Introduction to Astrophysics Neutrino flux The neutrino flu

Introduction to Astrophysics - Neutrino flux

The neutrino flux from SN 1987A was estimated to be 1.3 Times 10^14 m^-2 at the location of the Earth. If the average energy per neutrino was approximately 4.2 MeV, estimate the amount of energy released via neutrinos during the supernova explosion.

Solution

The flux ratio of neutrinos drops as the square of distance. The distance of 50 kpc to SN 1987A, and assuming a radius of about 1000 km for the neutrino-emitting core , we extrapolate to a flux of

=(1.3*10¹⁴)(50000*3.09*10¹³/1000)² =3.1*10⁴⁴ /m²

For this flux, the total number of neutrinos emitted over the surface of a sphere with radius of 1000 km is

=(3.1*10⁴⁴)*4pi*(1000*10³)² =3.9*10⁵⁷

If each neutrino carries 4.2 MeV of energy, the total energy carried by neutrinos is

=(3.9*10⁵⁷)(4.2*10⁶*1.6*10^-19)

=2.62*10⁴⁵ J