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Feynman-Diagramm: W+ or W-?

  1. Sep 17, 2011 #1
    Hi,
    lets say we have the scattering process [itex]e^- + \nu_{\mu} \rightarrow \mu^- + \nu_e[/itex] via the weak interaction. This would then be mediated by a W-Boson, right? Now my question ist: Which one?

    The Feynman-Diagramm should then look like the one on the right:

    http://accessscience.com/loadBinary.aspx?aID=9704&filename=450700FG0010.gif" [Broken]

    (in the line on top, one has to exchange [itex]e \rightarrow \mu[/itex]

    I know that this diagramm actually represents two processes (which could be drawn as slightly "skewed" diagramms):
    1) The electron emits a [itex]W^-[/itex] and thereby "changes" into an [itex]\nu_e[/itex]. This [itex]W^-[/itex] is then absorbed by the [itex]\nu_{\mu}[/itex] which therefore "changes into a [itex]\mu^-[/itex]
    2) The [itex]\nu_{\mu}[/itex] emits a [itex]W^+[/itex] and changes into a [itex]\mu^-[/itex]. The [itex]W^-[/itex] is then absorbed by the [itex]e^-[/itex] and changes into a [itex]\nu_e[/itex]

    So if I draw the Feynman-diagramm as in the link, is it ok to write [itex]W^+[/itex]? Because wouldn't that mean that the diagramm only represents option 2)? Or am I wrong to assume that [itex]W^+[/itex] and [itex]W^-[/itex] both have to do with the process and only one of the processes 1) and 2) is the "right" one?

    I have a big exam in particle physics soon. If the professor asks me to draw the Feynman-diagramm for this process I simply wouldn't know what to write ( [itex]W^+[/itex], [itex]W^-[/itex] or maybe forget about the charge (just write [itex]W[/itex]) and hope the professor doesn't ask which of the W's it is?).

    I know that my question is not very precise, so here is the short version:

    What belongs in the diagramm?
    a) W+
    b) W-
    c) simply W, because W+ and W- both have to do with the process
    d) other

    Thank you very much,
    QuantumCosmo
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 17, 2011 #2
    The [itex]W^+[/itex] and [itex]W^-[/itex] are related by a particle/anti-particle relationship, so for a virtual W, wether its charge positive or negative depends on how you implement conservation of momentum. You've perhaps heard the Feynman idea that an anti-particle is equivalent to a particle moving backwards in time? Have you studied charge conjugation symmetry?

    What this means is that wether its a particle or anti-particle depends on which way you consider its momentum to be flowing. In the diagram you've drawn, the momentum of the internal W must be flowing towards the bottom node in the diagram for charge conservation to work.

    If you want to draw a [itex]W^-[/itex], you also will reverse the direction of the four momentum. The two diagrams are completely equivalent, so ultimately it doesn't matter.
     
  4. Sep 18, 2011 #3

    Hepth

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    Gold Member

    As a side note that may also be helpful, the same rules apply to internal fermion lines, but you just have to make sure that whatever you decide for a convention you keep it. I think the general rule is for internal fermions you draw the momentum lines in the same direction as the particle arrow on the propagator. This way all of your propagators are for the particle, not the anti-particle.
     
  5. Sep 24, 2011 #4

    Bill_K

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    Science Advisor

    Label the line W±.
     
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