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I Unified gravity with electroweak through SU(2)xSU(2)xU(1)

  1. Oct 22, 2016 #1
    Consider symmetry SO(1,3)xU(1)=SU(2)xSU(2)xU(1) and puting a local spontaneous broken symmetry:
    Φ→exp(iαaτa)exp(iαaτa)exp(i(β/2))Φ
    The field Φ acquires vacuum expectation value: <Φ>=(0,v).
    Then ΔL=½(0 v)(gCμaτa+gAμaτa+
    ½g'Bμ)2(0 v)T.Then we find out three massive boson,one massless photon and two massless vector boson corresponding with 8 components of gravitation field hμν
    Is that correct?
     
  2. jcsd
  3. Oct 22, 2016 #2

    PeterDonis

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    Do you have a reference for this? PF does not allow discussion of personal theories.
     
  4. Oct 22, 2016 #3
    This is my theory.
     
  5. Oct 22, 2016 #4

    PeterDonis

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    Staff: Mentor

    Then it's off topic here. Thread closed.
     
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