Estimating Neutrino Flux Through Your Body?

AI Thread Summary
The discussion centers on estimating the flux of neutrinos passing through a human body, given a neutrino energy density of 0.2 MeV/m3 and a surface area of 0.01 m2. Participants debate whether to treat neutrinos as relativistic or non-relativistic, noting that assuming they are ultra-relativistic yields a flux of approximately 1.19 x 10^17 m-2 s-1, which is significantly higher than the solar neutrino flux. There is confusion about the surface area measurement, with suggestions that it should be 0.1 m2 instead of 0.01 m2. The conversation also touches on the challenges of detecting neutrinos due to their low energy and the implications of their speed and mass after decoupling from the Big Bang. Overall, the estimation process involves complex calculations and assumptions about neutrino behavior.
Joeseye
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1. Problem

"Estimate the flux of neutrinos passing through your body per second if the present energy density of neutrinos from the Big Bang is 0.2 MeV/m3. Assume that you are a standard size covering 0.01 m2".

Homework Equations



nv = Uv(T) / <Ev>

The Attempt at a Solution



I've assumed that the neutrinos have a temperature of 1.95 K. Now I'm not sure whether to presume that the neutrinos are relativistic (hence, zero mass and velocity of c) or non-relativistic (i.e. mv < 1 eV), since the question does not specify. Although I believe the Tv = 1.95 K value comes from assuming neutrinos are massless (I think).

I've attempted both and have different answers (although I doubt whether they are correct). Regardless, I've not had much success converting the neutrino density to a flux density. I assume that the neutrinos are traveling in all directions with the same velocity.
 
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You can assume that the neutrinos are ultra-relativistic, I think.

A human body with 0.01m^2 surface area is... strange.
 
mfb said:
You can assume that the neutrinos are ultra-relativistic, I think.

A human body with 0.01m^2 surface area is... strange.

I thought that 0.01 m2 was quite low, too. Perhaps he meant 0.1 m2.

Assuming the neutrinos are ultra-relativistic I got a flux of 1.19 x 1017 m-2 s-1... which I'm pretty sure is higher than the solar neutrino flux. o_O

I used:

Flux ϕ = (c . uv(T)) / (3 . <Ev>) = (c . uv(T)) / (3 . kB . T)

The factor of 1/3 comes from assuming the neutrinos are isotropic. Essentially, this is the power density (which is c/3 times the radiation pressure uv(T)) divided by the average energy of a neutrino <Ev>.

Do you think this is correct?
 
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Flux should be something "per time*area".

2K correspond to about .5meV, therefore we have ρ=0.4*10^9 primordial neutrinos per m^2, moving at nearly the speed of light. Using only one direction, the flux is 1/2 ρ c or about 10^17/(m^2*s).
Looks good.

Those neutrinos are hard (or even impossible) to detect as they have a very low energy.
 
mfb said:
Flux should be something "per time*area".

Ah yeah. Sorry I meant neutrinos per meter squared per second - I'll edit my post.

Thanks for your reply. Is it appropriate to assume the neutrinos are traveling at a velocity of c and are massless? I thought that when neutrinos decoupled (2s after the Big Bang) they had a velocity close to c, but have since slowed to approximately 105 - 106 m s-1?
 
If they are slow, they are not relativistic - with 2K, they would need some significant mass to be so slow.
 
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