Clarifying Black Hole Horizons: An Examination of Observer Perspectives

[Asher Weinerman]

By your (il)logic, a ball dropped from a uniformly accelerating rocket 'never' gets more than 1/a (with 'a' being the acceleration in natural coordinates) in distance from the rocket, because that is what is true in Rindler coordinates, which are the coordinates analogous to SC coordinates for a uniformly accelerating rocket.

That is precisely my point - be wary of transforming to alternate coordinate systems and then misinterpreting them because the new coordinates have obscure meanings.

As for the postulates of General Relativity, they are just the equivalence principal and that physical laws must be expressed in tensor form so they have the same form in all coordinate systems. Don't distract from the real issue.

You are confusing GR with quantum mechanics. Meaningful quantities must be able to be measured. They do not have to be scalars. So yes, time has meaning although it is different in different coordinate systems. Same with say, four-velocity, or energy or whatever. Try not to get distracted with too many complications that you have learned, and focus on the issue at hand.

 

[wabbit]

And what is your criteria for physically meaningful coordinates? Who chooses and how?

I don't know what Asher Weinerman's criteria are but I can tell you which distinction I make between coordinate systems relative to an observer: those that correspond to physically measurable quantities (times and distances measured by that observer in his worldline) are indeed special - for the hoverer, Kruskal coordinates do not belong to that class (not the whole chart at least) but I believe that Schwarzschild coordinates do, and the fact that they are singular at the Schwarzschild horizon reflects the fact that this is also a relative (apparent? ) horizon with respect to this observer in the sense that he cannot see past it.

 

[PeterDonis]

my perhaps idiosyncratic use of "an event that never happen for an observer" is "an event outside of any of that observer's future past lightcones". Obviously this is not the accepted meaning of the term

It's also not correct as you state it; the horizon is outside of all of the hovering observer's past light cones (because he can't ever see light from it), but it is not outside of all of his future light cones--he can quite easily send light signals into the black hole (just shine light radially inward).

 

[wabbit]

It's also not correct as you state it; the horizon is outside of all of the hovering observer's past light cones (because he can't ever see light from it), but it is not outside of all of his future light cones--he can quite easily send light signals into the black hole (just shine light radially inward).

Sorry I hesitated about that formulation for the reason you say but could not find a better one. I meant "all his future past light cones" as "all the past light cones he will have in the future", i.e. the event will never be in his past light cone, however long he waits.

I wish I knew a better way to say this.

 

[PeterDonis]

be wary of transforming to alternate coordinate systems and then misinterpreting them because the new coordinates have obscure meanings.

You don't have to transform to an alternate coordinate system to misinterpret the meaning of coordinates. You can do it quite easily with the original coordinates--Schwarzschild coordinates, in this case--as you have done. See below.

As for the postulates of General Relativity, they are just the equivalence principal and that physical laws must be expressed in tensor form so they have the same form in all coordinate systems. Don't distract from the real issue.

We aren't. You are, by focusing on a particular coordinate chart instead of focusing on invariants. You are entirely correct about what the postulates of GR are; but you are entirely incorrect about their implications for this particular scenario. Those implications are not "Schwarzschild coordinates tell us about the real physics". They are "forget about coordinates; look at invariants". And when you look at invariants, you find that objects can fall into black holes, and black holes can form, regardless of the fact that Schwarzschild coordinates cannot cover those events.

Meaningful quantities must be able to be measured.

Correct.

They do not have to be scalars.

Incorrect. Scalars are the only things we can measure. We can't measure vectors and tensors; we can only measure their components, which are scalars formed by contracting them with other vectors and tensors. The energy we measure for an object, for example, is the contraction of its 4-momentum with our 4-velocity--i.e., a scalar. We view it as one component of a 4-vector because, if we combine it with three other scalars--the contractions of each of our three spatial basis vectors with the object's 4-momentum--we can treat those four numbers as a single geometric object that transforms appropriately between frames. But "transforming between frames" means changing which set of 4 basis vectors (one timelike and three spacelike) we contract the 4-momentum with to get its components--i.e., it's just switching one set of 4 scalars for another in a particular way that we call a "coordinate transformation". None of this in any way implies that we actually measure anything but scalars.

 

[PeterDonis]

I hesitated about that formulation for the reason you say but could not find a better one. I meant "all his future (past light cones)" i.e. the reunion of all the past light cones corresponding to a given future point in his wordline.

Ah, I see. Yes, this is correct.

 

[PeterDonis]

the fact that they are singular at the Schwarzschild horizon reflects the fact that this is also a relative (apparent? ) horizon with respect to this observer in the sense that he cannot see past it.

This is fine as long as you remember that the words "relative" and "apparent" are technical terms in this connection--i.e., they don't have the implications that those words would normally be taken to have. The horizon isn't "relative" in the usual sense of the term because all observers share the same one--it doesn't change from observer to observer. The horizon also isn't "apparent" because it's not locally detectable; it's a global property of the spacetime.

There is, btw, another kind of horizon which is locally detectable and which is called an "apparent horizon"--Hawking originally came up with the term. This is a surface at which, locally, radially outgoing light does not move outward, but stays "in the same place" (I'm waving my hands here because it's hard to express this in ordinary language--the mathematics gives a much more precise definition of what this means). For a stationary black hole, i.e., one whose mass is the same forever, the apparent horizon coincides with the absolute horizon (the global property of the spacetime). But this is not always the case for black holes which gain mass (by objects falling in) or lose mass (by Hawking radiation).

 

[PAllen]

That is precisely my point - be wary of transforming to alternate coordinate systems and then misinterpreting them because the new coordinates have obscure meanings.

except that the coordinates you label as meaningful for an external observer of a BH correspond precisely to Rindler coordinates for a uniformly accelerating rocket. If you believe the one 'never' you must believe the other if you are logically consistent. The other logically consistent position - accepted by almost all physicists - is that both 'nevers' are coordinate artifacts, not something physically meaningful. Both 'never's are related to something physical though:

1) A transmitter 1/a behind a uniformly accelerating rocket cannot send a signal to the rocket.
2) A transmitter at a BH horizon can't send a signal to an external observer.

Note, there is no 'never' in the description of measurements, and no statement dropped balls don't reach the (Rindler/BH) horizon.

As for the postulates of General Relativity, they are just the equivalence principal and that physical laws must be expressed in tensor form so they have the same form in all coordinate systems. Don't distract from the real issue.

No, the physical content of general covariance is that the physics is what is independent of the coordinate choice.

You are confusing GR with quantum mechanics. Meaningful quantities must be able to be measured. They do not have to be scalars. So yes, time has meaning although it is different in different coordinate systems. Same with say, four-velocity, or energy or whatever. Try not to get distracted with too many complications that you have learned, and focus on the issue at hand.

All measurement processes in GR are defined by contractions, integrals of contracted quantities, or equivalent (e.g. tetrad basis of a given instrument).

 

[Asher Weinerman]

Scalars are the only things we can measure.

Incorrect. We can set up two clocks that are synchronized in our reference frame a certain distance apart, and measure the time between events that happen at each clock, say an infalling observer passing clock 1 and then clock 2. This is not a scalar and will be different in some other reference frame, but who cares? By the way, energy is not a scalar, but one component of a 4-vector. The contraction you described is a scalar which is the energy measured by a specific observer. All reference frames will observe Fred measuring an invariant energy E, but E itself is not invariant.

Anyway, back to crossing event horizons. Yes, I think it occurs in finite time because it is a two-body problem and breaks spherical symmetry, and the horizon engulfs the infalling observer. Appreciate if someone can substantiate that.

 

[PAllen]

Incorrect. We can set up two clocks that are synchronized in our reference frame a certain distance apart, and measure the time between events that happen at each clock, say an infalling observer passing clock 1 and then clock 2. This is not a scalar and will be different in some other reference frame, but who cares? By the way, energy is not a scalar, but one component of a 4-vector. The contraction you described is a scalar which is the energy measured by a specific observer. All reference frames will observe Fred measuring an invariant energy E, but E itself is not invariant.

Anyway, back to crossing event horizons. Yes, I think it occurs in finite time because it is a two-body problem and breaks spherical symmetry, and the horizon engulfs the infalling observer. Appreciate if someone can substantiate that.

But a frame is not a measurement. It is (at most) a convention applied to a series of measurements. A clock measuring when something passes by is an invariant by being the integral of interval up to a particular interaction. The non-invariant construct leading to coordinates is how you correlated the various measurements of individual clocks into a system. This latter is not a a measurement at all, but a convention. Any measurement by any instrument is an invariant. The (totally non-unique) way you assemble these produces a coordinate system.

 

[wabbit]

. Anyway, back to crossing event horizons. Yes, I think it occurs in finite time because it is a two-body problem and breaks spherical symmetry, and the horizon engulfs the infalling observer. Appreciate if someone can substantiate that.

While I have no issue with this for a large mass, I find it hard to believe for something of negligible mass relative to the black hole. The increase in surface horizon is tiny and for a large black hole the increase in radius is even tinier. There must be something that is small enough it does behave like a test particle, no?

 

[PAllen]

Anyway, back to crossing event horizons. Yes, I think it occurs in finite time because it is a two-body problem and breaks spherical symmetry, and the horizon engulfs the infalling observer. Appreciate if someone can substantiate that.

A test clock reaches a BH horizon in finite time measured by that clock (an actual measurement). This is a trivial calculation, no need for two body problem. Another (composite) measurement is that for any external observer, there is a precise time such that a flash occurring then is observed by an infaller at the moment of horizon crossing.

It is true that for a body of finite mass merging with a BH, the horizon expands to 'engulf' the object, and there is then a 'ring down' of the BH before the horizon settles down. However, if you believe coordinate artifices, the exact same approach that claims a test body 'never' reaches the horizon would say that infalling body was 'never' quite engulfed by the expanding horizon. This relates to the actual physical statement that no part of the infalling body can send a message to an external observer from the event of the horizon crossing it (whatever its shape). Coordinate systems that label such events (with past causal connection to external observers, but no future causal connection) , as t 'beyond infinity', will break down - simply not cover these events. This means nothing more than that they ill suited (for any observer) as a means of describing such events (that do occur).

 

[wabbit]

This is fine as long as you remember that the words "relative" and "apparent" are technical terms in this connection--i.e., they don't have the implications that those words would normally be taken to have. The horizon isn't "relative" in the usual sense of the term because all observers share the same one--it doesn't change from observer to observer. The horizon also isn't "apparent" because it's not locally detectable; it's a global property of the spacetime.

There is, btw, another kind of horizon which is locally detectable and which is called an "apparent horizon"--Hawking originally came up with the term. This is a surface at which, locally, radially outgoing light does not move outward, but stays "in the same place" (I'm waving my hands here because it's hard to express this in ordinary language--the mathematics gives a much more precise definition of what this means). For a stationary black hole, i.e., one whose mass is the same forever, the apparent horizon coincides with the absolute horizon (the global property of the spacetime). But this is not always the case for black holes which gain mass (by objects falling in) or lose mass (by Hawking radiation).

Yes, I guess the elusive relative meaning of horizon I am circling - one we can apply operationally - is in essence indistinguishable from "infinity". For a given observer, the region of spacetime he can (or will) receive signals from is I think, at least in some cases such as Schwarzschild or Rindler, a well defined (?) open submanifold of the full spacetime, and this could as well be the whole of spacetime from his viewpoint - ripping off the rest or transforming it in anyway would be unknowable to him (though he could make reasonable inferences about it by for instance assuming that his spacetime is part of a globally homogeneous one etc). This submanifold, "his spacetime", goes off to infinity in various ways e,g. for us a cosmological horizon and various black hole horizons.

Hawking's definition is a little different however, but how much so I am not sure. What does the light emitted towards us from a galaxy right at the Hubble horizon do? It certainly isn't "not moving" wrt that galaxy but it is not moving towards us either - not getting closer to us that is, so we might say that it "stays at the same place" relative to us, though this doesn't sound very meaningful (nor does "light stays in the same place" in Hawking's definition to be honest, it seems more of a wording aiming to capture something similar)

 

[JoeMath]

I see a fundamental problem with current discussion of black hole physics. Everyone starts with the assumption that there exists a theoretical black hole that is static and unchanging. This cannot be true. The only realistic origin of a black hole is through stellar collapse. If we observe this collapse from a safe distance (whatever that may be) we would observe a collapsing mass that appears to be cooling due to the gravity gradient and resulting time dilation. The surface would appear to slow down and stop before “crossing” the Schwarzschild radius. The usual response involves a coordinate change to a world line consistent with the surface of the collapsing core. But now we have a big problem. Because of the intense gravity just above the Schwarzschild radius and the surface of the core, Hawking radiation would bombard the surface of the core and knock particles away. There is also the cosmic background radiation that would gain tremendous energy at the surface of the core and because of time dilation, greatly increased photon flux. The result would be the same: material on the surface of the core would be knocked away or annihilated and the total mass of the core would decrease. This would result in a decrease in the Schwarzschild radius. The surface of the core would never catch up with the decreasing Schwarzschild radius. To an outside observer, the core would appear as we would expect a black hole to look because the surface of the core would be only very slightly larger than the Schwarzschild radius and this process would take gazillions of years. But the actual surface would always be just a hair above the Schwarzschild radius.

 

[wabbit]

I see a fundamental problem with current discussion of black hole physics. Everyone starts with the assumption that there exists a theoretical black hole that is static and unchanging. This cannot be true. The only realistic origin of a black hole is through stellar collapse. If we observe this collapse from a safe distance (whatever that may be) we would observe a collapsing mass that appears to be cooling due to the gravity gradient and resulting time dilation. The surface would appear to slow down and stop before “crossing” the Schwarzschild radius. The usual response involves a coordinate change to a world line consistent with the surface of the collapsing core. But now we have a big problem. Because of the intense gravity just above the Schwarzschild radius and the surface of the core, Hawking radiation would bombard the surface of the core and knock particles away. There is also the cosmic background radiation that would gain tremendous energy at the surface of the core and because of time dilation, greatly increased photon flux. The result would be the same: material on the surface of the core would be knocked away or annihilated and the total mass of the core would decrease. This would result in a decrease in the Schwarzschild radius. The surface of the core would never catch up with the decreasing Schwarzschild radius. To an outside observer, the core would appear as we would expect a black hole to look because the surface of the core would be only very slightly larger than the Schwarzschild radius and this process would take gazillions of years. But the actual surface would always be just a hair above the Schwarzschild radius.

Wow. Is this borne out by calculations? I mean, not the fact that we see a fading shell, but the Hawking radiation bombardment and core being materially affected by that, this sounds rather dramatic. But I'm probably misunderstanding your description.

Hawking radiation from a stellar mass black hole is a very tiny effect, overwhelmed by absorption of CMB photons - well, as I understand it - so this intense gravity doesn't seem relevant. The radiation comes from the horizon and is determined by its surface, so againn its hard to picture what you describe.

 

[PAllen]

I see a fundamental problem with current discussion of black hole physics. Everyone starts with the assumption that there exists a theoretical black hole that is static and unchanging. This cannot be true. The only realistic origin of a black hole is through stellar collapse. If we observe this collapse from a safe distance (whatever that may be) we would observe a collapsing mass that appears to be cooling due to the gravity gradient and resulting time dilation. The surface would appear to slow down and stop before “crossing” the Schwarzschild radius. The usual response involves a coordinate change to a world line consistent with the surface of the collapsing core. But now we have a big problem. Because of the intense gravity just above the Schwarzschild radius and the surface of the core, Hawking radiation would bombard the surface of the core and knock particles away. There is also the cosmic background radiation that would gain tremendous energy at the surface of the core and because of time dilation, greatly increased photon flux. The result would be the same: material on the surface of the core would be knocked away or annihilated and the total mass of the core would decrease. This would result in a decrease in the Schwarzschild radius. The surface of the core would never catch up with the decreasing Schwarzschild radius. To an outside observer, the core would appear as we would expect a black hole to look because the surface of the core would be only very slightly larger than the Schwarzschild radius and this process would take gazillions of years. But the actual surface would always be just a hair above the Schwarzschild radius.

There are two responses to this. First, if you assume classical physics (thus no Hawking radiation - a purely quantum effect), there are analytic solutions and general proofs that the horizon forms (and so does some type of singularity). CMB only hastens the effects (infinitesimally). It does not knock anythying away from the horizon, instead it simply makes horizon formation slightly faster as it adds blueshifted energy to the collapsing body.

As for accounting for Hawking radiation, a rigorous treatment is indeed very complex. However, a substantive consensus is the that the horizon forms anyway (whether it produces a firewall at or before the Page time is a separate issue). A review of the consensus and example of a detailed computation is:

http://arxiv.org/abs/0906.1768

 

[PeterDonis]

We can set up two clocks that are synchronized in our reference frame a certain distance apart, and measure the time between events that happen at each clock, say an infalling observer passing clock 1 and then clock 2. This is not a scalar and will be different in some other reference frame

Incorrect. It is a scalar and it will be the same in all reference frames. This should be obvious (what you've described is simply the invariant length along a particular timelike curve between two given events). If you disagree, please show your work: give the explicit calculations in two different frames and demonstrate how they are different.

The contraction you described is a scalar which is the energy measured by a specific observer.

Yes, because that's what is actually measured. "Energy" is only measurable by a specific observer, and the energy he measures is a scalar. There is no such thing as "measured energy that is not measured by any specific observer".

 

[PeterDonis]

For a given observer, the region of spacetime he can (or will) receive signals from is I think, at least in some cases such as Schwarzschild or Rindler, a well defined (?) open submanifold of the full spacetime

Yes; in fact this is always true, with the one caveat that the open manifold may be the full spacetime (for example, it is the full spacetime for an inertial observer in Minkowski spacetime).

this could as well be the whole of spacetime from his viewpoint

Not really. For cases where it is not the whole of spacetime, the observer can tell that it isn't by the fact that there are incomplete geodesics in the region of spacetime that he can see. For example, a Rindler observer in Minkowski spacetime can tell that the geodesic paths of objects that he releases into free fall are incomplete with respect to his region of spacetime: the portion of them within his region (i.e., up to his Rindler horizon) has a finite length, but there is nothing physically preventing the object from going further (i.e., no infinite spacetime curvature or other barriers). Another way of saying this is that the Rindler observer can tell that the portion of spacetime he can see can be analytically extended; he can in fact compute that the maximal analytic extension is the full Minkowski spacetime.

What does the light emitted towards us from a galaxy right at the Hubble horizon do?

It stays at the same proper distance from us, which could be thought of as "staying in the same place" relative to us. But it does not "stay in the same place" in comoving coordinates; its comoving spatial coordinate relative to us gets smaller (whereas the comoving coordinate of the galaxy emitting the light stays the same).

However, neither of these senses of "staying in the same place" are the one I was trying to describe. Let me give the more technical definition so I can make clear why it's not the same. The technical definition of an apparent horizon uses the notion of a "trapped surface". A trapped surface (more precisely, an "outer marginally trapped surface") is a spherical spacelike 2-surface at which the expansion scalar of the congruence of outgoing null geodesics is zero. "Congruence of null geodesics" means the family of light rays that are all moving radially outward from the surface, and the "expansion scalar" is a scalar invariant that can be computed for any congruence of timelike or null geodesics; for details, see, for example, here. An "apparent horizon" is then a 3-surface formed by a continuous collection of trapped surfaces. This 3-surface may be null, spacelike, or timelike, and may even switch from one to the other, depending on the details of the specific spacetime.

Associated with any black hole, there will be an apparent horizon in the above sense, which may or may not coincide with the absolute horizon (in general, it won't whenever the hole is gaining or losing mass). But there is no apparent horizon at all associated with the Hubble radius; there are no spherical 2-surfaces there that even come close to being trapped surfaces (in fact, since as far as we know our universe is spatially flat, spherical 2-surfaces are indistinguishable, with regard to outgoing light rays, from spherical 2-surfaces in Minkowski spacetime).

 

[PeterDonis]

Because of the intense gravity just above the Schwarzschild radius and the surface of the core, Hawking radiation would bombard the surface of the core and knock particles away.

The term "intense gravity" is misleading; spacetime curvature at the Schwarzschild radius gets smaller as the mass of the collapsing object gets bigger, and Hawking radiation gets less intense. For a black hole of a few solar masses, Hawking radiation at the Schwarzschild radius is negligible, with a temperature orders of magnitude less than the temperature of the CMB.

There is also the cosmic background radiation that would gain tremendous energy at the surface of the core and because of time dilation, greatly increased photon flux.

No, this is not correct. The incoming CMB radiation would appear blueshifted to an observer hovering just above the Schwarzschild radius; but the collapsing object is not hovering, it's free-falling inward. To an observer free-falling inward through the Schwarzschild radius, incoming CMB radiation actually appears redshifted, not blueshifted. (It will still add mass to the collapsing object, as PAllen says, and so hasten the collapse a bit.)

The surface of the core would never catch up with the decreasing Schwarzschild radius.

Incorrect. See above.

 

[Asher Weinerman]

Incorrect. It is a scalar and it will be the same in all reference frames. This should be obvious (what you've described is simply the invariant length along a particular timelike curve between two given events). If you disagree, please show your work: give the explicit calculations in two different frames and demonstrate how they are different

Sure. Let the proper time between two events at a fixed location r₁ be τ. Trivial integration of the Schwartzchild metric yields the result that an observer at radial coordinate r₂ will measure the time between the two events to be τ√(1-2GM/r₂)/√(1-2GM/r₁). Similarly an observer at position r₃ will measure the time between the two events to be τ√(1-2GM/r₃)/√(1-2GM/r₁). These observers measure time at location r₁ by having clocks synchronized to their wristwatch at location r₁. Obviously all these observers are measuring different times because time is not a scalar. Nor is energy. Why are we arguing over these nit-picky details and avoiding discussion of the relevant ideas. It's like that Monty Python skit where we just contradict each other and get nowhere.Incorrect, Incorrect, Incorrect.

 

[PeterDonis]

Obviously all these observers are measuring different times because time is not a scalar.

No, they are measuring different times because they are measuring different scalars; each one measures the scalar invariant length between two events on his own worldline, i.e., his own proper time, which is a scalar. The scalars are different along different worldlines because of the curvature of spacetime.

Why are we arguing over these nit-picky details

You're free to concede that scalars are the only things we can measure, which was what you originally objected to in this sub-thread, at any time.

and avoiding discussion of the relevant ideas.

You're free to respond to the other things that have been said in this thread--for example, the other things I said in post #65 that you didn't respond to when you picked up on "scalars are the only things we can measure". Or the responses PAllen has given.

 

[PAllen]

Sure. Let the proper time between two events at a fixed location r₁ be τ. Trivial integration of the Schwartzchild metric yields the result that an observer at radial coordinate r₂ will measure the time between the two events to be τ√(1-2GM/r₂)/√(1-2GM/r₁). Similarly an observer at position r₃ will measure the time between the two events to be τ√(1-2GM/r₃)/√(1-2GM/r₁). These observers measure time at location r₁ by having clocks synchronized to their wristwatch at location r₁. Obviously all these observers are measuring different times because time is not a scalar. Nor is energy. Why are we arguing over these nit-picky details and avoiding discussion of the relevant ideas. It's like that Monty Python skit where we just contradict each other and get nowhere.Incorrect, Incorrect, Incorrect.

These details are important, and the idea that measurements in GR are invariants is fundamental. You have described two different measurements along two different world lines. Each is invariant. Once you specify the simultaneity used to define the corresponding events on the different world lines, then you can compute these two different time interval measurements in Kruskal coordinates and get the same result.

 

[Haelfix]

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.

Yea, this is not what happens. Unfortunately I had forgotten how involved the actual analysis is regarding this particular problem and was not able to find an easy reference and was unsuccessful when I tried my hand at solving it just now. I remember that the answer depends on details of how quickly you lower the pole into the hole, and the material properties of the rigid rod. The point I wanted to get across was at some place near the horizon (likely slightly before) one calculates (when done correctly) that the speed of sound within the material exceeds the speed of light and therefore the rod would have to break. This removes any paradox one might have about whether the head and feet might agree.

 

[wabbit]

Not really. For cases where it is not the whole of spacetime, the observer can tell that it isn't by the fact that there are incomplete geodesics in the region of spacetime that he can see. For example, a Rindler observer in Minkowski spacetime can tell that the geodesic paths of objects that he releases into free fall are incomplete with respect to his region of spacetime: the portion of them within his region (i.e., up to his Rindler horizon) has a finite length, but there is nothing physically preventing the object from going further (i.e., no infinite spacetime curvature or other barriers). Another way of saying this is that the Rindler observer can tell that the portion of spacetime he can see can be analytically extended; he can in fact compute that the maximal analytic extension is the full Minkowski spacetime.

Right, although this is not a measurable property I think : there would be many possible extensions compatible with the finite amount of information he has about his observable patch.
But this doesn't really matter : what you are saying is that he can determine that some extension is possible, and perhaps some properties of spacetime at least "a little bit beyond" his horizon. In the black hole case, we infer that something is (or was) there by the shape of the horizon etc. - this certainly looks like a more reasonable explanation than "spacetime just stops at that unreachable limit".
I am not sure however that the distinction is strictly objective - using intelligent observers using reasonable inference etc. was not a good idea in my example, the definition should rest on the physical information alone. In that sense, a spacetime that consists exactly of the Schwarzschild exterior region and nothing else is a possibility and it would appear the same (provide the exact same information) to the observer.
It does seem like a weird possibility I must agree, but I don't see (yet) that it would contradict GR.

It stays at the same proper distance from us, which could be thought of as "staying in the same place" relative to us. But it does not "stay in the same place" in comoving coordinates; its comoving spatial coordinate relative to us gets smaller (whereas the comoving coordinate of the galaxy emitting the light stays the same).

However, neither of these senses of "staying in the same place" are the one I was trying to describe. Let me give the more technical definition so I can make clear why it's not the same. The technical definition of an apparent horizon uses the notion of a "trapped surface". A trapped surface (more precisely, an "outer marginally trapped surface") is a spherical spacelike 2-surface at which the expansion scalar of the congruence of outgoing null geodesics is zero. "Congruence of null geodesics" means the family of light rays that are all moving radially outward from the surface, and the "expansion scalar" is a scalar invariant that can be computed for any congruence of timelike or null geodesics; for details, see, for example, here. An "apparent horizon" is then a 3-surface formed by a continuous collection of trapped surfaces. This 3-surface may be null, spacelike, or timelike, and may even switch from one to the other, depending on the details of the specific spacetime.

Associated with any black hole, there will be an apparent horizon in the above sense, which may or may not coincide with the absolute horizon (in general, it won't whenever the hole is gaining or losing mass). But there is no apparent horizon at all associated with the Hubble radius; there are no spherical 2-surfaces there that even come close to being trapped surfaces (in fact, since as far as we know our universe is spatially flat, spherical 2-surfaces are indistinguishable, with regard to outgoing light rays, from spherical 2-surfaces in Minkowski spacetime).

Thanks for the detailed explanation, I need to digest this now - juggling horizons is a delicate business : )

 

[PeterDonis]

this is not a measurable property I think : there would be many possible extensions compatible with the finite amount of information he has about his observable patch.

More precisely: there would be many possible extensions, each of which would correspond to a different stress-energy tensor in the region of spacetime outside his observable patch. The analytic extension I described is the one the Rindler observer can make with the assumption that the SET is the same in the rest of spacetime as it is in his observable patch (i.e., vacuum). With that assumption, and with the requirement that the spacetime curvature at the boundary must match (i.e., for the Rindler observer case spacetime must be flat), the extension is unique.

In the black hole case, we infer that something is (or was) there by the shape of the horizon etc.

Not just that. We infer it by the fact that geodesics reach the boundary (the horizon) after a finite length. Where do they go after that? When an object falls to the horizon in a finite amount of its own proper time, what happens to it after that?

this certainly looks like a more reasonable explanation than "spacetime just stops at that unreachable limit".

I think "more reasonable" is an understatement, but it's true that there is judgment involved here. See below.

a spacetime that consists exactly of the Schwarzschild exterior region and nothing else is a possibility and it would appear the same (provide the exact same information) to the observer.

Correct. The Einstein Field Equation is local, so it is always mathematically consistent to claim that the solution on any open submanifold is a "complete" solution and ignore any possible extension. The question is whether such a claim is physically reasonable.

I don't see (yet) that it would contradict GR.

It wouldn't, in the sense that any open submanifold (such as the Schwarzschild exterior region) can in principle be viewed as a valid spacetime in its own right, without regard to any possible extension. Again, the question is whether such a model is physically reasonable. Is it physically reasonable to say that objects that fall to the black hole's horizon just "stop" after a finite amount of their own proper time? Most physicists believe that it isn't; that we should expect any physically reasonable spacetime to have geodesics that extend indefinitely, unless there is a curvature singularity--a place like $r = 0$ in Schwarzschild spacetime, where scalar invariants increase without bound. That doesn't happen at the horizon.

 

[wabbit]

With that assumption, and with the requirement that the spacetime curvature at the boundary must match (i.e., for the Rindler observer case spacetime must be flat), the extension is unique.

Right but he doesn't know that his spacetime is analytic, so this is not really an observation.

We infer it by the fact that geodesics reach the boundary (the horizon) after a finite length. Where do they go after that? When an object falls to the horizon in a finite amount of its own proper time, what happens to it after that?

We can infer it but he can't - he only sees objects receding farther and farther away, sthe proof of the finiteness is in the crossing, which he cannot observe nor prove.
One thing is, what a physicist could reasonably infer from plausible assuptions, but this is different from the information he objectively has as a physical system observing another. I confused things by not making this distinction - the two kind of conclusions are different.

Correct. The Einstein Field Equation is local, so it is always mathematically consistent to claim that the solution on any open submanifold is a "complete" solution and ignore any possible extension. The question is whether such a claim is physically reasonable.

While it may well be so in some cases, I am not claiming that it is in general a good assumption to make - though assumptions about things in principle unobservable really serve more I think as a useful and probably necessary tool to help model the observable - in themselves they can be highly plausible but they are I think more metaphysical than physical.

It wouldn't, in the sense that any open submanifold (such as the Schwarzschild exterior region) can in principle be viewed as a valid spacetime in its own right, without regard to any possible extension. Again, the question is whether such a model is physically reasonable. Is it physically reasonable to say that objects that fall to the black hole's horizon just "stop" after a finite amount of their own proper time? Most physicists believe that it isn't; that we should expect any physically reasonable spacetime to have geodesics that extend indefinitely, unless there is a curvature singularity--a place like $r = 0$ in Schwarzschild spacetime, where scalar invariants increase without bound. That doesn't happen at the horizon.

Agreed, I much prefer the assumption that the interior does exist. But the difference between a worldline ending in finite proper time at the horizon (actually in that view, never reaching it, or equivalently, accelerating away and "reaching" the border which is not part of the manifold, in finite time), and one ending in finite time at the singularity, is perhaps not that drastic.

Even if the usual description does sound more reasonable, I find it interesting to pursue the implications of the alternative view beyond that of physical information, to a what if scenario where for instance spacetime does stop before the Schwarzschild horizon.

One might also argue that singularities are unphysical and that the interior solution is just as realistic as the white hole in the analytic extension of that solution. It is not clear to me the case can be settled beyond arguments that either one is more unreasonable.

Also, if the continuation argument here and in the Rindler case is very strong, then why is the white hole part of the extended black hole solution often described as unphysical?

 

[PAllen]

Also, if the continuation argument here and in the Rindler case is very strong, then why is the white hole part of the extended black hole solution often described as unphysical?

Because it doesn't occur as part of any evolution from reasonable initial conditions. It only occurs for an eternal BH and a spacetime which is cosmologically implausible (eternal, static).

A BH is the natural evolution of collapse under a wide range of initial conditions. Lopping off part of the manifold (in the theoretical description of the 'reality') simply means refusing to specify what happens to a given piece of matter following locally SR laws, after a locally arbitrary point. This is fundamentally different from from saying we can't predict, with current theory, what happens to matter at Planck scale energies (that would arise even before the singularity). At the horizon, local physics is completely ordinary, so you are just saying you will stop the evolution at an arbitrary moment. Note, there is no possible white hole extension for a BH arising from collapse.

 

[PeterDonis]

he doesn't know that his spacetime is analytic

I'm not sure what you mean by this. "Spacetime is analytic" is a statement about the mathematical model, not about the physics. Physically, it basically corresponds to the statement that spacetime is a continuum. That is something that can be tested by observation (and so far, all observations support it--there are speculations that spacetime might become "granular" at the Planck scale, but those are just that: speculations).

We can infer it but he can't

Um, what? "We" don't have any information that "he" doesn't have; we're all outside the horizon. Anyone outside the horizon can run the same computations we are running that show that infalling geodesics reach the horizon in a finite proper time.

sthe proof of the finiteness is in the crossing

Incorrect; you can show that the proper time is finite by taking limits as the horizon is approached, without ever actually reaching it.

which he cannot observe nor prove.

Neither can "we"; "we" are outside the horizon just as "he" is.

the difference between a worldline ending in finite proper time at the horizon (actually in that view, never reaching it, or equivalently, accelerating away and "reaching" the border which is not part of the manifold in finite time), and one ending in finite time at the singularity, is perhaps not that drastic

No, there is a big difference: spacetime curvature increases without bound as $r = 0$ is approached. Spacetime curvature at $r = 2M$ is finite. Again, this can be observed physically: tidal gravity (which is what spacetime curvature is, physically) is perfectly well-behaved at the horizon (or as the horizon is approached--again, you can make measurements arbitrarily close to the horizon and take limits to show that tidal gravity is finite at the horizon).

One might also argue that singularities are unphysical and that the interior solution is just as realistic as the white hole in the analytic extension of that solution

The interior is more than just the singularity; even if the singularity at $r = 0$ is considered unphysical (which I personally think it is--to me, spacetime curvature increasing without bound at $r = 0$ really just means new physics, such as quantum gravity, will come into play before $r = 0$ is reached), there is a whole interior region from $r = 2M$ down to $r = 0$ (or down to some very small value of $r$ where new physics comes into play) which is a perfectly good spacetime without any "unphysical" properties.

The above is also true of the white hole interior region, but there is a big difference between the black hole interior and the white hole interior: the former is still present in a realistic model where a black hole forms by gravitational collapse. The latter is not present in any such model. See below.

why is the white hole part of the extended black hole solution often described as unphysical?

Because it would require the white hole singularity to be "built in" to the universe from the start; there is no way for it to arise from some ordinary process, the way a black hole singularity can arise from the ordinary process of gravitational collapse.

In a more realistic model in which a black hole is formed by gravitational collapse, there is still a vacuum exterior region (outside the collapsing object and outside the horizon) and a vacuum black hole interior region (outside the collapsing object but inside the horizon--the horizon forms during the collapse and the collapsing object falls through it, leaving behind an interior vacuum), plus a non-vacuum region containing the collapsing matter. But in this model, those three regions are the entire spacetime: there is no possible analytic extension. (This is because the portion of spacetime where the white hole would be in the maximally extended Schwarzschild vacuum model is "covered up" by the non-vacuum region; geodesics which would have been extendible into the white hole region instead end up inside the collapsing matter and are fully extendible within that region.)

 

[wabbit]

About the analyticity yes this was a confusion on my part, analyticity (in the mathematical sense) is not necessary here to reach the conclusion, only that spacetime is a differential manifold and that the metric is twice differtentiable or something like that I suppose. It is still a strong assumption, that we know from experience to be valid as an effective description well above Planck scale, but this is rather tangential to the case here : we are in any case discussing within the domain of validity of that assumption, and I was not envisioning that kind of breaking down of spacetime at the horizon.

For the singularity what I meant is that the viewpoint of spacetime stopping at the horizon would mean that no black hole singularity exists, which could be an appealing conclusion within GR. But this is in any case a weak argument, accepting it at the GR level and investigating its quantum properties certainly looks more productive.

As for the white hole I wasn't aware that there was no counterpart to it in a collapse model - so the solution can be extended only after the collapse ends, not before - weird and interesting, thanks for the explanation.

 

[Haelfix]

For the singularity what I meant is that the viewpoint of spacetime stopping at the horizon would mean that no black hole singularity exists, which could be an appealing conclusion within GR. But this is in any case a weak argument, accepting it at the GR level and investigating its quantum properties certainly looks more productive.

As for the white hole I wasn't aware that there was no counterpart to it in a collapse model - so the solution can be extended only after the collapse ends, not before - weird and interesting, thanks for the explanation.

The notion that you could simply excise the interior in classical GR doesn't make sense. This violates the equivalence principle, b/c the near horizon region is always Rindler space which is physically traversable. There are also very powerful theorems due to Hawking that guarantee's the existence of a singularity.

There is a big physical difference between using a singular coordinate system (Schwarschild coordinates) which have coordinate singularities (the time coordinate is not well behaved at the horizon) which is readily fixed by using Kruskal coordinates, and where we know that there are no coordinate independent quantities that diverge (proper time to the horizon is finite, curvature invariants are well behaved etc) and the notion that we must require quadrant III and quadrant IV of a maximally extended Schwarzschild spacetime to be included. These quadrants have thermodynamic properties that are physically unstable and as has been said, there is no well posed initial value problem that leads to their existence*, therefore it is perfectly reasonable to exclude these solutions on physical grounds.

*I'm actually not entirely sure this is true, I believe there are solutions within eternal inflation theory that output eternal black holes, and I am unsure what happens to the quadrants there, but this is perhaps beyond the scope.