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It is often quoted that most of the solar neutrinos passing through earth do just that – pass through – without interacting.

What's generally not said is what fraction of them do interact, and I've just noticed I never questioned this.

Wikipedia tells me that the characteristic electroweak cross-section $\sigma$ of electron neutrinos is 3.2 nanobarn. From this I would have estimated the differential loss fraction as $$ \frac{1}n \frac{\partial n}{\partial s} = \frac{\sigma\times\sum\text{atomic number}}{(\text{lattice constant})^3} \equiv \frac{\sigma\cdot k}{a^3} $$ Well, if I put in some rough numbers corresponding to $\mathrm{SiO_2}$, WolframAlpha gives me a characteristic length of $10^7\ \mathrm{m}$.

Not that much, is it? It's certainly not compatible with the statement that the vast majority of neutrinos pass through earth unhindered.

What did I do wrong in the calculation? What fraction of neutrinos actually interacts with earth, or what's the interaction half-life of a neutrino passing through rock?

leftaroundabout
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1 Answers1

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The cross-section of a nucleon to a solar (few MeV) neutrino is about $10^{-46}$ m$^2$ and higher than the cross-section for leptonic reactions (see here).

The mean density of the Earth is 5500 kg/m$^3$, and this is essentially made up of protons and neutrons with a number density of $3.3\times 10^{30}$ m$^{-3}$.

The mean free path is $(n\sigma)^{-1} \simeq 3\times 10^{15}$ m. This is the origin of the oft-quoted "light year of lead" to block neutrinos.

The Earth is about 13,000 km thick, so the fraction that interact is roughly $1.3\times 10^{7}/3\times 10^{15} = 4.3\times 10^{-9}$.

ProfRob
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