5
$\begingroup$

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)

$\endgroup$
12
  • 2
    $\begingroup$ Although the gravity is 0 at the center, the pressure will be maximum, I think. $\endgroup$
    – rodrigo
    Commented Sep 5, 2016 at 19:45
  • $\begingroup$ 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. $\endgroup$ Commented Sep 5, 2016 at 19:49
  • 1
    $\begingroup$ 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. $\endgroup$
    – rodrigo
    Commented Sep 5, 2016 at 19:55
  • $\begingroup$ 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. $\endgroup$ Commented Sep 5, 2016 at 20:03
  • 3
    $\begingroup$ Your understanding of the mechanism of degenerate gasses is faulty. To what extent does physics.stackexchange.com/questions/63383/…? help you out? $\endgroup$ Commented Sep 5, 2016 at 20:27

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
14
  • $\begingroup$ 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. $\endgroup$ Commented Sep 13, 2016 at 5:48
  • $\begingroup$ 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 . $\endgroup$ Commented Sep 13, 2016 at 5:59
  • $\begingroup$ 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. $\endgroup$ Commented Sep 13, 2016 at 6:04
  • 1
    $\begingroup$ @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. $\endgroup$
    – ProfRob
    Commented Sep 13, 2016 at 6:30
  • 1
    $\begingroup$ @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. $\endgroup$
    – ProfRob
    Commented Sep 13, 2016 at 8:42
0
$\begingroup$

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.

$\endgroup$

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.