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.)