# Pion+ and proton make kaon+ and another strange particle, X. Why is this the strong interaction?

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?

• 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. Jan 6 '20 at 16:34
• 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. Jan 6 '20 at 17:39
• 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. Jan 6 '20 at 17:43