Clarifying Black Hole Horizons: An Examination of Observer Perspectives

[wabbit]

I am confused about black hole horizons and such common statements as "light cannot escape from inside the horizon".

The way I currently understand it is as follows :

1. Horizons are always relative to an observer, and what is called "the black hole horizon" is just a shorthand for "the black hole horizon relative to a hovering observer", i.e. one with fixed spatial Schwarzschild coordinates. For a large back hole, this is essentially his Rindler horizon.

2. For other observers, the black hole horizon is different and may not exist.

3. For a free falling observer headed directly towards the singularity (constant angular Schwarzschild coordinates), I am guessing there is a naked singularity - and that light can in fact escape from the Schwarzschild interior region (as defined by the hovering observer) from his viewpoint, as the usual descrption of freefall across the horizon suggests.

4. For an oberver in free fall on a circular orbit, I wonder if there is a horizon, perhaps the same as for the hovering observer ?

I am looking for clarification as to which of the above statements are correct, and if not in which way they are wrong.

Thanks for your help.

 

[ShayanJ]

The horizon of a BH is different from the Rindler Horizon. The Rindler Horizon is not a property of spacetime but an effect of being in an accelerated frame. A BH horizon is a property of spacetime and its existence is not depended on the frame of reference. Its just that in Schwarzschild coordinates, the horizon appears as a singularity but in other coordinates, its not a singularity.

 

[Nugatory]

The horizon is a property of the spacetime geometry, so it is not observer-dependent. Some events can be connected by lightlike and/or timelike geodesics and others cannot, and all observers in all frames will agree about which are which. So the answer to #1 and #2 is "no".

For #3, you have to remember that in the interior region (region II in Kruskal coordinates, which are way more convenient than Schwarzschild coordinates for understanding the geometry here) the Schwarzschild $r$ coordinate is continually decreasing along any timelike geodesic. Thus, the singularity is in the future of all observers and there are no hovering observers in this region - everyone is an infaller. A light signal from the singularity cannot reach any observer (although light signals from some infallers can reach other infallers in region II).

 

[Mentz114]

4. For an oberver in free fall on a circular orbit, I wonder if there is a horizon, perhaps the same as for the hovering observer ?

The worldline of this observer is, for some constant $r$

$\frac{r-2\,M}{\sqrt{r}\,\sqrt{r-3\,M}}\partial_t + \frac{r\,\sqrt{M}}{\sqrt{r-3\,M}}\partial_\phi$

from which one may make certain inferences about the horizon.

From

Gravity: the inside story
T. Padmanabhan
Gen Relativ Gravit (2008) 40:2031–2036

Horizons are inevitable in such a theory and they are always observer dependent.

 

[Nugatory]

The horizon of a BH is different from the Rindler Horizon. The Rindler Horizon is not a property of spacetime but an effect of being in an accelerated frame.

I would prefer to say that it is a property of spacetime, just one that's not very interesting to anyone but the accelerated observer. There is a region of spacetime that has the the property that a lightlike geodesic from that region will not intersect the worldline of the accelerating observer, and the Rindler horizon is the edge of that region.

 

[wabbit]

The horizon of a BH is different from the Rindler Horizon. The Rindler Horizon is not a property of spacetime but an effect of being in an accelerated frame. A BH horizon is a property of spacetime and its existence is not depended on the frame of reference. Its just that in Schwarzschild coordinates, the horizon appears as a singularity but in other coordinates, its not a singularity.

This is what I am doubting. I may well be wrong but it seems to me that horizon is very real for some observers, and inexistent for others.

 

[wabbit]

The horizon is a property of the spacetime geometry, so it is not observer-dependent. Some events can be connected by lightlike and/or timelike geodesics and others cannot, and all observers in all frames will agree about which are which. So the answer to #1 and #2 is "no".

For #3, you have to remember that in the interior region (region II in Kruskal coordinates, which are way more convenient than Schwarzschild coordinates for understanding the geometry here) the Schwarzschild $r$ coordinate is continually decreasing along any timelike geodesic. Thus, the singularity is in the future of all observers and there are no hovering observers in this region - everyone is an infaller. A light signal from the singularity cannot reach any observer (although light signals from some infallers can reach other infallers in region II).

But then this means that as I cross the horizon feet first in free fall, I cannot see my feet for at least a brief time. Weird given that it is said that nothing special happens.

 

[Nugatory]

This is what I am doubting. I may well be wrong but it seems to me that horizon is very real for some observers, and inexistent for others.

You can save yourself much grief by identifying the invariant properties of the spacetime first... and then deciding what if anything they mean to various observers.

 

[ShayanJ]

I would prefer to say that it is a property of spacetime, just one that's not very interesting to anyone but the accelerated observer. There is a region of spacetime that has the the property that a lightlike geodesic from that region will not intersect the worldline of the accelerating observer, and the Rindler horizon is the edge of that region.

I think we're using different definitions of "property of spacetime". I meant its only the accelerated observer that sees such an effect and if there is no accelerated observer, there is no Rindler horizon. But I somehow see what you mean.

 

[CalcNerd]

You can think of the event horizon as a (Black) balloon surface. Imagining this balloon as the actual surface of the event horizon of a non rotating black hole. On the outside, we can all look down and we will see something (not actually in or from the hole though). We will see matter as it falls in and if the tidal forces are great, the material will be ripped apart. If it doesn't fall straight in, it will begin a spiral in fall into the hole with some of it emitting energy (estimates are as high as 40%) via radiation (making this a very efficient way to generate energy, more so than nuclear fusion, in fact and why Quasars are so bright).

However, as the material approaches the surface (of this event horizon, it is NOT a membrane like our visual black balloon), it becomes closer to a higher gravitational field ie it slows down it TWO distinct ways from us. First it is in a much higher gravitational field (which does actually inflict a time dilatation on the objects) and second, the light coming is actually being red shifted and dimmed by this same gravitational field as well.

If we jumped in and if we could avoid spaghettification, (we would probably choose a huge hole), we would probably see at the horizon, a large radiation flash right at that imaginary membrane that hasn't actually fallen into the hole, but can't escape either. As we pass through, we are now at a gravitational well that would affect our own time dilatation as well. And we would be falling towards a naked singularity. It would likely be black, because everything would be moving at it, nothing coming out (nothing, meaning nothing, not even light). If we could fire rockets, strong enough to stabilize us, which isn't feasible and most texts indicate any energy expended would actually propel us faster down into the singularity at this point, NO MATTER which way you aim your thrusters, technically, there is only one direction, DOWN; we would see light coming at us, being sucked in too. But since we can't slow down, we probably would see darkness all the way down to our spaghettification point and hit the singularity sometime thereafter. We would be in an extreme relativistic environment, extreme speeds and gravity in a hole, headed for a singularity.

I am not thinking we are going to see anything like "Interstellar". I was disappointed with that movie.

 

[wabbit]

You can save yourself much grief by identifying the invariant properties of the spacetime first... and then deciding what if anything they mean to various observers.

But then what is this invariant definition of the horizon ? After all, horizons are observer dependent, and the Rindler horizon occurs in Minkowski spacetime ?

 

[Nugatory]

But then this means that as I cross the horizon feet first in free fall, I cannot see my feet for at least a brief time.

That's not what happens. There is a never a moment when light that's reflected off your feet isn't reaching your eyes. It's just that the last light to reach your eyes before they fall through the horizon came from your feet just before they passed through the horizon, and the first light to reach your eyes after they pass through the horizon came from your feet just after they passed through the horizon. (We have to be very careful with the word "after" in this context - it's best to identify the the points on the worldlines of your feet and of your eyes at which the light is emitted and absorbed, speak about the relationships of these points).

 

[wabbit]

That's not what happens. There is a never a moment when light that's reflected off your feet isn't reaching your eyes. It's just that the last light to reach your eyes before they fall through the horizon came from your feet just before they passed through the horizon, and the first light to reach your eyes after they pass through the horizon came from your feet just after they passed through the horizon. (We have to be very careful with the word "after" in this context - it's best to identify the the points on the worldlines of your feet and of your eyes at which the light is emitted and absorbed, speak about the relationships of these points).

Hmmm... I was thinking of this solution but could not muster the courage to calculate and check it. But this seems hard to reconcile with the near-flat spacetime across the horizon of a very large black hole, and the "nothing special happens". After all if the horizon gravity is 1g and I just jumped off my hovering spacecraft , right above the horizon, I must be passing through it at well below the speed of light, no ?

But that "after" and the changing nature of the radial coordinate is perhaps throwing my intuition off here. Still, don't both feet and head have decreasing $r(\tau)$ where $\tau$ is my proper time - both for feet and head since I am in nearly flat spacetime. But even if not, my feet still are essentially stationary wrt my head, in my frame.

Maybe what I am missing is that the horizon must be moving towards me at light speed, whatever my velocity relative to say that spacecraft .

There is something similar in the Rindler case. The horizon suddenly disappears when he jumps off his spacecraft , although he then sees that spacecraft accelerating slowly away say at 1g or less. Something drastic happens to the horizon with no change of velocity, just a change in acceleration.

The Rindler analogy (and nearly indistiguishable spacetime metric from that of a small region crossing a very large black hole horizon) is what leads me to imagine that the back hole horizon simply disappears (or at least recedes away very fast) for me when I jump off my spacecraft ...

 

[wabbit]

The worldline of this observer is, for some constant $r$

$\frac{r-2\,M}{\sqrt{r}\,\sqrt{r-3\,M}}\partial_t + \frac{r\,\sqrt{M}}{\sqrt{r-3\,M}}\partial_\phi$

from which one may make certain inferences about the horizon.

I wish I knew how to do it :) though the discussion of the observer jumping straight into the black hole is already quite tough so I'm happy to leave the orbiting one alone for now...

Gravity: the inside story
T. Padmanabhan
"Horizons are inevitable in such a theory and they are always observer dependent."

This "always" would seem to support my assumption that like any other, the black hole horizon is observer dependent.

 

[Mentz114]

I wish I knew how to do it :) though the discussion of the observer jumping straight into the black hole is already quite tough so I'm happy to leave the orbiting one alone for now...

The orbiter must have $r>3M$ so they never can get to the horizon. But what is important is what happens to light sent by them to the center.

This "always" would seem to support my assumption that like any other, the black hole horizon is observer dependent.

Padmanabhan argues the proper acceleration causes a local entropy change that must be balanced by a horizon. Laws of thermodynamics.

I am not 100% convinced.

 

[wabbit]

@CalcNerd, so you are saying that despite the nearly flat metric, crossing the event horizon of a very large black hole is in fact very eventful and not like cruising in Minkowski spacetime ?

 

[wabbit]

Padmanabhan argues the proper acceleration causes a local entropy change that must be balanced by a horizon. Laws of thermodynamics.

Seems reasonable to me. In any case the entropy is also relative to the observer so it may well change between hovering and free fall. I believe this does happen in the Rindler case, no ?

 

[bcrowell]

But then this means that as I cross the horizon feet first in free fall, I cannot see my feet for at least a brief time. Weird given that it is said that nothing special happens.

No, that doesn't happen. One way of stating the equivalence principle is that spacetime is always locally Minkowski. Therefore a local experiment like the one you're describing would not give an unusual result. As a free-falling observer, you see nothing special happening. You can see your feet at all times. An observer using Schwarzschild coordinates would explain this by saying that although the light from your feet didn't escape the event horizon, your eyes crossed the horizon in time to see that light.

 

[bcrowell]

3. For a free falling observer headed directly towards the singularity (constant angular Schwarzschild coordinates), I am guessing there is a naked singularity - and that light can in fact escape from the Schwarzschild interior region (as defined by the hovering observer) from his viewpoint, as the usual descrption of freefall across the horizon suggests.

There is a local definition of a naked singularity and a global definition. The local definition is basically that it's timelike, http://physicsforums.com/showthread.php?t=635641 , and this is observer-independent. The global definition is defined in terms of intrinsic properties of the whole spacetime, and is therefore also observer-independent.

The Schwarzschild singularity is spacelike, and this is an observer-independent property. It's also surrounded by an event horizon, and this is also an observer-independent property. Therefore by both definitions it's not a naked singularity.

 

[wabbit]

No, that doesn't happen. One way of stating the equivalence principle is that spacetime is always locally Minkowski. Therefore a local experiment like the one you're describing would not give an unusual result. As a free-falling observer, you see nothing special happening. You can see your feet at all times. An observer using Schwarzschild coordinates would explain this by saying that although the light from your feet didn't escape the event horizon, your eyes crossed the horizon in time to see that light.

Right... This makes sense. From his viewpoint, it does. But from my viewpoint, I see my feet just fine, light travels from them to my eyes unimpeded. Which I still find hard to interpret otherwise than, from my viewpoint, I amnot crossing a horizon, just cruising in nearly flat space. And if I am still accelerating sightly (below what's required for hovering), then I "see" yet another horizon, which neither freefalling nor hovering observers will agree with me "exists" - all three are correct, within their own frame/perspective.

 

[CalcNerd]

Not saying that. The larger the whole, the greater the surface area. The colder the hole. A cold hole probably doesn't have nearly as much radiation at the horizon as a small "HOT" hole.
So, passing through the horizon of a large hole may give you an experience of a radiation flash, but it may also be uneventful. If it were a hole from an active galactic center, I suspect you would see a radiation flash, even though the horizon would be large, there would energy bound at the horizon, decaying outward or falling inward. A non-active hole would be a dull place as all the energy at the horizon would have dissipated away, leaving the hole at its natural (near zero K radiation) temp. Small holes (with huge tidal forces, the mean nasty little ones) with small surface areas (and higher temperatures) would most likely give you that Radiation Flash as you pass through the horizon.

I confess, I haven't made the trip.

 

[wabbit]

There is a local definition of a naked singularity and a global definition. The local definition is basically that it's timelike, http://physicsforums.com/showthread.php?t=635641 , and this is observer-independent. The global definition is defined in terms of intrinsic properties of the whole spacetime, and is therefore also observer-independent.

The Schwarzschild singularity is spacelike, and this is an observer-independent property. It's also surrounded by an event horizon, and this is also an observer-independent property. Therefore by both definitions it's not a naked singularity.

OK Thanks for that clarification, this is helpful. So each observer does "see" a horizon, though which horizon depends on the observer.

The infalling observer is a little special though since his wordline reaches the singularity in finite proper time, so even the definition of a horizon for him might be a little tricky (at least for the global definition).

 

[wabbit]

Not saying that. The larger the whole, the greater the surface area. The colder the hole. A cold hole probably doesn't have nearly as much radiation at the horizon as a small "HOT" hole.
So, passing through the horizon of a large hole may give you an experience of a radiation flash, but it may also be uneventful.

OK I think I get that. If there is a flash (not convinced yet of that part but I ll assume it does happen here), then it will be fainter and fainter for a larger black hole and this is how the experience of the free falling observer converges with his flat space experience. Thanks.

I confess, I haven't made the trip.

Yeah, I was kinda hoping to understand this without having to do that either : )

 

[pervect]

I am confused about black hole horizons and such common statements as "light cannot escape from inside the horizon".

The way I currently understand it is as follows :

1. Horizons are always relative to an observer, and what is called "the black hole horizon" is just a shorthand for "the black hole horizon relative to a hovering observer", i.e. one with fixed spatial Schwarzschild coordinates. For a large back hole, this is essentially his Rindler horizon.

There are several sorts of horizons, which may be part of the confusion. Absolute horizons are not observer dependent. They are the region in which light will never reach the region of the outside of the horizon if it starts inside the horizon. You do need a notion of "inside" and "outside", the later is usually taken to be "at infinity", but you don't really need an observer.

Because of their definition, absolute horizons won't exist for evaporating black holes.

Another type of horizon, which is observer dependent, is the apparent horizon. The exact definition is a bit technical, I believe the wiki article http://en.wikipedia.org/w/index.php?title=Apparent_horizon&oldid=650874054 captures the essence when it points out that inside the apparent horizon, both outgoig and ingoing light rays converege, while outside the apparent horizon, ingoing light rays converge and outgoing light rays diverge.

It's mentioned that the apparent horizons are observer dependent, but while I've read about them I haven't calculated them so I'm not sure of all the details, such as where they'd be for an orbiting observer.

For completeness, there are other sorts of horizons as well, such as Killing horizons.

 

[bcrowell]

OK Thanks for that clarification, this is helpful. So each observer does "see" a horizon, though which horizon depends on the observer.

The infalling observer is a little special though since his wordline reaches the singularity in finite proper time, so even the definition of a horizon for him might be a little tricky (at least for the global definition).

I think you need to decide on what definition you want to use of horizon. Until you do that, we can't really discuss this kind of thing meaningfully. As others have pointed out, the normal definition is a global definition in terms of intrinsic quantities, and is therefore observer-independent. If you have some other definition in mind, you need to tell us what it is.

 

[bcrowell]

But from my viewpoint, I see my feet just fine, light travels from them to my eyes unimpeded. Which I still find hard to interpret otherwise than, from my viewpoint, I amnot crossing a horizon, just cruising in nearly flat space. And if I am still accelerating sightly (below what's required for hovering), then I "see" yet another horizon, which neither freefalling nor hovering observers will agree with me "exists" - all three are correct, within their own frame/perspective.

The fact that spacetime is locally Minkowski is a fact that holds for free-falling observers. You seem to be imagining an observer who is hovering with his feet inside the horizon and his head outside the horizon. It's not physically possible for the feet to hover in this situation. If you tried to do this, your body would be shredded at or before the moment when your feet cross the horizon.

 

[wabbit]

I think you need to decide on what definition you want to use of horizon. Until you do that, we can't really discuss this kind of thing meaningfully. As others have pointed out, the normal definition is a global definition in terms of intrinsic quantities, and is therefore observer-independent. If you have some other definition in mind, you need to tell us what it is.

This may indeed be the source of my confusion, thanks for pointing this out. I was using the "light doesn't escape" definition (e.g. in my feet to head example) but also probably others without making a clear distinction. (I also referred to the Schwarzschild horizon as "the black hole horizon", distinct from the no-light-escape horizon defined by a given observer. )

Let's stick with light doesn't escape then. I will try to be more specific and to always say "Schwarzschild horizon" for the usual one, and "observer X horizon" for the light-doesn't-escape horizon as defined by observer X.

 

[wabbit]

The fact that spacetime is locally Minkowski is a fact that holds for free-falling observers. You seem to be imagining an observer who is hovering with his feet inside the horizon and his head outside the horizon. It's not physically possible for the feet to hover in this situation. If you tried to do this, your body would be shredded at or before the moment when your feet cross the horizon.

No, I am imagining an observer jumping off a hovering spacecraft and free-falling in across the horizon.

But that shredding, I do find hard to reconcile with "nearly Minkowski", it means that if I fall across infalling, and start a 1g rocket just as my head is above but my feet inside(*), I will be torn apart ?

(*)above/inside the Schwarzschild horizon as usually defined, corresponding to a fixed radial Schwarzschild coordinate.

 

[Haelfix]

Your problem, as others have told you is that you are thinking of the horizon as some sort of local surface, like a barrier or something of that nature. So then you are worried that the barrier has paradoxical properties inconsistent with special relativity. But this is not what happens.

The event horizon is a global property of spacetime, no local observer can make a measurement and ascertain its properties (he/she would need to know the entire future history of the universe). In the same way, when viewed in a frame and coordinate system that has access to this information, one sees that the event horizon actually hits the free falling observer at the speed of light. So it doesn't make sense to worry about whether his toes and his head 'look' differently.

 

[wabbit]

There are several sorts of horizons, which may be part of the confusion. Absolute horizons are not observer dependent. They are the region in which light will never reach the region of the outside of the horizon if it starts inside the horizon. You do need a notion of "inside" and "outside", the later is usually taken to be "at infinity", but you don't really need an observer.

This is (I think) what I was using (more precisely, I was using the property that light doesn't cross the horizon outward, as in feet vs head, but as defined / experienced by a given observer.)

Because of their definition, absolute horizons won't exist for evaporating black holes.

I should have said, I want to consider strictly classical, simple situations here, until I get this right I d rather not go into evaporation.

For completeness, there are other sorts of horizons as well, such as Killing horizons.

I probably need to read more about these distinctions.

 

[wabbit]

Your problem, as others have told you is that you are thinking of the horizon as some sort of local surface, like a barrier or something of that nature.

Not really, I was actually saying the Schwarzschild horizon doesn't exist for an infalling observer. But yes I was referring to the specific surface defined by a value of the Schwarzschild radial coordinate (calling this Schwarzschild horizon now to avoid ambiguity) and trying to figure out how different observers experience or describe this surface, especially as they fall across it.

Well, anyway, I should probably just calculate some trajectories to lay to rest those head and feet misgivings.

 

[bcrowell]

This may indeed be the source of my confusion, thanks for pointing this out. I was using the "light doesn't escape" definition (e.g. in my feet to head example) but also probably others without making a clear distinction. (I also referred to the Schwarzschild horizon as "the black hole horizon", distinct from the no-light-escape horizon defined by a given observer. )

Let's stick with light doesn't escape then. I will try to be more specific and to always say "Schwarzschild horizon" for the usual one, and "observer X horizon" for the light-doesn't-escape horizon as defined by observer X.

You need to spell out in more detail what you mean by "light doesn't escape." The standard definition is that light doesn't escape to null infinity. That's not an observer-dependent definition.

 

[bcrowell]

No, I am imagining an observer jumping off a hovering spacecraft and free-falling in across the horizon.

But that shredding, I do find hard to reconcile with "nearly Minkowski", it means that if I fall across infalling, and start a 1g rocket just as my head is above but my feet inside(*), I will be torn apart ?

(*)above/inside the Schwarzschild horizon as usually defined, corresponding to a fixed radial Schwarzschild coordinate.

In order to hover, you need some external force, such as the force from the deck of a spaceship pressing upward on your feet. That force is the force that destroys your body.

 

[wabbit]

You need to spell out in more detail what you mean by "light doesn't escape." The standard definition is that light doesn't escape to null infinity. That's not an observer-dependent definition.

Right. Actually as I realized after that I was using something slightly different, and this may simply be incorrect : that light doesn't travel from inside to outside. This is what the feet to head contradicts from the viewpoint of the infalling observer, but if this property doesn't hold (locally) than there is no issue, and this looks like the source of my confusion.

 

[wabbit]

In order to hover, you need some external force, such as the force from the deck of a spaceship pressing upward on your feet. That force is the force that destroys your body.

I understand. It's just hard to imagine a 1g force ripping my body apart in near-flat spacetime (well, not quite flat in that example, but something I am accustomed to here on earth) - like jumping off into free fall from 50cm above the Schwarzschild horizon, moving apart at 1g from said spacecraft , and then quickly (after just 60cm fall) , grabbing hold of the spacecraft with my hand, my shoulder will be torn apart. Maybe I just need to get used to it.

Actually 1g is rather strong, we might as well say 0.1g

Unless the g-field is highly inhomogenous there, so that my feet are pulled in much more strongly - but I don't think that's the case near the Schwarzschild horizon of a large black hole (I am relying on the EP/"nothing special happens as a free falling observer crosses the horizon" here. This is valid I presume if the spacetime curvature can be approximated by a constant g field in a small (a few meters) region that crosses the horizon.)

 

[wabbit]

One last question, hopefully specific enough to be either confirmed or infirmed unambiguously: I thought that the metric in a small region crossing the Schwarzschild horizon of a very large black hole is very close to corresponding to a constant g field. So a hovering observer at constant Schwarzschild radial and angular coordinates in this region is equivalent to a Rindler observer accelerating at that same g in Minkowski spacetime. I am assumimg that (locally) his Rindler horizon coincides with the black hole Schwarzschild horizon. Is this wrong?

 

[Khashishi]

I understand. It's just hard to imagine a 1g force ripping my body apart in near-flat spacetime (well, not quite flat in that example, but something I am accustomed to here on earth) - like jumping off into free fall from 50cm above the Schwarzschild horizon, moving apart at 1g from said spacecraft , and then quickly (after just 60cm fall) , grabbing hold of the spacecraft with my hand, my shoulder will be torn apart. Maybe I just need to get used to it.

If you are stationary near the Schwarzschild horizon, then you are extremely height-contracted. Or, in your perspective, the event horizon will be very far away. So you need to be more clear on what you mean by 50cm.

 

[wabbit]

If you are stationary near the Schwarzschild horizon, then you are extremely height-contracted. Or, in your perspective, the event horizon will be very far away. So you need to be more clear on what you mean by 50cm.

Yes ! this was ambiguous in my formulation, perhaps this is the key.
As I see it, from the viewpoint of the hovering observer, the Schwarzschild horizon is at an infinite proper distance - an object free falling away from him will take an infinite time (proper time of the hoverer) to reach the horizon. This is correct?
From the free falling observer after he jumps off, that Schwarzschild horizon is however a finite proper distance away. This is the distance (measured by the falling observer in his own frame, by timing light signals or even using a ruler I guess), that I am using to define those 50 cm. Those 50 cm are a very long (even infinite if it crosses the horizon ) distance from the viewpoint of the hovering observer.

So I should not have said "hovering 50cm above the horizon". This is meaningless, no one can hover at a finite distance (as measured by him) over the horizon. This should be instead "hovering at a point where the proper distance of a free falling observer to the horizon would be 50cm".

Maybe this changes things, I need to think about this point. Thanks for mentionning that.

 

[Haelfix]

Again, in this business it's really important to only ask questions that make sense based on what an observer has access too. A free falling observer does not have access to the location of the horizon. Only the observer at infinity, using charts that cover the whole spacetime can make that analysis and decide what did or did not happen. Of course that observer sees a large amount of red shift as the infalling observer is squashed against the horizon.

Nevertheless he can deduce based on his knowledge of GR that the observer must have crossed the horizon, where it met him at the speed of light and then passed him and that from this observers point of view, nothing special happened.

Now, a slightly more refined problem often given to students as homework, is the following experiment. Take the point of view of a hovering 'static' observer, on his rocket ship. Now give this person a long fishing pole (lets make it a light year across) and let him dip the pole down through the horizon (that observer won't know exactly where the horizon is, but let's say we arrange it so that he gets lucky). What happens?

 

[wabbit]

OK I would say it must take an infinite (static observer's proper) time for the pole to reach the horizon, so the static hoverer will just see the end of the pole moving farther and farther away as he dips it assuming the "dipping" is just letting the pole slip in his hand in free fall, or that he exerts a force to slow down the dipping.

But regarding the previous case I don't agree that the observer "at infinity" can know that the free falling observer has crossed the horizon since this only happens at infinite proper time for him, which is to say never. Is this incorrect?

 

[Nugatory]

But regarding the previous case I don't agree that the observer "at infinity" can know that the free falling observer has crossed the horizon since this only happens at infinite proper time for him, which is to say never. Is this incorrect?

The event which is the infaller crossing the horizon is not on the worldline of the observer at infinity, so there is no meaningful way to associate it with any proper time on that worldline. It's no more correct to say that it happens "at infinite proper time" than any other time.

However, there are two interesting events on the worldline of the observer at infinity:

One is the last point at which the observer at infinity will be able to send a radio message to the infaller and expect a reply. If sent before that point the message will reach the infaller before he has passed the horizon, so the infaller's reply will eventually (possibly many millennia later) make it back to the observer at infinity. Messages sent after that point will still reach the infaller, but the infaller will already be inside the horizon and the infaller's reply will not make it back out.

The second interesting point is the last point at which a message sent by the observer at infinity will reach the infaller at all; messages sent after that point will not reach the infaller before the infaller reaches the singularity. Imagine that the infaller is your beloved spouse, thrown into the black hole by some appalling mechanical failure on your spaceship hovering outside the black hole. There is no saving your spouse once the horizon is crossed, but you would wish for him/her to hear your promise of undying love before their inevitable death in the singularity... You do not have infinite time to get that message transmitted, in fact you don't have very long at all before it is too late and your spouse will die alone and uncomforted by any parting words from you.

You could reasonably choose any point on your worldline after the first to be the time when the horizon crossing "happens" and any point after the second to be when the infaller's meeting with the central singularity "happens". There's no need to wait for infinite time to pass on your worldline.

 

[PeterDonis]

Because of their definition, absolute horizons won't exist for evaporating black holes.

That's not correct--at least, it's not correct for the simplest model of an evaporating black hole, the one Hawking originally constructed. A Penrose diagram for this model is here:

http://en.wikipedia.org/wiki/Black_hole_information_paradox#Hawking_radiation

The triangular region at the upper left, with the singularity (sawtooth line) at the top, is all a black hole region in the absolute sense; no event in this region can send light signals to future null infinity. The boundary of this region (the 45 degree line going up and to the right from r = 0 to the right end of the singularity) is the absolute horizon.

There are other models of "black hole" evaporation where there is no absolute horizon--but putting the term "black hole" in quotes emphasizes the fact that in these models, there is no black hole--no singularity as well as no absolute horizon. In these models, there are only apparent horizons that exist temporarily but later disappear. However, we don't know yet which kind of model is the correct one.

 

[PeterDonis]

If you are stationary near the Schwarzschild horizon, then you are extremely height-contracted. Or, in your perspective, the event horizon will be very far away.

This is not correct. The proper distance to the horizon for a stationary observer is finite, and goes to zero as the $r$ coordinate of the stationary observer goes to $2M$. Don't misinterpret the meaning of the $(1 - 2M / r)^{-1}$ in the line element; it doesn't mean that distances get "infinitely stretched" as the horizon is approached, even though it looks that way on the surface. You need to actually compute the integral from some finite $r > 2M$ to $r = 2M$; you will see that it has the properties I described above.

 

[PeterDonis]

from the viewpoint of the hovering observer, the Schwarzschild horizon is at an infinite proper distance - an object free falling away from him will take an infinite time (proper time of the hoverer) to reach the horizon. This is correct?

As Nugatory said, there is no invariant meaning to "how much time" it takes for the infalling object to reach the horizon (though he also pointed out a very reasonable criterion for assigning a finite time to that event--what proper time the hovering observer would have to send a light signal in order for it to reach the infalling object before it crosses the horizon). There is, however, an invariant meaning to the proper distance from the hovering observer to the horizon; it's the integral I described in my previous post, in response to Khashishi. As I noted in that post, the proper distance is not infinite.

 

[wabbit]

As Nugatory said, there is no invariant meaning to "how much time" it takes for the infalling object to reach the horizon (though he also pointed out a very reasonable criterion for assigning a finite time to that event--what proper time the hovering observer would have to send a light signal in order for it to reach the infalling object before it crosses the horizon). There is, however, an invariant meaning to the proper distance from the hovering observer to the horizon; it's the integral I described in my previous post, in response to Khashishi. As I noted in that post, the proper distance is not infinite.

For the time, I was not assuming an invariant meaning, but a coordinate chart defined by the observer using as time his proper time. If I understand what you and Nugatory are saying, there is no matural choice of such a chart. I was implicitly thinking of time and distance as measured by the observer, extending his local chart as far as he can, and presuming that for the hovering observer this chart would extend to the Schwarzschild horizon and no further, and would (unambiguously) assign an infinite time (which I was calling "proper time of the hovering observer") to the event of the infalling observer crossing the Schwarzschild horizon.

I prefer to stay with a hovering observer than using the observer at infinity, for concreteness, though my understanding is that Nugatory's statements about the latter would also apply to the former.

Annway, at least now I understand that I don't understand anything here. Oh well.

 

[PeterDonis]

If I understand what you and Nugatory are saying, there is no natural choice of such a chart.

Correct.

I was implicitly thinking of time and distance as measured by the observer, extending his local chart as far as he can

This is one possible way of doing it, yes. But it has limitations; see below.

presuming that for the hovering observer this chart would extend to the Schwarzschild horizon and no further

To be precise, it is singular at the horizon, so it only covers the region outside the horizon.

would (unambiguously) assign an infinite time (which I was calling "proper time of the hovering observer") to the event of the infalling observer crossing the Schwarzschild horizon.

No, that's not what this chart does. As above, this chart is singular at the horizon, which means it cannot assign a "time" to events on the horizon at all. The chart is simply not well-defined there. To assign "time" to events on the horizon at all, you must switch to a chart that is not singular there.

Pop science presentations often are sloppy about this, saying that "infinite time" is assigned to events on the horizon. This just illustrates why you can't learn science from pop science presentations.

(Note, btw, that even though this chart is singular at the horizon, we can still take limits as the horizon is approached, and so we can still verify some calculations. For example, we can compute the proper time for a free-falling object to go from some radius $r_1$ to some smaller radius $r_2$; then we can take the limit of this as $r_2 \rightarrow 2M$ and find that the limit is finite, indicating that the free-falling object will reach the horizon in a finite proper time by its own clock. Or we can compute the proper distance between two hovering observers at radius $r_1$ and $r_2$, and take the limit of that as $r_2 \rightarrow 2M$, and find that the limit is finite, indicating that any hovering observer has only a finite proper distance to the horizon. So we can justify the statements made earlier in this thread about such things even without switching charts, strictly speaking. But of course it's a lot easier to see how things work at the horizon if you switch to a chart that's not singular there.)

 

[jartsa]

I am confused about black hole horizons and such common statements as "light cannot escape from inside the horizon".

The way I currently understand it is as follows :

1. Horizons are always relative to an observer, and what is called "the black hole horizon" is just a shorthand for "the black hole horizon relative to a hovering observer", i.e. one with fixed spatial Schwarzschild coordinates. For a large back hole, this is essentially his Rindler horizon.

2. For other observers, the black hole horizon is different and may not exist.

3. For a free falling observer headed directly towards the singularity (constant angular Schwarzschild coordinates), I am guessing there is a naked singularity - and that light can in fact escape from the Schwarzschild interior region (as defined by the hovering observer) from his viewpoint, as the usual descrption of freefall across the horizon suggests.

4. For an oberver in free fall on a circular orbit, I wonder if there is a horizon, perhaps the same as for the hovering observer ?

I am looking for clarification as to which of the above statements are correct, and if not in which way they are wrong.

Thanks for your help.

When observer hovers far from event horizon, far away from him there seems to be a place where things seem to freeze. When the observer starts free falling down, he sees those frozen things stay frozen, except for things that are hovering, observer sees those things start to slowly unfreeze when he starts to free fall.

When observer hovers near an event horizon, he sees the place where everything seems to be frozen to be close to him. When said observer starts to free fall down, he sees the frozen things start to unfreeze quickly. Nothing seems to stay frozen in this case, for some odd reason ... probably has something to do with frozen light unfreezing.

 

[wabbit]

The event which is the infaller crossing the horizon is not on the worldline of the observer at infinity, so there is no meaningful way to associate it with any proper time on that worldline. It's no more correct to say that it happens "at infinite proper time" than any other time.

However, there are two interesting events on the worldline of the observer at infinity:

One is the last point at which the observer at infinity will be able to send a radio message to the infaller and expect a reply. If sent before that point the message will reach the infaller before he has passed the horizon, so the infaller's reply will eventually (possibly many millennia later) make it back to the observer at infinity. Messages sent after that point will still reach the infaller, but the infaller will already be inside the horizon and the infaller's reply will not make it back out.
.

I did a calculation for the round trip of a photon from the hovering observer to some point which can be the inflalling observer reflecting that photon, and I find that time (proper time of the hoverer) is infinite. So if this is correct, any time in his future, the hoverer will still be receiving messages saying "not crossed yet, getting closer".

There may be errors here of course - here it goes :

Taking $c=1$ and considering only trajectories with constant angular Schwarzschild coordinates, r and t being the remaining Schwarzschild coordinates,
$$d\tau^2=(1-\frac{r_s}{r})dt^2-(1-\frac{r_s}{r})^{-1}dr^2$$
For a photon moving radially at $r>r_s$, this gives $$\frac{dt}{dr}=\epsilon\frac{r}{r-r_s}$$where the $\epsilon=+1$ is for a photon moving toward infinity, $\epsilon=-1$ for one moving toward the hole.
Integrating this yields
$$t-t_0=\epsilon\left(r-r_0+r_s\ln\frac{r-r_s}{r_0-r_s}\right)$$
For a photon emitted at $(r_0,t_0)$ toward the hole, reflected at $(r_1,t_1)$ and received back at $(r_0,t_2)$ this gives
$$\Delta t=_{def}t_1-t_0=t_2-t_1=\frac{t_2-t_0}{2}=r_0-r_1+r_s\ln\frac{r_0-r_s}{r_1-r_s}$$
And $lim_{r_1\rightarrow r_s^+}\Delta t=+\infty$

On the other hand, for a hovering observer at $r_0$, $$\frac{d\tau_{hov}}{dt}=\sqrt{1-\frac{r_s}{r_0}}$$ so $\Delta \tau_{hov}=\sqrt{1-\frac{r_s}{r_0}}\Delta t$ which also tend to infinity as $r_1\rightarrow r_s$

 

[PeterDonis]

I did a calculation for the round trip of a photon from the hovering observer to some point which can be the inflalling observer reflecting that photon, and I find that time (proper time of the hoverer) is infinite.

Yes, but think about what this calculation is telling you. What it is telling you is that, if the hovering observer sends light signals down to the infalling observer, who reflects them back, then as the infalling observer approaches the horizon, the time the hovering observer has to wait to receive the return signal increases without bound.

But now, suppose that the hovering observer sends light signals which are "time stamped" with the time of emission; and suppose that, when the infalling observer reflects the signals, he adds a second "time stamp" giving the time on his clock at the instant of reflection. Then, as the return signals come back, the hovering observer will see the two time stamps approach finite values, even though the time at which he receives the return signals increases without bound. The finite value approached by the first time stamp (the time of emission) is what Nugatory called "the last point at which the observer at infinity will be able to send a radio message to the infaller and expect a reply". The finite value approached by the second time stamp (the time of reflection) is the proper time on the infaller's clock at which he crosses the horizon.

 

[Nugatory]

did a calculation for the round trip of a photon from the hovering observer to some point which can be the infalling observer reflecting that photon, and I find that time (proper time of the hoverer) is infinite. So if this is correct, any time in his future, the hoverer will still be receiving messages saying "not crossed yet, getting closer".

The round trip time is always finite, although it becomes arbitrarily large as the infaller gets arbitrarily close to the horizon. When the infaller reaches the horizon, the round trip time isn't infinite - it's undefined, because the light signal doesn't make the round trip at all. But it is a great stretch, and one that I doubt you would accept in any other context, to say that because light from an event never reaches your eyes that event never happens (or doesn't happen until an infinite amount of time has passed).

Suppose that every five seconds you send a time stamped radio message to the infaller, and when he receives it he replies "I saw your message dated <whatever>". You send a sequence of messages reading "12:00:05", "12:00:10", "12:00:15", ... and you receive (perhaps millennia later) messages reading "I saw your 12:00:05 message", and "I saw your 12:00:10 message", but no matter how long you wait you will never ever receive a reply to your "12:00:15" message. I can reasonably describe this as the infaller passing through the horizon sometime between receiving my 12:00:10 message and my 12:00:15 message.

As for when these three events (infaller received 12:00:10, infaller passed horizon, infaller received 12:00:15) "happened"? None of them are on my worldline, so I have to map them to events that are on my wordline and to which I can assign proper times. So I choose a coordinate system that covers both my worldline and the infaller's worldline; note the time coordinate of the event on the infaller's worldline; then identify the point on my worldline that has the same time coordinate so happened "at the same time". Whatever my wristwatch reads at that point... that's "when" the event happened.
1) This is equivalent to the procedure that we follow in ordinary flat Minkowski spacetime for assigning times to events off our worldline. The only difference is that in the flat spacetime case, there is an obvious and natural choice of coordinate system, so we don't notice the extent to which "at the same time" is just a convention based on our choice of coordinate system. In curved spacetime, we don't have this luxury.
2) Schwarzschild coordinates are unusable for this purpose because they don't cover both worldlines.
3) The Kruskal time coordinate works just fine however, and there is no difficulty converting the Kruskal time coordinate of a point on the observer's worldine to the time displayed by the observer's wristwatch at that point. Of course, a different choice of coordinates would produce different results... which is why we say that simultaneity is a convention.

There is, however, a very great difference between saying that there is no single universally accepted definition of when the infaller's crossing the horizon "happens", and saying that it never "happens" or that it "happens" after infinite time.