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Z boson

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Just a quick question, the Z can couple to leptons and there antiLeptons and quarks and there corresponding antiquark. Is the coupling constant just Gz for all these interactions? Or is it Gz.Sin(z) for some and not for others?

    Iv seen different notation and am wondering why?
     
  2. jcsd
  3. Dec 7, 2009 #2
    What is the solution of the photon?

    ie

    http://upload.wikimedia.org/wikipedia/commons/8/8c/Standard_deviation_diagram.svg

    In SR?

    ie

    [tex]\bar{x}=x[/tex]

    and

    [tex]\bar{z}=z[/tex]

    The coupling constant is constant. so [tex]G_z=Gz.Sin(z)[/tex] if there are no CPT violations.

    Charge Parity Time (CPT)

    Thus CPT symmetry.

    And thus QFT:

    http://en.wikipedia.org/wiki/Quantum_field_theory
     
    Last edited by a moderator: Apr 24, 2017
  4. Dec 7, 2009 #3
    Cheers for the response :)

    So for a Z --> e-e+ and Z--> qq" the coupling constants would be the same? Gz = gz*Sin(z)? I know that it changes for the weak depending on if its interacting with quarks or leptons, im still unsure of why the difference?

    Also out of curiosity... i thought it cant be constant because of the 'running of the coupling constants' idea?
     
  5. Dec 7, 2009 #4
    Exactly weak has CPT violations inherent in it thus The Nobel prize for electro/weak theory.
     
  6. Dec 7, 2009 #5

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Electroweak theory has CP violations. No theory has CPT violations.
     
  7. Dec 8, 2009 #6
    Subsequently i have found it to be a little different then mentioned above:

    (z --> uu') has a coupling constant GzCos(z)
    (z --> dd') has a coupling constant GzCos(z)

    (z --> cc') has a coupling constant GzSin(z)
    (z --> ss') has a coupling constant GzSin(z)

    Can i confirm that this is correct? Also i am still unsure for leptons, am i to assume its just Gz like for W bosons?

    Cheers
     
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