is total number of scattering centers.

Why isn't total number of scattered particles equal to the number of collisions? Is it considered that there's more than one collision per scattered particle? If yes, then why should it appear that each scattered particle collides with each of scattering centers (only in this case would we just multiply by number of scattering centers)?

1. In English version which I cite the formula lacks a $dV$ factor. I put it here according to original Russian book.

• "$\sigma v_{\text{rel}}n_1dt$ is total number of scattered particles in time $dt$" Why do you think that? As you say, total number of scattered particles in time $dt$ in volume $dV$ equals number of collisions during that time there - only one collision per particle is considered. Jun 12 '14 at 22:23
• Because, as I understand it, $n_1 v_{\text{rel}}=n$ is number of particles passing in unit time through unit area of the beam cross-section, and then $\sigma n=N$ is number of particles scattered per unit of time through angles between $0$ and $\pi$ (i.e. all scattered particles), then multiplying this by $dt$ will give us total number of scattered particles. (Here I'm using variable names from "Mechanics" $\S18$). Jun 13 '14 at 6:56
• $\sigma n$ is number of particles scattered per unit of time per unit volume $by~one$ particle of type 2. To obtain $total$ number of particles scattered per unit time per unit volume, we have to multiply by number of scatterers in $dV$, which is $n_2dV$. Jun 13 '14 at 7:06
• This is because the cross-section $\sigma$ is meant to be "sensitive area per scatterer", which when it is hit by the particle of type 1, scatters it. Jun 13 '14 at 7:07
• Aaah, so $\sigma$ is a value for a single scattering center! It seems I now understand. This then explains why I got $\text{items}^2$ as the dimension of $d\nu$ instead of $\text{items}$, which, although is a non-dimensional thing, still made me bother. I think you could put this into an answer so that this question wasn't left in unanswered list. Jun 13 '14 at 8:09

As you say, total number of scattered particles in time $dt$ in volume $dV$ equals number of collisions during that time there - only one collision per particle is considered.