<p>The idea of a hard surface on a black hole is very much a minority theory, bordering on a fringe theory. The reason is that general relativity is fairly well tested, and does not just predict event horizons around black holes but also the <a href="https://en.wikipedia.org/wiki/Buchdahl%27s_theorem" rel="nofollow noreferrer">Buchdahl bound</a> on how small objects in hydrostatic equilibrium can be. Still, it is always worth testing the theory, <a href="https://arxiv.org/abs/1703.00023" rel="nofollow noreferrer">which is what the paper proposes</a>.</p>
<p>If you drop an object from infinity onto an object of radius <span class="math-container">$R$</span> the Newtonian kinetic energy at that moment would be <span class="math-container">$GM/R$</span> J/kg, much of which would be released as a flash. This is why accretion onto white dwarfs and neutron stars produce a lot of radiation beside the accretion disk: for white dwarfs about 0.0003 of the mass-energy would be released, while for a neutron star 0.147 of the mass energy would be released. For a black hole-radius object this crude estimate produces 50% efficiency of converting mass into energy.</p>
<p>However, the energy of that flash is reduced by the <a href="https://en.wikipedia.org/wiki/Gravitational_redshift" rel="nofollow noreferrer">gravitational redshift</a>. At the Buchdahl limit the redshift is z=2, halving the energy received (for neutron stars you only get 74% of the energy). But for a black hole horizon general relativity  gives infinite redshift, making the flash disappear altogether.</p>
<p>It does not make sense to see black holes as solid objects since the structure of spacetime inside the event horizon makes the inward direction behave like time does outside: you can only move inward, just as you can only move forward in time. Thinking of the singularity as a material object with mass is very misleading: <a href="https://arxiv.org/abs/1907.05292" rel="nofollow noreferrer">it is not even point-like</a>. Hence talk about the density of a black hole or imagining it as a sphere of something dense tends to end in confusion. Similarly talking about escape velocity is also tricky: it does not define the horizon as people usually say. The horizon is defined by being a trapped surface that can only be crossed in one direction.</p>

CC BY-SA 4.0
2021-03-04T13:40:22.173Z
2021-03-04T13:59:12.353Z
2021-03-04T13:59:12.353Z
Anders Sandberg
<p>This seems to be a fringe idea that appears inconsistent with General Relativity. For a spherically symmetric mass distribution it is easy to show, from the Schwarzschild metric, that it is impossible for anything (with or without mass) to avoid moving towards smaller values of <span class="math-container">$r$</span> if <span class="math-container">$r$</span> is less than the Schwarzschild radius.</p>
<p>Whilst GR breaks down as a theory as <span class="math-container">$r \rightarrow 0$</span>, it is safe to say that anything &quot;odd&quot; that happens will be well inside the Schwarzschild radius and so evidence of it in the form of a burst of light can never reach an outside observer.</p>
<p>A solid surface cannot exist outside, but close-to (within about 1.2 times), the Schwarzschild radius either. This is because of the self-defeating nature of pressure in General Relativity, which contributes to the curvature of spacetime. Any self-supporting structure close to the event horizon cannot be in equilibrium because of the increase in spacetime curvature that its internal pressure would cause.</p>

CC BY-SA 4.0
2021-03-09T10:07:07.440Z
2021-03-09T10:07:07.440Z
ProfRob
<p>To answer your questions: citing the same article you refer to:</p>
<blockquote>
<p>But if event horizons do exist, there wouldn't be flash of light. Instead, matter would just completely vanish when pulled in.</p>
</blockquote>
<p>So, the existence of an event horizon (where even light can not escape the gravitational pull) would avoid seeing any burst of light on this theoretical collision with a surface.</p>
<blockquote>
<p>Black holes are incredibly dense wouldn't it make more sense from them to be solid?</p>
</blockquote>
<p>It would not. Take as an example neutron stars, which are surely more dense than anything in our solar system. The star is so dense, that matter becomes <em>neutron-degenerate</em>. This is a special kind of plasma state, a ultra compressed one where the fermions are the closest possible without breaking the Pauli Exclusion Principle. The rules that govern this state are  determined by quantum mechanics rather than classic mechanics. I am just giving you this example to show that, under extreme densities, &quot;normal&quot; rules (such as the ones that follow for solids) no longer apply and things become more complex.</p>

CC BY-SA 4.0
2021-03-04T15:24:46.943Z
The PEP does not forbid fermions getting arbitrarily close together. That isn't the issue. The issue is that inside the event horizon, nothing can stand still.
ProfRob
CC BY-SA 4.0
2021-03-04T15:42:09.677Z
I only stated that the fermions get as close as possible without sharing the same configuration inside a NS. What exactly happens inside the EH of the BH beats me, wanted only to show the "strange" behaviour of matter under extreme conditions.
Wilhelmroentgen
CC BY-SA 4.0
2021-03-04T14:01:38.933Z
2021-03-04T14:43:34.057Z
2021-03-04T14:43:34.057Z
Wilhelmroentgen
<p>I was reading <a href="https://www.space.com/37075-how-stars-fall-into-black-holes.html" rel="nofollow noreferrer">Do Stars Vanish Into a Black Hole or Crash Against a Surface? A New Test Answers</a> on space.com.</p>
<p>It says the following:</p>
<blockquote>
<p>Another proposed alternative to event horizons is the &quot;hard surface
theory,&quot; which suggests that matter within a black hole is destroyed
by smashing into a solid surface. Instead of a singularity with no
surface area, the black hole is a giant mass with a hard surface, and
material being pulled closer — such as a star — would not actually
fall into a black hole, but hit this hard surface and be destroyed. If
that were the case, the collision should create a large burst of
light.</p>
</blockquote>
<p>If we focus on the last sentence:</p>
<blockquote>
<p>If that were the case, the collision should create a large burst of
light.</p>
</blockquote>
<p>If the escape velocity of black hole becomes at certain altitude above the say solid surface of a blackhole more than that of speed of light, would we still see a large burst of light when the collision happens? Black holes are incredibly dense wouldn't it make more sense from them to be solid?</p>
<p>Something like a <a href="https://en.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics)" rel="nofollow noreferrer">dark star</a>.</p>
<p>No one really knows what happens to the matter at the singularity. Wouldn't it make sense that the reason that a black hole gets larger is due to more matter being smashed into the singularity making it gain mass. It also seems that black holes can be hurled out their galaxies, it must be in some form of matter that it stays together even when hurled out?</p>

CC BY-SA 4.0
2021-03-04T13:47:48.080Z
If the escape velocity is greater than that of light, then that would be an event horizon.
ProfRob
<black-hole>
Why are blackholes generally not considered as solid spheres with an escape velocity FTL?
O S
# Why are blackholes generally not considered as solid spheres with an escape velocity FTL?

(Asked by: username_0 on 2021-03-04)

I was reading [Do Stars Vanish Into a Black Hole or Crash Against a Surface? A New Test Answers](https://www.space.com/37075-how-stars-fall-into-black-holes.html) on space.com.

It says the following:

> Another proposed alternative to event horizons is the "hard surface
> theory," which suggests that matter within a black hole is destroyed
> by smashing into a solid surface. Instead of a singularity with no
> surface area, the black hole is a giant mass with a hard surface, and
> material being pulled closer — such as a star — would not actually
> fall into a black hole, but hit this hard surface and be destroyed. If
> that were the case, the collision should create a large burst of
> light.

If we focus on the last sentence:

> If that were the case, the collision should create a large burst of
> light.

If the escape velocity of black hole becomes at certain altitude above the say solid surface of a blackhole more than that of speed of light, would we still see a large burst of light when the collision happens? Black holes are incredibly dense wouldn't it make more sense from them to be solid?

Something like a [dark star](https://en.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics)).

No one really knows what happens to the matter at the singularity. Wouldn't it make sense that the reason that a black hole gets larger is due to more matter being smashed into the singularity making it gain mass. It also seems that black holes can be hurled out their galaxies, it must be in some form of matter that it stays together even when hurled out?

**Tags:** black-hole

> **Comments:**
> **username_1**: If the escape velocity is greater than that of light, then that would be an event horizon.  

---

## 3 answers

### Answer by username_2 on 2021-03-04. Score: 4

The idea of a hard surface on a black hole is very much a minority theory, bordering on a fringe theory. The reason is that general relativity is fairly well tested, and does not just predict event horizons around black holes but also the [Buchdahl bound](https://en.wikipedia.org/wiki/Buchdahl%27s_theorem) on how small objects in hydrostatic equilibrium can be. Still, it is always worth testing the theory, [which is what the paper proposes](https://arxiv.org/abs/1703.00023).

If you drop an object from infinity onto an object of radius $R$ the Newtonian kinetic energy at that moment would be $GM/R$ J/kg, much of which would be released as a flash. This is why accretion onto white dwarfs and neutron stars produce a lot of radiation beside the accretion disk: for white dwarfs about 0.0003 of the mass-energy would be released, while for a neutron star 0.147 of the mass energy would be released. For a black hole-radius object this crude estimate produces 50% efficiency of converting mass into energy.

However, the energy of that flash is reduced by the [gravitational redshift](https://en.wikipedia.org/wiki/Gravitational_redshift). At the Buchdahl limit the redshift is z=2, halving the energy received (for neutron stars you only get 74% of the energy). But for a black hole horizon general relativity gives infinite redshift, making the flash disappear altogether.

It does not make sense to see black holes as solid objects since the structure of spacetime inside the event horizon makes the inward direction behave like time does outside: you can only move inward, just as you can only move forward in time. Thinking of the singularity as a material object with mass is very misleading: [it is not even point-like](https://arxiv.org/abs/1907.05292). Hence talk about the density of a black hole or imagining it as a sphere of something dense tends to end in confusion. Similarly talking about escape velocity is also tricky: it does not define the horizon as people usually say. The horizon is defined by being a trapped surface that can only be crossed in one direction.


---

### Answer by username_1 on 2021-03-09. Score: 2

This seems to be a fringe idea that appears inconsistent with General Relativity. For a spherically symmetric mass distribution it is easy to show, from the Schwarzschild metric, that it is impossible for anything (with or without mass) to avoid moving towards smaller values of $r$ if $r$ is less than the Schwarzschild radius.

Whilst GR breaks down as a theory as $r \rightarrow 0$, it is safe to say that anything "odd" that happens will be well inside the Schwarzschild radius and so evidence of it in the form of a burst of light can never reach an outside observer.

A solid surface cannot exist outside, but close-to (within about 1.2 times), the Schwarzschild radius either. This is because of the self-defeating nature of pressure in General Relativity, which contributes to the curvature of spacetime. Any self-supporting structure close to the event horizon cannot be in equilibrium because of the increase in spacetime curvature that its internal pressure would cause.


---

### Answer by username_3 on 2021-03-04. Score: 0

To answer your questions: citing the same article you refer to:

> But if event horizons do exist, there wouldn't be flash of light. Instead, matter would just completely vanish when pulled in.

So, the existence of an event horizon (where even light can not escape the gravitational pull) would avoid seeing any burst of light on this theoretical collision with a surface.

> Black holes are incredibly dense wouldn't it make more sense from them to be solid?

It would not. Take as an example neutron stars, which are surely more dense than anything in our solar system. The star is so dense, that matter becomes *neutron-degenerate*. This is a special kind of plasma state, a ultra compressed one where the fermions are the closest possible without breaking the Pauli Exclusion Principle. The rules that govern this state are determined by quantum mechanics rather than classic mechanics. I am just giving you this example to show that, under extreme densities, "normal" rules (such as the ones that follow for solids) no longer apply and things become more complex.

> **Comments:**
> **username_1**: The PEP does not forbid fermions getting arbitrarily close together. That isn't the issue. The issue is that inside the event horizon, nothing can stand still.  
> **username_3**: I only stated that the fermions get as close as possible without sharing the same configuration inside a NS. What exactly happens inside the EH of the BH beats me, wanted only to show the "strange" behaviour of matter under extreme conditions.  

---

