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Baryon asymmetry and conservation of charge

  1. Dec 13, 2008 #1
    I am confused with baryon asymmetry and conservation of charge..we say that there is baryon asymmetry but the universe is neutral..how is that possible..if there is baryon asymmetry then there should be charge asymmetry also? I think i am missing some point..can someone explain these two in detail?

    I got the Sakhorov conditions for baryon asymmetry from wikipedia:

    1. baryon number violation
    2. c or cp violation
    3. interactions out of thermal equilibrium

    The second point says that there is cp violation for baryon asymmetry to happen.which means the conservation of charge should be violated..so universe cannot be neutral..am i wrong?
  2. jcsd
  3. Dec 14, 2008 #2

    Vanadium 50

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    Not necessarily. Imagine a universe made up only of neutrons.

    CP violation does not mean charge isn't conserved. Why do you think it might?
  4. Dec 17, 2008 #3
    we distinguish matter and antimatter based on their charge. what i think as cp violation is, rate at which matter decay is not same as the rate at which antimatter decay. supposing we have 10 matter and 10 antimatter at time t=0. then after some time, 10 matter decays to 10 anti matter but only 5 antimatter decays to matter because of cp violation.so,at last,we 15 antimatter and 5 matter.charge is not conserved.is this calculation correct?
  5. Dec 17, 2008 #4

    George Jones

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    To illustrate Sakharov's conditions, let me use a silly toy example that probably violates all kinds of stuff.

    Consider the reaction where a positron (anti-electron) and a photon combine to form a proton,

    Code (Text):
    e[SUP]+[/SUP] + gamma --> p[SUP]+[/SUP]
    Baryon number conservation goes form zero to +1, so condition 1. is satisfied, but electric charge is conserved.

    Suppose this reaction in invariant under C. Then,

    Code (Text):
    e[SUP]-[/SUP] + gamma --> p[SUP]-[/SUP]
    is equally likely to happen. Here, baryon number goes from zero to -1.

    For condition 2. to be satisfied (and thus 1.), one of these reaction has to be more likely than the other. This gives baryon number non-conservation while conserving electric charge.
  6. Dec 17, 2008 #5
    Thanks..i got the difference..
  7. Dec 17, 2008 #6

    Vanadium 50

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    Are you arguing that anti-neutrons don't exist as separate particles? I assure you that they do. (You can tell the difference between a neutron and an anti-neutron by their magnetic properties)

    First, that's CPT violation, not CP violation. There is a a theorem called the "CPT theorem" which states that all theories that can be written down with a finite number of derivatives and respect relativity conserve CPT.

    Second, and more fundamentally, if the individual decays themselves don't violate charge conservation, no amount of changing relative rates will cause a charge nonconservation to appear.

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