# Neutrino flux through Earth

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1. Apr 21, 2013

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

Last edited: Apr 22, 2013
2. Apr 22, 2013

### Staff: Mentor

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.

Last edited: Apr 22, 2013
3. Apr 22, 2013

### Astronuc

Staff Emeritus
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.

4. Apr 22, 2013

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:

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 $\frac{m}{ρσ}$ where m = mp, but I am lost on how the problem makes the leap to their mean free path value. I then calculated σ/nu and got $\frac{1}{ρ}$ (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 tdiff = 0.4 seconds. For the first part of c), am I just plugging in the l$\nu$ equation into the tdiff one? I seem to be missing terms when I do this.

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

Last edited: Apr 22, 2013
5. Apr 22, 2013

### Staff: Mentor

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.

6. Apr 22, 2013

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/mp = ρ/mp
Note: I followed someone's steps for the above calculations, here is their note:
By the above equation, Volume/nucleon = VN = mp

So, VS = VN
So, Lσ = mp
Finally, L = mp/ρσ

My result in the above (b) I found by using the given σ$\nu$ equation using the given σ0 and E$\nu$ = 10 MeV (I chose this because I assumed E$\nu$ was referring to the energy per neutrino.

Last edited: Apr 22, 2013