# Why is the density of the Fermi gas in a neutron star not changing the potential depth caused by the strong nuclear interaction?

In some textbooks, the neutron star is explained as a degenerate Fermi gas. To calculate the degenerate pressure of the neutron fermi gas the average Energy of a neutron, U is calculated when the Volume V is changed. p = dU/dV. However, this does not consider that the changing neutron density under compression also changes the potential caused by strong interaction between the neutrons. At least the (attractive) Hartree-Potential in the Hartree-Fock Formalism grows linearly with nulceon density (which grows by 1/V). This gain in 1/V would win over the (repulsive) increase in the kinetic energy under compression, which grows only by V^-(2/3). Thus, the negative pressure caused by the increasing binding energy grows faster than the positive Fermi gas pressure due to the kinetic energy U. It is energetically advantageous to collapse.

What will certainly prevent a collapse is the "hard core" of the neutron, i.e. the strongly repulsive nuclear force for distances below 1 fm, as dicussed in question "Why is the central density of the nucleus constant". The density of the nuclei is not too far away from tightly packed spheres with 1 fm radius. However, this is totally different physics, and the textbooks would wrongly explain the neutron star as a Fermi-Gas.

I suspect that the constant density may have something to do with the Fock-term (exchange potential) in the Hartree-Fock equation, which in fact is another representation of Pauli-exlusion principle, compensating the increase in the Hartree-Term.

What is the nature of a neutron star?

• This phenomena is called the saturation of Nuclear forces. It is already observable for heavy nuclei and is the same physics that operates in neutron stars. The tightly packed spheres model discussed in the question you cited is not really a good explanation for this effect. Nucleons in a nucleus are a quantum fluid and exhibit strong shell effects just like electrons in atoms. I'll attempt to answer your question when I have time. Nov 15, 2016 at 21:53
• @Rob Jeffries gave a nice answer on the specific question you asked. If you are interested in some details on the degenerate Fermi gas EoS and some more general points on compact stars I can recommend this paper: arxiv.org/abs/astro-ph/0506417. In that paper you can also find some plots and equations directly related to Rob's answer.
– N0va
Nov 15, 2016 at 22:39
• I also endorse @RobJeffries answer. I will try to give you an indication of where his "hard core" comes from when I get a chance. As a prelude I invite you to read my answer to this question: physics.stackexchange.com/questions/288357/… Nov 16, 2016 at 4:57
• Though the real issue is not just the nucleon-nucleon potential, it is the many-body problem for a neutron fluid with a small fraction of protons. I Nov 16, 2016 at 7:02
• I want to mention that I probably used wrong terminology: in german text books, "Fermi-Gas" is used for what is apparently called "Fermi liquid" in english. Jan 3, 2017 at 13:52

You are quite correct that a neutron star is not supported by ideal neutron degeneracy pressure. Any book or web source that claims so should be given a wide berth.

As long ago as 1939 Oppenheimer & Volkhoff showed that a neutron star supported by ideal NDP became unstable at finite density, with a maximum mass of around $0.7 M_{\odot}$. All measured neutron star masses are much higher than this.

The repulsive core of the strong nuclear force in asymmetric nuclear matter is almost certainly what supports neutron stars. The polytropic index of the pressure can exceed 2, as opposed to somewhere between 4/3 and 5/3 for ideal NDP, so is a much harder equation of state.

The review by Lattimer (2013) does a good job of describing how observations of neutron star masses and radii provide constraints on the equation of state and uncertain parameters in the symmetry energy beyond nuclear densities.

My answer concerns the origin of the "repulsive hard core" referred to in @RobJeffries answer (with which I totally concur). This mechanism is responsible for the relative constancy of the central nuclear density of large nuclei as well as the stability of neutron stars below a certain mass limit. If you have read my answer to this question: Why is nuclear force spin dependent? you will see that within the relativistic mean field (or Hartree) approach to the nuclear many-body problem the central nuclear potential has two terms: 1) $U_s$, an attractive term arising from the exchange of the sigma (scalar, isoscalar) meson, and 2) $U_0$, a repulsive term arising from the exchange of the omega (vector, isoscalar) meson. These empirically determined potentials are the result of an elaborate self-consistent field (Hartree or Hartree-Fock) procedure that involves solving by numerical integration the Dirac equation $$[c\vec{ \alpha } \cdot \vec{ p}+\gamma^0(m+U_s+\gamma^0U_0)]\psi=E\psi$$for single particle orbitals for doubly magic nuclei over the entire periodic table. The wave function $\psi$ is a Dirac 4-component spinor that takes the form: $$\psi(\vec r)=\frac{1}{r}\binom {F(r)Y^{\omega}_{jm}(\hat r)}{iG(r)Y^{-\omega}_{jm}(\hat r)}$$ when the potentials are spherically symmetric. The functions $F(r)$ and $G(r)$ are the large and small component radial wave functions and $Y^{\omega}_{jm}(\hat r)$ is the Pauli central field spinor. Once a set of trial orbitals for a particular nucleus are found, the nuclear potentials are obtained by folding the relevant nucleon density functions with the associated Yukawa function. Iteration continues until two successive sets of orbitals and potentials are found to be in agreement (self-consistency).

The masses and couplings of these two mesons are the primary parameters in the model. When the work was done initially only the omega mass was known so the other parameters were adjusted to yield agreement with the experimental binding energies, charge distributions, and single particle separation energies of these doubly magic nuclei. The single particle eigenvalues turned out to have the correct ordering needed to reproduce the known magic number sequences for neutron and proton shells. This was a result of the large spin-orbit interaction inherent in this combination of attractive and repulsive potentials. The nuclear potentials resulting from this empirical procedure can be roughly described as follows: the attractive (scalar) potential ($U_s(r)$) has a depth between 500-600 MeV while the repulsive (vector) potential ($U_0(r)$) has a height between 400-500 MeV. In a zeroth order non-relativistic approximation the nuclear potential would be the sum of these two ($U_s(r)+U_0(r)$, a well with a depth of 50-100 MeV).

So what negates the argument:

"At least the (attractive) Hartree-Potential in the Hartree-Fock Formalism grows linearly with nulceon density (which grows by $\frac{1}{V}$). This gain in $\frac{1}{V}$ would win over the (repulsive) increase in the kinetic energy under compression, which grows only by $V^{-\frac{2}{3}}$). Thus, the negative pressure caused by the increasing binding energy grows faster than the positive Fermi gas pressure due to the kinetic energy U. It is energetically advantageous to collapse."

and prevents collapse (even in finite nuclei as well as neutron stars)? The answer comes from studying the nucleon densities that are folded with Yukawa functions to yield the scalar ($U_s$) and vector ($U_0$) potentials. The vector nucleon density is the same as the probability density so the above argument applies, except that the vector potential is repusive, not attractive. The scalar density, on the other hand, differs from the probability density (the square of the small component of the radial wave function is subtracted from the square of the large component ($F^2-G^2$) rather than added ($F^2+G^2$) in the case of the vector density. This means that if the nucleon density is squeezed beyond the normal equilibrium value, the repulsive potential indeed grows but the attractive scalar potential will not grow as much since the squeezing will also enhance the size of the small component ($G$) relative to the large component ($F$) ($G$ is the order of $\frac{v}{c}$ compared to F). This is the heart of the nuclear saturation mechanism in this approach. It is more a subtle relativistic effect that enhances the repulsive omega meson exchange relative to the attractive sigma meson exchange as the nucleon density increases. In the end a very stiff EOS results.

References: This work was my dissertation research. The first published calculation (Hartree approximation) was Phys. Rev C5 (1972) 241. The full Hartree-Fock calculation was Phys Rev C9 (1974) 537. My calculations were restricted to finite nuclei. Walecka published results with a similar model for infinite nuclear matter in the Fermi gas approximation (for balanced as well as unbalanced nuclear matter, applicable to neutron stars): Ann of Physics (NY) 83 (1974) 491.

• Thank you; while Rob Jeffrie's certainly answered my initial question, you just gave the answer to the next question I would have asked. I do experminetal and pllied research, but since this semester I give a course on basic nuclear physics, and I also want to have the theory correct. Jan 3, 2017 at 13:49
• Thanks. I've noticed a fair amount of confusion about the saturation of nuclear forces on PSE and even Wikipedia, so led me to expand on Rob's answer. Jan 3, 2017 at 15:30