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Do Gauge Bosons have anti-particles?

  1. Aug 9, 2011 #1
    1. Can two Bosons 'collide' in the same sense as the Fermions (Since Pauli's exclusion principle is not applicable for Bosons)?
    2. The Leptons have anti-leptons (positron, anti-muon, anti-tau and three anti-neutrinos). Each of the 6 Quarks have their corresponding anti-quark. So, do the gauge bosons (photon, gluon, Z & W, graviton) have corresponding anti-particles?
    3. If so, what happens when a gauge boson and it's anti-particle meet up at the same point in space?
     
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  3. Aug 9, 2011 #2

    Ben Niehoff

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    1. No, gauge bosons are bosons, they can share the same quantum state.

    2. Yes, all gauge bosons have antiparticles. The photon is its own antiparticle. The [itex]W^+[/itex] and [itex]W^-[/itex] are antiparticles of each other. The Z is its own antiparticle. A "red x anti-blue" gluon is the anti-particle of an "anti-red x blue" gluon, and similar for the other colors. The two diagonal gluons are each their own anti-particle.

    3. When a gauge boson and its antiparticle meet up, they annihilate and produce something else. The most commonly-known example is pair-production from two photons: [itex]\gamma + \gamma \rightarrow e^+ + e^-[/itex].

    And to answer the obvious follow-up question, "If photons are their own anti-particles and can annihilate against each other, why doesn't it happen all the time?", the answer is two-fold:

    First, there is not always enough energy to pair-produce. The photon is massless, but the electron and positron are massive. So the photons need enough momentum in order to annihilate and produce massive particles.

    Second, the process is reversible, as the electron and positron can also annihilate to produce photons. In fact, it's much more likely to happen in this direction, since the electron and positron have rest mass, and so always have enough energy to produce photons. So, in reality this process reaches an equilibrium where it is happening in both directions, and at low energies you will have many more photons around than pairs of [itex]e^-[/itex] and [itex]e^+[/itex].
     
    Last edited: Aug 9, 2011
  4. Aug 9, 2011 #3
    Thanks!

    I actually wanted to jokingly ask if the Gauge Boson and its anti-particle would annihilate and do things in the reverse and create a particle-antiparticle pair. What do I know...it turns out to be true :tongue2:

    >> So, in reality this process reaches an equilibrium where it is happening in both directions, and at low energies you will have many more photons around than pairs of e− and e+.

    • Can you give an idea about the energy level involved? Does a Gamma Ray Burst from a supernova provide enough energy for the reaction to happen in both directions equally?
    • Since photon is it's own anti-particle, do we have to limit our thinking that they have to annihilate in pairs? Is it impossible for 3 or 5 or more low energy photons come together and annihilate to create this particle-antiparticle pair?
     
  5. Aug 10, 2011 #4

    Ben Niehoff

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    a. Assuming a head-on collision of equal-energy photons, each photon must have at least as much energy as the rest mass of an electron. In general, you need conservation of energy and momentum.

    b. Yes, any number of photons [itex]\geq 2[/itex] can collide to produce an electron-positron pair. But the more photons you expect to collide, the less likely it will happen.
     
  6. Aug 10, 2011 #5
    Just to add, the rest mass of an electron is 511 keV, so you need at least 511 keV photons, which is solidly in the gamma-ray range. So these are reasonably serious photons. Sure, gamma ray burst photons are plenty high enough energy, but you'll have to wait until they collide with something for a chance they will produce some positrons. I am sure plenty of positrons are produced during gamma ray bursts for other reasons though.
     
  7. Aug 10, 2011 #6
    You mean higher order Feynman diagrams with >2 photons and an electron-positron pair?
     
  8. Aug 10, 2011 #7

    Ben Niehoff

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    Yes.
     
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