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W Boson Energy (Mass)

  1. May 11, 2015 #1
    I have read that W boson get their energy (mass) by temporarily breaking the energy conservation laws in accordance with the Uncertainty principle. I have also read that the W bosons get their mass through the Higgs mechanism. How are these two things reconciled? Is this the distinction between a virtual W and a real W?
     
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  3. May 11, 2015 #2

    mfb

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    Where did you read that? It is completely wrong.
    That is right.
     
  4. May 11, 2015 #3
  5. May 11, 2015 #4

    mfb

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    That is a problematic way to try to put formulas into words.
    Energy conservation is not violated. The W mass can be "wrong" (lower than the mass of a real W) for virtual particles. That is a violation of the energy/momentum relation, but both energy and momentum are conserved exactly and everywhere. They just don't match the properties a real W would have.

    Note that he describes it as "apparent violation" (44:40), it is not an actual violation.
     
  6. May 11, 2015 #5

    ChrisVer

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    I think his comment is exaggerated.
    There is no energy violation, and the energy is conserved in all vertices individually.
    The virtual particle however is an off-shell particle, which means that its 4-momentum squared is not equal to the particle's rest mass. In formula what is violated is the equality : [itex]E^2 - |p|^2 c^2 = m^2 c^4 [/itex].
    The mass would have to be replaced by [itex] s= q_w^\mu q_{w \mu} \ne m_W^2 [/itex].

    If someone wants to use the Uncertainty principle to describle virtual particles he generally says that there can be fluctuations in energy [itex]\Delta E[/itex] within some time [itex]\Delta t[/itex], that obeys the [itex]\Delta E \Delta t \sim \hbar[/itex]. This can help you get an insight about how the particle pops up into being. Of course it's not a real particle, meaning that you cannot observe it. In order to observe it you have to use higher energies and make it pop up into being real (that's a resonance)
     
  7. May 11, 2015 #6
    OK, so the Higgs mechanism actually generates the mass and the Uncertainty Principle approach is not quite true but gives some insight as to what is going on. So in the case of neutron to proton decay, where a down quark switches to up and emits a W-, what actually happens at the vertex? The Higgs generates the mass and there is a charge transference (weak isospin/weak hypercharge) to the W-. Is this also through the Higgs field or a different mechanism?
     
  8. May 11, 2015 #7

    ChrisVer

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    The higgs mechanism gives the W boson its mass by letting it interact with the Higgs field's vev... The virtual particle doesn't have the W boson's mass although it's a W.

    Also what do you mean by what happens at the vertex? That W boson which appears does not have 90GeV mass (that's impossible from the kinematics). So it should be lighter than 90GeV - should be virtual. What it does is to make the transition of the "quark" and also transfer some momentum.
     
  9. May 11, 2015 #8
    OK, I think I am getting more confused. The Higgs gives mass to a real W correct? The virtual W has less mass and mediates the down to up quark transition. It has some mass. Where does that mass come from? Apologies in advance for my lack of understanding. I am a QM guy from years ago!
     
  10. May 12, 2015 #9

    ChrisVer

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    Nowhere, there is no fixed mass.. It's a whole spectrum. It can be any value and that depends on probabilities.
    And it comes from the fact that the virtual particle carries some 4mommentum from the quarks to the electron/neutrino.
     
  11. May 12, 2015 #10

    Orodruin

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    I believe you need to rethink your perception of what a virtual particle is. It is not something that you can measure. The mass of a field is what it is and is related to the pole of the propagator. Now, Feynman diagrams are a graphical representation of terms in a mathematical expression. They do not "mean" that a reaction occurs in that way, just as little as an electron goes through the left slit in a double slit experiment. What suppresses the terms for weak interactions is that the momentum in that term is way smaller than the value needed to be at the propagator pole, this is what we call an off shell propagator.
     
  12. May 12, 2015 #11
    OK. I think I got it. Thanks for you help and patience.
     
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