$\mu^+ \rightarrow e^+ + \nu_e + \bar \nu_\mu$

Consider the above decay, it is given in my lecture notes as a weak force interaction. I thought the weak force was only involved when quarks change flavours. Here we have no quarks. Why are pure leptonic decays a weak force interaction?

  • $\begingroup$ beta decay is probably the first process where the weak force is mentioned in school. The two vertices of the W boson do both - interact with quarks and with leptons, see e.g. en.wikipedia.org/wiki/Beta_decay#/media/… $\endgroup$
    – Vangi
    Commented May 24, 2019 at 14:35

1 Answer 1


"Only the weak force can change quark flavours" is probably what you heard. This is not the same as "the weak force is only involved when quark flavours change". All fermions couple to the weak force. In fact, neutrinos only couple to the weak force. If an interaction involves neutrinos, then the weak force has to be involved.


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