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Relative Angular Momentum in 2 Body Decay at Detector Level

  1. Apr 29, 2015 #1
    How the relative angular momentum of two particles can be detect by detector in two particle decay (center of mass frame)? I am curious about the signatures/differentiation between different relative momenta, means how one can decide that it is L=0, L=1,2,3,....?

    Of course the distribution would be different, but what kind of difference exist exactly? Looking for some pictorial spirit to understand the difference.

    Thanks!
     
  2. jcsd
  3. Apr 29, 2015 #2

    ChrisVer

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    I think it has to do with the spherical harmonics functions [itex]Y^m_{l}[/itex].
    There will be some particular dependence on the [itex]\theta[/itex] coordinate and this will help you find the [itex]L[/itex].
    http://www.trinnov.com/wp-content/uploads/2011/11/sphericalHarm.jpg

    For example an [itex]L=0[/itex] particle will be independent from [itex]\theta= - \pi [/itex] to [itex]\pi[/itex] (or [itex] \cos \theta \in [-1,1][/itex].
    An [itex]L=1[/itex] will have some particular dependence on theta..

    I don't know maybe there are other more effective ways to do that in a detector.
     
  4. Apr 29, 2015 #3
    Ok got it. Thanks!

    But in theoretical calculation I am not clear what is the major difference I should have to keep in mind in calculating these partial decay widths. I know in the final result there must be projection on some [itex]Y^m_{l}[/itex] but I am not able to get the difference from start. Do you have any idea?
     
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