1
$\begingroup$

Is the following reaction allowed and why? $$ \nu_e \to e^- + \mu^+ + \nu_{\mu} $$

I would say it is allowed since individual lepton number and charge are conserved.

$\endgroup$

2 Answers 2

6
$\begingroup$

It is by lepton number and charge, but you can't get energy/momentum to balance. In the $\nu_e$ rest frame there isn't enough energy to make the products. If there is a nucleus around, you can imagine the $\nu_e$ emitting a virtual $W^+$ making the $e^-$, the $W^+$ scattering electromagnetically off a nucleus to deal with the momentum, then decaying into $\mu^++\nu_\mu$

$\endgroup$
1
  • 1
    $\begingroup$ Yeah, decays are always constrained by this. Collisions are easier--you can always imagine putting more energy into the collision. But a decay has to be possible in the rest frame of the single particle, and so is not so free. $\endgroup$
    – zeldredge
    Commented Feb 25, 2015 at 15:31
-2
$\begingroup$

Also : Lepton number conservation law, respectively Barion number conservation law, are laws which explain if any reaction can occur or not. Every particle has its own leptonic (barionic) number, and for the reaction to occur, the sum of those numbers in the right side of the equation MUST be equal to the sum of leptonic numbers in the left side of the equation. So if you have any doubt in any of these reactions, just look at the leptonic respectively barionic number of the particle.

$\endgroup$
2
  • $\begingroup$ Unfortuantly this completely fails to answer the question: there are no baryons involved, so baryon conservation is automatically assumed and the proposed reaction conserves both the overall lepton number and the lepton flavor numbers. The answer to the question is the one that Ross gave: it is disallowed by energy conservation. $\endgroup$ Commented Mar 14, 2015 at 21:25
  • $\begingroup$ I know that barions are not involved, however that is one of the conservation laws overall among with energy conservation law.All of those laws help to understand these kind of reactions so i wouldn't say my answer failed completely :) $\endgroup$ Commented Mar 14, 2015 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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