<p>You have the wrong idea about degeneracy pressure. There is no limit in principle on how closely together you can squeeze electrons (or other fermions) and at no point is the Pauli Exclusion Principle violated. All that happens is that their RMS momentum must increase - in accordance with the Heisenberg uncertainty principle.</p>
<p>Thus, the more densely you pack the electrons, the higher becomes their average momentum and it is this momentum that leads to &quot;degeneracy pressure&quot;.</p>
<p>Why then are &quot;neutron stars&quot; not held up by electron degeneracy pressure? The answer is because they are (mostly) made of neutrons!</p>
<p>The composition of a compact object will itself be density dependent. The reason for this is that the matter will attempt to convert itself into whatever form has the <em>minimum</em> energy density, where energy in this case includes the rest mass of the constituents and their kinetic energy.</p>
<p>In the cases of white dwarfs and the cores of massive stars that are supported by electron degeneracy pressure, once the density reaches a threshold, the most energetic electrons are capable of &quot;neutronising&quot; the protons in the gas (there must be an equal number of protons for charge neutrality). The electrons combine with the protons to form neutrons.</p>
<p>In both cases, the removal of free electrons means that electron degeneracy pressure does not increase as the star becomes denser - an unstable situation that can lead to collapse or possibly to a thermonuclear explosion (in the case of a white dwarf when the density becomes high enough to trigger pycnonuclear reactions). Note that it <em>is</em> possible for a white dwarf to collapse to a neutron star (see <a href="https://astronomy.stackexchange.com/a/25907/2531">https://astronomy.stackexchange.com/a/25907/2531</a> ); the fate of massive white dwarfs and whether they collapse or blow-up is a complex problem (see <a href="https://astronomy.stackexchange.com/a/14747/2531">https://astronomy.stackexchange.com/a/14747/2531</a>).</p>
<p>During the collapse, temperatures become so hot that photons have enough energy t break up any nuclei and produce a gas that is mostly neutrons with about 1% protons and electrons. The collapse <em>may</em> be arrested by the combined effects of neutron degeneracy pressure (those occurs at higher densities than for electrons, because neutrons are much more massive) and strong nuclear force repulsion once the neutrons are separated by <span class="math-container">$\sim 10^{-15}$</span>m. The electrons (and protons) are also degenerate, but because of their low densities (still much denser than in a white dwarf!) compared with the neutrons, they are a minor contributor to the pressure.</p>

CC BY-SA 4.0
2022-08-09T02:34:17.447Z
Neutronising is always possible.  The question is whether the reaction is endothermic or exothermic--collapse occurs when the pressure becomes high enough that it's exothermic.  We see it in the chemical world with ordinary water ice under pressure.
Loren Pechtel
CC BY-SA 4.0
2022-08-09T07:32:42.117Z
Neutronisation becomes possible when electrons have sufficient energy. (Fortunately) It does not occur for electrons with energy below the neutronisation threshold. - about 10 MeV for carbon, but lower for iron nuclei and only 1.3 MeV for protons @LorenPechtel
ProfRob
CC BY-SA 4.0
2022-08-09T15:06:23.910Z
and neutronisation is always "endothermic" in the sense that it takes kinetic energy from an electron and turns it into (mostly) rest mass. That's why it destabilises the star.
ProfRob
CC BY-SA 4.0
2022-08-10T15:24:00.960Z
But not all electrons will be at the same energy level.  Even when most aren't energetic enough the reaction would happen--just not at a meaningful rate.  And I'm looking at the big picture in saying it's exothermic.  Of course it takes energy to do it, but it yields energy in letting the star shrink a tiny bit.  When the yield from the shrink exceeds the energy to drive the reaction the star goes down the rabbit hole.
Loren Pechtel
CC BY-SA 4.0
2022-08-10T18:31:01.217Z
The uncertainty principle implies an increase in the standard deviation of the momentum, not its mean. It's more realistic to say each Cartesian component of momentum has mean zero, but increased root mean square. This also translates into a greater mean modulus of each component, and mean modulus of the momentum vector. And there's also an increase in the mean KE.
J.G.
CC BY-SA 4.0
2022-08-08T15:54:11.967Z
2022-08-10T19:13:00.390Z
2022-08-10T19:13:00.390Z
ProfRob
<p>A neutron star contains relatively few electrons.</p>
<p>In a white dwarf, electron degeneracy pressure does prevent further collapse. But if you add more matter to a white dwarf, it will shrink in volume, increasing the pressure.</p>
<p>Now if you keep adding matter, at some point, some kind of nuclear reactions will start. In a regular carbon-oxygen white dwarf, it is nuclear fusion and the white dwarf explodes as a type 1a supernova.  But in the iron core of a massive star a different reaction occurs. Protons combine with electrons to produce neutrons.  This removes electrons and allows further collapse. This happens extremely rapidly and releases an enormous pulse of neutrinos, which results in the outer parts of the star exploding in a type II supernova.</p>
<p>In a very large star the core passes through a state of election degeneracy very rapidly.  There will be an electron degenerate iron core, but it won't last long! Less than a day (?) So much iron is produced in such a short time that electron degeneracy is rapidly overcome and neutronisation occurs.  There is no visible &quot;iron white dwarf&quot;</p>
<p>An iron white-dwarf couldn't easily form.  Stars comparable to the sun will produce carbon-oxygen white dwarfs.  Larger stars will explode in supernovae.  There is no sweet spot at which a star will produce just enough iron to form an iron white dwarf supported by electron degeneracy.</p>

CC BY-SA 4.0
2022-08-08T12:07:52.993Z
2022-08-08T12:07:52.993Z
James K
<p>When an old star begins to collapse under its own gravity, one can in principle numerically solve the Schrödinger equation for the potential its electrons collectively experience. Although this is a linear equation, we are interested not in arbitrary solutions to it, but those which are antisymmetric under exchanging any two electrons. If the gravity-induced classical pressure is moderate, the quantum numbers of legal states are the same as with no gravity at all; we can work at say first order in perturbation theory to compute how the solutions shift, but the same discrete parameters describe them. But if the pressure is too high, the solutions are so unlike the zero-gravity case the solutions need to be differently labelled. Ultimately, the new low-energy states are numerous enough for electrons' marginal wavefunctions to overlap more than they used to, so electron degeneracy pressure no longer resists the collapse.</p>
<p>In a neutron star, after EDP fails the collapse continues to an extent, but we can repeat the above logic with neutrons. This time, it is their degeneracy pressure which wins, because the quantum numbers labelling multi-neutron states are the same as in the zero-gravity case. If we go to even more extreme gravity, of course, even this fails, so neutron degeneracy pressure no longer prevents collapse.</p>

CC BY-SA 4.0
2022-08-08T12:34:37.093Z
2022-08-08T12:34:37.093Z
J.G.
<p>You don't need quantum instrumentarium for a basic understanding of the compact star matters:</p>
<ol>
<li>The inverse beta decay does not &quot;violate&quot; the exclusion principle. It simply removes electrons from the whole system, reducing the pressure. The inverse beta decay consumes energy and this energy is usually delivered by the gravitational collapse of the star.</li>
</ol>
<p>If the star is not heavy enough, it can't deliver the needed energy for the inverse beta so it stays in the &quot;electron degeneracy state&quot; indefinitely.</p>
<ol start="2">
<li><p>There still is an electron degeneracy pressure in a neutron star. It is way higher than the pressure in a white dwarf. There is a proton degeneracy pressure in there as well. Both are just not this much important, compared to the neutron degeneracy pressure.</p>
</li>
<li><p>In order to get a neutron star, you start with a bigger star in the first place. Its core is hot and dense enough to burn everything all the way to iron before collapsing. Then the collapse and the inverse beta happens.</p>
</li>
</ol>

CC BY-SA 4.0
2022-08-08T15:23:57.660Z
2022-08-08T15:46:23.480Z
2022-08-08T15:46:23.480Z
fraxinus
<p>In a Fermi gas of electrons, the electron energy reflects the pressure. If the highest energy electrons are relativistic, the equation of state softens: a density increase doesn't produce as much pressure as it does in the nonrelativistic case. The softer equation of state is less able to resist gravitational collapse: thus we get the <a href="https://en.wikipedia.org/wiki/Chandrasekhar_limit" rel="nofollow noreferrer">Chandrasekhar Limit</a> for the mass of a white dwarf star.</p>
<p>It's more complicated for a neutron star: although the Fermi pressure of the neutrons is important, the nuclear force between them modifies the equation of state, making it stiffer.</p>

CC BY-SA 4.0
2022-08-08T13:45:51.767Z
2022-08-08T13:45:51.767Z
John Doty
<p>I was thinking a little bit, and never asked myself the following. If white dwarfs do not collapse, because electron degeneracy pressure stops the star from collapsing by its own gravity, and this is due to the Pauli exclusion principle, then why aren't neutron stars held by the same pressure? How does a star become a more compact object if the electrons &quot;cannot be more together&quot;, what is going on when the inverse beta decay process &quot;violates&quot; the Pauli exclusion principle (for electrons)? And why is neutron degeneracy pressure now the one that doesn't allows gravity to do its thing?</p>
<p>I think I have a wrong picture, like that the stars are first on a &quot;white dwarf state&quot; and then moves to a &quot;neutron star state&quot; and in reality a white dwarf is a result of a very specific final evolution phase of a low mass kind of star. And on the other hand a neutron star is the core remnant of a massive star.
I don't think this is the answer because why would the core go from an iron solid state to a degenerate gas of neutrons (and other stuff) without converting in a electron degenerate gas, but that it was what came to my mind.</p>
<p>Another thing that maybe answers my question, is quantum field theory. I mean, in this model there are particles that have some energy, properties and quantum numbers, and if you have the proper ones you can have any physical possible state, so at some point the question is not why something is violating the Pauli exclusion principle instead the question would be, which state is more probable to achieve given the conditions the object brings.</p>
<p>Thanks for reading, I'll be waiting for an answer.</p>

CC BY-SA 4.0
2022-08-07T23:48:00.337Z
+1. As a to brief thought on this: A degenerate election gas or whatever state describes matter in a neutron star also has a finite pressure. If gravity is stronger...
planetmaker
<gravity><neutron-star><quantum-mechanics>
Why does electron degeneracy pressure not stop massive star collapse?
martín canullán
# Why does electron degeneracy pressure not stop massive star collapse?

(Asked by: username_0 on 2022-08-07)

I was thinking a little bit, and never asked myself the following. If white dwarfs do not collapse, because electron degeneracy pressure stops the star from collapsing by its own gravity, and this is due to the Pauli exclusion principle, then why aren't neutron stars held by the same pressure? How does a star become a more compact object if the electrons "cannot be more together", what is going on when the inverse beta decay process "violates" the Pauli exclusion principle (for electrons)? And why is neutron degeneracy pressure now the one that doesn't allows gravity to do its thing?

I think I have a wrong picture, like that the stars are first on a "white dwarf state" and then moves to a "neutron star state" and in reality a white dwarf is a result of a very specific final evolution phase of a low mass kind of star. And on the other hand a neutron star is the core remnant of a massive star.
I don't think this is the answer because why would the core go from an iron solid state to a degenerate gas of neutrons (and other stuff) without converting in a electron degenerate gas, but that it was what came to my mind.

Another thing that maybe answers my question, is quantum field theory. I mean, in this model there are particles that have some energy, properties and quantum numbers, and if you have the proper ones you can have any physical possible state, so at some point the question is not why something is violating the Pauli exclusion principle instead the question would be, which state is more probable to achieve given the conditions the object brings.

Thanks for reading, I'll be waiting for an answer.

**Tags:** gravity, neutron-star, quantum-mechanics
> **Comments:**
> **username_1**: +1. As a to brief thought on this: A degenerate election gas or whatever state describes matter in a neutron star also has a finite pressure. If gravity is stronger...  

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## 5 answers

### Answer by username_2 on 2022-08-08. Score: 23

You have the wrong idea about degeneracy pressure. There is no limit in principle on how closely together you can squeeze electrons (or other fermions) and at no point is the Pauli Exclusion Principle violated. All that happens is that their RMS momentum must increase - in accordance with the Heisenberg uncertainty principle.

Thus, the more densely you pack the electrons, the higher becomes their average momentum and it is this momentum that leads to "degeneracy pressure".

Why then are "neutron stars" not held up by electron degeneracy pressure? The answer is because they are (mostly) made of neutrons!

The composition of a compact object will itself be density dependent. The reason for this is that the matter will attempt to convert itself into whatever form has the *minimum* energy density, where energy in this case includes the rest mass of the constituents and their kinetic energy.

In the cases of white dwarfs and the cores of massive stars that are supported by electron degeneracy pressure, once the density reaches a threshold, the most energetic electrons are capable of "neutronising" the protons in the gas (there must be an equal number of protons for charge neutrality). The electrons combine with the protons to form neutrons.

In both cases, the removal of free electrons means that electron degeneracy pressure does not increase as the star becomes denser - an unstable situation that can lead to collapse or possibly to a thermonuclear explosion (in the case of a white dwarf when the density becomes high enough to trigger pycnonuclear reactions). Note that it *is* possible for a white dwarf to collapse to a neutron star (see <https://astronomy.stackexchange.com/a/25907/2531> ); the fate of massive white dwarfs and whether they collapse or blow-up is a complex problem (see <https://astronomy.stackexchange.com/a/14747/2531>).

During the collapse, temperatures become so hot that photons have enough energy t break up any nuclei and produce a gas that is mostly neutrons with about 1% protons and electrons. The collapse *may* be arrested by the combined effects of neutron degeneracy pressure (those occurs at higher densities than for electrons, because neutrons are much more massive) and strong nuclear force repulsion once the neutrons are separated by $\sim 10^{-15}$m. The electrons (and protons) are also degenerate, but because of their low densities (still much denser than in a white dwarf!) compared with the neutrons, they are a minor contributor to the pressure.

> **Comments:**
> **username_3**: Neutronising is always possible.  The question is whether the reaction is endothermic or exothermic--collapse occurs when the pressure becomes high enough that it's exothermic.  We see it in the chemical world with ordinary water ice under pressure.  
> **username_2**: Neutronisation becomes possible when electrons have sufficient energy. (Fortunately) It does not occur for electrons with energy below the neutronisation threshold. - about 10 MeV for carbon, but lower for iron nuclei and only 1.3 MeV for protons @LorenPechtel  
> **username_2**: and neutronisation is always "endothermic" in the sense that it takes kinetic energy from an electron and turns it into (mostly) rest mass. That's why it destabilises the star.  
> **username_3**: But not all electrons will be at the same energy level.  Even when most aren't energetic enough the reaction would happen--just not at a meaningful rate.  And I'm looking at the big picture in saying it's exothermic.  Of course it takes energy to do it, but it yields energy in letting the star shrink a tiny bit.  When the yield from the shrink exceeds the energy to drive the reaction the star goes down the rabbit hole.  
> **username_4**: The uncertainty principle implies an increase in the standard deviation of the momentum, not its mean. It's more realistic to say each Cartesian component of momentum has mean zero, but increased root mean square. This also translates into a greater mean modulus of each component, and mean modulus of the momentum vector. And there's also an increase in the mean KE.  

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### Answer by username_5 on 2022-08-08. Score: 14

A neutron star contains relatively few electrons.

In a white dwarf, electron degeneracy pressure does prevent further collapse. But if you add more matter to a white dwarf, it will shrink in volume, increasing the pressure.

Now if you keep adding matter, at some point, some kind of nuclear reactions will start. In a regular carbon-oxygen white dwarf, it is nuclear fusion and the white dwarf explodes as a type 1a supernova. But in the iron core of a massive star a different reaction occurs. Protons combine with electrons to produce neutrons. This removes electrons and allows further collapse. This happens extremely rapidly and releases an enormous pulse of neutrinos, which results in the outer parts of the star exploding in a type II supernova.

In a very large star the core passes through a state of election degeneracy very rapidly. There will be an electron degenerate iron core, but it won't last long! Less than a day (?) So much iron is produced in such a short time that electron degeneracy is rapidly overcome and neutronisation occurs. There is no visible "iron white dwarf"

An iron white-dwarf couldn't easily form. Stars comparable to the sun will produce carbon-oxygen white dwarfs. Larger stars will explode in supernovae. There is no sweet spot at which a star will produce just enough iron to form an iron white dwarf supported by electron degeneracy.


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### Answer by username_4 on 2022-08-08. Score: 4

When an old star begins to collapse under its own gravity, one can in principle numerically solve the Schrödinger equation for the potential its electrons collectively experience. Although this is a linear equation, we are interested not in arbitrary solutions to it, but those which are antisymmetric under exchanging any two electrons. If the gravity-induced classical pressure is moderate, the quantum numbers of legal states are the same as with no gravity at all; we can work at say first order in perturbation theory to compute how the solutions shift, but the same discrete parameters describe them. But if the pressure is too high, the solutions are so unlike the zero-gravity case the solutions need to be differently labelled. Ultimately, the new low-energy states are numerous enough for electrons' marginal wavefunctions to overlap more than they used to, so electron degeneracy pressure no longer resists the collapse.

In a neutron star, after EDP fails the collapse continues to an extent, but we can repeat the above logic with neutrons. This time, it is their degeneracy pressure which wins, because the quantum numbers labelling multi-neutron states are the same as in the zero-gravity case. If we go to even more extreme gravity, of course, even this fails, so neutron degeneracy pressure no longer prevents collapse.


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### Answer by username_6 on 2022-08-08. Score: 4

You don't need quantum instrumentarium for a basic understanding of the compact star matters:

1. The inverse beta decay does not "violate" the exclusion principle. It simply removes electrons from the whole system, reducing the pressure. The inverse beta decay consumes energy and this energy is usually delivered by the gravitational collapse of the star.

If the star is not heavy enough, it can't deliver the needed energy for the inverse beta so it stays in the "electron degeneracy state" indefinitely.

2. There still is an electron degeneracy pressure in a neutron star. It is way higher than the pressure in a white dwarf. There is a proton degeneracy pressure in there as well. Both are just not this much important, compared to the neutron degeneracy pressure.
3. In order to get a neutron star, you start with a bigger star in the first place. Its core is hot and dense enough to burn everything all the way to iron before collapsing. Then the collapse and the inverse beta happens.


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### Answer by username_7 on 2022-08-08. Score: 3

In a Fermi gas of electrons, the electron energy reflects the pressure. If the highest energy electrons are relativistic, the equation of state softens: a density increase doesn't produce as much pressure as it does in the nonrelativistic case. The softer equation of state is less able to resist gravitational collapse: thus we get the [Chandrasekhar Limit](https://en.wikipedia.org/wiki/Chandrasekhar_limit) for the mass of a white dwarf star.

It's more complicated for a neutron star: although the Fermi pressure of the neutrons is important, the nuclear force between them modifies the equation of state, making it stiffer.


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