12
$\begingroup$

I think I might have a serious misunderstanding of some concepts to do with scattering cross sections and would really appreciate any help.

As far as I can tell the differential cross section is basically a measure of what fraction of particles get scattered into some solid angle. To get the total cross section all we need to do is then integrate over all solid angles. However I can't determine why this wouldn't just be equal to 1 i.e. that we are guaranteed to find the scattered particle at some solid angle. An attempt in my mind to fix this would be to not include the infinitesimal on axis solid angle. However, unless the differential cross section was infinite their, it wouldn't make any difference.

Could anyone tell me what's gone wrong with my reasoning? Much appreciated!

Improve this question
$\endgroup$
3
  • $\begingroup$ Note that a total cross-section need not be finite. $\endgroup$
    – Urgje
    Commented Feb 3, 2016 at 9:09
  • 3
    $\begingroup$ Indeed, scattering from 1/r$^{2}$ potentials (Rutherford scattering) has an infinite cross section. No matter the impact parameter, the particle will deviate from the original path. So, the confusion appears to be exactly what a scattering cross section represents. $\endgroup$
    – Jon Custer
    Commented Feb 3, 2016 at 13:38
  • $\begingroup$ Note the total integrated cross section has units of area, so cannot equal 1. Your intuition is incorrect. $\endgroup$
    – ComptonScattering
    Commented Mar 16, 2023 at 21:14

1 Answer 1

Reset to default
4
$\begingroup$

Scattering cross sections can have other parameters besides angle. For example you commonly have cross sections vs. angle and final energy, $d\sigma/d\Omega\, dE_\text{final}$. This might reflect the fact that in a generic elastic scattering process forward-scattered particles tend to retain most of the beam energy, while backwards-scattered particles must deposit a lot of energy and momentum in the target.

If you're doing purely elastic scattering from a crystal, for example Bragg scattering, the cross section $d\sigma/d\Omega$ will essentially vanish except for some very particular angles. It's the values of those magic angles that are useful, and the relative strengths of scattering at one angle versus another. The fact that the entire incident beam has to go somewhere can sometimes be a useful check, but more often the unscattered / forward-scattered beam gets sent to some beam dump and ignored.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ Ok thanks! So just to clarify performing the integral over all solid angles will in fact give 1? $\endgroup$
    – plzhelp
    Commented Feb 3, 2016 at 5:28
  • $\begingroup$ The probability of finding the incident particle somewhere (neglecting absorption or decay or whatnot) is 1. However the cross section isn't a probability --- for one thing probabilities are dimensionless, while cross sections have units of area. As Jon Custer points out, it's possible for a total integrated cross section to be infinite. $\endgroup$
    – rob ♦
    Commented Feb 3, 2016 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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