A Are neutrinos created by pair production

Cato
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Assume just for this question that a neutrino has a mass of 1 ev. That is about the energy of an infrared photon with a wavelength of 1 micron. Is it possible for a visible light photon, or a more energetic photon, to create a neutrino-antineutrino pair?
 
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Cato said:
Assume just for this question that a neutrino has a mass of 1 ev. That is about the energy of an infrared photon with a wavelength of 1 micron. Is it possible for a visible light photon, or a more energetic photon, to create a neutrino-antineutrino pair?
What do you think? Have you done any research on this?
 
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Cato said:
Is it possible for a visible light photon, or a more energetic photon, to create a neutrino-antineutrino pair
The photons don't (directly) couple to neutrinos, since the neutrinos are chargeless. So the fast answer is no...
Neutrino pairs can be produced by the Z-boson decays. The Z decay Z \rightarrow l \bar{l} can also have neutrinos in the place of leptons.
 
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It is possible, but as photons do not couple to neutrinos, it has to be an indirect process. It needs to involve the weak interaction and the electromagnetic interaction combined, which makes it ridiculously unlikely.
 
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mfb said:
It is possible, but as photons do not couple to neutrinos, it has to be an indirect process. It needs to involve the weak interaction and the electromagnetic interaction combined, which makes it ridiculously unlikely.

And it requires a bystander particle to shed some momentum. By itself, photon-to-neutrino-pair process would violate energy and momentum conservation.
 
Evidently, it does. Otherwise, no neutrino oscillation.
 
nikkkom said:
Evidently, it does. Otherwise, no neutrino oscillation.
nah, my reply was wrong... even if it's massless the problem comes from 1->2+3... so even if 2 is massless you can still go to a CoM frame in the final state.
But the neutrino oscillations indicate that at least 2 neutrinos are massive (we have 3).
 
nikkkom said:
And it requires a bystander particle to shed some momentum. By itself, photon-to-neutrino-pair process would violate energy and momentum conservation.
Assuming a neutrino has mass (something only demonstrated for two mass eigenstates out of three). But even if an eigenstate of neutrino is massless, a photon has no time to decay to a neutrino pair - same reason a graviton has no time to decay into photons.

Can an electronically excited state of an atom decay by emission of neutrino/antineutrino pairs? An electron on orbit around a nucleus does have a legal interaction with neutrino. And what are the selection rules for such emission?
 
snorkack said:
Assuming a neutrino has mass (something only demonstrated for two mass eigenstates out of three). But even if an eigenstate of neutrino is massless, a photon has no time to decay to a neutrino pair - same reason a graviton has no time to decay into photons.

I don't understand what you mean by this.
A photon _does_ decay into a pair (or more) or fermions, namely e+e-, if it has sufficient energy and has something to bump into.
 
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nikkkom said:
And it requires a bystander particle to shed some momentum. By itself, photon-to-neutrino-pair process would violate energy and momentum conservation.
Sure, same as "regular" pair production.

I guess the leading order process would be photon -> electron+positron, those two exchange a W boson to produce neutrinos. Then add a photon from a nucleus to the electron or positron to conserve energy and momentum.
Photon -> e+ e- -> Z -> neutrinos could also work, with the same additional photon as above.
 
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nikkkom said:
Evidently, it does. Otherwise, no neutrino oscillation.

oscillations and production are not the same things.
 
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That post was a reply to a (now deleted by the author) post about neutrino masses.
 
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malawi_glenn said:
oscillations and production are not the same things.
that was a reply to a deleted comment from me, in which I said that if the neutrino was massless then the energy-momentum conservation wouldn't be violated (which is obviously a wrong statement from my side).
 
  • #16
Thank you all for the discussion. It is helpful.
 

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