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A neutrino has rest-mass and travels at (near) $c$, why isn't its mass/ energy (nearly) infinite? Because it has too low rest mass or still too low speed. Neutrinos are very light particles: Their rest energy is comparable to energy of a hydrogen bond (weaker than typical chemical bound). So you can understand, that full energy must not be infinite. ...


It is more likely that the non-detection would be associated with statistics than SNR's not accelerating protons. Fermi-LAT has already shown that $\gamma$-ray emissions from 4 galactic supernova remnants (with molecular clouds nearby them) are coming from proton-proton collisions leading to neutral pions ($pp\to pp\pi^0$, $\pi^0\to2\gamma$): (source, ...


Solar neutrinos have energies under 10 MeV. There isn't enough energy to make heavy leptons. That's really the whole story.


Neutrino definitely have anti-particles. Another question whether an anti-particle differs from its "particle" or not. The charge conjugation operator may generally change a given "neutrino state" because a neutrino solution is not completely determined with just Dirac's equation.


It sounds like that site is discussing the Standard Model neutrino. Neutrinos were presumed to be massless for a long time and the SM still models them as massless. We now know that this is not true, that (at least two of the three flavors of) neutrinos do have a small mass, although the SM is still a good approximation. So, like other massive particles ...


Correction of typos and clarifications In the Friedmann equation, '$\rho$' is strictly speaking $\rho_m$, the mass density. Hence the presented Friedman equation has to be changed as follows: \begin{equation} H = \sqrt{\frac{8\pi}{3}G\rho_m}=\sqrt{\frac{8\pi}{3}\frac{G\rho}{c^2}}, \end{equation} in order to use the expression given below for the energy ...

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