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

I Decay process as 1 to n

  1. Dec 7, 2016 #1
    Decay processes are quite common in particle physics.

    Is the decay process always a ##1 \rightarrow n## process?

    In other words, can we call the reaction $$\mu^{-} + \mu^{+} \rightarrow \phi,$$

    where ##\phi## is some scalar particle, the decay of the muon?
  2. jcsd
  3. Dec 7, 2016 #2


    User Avatar
    Science Advisor

    As a matter of definition, decay processes start with one particle.
  4. Dec 7, 2016 #3
    Can we have the incoming particle also in the set of outgoing particles?
  5. Dec 7, 2016 #4


    User Avatar
    2017 Award

    Staff: Mentor

    That would violate energy/momentum conservation.
  6. Dec 7, 2016 #5
    I believe this would require an intermediate stage, plus an external energy source. Consider neutron → proton → neutron transformation through beta decay:

    β decay: when a free neutron decays into a proton

    n → p + e + -νe

    β+ decay: when a proton inside a nucleus decays into a neutron

    p → n + e+ + νe

  7. Dec 8, 2016 #6


    User Avatar
    Gold Member

    you call it annihilation of muon-antimuon...
    obviously you don't have 1 muon to call it decay of the muon.
    Can you have the incoming particle also in the outgoing particles? In vacuum as already mentioned no... but in other cases, yes, like Brehmstralung [itex] e \rightarrow e \gamma[/itex].
  8. Dec 8, 2016 #7
    Since the charged particle is only losing kinetic energy and its invariant mass remains unchanged, is this really considered decay?
  9. Dec 8, 2016 #8


    User Avatar
    Gold Member

    I didn't call it a decay- I gave that as an example to that you can have the same incoming and outgoing particle.
  10. Dec 9, 2016 #9


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Of course bremsstrahlung is not a decay process since you always need the electron to scatter with something since a free electron won't radiate. Only accelerated charges radiate. So the correct bremsstrahlung process is a scattering process like ##\mathrm{e}^-+X \rightarrow \mathrm{e}^- + X +\gamma##, where ##X## is some particle or atomic nucleus scattering with the electron.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted