I am starting to learn about Feynman diagrams and I wonder if this reaction could happen as a physical process on its own, as opposed to part of a more complicated diagram:
$e^- \to W^- + \nu_e$
Following the logic used by David Z here:
First-order EM Feynman diagram?
The question asked if a first order electromagnetic diagram could happen on its own. For example, an electron coming in and emitting a photon
$e^- \to \gamma + e^-$.
The answer given by David Z was that this was forbidden because in the rest frame of the initial state ($e^-$ on its own), the energy was $E=m_ec^2$, while the energy in the final state is at least $E=m_ec^2 + E_\gamma$. I am wondering why we need to look at the rest frame of the electron in the first place.
My logic tells me that if an electron highly energetic comes in, it should be able to emit a photon and end up with less kinetic energy. This is wrong but could someone tell me why?
Following David's logic, the reaction above could not happen on its own because the rest mass of the electron is obviously much smaller than the reduced mass of the $W^- + \nu_e$.
Could someone clarify my ideas?