The differential cross section of Rutherford scattering blows as $\theta\rightarrow 0$. People express this fact as "The Coulombic potential is of long range". But I am seeing the opposite: $\theta \rightarrow 0$ is when the incident particle is not affected by the potential, and therefore a blow when $\theta \rightarrow 0$ means short range potential.

That was first. Secondly, I cannot see how the differential cross section is related to (some kind) of probabilities when infinities are involved.

  • $\begingroup$ It is a well known fact that for Coulomb scattering the total scattering cross-section is infinite. Since the optical theorem connects the total cross-section to the imaginary part of the forward (θ=0) scattering amplitude the latter also becomes infinite. You can find more information on wikipedia, search for optical theorem and Coulomb scattering. $\endgroup$
    – Urgje
    May 3, 2014 at 17:11
  • $\begingroup$ Thank you, but that doesn't reconcile the difficulties I have which are described in the question. $\endgroup$ May 3, 2014 at 17:21

1 Answer 1


Since the Coulomb potential is long ranged, incident states with arbitrarily large impact parameters are scattered by small angles. Taking the impact parameter to infinity, the scattering angle goes to zero, so essentially there are an infinite number of forward scattering events for which the total cross section must account.

  • $\begingroup$ This is true, but in a sense uninteresting when you put real beams on real targets. In that cases there are many scattering centers in the target and the actual scattering is dominated by the closest scattering center, meaning that the small angle scattering fades smoothly into the multiple scattering regime. $\endgroup$ May 4, 2014 at 1:12
  • $\begingroup$ Actually, in a real life, there is also electron screening, that is much more important. $\endgroup$
    – jaromrax
    Feb 16, 2017 at 8:20

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