How to determine the force in hadron decays

  1. I was wondering if someone could explain to me how you determine how a Baryon/Meson decays, meaning Strong decay, Weak decay, and Electromagnetic decay. Or if you could simply link me to a chart with a list of all of the forces they use to decay (if such a thing exists, I can't find one) it would be much appreciated. I know that neutral Pion decays electromagnetically, and that the charged pions and the neutron and a few others decay weakly, but I would very much like the rest. Thank you.
     
  2. jcsd
  3. e.bar.goum

    e.bar.goum 643
    Science Advisor

    So the easiest way to do this is by looking at the particles involved, and knowing how different particles interact. For example, if you see a neutrino, you know immediately that it was a weak decay, eg:

    π+ -> π0 + e+ + ve

    Because neutrinos only interact with the weak force.

    Similarly, you know that this

    Z0 -> e+ + e-

    must be weak, because Z0 is neutral and non-composite, so no EM, and electrons don't feel the strong force.

    However,

    π0 ->e+ + e-

    May be EM or weak, but strong is forbidden (same as above), however weak is less probable in this instance, because of the timescales of the reactions - τEM=10^-16s, τweak=10^-10s. This is true in general. If it can go by strong, it will mostly go by strong, and so on.

    Anything that involves a γ must be a EM interaction, because γ feel the EM interaction.

    Anything with a flavour change must be from a weak interaction, because it is the weak interaction that changes flavour, for example

    D+-> K- + π+ + π-

    (The easiest way to see that is to write out the quarks).

    What about something like

    [itex]p + \overline{p} \rightarrow \pi^+ + \pi^-[/itex]?

    In this case, it is not forbidden by the strong force, therefore most of the time it will happen by the strong force.

    By the way, this is from something called the Totalitarian Priniciple that states that any decay that is not forbidden must occur. In this way, you can uncover new conservation rules - if a reaction doesn't happen, you have to figure out why.
     
  4. Thank You so much! This was very helpful, I hope that replying to this did not inconvenience you in any way.
     
  5. e.bar.goum

    e.bar.goum 643
    Science Advisor

    No worries at all. I'm glad it helped! This is something I teach, so it's pretty well stuck in my head.
     
  6. Orodruin

    Staff: Mentor

    It is weak by definition since the Z is a mediator of the weak force and therefore must have at least one weak vertex.
     
  7. mfb

    Staff: Mentor

    Or it just happens too rare for the current experimental precision. Prominent examples are the decays of neutral mesons to two muons. It is easy to write down Feynman diagrams for their decays, it is easy to search for two muons - but the decays are so extremely rare (a few decays in a billion, or even less) that it is hard to see them at all.
     
  8. e.bar.goum

    e.bar.goum 643
    Science Advisor

    Yeah, totally. That's why you do those rare searches - if you can't find the decay, you need a good reason.


    As a bit of an aside, another cool example is neutrinoless double beta decay. That's some hardcore experiments right there. Although I'm not sure I like the idea of spending my career not finding things!
     
  9. Orodruin

    Staff: Mentor

    Or, if we are on the topic of not finding things: proton decay.
     
  10. ChrisVer

    ChrisVer 2,271
    Gold Member

    is proton decay predicted by the Standard Model?
    I think it's not...
     
  11. e.bar.goum

    e.bar.goum 643
    Science Advisor

    Yeah, protons are stable in the Standard model due to Baryon number conservation.

    The search for proton decay is motivated by BSM models that predict the violation of baryon number conservation, but instead say that the Baryon-Lepton number (B-L number) is conserved.

    (The reason you invoke B-L conservation is that unlike B conservation, the symmetry would not be broken by chiral or gravitational anomalies, if the symmetry is global)
     
  12. Orodruin

    Staff: Mentor

    This also goes for neutrinoless double beta decay. In the SM, lepton number is also conserved (on the perturbative level).

    In the SM baryon number is broken by non-perturbative effects and the accidental symmetry is B-L. This is the basics underlying electroweak baryogenesis (ok, that does not work, but it would have been the basics if the parameters were more favorable ...) and part of the reason baryogenesis via leptogenesis is a possibility.
     
  13. ChrisVer

    ChrisVer 2,271
    Gold Member

    I guess that's why some people consider the Majorana nature of neutrinos to be an extension to the Standard Model (of course this is playing with words and definitions-what you define as SM-)... However double beta decay with neutrinos is a SM process.
     
  14. Orodruin

    Staff: Mentor

    Either Dirac or Majorana masses are typically considered an extension of the SM. Neither can actually be assumed as standard as we do not know which one is true. Double beta decay does not violate neither B or L and is therefore fine within the SM.
     
  15. ChrisVer

    ChrisVer 2,271
    Gold Member

    I stated for Majorana nature of neutrinos. In case we can measure a [itex]0v \beta \beta[/itex] decay this would imply that neutrinos are their own antiparticles, so they would satisfy the Majorana condition something the Dirac neutrinos don't have to...
    the [itex]2v \beta \beta[/itex] decay is indeed not violating any of those symmetries and thus is fine within SM...
     
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