# Is it possible that black holes do not exist?

#### JustinLevy

I agree about the singularity, but I also question the EH itself. Maybe you can help me find the problem with my logic.

1. Do you agree that all frames outside of the BH calculate that no mass ever crosses the EH (or, more specifically, mass crosses the EH at t=infinity)?

2. Do you agree that the EH does not expand until mass has crossed the EH (i.e. backreaction)?

If you concur with #1 and #2, then run the clock backwards in your mind and describe to me how this theoretical black hole formed in the first place. There are other problems that I have as well, but lets start here...
I think this discussion will take way too long if we approach it from this route. I suggest simplifying this as greatly as possible to get at the heart of the confusion here. We can discuss the underlying confusion without even needing GR. Quantum mechanics in flat spacetime + SR is enough.

Consider a Rindler horizon, and the associated Unruh radiation.

Just like in your openning paragraph, an observer far from the horizon and thus experiencing negligible proper acceleration (you could go to infinity to get your "infinity" observer as you call him) will see a object approaching the horizon take infinite time according to his watch to reach the horizon. He will also see the object bathed by an infinite amount of radiation before reaching the horizon.

Let's consider this scenario: described from an inertial frame, we see the global structure that is the rindler observer's event horizon. Imagine a baseball made of normal matter, and a baseball made of anti-matter. The baseball is on one side of the event horizon (inside the Rindler observer's past light cone), and the anti-baseball is on the other side. The baseball is thrown towards the horizon, and the antibaseball is in a trajectory so it is on the otherside to meet the baseball just as it crosses.

Alright, this is just good old flat-spacetime. What happens?

If the Rindler observer is always at constant acceleration, the baseball with never reach the horizon in any finite time on his watch. The baseball will be bathed with infinite amount of radiation.

From the baseball point of view, nothing spectacular happens as it crosses the horizon, and it reaches it in finite proper time. Upon crossing the horizon, it is annihilated, but the radiation from that anhihilation can never reach the Rindler observer.

Yes, to our intuition these sound contradictory.
But no, the math is consistent. Where our intuition fails is the expectation of "observer independence" of particle number.

Do we at least agree to this point?

The next issue of intuition failure is the fact that the Rindler observer's coordinate chart doesn't cover all of spacetime. So while what you are implying would be like claiming the "annihilation event" never occured according to the Rindler observer, that is not actually the case. He doesn't have a coordinate to assign to that event, according to his coordinate chart. That is completely separate from the claim that the event doesn't occur. He also cannot physically measure that it occurs, since the event is outside his light cone, but again, that is distinct from claiming the event didn't occur. For the event did, objectively occur.

So the second intuition issue is taking coordinate labels as too "physical". I have trouble describing this one well (it's come up in many other contexts with people asking questions on this site, and I'm often not able to help them see this clearly). So hopefully someone can word or describe it much better than I. Anyone? Please? I would like to learn how to describe this better.

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#### JustinLevy

bcrowell said:
This has been settled for a long time, in GR, and the answer is yes: http://en.wikipedia.org/wiki/Penrose...arity_theorems [Broken]
From the referenced Wiki article:
" ... It is still an open question whether time-like singularities ever occur in the interior of real charged or rotating black holes ..."
And
" ... Hawking's singularity theorem is for the whole universe, and works backwards-in-time: in Hawking's original formulation, it guaranteed that the Big Bang has infinite density. Hawking later revised his position in "A Brief History of Time" (1988) where he stated "There was in fact no singularity at the beginning of the universe" (p50). This revision followed from quantum mechanics, in which general relativity must break down at times less than the Planck time. Hence general relativity cannot be used to show a singularity."
You may be missing the point here.
bcrowell is correct. This has been settled for a long time, in GR, and the answer is yes.
There is no question whether, according to GR, time-like singularities form in the interior of blackholes formed from the infall of realistic starting conditions of stellar matter (ie. it isn't an artifact of very specific initial conditions).

The issue is instead, the real world clearly isn't completely classical as we have quantum mechanics. So in the real world, does a singularity form? Hawking isn't contradicting or revising, he is clarifying that his theorem was for GR and what happens in Quantum-Gravity is still not settled since we don't have a theory of quantum gravity that we can run the predictions with yet.

Maybe you already agree with all that, and we're on the same page. I just wanted to make sure.

That being said, most people wouldn't consider a theory of quantum-gravity that predicted singularities as being a consistent theory. GR's applicable domain does not cover short length scales as it is a classical theory, but quantum-gravity would have no excuse. Finding a theory predicts that it can't even predict further evolution (due to singularity), would make it not consistent and predictive. For this reason, there really isn't even debate amongst people looking for a theory of quantum-gravity whether there are black hole singularities "in the real world". They believe there exists a consistent theory of quantum-gravity (even if it hasn't been completely worked out / found yet), which without even getting details can make the prediction (just due to being predictive and consistent) as: no.

So there really isn't much debate on either side.

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#### DavidGTaylor

If there isn't debate, there needs to be. The border of a Black Hole/Schwarzschild object is not a wall, it is a marker where actual 'escape' requires light velocity. There is no reason why an object moving at orbital velocity for that S.O. would not simply pass the border and go into orbit. You maybe could make proclamations about relativistic distortion slowing things down so that normal physics doesn't apply, but think about this.
If our Universe has a mass (just a proposition) of 1.0E+53 kg, then the radius of an S.O with that mass would be about 1.535E+26m or 1.623E+10 light years. The gravity at that point would be 2.93E-10 m/s^2. Current theory says that distortion is infinite at that point?

#### JesseM

If there isn't debate, there needs to be. The border of a Black Hole/Schwarzschild object is not a wall, it is a marker where actual 'escape' requires light velocity. There is no reason why an object moving at orbital velocity for that S.O. would not simply pass the border and go into orbit.
What does S.O. stand for? Anyway, there is no "orbital velocity" at distances closer than the photon sphere, which is at a radius of 1.5 times the radius of the event horizon (i.e. there is a region between the photon sphere and the event horizon where it is possible to escape for an object traveling outward in the radial direction with sufficient speed, but it is impossible for anything to be in a circular orbit no matter how fast its speed in the nonradial direction) And once you are at or inside the "border" that is the event horizon, it's impossible for anything to escape (except for a hypothetical particle moving faster than light, but it's generally assumed these don't exist)

#### rjbeery

zhermes, thanks for the link. I read it all yet I still have problems.
We saw that the radial position of a test particle starting at radius r = 10m and t = 0 (for example) as a function of the particle’s proper time is a simple cycloid right down to r = 0, whereas if the same trajectory is described in terms of Schwarzschild coordinate time, the infalling object traverses through infinite coordinate time in order to reach the event horizon, and then traverses back through an infinite range of coordinate times until reaching r = 0 (in the interior) in a net coordinate time that is not too different from the elapsed proper time. In other words, the object goes infinitely far into the "future" (of coordinate time), and then infinitely far back to the "present" (also in coordinate time), and since these two segments must always occur together, we can "re-normalize" the round trip and just deal with the net change in coordinate time (for any radius other than precisely r = 2m).
I am not challenging the mathematical model of GR; I am applying common sense to what we can consider to be the real-world model of black holes. Be honest, does the above maneuver sit well with you? Claiming, basically, that infinity minus infinity gets us back to a finite proper time? It doesn't really matter because the author goes on to say
the "frozen star" interpretation does have the advantage of simplifying the topology, i.e., it allows us to exclude event horizons separating transfinite regions of spacetime. More importantly, by declining to consider the fate of infalling worldlines through the event horizon, we avoid dealing with the rather awkward issue of a genuine spacetime singularity at r = 0. Therefore, if the "frozen star" interpretation gave equivalent predictions for all externally observable phenomena, and was logically consistent, it would probably be the preferred view. The question is, does the concept of a "frozen star" satisfy those two conditions? We saw above that the idea of a frozen star as an empty region around which matter "bunches up" outside an event horizon isn't viable, because if nothing ever passes from the exterior to the interior of an event horizon (in finite coordinate time) we cannot accommodate infalling matter. Either the event horizon expands or it doesn't, and in either case we arrive at a contradiction unless the value of m inside the horizon increases, and does so in finite coordinate time.
I guess I don't know the strict definition of a "frozen star", but my "black hole region" contains no event horizon. It appears that the author, as I did, questions the process of growth of the event horizon, however his PRESUMPTION of the EH's existence forces him to reject the frozen star rather than rejecting the EH itself. Note that the author claims that, but for this "logical inconsistency", the frozen star would be the preferred view, and I am claiming that the inconsistency does not exist.

Furthermore, the author relies on the experience of the infalling frame as "evidence" that the frame is viable. Although admittedly speculative, I have suggested that the infalling masses are destroyed through the pre-Hawking radiation process.

#### JesseM

I am not challenging the mathematical model of GR; I am applying common sense to what we can consider to be the real-world model of black holes. Be honest, does the above maneuver sit well with you? Claiming, basically, that infinity minus infinity gets us back to a finite proper time?
But that's just in a particular coordinate system which is badly-behaved on the horizon, namely Schwarzschild coordinates. Other coordinate systems on the same spacetime don't say the coordinate time to reach the horizon is infinite--see Kruskal-Szekeres coordinates for example. And contrariwise, even in ordinary boring flat SR spacetime you can come up with coordinate systems that are badly-behaved at some boundary, in the sense that the coordinate time to cross that boundary is infinite--this is true of Rindler coordinates, an accelerating coordinate system in flat spacetime where it takes an infinite coordinate time for anything to cross the Rindler horizon (in fact there is a close analogy between these cases, if you plot lines of constant Rindler position and Rindler time in an ordinary Minkowski diagram where the vertical and horizontal axes are those of an inertial frame, it looks just like what you get if you plot lines of constant Schwarzschild position and Schwarzschild time in a diagram where the vertical and horizontal axes are those of Kruskal-Szkeres coordinates...compare the yellow diagram on the Kruksal-Szkeres wikipedia page with the second diagram on the Rindler horizon page).

Basically the lesson is, don't take "coordinate singularities", where some coordinate-based quantity like the time dilation of a clock relative to coordinate time goes to infinity, too seriously. The only singularities that are really physical are ones where some coordinate-invariant quantity goes to infinity, like singularities in the curvature of spacetime which occur at the center of a black hole. Likewise the only notion of "time" that's really physical is proper time along some worldline, and both Schwarzschild coordinates and Kruskal-Szekeres coordinates predict it only takes a finite proper time for an infalling observer to cross the horizon (though in terms of Schwarzschild coordinates, what you'd actually show is that in the limit as the distance between the falling observer and the horizon goes to zero as the coordinate time goes to infinity, the proper time approaches a particular finite value, the exact same value that in Kruskal-Szekeres coordinates is the proper time of the falling observer at the moment she crosses the horizon).

#### rjbeery

I respectfully disagree. Kruskal coordinates have problems with the OP also (remember, I am using the infinite observer's frame which presents Kruskal indeterminability). My OP is quite simple: if the black hole region evaporates in finite time, and mass crosses the EH in "infinite" time, then then black hole region is gone before any mass reaches it. It really doesn't need to be much more complicated than that, and the aesthetic advantages of this interpretation are indisputable (aren't they?).

#### JesseM

I respectfully disagree. Kruskal coordinates have problems with the OP also (remember, I am using the infinite observer's frame which presents Kruskal indeterminability).
What does "Kruskal indeterminability" mean? What exactly is indeterminable? Remember that Susskind's discussion of two different opinions on what happens to the falling observer had to do with speculations about quantum gravity, there are no such disagreements in classical general relativity (although certain observers may not be able to see certain events that others can see due to event horizons)

Also, note that observers don't really have "frames" in general relativity, except for locally inertial frames in regions of spacetime small enough that the curvature is insignificant. In SR we can talk about an observer's frame as just a shorthand for the inertial frame where they are at rest, since there is a unique way to construct an inertial frame for any inertial observer (aside from the position of the origin and the orientation of the spatial axes), and inertial frames are "preferred" in SR (the basic equations of SR like time dilation only hold in inertial frames, and the laws of physics obey the same equations in all inertial frames). But in GR the principle of "diffeomorphism invariance" means that the equations of GR hold in any arbitrary coordinate system, so there are an infinite number of different ways you could construct a coordinate system where a given observer in curved spacetime is at rest, and none of these different coordinate systems would be preferred over any other. So, talking about an observer's "frame" in a decent-sized region of curved spacetime does not pick out any unique coordinate system, and thus the term isn't really meaningful in this context, so physicists don't talk about the "frames" of GR observers.
rjbeery said:
My OP is quite simple: if the black hole region evaporates in finite time, and mass crosses the EH in "infinite" time, then then black hole region is gone before any mass reaches it.
What does "mass crosses the EH in infinite time" mean, though? Are you talking about coordinate time in some coordinate system, and if so which one? If the black hole evaporates in finite time, I don't think you can use Schwarzschild coordinates which are specific to the Schwarzschild spacetime which describes an eternal black hole. And even in this spacetime you can find coordinate systems (like Kruskal-Szkeres) where mass crosses the EH in a finite coordinate time.

Perhaps you're not talking about coordinate time at all, but rather what is seen visually by some observer outside the horizon (i.e. the proper time according to their own clock when they receive various light signals from events on the worldline of an object falling in). In the case of an eternal Schwarzschild black hole, it is true that if a falling clock is ticking as it falls in, the proper time for an external observer between seeing the light from successive ticks will get larger and larger, and the outside observer will never get to see the clock reading the exact time it read at the moment it crossed the horizon. However, for an evaporating black hole the situation is probably different, in theory the outside observer would see the light from the falling observer cross the horizon at the exact moment the outside observer sees the black hole evaporate to nothing (with the understanding that this theoretical answer would only hold if light waves were emitted continuously by the falling observer rather than in discrete photons, and if the outside observer could still see light at extremely high finite redshifts), as discussed in the last section of this entry from the Usenet Physics FAQ:
What about Hawking radiation? Won't the black hole evaporate before you get there?

(First, a caveat: Not a lot is really understood about evaporating black holes. The following is largely deduced from information in Wald's GR text, but what really happens-- especially when the black hole gets very small-- is unclear. So take the following with a grain of salt.)

Short answer: No, it won't. This demands some elaboration.

From thermodynamic arguments Stephen Hawking realized that a black hole should have a nonzero temperature, and ought therefore to emit blackbody radiation. He eventually figured out a quantum- mechanical mechanism for this. Suffice it to say that black holes should very, very slowly lose mass through radiation, a loss which accelerates as the hole gets smaller and eventually evaporates completely in a burst of radiation. This happens in a finite time according to an outside observer.

But I just said that an outside observer would never observe an object actually entering the horizon! If I jump in, will you see the black hole evaporate out from under me, leaving me intact but marooned in the very distant future from gravitational time dilation?

You won't, and the reason is that the discussion above only applies to a black hole that is not shrinking to nil from evaporation. Remember that the apparent slowing of my fall is due to the paths of outgoing light rays near the event horizon. If the black hole does evaporate, the delay in escaping light caused by proximity to the event horizon can only last as long as the event horizon does! Consider your external view of me as I fall in.

If the black hole is eternal, events happening to me (by my watch) closer and closer to the time I fall through happen divergingly later according to you (supposing that your vision is somehow not limited by the discreteness of photons, or the redshift).

If the black hole is mortal, you'll instead see those events happen closer and closer to the time the black hole evaporates. Extrapolating, you would calculate my time of passage through the event horizon as the exact moment the hole disappears! (Of course, even if you could see me, the image would be drowned out by all the radiation from the evaporating hole.) I won't experience that cataclysm myself, though; I'll be through the horizon, leaving only my light behind. As far as I'm concerned, my grisly fate is unaffected by the evaporation.

All of this assumes you can see me at all, of course. In practice the time of the last photon would have long been past. Besides, there's the brilliant background of Hawking radiation to see through as the hole shrinks to nothing.

(Due to considerations I won't go into here, some physicists think that the black hole won't disappear completely, that a remnant hole will be left behind. Current physics can't really decide the question, any more than it can decide what really happens at the singularity. If someone ever figures out quantum gravity, maybe that will provide an answer.)

#### DavidGTaylor

Lets get one thing straight. Velocity is not the exclusive element in escape velocity. There aren't Classic Gravitational theory speed traps at the border stopping object if it isn't going fast enough. The whole reason escape/orbital velocity works is because the momentum acquired by the escaping body allows it to coast completely out of the body's gravitational influence. But in a relativistic world a 1 kilo object does not need velocity of 299,792,458 m/s to escape. It needs a momentum of: 299,792,458 kg*m/sf. And if you were paying attention when Dr. Einstein was spouting that "Relativity" mass increase business in class last century, you would be able to calculate that a body with a rest mass of 1 Kg has momentum of 299,792,458 kg*m/s when it is going a little more than 2.11985E+08 m/s.

#### JesseM

Lets get one thing straight. Velocity is not the exclusive element in escape velocity. There aren't Classic Gravitational theory speed traps at the border stopping object if it isn't going fast enough. The whole reason escape/orbital velocity works is because the momentum acquired by the escaping body allows it to coast completely out of the body's gravitational influence.
The event horizon is not simply the distance where the "escape velocity" is the speed of light. "Escape velocity" is a concept from Newtonian physics which does not really explain what's going on in general relativity, although authors writing for a popular audience do sometimes simplify the concept of an event horizon in this way. In Newtonian physics, it's quite true that if the escape velocity at radius R is vR, an object traveling outward from just under R with v < vR can travel out past R, it just won't "escape" in the sense that it will eventually start falling back down. However in GR, where we are dealing with curved spacetime, the situation is totally different--for any event which occurs on or inside the event horizon, spacetime is curved in such a way that the entire future light cone of that event remains inside the horizon, so no worldlines passing through that event can even briefly stray outside the horizon. This can be illustrated nicely if we use Eddington-Finkelstein coordinates, where the future light cones of different events actually tilt over for events closer to the horizon (as discussed here), you can see visually that future light cones of events on or inside the horizon don't have any part outside the horizon:

[URL]http://www.etsu.edu/physics/plntrm/relat/eventho1.gif[/URL]

[URL]http://www.etsu.edu/physics/plntrm/relat/eventho2.gif[/URL]

(images from http://www.etsu.edu/physics/plntrm/relat/blackhl.htm)

DavidGTaylor said:
But in a relativistic world a 1 kilo object does not need velocity of 299,792,458 m/s to escape. It needs a momentum of: 299,792,458 kg*m/sf.
Again, you appear to be thinking in terms of classical escape velocity, but the event horizon cannot be understood in this way. Also, in relativity momentum is not given by (rest mass)*(velocity), it is given by (rest mass)*(velocity)*gamma, where gamma is the relativistic factor $$\frac{1}{\sqrt{1 - (v/c)^2}$$. So, in the limit as v approaches c, gamma (and thus the momentum of any object with nonzero rest mass) approaches infinity.

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#### rjbeery

JesseM said:
What does "mass crosses the EH in infinite time" mean, though? Are you talking about coordinate time in some coordinate system, and if so which one? If the black hole evaporates in finite time, I don't think you can use Schwarzschild coordinates which are specific to the Schwarzschild spacetime which describes an eternal black hole. And even in this spacetime you can find coordinate systems (like Kruskal-Szkeres) where mass crosses the EH in a finite coordinate time.
Pick any coordinate system you wish. Kruskal coordinates are not defined at the infinite observer mentioned in my OP. Are you aware of a coordinate system in which the infinite observer would calculate that a mass infalling toward an EH would do so in finite time? What else could I possibly mean when I say "crosses the EH in infinite time"?

Also, to say that Schwarzschild coordinates only apply to eternal black holes simply sounds like an ad hoc defense. Where did you come across this? Or is this personal supposition?
JesseM said:
Perhaps you're not talking about coordinate time at all, but rather what is seen visually by some observer outside the horizon (i.e. the proper time according to their own clock when they receive various light signals from events on the worldline of an object falling in). In the case of an eternal Schwarzschild black hole, it is true that if a falling clock is ticking as it falls in, the proper time for an external observer between seeing the light from successive ticks will get larger and larger, and the outside observer will never get to see the clock reading the exact time it read at the moment it crossed the horizon.
Light is Einstein's absolute system of measurement. There is no "trick" or "illusion" when light suggests that the mass does not cross the hypothetical event horizon. The proof of this is revealed by having the infalling mass accelerate itself back to the outside observer and compare clocks after any arbitrary (finite) amount of time has passed.

#### JesseM

Pick any coordinate system you wish. Kruskal coordinates are not defined at the infinite observer mentioned in my OP.
Kruskal coordinates cover observers at arbitrarily high finite radii, as do Schwarzschild coordinates. I don't think it is literally possible for a manifold with a metric defined on it (like a Riemannian manifold or a pseudo-Riemannian manifold edit: I seem to have gotten confused about the difference between a metric and a metric tensor, see this thread, but just replace 'pseudo-Riemannian manifold' with 'manifold with a pseudo-Riemannian metric') to actually include two points that are an infinite distance apart, see for example item (ii) on this page which I think is saying that any two points must be connected by a curve with a finite distance in the affine parameter along that curve. Anyway, even if a GR spacetime with a metric defined on it literally could include points at infinite distance from some other points (like points in the immediate neighborhood of the black hole), you'd still be able to represent these points in a Penrose diagram which for a Schwarzschild black hole is similar to a Kruskal-Szkeres diagram but with the entire exterior region compressed to a finite area on the diagram (and the Penrose diagram in this case can be defined in terms of a coordinate transformation from Kruskal-Szkeres coordinates, given on the link). Points on the right edge of the diagram would be points at infinite distance from the black hole in Schwarzschild or Kruskal-Szekeres coordinates, although as I said I don't think the edge is literally meant to be part of the spacetime, though points arbitrarily close to the edge at arbitrarily large distances in Schwarzschild/Kruskal-Szekeres coordinates are part of it (in other words I think the set of points in the diagram which are meant to be part of the spacetime form an open set)
rjbeery said:
Are you aware of a coordinate system in which the infinite observer would calculate that a mass infalling toward an EH would do so in finite time? What else could I possibly mean when I say "crosses the EH in infinite time"?
Falling from where? I assumed you were talking about falling from a finite distance (greater than the distance of the event horizon) in Schwarzschild coordinates, since in Schwarzschild coordinates objects falling in move slower and slower as they approach the horizon, never quite reaching it in any finite coordinate time. In Kruskal-Szekeres coordinates this problem doesn't occur, an object falling from a finite distance above the event horizon will pass the horizon in a finite coordinate time.

If you were instead talking about the time needed to literally fall from an infinite distance, then this would only be possible if the spacetime actually contained points at an infinite distance, which as I said I don't think is allowed. But even if it were, this would be a very odd thing to worry about, since it would have nothing to do with the fact that the central source of gravity is a black hole! After all, if we had a spacetime containing a single central planet with no event horizon, it would still take an infinite coordinate time in most coordinate systems (like Schwarzschild coordinates or Kruskal-Szkeres coordinates which could still be used in the vacuum region outside the planet's surface) for an object falling from an infinite distance to reach the planet. And in any case, the coordinate system used to draw a Penrose diagram would ensure that even such an infinite fall would only take a finite coordinate time.
rjbeery said:
Also, to say that Schwarzschild coordinates only apply to eternal black holes simply sounds like an ad hoc defense. Where did you come across this? Or is this personal supposition?
I'd only ever seen Schwarzschild coordinates defines on a Schwarzschild metric which is a static spacetime in the exterior region (the curvature, and thus the gravity, remains constant at every point for all eternity). However, I seem to have been wrong about the meaning of "Schwarzschild coordinates", reading the wikipedia article (along with this section of the textbook 'gravitation') it appears that they refer to a general type of coordinate system that can be constructed on any spherically symmetric spacetime, and that the formula for ds^2 (the line element) can in general be different in the dt^2 and dr^2 terms than the line element for the Schwarzschild metric in Schwarzschild coordinates. So, I guess I was wrong, if you have a star which collapses into a black hole while remaining spherically symmetric the entire time, it seems you'd be able to use Schwarzschild coordinates (this section of a page on deriving the Schwarzschild solution seems to support that) edit: But an evaporating black hole may be a trickier case, since a black hole which evaporates isn't even a valid solution to the equations of classical GR, so you'd probably need to use some sort of semiclassical approach to figuring out how curvature would vary as a function of time, which I'd guess would be necessary in figuring out the timing of when a distant observer would receive various light signals...
rjbeery said:
Light is Einstein's absolute system of measurement.
A rather vague statement. He defines inertial coordinate systems so that light has a constant speed in them, and defines clock synchronization in inertial systems using light signals, but beyond this what specific role do you think light plays in general relativity?
rjbeery said:
There is no "trick" or "illusion" when light suggests that the mass does not cross the hypothetical event horizon.
It only "suggests" this to a certain set of observers, those who remain outside the event horizon themselves. Similarly, to the set of accelerating observers whose positions are constant in Rindler coordinates defined in flat SR spacetime, no light from beyond the Rindler horizon will ever reach them, although if at any time they chose to stop accelerating they'd cross the horizon and see light from events beyond it. Here is a diagram drawn from the perspective of an ordinary inertial frame, showing both the worldlines of the accelerating Rindler observers (black hyperbolas) and the Rindler horizon (diagonal dotted line):

If you know something about spacetime diagrams drawn from the perspective of inertial frames, you can see that the Rindler horizon is just the future light cone of an event at the origin of the coordinate system, and that the accelerating Rindler observers will never enter this future light cone as long as they maintain the same constant acceleration (well, constant proper acceleration, their coordinate acceleration in this frame continually decreases as they approach c). So, it should be clear why they will never receive any light signals from beyond the horizon. But presumably you don't think that non-accelerating objects never really cross the horizon, since obviously they do from the perspective of the inertial frame (just draw a vertical worldline on that diagram and you can see it'll cross the dotted line at some finite t)

Assuming you agree in the case of the Rindler observers, why do you think the case of a black hole is so different? Just as the above diagram shows what curves of constant Rindler position coordinate (the black hyperbolas) and lines of constant Rindler time coordinate (the gray straight lines) look like when plotted in an inertial frame, so we can plot what curves of constant Schwarzschild radial coordinate and constant Schwarzschild time coordinate look like in a Kruskal-Szekeres diagram, and the result looks identical in the "exterior region" labeled with an I:

rjbeery said:
The proof of this is revealed by having the infalling mass accelerate itself back to the outside observer and compare clocks after any arbitrary (finite) amount of time has passed.
And what do you suppose happens in flat spacetime if an observer departs one of the accelerating Rindler observers and heads toward the Rindler horizon, gets close to it, then turns around and moves at high velocity to catch up with that same Rindler observer again so they can compare elapsed times on their clocks?

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#### bcrowell

Staff Emeritus
Gold Member
I don't think it is literally possible for a manifold with a metric defined on it (like a Riemannian manifold or a pseudo-Riemannian manifold) to actually include two points that are an infinite distance apart, see for example item (ii) on this page which I think is saying that any two points must be connected by a curve with a finite distance in the affine parameter along that curve.
Hmm...this is an interesting question. I don't think your interpretation can be technically quite right, but I'm afraid if we try to discuss it here as an OT side-discussion it will get swamped by the stuff about black holes. I'm going to start a separate thread on this.

#### rjbeery

JesseM: I don't have time to digest what you've written just yet, but I have to say that the general tone on this forum seems to be very mature and I appreciate the feedback of everyone.

#### Dale

Mentor
The point of Hawking radiation is that the black hole dissipates slowly, essentially shrinking.
Don't forget the CMB. A typical black hole of a few solar masses is much colder than the CMB, so it will absorb much more energy than it radiates and therfore grow, not dissipate. A supermassive one where tidal forces would be negligible would be very cold.

#### Chronos

Gold Member
I respectfully disagree. Kruskal coordinates have problems with the OP also (remember, I am using the infinite observer's frame which presents Kruskal indeterminability). My OP is quite simple: if the black hole region evaporates in finite time, and mass crosses the EH in "infinite" time, then then black hole region is gone before any mass reaches it. It really doesn't need to be much more complicated than that, and the aesthetic advantages of this interpretation are indisputable (aren't they?).
The problem is you are mixing coordinate systems in arriving at this apparent paradox [as has already been noted]. See Penelope for a simple explanation:
http://cosmology.berkeley.edu/Education/BHfaq.html#q4

#### Passionflower

The problem is you are mixing coordinate systems in arriving at this apparent paradox [as has already been noted]. See Penelope for a simple explanation:
http://cosmology.berkeley.edu/Education/BHfaq.html#q4
You mean question 9 not 4:

We've observed that, from the point of view of your friend Penelope who remains safely outside of the black hole, it takes you an infinite amount of time to cross the horizon. We've also observed that black holes evaporate via Hawking radiation in a finite amount of time. So by the time you reach the horizon, the black hole will be gone, right?

Wrong. When we said that Penelope would see it take forever for you to cross the horizon, we were imagining a non-evaporating black hole. If the black hole is evaporating, that changes things. Your friend will see you cross the horizon at the exact same moment she sees the black hole evaporate. Let me try to describe why this is true.

Remember what we said before: Penelope is the victim of an optical illusion. The light that you emit when you're very near the horizon (but still on the outside) takes a very long time to climb out and reach her. If the black hole lasts forever, then the light may take arbitrarily long to get out, and that's why she doesn't see you cross the horizon for a very long (even an infinite) time. But once the black hole has evaporated, there's nothing to stop the light that carries the news that you're about to cross the horizon from reaching her. In fact, it reaches her at the same moment as that last burst of Hawking radiation. Of course, none of that will matter to you: you've long since crossed the horizon and been crushed at the singularity. Sorry about that, but you should have thought about it before you jumped in.

#### Phrak

An issue to consider is whether a true 'singularity' can form in our spacetime. There is little doubt an event horizon can form.
What period of time would be required for this formation according to an outside observer?

#### JustinLevy

rjbeery:
I would appreciate it if you reread my post and respond about your thoughts on the Rindler case.

I really think you need to understand the Rindler case in flat-spacetime before you try to understand the GR case. It appears your misunderstanding stems from coordinate system issues which we can deal with in the much simpler flat-spacetime case.

Heck, we don't even need an accelerating observer to see these issues. Even in flatspace time with an inertial spatial origin (an "inertial observer" if you will), we could easily choose a coordinate chart which doesn't cover all spacetime and have coordinate time go to infinity before reaching relevant events. You are thinking of coordinates themselves too physically, and therefore trying to demand we consider coordinate artifacts as physical artifacts.

Also, let me remind posters here that while full quantum gravity may be needed to see what exactly occurs instead of a true spacetime singularity, semiclassical gravity is very capable of treating blackholes at the event horizon for large blackholes. That is because, once again, an event horizon is a global construct. Locally nothing spectacular happens there. No local measurements can tell. With large enough mass, the spacetime curvature at the event horizon can become very small. The Ricci curvature is already zero, and the tidal forces at the event horizon for blackholes at the center of galaxies would not harm you at all. This local patch of spacetime is not "extreme" in any sense. Semiclassical gravity handles this just find and people have already worked out the calculations. The analog of Hawking radiation and Unruh radiation gradually turns on before an event horizon forms, and this "quantum-gravity backreaction" has been calculated, and no, it doesn't prevent large blackholes (which again, semi-classical gravity should handle the formation of the event horizon just fine) from forming.

#### zhermes

I agree about the singularity, but I also question the EH itself. Maybe you can help me find the problem with my logic.
1. Do you agree that all frames outside of the BH calculate that no mass ever crosses the EH (or, more specifically, mass crosses the EH at t=infinity)?
2. Do you agree that the EH does not expand until mass has crossed the EH (i.e. backreaction)? .... run the clock backwards in your mind and describe to me how this theoretical black hole formed in the first place.
I've been thinking about this for a while. '2' is the problem. Mass in-falling into a BH causes the EH to expand before it crosses, from a distant observer's perspective. Hartle's gravity talks about it--I don't have my copy around, so I can't make a specific reference for a few days (sorry).
If I recall correctly, its because of gauss' law (slash the divergence theorem).... the details aren't coming to me however--can anyone back me up here?

#### rjbeery

JustinLevy and JesseM: I appreciate your reference to Rindler but I simply "not getting it". It sounds like you're trying to make the case that because apparent future null infinity events can be described without gravity (but rather constant acceleration) via Rindler horizons, then I should not have a problem accepting future full infinity events caused by an event horizon. My stance on both remains, and that is they "never" happen. In fact, I was being purely speculative when I said an infalling mass to an EH could be destroyed by infinitely more powerful radiation, but I think we would all agree that this is indisputably the case for an infinitely accelerating body in flat spacetime (due to CBR)!
JustinLevy said:
Also, let me remind posters here that while full quantum gravity may be needed to see what exactly occurs instead of a true spacetime singularity, semiclassical gravity is very capable of treating blackholes at the event horizon for large blackholes.
Ahh, I've seen this objection before. Be mindful, though, that black holes do not spontaneously appear; they have a creation event at the center of gravity of a mass. This creation event must begin on the atomic level - would you like to discuss the event horizon radius of two hydrogen atoms for example? It's on the order of 1E-20 smaller than the Planck length! My point is that one cannot point out that arbitrarily large black holes avoid unpleasant mathematical problems when you MUST accept arbitrarily small black holes before anything else can be discussed.
zhermes said:
I've been thinking about this for a while. '2' is the problem. Mass in-falling into a BH causes the EH to expand before it crosses, from a distant observer's perspective. Hartle's gravity talks about it--I don't have my copy around, so I can't make a specific reference for a few days (sorry).
If I recall correctly, its because of gauss' law (slash the divergence theorem).... the details aren't coming to me however--can anyone back me up here?
Well I'm glad you see there is a disconnect between my outlined logic and the existence of black holes (which to point no one else has even addressed - so far its been obfuscation and references to their lecture notes). I'm also glad you've given it some thought. I will research Hartle's gravity, thanks for the reference.

The rest of this thread is really academic, either there is a problem with one of the two logic points that I listed in post 22 (which is possible), or black holes do not exist as is currently being taught.

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#### JesseM

JustinLevy and JesseM: I appreciate your reference to Rindler but I simply "not getting it". It sounds like you're trying to make the case that because apparent future null infinity events can be described without gravity (but rather constant acceleration) via Rindler horizons, then I should not have a problem accepting future full infinity events caused by an event horizon.
I don't understand the term "future null infinity events", can you elaborate? My point was just that you seemed to be using the following two facts to argue against the idea that any observer can experience crossing the black hole event horizon:

1. An observer who remains outside the horizon will never see light signals from any events on or beyond the horizon

2. The time for an infalling observer to reach the horizon is infinite (which I took to mean the time in Schwarzschild coordinates for an observer falling from a finite distance about the horizon, although it's possible I misunderstood--see my requests for clarification on this point in my previous post)

So, I was just pointing out that if we replaced the observer outside the horizon by a Rindler observer, replaced the event horizon with a Rindler horizon, and replaced Schwarzschild coordinates with Rindler coordinates, then the same facts would be true in SR! So if you don't think the SR versions of the facts are sufficient to prove that no observer can ever experience crossing a Rindler horizon, that suggests that the black hole versions above aren't sufficient to show that no observer can ever experience crossing the event horizon.

#### rjbeery

To JesseM:
RJBeery said:
My stance on both remains, and that is they "never" happen.

#### JesseM

Are you saying that you think an observer "never" experiences crossing a Rindler horizon? Keep in mind that the position of a Rindler horizon just depends on where you choose to define your Rindler coordinate system, for any future light cone in a Minkowski spacetime, you can define a Rindler coordinate system such that the Rindler horizon coincides with one side of that future light cone. So absolutely any point in Minkowski spacetime is on the Rindler horizon of some Rindler coordinate system!

#### rjbeery

JesseM and JustinLevy: OK I still need to digest Rindler horizons because there is something I'm missing. I have a question on them but you must give me time to formulate it. My first thought was that the very postulate of Rinder horizons is an affront against Einstein's "constancy of the speed of light", as a light source beyond the horizon would not be seen. Now I'm wondering if the horizon is simply an equivalency to a spacelike separation? I want to explore this issue more, just give me some time.

Anyway, Rindler horizons, just like event horizons IMHO, are interesting mathematical structures with no physicality. We all agree that there are sufficient practical barriers to a Rindler horizon, correct (infinite fuel source, CMB annihilation, etc)? I will continue giving this topic some thought but I ask that the two of you give me feedback on the simple logic of post 22...

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