# Neutrino interaction with dense molecular structures

Would dense molecular structures on heavier planets (crystalline or other compounds which are generally unknown to us) allow neutrinos to pass through as easily as on Earth?

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One impressive thing about neutrinos is that you would need about one light year of lead to stop a neutrino about 50% of the time. Which is due to neutrinos interacting only weakly with matter.

So, to answer your question, it would make $X$ time difference in neutrino opacity if you take you planet $X$ times larger in radius or $X$ times denser. However, the net effect will be still vanishingly small.

Also, molecular-level structures do not affect the interaction with neutrinos, but rather the nuclear structure.

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Ok, so is the nuclear density of atoms in red or brown dwarf stars high enough to interact with neutrinos? –  Cymatical Jan 17 at 14:49
It depends on what do you mean by interact. The Earth also interacts with neitrinos. However, for both the Earth and brown dwarfs neutrino opacity is small and close to zero. –  Alexey Bobrick Jan 17 at 15:24
@AlexeyBobrick It might help to talk about interaction cross-section. You could be a bit more precise with your answer when you say "still vanishingly small". –  astromax Jan 19 at 5:44
@astromax, given that the question is "interacts or not", the answer is simply "not", on the argument that $10^{4}-10^{6} \textrm{km}$ of ordinary matter is by about ten orders of magnitude smaller than a light year. –  Alexey Bobrick Jan 19 at 12:15
Well, hold up a second. You're "impressive thing" in retrospect is incorrect. It's a light year of lead for it to interact 50% of the time. I'm editing your answer to reflect that. Secondly, I would still find it appropriate and informative to include a general discussion about interaction cross-section. Take it or leave it - no need to debate with me regarding this. –  astromax Jan 19 at 16:02

Neutrinos wouldn't pass through white dwarfs or neutron stars as easily as through Earth, measured by the scattering rate per distance, since these types of stars are very dense: white dwarfs about one million times denser than water, neutron stars about as dense as atomic nuclei.

To scatter a neutrino with a chance of > 50%, you "just" need the inner of a neutron star along 10 times the diameter of Earth's orbit around the sun:

The neutrino-nucleon scattering cross section is about $10^{-38}cm^2=10^{-42}m^2$ for neutrinos with energies above about 1 GeV, according to this paper. A proton has a diameter of about $1.75\cdot 10^{-15}m$. With $$1.75\cdot 10^{-15}m/10^{-42}m^2=1.75 \cdot 10^{27}$$ we need about $1.75\cdot 10^{27}$ nucleons in a row to scatter a neutrino. That's about $$1.75\cdot 10^{27}\cdot 1.75\cdot 10^{-15}m=3\cdot 10^{12}m=20\mbox{ a.u.}$$ as a rough estimate. (For simplicity I've neglected the exponentially decreasing intensity of the neutrino beam in the scattering medium.) Neutrinos will probably scatter a little more easily in a neutron star than in the same number of single traversed neutrons, but that's a guess based on the scattering cross section for heavier nuclei.

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Wow, tough subject for amateurs. Even with the math and emerging understanding the relative distance ratios, or rather spatial comparison, between quantum interaction compared to solar and planetary orbits it seems astounding that one form of "matter" wouldn't interact with other forms. –  Cymatical Jan 27 at 3:13
That said, check out the Thunderbolt Project on Youtube for a very very very fresh perspective on the universe. From the electric comet theory that pretty much proves that in the solar wind all that hydrogen missing an electron fuses with the +++++/----- anode cathode charged long orbit asteroid/meteor/comets –  Cymatical Jan 27 at 3:18