Why Does π+ Decay to Muon and Muonic Neutrino?

[Apashanka das]

I am having a question in mind that why π⁺ decays to muon and muonic neutrino ,why no electron and electronic neutrino?

 

[mfb]

It is possible, but extremely rare. The decay happens via the weak interaction, so all particles involved must be left-handed. If you consider the spin of the involved particles, then the charged lepton has to be right-handed: If the leptons would be massless it would be completely impossible. Thanks to the non-zero mass, left-handed for the weak interaction (required for the decay) and left-handed for the direction of motion (forbidden by spin) are not exactly the same. With the large muon mass the difference is large, so this decay is quite likely. The electron is very light compared to the pion, it is close to the massless case - this decay is very rare (~1/10,000).

Wikipedia has a longer discussion of it

 

[Apashanka das]

Am I right that since electron is left handed and neutrino is also left handed that's why
doesn't match their orientation ,but in place of neutrino if anti neutrino is produced doesn't it work?

 

[mfb]

That would violate lepton number conservation.

 

[ChrisVer]

The two decays:
$\pi^+ \to \mu^+ \nu_\mu$

$\pi^+ \to e^+ \nu_e$
are possible via weak interactions.
However, the 2nd is helicity-suppressed, due to the low mass of the electron. The width (if you follow some simple QFT calculations) is given by:
$\Gamma ( \pi \rightarrow \mu\nu) = \beta_\pi m_\mu^2 \Big(1 - \frac{m_\mu^2}{m_\pi^2} \Big)^2$
Where I have isolated some constants related to weak interactions and pions into the $\beta_\pi$.
A similar calculation for the electron (in fact just replacing the masses which is the only thing that differs) will give you:
$\Gamma ( \pi \rightarrow e \nu) = \beta_\pi m_e^2 \Big(1 - \frac{m_e^2}{m_\pi^2} \Big)^2$
Now you only have to divide the widths to see what's the ratio of decays to muons to those to electrons:
$R(\mu / e ) = \frac{\Gamma ( \pi \rightarrow \mu\nu)}{\Gamma ( \pi \rightarrow e\nu)} = \frac{m_\mu^2}{m_e^2} \dfrac{\Big(1 - \frac{m_\mu^2}{m_\pi^2} \Big)^2}{\Big(1 - \frac{m_e^2}{m_\pi^2} \Big)^2}= \frac{m_\mu^2 ( m_\pi^2 - m_\mu^2 )^2 }{m_e^2 (m_\pi^2 - m_e^2 )^2}$
Just putting in numbers you will get the result...
$R(\mu / e ) \approx 8000$ (or as they say ~10,000 times more probable to decay to muons).​

 

[arivero]

It is always amusing to consider that if the mass of muon and pion were the same, the pion would be an stable particle at tree level, wouln't it? Well, considering also that the mass of electron is practically zero.

 

[Vanadium 50]

It is always amusing to consider that if the mass of muon and pion were the same, the pion would be an stable particle at tree level, wouln't it?

Why? The sum of the daughter masses is above the parent mass, so the decay is forbidden on kinematic grounds. Like zillions of others.

Well, considering also that the mass of electron is practically zero.

If the muon mass were equal to the pion mass and the electron were massless, I think you would have $\pi^+ \rightarrow \pi^0 + e^+ + \nu$.

 

[mfb]

It is always amusing to consider that if the mass of muon and pion were the same, the pion would be an stable particle at tree level, wouln't it?

It would still decay to electron plus neutrino. Its lifetime would be a factor 10,000 larger, but that is still just 0.25 milliseconds.
If you also make the electron massless, then you still have the decay to neutral pions or higher order diagrams. But the world as we know it wouldn't exist with massless electrons.