[Physics] How much of neutrinos pass through a neutron star

Question

Neutrino rarely interact with matter however neutron star is extremely dense and I suppose only an insignificant amount of neutrinos can pass through the densest object in the universe second to black hole, right?

Answer 1

Quick answer: Neutron stars are transparent to their own thermally produced neutrinos and any other neutrinos of energies below a few MeV once they have cooled to highly degenerate states at temperatures of $\leq 10^{10}$ K. Now read on...

The interaction of neutrinos with matter in a neutron star will be heavily suppressed as dmckee suggests.

Both the baryonic (neutrons and protons) and leptonic (electrons and possibly muons) are highly degenerate. The Fermi energies are more than a GeV for neutrons and perhaps tens of MeV for electrons. On the other hand, the neutrinos that are produced inside the neutron star have thermal energies - at $\sim 10$ MeV when the neutron star is born, but decreasing rapidly.... due to neutrino emission!

I think there are then two things going on here. First, the cross-sections for neutrino interactions are energy dependent and decrease rapidly with neutrino energy as the neutron star cools.

Secondly, neutrinos can only interact with a particle if they can give the particle some of their energy (scattering) or if they can create new particles (e.g. inverse beta decay), but this is difficult if all the available energy and momentum states for particles are full, as in the case of degenerate fermion gases. Essentially, only particles within $\sim kT$ of the Fermi energy are able to participate, which is a very small fraction in a highly degenerate gas where $E_F \gg kT$ by definition. Hence the rates are even lower than the already low interaction rates between neutrinos and baryons/leptons.

In addition, neutrino absorption events on single neutrons become blocked because energy and momentum cannot be simultaneously conserved by the electron and proton that are created at the top of their respective Fermi seas. Instead such reactions require an additional third-body bystander neutron, which greatly decreases the efficiency and since all reactants and products must be near their Fermi energies, makes the cross-section decrease rapidly with temperature.

The net outcome is whereas the proto-neutron stars and very young neutron stars are significantly opaque to the $>$ MeV neutrinos they produce at $T>10^{10}$ K (the process of neutrino trapping is very important in the core of a supernova), this is because the neutrinos have relatively high energies and the various fermion species are not fully degenerate (plus in the core of a supernova there are more electrons with higher neutrino cross-sections). But as the neutron star cools, such that the neutrino energies fall below an MeV (even after a minute) and the constituent fermions become highly degenerate, then neutrino transparency can be safely assumed.

Neutron stars do have an outer crust with non-degenerate nuclei and I would expect these to be a source of neutrino opacity. The thickness is about 1km and the density about $10^{15}$ kg/m$^3$. This provides a column density of nucleons of around $10^{45}$ m$^{-2}$. MeV neutrino-nucleon cross-sections are about $10^{-48}$ m$^2$, so the crust material is also nearly transparent even to MeV neutrinos.

Some nice calculations that support what I have outlined above appear in Haensel & Jerzak (1987). They consider both absorption and scattering processes. At $5\times 10^{10}$K, the mean free path to scattering is about 1 km for few MeV neutrinos (cf a neutron star radius of 10 km), but with a strong inverse energy dependence. As the neutron star cools, the mean free path for neutrinos with a similar $E_{\nu}/kT$ increases strongly as $T^{-3}$ such that the star is completely transparent to its own thermal neutrinos by the time it reaches $10^{10}$ K.

Absorption is slightly more important than scattering at 1 MeV energies and $T= 5\times10^{10}$ K - producing a mean free path of about 10 km, but its importance declines on cooling as $T^{4}$, so again, complete transparency to thermal neutrinos by $10^{10}$K.