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I have read this:

Because the spacetime curvature at the horizon is so great that there is no light-like world line the extends beyond the horizon.

Why does time stop in black holes?

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If the curvature is extreme, that would that mean that laser beams (for the sake of argument let's assume they are visible because of some scattering like from the atmosphere) appear to be bent?

Question:

  1. Is the curvature so extreme at the event horizon, that you could see curved laser beams?
Árpád Szendrei
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  • small vs large scale curvature What does that mean? The curvature is the curvature, and it can be arbitrarily small at the horizon. A supermassive black hole, for example, does not have extreme curvature at its horizon. – Ghoster Mar 28 '24 at 23:28
  • @Ghoster I mean when they claim that spacetime is "locally flat", although there is extreme curvature at the black hole. But I will edit. – Árpád Szendrei Mar 28 '24 at 23:31
  • Near the singularity, yes. Near the event horizon, not necessarily. As I said, the curvature at the horizon can be arbitrarily small for large holes. Try estimating it using dimensional analysis. – Ghoster Mar 28 '24 at 23:38
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    they claim… It’s not a “claim”. Manifolds are locally flat by definition (except at singularities, but those aren’t part of the manifold). However… local flatness is something of a misnomer: it only means that coordinates can be chosen to make the Christoffel symbols vanish at any point; it doesn’t imply anything about the curvature, which in general can’t be made to vanish. It doesn’t make sense to talk about small-scale and large-scale curvature for black holes; this notion was based on a misunderstanding of local flatness. – Ghoster Mar 29 '24 at 00:42
  • This might help - How to Understand What Black Holes Look Like – mmesser314 Mar 29 '24 at 02:25
  • @Ghoster thank you I edited. – Árpád Szendrei Mar 29 '24 at 03:15
  • I have BH photon trajectory diagrams here: https://physics.stackexchange.com/a/805213/123208 But bear in mind that those diagrams use Schwarzschild coordinates, so they aren't exactly what a local observer would see. – PM 2Ring Mar 29 '24 at 03:22
  • The curvature at the horizon is finite, see the square root of the Kretschmann scalar, it only gets infinite at the singularity. – Yukterez Mar 29 '24 at 16:02
  • If you look along the beam of the laser you are holding, it would look “straight” to you (despite actually being curved), because your “line of sight” would be equally curved. However, if you look across the (non-radial) beam of the laser I am holding, then absolutely, you would see it bent toward the black hole +1 – safesphere Mar 30 '24 at 03:07
  • Note that “curvature” is not uniquely defined. Everyone here repeats the textbook mantra that “the spacetime is locally flat near the horizon”. This refers only to the curvature scalars. However the metric components diverge (become infinite) at the horizon and this exactly what defines the curved shape of the laser beam. So in this sense the metric curvature at the horizon is infinite. If you draw a straight line (“laser beam”) on a flat sheet of rubber and stretch this sheet unevenly, then the line (“laser beam”) becomes curved while the sheet remains flat (but not the same as it was). – safesphere Mar 30 '24 at 03:33

2 Answers2

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Yes, the curvature is enough to bend lasers. In fact, there is an interesting feature in Schwarzschild spacetime that might answer your question well. If a non-rotating, uncharged black hole (i.e., a Schwarzschild black) has mass $M$, then the event horizon is at $R = 2M$. At $r = 3M$ one has what is called the photon ring. At $r = 3M$, light rays can follow (unstable) circular orbits, meaning it is possible for a light ray to stay indefinitely orbiting the black hole at constant $r = 3M$. This orbit is unstable, so if it is slightly perturbed the light ray will either escape to infinity or fall into the black hole, but in principle you can arrange light to orbit the black hole indefinitely.

I should add that this does not necessarily means the curvature is large. For very big black holes, the curvature at the horizon can be as small as the curvature in your living room (or smaller).

Níckolas Alves
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  • Thank you so much! Does it make even sense to distinct the small vs large scale curvature at black holes (some claim that spacetime is locally "flat")? – Árpád Szendrei Mar 28 '24 at 23:35
  • @ÁrpádSzendrei Spacetime is locally flat. That is a basic assumption on general relativity and it can be understood as "locally, spacetime can be linearly approximated by a flat spacetime". Curvature is precisely a measure of how this linear approximation fails (you can think of it as a measure of the quadratic approximation) – Níckolas Alves Mar 29 '24 at 12:50
  • I wouldn't say it makes sense to speak of the small scale or large scale curvature. Curvature is too a local property (it has a value at each point in spacetime) – Níckolas Alves Mar 29 '24 at 12:50
  • Thank you. One thing I do not comprehend is you say the curvature is small (for large black holes, curvature is like in the living room), but light still cannot escape (it will orbit). So the fact that light cannot escape is not connected to the extreme curvature itself? Then what is it connected to and how can you have small curvature but still orbiting light? Naively said, my living room has no bent light rays (and none orbiting Earth), so what is it exactly that is different in my living room and the event horizon that makes light orbit (I assume now it is not the extreme curvature)? – Árpád Szendrei Mar 29 '24 at 16:36
  • If you want, I can ask this as a separate question. – Árpád Szendrei Mar 29 '24 at 16:36
  • @ÁrpádSzendrei I might have phrased it badly. Since curvature is a tensor, and not a scalar, it is a bit difficult to quantify the value of the curvature. Certain components of the curvature are really small at the horizon (for example, you have very small tidal forces for very big black holes), but you still have enough curvature in some other senses. For example, you need infinite acceleration to hover just above the event horizon at a constant radius. It was perhaps a poor choice of words (although I've read similar statements elsewhere) – Níckolas Alves Mar 29 '24 at 17:02
  • I have my new question here : https://physics.stackexchange.com/questions/808279/are-light-rays-always-bent-in-any-gravitational-field-or-just-in-a-strong-one-l – Árpád Szendrei Mar 30 '24 at 02:33
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Yes, this is true. Whilst laser light differs from ordinary light in being coherent, it is still light and so the effect of black hole still holds.

Its worth noting that close to black hole we will have light orbiting the black hole. But just away from this orbit the light will be able to escape. This means when you look at a closeby black hole with an accretion disk, then you will see a ring of light. This is an optical illusion, no matter where you stand you will still see this ring. It is this optical illusion that was famously photographed recently.

Mozibur Ullah
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