# Neutrino mass

I tried and I got a ridiculous large number for the neutrino mass. I basically used $$E=\gamma m c$$ and then, for 10 MeV neutrinos, time taken is t + 10 seconds to reach Earth, for 50 MeV neutrinos, time taken is t seconds.
$$Speed = \frac{100 000}{(t+10)}$$ for 10 MeV and $$Speed = \frac{100 000}{t}$$ for 50 MeV. Here's the question:

Neutrinos were detected from a supernova (distance ~ 100 000 light years) over a time period of ~ 10 seconds. The neutrinos have an energy range from 10 MeV to 50 MeV.
What's the mass of the neutrino, assuming all neutrinos were produced at the same time and their mass is small compared to their energy.

Last edited:

Meir Achuz
Homework Helper
Gold Member
Your speed is in LY/sec. Did you convert units?
Speed in LY/Y is so close to one that you have to use a Taylor expansion.

Your speed is in LY/sec. Did you convert units?
Speed in LY/Y is so close to one that you have to use a Taylor expansion.

Yes, I did unit conversions. A light year = $$9.46 \times 10^{15} meters$$

Here's what I did:

for 10 MeV neutrinos, $$10\;Me = \frac{m\;c}{\sqrt{1 - (\frac{d}{t+10})^2/c^2}}$$

and hence, $$t+10 = \frac{10\;M\;e\;d}{10\;M\;e\;c}(1-\frac{m^2\;c^2}{10^2\;M^2\;e^2})^{-1/2}$$

Binomial expansion, keep only first two terms yield:
$$t+10 = \frac{d}{c}(1+\frac{m^2\;c^2}{2\times10^2\;\M^2\;e^2})$$

do the same for 50 Mev neutrinos,

for 50 MeV neutrinos, $$50\;Me = \frac{m\;c}{\sqrt{1 - (\frac{d}{t})^2/c^2} }$$

simultaneous equations, cancelling t, yield:

$$10 = \frac{d\;m^2\;c}{2\;M^2\;e^2}(\frac{1}{10^2}-\frac{1}{50^2})$$

Solving for m (neutrino mass), I got 7.7 GeV !?

Last edited:
Meir Achuz
Homework Helper
Gold Member
I get: 10 sec=(10^5 yr X m^2/2)(1/100-1/2500), with m in MeV.
I don't see where your e^2 came from, why your c is i the numerator.