# Why can a neutron star implode? [duplicate]

In a book* I was reading, it said that a neutron star is formed when the pressure is so large that the electrons in a white dwarf interact with the protons, forming neutrons. The neutrons then collapse until the Pauli Exclusion Principle stops any further collapse.

However, my issue is with why, when the mass is even greater, how the neutron star collapses into a black hole:

The book says that the pressure is so high, and therefore the temperature is so high, that the neutrons start to move close to the speed of light. At this speed, the Pauli Exclusion Principle doesn't apply any more and the neutron star goes on to implode.

Why does the Exclusion Principle stop working at high speeds?

* The book was The Quantum Universe by Brian Cox and Jeff Forshaw. The subject of this question can be found in the chapter Epilogue: The Death of Stars

• @ACuriousMind The book is "The Quantum Universe" by Brian Cox and Jeff Forshaw. You can find the claim in the epilogue Aug 21, 2017 at 12:43
• Aug 21, 2017 at 18:38
• I don't think the authors meant to say the PEP does not apply any more. Rather, an argument based on PEP which would predict that a stable state can be reached at low temperature (non-relativistic behaviour) also predicts that that stable state is not available at high temperature. The PEP is valid throughout; it is just the relationship between momentum (and hence pressure) and energy that changes. Aug 28, 2021 at 10:08

Summary

In a Newtonian scheme, an increasing "weight" would be supported by ideal degeneracy pressure. An increased density leads to increased degeneracy pressure (provided by the PEP) and this can simply continue until an infinite density is approached (though at a finite mass).

In General Relativity this does not happen. Instability is reached at a finite density, because in GR, the increasing density and pressure contribute to the increasing curvature of space. Thus the increase in pressure ultimately becomes self-defeating and actually triggers the collapse.

There is no question of the PEP being "overcome" or dis-applying. The maximum neutron momentum and hence neutron degeneracy pressure, will continue to increase whilst the collapse proceeds but never by enough to re-establish an equilibrium.

Details

Your book is incorrect on a number of points - possibly for simplicity.

Neutron stars do indeed consist mainly of neutrons. They are packed together at separations of around a femto-metre and at these densities, the Pauli-Exclusion-Principle (PEP) does come into consideration It is not possible for the neutrons to be all placed in quantum states with low momentum whilst also being closely packed. Since the neutrons have a distribution of quite significant momenta, with speeds getting on for a good fraction of the speed of light, they do of course exert a degeneracy pressure.

However, degeneracy pressure is not what is (mostly) responsible for supporting neutron stars. It has been known since the late 1930s (Oppenheimer & Volkhoff 1939), that ideal neutron degeneracy pressure is only capable of supporting stars up to around 0.7 solar masses. All of the neutron stars where measurements have been performed have masses greater than 1.15 solar masses.

Neutron stars would not exist without the repulsive force between neutrons provided by the strong nuclear force at small nucleon separations. This hardens the equation of state in dense, neutron-rich matter, and is responsible for the support of neutron stars more massive than 0.7 solar masses and for the "core-bounce" that is ultimately responsible for core-collapse supernovae explosions.

OK, but even disregarding this important point, your question still remains - even in the approximation of ideal degeneracy pressure, there is a finite limit to the mass and density of a neutron star that can be built in general relativistic conditions. To see why, one examines the Tolman-Oppenheimer-Volkhoff (TOV) equation for hydrostatic equilibrium, which is the correct formulation under General Relativistic conditions: $$\frac{dP}{dr} = -\left(\frac{Gm(r) \rho}{r^2}\right)\, \frac{(1 + P/\rho c^{2})(1 + 4\pi r^{3}P/m(r)c^{2})} {(1 - 2Gm(r)/rc^{2})},$$ where $$m(r)$$ is the mass contained within radius $$r$$ and $$P$$ and $$\rho$$ are the pressure and density at radius $$r$$.

The TOV equation reverts to its Newtonian approximation when $$P \ll \rho c^2$$ and when $$Gm(r)/rc^2 \ll 1$$, which is equivalent to saying that the pressure is (relatively) small and the curvature of space given by the Schwarzschild metric can also be ignored.

The difference between the Newtonian and TOV formulations is that pressure appears on the right hand side of the TOV equation and the General Relativistic corrections act to increase the required pressure gradient. This means that increasing the pressure, by increasing the density ultimately becomes self-defeating because the required pressure gradient is increased even more. Detailed calculations (e.g. see section 9.5 of Black Holes, White Dwarfs and Neutron Stars by Shapiro & Teukolsky) show that whereas in a Newtonian scheme the density and degeneracy pressure can just keep increasing until instability is only reached at infinite density (exactly as you suppose in your question), in General Relativity the instability is reached at a finite density where increasing pressure cannot supply the increasing pressure gradient demanded by the TOV equation.

At what density this threshold is reached depends on the exact equation of state (relationship between pressure and density) governing the neutron star matter and the composition of the neutron star core. But even for the "hardest" equations of state the limit is reached at a little over 3 solar masses.

• I don't if your answer states this or not, but why exactly is the PEP overcome here? Is the answer due to the curvature of spacetime? Aug 21, 2017 at 16:17
• @BetaDecay The PEP is not a force. There is nothing to "overcome". As I describe above, it isn't even degeneracy that provides most of the pressure. It is simply that the RHS becomes larger than the LHS (in absolute terms) and they cannot be made to balance. But yes, in broad terms, both density and pressure increase the required pressure gradient in GR. Aug 21, 2017 at 18:30