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Original question and what I think the Feynman diagram is

The answer is the strong nuclear force.

After having looked at Feynman diagrams online for similar interactions I figured out an interaction for this one that I have drawn above. Is that correct? I believe I understand the interaction that I have drawn but I don't understand why it's impossible to be any other type of interaction.

As K+ has a strangeness of +1, X must have a strangeness of -1 to conserve strangeness. I know strangeness can change by +/-1 in weak interactions, but as we know X has strangeness, it must have a strangeness of -1 to satisfy strangeness not changing, or changing by +/-1.

X must be a baryon to conserve baryon number and it must have a charge of +1e to conserve charge.

So, it therefore has three quarks, one being a strange quark, and the other two must be either up, charm or top quarks to conserve charge.

Why must it be two ups (if it even must be that)? If a quark changed flavour from an up to something else the weak force would be in play. Is that a possibility?

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    $\begingroup$ When the leading contribution to a process dominates by orders of magnitude people tend to neglect the other contributions unless they plan to talk about the sub-dominate contributions in particular. The more so when the EM contribution is not separable as here so you are comparing strong to weak. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jan 6, 2020 at 16:34
  • $\begingroup$ Ah, so the decay that I have written above is much more probable than any other decay, so we are ignoring the other interactions. I'm not sure what you mean by the EM contribution is not separable, would you mind clarifying? Thanks. $\endgroup$
    – Kiran Govind
    Jan 6, 2020 at 17:39
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    $\begingroup$ I mean there is no experimental way to say "fraction $f_{EM}$ of the decays recorded can be attributed to the electromagnetic process because ...". As opposed to weak interactions, where you can (in principle) use parity viloation to extract the scale of the weak contribution. I say "in principle" because while we do that for processes which can only proceed by weak and EM channels ($G^0$, $q_\text{weak}$, etc...) it's still a bit impractical for cases where the leading contribution is strong. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jan 6, 2020 at 17:43

1 Answer 1

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As dmckee says in the comments, you have found a first order diagram.

What are Feynman diagrams? they are the diagrammatic representations of complicated integrals that will give the cross section of an interaction. These integrals are a series expansion , and each term has a diminishing constant allowing to use the first terms to first order, the way you have drawn. They are very useful because they simplify adding up quantum numbers that have to be accounted for.

It is the strong interaction because there is a gluon in first order, i.e the gauge particle of strong interactions. If you look at the ordering of interactions in terms of the coupling constant, you will find that a first order diagram with a photon exchange, which also creates quark antiquark , is suppressed by its coupling. The weak interaction is suppressed by its coupling even more.

Please note though that because of the coupling being 1 the series of perturbative diagrams for strong interactions do not converge and the first order diagram you have drawn cannot be calculated with perturbation expansion (There are innumerable diagrams with gluon exchanges in higher orders ).New mathematics are used depending on the calculation needed, dominant being QCD on the lattice. The diagram is useful for quantum number conservation display but not for calculating.

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  • $\begingroup$ Ah, I see. So a very similar Feynman diagram could be drawn with a photon turning the down-antidown pair into a strange-antistrange pair, but because the coupling constant/strength is 1/137 of the strong forces' that interaction will happen with a lower frequency. Also, a weak force diagram could be drawn with the same quark change mentioned in the previous sentence but it would be mediated by a Z boson? This would be much much less likely due to the coupling constant. Would that be a first-order interaction of the weak force? $\endgroup$
    – Kiran Govind
    Jan 8, 2020 at 10:32
  • $\begingroup$ @KiranGovind the weak force not only has even smaller couplings , but the large mass of Z is enormous and deoresses crossections more for energies lower than its mass. It contributes at LEP energies for example. $\endgroup$
    – anna v
    Jan 8, 2020 at 19:12

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