In a white dwarf, the star is prevented from collapsing due to the Pauli exclusion principle. If the star is heavy enough, the protons in the star will capture electrons, forming neutrons and bypassing the exclusion principle for electrons. The exclusion principle can come into effect again with the neutrons, forming a neutron star. If the star is heavy enough, it will again overcome this exclusion principle and become a black hole.

My question is, how is this possible? If the neutrons are in their absolute ground state, how can the star collapse down to a lower energy state? Do the neutrons themselves decay?

  • $\begingroup$ See this answer by ProfRob physics.stackexchange.com/a/672929/123208 $\endgroup$
    – PM 2Ring
    Feb 22 at 20:04
  • $\begingroup$ To summarise an answer I can't find that I wrote for a similar question, very intense gravity/pressure is such a big perturbation term in the Hamiltonian you can no longer give its eigenstates the same labels as you can in the zero-gravity limit. $\endgroup$
    – J.G.
    Feb 22 at 20:18
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    $\begingroup$ If the star is heavy enough, it will again overcome this exclusion principle and become a black hole” - (1) The Pauli Exclusion Principle for specific particles is never violated for as long as these particles exist. It can be overcome for neutrons only if neutrons split into quarks, but then it cannot be violated for elementary quarks. (2) The Pauli Exclusion Principle does not need to be violated in the gravitational collapse. (3) Conceptually a non-rotating galaxy can collapse to a supermassive black hole without the stars being destroyed or even compressed, as observed from outside. $\endgroup$
    – safesphere
    Feb 23 at 3:39
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    $\begingroup$ The answers, while generally good, assume the Euclidean geometry of space while calculating the volumes. However, in a gravitational collapse the geometry becomes Schwarzschild, which is very different. For example, once the horizon forms in the center, no matter how small, the black hole is formed, even if its mass is negligible compared to the mass of all the star matter outside. From this point on things like pressure, degeneracy, or Pauli Exclusion Principle no longer matter. The black hole will inevitably grow until it eats all the star matter by sucking it in at near the speed of light. $\endgroup$
    – safesphere
    Feb 23 at 3:59
  • $\begingroup$ Even if matter cannot collapse any further, spacetime can warp so that the relationship between radius and volume becomes non-Euclidian and the volume can contain more matter than the apparent radius would be able to in Euclidian space. $\endgroup$ Feb 23 at 14:59

2 Answers 2


In both white dwarfs and neutron stars, collapse occurs because hydrostatic equilibrium cannot be achieved by decreasing the radius. This process occurs even if the electrons or neutrons remain present and unchanged, but might be accelerated if some process removes them.


White dwarfs and Newtonian gravity

Here is a simple Newtonian argument for why a collapse must eventually occur in white dwarfs of increasing mass, governed by relativistic electron degeneracy pressure, even if no neutronisation occurs.

As the mass of a white dwarf increases, its radius decreases and its density increases. At low densities, the degenerate electrons have non-relativistic energies, and the pressure $P \propto \rho^{5/3}$. As the density increases, the degenerate electrons get pushed to relativistic energies and the relativistic degeneracy pressure goes as $P \propto \rho^{4/3}$.

What supports a star against gravity is the pressure gradient, $dP/dr$, through the equation of hydrostatic equilibrium $$ \frac{dP}{dr} = - \rho g\ . $$ If we just deal in proportionalities so that $dP/dr \propto P/R$ and $\rho \propto M/R^3$, then in the low density, non-relativistic case, with $g \propto M/R^2$, we have $$\frac{M^{5/3}}{R^6} = \frac{M^2}{R^5}\ .$$ For any given increase in mass it is then possible to decrease $R$ to keep the LHS equal the increasing RHS.

In the relativistic case, the hydrostatic equilibrium equation gives $$\frac{M^{4/3}}{R^5} = \frac{M^2}{R^5}\ . $$ In this case, the equation can only work for a single mass. Any increase in the mass above that means the RHS would become bigger than the LHS and the star will collapse. This limiting mass is the Chandrasekhar mass.

It is worth noting then that the pressure provided by fermion degeneracy will always be "overcome" at some finite mass threshold even in Newtonian physics. Consideration of General Relativity (see below) simply lowers the mass threshold.

In practice, the "real" Chandrasekhar mass is a little lower in typical carbon white dwarfs because indeed, electrons are captured by protons in the nuclei once they become highly relativistic; this removes electrons and lowers degeneracy pressure, leading to collapse.

Neutron stars and General Relativity

In neutron stars, the reason for the upper limit is also because hydrostatic equilibrium cannot be reached, either because of the increasing density (even if the neutrons remain intact) but possibly accelerated if the neutrons are removed.

We cannot use the Newtonian hydrostatic equilibrium equation in neutron stars. Its General Relativistic equivalent is the TOV hydrostatic equilibrium equation. $$\frac{dP}{dr}=-\left(P+\rho\right)\frac{m(r)+4\pi r^3P }{r\left(r-2m(r)\right)}\ .$$ A major difference is that the pressure appears on the RHS. This means that increasing the density at the centre of the star in order to increase the pressure gradient and support a more massive star also increases the pressure gradient required to support that star. Ultimately this is self-defeating and the RHS will always be bigger than the LHS, for any radius, and the star collapses. The mass threshold for collapse is lower than it would be if Newtonian hydrostatic equilibrium were considered.

There is considerable debate however as to whether this process might be accelerated by the disappearance of neutrons. This might happen because they have enough energy to create heavy hadrons - the hyperons - like $\Sigma$ and $\Lambda$ particles. This would have the effect of turning neutron kinetic energy, which is a source of pressure, into additional rest-mass energy and thus decreases pressure for a given density and might destabilise the star.

More catastrophic may be the production of mesons via strong force interactions - pions or kaons. These feel the strong nuclear force but are bosons, can form a condensate, and so that component of the pressure contributed by neutron (fermion) degeneracy would be removed and might trigger the collapse.

There are other possibilities too - like quark matter, but it is unclear whether that would hinder any collapse. The discovery of neutron stars of mass $2M_\odot$ possibly means that equilibrium structures featuring quark matter do not exist (e.g. Ozel & Freire 2016), but I'm sure you can also find papers that disagree.

  • $\begingroup$ Don't 2+ solar mass neutron stars rule out these fun ideas? Such heavy corpses are only stable if neutrons cannot squish into quarks or other exotica. $\endgroup$ Feb 23 at 8:51
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    $\begingroup$ @KevinKostlan - possibly quark matter is excluded by these, but not hyperonic matter. That does not mean those transitions do not occur, only that they do not lead to equilibrium structures. e.g. see annualreviews.org/doi/full/10.1146/annurev-astro-081915-023322 $\endgroup$
    – ProfRob
    Feb 23 at 9:06

In broad strokes, the answer is that beyond a certain mass limit, called the Tolman-Oppenheimer-Volkoff limit (TOV limit) the radius can decrease indefinitely without the outward forces ever exceeding the inward force due to gravity. But there are a few things wrong with your understanding of the situation that I'll address along the way.

This is the same thing that happens with white dwarves and electron degeneracy pressure, but that limit is called the Chandrasekhar limit. If we ignore for a moment that protons and electrons can combine to form neutrons, the Chandrasekhar limit says that if the mass is beyond: $$ M_C=3.098\left(\frac{\hbar c}{G}\right)^{3/2}\frac{1}{ m_H^2}, $$ The white dwarf's radius decreases indefinitely until it becomes a black hole. To briefly explain where this limit comes from, because fermions (electrons are fermions) cannot occupy the same quantum state as one another, as we add fermions to a system, each new fermion needs to have a higher energy than the previous one, because all the quantum states with lower energy are already occupied. On the other hand, because of gravitational potential energy, the energy of a system decreases when it is compressed. We find that beyond the Chandrasekhar limit, the radius can decrease indefinitely, and the increase of energy due to the increasing density of electrons is less than the decrease of energy due to gravity. It's kind of a common physicist's logical fallacy to assume that because there are two opposing effects, there must be an equilibrium - this is not always true.

In reality, if a white dwarf's mass is between the Chandrasekhar limit and the TOV limit, once the white dwarf has collapsed to a certain point, neutrons are formed. You may not have realized that if we apply the same degeneracy pressure argument to neutron stars, we actually arrive at a mass limit that is roughly half the Chandrasekhar limit. This would seem to predict that neutron stars cannot exist if we only consider degeneracy pressure.

The resolution is that nuclear forces (the strong force) is actually not negligible for the neutron star, and significantly increases the outward pressure. So the TOV limit is a bit higher than the Chandrasekhar limit, and it is still a subject of active research - both theoretical and in astronomical observations (see paper PSR J0952−0607: The Fastest and Heaviest Known Galactic Neutron Star).

Just like with the white dwarves and the Chandrasekhar limit, if the mass of a neutron star is beyond the TOV limit, it collapses indefinitely until it becomes a black hole. There is no equilibrium radius whatsoever if the mass is beyond the TOV limit.


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