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Homework Help: High energy electron positron branching ratios.

  1. Sep 12, 2013 #1
    The problem statement, all variables and given/known data

    Why does the ratio:

    [itex]\frac{σ(e^- + e^+ \rightarrow μ^- + μ^+) }{σ(e^- + e^+ \rightarrow τ^- + τ^+)}[/itex]

    tend to unity at high energies and would you expect the same for:

    [itex]\frac{σ(e^- + e^+ \rightarrow μ^- + μ^+) }{σ(e^- + e^+ \rightarrow e^- + e^+)}[/itex]

    The attempt at a solution

    So I have a good idea of the first part:

    An electron positron annihilation is going to lead to a photon. If the photon's energy is large compared to twice the rest mass of a tau then both channels are available and, if the mass difference between a muon and tau is small on the scale of the photon energy, approximately the same as far as the photon is concerned.

    Now for the second part. As far as I can tell my argument applies just as well to say that the second ratio should also tend to unity but the way the question is phrased implies it doesn't :tongue:.

    I guess this means something is wrong with my approach to justifying the first ratio tending to unity?

    Thanks in advance for any help :smile:
  2. jcsd
  3. Sep 12, 2013 #2


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    Staff: Mentor


    Your argument is valid only for the specific process you considered...
    Your feeling is right, the ratio does not go to 1.
  4. Sep 12, 2013 #3
    Thanks for the reply.

    But as soon as you've got the free photon I don't understand why it would favor (or disfavor) an electron positron pair rather than a muon or tau pair.

    The argument would just be if the energy is much greater than twice the muon rest mass the muon and electron are the same as far as the photon is concerned.

    The only thing I can think of is that, since the photon is created by an electron positron pair, is it at a sort of resonance energy for electron positron production? This would obviously favor electron pair production.
  5. Sep 12, 2013 #4


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    It does not, but getting a single (virtual) photon is not the only process.
    There are other processes for ee->ee only.
  6. Sep 13, 2013 #5
    Thanks for the hint! I think I understand:

    The ee -> ee interaction has a scattering channel as well as the annihilation channel. By the Feynman rules these matrix elements should be subtracted leading to a reduced ee -> ee cross section.
  7. Sep 13, 2013 #6


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    Staff: Mentor

    Are you sure?

    Especially for low momentum transfers, the cross-section is large.
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