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

Thinking about propagators

  1. May 28, 2008 #1
    i was thinking about propagators, and I've got bit of a silly question.

    The amplitude for any given decay takes the form:


    for EMAG decays, the Mx=0 (as we're dealing with photons), so the amplitude is just:


    Why do we not consider the propagator: [tex]\frac{1}{q^{2}}[/tex] when determining decay rates (we just say its proportional to the fine structure constant to power of the number of vertices)?
  2. jcsd
  3. May 28, 2008 #2
    In the case of EM interactions,
    g IS the fine structure constant.. where's your problem ?
  4. May 28, 2008 #3
    thanks Atakor.

    I understand that. However, the gauge boson (the carrier of the electromagnetic force), in this case a massless photon, transfers a momenmtum q. Why is the propagor 1/q^2, ignored when calculating the transition amplitude?
  5. May 28, 2008 #4
    Hello vertices!
    In feynman language:

    to each vertex we associate the couplig constant : -ig
    for each _internal_ line we associate the function you dubbed f(q)

    but f(q) is not the amplitude of the transition..
    we have to integrate over all momentum space (q) (with conservation of energy-momentum)
    after taking into account the functions of the particles (spin 1/2 ->spinors ...)

    I hope this will help you.
  6. May 28, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    We do take that factor into account! Do you have any specific example to discuss? That factor will be there with q having the value of some energy scale in the problem (of order the mass of the decaying particle).
  7. May 28, 2008 #6
    Yes, as it happens I do. Please look at the attached files which consider the decays of the pi and pi- mesons. The bit I have highlighted bit in the file "Propagator 2", basically says that, in the case of the neutral pion, which decays electromagnetically, "there is no propagator". I don't get this!

    Attached Files:

  8. May 28, 2008 #7

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    A propagator is used when a virtual particle propagates from one vertex to another. In the decay of the pi0, there are no virtual photons, so there are no photon propagators. In decays where you do have a virtual photon, e.g. [tex]J/\psi \rightarrow \mu^+ \mu^-[/tex], you have a photon propagator.
  9. May 28, 2008 #8
    i see. thanks for your reply Vanadium:)
  10. May 28, 2008 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you look at the Feynman diagram, you will see that there is no gauge boson propagator (no photon propagator or W or Z boson propagator) involved in the neutral pion decay.

    However, this is not the end of the story. And the excerpts you provided don't make much sense. You see right away that what they say does not make sense because a decay rate must always have the same dimensions no matter what process is computed. They seem to say that in one case the decay rate goes like 1/M^4 and in the other case it is dimensionless. This is clearly nonsense.

    two things are missing to the discussion. First, in the charged pion decay there will be spinor factors coming from the outgoing fermions taht are being produced. These will provide additional dimensionful factors that must be taken into account. These will produce an additional factor with dimension of energy.

    Also, in the neutral pion decay there is a propagator in the diagram: it's the propagator of the virtual quark between the two photon vertices! This must be taken into account as well. This wil produce a factor of dimension [tex] E^{-1} [/tex]. This way the two diagrams will have teh same dimensions (I am setting aside the factors coming from the initial quarks which are presumably of the same order in both diagrams so if we are taking a ratio of the two diagrams we don't have to think about those).

    What is the source of those excerpts you provided. I hope it's not the book by D McMahon!
  11. May 28, 2008 #10
    Yes indeed it is, thanks for pointing that out!

    aahhh. didn't think of that. here comes a really stupid question - a virtual quark isn't a gauge boson, so how does it/why can it mediate the process in question?

    Its "Nuclear and Particle Physics" by Brian Martin. I can overlook its complacency/lack of rigour because it does explain concepts quite well:)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook