I'm writing the diagrams for the following process in Standard Model: $$\nu_e + e^+\rightarrow \mu^++\nu_\mu+\gamma$$ I want to know if the W boson changes the flavor of $\mu^+$, for instance, necessarily to its anti-neutrino counterpart $\bar{\nu_\mu}$ or it can also change into $\nu_\mu$.
$\begingroup$
$\endgroup$
8
-
$\begingroup$ Muon and electrons are different flavors, not muons and muon neutrinos. As for what neutrino appears with a muon depends on muon number conservation. $\endgroup$– TriatticusCommented Jun 27, 2019 at 1:49
-
$\begingroup$ I'm confused. I thought that within the leptons category we could distinguish between 6 different flavors: electron, muon, tau, electron neutrino, muon neutrino and tau neutrino. Am I wrong? $\endgroup$– RicardoPCommented Jun 27, 2019 at 2:03
-
1$\begingroup$ you can distinguish, but each lepton number is conserved in their own right. The reaction you write annihilates the flavor number so just energy and charge remains, an annihilation, ( like e+e-, anything can happen). It is conservation of charge on the right that asks for a W+, , which then has to conserve whatever lepton it creates with antilepton. $\endgroup$– anna vCommented Jun 27, 2019 at 4:51
-
$\begingroup$ @annav I think the OP is considering a $t$-channel process. $\endgroup$– JEBCommented Jun 27, 2019 at 5:43
-
$\begingroup$ @JEB depending on charge conservation in the specific process. anything that is not forbidden by conservation laws can happen, though it might be of very low probability. $\endgroup$– anna vCommented Jun 27, 2019 at 6:00
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
1
I want to know if the W boson changes the flavor of μ+, for instance, necessarily to its anti-neutrino counterpart $\bar ν_μ$, or it can also change into $ν_μ$.
No it does not, in the Standard Model. All vertices there conserve lepton number, so the virtual $W^+$ involved will decay to just $\mu^+ ~ ν_μ$. To instead go to $\bar ν_μ$ would violate lepton number by 2. The photon in the reaction will attach to any charged line in your diagram.
Equivalently, only $\bar ν_μ$ can go to $W^- ~ \mu^+$ while $ν_μ$ can never do this.
When it comes to the (anti-)neutrino subscripts μ,e,..., in the standard model they merely define the special linear combinations of "real" neutrino mass eigenstates which couple the charged leptons μ,e,... to a W. (Some people like to call these definitions as enforcement of "lepton flavor".) They are fancy wavepackets mutating among themselves in all neutrino oscillation experiments, but that is a longer, recondite story, far outranged by this question.
answered Jun 27, 2019 at 14:44
-
$\begingroup$ The last paragraph is indispensable for the complete picture when neutrino mutation shall NOT be ignored. $\endgroup$– MadMaxCommented Jun 27, 2019 at 15:23