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Is cross section is proportional

  1. Oct 22, 2006 #1
    If I show something that a cross section is proportional to:

    [tex] 1 / (16p^2 sin^4( x) + 8mpsin^2 (x) + m^2) [/tex]

    does it imply that the cross section is proportional to

    [tex] 1 / sin^4 (x) [/tex] as well?
     
    Last edited: Oct 22, 2006
  2. jcsd
  3. Oct 22, 2006 #2

    OlderDan

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    No. But can you factor the first denominator?
     
  4. Oct 22, 2006 #3
    [tex] 1 / (4psin^2(x) + m)(4psin^2(x) + m) [/tex]

    Not sure what to do from there.
     
  5. Oct 22, 2006 #4

    OlderDan

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    The two factors in the denomnator are identical. What you have shown is that your thing is inversely proportional to the square of (4psin^2(x) + m). It is inversely proportiona to sin^4(x) only if m = 0. You could reach the same conclusion from your original expression, but it is a bit more evident when you factor it into a product of identical terms.
     
  6. Oct 22, 2006 #5
    That cause a problem then I was the scattering cross section is proportional to Mfi and

    [tex] |M_{fi}|^2 /propto 1 / (q^2 + m^2)^2 [/tex]

    Starting from the definition of q = [tex]p_{final} - p_{initial}.[/tex]

    Show that the angular dependance of the scattering is then given simply by the Rutherford formula:CS= scattering cross secion

    [tex] CS /propto 1 /(sin^4(\theta /2) [/tex]

    I found [tex]q = 2psin(\theta /2)[/tex]
    Since Mfi is proportional to the scattering CS I just tried sticking q into into the expression for [tex] |M_{fi}|^2 [/tex] as above but that clearly didn't work. I guess the wording angular dependance is key here.
     
    Last edited: Oct 22, 2006
  7. Oct 22, 2006 #6

    OlderDan

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    I had no context for your question, and I'm not completely following what you are trying to do. I have to log off now. See if this helps you any

    http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html
     
  8. Oct 22, 2006 #7
    Thanks, I've got it now m=0 because the exchange boson is a photon :).
     
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