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Why are leptons antibaryons in gut?

  1. Oct 24, 2014 #1
    Why does gut preserve B-L, not B+L?

    Also, does weak interaction see both baryons and leptons as particles? Is the helicity required for weak interaction the same (left) for both baryons and leptons and the opposite (right) for both antibaryons and antileptons?
     
    Last edited: Oct 24, 2014
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  3. Oct 24, 2014 #2

    arivero

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    In SU(4) lepton-colour unification it is pretty clear. The U(1) for "B L" must be a diagonal of this matrix group, and it must be traceless, and it must have the same charge for the three colours. So it is 1/3 for quarks, -1 for Lepton, or opposite. Now, if the full group keeps this assignment, then B-L survives. But you could engineer that the SU(4) leptocolour components are really no lepton and quarks, but leptons and antiquarks. Or that the full "B L" mixes with some othe U(1). So it is not a strong requisite.
     
    Last edited: Oct 24, 2014
  4. Oct 24, 2014 #3

    Vanadium 50

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    If (and only if) both B+L and B-L are conserved, B and L are conserved separately.

    If you ask why the conserved quantity has a minus sign and not a plus sign, the answer is "convention". By convention, the e- has L=1 and the e+ has L=-1. We could have defined it the other way, and then B+L would be conserved.
     
  5. Nov 7, 2014 #4
    Yes. As they are outside GUT.
    So, if B separately is not conserved, why should there be conservation of B-L, but not conservation of B+L without conserving B-L, or not conserving either B+L or B-L?
    But that´s a matter of sign definition. I am asking substantive reason for the underlying process. Why does GUT allow baryon decay to antileptons while forbidding baryon decay to leptons?
     
  6. Nov 8, 2014 #5

    ChrisVer

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    Doesn't the last concern of yours imply CP violation instead?
    It's totally different to sign convention.
     
  7. Nov 8, 2014 #6

    Orodruin

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    B+L is violated on a non-perturbative level in the standard model, while B-L is an accidental symmetry. That it is B-L rather than B+L that is conserved is of course only a sign convention of what leptonic and baryonic charges we assign to the fields.
     
  8. Nov 8, 2014 #7

    arivero

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    I will repeat the point: The U(1) generator must be traceless. So it has more sense a convention which calls it "B-L" than the contrary.
     
  9. Nov 12, 2014 #8

    ChrisVer

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    I think writing the fields' Lagrangian, in an SU(5) model for example, and look at the vertices can help you clarify your problem.
     
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