Is it possible that black holes do not exist?

In summary: No. According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no
  • #36
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.
 
Physics news on Phys.org
  • #37
rjbeery said:
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):

Coords.gif


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:

360px-Kruksal_diagram.jpg

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?
 
Last edited:
  • #38
JesseM said:
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.
 
  • #39
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.
 
  • #40
rjbeery said:
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 therefore grow, not dissipate. A supermassive one where tidal forces would be negligible would be very cold.
 
  • #41
rjbeery said:
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
 
  • #42
Chronos said:
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.
 
  • #43
Chronos said:
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?
 
  • #44
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 black holes at the event horizon for large black holes. 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 black holes 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 black holes (which again, semi-classical gravity should handle the formation of the event horizon just fine) from forming.
 
  • #45
rjbeery said:
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?
 
  • #46
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 black holes at the event horizon for large black holes.
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.
 
Last edited:
  • #47
rjbeery said:
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.
 
  • #48
To JesseM:
RJBeery said:
My stance on both remains, and that is they "never" happen.
 
  • #49
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!
 
  • #50
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...
 
  • #51
rjbeery said:
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.
This comes up almost as often as the twin "paradox".

The speed of light is c when an observer makes a local measurement! If an observer makes a distance measurement or an average speed measurement between two locations the speed of light is not necessarily c. It is not c and anisotropic in case the observer undergoes proper or inertial acceleration.

rjbeery said:
We all agree that there are sufficient practical barriers to a Rindler horizon, correct (infinite fuel source, CMB annihilation, etc)?
I do not agree.
Even for travelers on a bullet train slowly accelerating from 0 to a top speed of 200 miles/hour in a timespan of one hour there is a Rindler horizon. Only after the acceleration stops the horizon will disappear.

Did you read question and answer nine of Ted Bunn's article?
 
Last edited:
  • #52
Yes, I realized this, thanks; there is acceleration involved, by definition. I was just going through my initial thought process aloud. It also did not occur to me that a Rindler horizon could exist for an arbitrary acceleration speed and time but that makes sense - if a receding light source was "just" observable, then even a slight acceleration could redshift it beyond detection. As I said, please let me think about it more before I comment on the subject further. I will also read Bunn's article, thanks.

If you get a chance, please comment on post 22, because I do not believe an event horizon expands before mass comes "in contact" with it. Maybe we need to strictly define phrases like this, but I am open to that.
 
  • #53
rjbeery said:
If you get a chance, please comment on post 22, because I do not believe an event horizon expands before mass comes "in contact" with it.

Vaidya metric.

See the left diagram of Figure 5.7 on page 134 (pdf page 150) of

http://www.physics.uoguelph.ca/poisson/research/agr.pdf,

and the first full paragraph beneath this diagram for the reason that this happens.
 

Similar threads

Replies
46
Views
4K
Replies
4
Views
897
Replies
6
Views
2K
Replies
1
Views
724
Replies
4
Views
2K
Replies
2
Views
1K
Replies
4
Views
959
Back
Top