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enter image description hereIf one considers the Semi Empirical Formula (SEMF) in nuclear physics containing the Volume, Surface, Coulomb energy and Asymmetry Energy terms for the binding energy of a nuclide, then by replacing the Coulomb energy term by gravitational potential energy term, one obtains a minimum radius of 4.8 or around 5 km for a neutron star.

However, I believe the above is an ideal case-as, the complete and realistic formulation for computing the minimum radius of a neutron star is obtained by using the concepts of hydrostatic equilibrium (i.e., degenerate Fermi gas/ neutron degeneracy pressure models, as can be found in these StackExchange discussion threads: Size and density of neutron stars Now while the aforementioned sites (among others) discuss realistic minimum possible radius for the neutron star by invoking hydrostatic equilibrium-based and General Relativity-based concepts, my interest lies in an ideal case. Is the concept of using SEMF to determine minimum stable radius of the neutron star correct, ideally?

Puppeteer
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  • What did you use for the gravitational potential energy term? That is inextricably linked to the (mass and energy) density profile within the star? Or did you just assume a uniform density and use Newtonian physics? – ProfRob May 09 '23 at 12:58
  • I just assumed a uniform density, spherical neutron star, and Newtonian physics-as I wanted to find out what value of minimum radius I would get using this idealized approach, without using GR-based concepts. So, my question was-as a crude approximation, is it correct to use the SEMF and Newtonian physics-based approach to estimate the minimum radius? – Puppeteer May 09 '23 at 13:51
  • I don't really follow - gravitational potential energy has the opposite sign to the Coulomb energy in the SEMF. Maybe you should post some details. – ProfRob May 09 '23 at 14:45
  • At short distances(as one finds in neutron star), the strong nuclear force between neutrons acts repulsively. This behavior is similar to Coulomb repulsion in electrostatics. So Coulomb term in the SEMF acts outwards and prevents the star from collapsing onto itself due to the inward gravitational PE-hence the two balance. This is analogous to hydrostatic pressure acting outwards to prevent the outer layers of the star from collapsing onto itself, in hydrostatic equilibrium models. In any case, the signs cancel while calculating. – Puppeteer May 09 '23 at 18:58
  • And the strong nuclear force is short range, whilst the Coulomb force is not and the short range nuclear repulsion is already part of the SEMF. Show your working. – ProfRob May 09 '23 at 20:46
  • My question was motivated by a problem that I came across, in which it was stated that the "Coulomb term is replaced by gravitational potential energy " and to find the minimum radius of the neutron star (which they claim to be 4.345km as the answer). I have attached the screenshot of the question (please see below). I was trying to examine how this was possible, as in many research articles it is stated that the minimum radius is almost never less than 8 km-hence the doubt. So I was wondering if the question is hypothetical, or possible in some ideal situation. – Puppeteer May 10 '23 at 07:37
  • I cannot see how the question works or where an answer to 4 significant figures comes from. Gravitational energy would be a positive contributor to binding energy. The total binding energy would just be a monotonically increasing function of $A$. Maybe show how the answer arises and comments might be possible. – ProfRob May 10 '23 at 12:43

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