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How about this:

In an $n,p,e$ gas the ratio of neutrons to protons decreases with density. For ideal degenerate gases the Fermi energies are related by $E_{F,n} = E_{F,p} + E_{F,e}$. In this situation, the largest the proton to neutron ratio can become is 1/8 when all the particles are ultra-relativistic at infinite density.

To conserve momentum in the direct URCA process $p_{F,n} = p_{F,p} + p_{F,e} = 2p_{F,p}$ (by charge neutrality) and when the reaction occurs in "a straight line". Hence because $p_F$ is proportional to $n^{1/3}$, then $n_n = 8 n_p$

For all other reaction geometries the proton number density needs to be proportionately higher. Hence in completely degenerate, ideal $n,p,e$ gases, the direct URCA process is blocked.

At densities above about $8\times 10^{17}$ kg/m$^3$, the Fermi energy of the electrons equals the rest-mass energy of muons (105 MeV) and muons can be produced (NB In more realistic models including nucleon interactions, this occurs at about $3\times 10^{17}$ kg/m$^3$). The charge neutrality equation changes and the net result is that the number of protons in the gas is able to increase slightly - it could get above 10% of the neutron numbers, or even higher if negatively charged Hyperons are produced at even higher densities. This opens up a channel for the direct URCA process to occur.

The image below (taken from a thesis by Stephen Portillo) illustrates the proton (gold), electron (purple) and muon (blue) fractions as a function of baryonic density (in units of the nuclear saturation density $2.8\times10^{17}$ kg/m$^3$). This has been calculated incorporating nucleon interactions. The muons appear at about the saturation density and the proton fraction exceeds 1/8 at about $10^{18}$ kg/m$^3$.

Proton (gold), electron (purple) and muon (blue) fractions as a function of baryonic density in units of 2.8e17 kg/m^3

These reactions must involve neutrons, protons and electrons that are close to their Fermi energies (within about $kT$). Therefore this requirement on each species introduces an extra dependence of $\sim kT/E_F \ll 1$ per species into the reaction rate.

In the modified URCA process "bystander particles" are available to balance the momentum and energy equations - i.e. their role is just to provide sources/sinks of energy/momentum. However, the requirement for two more bystander nucleons (also within $kT$ of their Fermi energies) suppresses the reaction rate considerably (two more factors of $kT/E_F$). So, even though it dominates over direct URCA at low densities, if densities are high enough for direct URCA to be possible, it dominates.

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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. –  spencer nelson Mar 11 '11 at 1:25

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