# Neutrinos from supernova

Observations of neutrinos emitted by the supernova SN1987a place an upper limit of 20 eV on the rest energy of the electron neutrino. If the rest energy of the electron neutrino were, in fact 9.8 eV, what would be the speed difference (in terms of cm/s) between light and a 3.1 MeV electron neutrino?

Answer

$E = m*c^2$

$m = m0/\sqrt{(1-v^2/c^2)}$

$E = m0c^2/\sqrt{(1-v^2/c^2)}$

$E = E0/\sqrt(1-v^2/c^2)$

$\sqrt{(1-v^2/c^2)} = E0/E = 9.8/3.1*10^6 =3.16*10^{-6}$

$1-v^2/c^2 =9.994*10^{-12}$

$V^2 = 0,9999999999950031217481664958328*C^2$

$C-V =1.5*10^{-3} m/s =0.15 cm/s$