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I Baryon number nonconservation in the early universe

  1. Sep 10, 2017 #1

    stevendaryl

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    It is said that the imbalance of matter versus antimatter in the present universe implies CP violations at very high energy. It seems to me that it most directly implies baryon number nonconservation: If we assume (and I'm not exactly sure why this is a necessary assumption) that immediately after the Big Bang, the universe was electrically neutral (and had zero lepton number, baryon number, etc.), then the fact that the baryon number of the universe is nonzero today means that it's not conserved.

    But what is the relationship between CP violation and baryon number nonconservation? Can you have one without the other?
     
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  3. Sep 10, 2017 #2

    Orodruin

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    Assuming that you start with everything in thermal equilibrium, you need three conditions satisfied:
    1. Baryon number violation.
    2. C and CP violation.
    3. Departure from thermal equilibrium.
    These are the so-called Sakharov conditions.

    Clearly, you need baryon number violation or you will never be able to create a baryon asymmetry. However, if you do not violate C, then the rate of the process ##X \to Y + B## is the same as its C conjugate process ##\bar X \to \bar Y + \bar B##, which gives you no net baryon number violation even if baryon number is violated in each individual interaction. The argument for the necessity of CP-violation is similar. Finally, if you do not have departure from thermal equilibrium, both baryons and anti-baryons have the same distribution and therefore the same numbers.

    A priori, there is no relation and you can have one without the other. You can have baryon number conserving CP-violation (weak interactions being a good example) or you can have CP-violating processes that violate baryon number (such as the one in the example above).
     
  4. Sep 10, 2017 #3

    stevendaryl

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    I'm not up on particle physics, but as far as I know, there has never been an experimental demonstration of baryon number non-conservation, right?
     
  5. Sep 10, 2017 #4

    mfb

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    Right.

    We don't have the energies available where we expect these processes to be relevant. Searches for proton decays are the only tool to indirectly probe these regions.
     
  6. Sep 10, 2017 #5

    Orodruin

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    Theoretically however, all the ingredients are there in the SM. However, the CP violation is too small to give you enough baryons. The B violation in the SM is due to non-perturbative processes that violate B+L while keeping the accidental SM symmetry B-L.

    There are also things such as searches for ##n-\bar n## oscillations.
     
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