Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Non-leptonic hadron decays: preferred paths?

Tags:
  1. Jun 13, 2015 #1
    My background of some introductory courses in particle physics has left me with severe shortcomings.

    Say we start from a hadron, which decays purely to other hadrons. My question is this: through which interaction does the process take place? Is there a preferred interaction, and why/why not?
    Let me put my thoughts/guesses/current understanding here.
    A process in which the flavour of one or more quarks in the original hadron changes, must happen through weak interaction, and there are no other options.
    Processes in which the flavours don't change could happen either through strong, weak or electromagnetic interaction. Right? Now, my guess is that there is a preference. How should I see this exactly? I thought about comparing the coupling constants, but I've understood that the coupling constant of the strong force depends on the distance of the quarks. Then again, how far can I get the quarks from each other depends on the energy they have. So once I fix the energy scale, I would be able to compare the couplings? Or is there an easier approach? Or am I missing something?
     
  2. jcsd
  3. Jun 13, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.

    It depends on energy, but binding energies in hadrons are always at the QCD energy scale. "Distance of quarks" is not a meaningful concept in hadrons.
    The strong interaction is so strong that it nearly always dominates.
    The electromagnetic interaction is significantly weaker.
    The weak interaction is negligible if one of the two other interactions is available.

    Only in rare cases like the J/Psi the strong interactions gets suppressed enough to see electromagnetic decays.
     
  4. Jun 14, 2015 #3
    I see, thank you for your insight. I take that by the qcd scale you mean the scale at which the coupling constant diverges, and as such perturbation theory cant be used.
     
    Last edited: Jun 14, 2015
  5. Jun 14, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    There is a parameter called QCD scale, it is important for hadron binding energies.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Non-leptonic hadron decays: preferred paths?
Loading...