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Beta Decay

  1. Feb 28, 2012 #1
    From what I understand, which is kind of limited, the neutron (939Mev) decays into a proton (938Mev) giving off a HUGELY massive particle called a W boson (80,000Mev). The W boson exists for 3*10^-25 seconds then gets transformed into an electron (.511Mev) and an electron anti-neutrino (.28ev). My question is where does all that mass/energy of the boson go, and where does it get it from? The electron and neutrino are tiny but has lots of kinetic energy, surely it didn't convert all that 80,000Mev to the kinetic energy of the electron and neutrino? Another question: in the Feynman diagram, why is the anti-neutrino shown going backward in time?
     
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  3. Feb 28, 2012 #2

    jtbell

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    The W is a virtual particle, which means it doesn't behave like a "real" particle. In many popular-level treatments you find descriptions of virtual particles being able to violate energy conservation or "borrow energy from the vacuum" for a short period of time via the Heisenberg uncertainty principle. Many physicists (including most particle physicists AFAIK) prefer to think of them as conserving energy and momentum, but not having the same mass as "real" particles. (The jargon for this is that virtual particles are "off the mass shell.")

    Going further, many (but not all) physicists consider virtual particles to be mathematical artifacts of the intermediate stages of a particular method of calculating experimentally observable quantities in quantum field theory, that is, they don't "really exist," and are basically pictorial devices to help keep track of the steps in a calculation. There have been many contentious arguments on this point in these forums, which tend to devolve into debates about what "real" means, with "dueling quotations" from prominent physicsts who take one side or the other.
     
  4. Feb 29, 2012 #3
    But isn't the photon a virtual particle also, and if so, woudn't it behave similarly to a W boson (never took QFT so am wondering about this)? Yet we have evidence the hoton really exists (unless this is what you mean by "real").
     
  5. Feb 29, 2012 #4

    jtbell

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    There are virtual photons and real photons. Virtual photons mediate the interaction between (say) two colliding electrons; real photons are what you see by, and give you sunburn.

    Similarly there are virtual W-bosons and real W-bosons. Virtual W-bosons mediate the weak interaction in things like beta decay. Real W-bosons are very rare because they have so much mass and require so much energy to produce; as far as I know we've "seen" them only in experiments at high-energy particle colliders like the one at CERN (where they were first observed in the mid 1980s, for which Carlo Rubbia got his Nobel Prize).

    When I wrote, "The W is a virtual particle" in post #2, I should have written "The W in beta decay is a virtual particle."
     
    Last edited: Feb 29, 2012
  6. Feb 29, 2012 #5

    DrChinese

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    In many ways, I think the same debate does extend to photons. Some say that a photon is similarly a mathematical artifact, others saying that it is "real". Clearly, the reason we know that photons "exist" it because they are emitted and absorbed. Which is equivalent to the mathematical viewpoint as much as any other (which asserts they are real).

    One question I have asked and never seen a satisfactory (or definitive) answer to is: do free photons exist? Specifically, are there photons from the CMBR which will forever go on without interacting with matter?
     
  7. Feb 29, 2012 #6
    Thanks, that clears it up.
     
  8. Feb 29, 2012 #7
    So, it 'borrows' 80,000Mev from the vacuum, exists virtually for 3*10^-25 seconds then most of that 80,000Mev disappears back into the vacuum. Some of it is transformed into the mass of the electron and anti-neutrino, giving them their kinetic energy. Is that really what is going on in beta decay? I also want to know, why is the anti-neutrino shown in the Feynman diagram to be going backward in time before interacting withe the W boson? Or is it the neutrino going back in time, getting changed into a forward in time anti-neutrino when it interacts with the W boson?
     
  9. Feb 29, 2012 #8

    jtbell

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    No. The energy of the original nucleus equals the sum of the energies of the final nucleus, the outgoing electron and the outgoing antineutrino, where in each case "energy" includes the object's rest-mass energy. In this picture the energy "borrowed" from the vacuum serves only to provide the supposed mass-energy of the intermediate W-boson.

    As I noted before, most particle physicists (at least AFAIK) consider the intermediate (virtual) W-boson not to have 80 GeV of mass-energy to begin with, because it's "off the mass shell," so there's no need to explain where that energy comes from.
     
  10. Feb 29, 2012 #9
    No energy is ever borrowed from the vacuum. The W boson does not have an energy of 80,000 MeV, only 939 MeV-938 MeV = 1 MeV. This 1 MeV then goes into the mass and kinetic energy of the electron and antineutrino. You might object that the W boson is supposed to have a mass of 80,000 MeV, so it must have at least that much energy, but virtual particles by definition violate the equation E^2 = p^2 + m^2.

    The backwards arrow on the antineutrino line is a convention that is used to distinguish particles from antiparticles. This is a useful convention because this way if you follow a fermion line through any Feynman diagram, the arrows on that line should always keep pointing the same way along the line. The arrow does not indicate anything going backward in time.
     
  11. Feb 29, 2012 #10
    Are you sure about that? I read that in Feynman diagrams, time progresses from the bottom to the top or from the left to the right. And that, if the arrow is pointing in the opposite direction, it means that the particle could be thought of as going backwards in time. I could be wrong though, it was in that Quantum World for Everyone by Ford.
     
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