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[–] Devilbiter 0 points 1 point (+1|-0) ago 

If we can't determine momentum and can only discern energy, we can compare the measurements of having the facility face the sun (noon) and face opposite of the sun (midnightish). Using the difference in the measurements, knowing the decay rate through the earth of the neutrinos and the effective width of the earth, we should know what fraction come from the sun. That is, if the earth is any good at shielding neutrinos.

a light-year's worth of lead, equal to about 5.8 trillion miles (9.5 trillion kilometers) would only stop about half of the neutrinos flying through it.

The earth probably can't stop too many neutrinos. Then again, I thought we had a good idea of how many neutrinos are coming from the sun since we know what nuclear reactions are occurring and the mass of the sun. If you assume no (or almost no) neutrinos are absorbed by the earth from the sun, then we can simply subtract the sun neutrinos from the number measured underground. That sounds more reasonable.

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[–] 146Dissident_Fulcrum ago 

Damn. That is one penetrative particle. Does it just not interact with matter?

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[–] Devilbiter ago 

I think it only interacts via the weak force, so in general its cross section is much smaller than other particles that interact via the strong or electromagnetic forces.

In a related note, we had t-shirts in udergrad that said "Four forces in four years? This'll be easy." Heh.