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How are W Bosons Formed in Beta Decay

  1. Jan 15, 2012 #1
    I understand how beta decay works on a fundamental level, in that either an up or down quark in the one of the nucleons decays into an up or down quark and in the process, a W boson is emitted which in turn decays into an electron and an electron antineutrino or opposites thereof depending on which quark decayed.
    What I don't understand is how the W boson is formed in the first place.

    Also, in Beta- decay, a neutron 'decays' into a proton which I can understand because a proton is smaller than a neutron. What happens to that excess mass, I also don't know - does it go towards the W boson?

    If that's the case, then how can a proton 'decay' into a neutron - gaining mass?

    Thanks for any insight...
     
  2. jcsd
  3. Jan 15, 2012 #2
    I am not entirely sure what you are asking. Are you asking where the W comes from? It is no different to asking where the photon comes from when two electrons repel each other. Perhaps thinking of them in the field theory picture will make more sense to you: there are quark fields and W boson fields permeating all of space and the physical particles are just quantised excitations of these fields; because the fields are coupled together through the charges of the weak nuclear force it is possible for energy to flow from one field to another, so the energy to make the W of course comes from the quark.
    Perhaps you are confused by the fact that the W is really heavy and the quark doesn't have that much energy to give away? Well you can either say that the uncertainty principle means you can make a W anyway so long as it doesn't last very long, or you can say that you make a really light W (which also doesn't last very long because it's not a "real" W). Energy is conserved in the end because the neutrino/electron pair the W turns into is very light.
    I like to think of it this way. The quark field doesn't couple directly to electrons and neutrinos, so you have to send the energy through some different field first. This is the W field. However, the W field has much heavier excitations so it takes a lot of effort to make the field "ring up" enough to transfer the energy through from the quark field to the electron/neutrino fields. But this is quantum theory, so you can send the energy through this channel anyway, but it just happens with a very small probability, sort of like you have to tunnel through the big potential barrier of the giant W mass.

    Well, for proton-neutron "decay" you have to remember we aren't talking about isolated protons. As long as anybody has watched them for, no-one has ever seen a free proton decay, basically for the very reason you mention. Inside a nucleus, however, the nucleus mass overall can be lowered by changing a proton to a neutron, due to the change in nuclear binding energy.

    As for the excess mass, it sort of goes towards the W I guess, in that if you have more of it you can make the W more easily (so the decay will occur more quickly), but it still won't be enough to make a real W so you can only make virtual ones. In the end the energy ends up in the mass of the product electron-neutrino pair, and in the kinetic energy of that pair and the daughter nucleus.
     
  4. Jan 15, 2012 #3
    Something related that I'm a bit confused about concerns the beta decay of charged pions, say for argument's sake π+. As I understand it, the (most common mode of) decay goes

    π+ (= u dbar) → W+ → μ+ + vμ

    But the π is spin 0 and the W spin 1, so how can the W be created?
     
  5. Jan 15, 2012 #4
    So, what about the energy required for the quark? If the energy transfers from the quark field to the W field, then what is left for the actual quark in the nucleon?
    And what about when a W boson decays into quarks? I now understand virtual particles properly because the final products conserve energy. But how can energy be conserved when you have turned one quark into another, but in the process, have ended up with an extra two quarks from the W boson decay?
     
    Last edited: Jan 15, 2012
  6. Jan 15, 2012 #5

    Bill_K

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    The W is a massive vector particle. Its propagator has a numerator gμν - pμpν/M2 which guarantees that on the mass shell there are only three polarization modes and the W has spin 1. However off the mass shell all four polarizations propagate including the 'timelike' one, which is spin 0.
     
  7. Jan 16, 2012 #6
    Neutron decay works this way:
    https://upload.wikimedia.org/wikipedia/commons/8/89/Beta_Negative_Decay.svg

    A single down-quark decays into an up-quark and sends out a W- boson to compensate for the charge difference (since the conservation of charge must be kept). Quickly afterwards, the W- boson decays into an electron and an electron anti-neutrino.

    At high energies, this process can be shown to be true. At low energies, it can be said that the energy to create the W- boson is temporarily borrowed.

    By using the Feynman diagram rule called crossing, we can change outgoing particles to incoming particles and change it into an antiparticle to keep a legit process. Thus, starting from neutron decay, we get this:

    n → p+ + anti-ve + e-
    p+ + anti-ve + e- → n
    p+ → n + ve + e+
     
  8. Jan 16, 2012 #7

    Bill_K

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    This is false. The W boson does not "borrow" energy, it is created off the mass shell.
     
  9. Jan 16, 2012 #8
    Thanks!
     
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