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Compare the number of scattered particles:
$N_s=Fa\int\sigma(\theta)d\Omega$
With the total number of incident particles:
$N_{in}=Fa$

Here, F is the flux of incoming beam, a the area. sigma the crossection and omega the solid angle.

Why isnt $N_s=N_{in}$? How does one define which particles are scattered and which are not, arent they all interacting with the target to some degree? Isnt particles conserved normally?

Do almost all the particles either pass right through almost undetected or are scattered signficiantly, so what we are really integrating over is a sphere surrounding the target except a spot of area $a$ where the beam exits?

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The answer may depend of the purpose for which the question is being asked. Is this a pedagogical inquiry meant to emphasize the unitarity of the process, part of a experimental design process, or something else entirely? –  dmckee Nov 12 '12 at 22:06
    
Well my book didnt define 'scattered particle', but if you say that there is no clear definition in wide use, then its OK. So I may be right that they are removing an area a then? The context is collisions in accelerators. With the definition they use then all outgoing particles are scattered. So there is some arbitrary definition of what counts as scattered normally? –  Hobo Nov 12 '12 at 22:10
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1 Answer

up vote 1 down vote accepted

With the exception of stopping targets, there is going to be a portion of the beam which continues down the beam pipe (i.e. misses even the innermost elements of the detector), which would usually be counted as "unscattered" for that purpose.

In that case the integration is often over the solid angle covered by active detector elements.

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Ok, but the differential cross-section has an unequivocal definition? –  Hobo Nov 12 '12 at 22:32
    
Sure, but if the scattering angle is sufficiently small you won't see it in your detector. –  dmckee Nov 13 '12 at 0:40
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