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Asymmetry in e+e- anihilation

  1. May 21, 2008 #1

    neu

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    I understand the following:

    The cross section [tex]\frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \propto 1 + cos^2 \theta [/tex] for a purely vectorial (electromagnetic) interaction. Hence [tex]\sigma[/tex] is expected to be symetric in [tex]cos \theta[/tex].

    The axial vector (weak) coupling of the Z boson violates parity and give an asymmetric contribution to the [tex]\sigma[/tex] distribution.

    So obviously the asymmetry give a measure of the Z exchange contribution.

    OK, so Q: I don't understand why this reasoning doesn't apply when you consider [tex]e^+ e^- \rightarrow e^+ e^- [/tex]
    Surely this behaves the same, since the photon and Z exchange is equally probable?
     
  2. jcsd
  3. May 21, 2008 #2

    neu

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    OK, well is it because
    [tex]
    e^+ e^- \rightarrow e^+ e^-
    [/tex]

    can occur via t-channel as well, since there's no species change. If so how does this affect angular distribution?
     
  4. May 21, 2008 #3

    blechman

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    well, first off, the photon and Z boson exchanges are not equally probable: they involve different coupling constants, and the Z-boson is heavy while the photon is not (more relevant at lower energies).

    I'm not quite sure the answer to your question, but the fact that there is a t-channel process might bury the relatively small asymmetry coming from the axial Z coupling. That's just a first-blush guess.
     
  5. May 22, 2008 #4

    neu

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    Yes, I realised I'd written that badly. I meant to say the photon and Z couplings are just as they are for muon pair production as electron pair production/scatter.

    So Why is it that

    [tex] \frac{d \sigma (e^+e^- \rightarrow e^+e^-)}{d\Omega} \neq \frac{d \sigma (e^+e^- \rightarrow \mu^+\mu^-)}{d\Omega} [/tex]

    My thinking is that [tex] \frac{d \sigma}{d\Omega} \propto \mid M_{z} +M_{\gamma}\mid ^2[/tex]

    where there is s, t and u channel scatter for [tex] (e^+e^- \rightarrow e^+e^-) [/tex] and only s channel for [tex] (e^+e^- \rightarrow \mu^+\mu^-) [/tex]

    And so : [tex] \frac{d \sigma (e^+e^- \rightarrow e^+e^-)}{d\Omega} > \frac{d \sigma (e^+e^- \rightarrow \mu^+\mu^-)}{d\Omega} [/tex]


    How does that sound?
     
  6. May 23, 2008 #5

    blechman

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    strange! i'm a little confused what you're asking. are you only asking why Bhabha scattering is not the same as muon production? That's straightforward - the processes are not topologically the same (that is, as you say, there is a t-channel diagram in bhabha but not mu-production), so there is no reason why they are the same. that one is "larger" than the other: that's not immediately obvious since (in principle) there could be DEstructive interference making bhabha scattering smaller.

    but none of this seems to have anything to do with your original question about why you can't measure the Z-contribution to Bhabha scattering but you can for mu-production. I think it depends on precisely what you are trying to measure.

    I'm not doing a good job answering your question because i'm confused. help me out: what precisely do you want to do with bhabha scattering that you claim you cannot do?
     
  7. May 24, 2008 #6

    neu

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    OK, I'm not being clear, thanks for sticking with me. I'll re-state my question.

    Question:
    Ignoring mass effects/ phase space arguments, why at leading order do we expect:

    [tex]
    \frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \neq \frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow e^+e^-)
    [/tex]

    And which should be larger.

    My thoughts:
    -only s-channel for muon production, but s, t and u-channel for bhabha,
    -so bhabha scatter/anihillation cross section should be larger (aside from phase space favouring low final state mass).

    Another question:
    For photon exchange we have
    [tex]
    \frac{d\sigma}{d(cos \theta )} (e^+e^- \rightarrow \mu^+ \mu^-) \propto 1 + cos^2 \theta
    [/tex]

    This is symmetric in cos, but measurement shows asymmetry due to the axial vector coupling of the Z boson. Why doesn't this reasoning apply (correct me if I'm wrong) for bhabha scatter/anihillation.


    My thoughts:
    Even though, as you say, the two processes are topologically different, they can still have Z/photon exhange for all processes.

    I'm thinking that u and t-channel processes don't posses the same symmetric (in cos) properties as the s-channel process so they wash out this method of measuring the Z contribution.
     
  8. May 24, 2008 #7

    Vanadium 50

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    Correct.

    Incorrect. You don't know if it is larger or smaller without doing the calculation, because you don't know the relative phase of the s and t channel contributions. The interference could be either constructive or destructive. (In QM you add amplitudes and then square)



    This has me totally confused. If you write down the angular distribution using photon exchange, you don't see the parity violating effect of the Z. That's true for muon exchange and it's true for Bhabbas. If you include the Z, both processes show parity violation.
     
  9. May 24, 2008 #8

    blechman

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    see vanadium 50's comments.

    you're saying it doesn't?! Naively I'd expect both processes should be sensitive to the axial Z-boson coupling. The only thing I can think of is that there are some experimental barriers. Can you give me a reference to this conflict?

    t-channel processes most assuredly do NOT have the same angular dependence as s-channel processes, that is true; you can see this by explicitly doing the calculation, or by expanding in partial waves and noticing that the t-channel has no s-wave component, and is therefore not isotropic. In any event, this might be true, but please clarify my previous question about where you heard Bhabha is such a bad boy.
     
  10. May 25, 2008 #9

    neu

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    Thanks for your responses, things are clear now.

    It seems I've misinterpreted things regarding the second question I asked. It's a question of how the amplitudes differ. Of course both have an axial vector contribution.

    It was to do with a conversation I had with a lecturer.

    Thanks again for your help.
     
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