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For certain processes with low momentum transfer, such as the Kaon decay shown below, quarks will form bound states of mesons.

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Whereas for higher momentum-transfer processes, such as the decay of an on-shell heavy particle to a quark-anti-quark pair, you expect the quarks to instead hadronize to form either a pair of hadronic jets or a single large-$R$ jet. My question is what precisely determines this behaviour. Consider as an example the decay of a new right-handed neutrino $N$, which occurs via an off-shell $W$. For the $W \rightarrow qq$ decay channel, when would we expect this to give $N \rightarrow l \pi$ and when would we expect $N \rightarrow ljj$ ($j$ here meaning jet)? I know that it involves some relation between the momentum transfer and the QCD scale, but the precise criteria are not clear to me.

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  • $\begingroup$ the kaon is a bound state of the quark antiquark pair, as far as our models in mainstream physics go, whereas the W is modeled as a point particle with the mass of W that decays into constituents., and neutrinos do not decay, as they have very small mass, and there would not be enough energy in their center of mass . $\endgroup$
    – anna v
    Commented Dec 2, 2023 at 5:56
  • $\begingroup$ also how could there be quantum number conservation? $\endgroup$
    – anna v
    Commented Dec 2, 2023 at 6:30
  • $\begingroup$ Jets are collections of mesons, and sometimes baryons, in a fast frame. What's your point? $\endgroup$
    – Cosmas Zachos
    Commented Dec 2, 2023 at 14:28
  • $\begingroup$ Review. $\endgroup$
    – Cosmas Zachos
    Commented Dec 2, 2023 at 14:41
  • $\begingroup$ @CosmasZachos maybe my question was not clear. Above a certain energy scale a pair of quarks will undergo a Parton shower and give rise to many hadrons in a jet. In low momentum transfer processes this will not happen and instead particles will decay to pairs of pions. The exact mechanism behind this is what I am asking $\endgroup$
    – Jackson Burzynski
    Commented Dec 3, 2023 at 6:08

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