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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?

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  • $\begingroup$ I was prompted to this question by a post over at Worldbuilding.SE. $\endgroup$
    – leftaroundabout
    Commented Jul 6, 2016 at 15:34
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    $\begingroup$ Wikipedia does not state what exactly it means by that "cross section" (and probably wrongly links ot the generic cross section article). "Cross sections" in particle physics are certain probabilities per unit spacetime volume for particle interactions, not the usual cross sections for what happens when you shoot a beam of something into a target of other stuff. (The CERN website has a short article about this usage of "cross section") $\endgroup$
    – ACuriousMind ♦
    Commented Jul 6, 2016 at 15:56
  • $\begingroup$ I think there is something wrong there. Solar neutrinos pass trough the entire sun without much interaction, otherwise they couldn't be used to probe the interior of the sun. What you have there is 20,000km, which is a fraction of the sun's radius. $\endgroup$
    – CuriousOne
    Commented Jul 6, 2016 at 16:02
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    $\begingroup$ Your scattering cross section is 16 orders of magnitude too big $\endgroup$
    – John Rennie
    Commented Jul 6, 2016 at 16:26
  • $\begingroup$ Not that it makes the kind of difference you're looking for, but isn't oxygen 8, not 10? :) $\endgroup$
    – hobbs
    Commented Jul 6, 2016 at 16:27

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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 $1.3\times 10^{7}/3\times 10^{15} = 4.3\times 10^{-9}$.

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