# What allows the modified Urca process to work at lower density than direct Urca in neutron star cooling?

The dominant method of neutron star cooling is neutrino emission. There are two regimes usually presented, the "direct Urca" and "modified Urca" processes, each of which are sequences of neutron decay and inverse reactions. The direct Urca looks like this: $$n\rightarrow p+l+\overline{\nu_l},\quad p + l \rightarrow n + \nu_l$$ where $l$ is a lepton - either an electron or a muon. These processes cause continuous emission of neutrinos which cools a neutron star relatively quickly.

But below a density of $\rho\approx 10^{15}\mathrm{\,g\,cm^{-3}}$ (about three times the nuclear density) this process is suppressed, which means that the direct Urca process only occurs in the core. This is the reason according to a review of neutron star cooling from Pethick and Yakovlev (2004):

The process can occur only if the proton concentration is sufficiently high. The reason for this is that, in degenerate matter, only particles with energies within ~$k_BT$ of the Fermi surface can participate in reactions, since other processes are blocked by the Pauli exclusion principle. If the proton and electron Fermi momenta are too small compared with the neutron Fermi momenta, the process is forbidden because it is impossible to satisfy conservation of momentum. Under typical conditions one finds that the ratio of the number density of protons to that of nucleons must exceed about 0.1 for the process to be allowed.

This makes some sense. But what surprises me is that this process can still work with a slight modification at lower densities. The modified Urca process can cool the star $$n+N\rightarrow p+N+l+\overline{\nu_l},\quad p + N + l \rightarrow n + N + \nu_l$$ where $N$ is a nucleon - a proton or a neutron.

This process, I'm told, can work at much lower densities, but produces 7 orders of magnitude less emissivity. As a result, it's the dominant process in the superfluid outer core.

My question is why does the additional nucleon permit lower densities? How does an additional neutron or proton get us out of the conservation of momentum problem with the direct Urca process?

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I am not an expert on this, but the two cases indicate a change of phase. The process $$n~\rightarrow~ p+l+\overline{\nu_l},\quad p + l \rightarrow n + \nu_l$$ might be thought of as two steps in a process where a neutron in a high energy state in the Fermi surface, we denote by $n^*$ falls to a low energy by $n^*~\rightarrow~n~+~\nu~+~\bar\nu$. For a lower mass density $<~10^{15}g/cm^3$ the neutron state, which is a form of liquid as I understand, changes to the cooler phase with nuclear matter. Using the shorter notion above a neutron scatters off a nucleus, so the $n^*$ neutron scatters off a nucleus $N$ so $$n^*~+~N~\rightarrow~n~+~N~+~\nu~+~\bar\nu$$ This process however competes for the process $n^*~+~N~\rightarrow~N’$ which is an isotope of the $N$. This is a nuclear interaction which is strong than weak interactions, and by that fact probably dominates the neutrino emitting process governed by weak interaction.

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Thanks. This seems to be in the right direction, but modified Urca is actually dominant in the neutron superfluid, so I don't think it's specifically about the phase transition to nuclear matter. –  kharybdis Mar 11 '11 at 1:25