Naked Science Forum
On the Lighter Side => New Theories => Topic started by: talanum1 on 18/09/2020 17:48:23
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I've been doing a nuclear model and realized that there is no force balancing the nuclear force for Neutrons. For Protons there is the opposing Coulomb force.
Is there a formula for Neutron Degenerate Pressure?
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Yes, neutron degeneracy pressure is what you are looking for. I don't know of any equation to calculate it, unfortunately.
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How can a nucleus have orbital angular momentum not equal to zero, if all the particles are close enough for degeneracy pressure to apply and supposing at least two nucleons are static (not orbiting with spin paired)?
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How can a nucleus have orbital angular momentum not equal to zero, if all the particles are close enough for degeneracy pressure to apply and supposing at least two nucleons are static (not orbiting with spin paired)?
Quantum spin is not macroscopic spin.
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Here is a term paper that derives the equation for neutron degeneracy pressure (pages 4-6):
http://www.physics.drexel.edu/~bob/Term_Reports/John_Timlin.pdf
- This is calculated in the context of a neutron star, which has few protons in the core.
- You may prefer to calculate it in the context of a Uranium nucleus, where protons make up a little over 1/3 of the nucleons....
The equations for neutron degeneracy are in the process of being analyzed with input from the NICER X-Ray telescope on the ISS. This telescope collects data on the rotational red shift of pulsars to estimate the relationship between mass and volume, which allows verification of the theoretical neutron degeneracy equation.
This analysis will help determine if there is something denser than a neutron star which can resist collapse into a black hole (a so-called quark star, held up by quark degeneracy pressure).
See: https://en.wikipedia.org/wiki/Neutron_Star_Interior_Composition_Explorer
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Thanks evan_au.