Why can't we detect neutrino-antineutrino annihilation?

  • #1
bcrowell
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Various astrophysical processes produce antineutrinos, which then fly off into outer space. I assume there are pretty accurate estimates of the production rates. I can imagine three possible fates for such an antineutrino: (1) annihilating with a neutrino, (2) interacting with baryonic matter, (3) ending up as the only particle inside its own cosmological horizon. It seems like we ought to have pretty good estimates of the rate of the neutrino-antineutrino annihilation process. Each such annihilation produces two back-to-back photons. If the neutrino masses are on the order of 0.1 eV, then these are infrared photons with wavelengths on the order of 10^4 nm. Why can't we detect these photons and thereby determine the neutrino mass spectrum? Are the peaks too weak? Too spread out by Doppler broadening?
 
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  • #2
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Too spread out by Doppler broadening?
Surely the Doppler broadening is so great that there would be nothing you could call a "peak" in the spectrum? Essentially all neutrinos are relativistic; the energy of the annihilation photons would be almost entirely determined by the neutrinos' kinetic energies and not their masses.
 
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  • #3
bcrowell
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Surely the Doppler broadening is so great that there would be nothing you could call a "peak" in the spectrum? Essentially all neutrinos are relativistic; the energy of the annihilation photons would be almost entirely determined by the neutrinos' kinetic energies and not their masses.
Makes sense!
 
  • #4
Vanadium 50
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The vv → γγ cross-section is truly microsocpic. (zeptoscopic?)

It's got to be photons, because everything else is kinematically blocked. It's got to go through the Z-pole, at 90 GeV when the neutrinos are at 1/40,000 of an eV. It also has to go through a loop which gives additional kinematic reduction, higher powers of the couplings, and a 1/16π2, which in light of the other factors, is almost too small to worry about.
 
  • #5
bcrowell
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The vv → γγ cross-section is truly microsocpic. (zeptoscopic?)

It's got to be photons, because everything else is kinematically blocked. It's got to go through the Z-pole, at 90 GeV when the neutrinos are at 1/40,000 of an eV. It also has to go through a loop which gives additional kinematic reduction, higher powers of the couplings, and a 1/16π2, which in light of the other factors, is almost too small to worry about.
The cross-section is small ... compared to what? Is the probability of this fate small compared to both of the other possibilities listed in #1? Negligibly small?
 
  • #6
Vanadium 50
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I would say "so small that I don't trust the estimation to be done correctly".

If you want a ballpark, I'd estimate that the vv cross-section divided by the vp cross-section is alpha x m(v)/m(p) x [m(v)/m(e)]^4 /16 pi^2.
 
  • #9
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Thanks Bill. I had acquired the notion that Majorana particles do not self annihilate. Upon further review, I find the consensus is that is only true for bosons, not leptons.
 
  • #10
ChrisVer
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? If neutrinos are majorana they can annihilate - we are supposed to find the Majorana nature by neutrinoless double beta decay.
 
  • #11
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Don't we know already that neutrinos can self-annihilate through a Z boson, because we observe that the Z can decay to neutrinos? ( simply the reverse process)
 
  • #12
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Thanks Bill. I had acquired the notion that Majorana particles do not self annihilate. Upon further review, I find the consensus is that is only true for bosons, not leptons.
It seems to me that this also isn't true, as two Z bosons ( which are there own antiparticles) can self-annihilate through a higgs boson.
 

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