Are there any fundamental rules of thumbs that can be used to identify the type of force interaction (weak, electromagnetic, strong) in a particle decay without drawing the Feynman diagrams at the beginning of the problem. At least to rule out some types of forces from a given decay reaction, before start drawing the Feynman diagrams.

  • If a particle changes flavor, it's a charged-current weak decay. Example: $n\to pe\bar\nu$.

  • If there's a neutrino in the final state, it's a weak interaction. Decay example: $\pi^+\to\mu^+\nu$. See also neutrino scattering.

  • If parity isn't conserved, it's a weak interaction. Examples: $K^0 \to 2\pi$ and $K^0 \to 3\pi$. Note that kaon decays and $K\leftrightarrow\bar K$ oscillations also change strangeness, a flavor quantum number.

  • If only hadrons are involved, and all the flavor quantum numbers are conserved, it's a strong interaction. Examples: $pp \to p\Delta^{++}\pi^-$, $pp \to p\Lambda^0 K^+$.

  • If photons are involved, it's an electromagnetic interaction. Examples: $\pi^0\to\gamma\gamma$, $e^+e^- \to \gamma\gamma$, $\gamma + {}^AZ \to {}^{A-1}Z + n$

This isn't an ironclad rulebook, because the separation of interactions into strong, weak, and electromagnetic is something we can do artificially since we live in a world where the intrinsic energy associated with each interaction is orders of magnitude different. All the fundamental interactions contribute, at some level, to all of the decays.

For example deuterium formation with cold neutrons ($np\to d\gamma$) is a transition from a strong state (unbound $np$, isospin 1) to a strong state (bound $np$, isospin 0). This transition liberates a magnetic dipole photon because no hadronic degree of freedom exists to carry away the binding energy; the strong matrix elements are presumably the same as in a process like $\pi^0d \to np$.. The photons in $np\to d\gamma$ have a tiny parity-violating asymmetry due to the weak neutral current acting between the neutron and proton.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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