# Special Relativity problem - proper time interval

The supernova 1987A explosion in the Large Magellanic Cloud 170 000 light years from Earth produced a burst of anti-neutrinos ν ̄e which were observed in terrestrial detectors. If the anti-neutrinos are massive, their velocity would depend on their mass as well as their energy. What is the proper time inter- val between the emission, assumed to be instantaneous, and the arrival on Earth? Show that in the limit of vanishing mass the proper time interval is zero. What information can be derived about the anti-neutrino mass from the observation that the energies of the anti-neutrinos ranged from 7 to 11 MeV, and the arrival times showed a dispersion of 7 s?

I'm having some troubles with the first part of this problem. When it says proper time interval I'm assuming that just means to integrate dt'=dt/sqrt(1-(v/c)^2)? But I need to find the velocity that the neutrinos are traveling at. I know that I can find the mass of the neutrinos using E=mc^2 but could I also use the fact that m=(hv)/c^2 where h is the plank constant and substitute that into E=mc^2 to get the velocity as v=E/h? Then using that in the equation for dt' just integrate with the limits of t_emission and t_arrival? Also, would I use the fact that the light from the supernova arrived at earth in 1987 and that it is 170000 ly away as the t_emission and t_arrival times?

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## 1 Answer

You have$\frac E{m_0}$, the energy divided by the rest mass is $\gamma=\sqrt{1-\frac{v^2}{c^2}}$. The proper time is lab time divided by $\gamma$. Since you have a fixed $E$, as $m_0 \to 0, \gamma \to \infty$ and the proper time goes to $0$. For the last part, you are supposed to assume that an $11$ MeV neutrino arrived $7$ seconds before a $7$ MeV neutrino. This is in the lab frame and you can use the $7$ second time difference out of $170,000$ years to assess the lab velocity difference. It will give you a measure of the rest mass because if the neutrinos were massless they would all arrive together after traveling at $c$. If they were rather heavy, the time difference would be larger.