<p>In general relativity the thickness is zero. In quantum gravity the thickness has a Planck unit thickness which is very thin.</p>
<p>On this sphere, or better, just outside it, Hawing radiation emerges. To avoid the information paradox some theorists have proposed a firewall to exist on the horizon.</p>
<p>Seen from us it is like all matter ends up frozen on the horizon, in a zero volume shell. In a falling frame all matter falls through. There is no horizon at all in this frame.</p>
<p>The Schwarzschild surface is a purely mathematical surface. On this surface light will not fall into the hole nor escape. For us it looks frozen on this sphere. For a falling observer the light on the surface (if tangentially moving) will circle the hole forever (in the classical picture). Untill the horizon gets smaller by vaporizing (Hawking radiation). The photons will be set free in that case.</p>
<p>So. All light that moves close to the horizon can circle it for a small distance and move away again. It can  go 180 degrees round the hole. Reflection! Once photons are only 0.00000000000000....1 meter over the horizon they are lost forever only to reappear again as Hawing photons, which are entangled with all particles that fall in (well, if they are entangled is still the question and this is one of the reasons a firewall is thought to exist).</p>

CC BY-SA 4.0
2021-07-15T12:16:06.003Z
@DaddyKropotkin That is true. The photons trace out a different path if the approach the hole at the top and sideways.
Deschele Schilder
CC BY-SA 4.0
2021-07-19T13:07:48.790Z
"In general relativity the thickness is zero. In quantum gravity the thickness has a Planck unit thickness which is very thin." The thickness of what? What are you referring to here? The event horizon? The accretion disk? It would likely help the OP if you were extra clear here, since, as the comments in the OP above show, they are kind of confused about what the event horizon is.
Daddy Kropotkin
CC BY-SA 4.0
2021-07-14T19:54:52.033Z
2021-07-14T20:11:20.400Z
2021-07-14T20:11:20.400Z
Deschele Schilder
<p>I would premit that I am just an enthusiast, so don't expect a good background or a deep mathematical knowledge about the topic.
I just read an article about how light is reflected by a black hole (<a href="https://www.sciencealert.com/we-now-have-precise-maths-to-describe-how-black-holes-reflect-the-universe/amp" rel="nofollow noreferrer">https://www.sciencealert.com/we-now-have-precise-maths-to-describe-how-black-holes-reflect-the-universe/amp</a>). That let me think because, as far as I can understand that imply that the geometrical place where gravitational pull is &quot;just strong enough&quot; to &quot;capture&quot; light has to be a sphere (or geoid if the black hole rotate). What is hard to grasp for me is how perfect that sphere should be in reality (mathematical place doesn't not exist in real world).In other words I am asking if that topic was studied and which is the trending hipotesys about real thickness and smoothness of event horizon (intended as the region of space in which gravitational pull is just strong enough to capture the light, but not too strong).</p>

CC BY-SA 4.0
2021-07-14T19:53:51.363Z
"...event horizon (intended as the region of space in which gravitational pull is just strong enough to capture the light, but not too strong)." That's an unhelpful definition of event horizon, since the tidal forces felt by a particle at a black hole's event horizon depends on the mass of the black hole. Black holes are not the only objects whose gravity can bend the path that light travels on, this is generally known as gravitational lensing and is done by galaxies too, where the shape of the lens (the galaxy) effects the types of images that are produced.
Daddy Kropotkin
CC BY-SA 4.0
2021-07-15T05:45:58.863Z
Sorry English is not my mother tongue so I may be misleading.What I mean is understand if there are theory about the thickness and smoothness of the black hole region in which photon start "orbiting" without falling (in a given moment, so assuming the gravitational pull does not change).So I would better understand the current theory about the "border" of a black hole and not it's interior.Is the border a infinitesimal thin film or has a thickness? Is smooth or has irregularities?
Skary
CC BY-SA 4.0
2021-07-15T12:15:15.850Z
That's okay! Thanks for clarifying. By "border" are you referring to the event horizon of the black hole, or are you referring to the inner most stable circular orbit (ISCO) where the accretion disk begins? These two things are not the same. It seems like you're asking about the event horizon, but then you ask about the thickness of an accretion disk which involves the ISCO. Which is it?
Daddy Kropotkin
CC BY-SA 4.0
2021-07-19T12:55:12.487Z
Sorry for the delay, I don't even know these are two separated region. From what I've understand if you are inside a black hole you start the fall, no stable orbit here. So I am referring to the event horizon "outer" face (that I suppose it's a 2D surface, but I am no sure, for that reason I have asked). I definitely not refer to accretion disk outside, where gravitational pull should not be so strong to make light orbiting (afaik)
Skary
CC BY-SA 4.0
2021-07-19T13:02:17.110Z
All good. What does "Inside a black hole" mean??? IT seems you might want to start reading a textbook before asking questions on here. For a non-rotating (Schwarzschild) black hole, the inner most stable circular orbit (ISCO) is at $r = 6m$ (assuming G=C=1 and using Schwarzschild coordinates) which is farther away from the center than the event horizon, which is at $r = 3m.$ So, as any textbook shows you, the ISCO is at a wider orbit than the event horizon. Light can orbit the black hole on at the ISCO as well, so the dichotomy you're making is not true.
Daddy Kropotkin
<black-hole><event-horizon><space-geometry>
What is the thickness of black holes event horizon?
Skary
# What is the thickness of black holes event horizon?

(Asked by: username_0 on 2021-07-14)

I would premit that I am just an enthusiast, so don't expect a good background or a deep mathematical knowledge about the topic.
I just read an article about how light is reflected by a black hole (<https://www.sciencealert.com/we-now-have-precise-maths-to-describe-how-black-holes-reflect-the-universe/amp>). That let me think because, as far as I can understand that imply that the geometrical place where gravitational pull is "just strong enough" to "capture" light has to be a sphere (or geoid if the black hole rotate). What is hard to grasp for me is how perfect that sphere should be in reality (mathematical place doesn't not exist in real world).In other words I am asking if that topic was studied and which is the trending hipotesys about real thickness and smoothness of event horizon (intended as the region of space in which gravitational pull is just strong enough to capture the light, but not too strong).

**Tags:** black-hole, event-horizon, space-geometry
> **Comments:**
> **username_1**: "...event horizon (intended as the region of space in which gravitational pull is just strong enough to capture the light, but not too strong)." That's an unhelpful definition of event horizon, since the tidal forces felt by a particle at a black hole's event horizon depends on the mass of the black hole. Black holes are not the only objects whose gravity can bend the path that light travels on, this is generally known as gravitational lensing and is done by galaxies too, where the shape of the lens (the galaxy) effects the types of images that are produced.  
> **username_0**: Sorry English is not my mother tongue so I may be misleading.What I mean is understand if there are theory about the thickness and smoothness of the black hole region in which photon start "orbiting" without falling (in a given moment, so assuming the gravitational pull does not change).So I would better understand the current theory about the "border" of a black hole and not it's interior.Is the border a infinitesimal thin film or has a thickness? Is smooth or has irregularities?  
> **username_1**: That's okay! Thanks for clarifying. By "border" are you referring to the event horizon of the black hole, or are you referring to the inner most stable circular orbit (ISCO) where the accretion disk begins? These two things are not the same. It seems like you're asking about the event horizon, but then you ask about the thickness of an accretion disk which involves the ISCO. Which is it?  
> **username_0**: Sorry for the delay, I don't even know these are two separated region. From what I've understand if you are inside a black hole you start the fall, no stable orbit here. So I am referring to the event horizon "outer" face (that I suppose it's a 2D surface, but I am no sure, for that reason I have asked). I definitely not refer to accretion disk outside, where gravitational pull should not be so strong to make light orbiting (afaik)  
> **username_1**: All good. What does "Inside a black hole" mean??? IT seems you might want to start reading a textbook before asking questions on here. For a non-rotating (Schwarzschild) black hole, the inner most stable circular orbit (ISCO) is at $r = 6m$ (assuming G=C=1 and using Schwarzschild coordinates) which is farther away from the center than the event horizon, which is at $r = 3m.$ So, as any textbook shows you, the ISCO is at a wider orbit than the event horizon. Light can orbit the black hole on at the ISCO as well, so the dichotomy you're making is not true.  

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## 1 answers

### Answer by username_2 on 2021-07-14. Score: 3

In general relativity the thickness is zero. In quantum gravity the thickness has a Planck unit thickness which is very thin.

On this sphere, or better, just outside it, Hawing radiation emerges. To avoid the information paradox some theorists have proposed a firewall to exist on the horizon.

Seen from us it is like all matter ends up frozen on the horizon, in a zero volume shell. In a falling frame all matter falls through. There is no horizon at all in this frame.

The Schwarzschild surface is a purely mathematical surface. On this surface light will not fall into the hole nor escape. For us it looks frozen on this sphere. For a falling observer the light on the surface (if tangentially moving) will circle the hole forever (in the classical picture). Untill the horizon gets smaller by vaporizing (Hawking radiation). The photons will be set free in that case.

So. All light that moves close to the horizon can circle it for a small distance and move away again. It can go 180 degrees round the hole. Reflection! Once photons are only 0.00000000000000....1 meter over the horizon they are lost forever only to reappear again as Hawing photons, which are entangled with all particles that fall in (well, if they are entangled is still the question and this is one of the reasons a firewall is thought to exist).

> **Comments:**
> **username_2**: @DaddyKropotkin That is true. The photons trace out a different path if the approach the hole at the top and sideways.  
> **username_1**: "In general relativity the thickness is zero. In quantum gravity the thickness has a Planck unit thickness which is very thin." The thickness of what? What are you referring to here? The event horizon? The accretion disk? It would likely help the OP if you were extra clear here, since, as the comments in the OP above show, they are kind of confused about what the event horizon is.  

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