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Consider the following decay equation:-

A Kaon(+) decays into a muon(-) and an antineutrino(?)

The question is as follows :- State and explain whether the anti-neutrino is an electron, muon or tau neutrino.

After practising a couple of Feynmann diagram I guess that the answer to the first part must be muon antineutrino, But I don't know the reason, could anyone please explain?

Thanks...

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Have you met the conservation rules of a feynmann diagram?

They basically say the following:

"At each vertex (that is every point on the feynmann diagram you see more than one branch touching that point) there are several quantities that must be conserved"

The rules slightly differ in QCD (that's roughly speaking the study of quarks and weak interaction) but always the following quantities are conserved

1.Electric Charge (ALWAYS CONSERVED; Total charge entering vertex = total charge leaving)

2.Baryon number (ALWAYS CONSERVED = Number of baryons you start with = number you end with)

  1. INDIVIDUAL Lepton number (that means electron lepton number, muon lepton number, tau lepton number)

4.4-Momentum (This is usually implicit in the diagram; for the purposes of your understanding just be aware of it)

5.Colour charge (this refers to gluons in QCD)

6.Quark Flavour* (The number of quarks entering a vertex= number leaving; however the FLAVOUR is allowed to change in WEAK interactions only! it is completely conserved however in EM and Strong Interactions)

In your case, you start of with

$K^+$ which is equivalent to $K^+=u\bar{s}$

So you start of with the following values before the vertex (only listing the important ones here)

Charge = +1

Baryon Number = 0 (since a Kaon is a meson)

Electron Lepton Number = 0

Muon Lepton Number = 0

Tau Lepton = 0

You then produce a antimuon $\mu^+$ (you can't have a negative charge coming out of nowhere!) in the future so on the other side of the vertex you have the following values:

Charge = +1

Baryon Number = 0 (Since a muon is also a meson)

Electron Lepton Number = 0

Muon Lepton Number = -1

Tau Lepton Number = 0

There are only two ways to reduce our muon lepton number to 0; either add a muon $\mu^-$ or a muon neutrino $\nu_\mu$ (since both of these are leptons/lepton neutrinos so they will contribute a muon lepton number of +1)

But if we choose to add the $\mu^-$ this will change our charge.

Therefore the obvious choice is to use the muon neutrino $\nu_\mu$

Therefore the interaction we are after is

$K^+ \longrightarrow \mu^+ + \nu_\mu$

Which is a well established decay mode. As a question - can you guess what the virtual particle should be? (Hint: our flavour of our quarks had to change in order for this interaction to even be possible - that should narrow down what type of interaction we are dealing with!)

Feynmann Diagram for $K^+$

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  • $\begingroup$ Analogously, you can simply replace this equation with each of the antiparticles $$K^- \longrightarrow \mu^- + \bar{\nu}_\mu $$ $\endgroup$ – MKF Nov 13 '15 at 13:41
  • $\begingroup$ Since the strangeness is not conserved, this implies that the interaction is a weak interaction. This leaves us with the W and the Z boson. But the W bosons carry charges whereas the charge does not change, which means that it must be Z boson. Please correct me if my reasoning is wrong. $\endgroup$ – model_checker Nov 13 '15 at 14:28
  • $\begingroup$ You're so close! so Weak interaction is correct! but it turns out to actually be the W boson because we need to "carry" the charge from the Kaon to the muon as you said! I'll draw a diagram to better emphasise this $\endgroup$ – MKF Nov 13 '15 at 14:44
  • $\begingroup$ Wow! that was insightful, and all the while I thought that the W boson would leave the particle 'chargeless'. Now I understand the real meaning of the term 'Exchange Particle'. It is as if the Kaon has exchanged its charge with a muon. Is it a W(-) boson? $\endgroup$ – model_checker Nov 13 '15 at 16:41

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