Determining the type of interaction

[maximus123]

Hi everyone,

I have the following question

When a positive pion interacts with a proton, a kaon can be produced, along with another strange particle, as shown in this equation

\pi^++\textrm{P}\longrightarrow \textrm{K}^++\textrm{X}

Circle the type of interaction shown

\textrm{Electromagnetic} \qquad\textrm{gravitational}\qquad\textrm{weak nuclear}\qquad\textrm{strong nuclear}

I know this is a strong interaction as I have gotten used to recognising them. Given a reaction equation, if I was not sure what the interaction type, I would be able to work it out by seeing if the quark composition was conserved. If it was then we have the strong nuclear force. However in the above example. Since we have not been given the quark composition of the X particle, or whether or not its strange particle is strange or antistrange, that is not possible in this case. So how else could I know by looking at the above (incomplete) equation, that it is an example of a strong interaction?

Thanks a lot for any help

 

[Meir Achuz]

It's true that if X is a proton, it would be a weak interaction. But the branching ratio to produce a proton is so small that it probably has never been seen experimentally. You could check on PDG to see if it has been seen, or what the upper limit for it to happen is.

 

[dukwon]

along with another strange particle

That should tell you that $X$ has non-zero strangeness. To conserve charge and baryon number, $X$ is going to have negative strangeness (regular baryons don't have valence antiquarks) and charge +1, like a $\Sigma^{+}$. The $K^{+}$ has strangeness +1, so if all the flavour numbers are conserved, this can go as a strong interaction.

I'm not sure why the first reply mentions branching ratios, since this is a $2 \to 2$ body scatter.

 

[SnoliF]

Just testing my current knowledge- would the particle $X$ be a $\Sigma^+$ ? And would this be the corresponding Feynman diagram?

Also, rather than a gluon, could the mediation particle be a photon or $Z^0$?

 

[ChrisVer]

I'm not sure why the first reply mentions branching ratios, since this is a 2→22→22 \to 2 body scatter.

because it was confusing. You are right since in that case there is no Branching ratio for proton production (proton is already there).

Also, rather than a gluon, could the mediation particle be a photon or Z0Z0Z^0?

quarks prefer interacting with gluons rather than photons (or even worse the massive Z bosons)... so to your question the answer is yes but not as likely.

W's apart from being heavy also have Cabibbo suppressed vertices.