How Is Neutrino Flux Calculated Through Earth?

[Darth Tader]

EDIT: I apologize for not noticing this earlier, but I realize this should be in the HW/Coursework section of the forum. If possible, would someone be able to move it over for me? Again, sorry for my error and any trouble/confusion.

I would like some help calculating the flux of neutrinos through the Earth.

I am given the energy released in neutrinos, 10 x 10⁵³ erg, and the mean energy 10MeV per neutrino. I am asked how many pass through a cm^2.

I know...
• Flux = N/(At) = E/(At) * N/E
• 10⁵³ erg = 6.24 x 10⁶⁴ eV, which leads to...
• 6.24x10⁵⁷ neutrinos.

I'm feeling lost on the next step as I believe I need the energy that passes through some area on Earth, but I'm not sure. It seems to simple and wrong to simply divide # of neutrinos by (1 x 10^-4 m)*s

Thanks for your help.

 

[mfb]

Only a tiny fraction of those neutrinos will hit a specific cm^2. Where do the neutrinos come from? Can you find some closed surface where those neutrinos will flow through (with the same rate everywhere)?

It is strange that the problem statement does not give erg/s, so it might be integrated over the whole lifetime. The number is larger than I would expect, however. Well, the average neutrino energy is wrong, too.

 

[Astronuc]

Considering that the energy output of the sun is ~3.86e33 ergs/sec, then the 10⁵³ ergs (I think that it should be erg/s as to which mfb alluded), must represent lots of stars, and at 20 orders of magnitude, it would be from all stellar objects in the universe.

One has the number of neutrinos coming at the Earth from all directions and they must pass through the surface (projected surface) in all directions.

It's similar to how neutron flux is calculated at point in a nuclear reactor.

 

[Darth Tader]

Sorry for the confusion, I should have posted the entirety of the question.

The question does say 10⁵³ erg, but it does make more sense to believe it means erg/s. I will post the full question though to try to alleviate any questions:

This problem considers the neutrino burst generated by SN1987A in the Large Magellanic Cloud (LMC) at a distance of 50kpc.

a. If 10⁵³ erg were released in neutrinos of mean energy 10Mev each, how many neutrinos passed throiugh a square centimeter on Earth?

b. For neutrino energies much less than 1 GeV, the cross section for scattering off of a nucleon is

σ_\nu = σ₀(E_v / m_ec³)

where σ₀ = 1.76 x 10^-44 cm²

and G_f is Fermi's constant. Show that the mean free path for neutrino scattering in matter with density ρ is

l_/nu = (2 x 10⁵ cm) (E_/nu / 10MeV)^-2 (p / 10¹² g cm^-3)^-1

c. If the neutrinos perform a random walk out of the proto-neutron star at the center of the supernova explosion, the time to diffuse a radial distance, R, is given by

t_diff = [itex]\frac{R*R}{l(nu)c}[/itex]Show that for a uniform density sphere of mass 1.4 M_{[itex]\odot[/itex]} the time for a neutrino to escape is

t_diff = (1.0 x 10^-2 s)(ρ/10¹²gcm^-2)^1/3(E_{[itex]\nu[/itex]}/10MeV)²

Evaluate this time for a typical (proto-) neutron star density of 1.0 x 10¹⁵ g cm^-3 and a neutrino energy of 20 MeV.

I don't expect much/any help on b) and c) as I have not shown any work on them. But I am unlear on them in general, so a slight push in the right direction would be greatly appreciated.

Does this help clear up any problems you may have had with part a) originally?

EDIT: I've worked out that the mean free path is equal to [itex]\frac{m}{ρσ}[/itex] where m = m_p, but I am lost on how the problem makes the leap to their mean free path value. I then calculated σ_/nu and got [itex]\frac{1}{ρ}[/itex] (2.59 x 10^-14 kg/m². This just seems wrong to me, but I'm not sure where I've made an incorrect calculation.

As for part c), I have solved the second part of the question and determined a t_diff = 0.4 seconds. For the first part of c), am I just plugging in the l_{[itex]\nu[/itex]} equation into the t_diff one? I seem to be missing terms when I do this.

Thanks again Astronuc and mfb, I really appreciate the help.

 

[mfb]

10⁵³ erg is the total energy released (as neutrinos) in the whole supernova then. That is fine. The given distance is important.

I don't understand what you did at (b). I think you skipped some steps in the explanation, so it is hard to follow.

For the first part of c: Use your result of (b). You have to find a relation between the stellar radius and its density, too.

 

[Darth Tader]

I am sorry for the ambiguity in (b), I will post my steps here.

The mass a 1m by 1m by 1m cube is M = ρV = ρ(1 m³)
So, the number of nucleons in this cube = # nucleons m^-3, N = M/m_p = ρ/m_p
Note: I followed someone's steps for the above calculations, here is their note:

For simplicity, let's say that the mass of a proton is equal to the mass of a neutron, and let's also ignore the mass of the electrons in the cube. With these assumptions, the number of nucleons in this cube of material is simply the mass of the cube divided by the mass of a proton.

By the above equation, Volume/nucleon = V_N = m_p/ρ

So, if the neutrino travels far enough, it will sweep out a volume equal to the typical volume occupied by one proton or neutron, and thus interact with the matter. We can therefore compute the typical distance the neutrino must travel before interacting L.

So, V_S = V_N
So, Lσ = m_p/ρ
Finally, L = m_p/ρσ

My result in the above (b) I found by using the given σ_{[itex]\nu[/itex]} equation using the given σ₀ and E_{[itex]\nu[/itex]} = 10 MeV (I chose this because I assumed E_{[itex]\nu[/itex]} was referring to the energy per neutrino.