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enter image description here

On this picture you can see that this $d$ quark turns into $u$ quark and makes this $W^-$ boson. $W^-$ then apparently turns the anti-electron neutrino, $\bar{\nu_e}$, into an electron, $e^-$.

My question is why can $d$ turn into $u$ and how is this $W^-$ produced?

And how does this $W^-$ boson turn into electron?

(what I found on Google didn't explain it really clearly to me)

Kyle Kanos
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zoran404
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  • You're reading the Feynman diagram wrong. Time goes upwards. The arrows (see this question) don't correspond to time. Here, the neutron turns into a proton and a W$^-$ boson, and the W$^-$ turns into an electron and an anti-electron neutrino. – Peter Shor Jul 21 '13 at 18:19
  • BTW--Most particle physics type prefer time horizontal to the right for these diagrams, but both conventions do occur so you have to stay alert. – dmckee --- ex-moderator kitten Jul 21 '13 at 18:33
  • @Peter Shor, why do you say I'm reading it wrong? – zoran404 Jul 21 '13 at 18:40
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    @zoran404: do you see that little t with an arrow? That means time goes up. So it is the case that the W$^-$ turns into an electron and an antineutrino, and not the case (as you say) that the W$^-$ turns an anti-electron neutrino into an electron. If that happened, lepton number would not be conserved. – Peter Shor Jul 21 '13 at 20:53
  • It's not the time arrow that confused me, it is that the arrow for Ve is pointing to W- not from it, like the arrow for e-. Why is it pointing in? – zoran404 Jul 21 '13 at 21:48
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    @zoran404 in Feynman diagrams, anti-matter particles have arrows that point "backwards" from their normal-matter counterparts. It doesn't represent "movement" or "time", it's just a convention to distinguish matter from anti-matter. The anti-electron neutrino and the electron formed at the same time and moved away from each other. – KutuluMike Jul 21 '13 at 22:23
  • That explains a lot – zoran404 Jul 21 '13 at 22:35

2 Answers2

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The $d$ quark can turn into the $u$ quark because they are particular quantum states of the quark (generally called flavours), in this case the quantum state the weak isospin state $T_3$.

The $W^-$ boson is the particle that changes this flavour, so it is generated as a result of the beta-decay of the neutron (in the process of $T_{3}=-\frac{1}{2}\to T_3=+\frac12$). The $W^-$ boson is unstable with a half life of $\sim10^{-25}$ seconds, where it decays into an anti-neutrino and an electron.

Kyle Kanos
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    please note that the W- is off mass shell, because certainly the neutron does not have the mass to generate an 80GeV+ real particle. Off mass shell particle have all the quantum numbers of the particles with their name, except their four vector is off the mass value they have as real particles. – anna v Jul 21 '13 at 18:22
  • I read on wikipedia and it is not clear to me what is this anti-neutron? What happens to it after it's made? – zoran404 Jul 21 '13 at 18:38
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    It's a neutrino, not a neutron, and it is another sub-atomic particle (in particular, a fermion). After being produced, the anti-neutrino will propagate onwards, possibly forever since it is not affected by gravity or E&M, but via the weak force (with a range $\sim10^{-17}$ meters). – Kyle Kanos Jul 21 '13 at 19:01
  • propagate onwards, what does that mean? Does it effect anything in anyway? – zoran404 Jul 21 '13 at 20:02
  • By "propagate onwards," I mean "move as sub-atomic particles do." The neutrino can pretty much pass through anything without stopping. – Kyle Kanos Jul 21 '13 at 20:39
  • Can it be detected? – zoran404 Jul 21 '13 at 21:50
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    Not directly. When neutrinos do interact with matter, they produce charged particles (either electrons or muons) that we can detect. Duke University has a good web page on neutrinos. – Kyle Kanos Jul 21 '13 at 22:10
  • I want to make clear that the neutrinos do have masses, even though very small, and thus are affected by gravity too, even at the Newtonian framework, and certainly by General Relativity, as the have a four vector. http://en.wikipedia.org/wiki/Neutrino#Mass – anna v Jul 22 '13 at 03:42
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I want to elucidate a bit that this diagram is a type of Feynman diagram. This means it guides us to calculate the integrals that will give us the lifetime of the particle decaying.

Calculable Feynman diagrams can be written for all three forces ( week strong electromagnetic). The W is the exchange particle of the weak force and the two vertices have very well determined values and functional forms.

So the down quark can decay to the up quark because it is energetically possible and the decay is mediated by the W-, is the way to read this diagram. All particles are off mass shell except the neutron, the proton , the electron and the antineutrino_e . Off mass shell means the particles have all the quantum numbers of the particles with their name, except their four vector is off the mass value they have as real particles

anna v
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