Is it possible with radioactive decay as pure radiation of neutrinos and antineutrinos? From a theoretical point of view?

I'm not asking for processes since I'm convinced there are no known processes resulting in pure neutrino radiation, that is decay without any other radiation than of neutrinos and antineutrinos.


2 Answers 2


All processes involving neutrinos are weak mediated because neutrinos have zero electromagnetic and color charge.

At tree level, all processes involving a $\nu + \bar\nu$ final state without other debris involve a time-like $Z^0$ (except neutrino NC scattering) which implies the annihilation of a particle/antiparticle pair (and will be suppressed relative outcomes with a photon or gluon mediator). That is not going to be a characteristic of "decay", though it is possible for reaction to have this final state.

Addendum: A beta-plus decay $^A_ZX \to \, ^A_{Z-1}\!X + e^+$ in an atomic context could, in principle see the positron annihilating one of the atomic electrons giving rise to the desired state. This will be suppressed by about $10^5$ relative the case where you get photons out.


I'm speculating here, basically looking at two considerations.

Considering you used the term "radioactive decay" I'm going to look at this from a gross nuclear viewpoint rather than a particle/subnuclear viewpoint.

If a nucleus is in it ground state, it would have no mode to lose energy without changing Z. So, in this case, no.

If a nucleus is not in its ground state, say a metastable state (or even a short-lived state), it could move to a lower state by emission of a photon. An alternative to a photon is internal conversion (inner shell atomic electron is pushed off the atom). Hmmm, since pair-production of massive leptons by photons is possible, why wouldn't pair-production of $\nu / \bar{\nu}$. Neutrinos definitely carry energy, lepton number is conserved, and cons. of angular momentum is possible. I can't think of an argument against the possibility of the process.

I strongly doubt with today's technology we could detect it above the noise. I'm unaware of theoretical predictions.

I admit this is not a comprehensive answer, but it's a starting point for discussion.

  • $\begingroup$ But if the source of the "decay" come from other kind of matter, as strangelets or something? Does the arguments about the ground state holds then. $\endgroup$
    – Lehs
    Jul 15, 2015 at 15:22
  • $\begingroup$ As I said, I'm limiting my answer to nuclei. I'm totally out of touch with strangelets. Someone else will need to address that possibility. $\endgroup$
    – Bill N
    Jul 15, 2015 at 15:25

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.