Before decaying into pions the Rho mesons can decay into a pair of charged rho mesons. Is this decay gluon mediated or does it happen just due to the quark exchange?
This isn't a decay process. A decay process means 1 meson is in the initial state, here there are 2. Of course, at higher order there will be gluons involved here.
The neutral rho is a superposition of quark-antiquark pairs. An additional quark-antiquark pair is created and the rho separates into two quark pairs, like in:
u ~u -> u ~u d ~d -> u ~d + d ~u (~p is the antiparticle of p)
The first process can be described as the decay of a gluon.
No-there doesn't exist any process that can be identified with gluon decay, i.e. that has a gluon in the initial state and the vacuum in the final state. However the diagram is misleading in the sense that the quarks that make up the rho mesons are in a bound state-and that bound state is possible through gluon exchange. The diagram just illustrates one possible transition amplitude between two states comprising u, ubar and d and dbar quarks.
The properties of the rho mesons are well known, cf. http://pdg.lbl.gov/2009/listings/rpp2009-list-rho-770.pdf
Incidentally, any such scattering process will involve gluons. Quark exchange is a misleading term, because there doesn't exist a fundamental four quark interaction-therefore the intersection point of the quark lines isn't a four-quark interaction vertex-the diagram is non-planar (so it's a disconnected diagram, that doesn't contribute to any scattering process) or there's a gluon exchange there.
no decay into charged rho mesons is possible because of energy conservation
Such a process is eventually possible inside a Quark-Gluon Plasma (QGP) via a recombination-coalescence mechanism (re-arrangement of quarks which are near in phase space). Of course gluons are involved in such a process.
The decay of a gluon gives a particle-antiparticle pair like for a photon. It must be a quark-antiquark pair since a gluon carries a colour. The pair is not created from nothing, neither form the vacuum since it would anihilate again immediately. For having two rho's instead of one, two additional valence quarks are needed.
The attached diagram is not related to the rho decay. This is inelastic scattering rho0 + rho0 -> rho+ + rho- Also the question "quark exchange" or "gluon mediation" is a misleading one. So called "quark exchange" is just a kind of visualization, hand waving approach. To CALCULATE this process, one should go into Feynman diagrams QCD formalism, then gluons are necessary. This scattering can be settled in the lowest order by one gluon emission decaying into q,qbar pair. In practice, te perform this calculation on the pure QCD level would be a real challenge :-(
The question about decay into (rho,rho) is sensible, assuming that one considers decay rho(1700) into rho(770) pair. This channel, however, is rather rare. Dominant channel is rho(1700)->(rho(770),pi,pi ) . The next one is rho(1700)-> 2(pi+,pi-)
Corresponding "quark exchange" or equivalent "gluon mediated" pictures (not diagrams!) are left as an exercise to the reader :-)
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Particle Physics
- Elementary Particle Physics