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Conservation Laws at a Vertex

  1. May 19, 2010 #1
    Is my understanding correct:

    Charge is conserved at a vertex. So the charges of particles pointing in towards a vertex and the charges of particles pointing out must be equal. Alternatively: The sum of the charges at a vertex must be zero.

    So if this is correct, can someone explain the attached diagram to me?

    This is for the decay process:

    [tex]\pi^- -> \mu^- + \overline{\nu_{\mu}}[/tex]

    To me, it should be a W^+ particle rather than W^-, since the charge at the first vertex is d = -1/3, anti-u = -2/3, and therefore if it's a W^- then the charge will be -2 rather than zero. If we put a W+ that solves the problem. Similarly on the other side, we have -1 from the muon and zero from the anti-muon-neutrino so we need a +1 from a W+ to conserve charge at the vertex.

    Attached Files:

  2. jcsd
  3. May 19, 2010 #2
    Forget to mention we use the x axis as time and the y axis as space.
  4. May 19, 2010 #3


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    If time goes to the right, think about it as charge flow. Whatever goes in must go out. If -1 goes in -1 goes out and so on. There is no "AT" the vertex.
  5. May 20, 2010 #4
    In that diagram, -1/3 goes in but -2/3 + (-1) goes out. Something is fishy there? Essentially this "flow change" you talk of means exactly that total charge should equal zero at a vertex, no? Otherwise it is implying charge is created..
  6. May 20, 2010 #5
    anti U + D gives -1 as charge

    that charge is entering the vertex

    what is leaving the vertex is a W, with charge -1

    thus charge is conserved at the vertex.

    The arrows on the solid fermion lines does NOT mean charge flow, but fermion number flow!
  7. May 20, 2010 #6
    If the arrows don't indicate charge flow, then how do you decide what charge is "entering" the vertex and what is "leaving" the vertex? Purely from the left-right nature of the time axis?
  8. May 20, 2010 #7
    OK I think I'm getting now, except how do I treat a particle at a vertex that is pointing vertically upwards ie.stationary in time?

    Example attached, can someone describe what the charges at each vertex would be for me here, using this flow analogy, because I don't see it, for instance in the bottom right vertex, it looks like charge -1/3 is flowing in, and a charge of 2/3 for the u particle, but then -1 via the W- flowing out?

    Attached Files:

    Last edited: May 20, 2010
  9. May 20, 2010 #8
    (you forgot bar on S)

    look at the lower left vertex the total charge is -1/3 from the d-quarks and -(+2/3) from the u quark since the arrow is reversed, he W then carries -1 charge

    i.e. -1/3 + -2/3 = -1

    another way to look at it is that the d-quark is going to an up quark and a W boson

    i.e. -1/3 = 2/3 + -1

    so when you have a vertical line, there are two ways to look at it.
  10. May 20, 2010 #9
    You just said that the arrows don't indicate charge direction! What you have just written is in contradiction there? So if an arrow points away from a vertex, it indicates that it's an anti-particle with respect to that vertex? If it is going towards it, it's a particle? When it is a vertical line I mean?
  11. May 20, 2010 #10
    well they do that SINCE the arrows indicate fermion number flow, I should have pointed that out ;)
  12. May 20, 2010 #11
    OK, now the top left vertex, +1/3 and +2/3 flowing in, but -1 flowing out? How does that work?
  13. May 20, 2010 #12
    no in comes +1/3 and out goes -2/3 (the u) and +1 (W+ .. why did you think that it should be a W-?)

  14. May 20, 2010 #13
    Because otherwise in the top right vertex you have +1 going in and -1 going out?
  15. May 20, 2010 #14
    that muon is +. the muon in the lower right is -
  16. May 20, 2010 #15


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    In the diagram the muons are unlabeled. The top one MUST have positive charge, from the diagram. Therefor it is an ANTI muon, and so the arrow should be the other way. Same goes for the lower.

    The diagram is wrong by 99% of people's conventions that a PARTICLE MUON has NEGATIVE charge.
  17. May 20, 2010 #16
    that was gonna be my second comment ;)
  18. May 20, 2010 #17
    OK thank you guys, ansgar sorry for not getting it as quick as you like, I'm new to particle physics and when the diagrams are labelled wrongly it doesn't help understanding. Now it makes sense to me though, thank you :)
  19. May 20, 2010 #18
    I was also new once and thought that this was difficult! Now Iam doing my PhD in theoretical elementary particle physics :)

    good luck and please do not hesitate to ask questions here
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