8
$\begingroup$

I have read this question (to Andrew's answer, in the comments):

What supports neutron stars is the repulsion provided by the strong nuclear force between closely-packed neutrons. The central pressure in a neutron star is an order of magnitude higher than ideal neutron degeneracy pressure. no, it's not quark degeneracy pressure, it's actual forces due to gluon exchange.

Will a neutron star always collapse into a black hole in the future?

Now as far as I understand, on this site (and wiki) it is said that neutron stars do not collapse because they are supported by neutron degeneracy pressure. Though, based on the comments it is at the core a different mechanism, being the residual strong force (repulsive at short distances) between neutrons.

One of the comments says it is mediated by gluons, but as far as I understand, the residual strong force is mediated by pions between neutrons.

Now the distinction is important, because even on this site, it is not clarified, whether it is neutron degeneracy pressure (which is explained differently, based on QM and the Pauli exclusion principle), or the repulsive (at short distances) residual strong force that actually supports the neutron star from further collapse.

So there are two main ideas:

  1. it is the neutron degeneracy pressure, between neutrons, and how they fill the QM energy levels (PEP)

  2. it is the repulsive (at short distances) residual strong force

Just to clarify, these are two completely different mechanisms, this is why I am asking the question.

Question:

What really supports neutron stars?

$\endgroup$
4
  • $\begingroup$ There isn't necessarily a contradiction - the energy-level structure is created by forces of interaction. Just like whether it is "electromagnetism" or Pauli exclusion that keeps ordinary material objects from sliding through each other. $\endgroup$ Commented Oct 15, 2021 at 23:50
  • 5
    $\begingroup$ Related: Why do nucleons feel a repulsive force when less than 1 fm? $\endgroup$ Commented Oct 16, 2021 at 0:30
  • 1
    $\begingroup$ The question may be answerable without knowing all of the details, but regarding the details, the abstract of arXiv:0812.4499 says "the incompressibility coefficient of neutron-rich matter remains an important open problem." $\endgroup$ Commented Oct 16, 2021 at 0:32
  • 2
    $\begingroup$ Related: physics.stackexchange.com/questions/292913/… $\endgroup$ Commented Oct 22, 2021 at 2:25

1 Answer 1

8
$\begingroup$

When someone says "degeneracy pressure", then I would assume that they mean the ideal degeneracy pressure felt in a gas of non-interacting, indistinguishable fermions, simply due to their non-zero kinetic energy.

The expression for ideal degeneracy pressure (e.g. for non-relativistic fermions) $$P = \frac{h^{2}}{20m}\left(\frac{3}{\pi}\right)^{2/3}\, n^{5/3}\ , $$ where $n$ is the number density of fermions of mass $m$, does not involve: charges, baryon numbers or any interaction constants associated with any type of force. It is a purely quantum mechanical effect that requires densely packed fermions to have non-zero momentum. The only force involved actually is gravity, which provides the potential that confines the fermions and hence quantises the momentum states.

It is a fair statement to say that white dwarfs are "supported by electron degeneracy pressure". That is because the Coulomb interactions between the electrons and nuclei, which are included in any proper treatment of a white dwarf's structure, are a very small perturbation to the equation of state, reducing the pressure by just a few per cent.

A neutron star however, or at least all the neutron stars found so far in nature, cannot be supported by ideal neutron degeneracy pressure (NDP). The interaction terms between the nucleons completely dominate the equation of state.

One of the first papers to discuss the possibility of neutron stars, by Oppenheimer & Volkoff (1939), showed that ideal neutron degeneracy pressure (NDP) can only support a stable ball of neutrons up to 0.75 solar masses. i.e. the "Chandrasekhar limit" (but using General Relativity) associated with NDP is only $0.75 M_\odot$; and all measured neutron stars are more massive than this (the least massive known is about 1.17 solar masses - Martinez et al. 2015).

To support more massive neutron stars or halt the core collapse in a supernova requires interactions between the neutrons, or the neutrons to turn into something else like a quark-gluon plasma. This interaction can be provided by the strong nuclear force, which (in broad terms) is attractive over ranges of $1-2 \times 10^{-15}$ m, but strongly repulsive if you try to squash nucleons closer together than this. The details of this interaction in a neutron star are still uncertain because of the relativistic many-body nature of the problem and that the nuclear matter is highly "asymmetric", in the sense of there being just 1 proton for every 100 neutrons.

The above terminology and use of language is entirely consistent with statements by the leading researchers in neutron star and core-collapse physics. e.g.

Lattimer & Prakash (2001) in "Neutron Star Structure and the Equation of State":

the pressure near the saturation density is primarily determined by the isospin properties of the nucleon-nucleon interaction, specifically, as reflected in the density dependence of the symmetry energy, Sv(n).

Woosley & Janka (2005) in "The Physics of Core-Collapse Supernovae":

Eventually the repulsive component of the short-range nuclear force halts the collapse of the inner core when the density is nearly twice that of the atomic nucleus, or 4–5 × 1014 g cm−3.

Ozel et al. (2016) in "The Dense Matter Equation of State from Neutron Star Radius and Mass Measurements":

Our understanding of the equation of state in the vicinity of the nuclear saturation density is firmly founded on nucleon-nucleon scattering experiments below 350 MeV and on the properties of light nuclei. An approach that makes use of these data most directly is based on describing the interactions between particles via static two- and three-body potentials at this density...

And so on...

$\endgroup$
1
  • $\begingroup$ Thank you so much! $\endgroup$ Commented Oct 22, 2021 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.