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Suppose we know differential cross-section for some type of scatterer particle. Now, consider a large number of such scatterers distributed randomly in some volume. If there's much space between these scatterers, I suppose the cross-sections of individual scatterers can be simply summed, because multiple scattering would be negligible(?).

But in general, it seems, multiple scatterings should distort differential cross-section of individual scatterers, so that simple sum won't work to get the cross-section for the collection. Is this correct? If yes, how can one calculate the differential cross-section of a large collection of randomly-distributed scatterers, given the cross-section for one scatterer?

Ruslan
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  • This is a very difficult problem without a general answer. For example, in radar for which land, rain, or sea scatter are very important they have quite different characteristics with the only common thread being that no theoretical analysis is good enough in practice. See the subject heading under "clutter". – hyportnex Jun 04 '18 at 13:59

2 Answers2

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As long as the target is "thin" (meaning the total odds of being scattered are small), then you are safe simply adding cross-sections together because the odds of being scattered more than once are negligible.

Once the total chance of a incident particle being scattered gets to be significant, then simply summing the cross-sections of scattering centers in the beam will over count—because a significant fraction of incident particles will be scattered by more than one center, but it is still only one particle.

As with other concerns the definition of "negligible" is driven by your precision goals for the measurement and by comparison to other uncertainties. To keep the frequency of individual incident particles experiencing multiple hard scattering events small enough to ignore when you want a 1% final precision (the goal in most of my dissertation work) you'll need a target that scatters less than $\approx 10\%$ of incident beam particles (we used targets up to 6% radiation length).

dmckee --- ex-moderator kitten
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  • well radiation lengths relates to electromagnetic scattering of an incoming beam or an out-going scattered (charged) particle. I think the OP is asking about the cross section at hand, say $ep\rightarrow ep$ where multiple scattering is a factor, but not multiple elastic scattering from a proton. – JEB Jun 04 '18 at 18:35
  • @JEB I'm not sure I understand your objection. The use of "radiation length" as a measure of target thickness certainly only applies to some processes, but most of the answer is coached in terms of scattering probability and not radiation length (my dissertation work of $H(e,ep)$ and $A(e,ep)$ so the term was sensible). – dmckee --- ex-moderator kitten Jun 04 '18 at 18:45
  • I picked a good example then. So pre scatter, you worry about radiation lengths so the $e$ has the beam energy when it scatters. Likewise the detected $e'$ (and also angle to some extent). The OP seems to be asking about the need to account for something like $ep \rightarrow ep$ followed by $ep'\rightarrow ep'$ and so on, and then detecting the electron. That scattering probability is so much smaller than multiple scattering (from atomic elections) that it is not relevant. – JEB Jun 04 '18 at 18:55
  • Ah. Perhaps I see. I've clarified the text to indicate that I am talking about individual; beam particles subjected to multiple hard scattering events, and not about "multiple scattering" in the many small-angle elastic events. – dmckee --- ex-moderator kitten Jun 04 '18 at 20:22
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I think you should write the scatterer, consisting of many centers, exactly in the Schroedinger (or a wave) equation and make some approximate treatment. If the wavelength of the incident "projectiles" (waves too) is much larger than the charateristic distance between scattering centers, you will not be able to add cross sections because you will have some sort of a "coherent scattering" from a compound target.

Vladimir Kalitvianski
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  • The coherent scattering your describe occurs in both laser scattering from a uniform white surface and radar-scattering (see a SAR image), where it is called speckle noise. For a given constant power $x$, instead of sampling from $\delta(x)$, you sample from $e^{-x}$ leading to a salt-n-pepper look (hence: speckle). – JEB Jun 04 '18 at 21:31