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Suppose there is a material that interacts with 50% of high-energy neutrinos. To what temperature such material will heat up naturally here on Earth? What power can be extracted from it?

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The obvious use of this material would be putting it around reactor cores, I presume. en.wikipedia.org/wiki/Neutrino#Artificial - 185 MW of neutrino energy loss out of 4000 MW thermal output (in a 1.3GW plant). – Deer Hunter Sep 7 '13 at 7:46
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@DeerHunter Well, reactor neutrinos are not "high energy" in the usual terminology. They are typically a few MeV. – dmckee Sep 7 '13 at 11:48
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I'm concerned that the question violates the "fictional physics" constraint. Above the Z mass, neutrinos do interact strongly matter, but that is uninteresting for the purposes of this question because the main neutrino flux through the Earth is at solar energies (which, like reactor energies are a few MeV). The cross-section there is low for very fundamental reasons relating to the mass of the weak bosons and the Fermi coupling constant. – dmckee Sep 7 '13 at 11:51
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Have a look at the neutrino slides of this talk: users.ictp.it/~smr2246/monday/gaisser-NUSKY.pdf . The numbers are very small there is no energy pool in the high energy neutrino flux and in addition, due to the weak interaction the by products take away most of the energy, except for electron neutrino interactions. – anna v Sep 7 '13 at 13:21
up vote 3 down vote accepted

If we indeed suppose that such an exotic material did indeed exist, then we just have to consider the flux of solar neutrinos from proton-proton fusion. This is about $10^{11} cm^{-2}s^{-1}$ at about 0.4 Mev each.

So a one-metre square sheet of annixxite, would generate, in Watts:

$$ {1\over2} \times 10^{11} \times 10^4 \times (0.4 \times 10^6) \times (1.602 \times 10^{-19}) = 32.04 $$

So about 30 Watts (a $1m^2$ photovoltaic panel is about 20 times more powerful).

You might like to add in the contribution from high-energy cosmic neutrinos... Their flux is about $100 cm^{-2}s^{-1}$ with energies all over the scale. Let's guess at 10 Gev:

$$ {1\over2} \times 100 \times 10^4 \times 10^{10} \times (1.602 \times 10^{-19}) = 8.01 \times 10^{-4} $$ So about 0.8 milliwatts and so negligible.

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I like this, the neutrino-to-light energy flux should (for steady state) approximately match the proportion of energy in the fusion process, taken away by neutrinos. We know that's not very much, and that it by all common sense is much below 1: both sources coming from the sun, this rule of thumb can tell you "it's much less than photovoltaic panels" :) – orion Jan 20 at 13:40

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