# Muon neutrino interacting with nucleon's quarks

I'm considering the muon neutrino/anti-quark interaction with the nucleons of ice in the IceCube experiment. The purpose is to decide which antiquarks are $$\bar{q}$$ and $$\bar{q}'$$. I have the following process: $$\nu_\mu+\bar{q}\rightarrow \mu^-+\bar{q}'$$. Since we are dealing with nucleons, I suppose in this case anti-nucleons, the $$q's$$ can only be either $$\bar{u}$$ or $$\bar{d}$$. In order to have charge conservation, I've drawn the following:

My problem concerns the $$W$$ boson. If I assume that the $$\bar{u}$$ decays into a $$W$$, the latter has to necessarily be a $$W^+$$ boson, in order for the $$\bar{u}$$ change its flavor to $$\bar{d}$$. However, the $$\nu_\mu$$ won't turn into muon by absorbing a +1 charged boson, correct? It had to be a $$W^-$$ boson then. However, if $$\bar{u}$$ is to transform into a $$\bar{d}$$, then it can't be a $$W^-$$ boson, right? Hence, my confusion. (vice-versa if it were the muon neutrino emitting the muon).

If you require that you have $$\nu_\mu\to\mu^-$$ during the reaction, then there must be a $$W^-$$ boson exchanged, which (if you are limited to $$\overline{u}$$ and $$\overline{d}$$ quarks) requires that you start with a $$\overline{d}$$ quark and end with a $$\overline{u}$$ quark.
If, on the other hand, you require that $$\overline{u}\to\overline{d}$$ during the reaction, then there must be a $$W^+$$ boson exchanged, which requires that you start with a $$\overline{\nu_\mu}$$ and end with a $$\mu^+$$.
• @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. – probably_someone Dec 29 '18 at 0:35
• 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. – dmckee Dec 29 '18 at 0:40
• @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! – RicardoP Dec 29 '18 at 0:41