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From what is given here free neutrons (neutrons outside of the atomic nuclei) are unstable and decay in about 15 minutes into proton, electron and an anti-neutrino (in most cases).

Also given that neutron stars exist, it would be the case that it is gravity that packs the neutrons close enough to remain stable.

As per my understanding, towards the center of the neutron star (the core), the net gravitational force should decrease (the mass at the middle and outer-core of the neutron star contributes to the gravitational force towards the surface however net force experienced due to gravity at the center itself would be zero given that the masses on the rest of the sphere would pull symmetrically at the core resulting in a net zero gravitational force).

With the above understanding, that there is nearly no gravitational pull at the center, the neutrons there would then be free to decay. Is this understanding correct or did I miss something fundamental? (articles linked do mention that there would be matter in other states but not for the above mentioned reason - rather the justification is due to the higher density - but my point is the exact opposite - that the density at the core would be less due to net gravitational force being zero)

Qmechanic
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Ravindra HV
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    Although the gravity is 0 at the center, the pressure will be maximum, I think. – rodrigo Sep 05 '16 at 19:45
  • But the only (or major) force that could drive pressure in case of a neutron star would be gravity which at the center would be not very significant. – Ravindra HV Sep 05 '16 at 19:49
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    Consider the Earth, for example, or Jupiter. Gravity is also zero at the center of the planet, but the pressure is enormous, because it it the weight of all the matter above, and above there is gravity. – rodrigo Sep 05 '16 at 19:55
  • Well what holds for neutron star should hold for earth and jupiter too. Given that the net gravitational force is zero near the center, effect of any pressure resulting out of liquids and gases in the other layers would be significantly less than what could have been. Besides a neutron star would mostly expected to consist of solid rather than any other in the layers other than towards the center. – Ravindra HV Sep 05 '16 at 20:03
  • What I am also trying to say is that the gravitational force acts outwards near the center (due to gravitational pull from the intermediate layers and surface) and it acts inwards near the surface - (due to the center and intermediate layers). The net gravitational force - for all heavenly bodies including earth, jupiter and any other would have to be near zero at the center. If there is pressure towards the center then it would have to be due to other forces such as gases. – Ravindra HV Sep 05 '16 at 20:09
  • I think that you did not understand my argument. Consider a column of section 1 cm^2 from the center of the Earth until the surface. Divide the column in little cubes 1 cm^3 each; then multiply the mass of each cube times the gravity at that cube; add up all these weights of cubes, that will be the weight per cm^2 a the center of the Earth. That at this very point gravity is 0 is irrelevant, pressure is maximum. – rodrigo Sep 05 '16 at 20:09
  • Let us continue this discussion in chat. – rodrigo Sep 05 '16 at 20:09
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    Your understanding of the mechanism of degenerate gasses is faulty. To what extent does http://physics.stackexchange.com/questions/63383/what-stabilizes-neutrons-against-beta-decay-in-a-neutron-star? help you out? – dmckee --- ex-moderator kitten Sep 05 '16 at 20:27
  • @dmckee - Thanks. I missed that one. I did look into degenerate gasses briefly but my question was based on an understanding which appears to contradict the shell theorem from what rodrigo indicated. So I'll need to start from there. Thanks again! – Ravindra HV Sep 05 '16 at 20:54
  • You have to reason this in terms of degenerate gasses, because it is the energy balance of such forces that controls this behavior. You're implicitly assuming that you can put the decay products into low energy states, but no such states are available. – dmckee --- ex-moderator kitten Sep 05 '16 at 20:55
  • @dmckee I was not focusing on degenerate gasses because my understanding was that the condition for their presence itself was not there - which is high pressure (and possibly other factors). But I guess I'll have to revisit the whole thing. – Ravindra HV Sep 05 '16 at 21:00
  • One final point, although the nuclear force between neutron and proton is stronger, the force between two neutrons is not zero. There have been searches for bound states between two neutrons and although they have not been found, they still might exist. – Lewis Miller Sep 05 '16 at 21:51

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Gravity is in balance with a pressure gradient, not the pressure. The equation of hydrostatic equilibrium is $$\frac{dP}{dr}=-\rho g\ ,$$ where $\rho$ and $g$ are the local density and gravity respectively.

You are correct that $g=0$ exactly at the centre of a neutron star. This also means that the pressure gradient is zero at the centre of the neutron star, which means the pressure and therefore the density are at a maximum.

What stops the neutrons decaying is not the local gravitational field, it is their extremely high number densities and the presence of a small fraction of degenerate protons and electrons.

Consider a thought experiment where you were able to confine a dense gas of pure neutrons. There would be an initial decay phase producing some protons and electrons. But the densities of these fermion species would build up until they were also degenerate. When the electron Fermi energy reaches the maximum possible from the neutron beta decay process, then further neutron decay is blocked. All lower electron energy states are already filled.

ProfRob
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  • From what I can understand from here , hydrostatic equilibrium would be a consequence of the interaction between gravity and pressure. The way you put it, it makes it appear that gravity would be a consequence rather than part of the cause. – Ravindra HV Sep 13 '16 at 05:48
  • Given that a neutron star can be considered as mostly homogenous (consisting of neutrons), from what I have understood from the shell theorem (following discussion with rodrigo), as we go towards the center of the sphere, the effect of gravity (therefore pressure acting towards the center) decreases . – Ravindra HV Sep 13 '16 at 05:59
  • So as we move towards the center, my understanding is that - due to the decreasing pressure, it would then be possible for the neutrons to decompose into protons and electrons, simply because there is no longer sufficient pressure to hold them close to each other, thus making them behave more like free neutrons. – Ravindra HV Sep 13 '16 at 06:04
  • It makes sense to me that the neutron star as a whole is in hydrostatic equilibrium where - gravity ensures the outer shell tends to collapse the structure (force acts towards the center) while the pressure from the decomposed protons and electrons at the core resists the same (force acts away from the center). Although wiki states that the degenerate particles in case of neutron star are neutrons themselves which contradicts the above understanding. – Ravindra HV Sep 13 '16 at 06:15
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    @RavindraHV Hydrostatic equilibrium balances gravity with a (negative) pressure gradient, not the pressure. Pressure of couse increases towards the centre. Pressure is dominated by the dominant species - neutrons. – ProfRob Sep 13 '16 at 06:30
  • Thanks for getting back! (a) If I understand correctly, by pressure-gradient, you mean that its not just pressure but its a pressure that varies along the axis the force due to pressure acts and in this case the pressure increases towards the center. If yes, then fine, I get that. (b) Given that neutrons are not known to repel each other, in this case it would not be the electrostatic force that is the source for the pressure but rather some other one. Is this correct? – Ravindra HV Sep 13 '16 at 06:49
  • @RavindraHV (a) $dP/dr = -\rho g$. At the very centre both $g=0$ and $dP/dr=0$. (b) Neutrons are affected by degeneracy (leading to degeneracy pressure), but also the strong nuclear force is repulsive between neutrons above the usual density of nuclear matter. – ProfRob Sep 13 '16 at 06:55
  • a) From Dictionary of Geophysics, Astrophysics, and Astronomy edited by Richard A. Matzner for hydrostatic-equilibrium : P and ρ are the local values of pressure and density at radius r. In the answer you say that both the pressure and density is highest at the center and at the very center its zero. b) Thanks. – Ravindra HV Sep 13 '16 at 07:24
  • Sorry, dP/dr being zero would be the gradient. So pressure would still be max at center. – Ravindra HV Sep 13 '16 at 07:32
  • To summarize the understanding, although the pressure caused by gravity is the highest at the surface and decreases as we move towards the center, the pressure still adds up as we move towards the center. As pressure increases so does density until the point (near the center) wherein due to the prevailing conditions, results in the formation of degenerate gas. (The misunderstanding was to not account for the shell theorem and therefore concluding that there was a gravitational force acting away from the center as a result of the surrounding mass). – Ravindra HV Sep 13 '16 at 08:10
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    @RavindraHV That would not be my summary. The pressure is zero at the surface (by definition). You appear not to be grasping that hydrostatic equilibrium is between the pressure gradient and gravity. The highest pressure gradient is near the surface. The neutrons (and protons and electrons) are degenerate in the vast majority of the neutron star, though the structure is more complex than you describe. – ProfRob Sep 13 '16 at 08:42
  • Right. I meant effect of pressure due to gravity is highest from below the surface on-wards. I do understand this : In any given layer of a star, there is a hydrostatic equilibrium between the outward thermal pressure from below and the weight of the material above pressing inward. Is that what you are trying to say? Also from what you say as well as from wiki it appears that there is no 'core'. – Ravindra HV Sep 13 '16 at 10:22
  • You write here “degenerate protons.” But isn’t it the case that the neutrons and electrons are degenerate, while the protons are not? The proton charge density should match the electron charge density, so the proton number density should be white-dwarf-ish. – rob Apr 01 '22 at 09:48
  • @rob neutron number density is of order $10^{45}$ m$^{-3}$, so proton number density is $\sim 10^{43}$ m$^{-3}$. In a white dwarf the electron number density is only of order $10^{35}$-$10^{36}$ m$^{-3}$, so more than a million times less. What matters for degeneracy is to compare $kT$ with the Fermi (kinetic) energy - which goes as $n^{2/3}/2m$ - so that is more than $10-100$ times bigger for protons. So even though neutron stars can be 10-1000 times hotter than white dwarfs, the protons in a neutron star are still degenerate. – ProfRob Apr 01 '22 at 13:00
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To summarize the understanding, although the pressure caused by gravity is the greatest at the surface and decreases progressively as we move towards the center, the pressure still adds up as we move towards the center (progressively increases). As pressure increases so does density until the point (near the center) wherein due to the prevailing conditions, results in the formation of degenerate gas.

The misunderstanding was to not account for the shell theorem and therefore concluding that there was a gravitational force acting away from the center as a result of the surrounding mass which also negated the pressure.

Ravindra HV
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