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I Confused about Feynman diagrams

  1. Jul 16, 2017 #1

    dyn

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    Hi. I'm self-studying particle physics.Just been looking at some questions where a reaction is listed and the questions asks to draw a Feynman diagram for the reaction and state which force is involved. I have the answers but they all seem so random and I would like to know how to decide which force is involved. If possible I would like any explanation for the following examples -

    1 - an electron and positron annihilate forming a photon which then decays to a muon and an antimuon
    This is an EM interaction but I have read that only the Weak Interaction changes particle flavour. But hasn't the flavour changed here from electron to muon ?

    2- an electron and positron move to the right (time axis) they are joined by a vertical line and 2 photons are emitted at the vertices. Is this electron-positron annihilation ? Why the vertical line ? Why do the electron and positron just not meet and annihilate to form a photon ?

    3 π+ → νμ + μ+ via a W+ boson.
    I have read that "the weak force cannot turn quarks into leptons and vice vera". But in this reaction hasn't quarks and antiquarks turned into leptons and antileptons ?

    Thanks
    PS I have the books by Martin & Shaw , Thompson , Barr & Devenish but none of them really explain how to construct Feynman diagrams just from a given reaction.
     
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  3. Jul 16, 2017 #2

    Orodruin

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    No. The total electron and muon numbers before and after the reaction are zero. The anti-particles count as negative.

    Also note that the photon is not an on-shell photon and does not "decay". It is just an s-channel interaction with a virtual photon.

    Forming a single on-shell photon would violate energy-momentum conservation.

    Same answer as for (1). The total baryon and lepton numbers before and after the decay are zero.
     
  4. Jul 16, 2017 #3

    ChrisVer

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    How? Initially your electron number is: [itex] L_e(e^-) + L_e(e^+) = -1 + 1 = 0[/itex], and at the final state is 0+0 =0 (conserved).
    Similarily for the muon lepton numbers (0+0 initially , +1-1 finally).

    Because you can find a reference frame (by a Lorentz transformation) where the total momentum of the electron+positron system is zero (center of mass frame)... In that frame, conservation of momentum would imply that the single photon would have to have 0 momentum... that's impossible for real photons (since they are massless and have energy E=pc). If you allow two (or more) photons in the final state then you don't have this problem.

    It's a [itex]u\bar{d} ( \rightarrow W^+ \rightarrow ) \mu^+ +\nu_\mu [/itex]
    What do you have initially and what you have finally? The lepton numbers are conserved, the baryon numbers (mesons have 0 baryon number since they are quark+antiquark pairs) are conserved.
    the statement "the weak force cannot turn quarks into leptons and vice vera" means that you cannot have something like:
    [itex] u \rightarrow l \nu_l[/itex]
    (which I think makes sense even from conservation of charge.)
    You can have:
    [itex] q \bar{q}' \rightarrow l \nu_l[/itex] (eg B-meson leptonic decays) and [itex] q \rightarrow q' l \nu_l[/itex] (eg top semileptonic decays)

    In general Feynman diagrams are like art when you want to construct them in a basic level (like a puzzle). They can give you an idea of conservation laws (just because conservation laws are always applied in them) and things like that in particle physics... the thing is how deep you wanna go into them, since afterall they only represent amplitudes for a given process (which allow you to calculate cross-sections/probabilities). That's why you can see mathematical formulas like: [itex] | \text{Feyn.Diag.1} + \text{Feyn.Diag.2} +... |^2[/itex]... With a QFT in particle physics, you are able to construct the Feynman diagram by reading the Lagrangian and using several rules that are there...
     
  5. Jul 16, 2017 #4

    dyn

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    Thanks both of you. That's a great help. But when just given the reaction and without resorting to QFT how do you know the most likely force involved ? eg. for
    Δ++ → p + π+ . the given answer was the Strong interaction but the reaction involves quarks so the force involved could be EM , Strong or Weak. What do I look for to know that the most likely force involved is the Strong one ?
     
  6. Jul 16, 2017 #5

    mfb

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    Go by interaction strength. If it is possible via the strong interaction, that is dominant. If not, but it is possible by the electromagnetic interaction, it is that. Otherwise it happens via the weak interaction.

    There are a few exceptions (e. g. Z bosons can be relevant even when there is a similar electromagnetic process), but they shouldn't be too relevant here.
     
  7. Jul 16, 2017 #6

    dyn

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    Thanks ! That makes a lot of sense but what about
    π+ → μ+ + νμ ?
    This involves quarks so could happen by the Strong force but the decay is via the Weak force and the W+ boson. How would I be supposed to know this without any QFT calculations ?
     
    Last edited: Jul 16, 2017
  8. Jul 17, 2017 #7

    ChrisVer

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    Could it happen with just the strong force? What's the difference between leptons and quarks (they are both fermions but...)?
    In your reaction you obviously have leptons in the final state (muon and neutrino). That answers your question.

    Also strong force (mediated by gluons) cannot change the flavor of the quarks, in the charged pion you have an up-flavor and down-flavor quarks interacting somehow (there is no vertex in which an u-->d by emitting/absorbing a gluon). W bosons can do that (they can be flavor changing), in fact in vertices with quarks they always do that and you can remember that by charge conservation...
     
  9. Jul 17, 2017 #8

    dyn

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    I'm not sure which difference you are referring to ? The only reason I can now see that the reaction couldn't proceed by the Strong Interaction is that quark flavours have changed ie. the number of up minus anti-up has gone from 1 to zero which can only happen with the weak force,
     
  10. Jul 18, 2017 #9

    ChrisVer

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    leptons are not seen by strong interactions. That's one distinction (even a definition) between leptons and quarks. So, leptons can't interact via strong interaction... So you can't see gluon+leptons vertices.
     
  11. Jul 18, 2017 #10

    dyn

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    Thanks for your help
     
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