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$$ e^- + p \rightarrow \Delta^{++} + e^- + \pi^- $$ Apparently this reaction is mediated by the EM force. My question is: how do you know it isn't the strong force? Yes, all the particles have charge, suggesting it could be the EM force. But there are quarks involved too, so why is it not the strong force?

More generally, could anyone please direct me to a recipe to follow to determine which force a reaction is mediated by? I can't find a solid explanation anywhere on how to indisputably determine the correct force.

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    $\begingroup$ The only recipe is to write down the elementary particle content, the possible Feynman diagrams that obey all the conservation laws , and quantum number conservation laws. Then look at the allowed diagrams and estimate from the coupling constants which of the strong or electromagnetic or weak interaction diagrams will give the higher contribution. $\endgroup$
    – anna v
    Commented Apr 17, 2023 at 12:04
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    $\begingroup$ Both photons and gluons are exchanged in the interaction, so both forces are involved. The electron can only interact through photon exchange; any hadronization process where quarks rearrange the hadrons involved is strong. $\endgroup$
    – Cosmas Zachos
    Commented Apr 17, 2023 at 13:52
  • $\begingroup$ the reaction you have shown conserves neither charge nor baryon number. $\endgroup$
    – JEB
    Commented Apr 17, 2023 at 14:24
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    $\begingroup$ @JEB The reaction correctly preserves both baryon number and charge, if the $\Delta^{++}$ is the baryon with quark content $\rm uuu$. $\endgroup$
    – rob ♦
    Commented Apr 17, 2023 at 17:58
  • $\begingroup$ @rob. charges says $(-1) + (+1) \rightarrow (+2) \rightarrow (-1) + (-1)$ which is $0 \rightarrow 2 \rightarrow -2$, while Baryon number is $(0) + (1) \rightarrow 1 \rightarrow (0) + (0)$ which is better, it has one conserving step: $ 1 \rightarrow 1 \rightarrow 0$. Unless we're looking at different things, or I've been working too hard. $\endgroup$
    – JEB
    Commented Apr 17, 2023 at 18:10

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If the reaction were mediated by the strong force only then the electron (which does not feel the strong force) would be unaffected by the reaction and so would have the same energy on both sides of the equation. In which case it is not clear how the proton gets sufficient energy to "decay" into the more massive $\Delta^{++}$.

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    $\begingroup$ So, are we agreed? - the reaction is mediated by both the EM force AND the strong force? (If so then a 2022 A-level mark scheme is wrong!) $\endgroup$
    – Bazley
    Commented Apr 17, 2023 at 23:08
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Yes, there is a method to determine which forces can be involved in a given reaction. The standard model admits 11 distinct types of interaction vertices that you can use to draw Feynman diagrams. You can find descriptions of them here. Aside from drawing all possible Feynman diagrams, which is often difficult for a reaction involving many particles, there are a few tricks you can use.

If a neutrino is present, the weak force must be involved. Leptons cannot interact via the strong force. Quarks can change color, not flavor, by interacting with a gluon (strong force). Quarks can only change flavor through weak force interactions.

There are probably other such broad statements you can make based on the allowed vertices, but these are quite useful. Note that there are generally many possible allowed diagrams, which can involve different forces. The dominant diagram, and thus the more dominant decay method, will be determined by the values of the matrix elements represented by the diagrams.

You can construct a valid Feynman diagram for this reaction which only includes electromagnetic interactions. However, you can also draw valid diagrams that involve the weak and strong force. It's generally a good rule of thumb that if a process can occur electromagnetically and weakly, the electromagnetic mode will dominate because weak vertex factors are suppressed by the masses of the W and Z bosons.

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