2
$\begingroup$

My professor told me that kaon producing reactions have a small cross section compared to other nucleon collision reactions. The reason for that is the production of a strange quark but he didn't elaborate further on why this actually is the case. I remember reading that reactions that involve quarks of different generations have lower probability of occuring but they mentioned this in the context of the weak interaction. Does the same still hold in reactions where kaons are produced via the strong interaction such as in nucleon collisions?

The Kaon producing reactions I'm talking about. B are nucleons, Y are hyperons and R resonances

$\endgroup$
7
  • $\begingroup$ If you google "feynman diagrams kaon production" you will see there is always a W involved, which means the weak coupling constant. $\endgroup$
    – anna v
    Commented Oct 6, 2021 at 5:39
  • $\begingroup$ Hello, i added a Feynman diagram for the reactions I mentioned. In this diagram no W is involved so I don't think that it is necessary that Kaon are produced only via the weak interaction $\endgroup$ Commented Oct 6, 2021 at 8:25
  • $\begingroup$ your diagrams are here researchgate.net/figure/… and have a caption "Kaon production processes in the resonance model. B i (i = 1, 2, 3), Y and R stand for respectively, either the nucleon or the ∆, either the Λ or the Σ hyperon, and the baryon resonances.". I think if your write down the quark interactions there will alwasy be a weak vertex in the final expression $\endgroup$
    – anna v
    Commented Oct 6, 2021 at 10:10
  • $\begingroup$ I read the paper these figures appear in. No lagrangian I see includes the weak coupling constant for the interaction between the Hyperon Kaon and nucleon. Shouldn't these lagrangians include the weak coupling constant if the resonances do decay via the weak interaction? $\endgroup$ Commented Oct 6, 2021 at 10:59
  • 1
    $\begingroup$ Ειμαι συνταξιουχος, (απο το 2000) ερευνητρια στο Δημοκριτο, ΙΠΦ, σωματιδιακη φυσικη, πειραματικος. $\endgroup$
    – anna v
    Commented Oct 6, 2021 at 12:39

1 Answer 1

0
$\begingroup$

I remember reading that reactions that involve quarks of different generations have lower probability of occuring but they mentioned this in the context of the weak interaction.

Flavor changing goes through the weak interaction.

Does the same still hold in reactions where kaons are produced via the strong interaction such as in nucleon collisions?

If there is no flavor change, it is the strong coupling that is involved

It depends on the particular production mechanism. What the diagrams you display show is the decay of hyperon resonances though the strong interaction, which has lifetimes of order $10^{-23}$ seconds. As you can see in the table here there are no lifetimes measured for the resonances. I expect the width is controlled by experimental errors which would mask any strong interaction width.

( this link might interest you)

$\endgroup$
2
  • $\begingroup$ I think the link you provided makes it clear why there exists such a big difference in lifetime between the weak and the strong interaction. But I still don't understand why the strangeness producing strong interactions are less probable than other reactions involving the strong interaction. $\endgroup$ Commented Oct 6, 2021 at 11:41
  • $\begingroup$ I think it is because they start from having an s quark in the quark content of the particular interactions, which means already there has been a weak vertex in the construction of the original s quark. $\endgroup$
    – anna v
    Commented Oct 6, 2021 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.