Are neutron star cores thought to be stiffer than photon gases?

The weak energy condition and with the dominant energy condition allow anything from w=-1 to 1 (meaning that the magnitude of the pressure can be at most equal to the total mass-energy density). Values outside this range are needed to make wormholes and other causality violations (note that the strong energy condition is violated by dark energy). Cosmic inflation and dark energy are very close to the limit of -1, but nothing seems to come close to +1 even though it is "allowed". Both photon gases and ultra-relativistic degenerate gases are +1/3, can we do better?

Neutron stars are thought to be stiffened by a repulsion in addition to the usual degeneracy pressure. Could this repulsion stiffen the equation of state enough to exceed 1/3? It takes energy to force repelling particles together (which can be interpreted as an the energy in the virtual particles that are mediating said repulsive force). This extra energy raises the mass, behaving much like kinetic energy in degenerate matter, which would seem to prevent rising above 1/3. Is there any way around this "problem"?

2 Answers

Repulsion between nucleons must stiffen the equation of state of a neutron star.

It was established in 1939 (well before neutron stars were discovered) by Oppenheimer & Volkoff, that ideal neutron degeneracy pressure was incapable of supporting a ball of neutrons with mass greater than $$0.75 M_{\odot}$$.

Since then, many neutron stars have been discovered and all of them have masses significantly greater than this, requiring stiffer equations of state.

Ultra-relativistic neutron degeneracy pressure has an equation of state of the form $$P \sim \rho/3$$, where $$\rho$$ is the total energy density. In order to get the neutron stars we see, the equation of state in the core needs to behave more like $$P \sim \rho$$. This stiffer equation of state is thought to be caused by the repulsion between neutrons in strongly asymmetric nuclear matter at densities greater than the nuclear saturation density.

NB: This assumes the size of $$w$$ indicates "stiffness".

NB2: In terms of pressure as a function of mass density, the "stiffening" of the equation of state must raise the adiabatic index from somewhere between 4/3 and 5/3 (appropriate for non-interacting fermions of an intermediate degree of "relativisticness") to $$\sim 2$$.

• Doesn't a degenerate gas that is not ultra-relativistic have 𝑃<𝜌/3? The confusion arises when people say "relativity makes things softer". A non-relativistic degenerate gas has 𝑃~𝜌^(5/3). 𝑤 increases with compression, but 𝑤 asymptotes to 1/3 when relativity kicks in and we get 𝑃 = 𝜌/3 ~ 𝜌_#^(4/3) in the limit. It never gets "softer" but it ceases to get harder fast enough to keep up with the gravity (and general relativity makes gravity worse). – Kevin Kostlan Sep 23 at 16:59
• @KevinKostlan I am now wondering whether there are two usages, one referring to w and one referring to the adiabatic index (which governs how compressible the gas is). – Rob Jeffries Sep 23 at 17:29
• Adiabatic index is gamma, not omega, and is usually used for "ordinary" conditions far from relativity. But gamma is also a dimensionless number that determines the equation of state in some sense. – Kevin Kostlan Sep 23 at 17:36
• @KevinKostian the adiabatic index is 4/3 for a relativistic fermion gas. It is a standard term used in NS astrophysics. If hardness just mean the size of w then a non-relativistic gas would always be softer (w ~ 0), but I have always taken "softer" to mean more compressible. However, I think my last paragraph is incorrect, so will delete. – Rob Jeffries Sep 23 at 17:45
• @KevinKostian I also appreciate that I haven't really answered your question, so much as saying that it must be true. – Rob Jeffries Sep 23 at 17:55

The idea of a short-range repulsion between nucleons was conceived to explain a paradox concerning the stability of nuclear matter, but it is not really grounded in particle physics -- neither in the meson-exchange nor in the gluon-exchange picture. (The paradox is that purely attractive two-body forces would predict an interaction energy per nucleon scaling as $$-{{n}^{1}}$$, whereas the kinetic energy of a free nonrelativistic Fermi gas scales as $$+{{n}^{2/3}}$$, where n denotes number density. The attraction would overwhelm the kinetic energy in the limit of infinite density, and neutron matter would collapse.) As a snotty former particle physicist, I regard the idea of short-range repulsion as a heuristic myth.

The meson-exchange picture predicts a bewildering variety of attractive and repulsive forces, depending on the quantum numbers of the meson involved. Single-pion exchange predicts the longest-range part of nuclear forces correctly, because pions are isovector and pseudoscalar. An isoscalar vector meson would mediate a consistent repulsion between nucleons, but there are many other mesons in nature’s catalog, and there is no proof of net short-range repulsion.

The quark-gluon picture seems most appropriate at extreme densities (probably higher than in any stable neutron star) where nucleons would overlap. The kinetic energy per light (hence ultrarelativistic) quark scales as $$+{{n}^{1/3}}$$, whereas gluon exchange predicts a net attraction that scales as $$-{{g}^{2}}{{n}^{1/3}}$$, but the running coupling fades as $${{g}^{2}}\propto 1/\log ({{p}_{F}})$$ where $${{p}_{F}}\propto {{n}^{1/3}}$$ , so kinetic energy ultimately dominates.

The bottom line is that nuclear matter at extreme densities, regarded as quark matter, should be about as “soft” as photon gas.

• Such a soft equation of state can only support a neutron star with a maximum mass of 0.75 solar masses. This was shown in 1939. If there is no short-range repulsive force between nucleons, why are nuclei the size they are? The observed masses and radii of neutron stars already put their cores at densities well above the nuclear saturation density. I agree that ultradense quark matter should be as soft as a photon gas and that is why the observations of 2 solar mass neutron stars basically rule that out. – Rob Jeffries Sep 19 at 13:54
• If the proof of the pudding is in the eating, then the phenomenological short-range repulsion is justified because it gets credible answers, but it isn't grounded in particle physics. Perturbative QCD cannot predict what happens at the densities of neutron stars, which are not extreme in the sense that I defined it. – Bert Barrois Sep 19 at 19:24
• This answer seems to interpret the failure of high-energy physics to predict inter-nucleon repulsion as an evidence that such repulsion does not occur. That seems backwards. If we have evidence in low-energy systems like stable nuclei and neutron stars for a repulsive inter-nucleon interaction, a theory which offers "proof" that there is no such repulsion without a better explanation is wrong. – rob Sep 23 at 18:24
• @rob -- I said nothing of the sort. I called short-range repulsion heuristic, and said that QCD has so far been unable to calculate properties of nuclear matter at normal density, not that it predicts absence of repulsion. Both approaches say that dP/dρ > 0 at extreme densities, but offer different explanations. Heuristic/effective theories (e.g. Ginzburg-Landau) are ultimately explained in terms of fundamental theories (viz. BCS). – Bert Barrois Sep 29 at 11:19
• @rob -- The effective potentials used in Nuc Phys are phenomenological, being calibrated upon low-energy scattering data and properties of nuclei. You cannot expect them to give accurate predictions of compressibility above normal density. – Bert Barrois Sep 29 at 11:19