<p>I have been made aware of the aforementioned 'missing important information' by way of a <a href="https://www.youtube.com/watch?v=8jVYe_g6y2E" rel="nofollow noreferrer">helpful video</a> I was referred to earlier. As I now understand, it is not the strength of atomic structures (strong force?) that keeps the singularity's formation at bay, but rather the formation of a 'proto neutron star core' during the star's final moments that essentially seeds the singularity (if it attains too much mass).</p>
<p>As I understand, the only significant relevance to nuclear fission in all of this is what happens when a star 'runs out of fuel', so to speak. In simple terms, the processed remains of this 'fuel' (iron atoms in the star's core) break apart due to temperature and pressure, and the star begins to collapse. During this process, a neutron star core begins to form, and it is this structure that can exceed critical mass and form a singularity.</p>
<p>If this sounds incorrect, add a comment to clarify.</p>

CC BY-SA 4.0
2023-11-14T04:16:31.557Z
2023-11-14T04:16:31.557Z
hamstar
<p>A different way of looking at the issue: what properties would matter need to have to avoid imploding into a black hole if enough matter was gathered in a static sphere?</p>
<p>The key thing about black holes is that they come from curved spacetime. Thinking about them in terms of force often produces confusion. The equations of general relativity predict that outside a spherical mass spacetime has the <a href="https://en.wikipedia.org/wiki/Schwarzschild_metric" rel="noreferrer">Schwarzschild metric</a>, which has an event horizon if the radius is smaller than <span class="math-container">$R_s = 2GM/c^2$</span>. That horizon has the property that particles with mass (and light) can only cross it inward, and this is enforced (sorry) by what directions they can move, not really any explicit force.</p>
<p>However, an object with radius greater than <span class="math-container">$R_s$</span> and constant density interior has a spacetime metric inside that is the <a href="https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric" rel="noreferrer">interior Schwarzschild metric</a>. The interesting aspect here is that the central pressure is <span class="math-container">$$p = \rho c^2 \frac{1-X}{3X-1}$$</span> where <span class="math-container">$X = \sqrt{1-\frac{R_s}{R}}$</span> where <span class="math-container">$R$</span> is the radius of the body. This goes off to infinity when <span class="math-container">$X=1/3$</span>, or <span class="math-container">$R=(9/8)R_s$</span>. This is known as the Buchdahl limit. Note that it ignores what the object is made from: unless you believe that there are materials that can withstand <em>infinite</em> pressure the sphere will start to compact, and then it will soon become a black hole.</p>
<p>How compact something can be without imploding has been studied a fair bit. <a href="https://arxiv.org/abs/2202.00043" rel="noreferrer">It turns out that for matter that has a speed of sound below the speed of light</a> (a <em>very</em> reasonable assumption!) must have <span class="math-container">$R&gt;1.29 R_s$</span> (about 14% larger than the Buchdahl limit).</p>
<p>These bounds don't care what the matter is made of, but as long as it doesn't have superluminal sound speeds and obeys the laws of general relativity it will implode if packed too densely. In this domain even having enormous resistance to pressure doesn't help since pressure itself is a component in the mass-energy tensor and contributes to gravity: you can literally implode due to the weight of pressure energy! And this implosion is less about matter moving as space itself giving way to form a black hole.</p>

CC BY-SA 4.0
2023-11-13T15:16:42.710Z
2023-11-13T15:16:42.710Z
Anders Sandberg
<h1>Gravity doesn't care at such extreme conditions.</h1>
<p>Yeah, I am definitely not joking.</p>
<p>Friction, nuclear fission and whatever forces are there in the core of a massive dying star have no chance against the intense gravity of a collapsing stellar core. We are talking about the pressure of 20 <strong>solar masses</strong> crushing down on you, and friction stands no chance.</p>
<ul>
<li>First, as soon as the star's core starts fusing iron, it basically starts losing energy. Not that iron cannot fuse or anything, but the fact that fusing iron-56 is endothermic, i.e. it consumes more energy than it releases<sup>1</sup>.</li>
<li>After the iron accumulates at the core, the radiation pressure of the stellar core decreases, as the endothermic reactions keep draining away more and more energy, until finally, the immense gravitational pressure of the electron-degenerate iron core causes it to collapse dramatically.</li>
</ul>
<p>If you are lucky, at some point, neutron-degeneracy pressure fights back and stops the collapse. The abrupt halting of the collapse triggers a <a href="https://youtu.be/XzaYNOKK6Xk?si=npRBHNOh6m1b8yJ7&amp;t=122" rel="nofollow noreferrer">rapid deceleration</a> on the surrounding gas falling onto it. This causes the star to explode in a Type-II supernova, and congrats, you have a neutron star.</p>
<p>The neutrons form because the iron nuclei are crushed so tightly that the protons and electrons ram into each other and form neutrons, emitting a neutrino. Here's the formula:
<span class="math-container">$$p + e \rightarrow n + \nu_e$$</span></p>
<p>However, get above 20 solar masses, and the core isn't so lucky.</p>
<p>We still do not have a clear picture of what exactly happens in the transition from <s>neutronium</s> neutron degenerate matter to a gravitational singularity. This source posits that there is a <a href="https://www.quora.com/What-happens-when-water-is-compressed-to-the-extreme#:%7E:text=We%E2%80%99ve%20crushed%20matter,possible%20with%20neutrons." rel="nofollow noreferrer">phase in which the neutrons transforms into bosons that can occupy the same space</a>, but that seems skeptical.</p>
<p>However, a certain image is clear. After a critical mass, the neutron degeneracy pressure cannot resist the force of gravity anymore, and the core collapses further. After it shrinks below the Schwarzschild radius, it's a goner. The escape velocity exceeds the speed of light, resulting in the formation of a event horizon, while the collapsing matter gets crushed into a infinitesimally small singularity with 0 volume. You have a black hole.</p>
<p>However, I think the premise of the question is hugely flawed:</p>
<blockquote>
<p>The energy needed to break the structure of an atom (fission) may not be the same as the energy needed to hold all its released pieces within the same relative vicinity. That is, if the gravity well can only break atoms, but not hold the collapsed atoms' material inside itself, it will lose mass with each broken atom, and never be able to cause a singularity (and thus a black hole).</p>
</blockquote>
<p>That isn't fission. You seem to be conflating nuclear fission with <a href="https://en.wikipedia.org/wiki/Photodisintegration" rel="nofollow noreferrer">photodisintegration</a>, a similar but an entirely different process that occurs in a collapsing star, where a heavy nucleus absorbs a high energy gamma ray and disintegrates.</p>
<p>Again mass is not lost in this reaction. Even if nuclear fission occurred, the mass would not have been lost, but converted to energy.</p>
<p>That's a W for the black hole. Black holes don't care about what enters their event horizons as long as it is a form of the <a href="https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence" rel="nofollow noreferrer">mass-energy equivalence</a>. A tank full of oxygen? A pair of socks? A lot of laserlight? a hydrogen bomb? Na na na... The black hole will simply get more massive. Since mass and energy are the same (W Einstein), the black hole will regard them as positive values and increase in mass.</p>
<p>That's exactly another reason why antimatter cannot annihilate a black hole. Antimatter and matter both have positive mass, and thus are positive values to a black hole. So antimatter, if dumped into a black hole, will simply make it more massive.</p>
<p><sub> Actually, certain fusion processes beyond Fe-56 are still exothermic, for e.g. <a href="https://astronomy.stackexchange.com/questions/7663/why-does-iron-consume-more-energy-in-the-fusion-process-than-it-produces#:%7E:text=Adding%20alpha%20particles%20to%2056Ni%20(and%20indeed%20to%20even%20heavier%20nuclei)%20is%20in%20fact%20still%20exothermic%20(in%20isolation).%20Claims%20that%20nuclear%20fusion%20in%20elements%20heavier%20than%20iron%20are%20%22endothermic%22%20are%20simply%20incorrect.">the fusion of Fe-56 and an alpha particle to form Zinc-60 releases more energy than it consumes (2.7 MeV, if I recollect correctly)</a>. However, it is important to know that there are very few, if any, alpha particles remaining in the stellar core at this stage, and so, the core remains relatively &quot;stagnant&quot;, collecting more and more iron till it collapses. </sub></p>
<blockquote>
<p>Fluctuations of the above 2 factors due to the tumultuous nature of hot things in general (think boiling kettle). There wouldn't be a nice uniform gravitational pressure environment to allow everything to collapse to singularity all at once.</p>
</blockquote>
<p>A star's core is not similar to a steaming kettle. Gravity is gravity, whether it's hot or cold. Even if it's BILLIONS of Kelvins hot, gravity doesn't care about it. No matter how chaotic and tumultous it is down there, gravity will crush it altogether like eggs against a rock.</p>
<p>Thus, black holes exist because of the aforementioned reasons.</p>

CC BY-SA 4.0
2023-11-13T13:47:11.053Z
The inner core isn't exactly stagnant. There's plenty of photodisintegration going on to supply the alpha particles required for fusion. BTW, photodisintegration *is* a form of fission, and (of course) fission of small nuclei is endothermic. Also, the whole [silicon-burning sequence](https://en.wikipedia.org/wiki/Silicon-burning_process) is quite brief: it lasts about a day before being impacted by the core collapse shockwave. ProfRob has more good info on this topic at https://astronomy.stackexchange.com/a/36725/16685
PM 2Ring
CC BY-SA 4.0
2023-11-13T12:55:22.030Z
2023-11-14T23:24:21.833Z
2023-11-14T23:24:21.833Z
Arcturus
<p>A black hole happens because enough material mass exists in an area to create a gravitational field strong enough that nothing escapes (including light). This gravity also exceeds the strength of atoms to hold their structure intact, which then collapse and release their subatomic composition to be further consolidated into a smaller point still, which happens to be infinitesimal (AKA a singularity). This is the most straightforward way I know of to explain a black hole.</p>
<p>Based on what I observe, I don't think this can happen. The energy needed to break the structure of an atom (fission) may not be the same as the energy needed to hold all its released pieces within the same relative vicinity. That is, if the gravity well can only break atoms, but not hold the collapsed atoms' material inside itself, it will lose mass with each broken atom, and never be able to cause a singularity (and thus a black hole).</p>
<p>Also, such singularity-denying atomic breakage (fission) may be made to happen from several sources of energy being released within same general vicinity:</p>
<ol>
<li>The intense friction of atoms pressing against each other under high temperature and gravitational pull.</li>
<li>The release of energy from adjacent natural fission and fusion of atoms.</li>
<li>Fluctuations of the above 2 factors due to the tumultuous nature of hot things in general (think boiling kettle). There wouldn't be a nice uniform gravitational pressure environment to allow everything to collapse to singularity all at once.</li>
</ol>
<p>To my mind, without bringing complex quantum or astro-physics concepts to bear, this seems too simple and sensical an explanation to ignore. Would I be misunderstanding anything important to have arrived at this conclusion?</p>

CC BY-SA 4.0
2023-11-13T10:20:22.367Z
Have you done the calculations to see how much energy you get from these processes?   It seems you are assuming that the energy liberated from the gravitational collapse is too small without showing that is the case.
BowlOfRed
CC BY-SA 4.0
2023-11-13T12:47:18.407Z
You've also seem to have assumed that energy is somehow different to mass? A black hole happens because enough material mass *and energy* exists ...
ProfRob
CC BY-SA 4.0
2023-11-13T14:03:41.173Z
I have been made aware of the aforementioned 'missing important information' by way of a [helpful video](https://www.youtube.com/watch?v=8jVYe_g6y2E) I was referred to earlier. As I now understand, it is not the strength of atomic structures (strong force?) that keeps the singularity's formation at bay, but rather the formation of a 'proto neutron star core' during the star's final moments that essentially seeds the singularity (if it attains too much mass). As I understand, the only significant relevance to nuclear fission in all of this is what happens when a star 'runs out of fuel', so to s
hamstar
<black-hole><gravity><singularity>
Can black holes even exist [if mass cannot be retained near the collapse threshold]?
hamstar
# Can black holes even exist [if mass cannot be retained near the collapse threshold]?

(Asked by: username_0 on 2023-11-13)

A black hole happens because enough material mass exists in an area to create a gravitational field strong enough that nothing escapes (including light). This gravity also exceeds the strength of atoms to hold their structure intact, which then collapse and release their subatomic composition to be further consolidated into a smaller point still, which happens to be infinitesimal (AKA a singularity). This is the most straightforward way I know of to explain a black hole.

Based on what I observe, I don't think this can happen. The energy needed to break the structure of an atom (fission) may not be the same as the energy needed to hold all its released pieces within the same relative vicinity. That is, if the gravity well can only break atoms, but not hold the collapsed atoms' material inside itself, it will lose mass with each broken atom, and never be able to cause a singularity (and thus a black hole).

Also, such singularity-denying atomic breakage (fission) may be made to happen from several sources of energy being released within same general vicinity:

1. The intense friction of atoms pressing against each other under high temperature and gravitational pull.
2. The release of energy from adjacent natural fission and fusion of atoms.
3. Fluctuations of the above 2 factors due to the tumultuous nature of hot things in general (think boiling kettle). There wouldn't be a nice uniform gravitational pressure environment to allow everything to collapse to singularity all at once.

To my mind, without bringing complex quantum or astro-physics concepts to bear, this seems too simple and sensical an explanation to ignore. Would I be misunderstanding anything important to have arrived at this conclusion?

**Tags:** black-hole, gravity, singularity

> **Comments:**
> **username_1**: Have you done the calculations to see how much energy you get from these processes?   It seems you are assuming that the energy liberated from the gravitational collapse is too small without showing that is the case.  
> **username_2**: You've also seem to have assumed that energy is somehow different to mass? A black hole happens because enough material mass *and energy* exists ...  
> **username_0**: I have been made aware of the aforementioned 'missing important information' by way of a [helpful video](https://www.youtube.com/watch?v=8jVYe_g6y2E) I was referred to earlier. As I now understand, it is not the strength of atomic structures (strong force?) that keeps the singularity's formation at bay, but rather the formation of a 'proto neutron star core' during the star's final moments that essentially seeds the singularity (if it attains too much mass). As I understand, the only significant relevance to nuclear fission in all of this is what happens when a star 'runs out of fuel', so to s  

---

## 3 answers

### Answer by username_0 on 2023-11-14. Score: 0 | Accepted answer

I have been made aware of the aforementioned 'missing important information' by way of a [helpful video](https://www.youtube.com/watch?v=8jVYe_g6y2E) I was referred to earlier. As I now understand, it is not the strength of atomic structures (strong force?) that keeps the singularity's formation at bay, but rather the formation of a 'proto neutron star core' during the star's final moments that essentially seeds the singularity (if it attains too much mass).

As I understand, the only significant relevance to nuclear fission in all of this is what happens when a star 'runs out of fuel', so to speak. In simple terms, the processed remains of this 'fuel' (iron atoms in the star's core) break apart due to temperature and pressure, and the star begins to collapse. During this process, a neutron star core begins to form, and it is this structure that can exceed critical mass and form a singularity.

If this sounds incorrect, add a comment to clarify.


---

### Answer by username_3 on 2023-11-13. Score: 6

A different way of looking at the issue: what properties would matter need to have to avoid imploding into a black hole if enough matter was gathered in a static sphere?

The key thing about black holes is that they come from curved spacetime. Thinking about them in terms of force often produces confusion. The equations of general relativity predict that outside a spherical mass spacetime has the [Schwarzschild metric](https://en.wikipedia.org/wiki/Schwarzschild_metric), which has an event horizon if the radius is smaller than $R\_s = 2GM/c^2$. That horizon has the property that particles with mass (and light) can only cross it inward, and this is enforced (sorry) by what directions they can move, not really any explicit force.

However, an object with radius greater than $R\_s$ and constant density interior has a spacetime metric inside that is the [interior Schwarzschild metric](https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric). The interesting aspect here is that the central pressure is $$p = \rho c^2 \frac{1-X}{3X-1}$$ where $X = \sqrt{1-\frac{R\_s}{R}}$ where $R$ is the radius of the body. This goes off to infinity when $X=1/3$, or $R=(9/8)R\_s$. This is known as the Buchdahl limit. Note that it ignores what the object is made from: unless you believe that there are materials that can withstand *infinite* pressure the sphere will start to compact, and then it will soon become a black hole.

How compact something can be without imploding has been studied a fair bit. [It turns out that for matter that has a speed of sound below the speed of light](https://arxiv.org/abs/2202.00043) (a *very* reasonable assumption!) must have $R>1.29 R\_s$ (about 14% larger than the Buchdahl limit).

These bounds don't care what the matter is made of, but as long as it doesn't have superluminal sound speeds and obeys the laws of general relativity it will implode if packed too densely. In this domain even having enormous resistance to pressure doesn't help since pressure itself is a component in the mass-energy tensor and contributes to gravity: you can literally implode due to the weight of pressure energy! And this implosion is less about matter moving as space itself giving way to form a black hole.


---

### Answer by username_4 on 2023-11-13. Score: 4

# Gravity doesn't care at such extreme conditions.

Yeah, I am definitely not joking.

Friction, nuclear fission and whatever forces are there in the core of a massive dying star have no chance against the intense gravity of a collapsing stellar core. We are talking about the pressure of 20 **solar masses** crushing down on you, and friction stands no chance.

* First, as soon as the star's core starts fusing iron, it basically starts losing energy. Not that iron cannot fuse or anything, but the fact that fusing iron-56 is endothermic, i.e. it consumes more energy than it releases1.
* After the iron accumulates at the core, the radiation pressure of the stellar core decreases, as the endothermic reactions keep draining away more and more energy, until finally, the immense gravitational pressure of the electron-degenerate iron core causes it to collapse dramatically.

If you are lucky, at some point, neutron-degeneracy pressure fights back and stops the collapse. The abrupt halting of the collapse triggers a [rapid deceleration](https://youtu.be/XzaYNOKK6Xk?si=npRBHNOh6m1b8yJ7&t=122) on the surrounding gas falling onto it. This causes the star to explode in a Type-II supernova, and congrats, you have a neutron star.

The neutrons form because the iron nuclei are crushed so tightly that the protons and electrons ram into each other and form neutrons, emitting a neutrino. Here's the formula:
$$p + e \rightarrow n + \nu\_e$$

However, get above 20 solar masses, and the core isn't so lucky.

We still do not have a clear picture of what exactly happens in the transition from ~~neutronium~~ neutron degenerate matter to a gravitational singularity. This source posits that there is a [phase in which the neutrons transforms into bosons that can occupy the same space](https://www.quora.com/What-happens-when-water-is-compressed-to-the-extreme#:%7E:text=We%E2%80%99ve%20crushed%20matter,possible%20with%20neutrons.), but that seems skeptical.

However, a certain image is clear. After a critical mass, the neutron degeneracy pressure cannot resist the force of gravity anymore, and the core collapses further. After it shrinks below the Schwarzschild radius, it's a goner. The escape velocity exceeds the speed of light, resulting in the formation of a event horizon, while the collapsing matter gets crushed into a infinitesimally small singularity with 0 volume. You have a black hole.

However, I think the premise of the question is hugely flawed:

> The energy needed to break the structure of an atom (fission) may not be the same as the energy needed to hold all its released pieces within the same relative vicinity. That is, if the gravity well can only break atoms, but not hold the collapsed atoms' material inside itself, it will lose mass with each broken atom, and never be able to cause a singularity (and thus a black hole).

That isn't fission. You seem to be conflating nuclear fission with [photodisintegration](https://en.wikipedia.org/wiki/Photodisintegration), a similar but an entirely different process that occurs in a collapsing star, where a heavy nucleus absorbs a high energy gamma ray and disintegrates.

Again mass is not lost in this reaction. Even if nuclear fission occurred, the mass would not have been lost, but converted to energy.

That's a W for the black hole. Black holes don't care about what enters their event horizons as long as it is a form of the [mass-energy equivalence](https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence). A tank full of oxygen? A pair of socks? A lot of laserlight? a hydrogen bomb? Na na na... The black hole will simply get more massive. Since mass and energy are the same (W Einstein), the black hole will regard them as positive values and increase in mass.

That's exactly another reason why antimatter cannot annihilate a black hole. Antimatter and matter both have positive mass, and thus are positive values to a black hole. So antimatter, if dumped into a black hole, will simply make it more massive.

Actually, certain fusion processes beyond Fe-56 are still exothermic, for e.g. [the fusion of Fe-56 and an alpha particle to form Zinc-60 releases more energy than it consumes (2.7 MeV, if I recollect correctly)](https://astronomy.stackexchange.com/questions/7663/why-does-iron-consume-more-energy-in-the-fusion-process-than-it-produces#:%7E:text=Adding%20alpha%20particles%20to%2056Ni%20(and%20indeed%20to%20even%20heavier%20nuclei)%20is%20in%20fact%20still%20exothermic%20(in%20isolation).%20Claims%20that%20nuclear%20fusion%20in%20elements%20heavier%20than%20iron%20are%20%22endothermic%22%20are%20simply%20incorrect.). However, it is important to know that there are very few, if any, alpha particles remaining in the stellar core at this stage, and so, the core remains relatively "stagnant", collecting more and more iron till it collapses.

> Fluctuations of the above 2 factors due to the tumultuous nature of hot things in general (think boiling kettle). There wouldn't be a nice uniform gravitational pressure environment to allow everything to collapse to singularity all at once.

A star's core is not similar to a steaming kettle. Gravity is gravity, whether it's hot or cold. Even if it's BILLIONS of Kelvins hot, gravity doesn't care about it. No matter how chaotic and tumultous it is down there, gravity will crush it altogether like eggs against a rock.

Thus, black holes exist because of the aforementioned reasons.

> **Comments:**
> **username_5**: The inner core isn't exactly stagnant. There's plenty of photodisintegration going on to supply the alpha particles required for fusion. BTW, photodisintegration *is* a form of fission, and (of course) fission of small nuclei is endothermic. Also, the whole [silicon-burning sequence](https://en.wikipedia.org/wiki/Silicon-burning_process) is quite brief: it lasts about a day before being impacted by the core collapse shockwave. username_2 has more good info on this topic at https://astronomy.stackexchange.com/a/36725/16685  

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