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Do Neutron stars have free quarks? Also Can Quark Stars be formed also due to this reason?

Because of Asymptotic freedom high energy causes quarks to be free but according to the mass-energy equivalence equation I.E $E=MC2$ the same should happen if it has a lot of mass thus it Led me to this question.

But however in order for a particle to have free quarks it also has to have less strong force and since the quanta for Strong force are Gluons which abides QCD/Quantum Chromodynamics, and QCD is due to the color so it also has to have the necessary colors to have less strong force to have free quarks and since Neutron stars are composed of Neutronium i.e in the core and rarely a few protons and neutrons thus it is important to check the Gluons And Quarks of an neutron with QCD.

While researching the question i found that there is an article in PhysicsWorld which talks about the same (however it does not mention the same reasons)

Blackholes might not have the same though because anything above the Tolman-Oppenheimer-Volkov limit don't have Quarks.

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    $\begingroup$ have a look nature.com/articles/d41586-020-00590-8 It ends with "Even further down, the pressure is so intense that almost all the protons combine with electrons to turn into neutrons, but what occurs beyond that is murky at best (see ‘Dense matter’). " $\endgroup$
    – anna v
    Commented Jan 5, 2023 at 18:31
  • $\begingroup$ Thanks, @annav. $\endgroup$
    – user348071
    Commented Jan 6, 2023 at 3:54
  • $\begingroup$ Quark stars are formed due : However, it is hypothesized that under even more extreme temperature and pressure, the degeneracy pressure of the neutrons is overcome, and the neutrons are forced to merge and dissolve into their constituent quarks, creating an ultra-dense phase of quark matter based on densely packed quarks. In this state, a new equilibrium is supposed to emerge, as a new degeneracy pressure between the quarks, as well as repulsive electromagnetic forces, will occur and hinder total gravitational collapse. $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 6, 2023 at 7:33
  • $\begingroup$ @AgniusVasiliauskas Thanks. However "Is this is another plausible reason?" $\endgroup$
    – user348071
    Commented Jan 6, 2023 at 7:36
  • $\begingroup$ Have a look at "strange star" en.wikipedia.org/wiki/Strange_star and the "strange matter hypothesis" (aka "Bodmer-Witten hypothesis"): when a large number of quarks are concentrated together, the lowest energy state has roughly equal numbers of up, down, and strange quarks. $\endgroup$
    – Quillo
    Commented May 2, 2023 at 12:11

1 Answer 1

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"Free" quarks may appear in the cores of neutron stars as a result of asymptotic freedom.

At low densities and in "normal" matter, quarks are confined by the strong force and cannot be separated from nucleons. Effectively, the force between them grows with separation. At very high densities, perhaps ten times that of normal nuclear matter, the force between quarks may become weak enough that they can be treated as free fermions.

At densities of $\sim 3 \times 10^{18}$ m$^{-3}$, any free neutron gas exists in equilibrium with protons and electrons and muons. It is easy to show that at these densities the ratio of neutrons to protons is about 10:1 and that the electrons have a relativistic Fermi energy of order 100-200 MeV. If asymptotic freedom is achieved, then a weak force equilibrium will be setup between quarks and leptons (muons must also be considered at these energies). The available energies are sufficient to create up, down and strange quarks.

The resultant material is known as a quark gluon plasma in which the quarks might be treated as effectively non-interacting and almost massless particles (certainly the u and d quarks).

Neutron stars may achieve high enough densities for quark gluon plasmas to exist near their centres. Such neutron stars may be termed quark stars, though that term is often used for stars that have been entirely transformed into quark matter. However, there are other possibilities that depend on the uncertain behaviour of the strong nuclear force in many particle systems at high densities - e.g. the creation of heavy hadrons or a pion/kaon condensation, that might lead to instability at densities too low to achieve quark freedom.

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  • $\begingroup$ Thanks, However What is the equilibrium term referring to in the "free neutron gas" case? $\endgroup$
    – user348071
    Commented Jan 7, 2023 at 10:23
  • $\begingroup$ @ScienceAJ the equilibrium between neutrons, protons, electrons and muons. I don't understand your question. $\endgroup$
    – ProfRob
    Commented Jan 7, 2023 at 10:25
  • $\begingroup$ I mean what type of equilibrium is between neutrons, protons, electrons and muons? $\endgroup$
    – user348071
    Commented Jan 7, 2023 at 10:26
  • $\begingroup$ @ScienceAJ it is a weak force equilibrium. Neutrons decay into protons, electrons (and neutrinos, which escape). $\endgroup$
    – ProfRob
    Commented Jan 7, 2023 at 10:45
  • $\begingroup$ Thanks a lot !. However I have 1 question more: According to The Asympotic freedom equation and then conversion of the energy value to mass by mass-energy equivalence i.e M=E/C2 , does it give 2.5 M☉? because 2.5 M☉ is the value for the mass of Quark Stars. $\endgroup$
    – user348071
    Commented Jan 7, 2023 at 12:21