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Do boson's have anti-particle versions of themselves?

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  1. Apr 28, 2015 #1
    i have been researching this question and have so far not found a conclusive answer. i have read in a few places that bosons may even be their own anti particles, but a) i don't understand what this means, and b) this idea doesn't seem to be consistent between articles. hopefully someone on this forum could shed some light on the matter?
     
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  3. Apr 28, 2015 #2

    e.bar.goum

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    Yes, some bosons, like photons, Z bosons, Higgses, gluons, neutral pions, are their own antiparticle. If the graviton exists, it should also be its own antiparticle. By the definition of antiparticles, the electromagnetic charge of such a boson must be 0.

    This means that two of these bosons may annihilate against each other, much like any antimatter collision.
     
  4. Apr 28, 2015 #3
    so, if a gluon is its own anti particle, that means that if it were to come in contact with another gluon the two would annihilate each other? wouldn't this happen very often then?
     
  5. Apr 29, 2015 #4

    e.bar.goum

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    Sometimes. Just like all antimatter-matter collisions, the annihilation isn't automatic -- if the particles get close to each other, they can just scatter. The cross-section (probability) for annihilation doesn't have to be large. For instance, two photon reactions are incredibly rare, even though a photon is it's own antiparticle.

    There are plenty of studies out there about gluon-gluon collisions. E.g. Strangeness production in Quark-gluon plasmas. http://en.wikipedia.org/wiki/Strangeness_production or charm production http://www.sciencedirect.com/science/article/pii/0003491678902701

    The precise physics of gluon-gluon interactions is a part of Quantum Chromodynamics (QCD) which is rather out of my league. Perhaps a more knowledgeable member will come along if you have more questions about gluons.
     
  6. Apr 29, 2015 #5
    well if its out of your league it must be way above my head! doesn't mean i won't look into it and learn what i can. thanks for the information it was very helpful!
     
  7. Apr 29, 2015 #6

    ChrisVer

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    Well yes, depends on what you have in your mind as annihilation.
    The gluons can be annihilated into quarks : [itex]gg \rightarrow qq[/itex] by the mediation of another gluon, the interaction is a strong interaction.
    The two gluons can also interact at a point interaction and give two gluons [itex]gg \rightarrow gg [/itex] (they have a 4 point interaction vertex) as well as mediated 2 gluons to mediate another gluon and give 2 gluons.



    Some other bosons are not their own antiparticles. In any case you have to check if your particle is an eigenstate of the charge conjugation operator, if yes it's its own antiparticle, if no then it's not. An example the [itex]\pi^+ ~(\text{Spin}=0)~ , W^\pm ~(\text{Spin}=1)[/itex] are not their own antiparticles. For the pion for example you have:
    [itex]\hat{C} | \pi^+ > = a | \pi^->[/itex]
     
  8. Apr 29, 2015 #7

    mfb

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    I'm not sure how useful it is to call a gluon-gluon interaction to two new gluons "annihilation". At low energy, this is not even a countable process (in the sense of "123 interactions happened") as perturbation theory does not work.

    There are 8 types of gluons, it is possible to see two of them as being their own antiparticle, then the other 6 form 3 pairs.
     
  9. Apr 30, 2015 #8
    Gluons have a nonzero conserved color charge. Thus they are not their own antiparticle.
     
  10. Apr 30, 2015 #9

    ChrisVer

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    At low energy it's not even useful to talk of strong interactions as an SU(3) symmetry because it's not perturbative.

    For that I need some help... Isn't the [itex]\bar{8}[/itex] (you could say anti-gluons) equivalent to [itex]8[/itex] (gluons) for the adjoint representation of SU(3) ?
     
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