An observer passing the event horizon of a black hole (Part 2)

[DennisN]

TL;DR Summary

Is this scenario regarding an observer passing a black hole event horizon reasonable or incorrect?

Hi all, this is a new scenario I got thinking about after having received great feedback and corrections from other PF'ers in this thread. Thanks again for the great help! This new scenario is similar to the previous one, but with a twist including a mirror. And as I said in that thread, I am not particularly well aquainted with GR.

First, the assumptions:

1. The black hole is massive enough that there are no significant tidal forces near nor at the event horizon.
2. There is no firewall nor quantum effects at the event horizon, i.e. only standard GR is assumed.
3. The container and its contents is in free fall and is doomed to pass the event horizon.

Scenario: Laser, Mirror and Observer.

An observer is sitting at the left side in a container which is heading left towards an event horizon.
The observer is holding a turned on laser pointer that is directed at a vertical mirror that is located at the right side of the container. The observer can thus observe the laser light since it is reflected by the mirror back towards the observer.

Picture 1. Laser, observer & mirror are outside the event horizon.

The observer observes the laser light reflected by the mirror.

Picture 2. Laser & observer are inside the event horizon, mirror is still outside the event horizon.

As the observer and laser passes the event horizon, I am pretty sure that no more (new) light from the laser can reach the mirror and get reflected back to the observer. Thus it seems reasonable that the observer sees no more light. Now, of course light travels at c, so there may be a very brief moment the observer sees the remaining light after the laser passes the event horizon, but after this it seems to me the observer will not see any more light from the laser.

3. (no picture) Laser, observer & mirror are inside the event horizon.

As the right side of the container and thus the vertical mirror finally passes the event horizon, it seems to me that the observer will still not see any laser light, since the light going towards the right can't overcome the spacetime curvature.

Is this scenario reasonable, and if not, where am I thinking wrong?
Cheers!

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(A personal note from me:

I'd like to add that I don't have an agenda to show that an observer notices passing an event horizon, I have been and still am perfectly fine with the opposite. The reason I'm posting the scenario is because (1) for some reason I've been thinking about event horizons lately and (2) I don't trust myself to think correctly about the scenario and (3) I'm always happy to get a better understanding of things I don't understand so well.

To add some spice to the scenario, the poor observer can be called "Muerte", sitting in a container called "Pasta" since it will eventually be turned into spaghetti. :) Figuratively, at least. And thus the dish that is served closer to the singularity is called "Pasta Muerte". Bon appetit!

(I actually have one more scenario I can't figure out with regards to the event horizon, but it's
completely different and hasn't anything to do with if an observer notices passing it. Therefore
I may post it in another thread later when I've understood this one.)

 

[DennisN]

3. (no picture) Laser, observer & mirror are inside the event horizon.

As the right side of the container and thus the vertical mirror finally passes the event horizon, it seems to me that the observer will still not see any laser light, since the light going towards the right can't overcome the spacetime curvature.

Hmm, when I read my post I just realized I may be wrong here, for the same reason I was wrong in the previous thread. The laser light may "freeze" due to spacetime curvature, but then the mirror will reach it... and probably get reflected back to the observer. My brain hurts, I don't know what really happens... gosh, black holes are hard to think about .

 

[Nugatory]

Hmm, when I read my post I just realized I may be wrong here, for the same reason I was wrong in the previous thread. The laser light may "freeze" due to spacetime curvature, but then the mirror will reach it... and probably get reflected back to the observer.

Yes

 

[DennisN]

Yes

And when I think about it the same ought to apply to picture 2. But looking at my picture 2 it seems the laser light will hit the the mirror at a different (lower) vertical position (y) than it would otherwise, and thus eventually reach the observer, but via another path than it would compared to picture 1.
Edit: No, I don't know, I'm confused now, haha. And a bit tired.

 

[DennisN]

No, I don't know, I'm confused now, haha. And a bit tired.

I rethought it, and it now seems to me that exactly the same thing will happen in picture 2. Which means the observer will not see any difference compared to picture 1. But I had to think very hard about that (if it's correct)!

 

[pervect]

Hmm, when I read my post I just realized I may be wrong here, for the same reason I was wrong in the previous thread. The laser light may "freeze" due to spacetime curvature, but then the mirror will reach it... and probably get reflected back to the observer. My brain hurts, I don't know what really happens... gosh, black holes are hard to think about .

You've basically got the idea, though I would not blame what you call the freezing of the light on the space-time curvature, but on the use of Schwarzschild coordinates.

Schwarzschild coordinates ARE known to be ill-behaved at the horizon. It's technincally called a coordinate singularity.

Your first assumption , "1`) The black hole is massive enough that there are no significant tidal forces near nor at the event horizon. ", basically assumes that there isn't any significant space-time curvature, as tidal forces are equivalent to one component of the mathematical entity that represents space-time curvature, the Riemann curvature tensor.

So, it's not the space-time curvature that's actually confusing you - it's the use of Schwarzschild coordinates. They are inherently confusing at the event horizon due to the coordinate singularity there.

 

[PeterDonis]

Is this scenario reasonable, and if not, where am I thinking wrong?

No. You are making the same mistakes as in the previous thread.

Also, your diagrams are not correct, because the event horizon is not a spatial location. It's an outgoing light ray. There is *no* way to draw a proper diagram of your scenario in which it is "standing still". You need to be thinking of it as moving from left to right, relative to your container, with the speed of light. If you do that, you will see for yourself the proper resolution of your scenarios. This was already pointed out in the previous thread.

 

[PeterDonis]

The laser light may "freeze" due to spacetime curvature

No, it doesn't. Spacetime curvature has nothing whatever to do with the resolution of the issues you are raising. They can be resolved entirely by viewing things from the standpoint of the local inertial frame in which the freely falling object is at rest, just as it crosses the horizon (although from the viewpoint of this frame, the process is better described as the horizon flying outward past the object at the speed of light).

 

[pervect]

No, it doesn't. Spacetime curvature has nothing whatever to do with the resolution of the issues you are raising. They can be resolved entirely by viewing things from the standpoint of the local inertial frame in which the freely falling object is at rest, just as it crosses the horizon (although from the viewpoint of this frame, the process is better described as the horizon flying outward past the object at the speed of light).

I believe that the OP is thinking that light freezes at the event horizon because outgoing light takes the path of a constant Schwarzschld coordinate of r = r_s, where r_s is the Schwarzschild radius, at the event horizon.

In those coordinate-dependent terms, the outgoing light reflected off the mirror just when the mirror reaches the event horizon, has a constant radial schwarazschild coordinate, which could be regarded as "freezing".

 

[DennisN]

You've basically got the idea, though I would not blame what you call the freezing of the light on the space-time curvature

No, I understand that. I was tired and wrote wrong. My thought about the "freezing" of light was based on a thought suggestion in the previous thread by @Dale (to view the event horizon as moving at c):

Another way to think about it is to consider the infalling observer’s local inertial frame. In that frame the event horizon is moving at c. Light emitted by the hand after it passes the horizon cannot catch up to the horizon, but after the horizon passes the observer’s eyes the light from the hand will reach it as normal.

...so I didn't base it on spacetime curvature.

I see there are more replies, which makes me very happy, but I will have to wait until tomorrow. My poor, tired brain simply can't handle black holes at the moment . I will definitely read through this thread thoroughly later! Now I'm heading for the event horizon of my bed, and hopefully I will manage to return from that one.

 

[PeterDonis]

I believe that the OP is thinking that light freezes at the event horizon because outgoing light takes the path of a constant Schwarzschld coordinate of r = r_s, where r_s is the Schwarzschild radius, at the event horizon.

And if the OP is thinking of it that way, he's wrong, and that error is what's causing the problem. See below.

In those coordinate-dependent terms, the outgoing light reflected off the mirror just when the mirror reaches the event horizon, has a constant radial schwarazschild coordinate, which could be regarded as "freezing"

But the Schwarzschild $r$ coordinate is not the same as "space" in the diagrams the OP drew: it can't be, because at the horizon it is null, and inside the horizon it is timelike, not spacelike. So what you are describing is not the solution of the problem, it is the problem.

 

[Orodruin]

My poor, tired brain simply can't handle black holes at the moment .

The main problem people have with black holes is that they try to think of them as just spheres in ordinary Euclidean space. In particular when imagining the event horizon. Nothing could be further from the truth. The event horizon is, as @PeterDonis has pointed out several times, that the event horizon is an outgoing null surface, which means that it will move at the speed of light in any local inertial frame. It also means you cannot think of it as a regular sphere in space. It actually has more in common with a sphere that is expanding at the speed of light than a sphere of fixed radius in Minkowski space. You need to think of it as a surface that is moving at the speed of light, you cannot draw "space" in any reasonable way such that it is standing still.

 

[pervect]

And if the OP is thinking of it that way, he's wrong, and that error is what's causing the problem. See below.
But the Schwarzschild $r$ coordinate is not the same as "space" in the diagrams the OP drew: it can't be, because at the horizon it is null, and inside the horizon it is timelike, not spacelike. So what you are describing is not the solution of the problem, it is the problem.

I'd agree that the r coordinate is null, aka "lightlike". I assume you'd agree that it's also true that an outgoing light beam emitted or reflected just at the horizon would have a constant r coordinate. The fact that the coordinate is constant along a lightbeam indicates that the coordinate can't be a spatial coordinate.

 

[DennisN]

@Peter Donis and @pervect:

Regarding spacetime curvature:

Please forget what I wrote about spacetime curvature in post #1 and #2, it was not what I meant. What I was thinking about was @Dale's suggestion of thinking of the event horizon as moving at c. So I have absolutely no idea why I wrote "spacetime curvature" in those two posts .

Schwarzschild coordinates:

It wasn't about this either, though what @pervect wrote about it sounds very interesting. I know what a Schwarzschild black hole is, but I have no experience calculating coordinates in GR.

Regarding my pictures:

I also understand that things like these can't be easily pictured in 2D, but it was the best I could think of at the moment to illustrate my scenarios. I did the pictures as an illustration to help convey the idea of the scenarios. They were not intended to be exactly scientifically accurate.

Regarding the previous thread versus this thread:

I understood the feedback in the previous thread regarding those three scenarios, but I couldn't initially apply the knowledge gained to this scenario.

This is how I thought up the scenario:

In the previous scenarios the observer was looking towards the event horizon, but what does an observer see when he is initially looking away from the event horizon, but then finds himself looking towards it after he has passed it? I realized it wasn't particularly interesting to put a simple light source to the right of him, because it was clear to me that he would see that light both before and after passing the event horizon. So I would have to find a way for the light to be first emitted away from him, and then turn back to reach him again. And then I came up with the laser and mirror system.

And it got too complicated for my brain, that's why I started the thread, but I started to realize short after I posted the scenario that similar things seemed to be going on in this scenario. In short, it was the event horizon, the mirror and the associated reflection which confused me in this scenario. And the knowledge gained from the previous thread eventually helped me figuring things out in this scenario.

@Orodruin:

Even though my scenario pictures are feeble :), I am aware of that black holes can't be thought of as embedded in 3D space (I learned that from another PF thread posted by some other PF'er long ago). And that is one of my intentions with my scenarios, to get a better way of thinking about black holes. So I am very thankful for your and the other replies!

that the event horizon [...] will move at the speed of light in any local inertial frame.

Understood.

the event horizon is an outgoing null surface

And if I'm not mistaken a null geodesic is a geodesic with no proper time (light-like). And a null surface is a surface where all the normal vectors are null geodesics, or?

 

[PeterDonis]

I assume you'd agree that it's also true that an outgoing light beam emitted or reflected just at the horizon would have a constant r coordinate.

Yes.

The fact that the coordinate is constant along a lightbeam indicates that the coordinate can't be a spatial coordinate.

Yes, and it also indicates that something that has a constant $r$ coordinate at the horizon can't be a "place in space", which means the diagrams @DennisN drew that treat the horizon as a place in space can't be right.

 

[PeterDonis]

I did the pictures as an illustration to help convey the idea of the scenarios. They were not intended to be exactly scientifically accurate.

The problem isn't that they're not "scientifically accurate". The problem is that they're misleading, because they show the horizon as standing still while the container falls past it, when what they should be showing is the container standing still while the horizon flies upward past it at the speed of light. Or you could think of the container as "falling" leftward in the diagram, that would be OK, but you cannot think of the horizon as standing still in your diagram--you have to think of it as moving upward (rightward in the diagram) at the speed of light. That is the crucial correction that needs to be made.

if I'm not mistaken a null geodesic is a geodesic with no proper time (light-like).

"No proper time" is not really correct terminology. Zero spacetime interval along the geodesic is correct. The concept of "proper time" is not well-defined for any curve that is not timelike.

a null surface is a surface where all the normal vectors are null geodesics

It's defined as a surface that has null tangent vectors (but the tangent vectors can't all be null--in the case of the horizon there are also spacelike tangent vectors, in the "tangential" directions, only the tangent vector in the radial direction is null). But since null vectors are orthogonal to themselves (their dot product with themselves is zero, since they have zero spacetime interval), a surface with null tangent vectors will also have null normal vectors (the same vectors). This is one of the counterintuitive properties of null vectors.

 

[Dale]

a surface with null tangent vectors will also have null normal vectors (the same vectors)

Cool! I never thought of that.

 

[pervect]

Cool! I never thought of that.

If we consider a 4-velocity u, it's null if and only if $u \cdot u$ is zero. So a null vector is always orthogonal to itself.

I believe it's also true that if u and v are null vectors pointing in different directions, then they are not orthogonal. If we let u = (1, $\vec{a}$) and v = (1, $\vec{b}$), then with a diag(-1,1,1,1) metric in flat space-time, $u \, \cdot v$ is $-1 + \vec{a} \cdot \vec{b}$. The null length condtion for u and v gives us that $\vec{a} \cdot \vec{a} = 1$ and $\vec{b} \cdot \vec{b} = 1$, i.e. they are unit 3-vectors.

So it is necessary that the dot product of the unit 3-vectors $\vec{a}$ and $\vec{b}$ be equal to 1 for the null 4-vectors u and v to be orthogonal, which implies that $\vec{a} = \vec{b}$.

A good set of 4 basis vectors at the event horizon would be the ingoing null vector, the outgoing null vector, and the two spacelike vectors in the $\hat{\theta}$ and $\hat{\phi}$ directions.

 

[PeterDonis]

A good set of 4 basis vectors at the event horizon would be the ingoing null vector, the outgoing null vector, and the two spacelike vectors in the $\hat{\theta}$ and $\hat{\phi}$ directions.

In fact, this is what you get at the horizon if you use ingoing Eddington-Finkelstein coordinates.

 

[DennisN]

Regarding the event horizon moving at c:

Another way to think about it is to consider the infalling observer’s local inertial frame. In that frame the event horizon is moving at c. Light emitted by the hand after it passes the horizon cannot catch up to the horizon, but after the horizon passes the observer’s eyes the light from the hand will reach it as normal.

You need to be thinking of it [the event horizon] as moving from left to right, relative to your container, with the speed of light.

You need to think of it [the event horizon] as a surface that is moving at the speed of light, you cannot draw "space" in any reasonable way such that it is standing still.

Beyond the event horizon (towards the singularity), can it be useful to view space as moving faster (outwards) than the speed of light in any local inertial frame? And increasingly faster as the frame gets closer to the singularity?

It's defined as a surface that has null tangent vectors (but the tangent vectors can't all be null--in the case of the horizon there are also spacelike tangent vectors, in the "tangential" directions, only the tangent vector in the radial direction is null). But since null vectors are orthogonal to themselves (their dot product with themselves is zero, since they have zero spacetime interval), a surface with null tangent vectors will also have null normal vectors (the same vectors). This is one of the counterintuitive properties of null vectors.

I have tried to digest this, and I am doing a picture which I will post later in this thread to see if I understand it correctly...

A good set of 4 basis vectors at the event horizon would be the ingoing null vector, the outgoing null vector, and the two spacelike vectors in the ^θ\hat{\theta} and ^ϕ\hat{\phi} directions.

...I am using 4 vectors in the picture I am doing, and I actually think they are the vectors you are describing here. We'll see soon...

 

[PeterDonis]

Beyond the event horizon (towards the singularity), can it be useful to view space as moving faster (outwards) than the speed of light in any local inertial frame?

No. Space does not "move" at all. We specifically said the horizon moves at the speed of light; that's because it's a lightlike surface. The horizon is not "space".

I have tried to digest this, and I am doing a picture

Try the diagram towards the end of this Insights article:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-2/

Also the accompanying explanation of what the diagram is telling you about the "grid lines" of Schwarzschild coordinates (the light gray lines in the diagram) and how they are distorted near the horizon.

 

[PeterDonis]

A good set of 4 basis vectors at the event horizon would be the ingoing null vector, the outgoing null vector, and the two spacelike vectors in the $\hat{\theta}$ and $\hat{\phi}$ directions.

In fact, this is what you get at the horizon if you use ingoing Eddington-Finkelstein coordinates.

Even better, if you use the null version of Kruskal coordinates (which you can obtain by taking the timelike/spacelike Kruskal coordinates described in the Insights article I just linked to, and defining null coordinates in the standard way as $U = T + X$ and $V = T - X$), you have these as the coordinate basis vectors everywhere in Schwarzschild spacetime.

 

[DennisN]

Beyond the event horizon (towards the singularity), can it be useful to view space as moving faster (outwards) than the speed of light in any local inertial frame? And increasingly faster as the frame gets closer to the singularity?

No. Space does not "move" at all. We specifically said the horizon moves at the speed of light; that's because it's a lightlike surface. The horizon is not "space".

I was very hesitant using the word "space" there , but I didn't know what word to use to convey what I meant. I will rethink this later.

Try the diagram towards the end of this Insights article:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-2/

Also the accompanying explanation of what the diagram is telling you about the "grid lines" of Schwarzschild coordinates (the light gray lines in the diagram) and how they are distorted near the horizon.

Thanks, I will read the article!

 

[PeterDonis]

I was very hesitant using the word "space" there , but I didn't know what word to use to convey what I meant. I will rethink this later.

If you absolutely must insist on thinking of "space moving" around a black hole, there is in fact an interpretation called the "river model" that works this way. You can read about it in this paper:

https://arxiv.org/abs/gr-qc/0411060

However, you should note carefully that in this interpretation, "space" flows inward, not outward. Also, this interpretation is based on a different coordinate system, Painleve coordinates, which you might not be familiar with; so you can't interpret "space" in this interpretation as the "space" of Schwarzschild coordinates. (In fact, "space" in this interpretation is flat everywhere.) Also, "time" in this interpretation is not the same as Schwarzschild coordinate time; it is in fact the proper time of an observer who is always falling towards the hole at the same rate that space flows inward.

 

[Dale]

Also, that model is only valid for Schwarzschild and Kerr spacetimes.

 

[DennisN]

It's [null surface] defined as a surface that has null tangent vectors (but the tangent vectors can't all be null--in the case of the horizon there are also spacelike tangent vectors, in the "tangential" directions, only the tangent vector in the radial direction is null). But since null vectors are orthogonal to themselves (their dot product with themselves is zero, since they have zero spacetime interval), a surface with null tangent vectors will also have null normal vectors (the same vectors). This is one of the counterintuitive properties of null vectors.

A good set of 4 basis vectors at the event horizon would be the ingoing null vector, the outgoing null vector, and the two spacelike vectors in the ^θ\hat{\theta} and ^ϕ\hat{\phi} directions.

Here's the picture I did to check if I got this info reasonably correct. Is it something like this?
(I put the black hole at the bottom in the picture instead of to the left)

 

[PeterDonis]

Is it something like this?

I can't tell; your picture is incomprehensible to me. But the fact that you say your layout is "like a standard 3-D diagram" is not encouraging: if there is one thing we can say about black holes, it is that it is impossible to describe what is going on near, at, and below the horizon using anything that is "like a standard 3-D diagram".

I would strongly suggest that you take the time to learn the conceptual tools that have already been invented for understanding black holes (such as the Kruskal diagram I showed in the Insights article), instead of trying to invent your own. You can't invent useful conceptual tools for something you don't understand.

 

[DennisN]

But the fact that you say your layout is "like a standard 3-D diagram" is not encouraging

I said the axis layout, so the viewer can see where the axises are pointing in the picture which is 2D (there is no actual x,y,z in the picture).

I would strongly suggest that you take the time to learn the conceptual tools that have already been invented for understanding black holes (such as the Kruskal diagram I showed in the Insights article), instead of trying to invent your own.

I am.

You can't invent useful conceptual tools for something you don't understand.

My picture was to see if I got the idea of a null surface which you described reasonably correct. Instead of writing it in text, I did a picture where possible errors are more easily seen.

 

[PeterDonis]

I said the axis layout

But at least one of your axes is null, and a null axis is not spacelike and does not behave like a spacelike axis, so again you can't possibly describe what is going on with a null axis using something that works like a standard 3-D diagram.

(Btw, one obvious error in the text in your diagram is the statement that null vectors have no direction. That's not correct. They have zero length, but they still have direction. This is one obvious way in which they don't work like spacelike vectors.)

I am.

No, you're not. You're trying to draw your own diagram. Don't do that. You're just going to further confuse yourself.

My picture was to see if I got the idea of a null surface which you described reasonably correct.

And I've given reasons why your diagram can't possibly be a reasonably correct description of the idea of a null surface. Rather that try to draw your own diagrams, you should, as I have already said, be using the ones physicists who understand how all this works have already invented.

 

[PeterDonis]

The OP question has been sufficiently discussed. Thread closed.