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jossives
Sep16-08, 12:14 PM
Let's take for example the decay K+ -> pi+,pi0,gamma

This can proceed via direct emission or inner bremsstrahlung. I have questions for each of these modes

1. For inner bremsstrahlung, it is a final charged product that radiates, correct? In this case the pi+. What is causing the deceleration of the pi+ such that it radiates?

2. Does direct emission only come from from virtual particles or can it come from (in this example) one of the 4 quarks in the final state?

Thanks for any help in advance

hamster143
Sep16-08, 06:39 PM
I don't think it's appropriate to talk about "deceleration" in this context. IB is simply a radiative QED effect.

By definition, DE occurs before pions are fully formed. Fully formed pi0 is neutral and can't radiate, gammas emitted from fully formed pi+ are part of IB.

jossives
Sep16-08, 09:44 PM
So the DE photons come only from the virtual particles and IB from the fully formed ones? How do you differentiate between these two modes experimentally?

hamster143
Sep17-08, 12:29 AM
IB is the dominant mechanism and DE is a tiny correction to that. IB width is easily calculated using QED, DE is much more nontrivial because you can't fully explain processes that go on during the decay of K in terms of virtual particles and perturbative QCD. I'm not an expert, but I'd imagine that spectra would be different.

See this presentation, hope it helps

http://www.lnf.infn.it/wg/vus/workshop07/Talks/Raggi.ppt