Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we assume that electrons (just like neutrinos) are massless, why can’t the decay $\pi^- \rightarrow e^- + \bar{\nu}_e$ occur under the weak interaction?

share|cite|improve this question
    
I believe a pion can decay in an electron and an antineutrino check wikipedia – SaudiBombsYemen Jan 25 at 15:08
    
@Pablo Not when the electron is massless. – pfnuesel Jan 25 at 15:12
1  
It's a helicity-suppressed decay, and massless particles can't take helicity values opposite to their chirality. – dukwon Jan 25 at 15:19

Since the spin of the charged $\pi$ is $0$, the spins of the daughter particles need to add up to $0$ as well, i.e., their spins need to be anti-parallel. That's nothing else than the conservation of angular momentum.

Assuming the anti-neutrino to be massless, it is always right-handed. Right-handed means that the momentum vector and the spin vector are parallel, while left-handed means that the momentum vector and the spin vector are anti-parallel. This is well defined for a massless particle, since it travels at the speed of light, and there is no inertial frame in which the momentum vector would switch direction.

Since the anti-neutrino is right-handed, the electron would need to be right-handed as well to conserve linear momentum and angular momentum (spin). But the decay of the $\pi$ happens via the weak interaction, i.e. via the $W$ boson. Since the $W$ boson is known to couple only to left-handed particles, the decay would be forbidden.

Since the electron is not massless, it has a small left-handed component. The decay is suppressed, but not forbidden. The heavier muon has a larger left-handed component, and its decay is less suppressed. Hence, pions usually decay into muons, although they have less phase space available.

share|cite|improve this answer

pfnuesel's answer is absolutely correct and very satisfying to understand, and you should read it before this one.

There is, however, a route which would permit $\pi\to e+\nu$ even if the electron were massless. The conserved quantity which suppresses that decay is not spin, but total angular momentum. There exists an electron+neutrino wavefunction with spin $S=1$, orbital angular momentum $L=1$, and total angular momentum $\vec J = \vec L + \vec S$ of magnitude $J=0$; this wavefunction is available for pion decay and competes with the $S=0, L=0$ channel described by pfnuesel.

However, like all high-$L$ decays, this one is suppressed because the overlap of a wavefunction with nonzero $L$ and the parent particle with radius $R$ is proportional to $R^L$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.