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Why gluon cannot decay to quark and photon

  1. Dec 18, 2011 #1
    What are flavor arguments that prevent a gluon from decaying into quark and photon, or anti-quark and photon, etc?
     
  2. jcsd
  3. Dec 18, 2011 #2
    1) Particles can only decay to 2+ body states with less mass than them. Since gluons are massless, it follows that they are stable.

    2) Pure gluons don't exist; QCD is confining, so a single gluon will hadronize into color singlets.

    3) The photon doesn't carry color charge, so it doesn't couple to the gluon. Conversely, the gluon doesn't carry electric charge, so it doesn't couple to the photon.

    4) Color is not conserved; you have a color 8 going to a color 3 x color 1. This is no good, as color symmetry is unbroken.
     
  4. Dec 18, 2011 #3

    tom.stoer

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    Let's assume that elduderino does not mean 'decay' but 'couple'; there is no gluon - quark - photon vertex b/c

    the guon lives in the SU(3) color octet, whereas the quark lices in the triplet representation; b/c the photon is a color singulet this vertex would violate color conservation (this is chrispb's 4)

    looking at the SM lagrangian there is simply no such vertex (there must not be such a vertex as the theory generated by ít would be inconsistent; this is chrispb's 3)

    chrispb's 2) is slightly misleading b/c "a single gluon will hadronize into color singlets" is wrong; a color octet cannot hadronize in color singulets

    chrispb's 1) is correct for free particles, ruling out a physical decay, but not a vertex.
     
  5. Dec 18, 2011 #4
    That question is breaking down the conservation of charge, Baryonic number (even the difference of Baryonic and Leptonic numbers).
     
  6. Dec 18, 2011 #5

    tom.stoer

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    baryon and lepton number need not be good quantum numbers as they are not related to local gauge symmetries; that's why I think it makes more sense to argue with electric charge and color charge conservation
     
  7. Dec 18, 2011 #6
    Another reason not specifically mentioned is that, because the gluon has no electric/electroweak charges but the quarks do, gluons can only take part in interactions that produce a quark and a matching anti-quark so that the total of these charges still adds up to zero.
     
  8. Dec 18, 2011 #7
    tom.stoer for what I know in StandarModel B and L numbers are totally correct. Maybe in theories beyond the SM don't need them, or need their differences or (blah blah blah).
     
  9. Dec 19, 2011 #8

    tom.stoer

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    Aren't their non-pertubative effects like "electroweak instantons" which can induce baryon- and lepton-number violation?
     
    Last edited: Dec 19, 2011
  10. Dec 19, 2011 #9
    I'm having a slight crisis of basic quantum mechanics right now, but it seems to me like there is a conservation of spin problem too. But when I continue that reasoning I can't see how gamma->W+W- could be allowed either (how can we start with 1 unit in some direction and end up with 2 units, or zero?), or is this one of those things that we can violate for a virtual process? I didn't think it was but perhaps I am wrong. Or perhaps I am forgetting something more fundamental than that. Can someone fix this embarrassing situation for me?
     
  11. Dec 19, 2011 #10
    i guess but they are not part of StandarModel. One coming in my mind almost immediately is the neutrinoless double beta decay- however it has not being observed.
     
  12. Dec 19, 2011 #11
    What do you mean? A complete mathematical description of the standard model is necessarily non-perturbative, since perturbation theory is just an approximation to the theory. The standard model doesn't only include perturbative effects. If it did, it would conflict with experiment.
     
  13. Dec 19, 2011 #12

    tom.stoer

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    Yang-Mills instantons and theta-vacuum tunneling are part of the SM, neutrinoless beta decay isn't.
     
  14. Dec 19, 2011 #13
    @tom.stoer: A vertex of the form gluon-photon-quark is excluded by symmetry arguments. I'm assuming one would get this external gluon as ISR or FSR off a pp collision or something of the sort; the initial state was a color singlet, so the final state had better be one too. The gluon will indeed hadronize and turn into a jet in this case. Either way, the point is that this vertex is not allowed.

    Yes, B and L can be violated via sphalerons, which are couplings to EW instantons, though I seem to recall B-L is conserved by these processes. The ability to interact with them is temperature-dependent; it's very hard to move into a different theta-vacuum at low temperatures (like ours). However, early in the universe, the potential energy one needed to move from theta-vacuum to theta-vacuum was small compared to the temperature of the universe, and so these processes could happen. In fact, this is one of the more popular avenues to explore in terms of baryogenesis.

    Neutrinoless double beta decay happens if neutrinos are Majorana. We don't know whether their masses are Majorana or Dirac. That's why we're looking for NDBD. It's up to you what you want to call the "Standard Model".

    Gamma -> W+ W- is indeed an induced coupling in the SM. Alternatively, you could imagine a W+ radiating off a Gamma; W+ -> W+ Gamma. The W+ carries electric charge, and consequently it couples to the photon.
     
  15. Dec 19, 2011 #14
    Yes I know it is allowed, I was just screwing up my conservation of angular momentum. I think I remember how this goes now: So we start with a gamma, say +/-1 unit of angular momentum in the z direction. This must get distributed between the W+ W- pair somehow. I was silly before though and forgot that these were massive and so the s_z=0 state is perfectly existing. So obviously one W has s_z=0 and the other has s_z=+/-1.

    So anyway, any vertex with two bosons and a fermion is therefore forbidden for this spacetime reason. You cannot start from s_z=-1,0,+1, subtract +/-0.5 units of angular momentum (spit out a fermion), and end up with another integer.

    At least this seems to make sense to me. I never thought about it before this though. Also I can't think of a counter-example :p.
     
  16. Dec 23, 2011 #15

    blechman

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    EXACTLY! I'm not sure why it took so long for someone to say this. Lorentz invariance requires that you have an EVEN number of fermions in any vertex.

    Gluons and photons certainly can couple to each other when going to higher order in the loop expansion. But it would require AT LEAST 2 gluons (or a gluon and at least 2 quarks) in order to get a color singlet.
     
  17. Dec 24, 2011 #16

    tom.stoer

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    @blechman: thanks, that's it
     
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