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Dec 29, 2018 at 0:41 comment added RicardoP @probably_someone Ah, of course! Considering all the momenta that the $W$ boson can have is equivalent to consider all the time orderings of events!
Dec 29, 2018 at 0:40 comment added dmckee --- ex-moderator kitten As usual, there is no unique charge identity for space-like $W$s. The vertex could be either $ \nu_\mu + W^- \to \mu^-$ or $\nu_\mu \to \mu^- + W^+$ (this is usually depicted in diagrams by tilting the exchange line), and even if you could determine which it is in some particular frame there exist other frame where it has the other character.
Dec 29, 2018 at 0:35 comment added probably_someone @RicardoP The diagram you have drawn integrates over all possible momenta of the $W$ boson. This includes momenta that make the $W$ boson propagate from the quark to the neutrino, and also includes momenta that make the $W$ boson propagate from the neutrino to the quark. This is why it is drawn as a vertical line.
Dec 29, 2018 at 0:15 vote accept RicardoP
Dec 29, 2018 at 0:15 comment added RicardoP By the way, another question I had was if it is possible for the muon neutrino to decay into a W+- boson or must I necessarily have the quarks/anti-quarks decay into the boson instead?
Dec 29, 2018 at 0:12 comment added RicardoP Thank you for the answer! I agree and I wonder if the statement of the exercise is wrong since they clearly wrote the process with a muon- in the end state. I'll ask my professor about his possible mistake,
Dec 29, 2018 at 0:00 history answered probably_someone CC BY-SA 4.0