Are thermalized neutrinos stable at temperatures their rest energy?

AI Thread Summary
The discussion centers on the stability of thermalized neutrinos at temperatures near their rest energy, particularly at the last scattering surface (LSS). It is argued that neutrinos, with a low interaction cross section, may remain stable despite high temperatures, contrary to particles with significant cross sections that become unstable. The stability of the electron neutrino is emphasized due to lepton number conservation, while the decay of mu and tau neutrinos is complicated by neutrino oscillation, preventing energy conservation violations. The implications of neutrino mass on energy density at the LSS suggest that if neutrinos have mass, it could affect the expected ratios of neutrino to photon energy density, potentially requiring adjustments in baryon-photon ratios. The conversation highlights ongoing questions in neutrino physics and the need for further research on their properties and interactions.
BillSaltLake
Gold Member
Messages
183
Reaction score
0
At the last scattering surface (LSS), the energy density of neutrinos is argued to be ~0.68 of the energy density of blackbody photons, based on a thermodynamic equilibrium argument. This is also required to obtain the correct total energy density for stable expansion. At that time, the average BB photon energy (2.7kT) was ~0.7 ev with the neutrino energy a little lower. If neutrinos have mass, there is evidence that the combined mass of the 3 varieties is < 0.3 eV.
I believe that particles with a significant interaction cross section are unstable (or exist in reduced numbers) at temperatures much higher than the energy eqivalent of their rest mass. Because of the low cross section of neutrinos, will this heat-instability effect not reduce the ~0.68 factor, or are neutrinos intrinsically stable with heat simply because they're leptons?
 
Space news on Phys.org
BillSaltLake said:
At the last scattering surface (LSS), the energy density of neutrinos is argued to be ~0.68 of the energy density of blackbody photons, based on a thermodynamic equilibrium argument. This is also required to obtain the correct total energy density for stable expansion. At that time, the average BB photon energy (2.7kT) was ~0.7 ev with the neutrino energy a little lower. If neutrinos have mass, there is evidence that the combined mass of the 3 varieties is < 0.3 eV.
I believe that particles with a significant interaction cross section are unstable (or exist in reduced numbers) at temperatures much higher than the energy eqivalent of their rest mass. Because of the low cross section of neutrinos, will this heat-instability effect not reduce the ~0.68 factor, or are neutrinos intrinsically stable with heat simply because they're leptons?
I'm not sure why you'd think they would be unstable. They pretty much can't be. My argument goes like this:

First, consider the electron neutrino. This is necessarily stable because it is (most likely) the lightest neutrino, and thus the lightest particle with lepton number = 1. It just can't decay because lepton number is conserved and so there's nothing lighter to decay into.

One might, at first, think the mu and tau neutrinos may potentially decay. But they could only decay into lighter neutrinos. This is because they don't interact with photons (having no charge), and because those are the only lighter particles around. But I don't think this can work due to neutrino oscillation. My argument starts with the example of a mu neutrino decaying into two electron neutrinos and an electron anti-neutrino. Here you have three neutrinos where before you had just one. This becomes a problem when you take into account neutrino oscillations, because now these three new neutrinos can oscillate back into mu neutrinos and decay again! So a universe where neutrinos both oscillate and decay into one another would also be a universe where neutrinos continue to multiply through division after division, producing a massive violation of conservation of energy. I just don't think that's possible. From this argument, I suspect that if you delved deeply into the details of the standard model, the fact that neutrinos don't decay would leap out clear as day. But my specific knowledge of the standard model isn't good enough to say for certain.
 
This is not the way to think about it.

For decays, you need to think about mass eigenstates, not flavor eigenstates. In the SM, the lightest neutrino, nu_1, being the lightest fermion, is absolutely stable. The heavier ones n_2 and n_3 can decay only if they are at least three times as heavy as nu_1, and decay via your three neutrino mode. A world where nu_3 weighs about 0.05 eV and the other two are much lighter is compatible with experiment.

However, the lifetime of such a neutrino would be ~1033 years.
 
Thanks. The flip side of this type of question concerns the effect of any neutrino mass on the total neutrino energy density. At the LSS, the average neutrino temperature was ~2150K, so 3/2(kT) = 0.28eV. If the average mass were as high as .05 eV, there could be a discrepancy.
Equilibrium was established much earlier than the LSS. In early times, ρneuρhot = 0.68 and both of the ρ scale as 1/a4. If mass becomes significant, ρ decreases more slowly, so that at the LSS, we might expect ρneuρhot > 0.68, which causes ρtotal to be too high unless the baryon:photon ratio is adjusted downward. Are there any papers that discuss this subject?
 
I am not an expert, but such a mass spectrum would give a mix of hot and cold dark matter at the LSS, a mix that would be different for the normal and neutrino hierarchies. Since that's still an open question, I can only assume that the observables from both models are too close for us to tell them apart.
 
https://en.wikipedia.org/wiki/Recombination_(cosmology) Was a matter density right after the decoupling low enough to consider the vacuum as the actual vacuum, and not the medium through which the light propagates with the speed lower than ##({\epsilon_0\mu_0})^{-1/2}##? I'm asking this in context of the calculation of the observable universe radius, where the time integral of the inverse of the scale factor is multiplied by the constant speed of light ##c##.
Why was the Hubble constant assumed to be decreasing and slowing down (decelerating) the expansion rate of the Universe, while at the same time Dark Energy is presumably accelerating the expansion? And to thicken the plot. recent news from NASA indicates that the Hubble constant is now increasing. Can you clarify this enigma? Also., if the Hubble constant eventually decreases, why is there a lower limit to its value?
Back
Top