Why is the typical momentum transfer in ultrarrelativistic Coulomb scattering of the same order as electron mass? I am reading the Course of Theoretical Physics, Vol. 4, by Beretevskii, Lifshitz and Pitaevskii and on $\S$ 39 there is a claim that the impact parameter that most contributes to the scattering is of the same order of the mass of the electron, i.e.,
$\rho \sim \frac{1}{m}$
I have been trying to find an explanation for this and I can't find it.
 A: I found the answer in $\S$93 which I was trying too read too fast and tired before.
I will post it here for future reference.
Verba volant, scripta manent.
It turns out that $\rho \sim \frac{1}{q} \sim \frac{1}{m}$ is not necessarily the case. This is just a regime that gives the larger contribution in ultra-relativistic bremsstrahlung.
We find that the denominators of the squared amplitude of this process have the following terms:
$\kappa=\omega(E-p\cos(\theta))$ and $\kappa'=\omega(E'-p'\cos(\theta'))$
where $E,p,\theta$ are the initial electron energy, 3-momentum and angle with the photon, respectively. The primed variables correspond to the same, but for the final electron. Using the expansion valid in ultra-relativistic case:
$p \sim E-\frac{m^2}{2E}$ and $\cos(\theta) \sim 1-\frac{\theta^2}{2}$
The denominators become:
$\kappa \sim \frac{1}{2}\left( \frac{m^2}{E^2}+\theta^2 \right)$
and
$\kappa' \sim \frac{1}{2}\left( \frac{m'^2}{E'^2}+\theta'^2 \right)$
This favors lower angles and therefore favors $q \sim \frac{1}{m}$.
To summarize, it is not the case that this regime is enforced in Coulomb scattering. But it is an useful regime in the case of ultra-relativistic bremsstrahlung.
