A question about "nothing" [Archive] - Page 2

Agent : Orange

27th December 2007, 03:40 PM

Actually no. If you're talking about a black hole, it's true our current understanding of physics does not apply in the vicinity of a singularity, BUT... around the singularity is an event horizon where time stops. A black hole happens when a star collapses to within the event horizon. By definition, this can never happen since time stops at the event horizon. So when we see a black hole, what we're seeing is actually a star that has almost collapsed to the point of crossing the event horizon but hasn't done it yet. It's probably so redshifted it's hard to distinguish it from an actual black hole. But I think this point spares us from having to allow for singularities in our universe, except for maybe the big bang.

Hi, just reading through the thread and I found this post. Time doesn't stop at the event horizon of a black hole. In the case of a simple Schwarzschild metric, the event horizon is only a coordinate singularity. You can prove this by finding the Riemann tensor and then the Kretschman scalar. The event horizon is like any other point in space and things are perfectly well behaved except for the fact that it represents the "point of no return" from the universe, and that all observers within this mathematical boundary have to continue on to the singularity. You can put the Schwarzschild metric into Eddington-Finkelstein or Kruskal coordinates and have no problem at the event horizon. The Schwarzschild radius is just a coordinate singularity, like the origin in polar coordinates.

Agent : Orange

27th December 2007, 03:42 PM

Clearly my message above has already been covered and I decided to throw in my $0.02 before finishing my reading. For what it's worth in Canadian currency. :)

Yllanes

27th December 2007, 04:09 PM

ETA: you said "because the collapse does take place for other observers".
Given that time runs slower and slower for a clock with respect to the rest of the universe the further into a gravity well the clock goes, until time actually stops at the event horizon, can you explain how observers outside the event horizon can ever see the star completely collapse?

Nobody said an observer outside the event horizon would see a star completely collapse. I said an observer falling with the star would. Nobody claims the presence of the black hole singularity has any effect on the outside observers (classically). But I think understanding what happens with something that falls into the black hole would be interesting, even if it doesn't affect us in the outside.

DrBaltar

27th December 2007, 06:15 PM

Hi, just reading through the thread and I found this post. Time doesn't stop at the event horizon of a black hole. In the case of a simple Schwarzschild metric, the event horizon is only a coordinate singularity. You can prove this by finding the Riemann tensor and then the Kretschman scalar. The event horizon is like any other point in space and things are perfectly well behaved except for the fact that it represents the "point of no return" from the universe, and that all observers within this mathematical boundary have to continue on to the singularity. You can put the Schwarzschild metric into Eddington-Finkelstein or Kruskal coordinates and have no problem at the event horizon. The Schwarzschild radius is just a coordinate singularity, like the origin in polar coordinates.

Again, I am talking about what an observer sees from outside the event horizon and looking at objects entering the event horizon. I agree that someone entering the event horizon notices nothing special aside from the increasing gravitational gradient. This is the third time now that I have acknowledged this. How many more times must we go through this?

ETA: Sorry, I see now that you saw this was already covered. As for time stopping, see 2 posts down...

DrBaltar

27th December 2007, 06:23 PM

Nobody said an observer outside the event horizon would see a star completely collapse. I said an observer falling with the star would. Nobody claims the presence of the black hole singularity has any effect on the outside observers (classically). But I think understanding what happens with something that falls into the black hole would be interesting, even if it doesn't affect us in the outside.

I agree with all that. With that finally behind us, what did you think of the penrose diagrams? Do you see that the region inside the event horizon is something separate from the rest of the universe?

DrBaltar

27th December 2007, 06:34 PM

Time does stop at the event horizon (as seen from the outside)

In case there is any doubt I wanted to clear this up. There is an online version of the book I studied from in a General Relativity course available here (http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html).

In the section on Black Holes, there is this:
This should be clear from the pictures, and is confirmed by our computation of delta-tau1/delta-tau2 when we discussed the gravitational redshift (7.61). As infalling astronauts approach r = 2GM (the schwarzchild radius), any fixed interval delta-tau1 of their proper time corresponds to a longer and longer interval delta-tau2 from our point of view. This continues forever; we would never see the astronaut cross r = 2GM, we would just see them move more and more slowly (and become redder and redder, almost as if they were embarrassed to have done something as stupid as diving into a black hole).

Scott1972

27th December 2007, 07:20 PM

Wow, this thread really took off. I've been out of town for the holidays and wasn't able to keep up.

Thank you everyone for taking the time to post your ideas. I'll try to make time to read through everything and see if I can come up with any useful comments.

DrBaltar

27th December 2007, 11:13 PM

sol invictus

27th December 2007, 11:55 PM

Given that time runs slower and slower for a clock with respect to the rest of the universe the further into a gravity well the clock goes, until time actually stops at the event horizon, can you explain how observers outside the event horizon can ever see the star completely collapse?

You're missing a crucial point here. Try calculating how long it takes for something to get within one Planck length of the horizon. It's not very long at all. At 1 Planck length from the horizon, the Hawking temperature is the Planck temperature. So no matter what you drop in to the hole, it will have been burned to a crisp by Hawking radiation by the time it arrives there. All of this is as seen by an outside observer, and believe it or not it is perfectly consistent with the statement that infalling observers see nothing at the horizon.

From the outside, a black hole is indistiniguishable from a hot, impermeable membrane. The size of that membrane can change - it can grow or shrink depending on whether matter is accreting or not. Things falling in approach the surface, thermalize, and are re-radiated as Hawking particles. Eventually if nothing keeps falling in to maintain the mass, the whole thing evaporates.

By the way, that conformal diagram for the rotating black hole you posted is very misleading. The inner horizon is unstable to form a space-like singularity, so the slightest perturbation will turn that diagram into one looking essentially identical to a Schwarzschild black hole. The other universe is unreachable.

DrBaltar

28th December 2007, 12:52 AM

sol invictus,

In "empty" space, virtual particle pairs pop in and out of existence all the time. It is my understanding that in Hawking radiation, there is a chance that one of the particles of the pair falls into the black hole, while the other particle is able to escape the black hole. So rather than being temporary particles, they become permanent. And so you would have a net flux of particles leaving the black hole, which is where the black body radiation comes from.

The part I don't get is if it takes an infinite amount of time for the star to completely collapse, then the region outside the black hole is not a vacuum. So no vacuum fluctuations and no virtual particles. So how does Hawking radiation happen? I am skeptical on this, especially since it has never been observed. I thought this theory applied to a 'classic' black hole in which the collapse has already happened and the region outside the event horizon was a vacuum.

As for the rotating black hole diagram, I agree it would be extremely difficult to navigate such that you can avoid the singularity and make it to an alternate universe, but it does seem to allow this in theory.

Yllanes

28th December 2007, 01:37 AM

I agree with all that. With that finally behind us, what did you think of the penrose diagrams?

That was certainly not the first time I saw a Penrose diagram...

jmercer

28th December 2007, 02:56 AM

In "empty" space, virtual particle pairs pop in and out of existence all the time. It is my understanding that in Hawking radiation, there is a chance that one of the particles of the pair falls into the black hole, while the other particle is able to escape the black hole. So rather than being temporary particles, they become permanent. And so you would have a net flux of particles leaving the black hole, which is where the black body radiation comes from.

That's my understanding as well, although it's my understanding that one of the pair actually appear at or inside the horizon, while one does not.

The part I don't get is if it takes an infinite amount of time for the star to completely collapse, then the region outside the black hole is not a vacuum. So no vacuum fluctuations and no virtual particles. So how does Hawking radiation happen? I am skeptical on this, especially since it has never been observed. I thought this theory applied to a 'classic' black hole in which the collapse has already happened and the region outside the event horizon was a vacuum.

As I understand it, two fluctuations can't appear at the same spacetime coordinates. So at the moment the two imaginary particles appear, that's all that will happen - at least until they cancel each other out and a "true vacuum" exists once more.

In the case of a black hole, you only get Hawking radiation when one particle appears inside or exactly at the (two dimensional?) horizon and the other does not - and then only if the outside particle is moving at the proper angle to escape the horizon. So, for the period of time while the new "outside" particle physically remains at the spacetime location, no fluctuations occur - but as it vacates that particular place by moving away (or falling into the horizon), a vacuum resumes and more random fluctuations occur.

At least that's my understanding.

DrBaltar

28th December 2007, 08:01 AM

That was certainly not the first time I saw a Penrose diagram...

That's nice. What did you think about them? Do you agree that they show the inside of the event horizon is a region separate from the universe or not?

DrBaltar

28th December 2007, 08:11 AM

That's my understanding as well, although it's my understanding that one of the pair actually appear at or inside the horizon, while one does not.Yes, I did mention that.

The part I don't get is if it takes an infinite amount of time for the star to completely collapse, then the region outside the black hole is not a vacuum. So no vacuum fluctuations and no virtual particles. So how does Hawking radiation happen? I am skeptical on this, especially since it has never been observed. I thought this theory applied to a 'classic' black hole in which the collapse has already happened and the region outside the event horizon was a vacuum.

As I understand it, two fluctuations can't appear at the same spacetime coordinates. So at the moment the two imaginary particles appear, that's all that will happen - at least until they cancel each other out and a "true vacuum" exists once more.

In the case of a black hole, you only get Hawking radiation when one particle appears inside or exactly at the (two dimensional?) horizon and the other does not - and then only if the outside particle is moving at the proper angle to escape the horizon. So, for the period of time while the new "outside" particle physically remains at the spacetime location, no fluctuations occur - but as it vacates that particular place by moving away (or falling into the horizon), a vacuum resumes and more random fluctuations occur.

At least that's my understanding.

You haven't addressed my point though. If it takes an infinite amount of time for a star to completely collapse, as seen from an outside observer, then the region above the event horizon is occupied by star matter. That is what I mean when I say there's no vacuum outside the event horizon. And I believe the virtual particles must appear just outside the event horizon. One particle goes in with negative energy, and the other escapes with positive energy.

Yllanes

28th December 2007, 08:55 AM

That's nice. What did you think about them? Do you agree that they show the inside of the event horizon is a region separate from the universe or not?

I never denied that. I don't know what you are trying to say.

sol invictus

28th December 2007, 08:58 AM

jmercer

28th December 2007, 09:46 AM

You haven't addressed my point though. If it takes an infinite amount of time for a star to completely collapse, as seen from an outside observer, then the region above the event horizon is occupied by star matter. That is what I mean when I say there's no vacuum outside the event horizon. And I believe the virtual particles must appear just outside the event horizon. One particle goes in with negative energy, and the other escapes with positive energy.

Ah - ok, I see what you're driving at. Here's my thinking on it, albeit this is strictly coming from me and not any kind of authoritative source. First, let me acknowledge that you are correct regarding where the virtual particles appear - both particles must appear outside the horizon, and not one inside and one outside. :)

Let me define the problem you're citing as I understand it:

Once the star reaches the point where a horizon is created, all star matter inside (or at) the horizon is cut off from the universe. This leaves star matter outside when the horizon appears. This star matter prevents a "true vacuum" from existing, and - according to your objection - under these circumstances, quantum fluctuation and Hawking particle pairs will not occur, since they must occur in a vacuum.

However, when we talk about quantum fluctuation in a vacuum, we're not speaking of a classical vacuum. We're speaking of a quantum vacuum, which exists below the planck length where classical matter and energy cannot exist. Quantum fluctuation supposedly happens in a space smaller than a planck, and it's duration is less than a planck according to Hawking's theory. (AFAIK) That's why the "quantum foam" can be said to exist literally everywhere - regular matter and energy exist above the planck level, while all of this stuff happens below the planck level. IRRC, I believe that the particle twins emerge out of the sub-planck level and then cancel each other out, returning to the sub-planck level in less than a planck time... preventing any possible detection of their existence.

So in essence, the presence of classical matter doesn't affect the existence of a quantum vacuum... any star matter lingering outside the horizon wouldn't prevent the creation of the virtual particle pairs. What does play a critical role, however, are the tidal forces of gravity at or near the event horizon - and that's why (theoretically!) the virtual pairs can be broken up... with one becoming "real" and the other disappearing into the black hole. The "new" particle exists for longer than a planck, making it detectable.

(BTW, supposedly these virtual pairs are created constantly everywhere - but are undetectable because they exist for less than a planck time - unless one is captured, as described above.)

Regarding the existence of star matter in the proximity of an event horizon... in order for the star matter to simply remain in the near vicinity of the event horizon, it would have to neither fall in or move away - in other words, "hover" just above the horizon.

Given the nature of how black holes supposedly form, I don't see how that would be possible. Certainly at the initial stages of formation, a lot of matter and energy is driven outward violently - so we can assume that star matter outside of where the horizon ultimately forms would have an outward momentum. Once the horizon forms, there's a steep gravity well to contend with... and any star matter not moving with sufficient velocity to escape the well would be recaptured and fall into the event horizon.

In summary:

1) Quantum fluctuation happens below the planck length and for less than a planck time, and is therefore not affected by the presence (or absence) of classical matter and energy. It is not necessary for the star to finish collapsing for Hawking radiation to occur; all that is required is that an event horizon be formed with a gravity gradient sufficient to prevent the virtual pairs from self- annihilating in less than a planck time.

2) Any remaining star matter in the vicinity of a newly formed black hole would either radiate away or fall into the horizon... clearing the area above the surface of the horizon and creating a classical vacuum in any case.

DrBaltar

28th December 2007, 12:03 PM

Regarding the existence of star matter in the proximity of an event horizon... in order for the star matter to simply remain in the near vicinity of the event horizon, it would have to neither fall in or move away - in other words, "hover" just above the horizon.
Imagine 3 particles of this star matter. Particle 1 (p1) is 1m from the event horizon. p2 is 500m from the event horizon. p3 is 1000m from the event horizon. From p2's frame, it is hurdling towards the event horizon at relativistic speeds. From p2's frame, p1 appears to be at almost a stand still. This is because time at p1 is going much slower than time at p2. In p1's frame, p1 is also hurdling towards the event horizon at relativistic speeds. From p3's frame, it also seems to be hurdling towards the event horizon at relativistic speeds, but not quite as fast as p1 or p2 think they are moving. From p3's frame, p2 seems to be moving very slowly and p1 appears to be almost stopped, or as you say, 'hovering'. This is again because time 'flows' slower the further in you go. I agree this seems to pose logical problems. But this is solved when you take into account the changing length scales and time scales.

As for the first part of your posting, it is true that atoms are mostly empty space. The nucleus is 1/100,000th the diameter of the atom. But in the case of star matter falling into the back hole things are very different. Remember the stages a star goes through before becoming a black hole. First it overcomes the electromagnetic force between the electrons and the nucleus - compressing the atomic components together. Then gravity overcomes the short range nuclear forces and it becomes so dense that electrons and protons are smashed together and become neutrons, so all there are are neutrons. That's the neutron star stage. This star matter is destined to become even denser than that, and it is theorized that a quark star state exists between a neutron star and a black hole, where neutrons are further smashed into quarks. For a star becoming a black hole, it doesn't stop there and the matter collapses into a singularity.

So basically I can't imagine there would be much empty space in matter that dense, and even if there were, I certainly cannot imagine any of the virtual particles flying through matter that dense that is hurdling towards the event horizon at relativistic speeds. Any virtual particles speeding away from the black hole would easily get knocked back towards the event horizon with the infalling star matter.

sol invictus

28th December 2007, 12:21 PM

You're very confused.

All three of your particles simply fall in, and none of them see any slowing of the others. That would only happen to an observer that accelerated like crazy to stay out of the horizon, and the redshift and time-dilation (and Hawking particles) are then attributable to the acceleration. The equivalence principle at work.

In any case, for matter collapsing into a black hole, all of it crosses the horizon and is gone quite rapidly, leaving nothing much behind.

jmercer

28th December 2007, 12:45 PM

Imagine 3 particles of this star matter. Particle 1 (p1) is 1m from the event horizon. p2 is 500m from the event horizon. p3 is 1000m from the event horizon. From p2's frame, it is hurdling towards the event horizon at relativistic speeds. From p2's frame, p1 appears to be at almost a stand still. This is because time at p1 is going much slower than time at p2. In p1's frame, p1 is also hurdling towards the event horizon at relativistic speeds. From p3's frame, it also seems to be hurdling towards the event horizon at relativistic speeds, but not quite as fast as p1 or p2 think they are moving. From p3's frame, p2 seems to be moving very slowly and p1 appears to be almost stopped, or as you say, 'hovering'. This is again because time 'flows' slower the further in you go. I agree this seems to pose logical problems. But this is solved when you take into account the changing length scales and time scales.

As for the first part of your posting, it is true that atoms are mostly empty space. The nucleus is 1/100,000th the diameter of the atom. But in the case of star matter falling into the back hole things are very different. Remember the stages a star goes through before becoming a black hole. First it overcomes the electromagnetic force between the electrons and the nucleus - compressing the atomic components together. Then gravity overcomes the short range nuclear forces and it becomes so dense that electrons and protons are smashed together and become neutrons, so all there are are neutrons. That's the neutron star stage. This star matter is destined to become even denser than that, and it is theorized that a quark star state exists between a neutron star and a black hole, where neutrons are further smashed into quarks. For a star becoming a black hole, it doesn't stop there and the matter collapses into a singularity.

So basically I can't imagine there would be much empty space in matter that dense, and even if there were, I certainly cannot imagine any of the virtual particles flying through matter that dense that is hurdling towards the event horizon at relativistic speeds. Any virtual particles speeding away from the black hole would easily get knocked back towards the event horizon with the infalling star matter.

I'm sorry, DrBaltar, but this is inherently incorrect.

First of all, your original contention was that quantum fluctuation wouldn't happen due to the presence of star matter; now you have compressed star matter knocking virtual particles around like billiard balls on a pool table into the event horizon. Regular matter cannot interact with virtual particles because they exist for less than a planck... plus you're changing the goal posts.

Secondly, the black hole will eventually clear the region of all content outside the horizon, like a giant vacuum cleaner, rendering infalling matter a moot point.

Thirdly, the event horizon of a non rotating black hole is a sphere; you would have to have matter falling in toward every point of the horizon simultaneously and constantly, for eternity. You may debate how long it will take a black hole to completely collapse, but outside of the horizon Einsteinian and quantum physics are in control - therefore, the horizon and activities near it's surface are still subject to the rules of spacetime. Nothing is going to fall into the horizon eternally, and certainly not at every mathematical point of the horizon at all times.

Fourthly, as I said, classical matter has nothing to do with interfering with a quantum vacuum whatsoever.

Lastly, planck length refers to thing smaller than quarks, not simply atoms. I don't care how compressed the matter is outside of the event horizon - it simply will not - and cannot - approach planck size or affect things that exist for less than a planck.

Sorry, but Sol is correct - I'm seeing a bit of confusion here.

DrBaltar

28th December 2007, 03:20 PM

DrBaltar

28th December 2007, 03:31 PM

jmercer

1) if the virtual particles only 'exist lest than a plank' then Hawking radiation couldn't happen. It is because one falls into the black hole, and the other escapes the region around the black hole that it can last.

2) See my previous post about matter falling into the black hole.

3) If a spherical star is collapsing into a black hole, then yes, you would have extremely dense star matter falling toward every point of the horizon simultaneously and constantly, for eternity.

4) Even neutrinos can interact with ordinary matter. It's rare, because atoms are mostly empty space, but it can happen. As I said the star matter of a star collapsing into a black hole is much denser than ordinary matter and it would certainly react.

sol is not correct. He is completely ignoring the fact that time slows down in a gravity well, and stops at the event horizon. Again see my previous posting.

sol invictus

28th December 2007, 04:53 PM

I have stated that matter, as seen from an outside observer, will never be seen to cross the event horizon in as many ways as I can, as clearly as I can, without being overly redundant. From reputable sources there is plenty of physics to back me up. I have shown links to actual text books which back up what I say. All I get from you guys is either 'no it's not like that' without any actual sources to back you up (other than the ones I have discredited), or you only read half of what I say. I don't know how else to say it other than face to face with a piece of paper and a book on General Relativity.

It's not a question of what the outside observer sees. You are correct that photons or other particles emitted by something falling into a black hole will be redshifted infinitely relative to an asymtotic observer as the source crosses the horizon. So what?

What matters is what actually happens, not what some specific observer can see with her eyes. And what actually happens is that matter falls into the black hole in finite proper time. You can see that from the conformal diagram, from a simple calculation, or by reading any book on GR.

You are the one that brought up Carter-Penrose diagrams - just look at the one in my post above, the one which shows the formation of a black hole from a collapsing shell of radiation. Look at the region of the diagram above the blue region. The metric in that region is precisely the vacuum Schwarzschild solution. End of story.

Matteo Martini

28th December 2007, 05:18 PM

The universe has a radius of 13.7 billion light years, plus or minus 137 million (see reference (http://map.gsfc.nasa.gov/m_mm/mr_age.html)).

What do we have outside there?

jmercer

28th December 2007, 05:44 PM

What do we have outside there?

There is no "outside".

sol invictus

28th December 2007, 05:44 PM

sol is not correct. He is completely ignoring the fact that time slows down in a gravity well, and stops at the event horizon. Again see my previous posting.

Let me give you an analogy that may help here. Imagine you are a creature that lives underwater, in a river. You're blind, and navigate solely by hearing and sonar, like a bat.

Now suppose the river you're in has a waterfall somewhere downstream from where you are, and at some point above the waterfall the speed of the river water becomes greater than the speed of sound. Let's call that the sound horizon, because a sound emitted there will not propagate upstream.

Now suppose your friend drifts downstream towards the horizon. You keep track of her location with sonar pulses. What will you conclude is her trajectory if you reconstruct it from the sonar echoes you hear?

The answer is that it will sound to you as though she is drifting more and more slowly, gradually approaching the horizon, but never quite reaching it. Not only that, but if she is emitting sounds at regular intervals you will hear them separated by longer and longer lengths of time - as though time were slowing down there at the horizon. And if you defined a time coordinate by those intervals, time would indeed stop at the sound horizon - in that coordinate system.

So what would you conclude from this - that time stops and nothing ever crosses the sound horizon? That your friend and everything else that drifted downstream is all piled up on top of itself at the horizon? That would be pretty remarkable, wouldn't it?

jmercer

28th December 2007, 05:49 PM

jmercer

[quote=DrBaltar;3283208]
1) if the virtual particles only 'exist lest than a plank' then Hawking radiation couldn't happen. It is because one falls into the black hole, and the other escapes the region around the black hole that it can last.

They exist for less than a planck; they are not less than a planck in size.

2) See my previous post about matter falling into the black hole.

I saw it. You're talking about observation from an external viewpoint. As Sol has stated, what we see isn't what's happening, any more than looking at a star 4 light years distant is showing you what the star is doing real-time.

3) If a spherical star is collapsing into a black hole, then yes, you would have extremely dense star matter falling toward every point of the horizon simultaneously and constantly, for eternity.

First I've heard of it. If you have references citing that interesting phenomenon, please share it.

4) Even neutrinos can interact with ordinary matter. It's rare, because atoms are mostly empty space, but it can happen. As I said the star matter of a star collapsing into a black hole is much denser than ordinary matter and it would certainly react.

I disagree.

sol is not correct. He is completely ignoring the fact that time slows down in a gravity well, and stops at the event horizon. Again see my previous posting.

Time slows down relative to the frame outside of the gravity well. It doesn't slow down inside the frame of the traveler.

Agent : Orange

28th December 2007, 05:59 PM

3) If a spherical star is collapsing into a black hole, then yes, you would have extremely dense star matter falling toward every point of the horizon simultaneously and constantly, for eternity.

I disagree with this point. The proper time in Schwarzschild coordinates is well-behaved, so the star matter falls right through the horizon with no ill-effect. Because the coordinate time diverges at the horizon, a distant observer watching a star collapse will see an image of the star as it appeared at the event horizon, but this image of the star will quickly fade because of the large gravitational redshift near the horizon. The spherical mass distribution of the star does NOT remain eternally falling toward the singularity, frozen near the event horizon because the proper time is well-behaved. An external observer watching the collapse will see the star slowly fade from view, and the periodicity of a regular signal sent from the surface of the star will grow to infinity as the horizon is approached.

To a distant observer, it takes a test body an infinite amount of time to cross the event horizon. To the test body itself it takes a finite amount of time to pass through r=2m and reach the singularity at r=0. The behavior of the line element at r=2m tells us that the Schwarzschild coordinates are inappropriate for describing the motion of an infalling particle. By adopting Eddington-Finkelstein coordinates we can describe this motion without any problems arising from the coordinate system itself.

jmercer

28th December 2007, 06:01 PM

Very cogently put. Thank you, Agent : Orange.

Agent : Orange

28th December 2007, 06:02 PM

It's not a question of what the outside observer sees. You are correct that photons or other particles emitted by something falling into a black hole will be redshifted infinitely relative to an asymtotic observer as the source crosses the horizon. So what?

What matters is what actually happens, not what some specific observer can see with her eyes. And what actually happens is that matter falls into the black hole in finite proper time. You can see that from the conformal diagram, from a simple calculation, or by reading any book on GR.

You are the one that brought up Carter-Penrose diagrams - just look at the one in my post above, the one which shows the formation of a black hole from a collapsing shell of radiation. Look at the region of the diagram above the blue region. The metric in that region is precisely the vacuum Schwarzschild solution. End of story.

Exactly. Well said.

Agent : Orange

28th December 2007, 06:08 PM

Again, I am talking about what an observer sees from outside the event horizon and looking at objects entering the event horizon. I agree that someone entering the event horizon notices nothing special aside from the increasing gravitational gradient.

Okay, so then what are you arguing? Clearly from what you've just said, time does not actually stop for a test particle crossing the event horizon from the point of view of that particle. To an outside observer, it does look like time slows down however. If you agree that someone who crosses the event horizon does not notice anything special happening, your argument about time stopping at the event horizon and stellar material being suspended there gets knocked down.

EDIT: I recognize the point you were making is about the collapse of a star as seen by an outside observer. Neglecting gravitational redshift, such an observer would see the star eternally collapsing, and would notice the period of a regular signal increasing to infinity as r->2m. However, in practice the wavelength of the signal becomes infinitely stretched out due to gravitational redshift. So as the collapsing stellar surface approaches the event horizon, an outside observer sees signals from the star being sent with longer and longer delays between and simultaneously sees the signals become redder and redder, until there's simply nothing to detect. This is all well known physics, can be found in any introductory book on GR and it certainly doesn't allow us to sidestep the existence of a singularity at r=0. The Kretschman scalar vanishes at this point, and that means that the r=0 singularity can't be transformed away like the event horizon singularity can be. This is the real deal - an actual, physical singularity and not simply a coordinate singularity.

DrBaltar

28th December 2007, 07:42 PM

Agent : Orange

28th December 2007, 08:51 PM

I am currently working on my Ph.D. in astrophysics. My thesis is based around modeling gravitational lens systems using a genetic algorithm to optimize a parametric mass model for the lens.

Yllanes

29th December 2007, 12:57 AM

This is all well known physics, can be found in any introductory book on GR and it certainly doesn't allow us to sidestep the existence of a singularity at r=0. The Kretschman scalar vanishes at this point, and that means that the r=0 singularity can't be transformed away like the event horizon singularity can be.

I'm sure it's just a typo, but the Kretschmann scalar diverges (is infinite) at r = 0.

sol invictus

29th December 2007, 02:34 AM

Just so we know where everyone's at, could everyone following this discussion please state what their educational background is relating to black holes and general relativity?

My background is a masters in physics. I've had courses in plasma physics, statistical mechanics, relativity, general relativity, and cosmology. For my thesis I did a paper and simulation on the effects of gravitational waves on primordial matter after the inflationary period following the big bang. I'm just trying to figure out what kind of education is behind the statements everyone is making.

Were all of us to reveal our credentials, the discussion would probably devolve into a question and answer session to the one with the most authority. I'm not sure anyone wants that, least of all that person.

Did you read my analogy of the river? It's a rather good one (I didn't invent it) and it captures a lot of the physics of black holes, including the issue that has you confused.

jmercer

29th December 2007, 03:19 AM

Well, as I said, I'm not an authority on the subject - I am, however, widely read on the subject, have attended various lectures, and have conversed here (and in other places) at length with people who hold doctorates in physics, astrophysics and related disciplines on this and similar topics. I myself have a Masters in Computer Science.

Further, all the things I've stated exist in various books on the subject - not to mention all over the internet from various universities. Conceptually, I think I'm pretty well informed - even though I can't do the math myself.

jmercer

29th December 2007, 03:31 AM

I am currently working on my Ph.D. in astrophysics. My thesis is based around modeling gravitational lens systems using a genetic algorithm to optimize a parametric mass model for the lens.

Now THAT has to be one of the coolest thesis subjects ever. :) My Masters dissertation was on the potential effects on society of quantum computing. (If it should become a commercial reality)

DrBaltar

29th December 2007, 08:30 AM

Did you read my analogy of the river? It's a rather good one (I didn't invent it) and it captures a lot of the physics of black holes, including the issue that has you confused.There are no relativistic effects in rivers.

DrBaltar

29th December 2007, 08:38 AM

Were all of us to reveal our credentials, the discussion would probably devolve into a question and answer session to the one with the most authority. I'm not sure anyone wants that, least of all that person.

The reason I ask is because especially in the last page of posts, there has been an onslaught of misinformed statements, and I'm just trying to figure out where it's all coming from. I feel like in order to get us all on the same page, I would have to go through at least 2 semesters worth of material. Or if you are trained in this subject, I could simply show the math, which would be irrefutable.

Yllanes

29th December 2007, 08:56 AM

You can show whatever math you want, but I don't know what use it would be. Nobody denies that, for example, if you write the Schwarzschild metric as

\footnotesize
\[
\mathrm{d}s^2 = -\left(1-\frac{2GM}{r}\right) \mathrm{d}t^2 +\left(1-\frac{2GM}{r}\right)^{-1} \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2_2,
\]

Then the timelike geodesics are

\footnotesize
\[
-1 = -\left(1-\frac{2GM}{r}\right) \dot t^2 +\left(1-\frac{2GM}{r}\right)^{-1} \dot r^2.
\]

This means that as the surface of the star approaches the Schwarzschild radius, \footnotesize$\dot t \to \infty, \dot r \to0\text{ and } \mathrm{d}r/\mathrm{d}t \to 0$. So in these coordinates the star never reaches its Schwarzschild radius. I believe everyone on this thread agrees on this.

What all of us are saying is that this does not mean that the collapse takes infinite time. This is easy to see, because the radial velocity behaves as \footnotesize $ \dot r \sim \left|1-\frac{2GM}{r}\right|^{1/2}$ and

\footnotesize
\[
\tau = \int^{2GM}\mathrm{d}r/\dot r \sim \int^{2GM} \mathrm{d}r\
\left|1-\frac{2GM}{r}\right|^{-1/2} < \infty
\]

So the proper time to reach r = 2GM is finite, and it is possible to prove that the time from r = 2GM to r = 0 is also finite. If one does things properly, the time of the collapse can be estimated as ~ 10-5 M/MSun seconds.

Agent : Orange

29th December 2007, 09:57 AM

I'm sure it's just a typo, but the Kretschmann scalar diverges (is infinite) at r = 0.

Whoops! Excuse me! :)

Agent : Orange

29th December 2007, 10:06 AM

The reason I ask is because especially in the last page of posts, there has been an onslaught of misinformed statements, and I'm just trying to figure out where it's all coming from. I feel like in order to get us all on the same page, I would have to go through at least 2 semesters worth of material. Or if you are trained in this subject, I could simply show the math, which would be irrefutable.

Instead of going through 2 semesters worth of material, why not just point out where the mistakes were made, and why you feel your interpretations are correct on that topic? Where are these misinformed statements? You should go ahead and show the math regardless of whether or not all posters are "on the same page" because that would clear it all up straight away. Yllanes did this and I think we all agree on the interpretation of the situation that he posted.

DrBaltar

29th December 2007, 11:11 AM

Yllanes
Thanks for posting the math. That's good stuff. You say
>So in these coordinates the star never reaches its Schwarzschild radius. I believe everyone on this thread agrees on this.

But I'm seeing:
You're very confused.

All three of your particles simply fall in, and none of them see any slowing of the others. That would only happen to an observer that accelerated like crazy to stay out of the horizon, and the redshift and time-dilation (and Hawking particles) are then attributable to the acceleration. The equivalence principle at work.

In any case, for matter collapsing into a black hole, all of it crosses the horizon and is gone quite rapidly, leaving nothing much behind.It seems that sol invictus claims that there is no time dilation between points entering the event horizon at different radii.

And this:
Secondly, the black hole will eventually clear the region of all content outside the horizon, like a giant vacuum cleaner, rendering infalling matter a moot point.

Thirdly, the event horizon of a non rotating black hole is a sphere; you would have to have matter falling in toward every point of the horizon simultaneously and constantly, for eternity.
...
Nothing is going to fall into the horizon eternally, and certainly not at every mathematical point of the horizon at all times.This disagrees with your and my statement that the outer parts of the star never reach the schwarzchild radius as seen from the outside.

sol invictus continues to concentrate on what happens inside the event horizon. And I have continued to acknowledge that in that frame it takes a finite time for the star to collapse. The region inside an event horizon is NOT causally related to the rest of the universe. Yes information comes in, but nothing goes out. That fact separates it from the rest of the universe. Which makes it irrelevant to the discussion of a singularity in THIS universe. Hawking does say that Hawking radiation allows for information to escape. I know I am pond scum when it comes to his intellect, but I am wondering how Hawking radiation happens when extremely dense star matter from further outside the virtual particle's position entering the black hole at relativistic speeds from the virtual particle's POV. I believe this prevents any virtual particles near the event horizon from leaving the vicinity of the black hole, therefore how can Hawking radiation work? If Hawking radiation doesn't work, then no information gets out. If no information gets out, then the inside of the event horizon is separated from the universe. If the inside of the event horizon is separated from our universe, then there is no singularity in our universe, and I don't dispute the fact that there is a singularity inside the event horizon.

I realize that I don't have the credentials to dispute Hawking, but please don't make arguments against my objections that counter what every physicist accepts about black holes like: All three of your particles simply fall in, and none of them see any slowing of the others.

Then there's this:
You're talking about observation from an external viewpoint. As Sol has stated, what we see isn't what's happening, any more than looking at a star 4 light years distant is showing you what the star is doing real-time.
As far as this universe goes (i.e. from an external view point) what you see is what is happening. Yes, in the star's frame it does enter the event horizon in finite time, but from the outside, when does this happen? Never.

sol invictus

29th December 2007, 11:17 AM

There are no relativistic effects in rivers.

http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1981PhRvL..46.1351U

sol invictus

29th December 2007, 11:24 AM

It seems that sol invictus claims that there is no time dilation between points entering the event horizon at different radii.

I don't know what "event horizon at different radii" means, so I can't comment on that.

sol invictus continues to concentrate on what happens inside the event horizon.

Nope - please re-read my posts.

I realize that I don't have the credentials to dispute Hawking, but please don't make arguments against my objections that counter what every physicist accepts about black holes like: All three of your particles simply fall in, and none of them see any slowing of the others.

Your three particles were freely falling in. None of them sees any significant time dilation relative to the others (modulo tidal effects which will be significant only if the particles are far apart compared to the size of the hole). Any physicist that understands GR will tell you the same.

Again, please look at the conformal diagram I posted several times, the one which describes precisely the situation you're interested in (a collapsing star or other spherical matter distribution). Look at the region above the blue shaded area and outside the event horizon. What do you think the metric is in that region?

By the way, at least three of the posters on this thread are pretty clearly phycisists with credentials that equal or exceed yours (not that that should matter). If you want to use math, please go ahead.

jmercer

29th December 2007, 11:33 AM

By the way, at least three of the posters on this thread are pretty clearly phycisists with credentials that equal or exceed yours (not that that should matter). If you want to use math, please go ahead.

:jedi:

:popcorn3

Agent : Orange

29th December 2007, 03:28 PM

:jedi:

:popcorn3
Seconded:

:popcorn3

By the way, I love your icon. The Question kicks all kinds of ass.

Agent : Orange

29th December 2007, 04:22 PM

It seems that sol invictus claims that there is no time dilation between points entering the event horizon at different radii.
No, I think Sol is saying that test particles dropped from r=infinity at different angles will pass through the schwarzschild radius at the same proper time. This is just a consequence of the spherical symmetry of the metric. Think about a spherical shell of matter impinging on the Schwarzschild black hole at the origin. All of the particles that make up that infalling shell are equidistant from the origin and so will see no time dilation between them.

You are right, however, in saying that a "chain" of test particles released periodically from r=infinity would see time dilation of the previously released particle. But this is just the statement that time dilation varies with distance from the origin, which is obvious since the effect vanishes completely at r=infinity. In this second example, picture concentric shells of material falling into the black hole. Then there will be time dilation between separate shells of matter.

Is this what you were talking about?

That fact separates it from the rest of the universe. Which makes it irrelevant to the discussion of a singularity in THIS universe.
"There is no such thing as a naked singularity". It seems that all such singularities come packaged with event horizons. I guess it sort of comes down to how you want to define the "limits" of the universe itself. If you say that anything inside the event horizon is not a part of our universe anymore then I suppose that means the singularity itself is actually outside of the universe. However, we are still left with "something" anomalous (the horizon itself) in our universe. How is this accounted for given the fact that the horizon is a coordinate singularity and not a physical singularity?

The gravitational influence of the singularity actually can be observed in our universe and the horizon grows with each meal the black hole has. It just doesn't mean much to me to say the horizon itself is "separate" from the universe since it's an artifact of using a specific coordinate system in the first place.

As far as this universe goes (i.e. from an external view point) what you see is what is happening. Yes, in the star's frame it does enter the event horizon in finite time, but from the outside, when does this happen? Never.
So what? You *will* lose contact with a signal sent from the surface of the star because of a combination of time dilation and gravitational redshift. The image of the collapsing star doesn't exist forever around the black hole because it's intensity dies off quickly due to the red shift at the event horizon. Once this residual image fades, all that's left is a black hole. So any way you slice it the end product is the same.

DrBaltar

29th December 2007, 05:06 PM

No, I think Sol is saying that test particles dropped from r=infinity at different angles will pass through the schwarzschild radius at the same proper time.

Sol was responding to what I said here:
Imagine 3 particles of this star matter. Particle 1 (p1) is 1m from the event horizon. p2 is 500m from the event horizon. p3 is 1000m from the event horizon. From p2's frame, it is hurdling towards the event horizon at relativistic speeds. From p2's frame, p1 appears to be at almost a stand still. This is because time at p1 is going much slower than time at p2. In p1's frame, p1 is also hurdling towards the event horizon at relativistic speeds. From p3's frame, it also seems to be hurdling towards the event horizon at relativistic speeds, but not quite as fast as p1 or p2 think they are moving. From p3's frame, p2 seems to be moving very slowly and p1 appears to be almost stopped, or as you say, 'hovering'. This is again because time 'flows' slower the further in you go.

You are right, however, in saying that a "chain" of test particles released periodically from r=infinity would see time dilation of the previously released particle. But this is just the statement that time dilation varies with distance from the origin, which is obvious since the effect vanishes completely at r=infinity. In this second example, picture concentric shells of material falling into the black hole. Then there will be time dilation between separate shells of matter.

Is this what you were talking about?Not quite, but close enough. Thank you for seconding that.

"There is no such thing as a naked singularity". It seems that all such singularities come packaged with event horizons. I guess it sort of comes down to how you want to define the "limits" of the universe itself. If you say that anything inside the event horizon is not a part of our universe anymore then I suppose that means the singularity itself is actually outside of the universe. However, we are still left with "something" anomalous (the horizon itself) in our universe. How is this accounted for given the fact that the horizon is a coordinate singularity and not a physical singularity?The penrose diagrams seem to show that the region inside the event horizon is separate from our universe. Since it is not completely causally linked to the outside universe I would agree with that.

The horizon may be anomalous, but it is still something we can mathematically deal with using an appropriate coordinate system, such as the Kruskal-Szekeres coordinate system.

The gravitational influence of the singularity actually can be observed in our universe since the horizon grows with each meal the black hole has. I think this amounts to confusion from definitions. It just doesn't mean much to me to say the horizon itself is "separate" from the universe since it's an artifact of using a specific coordinate system in the first place.No, that's not why I'm saying the interior of the event horizon is separate from our universe. I'm saying it's separate because I'm thinking it is not causally connected to our universe.

So what? You *will* lose contact with a signal sent from the surface of the star because of a combination of time dilation and gravitational redshift. The image of the collapsing star doesn't exist forever around the black hole because it's intensity dies off quickly due to the red shift at the event horizon. Once this residual image fades, all that's left is a black hole. So any way you slice it the end product is the same.Yes, the luminosity rapidly decays, proportional to Exp(-(2/3)*(1/sqrt(3))*(t/2M)). And because of this, black holes will appear to be black. But nevertheless there is still a layer of star matter eternally wrapped around the event horizon, as seen from the outside. It may be black, but would you agree with that?

sol invictus

30th December 2007, 12:24 AM

No, I think Sol is saying that test particles dropped from r=infinity at different angles will pass through the schwarzschild radius at the same proper time. This is just a consequence of the spherical symmetry of the metric. Think about a spherical shell of matter impinging on the Schwarzschild black hole at the origin. All of the particles that make up that infalling shell are equidistant from the origin and so will see no time dilation between them.

No - that's not what I was saying. Freely falling particles at different radii will NOT see the others as red-shifted, so long as the distance between them is small compared to the radius of curvature of the spacetime (which near the horizon means the radius of the hole). That's an immediate consequence of the equivalence principle - detecting a redshift would constitute a local experiment for which a freely falling observer can detect a grav. field - and it also follows from the fact that the near-horizon metric is Rindler.

Furthermore it is NOT true that the particles will see the others as hovering near the horizon - that's essentially the same thing as a redshift. By the way, another way to see this is simply to draw some 45 degree lines on one of those conformal diagrams. Those lines are the trajectories of radial photons, and you can see immediately why there is an infinite redshift to an asymptotic observer, and why there is no redshift between two freely falling observers.

Again, I stress that what I am saying only holds true if the distances separating the particles are small compared to the radius of curvature. One particle will reach the singularity eventually, and then its redshift will be large (the radius of curvature goes to zero there).

sol invictus

30th December 2007, 07:01 AM

By the way, here's the math associated with this. If we're interested in the near-horizon region, we can expand $f(r)=1-2GM/r=1-r_h/r$ around the horizon radius, $r=r_h(1 + \epsilon)$. This gives $ds^2 = -\epsilon dt^2 + r_h^2 d\epsilon^2/\epsilon + r_h^2 d\Omega^2$ to leading order. The approximation is accurate when $\epsilon = (r-r_h)/r_h \ll 1$ up to quadratic order in epsilon (there is a linear correction to the sphere term in the metric, but that doesn't concern us here).

But this metric is simply flat space in non-standard coordinates (I can show that explicitly if anyone doubts it). So we see that the metric near the horizon is nothing but flat Minkowski space up to corrections quadratic in the distance from the horizon divided by the horizon radius, as expected from the equivalence principle. Therefore particles following geodesics (freely falling particles) in the near horizon region have no significant redshift relative to each other, and do not stop (or appear to stop) at the horizon, so long as the distances involved are less than the horizon radius.

If you find this confusing, remember that these freely falling particles will inevitably fall into the horizon. In order to stay outside, they must accelerate, and it is that acceleration that causes the redshift we were discussing. If they do not accelerate and continue to fall freely, they disappear into the horizon in finite time. An observer who accelerates and remains outside the horizon will never see this occur with photons, but that is no more indicative of something remaining frozen on the horizon, or time stopping, than not hearing sound from something crossing an acoustic horizon is.

jmercer

30th December 2007, 08:16 AM

SI, let me recap what you just said to make certain I understand it, please. (Please forgive my layman's inexactitude in phrasing, and feel free to correct anything I'm missing.)

You're saying:

1) An observer outside the horizon (who is not falling in - essentially, accelerating out of the gravity gradient by holding station-keeping off the black hole) will perceive the particles as NOT reaching the horizon because the photons showing the event would take an infinite amount of time to reach the observer.

2) This would give the invalid perception that the particles were "frozen in time" at (or near) the horizon based on the last photons able to escape the area and reach the observer.

3) The particles actually do fall through the horizon, even though the photonic evidence is never reported.

sol invictus

30th December 2007, 01:25 PM

Yes, precisely.

As a parenthetical comment, if we include Hawking radiation things get much more interesting. In that case, so long as we only ask about observers that remain only outside the horizon, it does make sense to regard the horizon as an impermeable membrane. However time does not stop there (because the membrane is "stretched" a little outside where the horizon would be), the membrane is extremely hot, and anything that gets close to it melts and is re-radiated. This is really another story, and the classical physics (which is all we've been discussing so far) must be understood thoroughly before the effects of Hawking radiation can be tackled.

jmercer

30th December 2007, 01:56 PM

Excellent - thanks for the clarification. Once we're past this bump with everyone, I'd love to discuss Hawking radiation effects. :)

I was also hoping at some point for some dialogue on how Kerr holes might have two horizons... but again, I don't want to derail or jump ahead until this part is resolved to everyone's satisfaction. (Or at least, to the limits of their patience. :D)

Agent : Orange

30th December 2007, 02:29 PM

Therefore particles following geodesics (freely falling particles) in the near horizon region have no significant redshift relative to each other, and do not stop (or appear to stop) at the horizon, so long as the distances involved are less than the horizon radius.

If you find this confusing, remember that these freely falling particles will inevitably fall into the horizon. In order to stay outside, they must accelerate, and it is that acceleration that causes the redshift we were discussing. If they do not accelerate and continue to fall freely, they disappear into the horizon in finite time. An observer who accelerates and remains outside the horizon will never see this occur with photons, but that is no more indicative of something remaining frozen on the horizon, or time stopping, than not hearing sound from something crossing an acoustic horizon is.

Yeah, I realized that within a small enough region they'd have to not see any redshift because of the equivalence principle. I agree, and thanks for all the extra details as well.

Yllanes

30th December 2007, 04:38 PM

I was also hoping at some point for some dialogue on how Kerr holes might have two horizons...

That's quite interesting, but I am afraid it may complicate the thread. Remind us later to write about it, or you can start a different thread. Although charged black holes, which also have two horizons, are easier.

jmercer

30th December 2007, 05:19 PM

That's quite interesting, but I am afraid it may complicate the thread. Remind us later to write about it, or you can start a different thread. Although charged black holes, which also have two horizons, are easier.

Will do. :)

Plus I'm more interested in Hawking radiation and how it works at the moment. :)

DrBaltar

30th December 2007, 06:16 PM

I'm not sure yet if this is the right way to post images of equations, but here it goes...

This is in response to sol's objections to time dilations between infalling particles into a black hole. The particles I mentioned before were at 1m, 500m, and 1000m from the event horizon. I changed the 2nd one to 2m to make it more interesting. In the equation attachment, In[1] is the mass of the star, which is 10 times the sun's mass. In[2] - In[6] set up constants including the positions of the particles. In[7,8,9] produce the ratios between the frequencies between particles at r1,r2 and r3.

And for something interesting, In[10] is the frequency of light from a blue light at r1. In[11] multiplies it by the redshift factor .707119 with the result output at Out[12]: 678nm. At 2m from the event horizon, a blue light looks red.

DrBaltar

30th December 2007, 08:46 PM

That equation image came out way too small. How do I upload the image in its full resolution?

sol invictus

30th December 2007, 10:31 PM

That equation image came out way too small. How do I upload the image in its full resolution?

I can't read the image (and I have limited internet access from here, or I'd download it and try to blow it up).

One comment though - it's not sufficient to show that there can be a redshift between particles close to the horizon and those far away. What you need to show is that there must be - or better, that as the particles fall along their geodesics, the redshift increases as one approaches the horizon.

The reason is that in flat space particles in relative motion will have a redshift between them (which is just the standard Doppler shift). So if you pick two geodesics, there will generically be a redshift. However one thing that distinguishes flat space is that the redshift will not change with time, unlike what you were claiming for the hole. So you'd need to demonstrate that the redshift between two inertial particles changes with time as they fall.

Yllanes

31st December 2007, 01:49 AM

That equation image came out way too small. How do I upload the image in its full resolution?

You attached the image. Attachments have strict size limits. The way to post a full size image is to host it online (you can do it in the forum here (http://forums.randi.org/vbimghost.php?do=upload)) and then link to it with ... tags.

69dodge

31st December 2007, 06:06 AM

It doesn't spare us, because in the reference frame of the star the collapse takes only a fraction of a second.

But by then an infinite amount of time has passed for the rest of the universe. The show is over.

"by then"?

Can we really compare the times of distant events like that? I don't think so. Simultaneity is relative. Someone who falls into a black hole and someone who stays out of it never meet again to compare their watches. In what sense can it be said that A's watch reads 10:00 "at the same time" that B's watch, far away, reads 11:00?

DrBaltar

31st December 2007, 07:23 AM

Ok, here's plan B. And I'll repeat the text I wrote for the image from above...

This is in response to sol's objections to time dilations between infalling particles into a black hole. The particles I mentioned before were at 1m, 500m, and 1000m from the event horizon. I changed the 2nd one to 2m to make it more interesting. In the equation attachment, In[1] is the mass of the star, which is 10 times the sun's mass. In[2] - In[6] set up constants including the positions of the particles. In[7,8,9] produce the ratios between the frequencies between particles at r1,r2 and r3.

And for something interesting, In[10] is the frequency of light from a blue light at r1. In[11] multiplies it by the redshift factor .707119 with the result output at Out[12]: 678nm. At 2m from the event horizon, a blue light looks red.
http://www.geocities.com/quarkburger/pics/redshift.gif

ETA: sol, I'll try and modify this to free falling particles, but I'll have to squeeze that in between remodeling the bathroom and fixing cars while I have time off.

jmercer

31st December 2007, 12:45 PM

:bump4

sol invictus

31st December 2007, 06:49 PM

This is more naive than I had guessed. Disregard my comment from above.

ETA: sol, I'll try and modify this to free falling particles, but I'll have to squeeze that in between remodeling the bathroom and fixing cars while I have time off.

So you do understand that this has no bearing whatsoever on the issue we were discussing?

This is the redshift for accelerating particles at fixed radius in the Schwarzschild metric, which everyone agrees is there, and which goes to infinity when one particle is at the horizon. Precisely the same effect is there in flat space (for accelerating particles). Freely falling particles do not experience such a redshift.

Anyway, gotta go - happy new year!

DrBaltar

1st January 2008, 07:50 AM

Agent : Orange

1st January 2008, 09:32 AM

Happy New Year everyone!

I had left a comment/question for Agent:Orange or for anyone else who wants to answer:

Yes, the luminosity rapidly decays, proportional to Exp(-(2/3)*(1/sqrt(3))*(t/2M)). And because of this, black holes will appear to be black. But nevertheless there is still a layer of star matter eternally wrapped around the event horizon, as seen from the outside. It may be black, but would you agree with that?

Happy new year!

No! I would totally disagree with what you've written. It's been shown and can easily be shown that the proper time of the infalling particles is well-behaved. Stellar matter falls right through the event horizon though an asymptotic observer will see time slow to a crawl at the event horizon. This has been described many times on this thread and you yourself have acknowledged this fact. How many times does it need to be repeated?

Think about what you are arguing for here: From the perspective of a test particle falling into the hole the path of the particle crosses the horizon and reaches the r=0 singularity. In one frame the matter passes right through, but in all others it's wrapped around the horizon? That's completely unphysical and paradoxical.

You're bringing up stuff that we've already discussed and put to bed pages ago. You should go back to the beginning of the thread and read closely. You may find your problems have already been solved.

DrBaltar

2nd January 2008, 07:11 AM

Happy new year!

No! I would totally disagree with what you've written. It's been shown and can easily be shown that the proper time of the infalling particles is well-behaved. Stellar matter falls right through the event horizon though an asymptotic observer will see time slow to a crawl at the event horizon. This has been described many times on this thread and you yourself have acknowledged this fact. How many times does it need to be repeated?

Think about what you are arguing for here: From the perspective of a test particle falling into the hole the path of the particle crosses the horizon and reaches the r=0 singularity. In one frame the matter passes right through, but in all others it's wrapped around the horizon? That's completely unphysical and paradoxical.

You're bringing up stuff that we've already discussed and put to bed pages ago. You should go back to the beginning of the thread and read closely. You may find your problems have already been solved.

Why are you rehashing that the proper time of an infalling particle is well-behaved again? Did I call that into question? No. As you say this has been rehashed over and over (for reasons I cannot fathom since we are concerned with what an asymptotic observer sees on the outside), and as I have said, and as you have seen me say, I agree with this.

You are also contradicting yourself. You are saying both that "an asymptotic observer will see time slow to a crawl at the event horizon", and that it's completely unphysical and paradoxical to say that star matter collapsing into a black hole will be eternally wrapped around the event horizon as seen from the outside.

You can't have it both ways. Star matter collapsing into a black hole will appear to be eternally wrapped around the event horizon as seen from the outside because an asymptotic observer will see time slow to a crawl at the event horizon.

In one frame the matter passes right through, but in all others it's wrapped around the horizon? That's completely unphysical and paradoxical.
Anyone studying relativity should have the subject of simultaneity drilled into their heads. Relativity shows that no two observers can agree on simultaneous events.

BTW, the radius of the outside shell of the star, as a function of t (time in an asymptotic observer's frame) is:

R(t) = Rs + (R0 - Rs)*exp(-t/Rs)

where Rs is the Schwarzschild radius, and R0 is the radius of the shell before collapse.

For all t, R(t) > Rs.

Agent : Orange

2nd January 2008, 09:54 AM

The "shell of matter" is essentially a time dilated image of the star's surface as the stellar material crosses the horizon. The gravitational redshift will cause this image to rapidly fade. I apologize if my last post was a bit gruff. I've mentioned the effect of the gravitational redshift before and I should have brought it up again.

For an asymptotic observer and an observer falling through the horizon the end result in practice is the same - a black hole. There's no physical material structure that encases the horizon around a black hole. Because the proper time of an infalling particle is well-behaved this stuff crosses the horizon without incident. That's what I'm trying to say.

sol invictus

2nd January 2008, 09:55 AM

You continue to make the elementary mistake of confusing seeing with observing. Just because an outside observer sees matter slowing down does not mean it actually slows down.

Please think about the acoustic horizon example I gave above. It has all the relevant features of a black hole, including time stopping (if you define time in the way you insist on doing). But no reasonable person - certainly not one that understood the physics - would conclude from what they were hearing that matter slows down and accumulates at the acoustic horizon.

69dodge

2nd January 2008, 03:23 PM

You continue to make the elementary mistake of confusing seeing with observing. Just because an outside observer sees matter slowing down does not mean it actually slows down.

I think I know what each of those terms means in special relativity, but observing seems fuzzy in general relativity. In special relativity, a unique reference frame can be associated with an inertial observer, namely, the inertial reference frame in which he's motionless. But in general relativity, there's a great deal of freedom in choosing a coordinate system, and less reason to prefer one to another.

What's the "right" coordinate system for a distant observer to use, whose time coordinate will tell him "when" some bit of matter really crosses the event horizon?

The question isn't meaningful, is it?

The only thing that matters to a distant observer is what he sees, as far as I can tell, because nothing else can affect him in any way. Am I wrong here?

sol invictus

2nd January 2008, 03:35 PM

The only thing that matters to a distant observer is what he sees, as far as I can tell, because nothing else can affect him in any way. Am I wrong here?

Yes and no. The outside observer could choose to dive into the black hole, either pulling up at the last moment and remaining outside the horizon, or actually falling in. Either way she will learn something she would have had to wait a long time to see if she remained far away. It's true that if she comes very close to the horizon (but stays out), a lot of time will pass for the distant observer before she can rejoin him (that's essentially nothing but the familiar twin "paradox" from SR).

Note that all these statements are perfectly true also for an acoustic horizon, so long as we posit that the creatures investigating it don't posses supersonic submarines.

But the key question under discussion here was whether the matter "really" accumulates and remains forever outside the horizon. We can give that question a precise meaning - we can ask, is there a region of the spacetime, close to the horizon, which is empty of matter (because it fell in) and is just described by the vacuum black hole metric? The answer is yes, as can be seen from the conformal diagram, or from the fact that the proper time is finite.

Good question, though. And as I've said, there is a sense in which we can consistently regard the horizon as impermeable. But it's rather subtle, and it does NOT involve the infalling matter slowing down and freezing - just the opposite!

DrBaltar

2nd January 2008, 06:08 PM

You continue to make the elementary mistake of confusing seeing with observing. Just because an outside observer sees matter slowing down does not mean it actually slows down.

I'm not talking about the light. I'm talking about the position of the outer shell of stellar matter as a function of time, t, in an asymptotic observer's frame:

R(t) = Rs + (R0 - Rs)*exp(-t/Rs)

where Rs is the Schwarzschild radius, and R0 is the radius of the shell before collapse.

For all t, R(t) > Rs.

Please think about the acoustic horizon example I gave above. It has all the relevant features of a black hole, including time stopping (if you define time in the way you insist on doing). But no reasonable person - certainly not one that understood the physics - would conclude from what they were hearing that matter slows down and accumulates at the acoustic horizon.

I read the abstract that you linked to. I think it is dangerous to use analogies to explain a physical system. The analogy may capture some, but not all of the effects. And second, it seems that you have drawn the wrong analogy from it the link you gave which says this:

Attention is called to the existence of a physical system which has all the properties of a black hole as far as the quantum thermal radiation is concerned but in which all of the basic physics is completely understood. This model of the behavior of a quantum field in a classical gravitational field is the motion of sound waves in a convergent fluid flow. This system is considered to form an excellent theoretical laboratory where many of the unknown effects that quantum gravity could exert on black-hole evaporation can be modeled. It is noted that the low-energy fluid equations which have led to quantum thermal sonic emission by a transonic background flow break down at high frequencies because of the atomic nature of the fluid. The phonons emitted are quantum fluctuations of the fluid flow and therefore affect their own propagation in exactly the same way that graviton emission affects the space-time on which the various relativistic fields propagate. It is shown that the very arguments which lead to black-hole evaporation also predict that a thermal spectrum of sound waves should be given out from the sonic horizon in transonic fluid flow.

Both the title and the abstract discuss quantum thermal radiation, or black-hole evaporation. It may be relevant to black-hole evaporation, or may not. As I say, I am reluctant to draw too many similarities from such different physical processes. In any case, I'm not sure how this is relevant to the collapse of a star into its event horizon.

sol invictus

2nd January 2008, 08:05 PM

I'm not talking about the light. I'm talking about the position of the outer shell of stellar matter as a function of time, t, in an asymptotic observer's frame:

No one disagrees with you on that. We do disagree (or at least I do) with your statement that matter accumulates forever (in any meaningful sense) on the horizon, particularly when you use it to argue that Hawking radiation will therefore be affected since there is "always" matter near the horizon. That's just not the case - there are large regions of the spacetime where all the matter has passed through the horizon, leaving behind a vacuum black hole solution. That is the region where nearly all of the Hawking particles originate, and that part of the spacetime is not well described by the coordinates you are using (because they are singular on the horizon).

I've asked you at least four times to look at the Penrose diagrams you yourself brought up - only look at the ones showing the formation of a black hole by collapse - and you have not responded. Those diagrams show clearly what I keep telling you, and until you look at them, there is no point in discussing this further.

I read the abstract that you linked to. I think it is dangerous to use analogies to explain a physical system. The analogy may capture some, but not all of the effects.

All of the effects you have been using to try to argue for your case are identical for an acoustic horizon, including the "stopping" of time. In fact the Navier-Stokes equations (in a certain limit) can be put in precisely the same form as the equations of motion for a scalar field in a black hole spacetime, by choosing the time coordinate you like. That's what he shows in that paper - he's interested in using it to study Hawking radiation, but the classical effects we're talking about are there too (they're just not very interesting because they are very well understood).

DrBaltar

2nd January 2008, 11:54 PM

I'm not talking about the light. I'm talking about the position of the outer shell of stellar matter as a function of time, t, in an asymptotic observer's frame:
No one disagrees with you on that. We do disagree (or at least I do) with your statement that matter accumulates forever (in any meaningful sense) on the horizon
You don't disagree with what exactly? What you quoted from me talked about the position of the outer shell of stellar matter as a function of time. The equation after what you quoted, and what the quote was about shows that matter accumulates forever (as seen from the outside) on the horizon.

I ran the numbers, and admittedly, the shell becomes very thin pretty quickly. With a 10 solar mass star, with an initial diameter of 1AU, assuming the outer 1/2 of the star free falls to the event horizon, after 10 days it would become a 1mm thick shell around the event horizon as measured from the outside. It continues to shrink rapidly but never completely goes away.

Virtual particles can form outside this shell, and some pairs of particles will have one fall in behind the shell, with the other particle leaving the black hole. The particle that falls in behind the shell will follow the shell into the event horizon (see I said it - I don't want any comments saying that in the particle's frame it passes into the event horizon without any problem - I know this). The problem is, that from an outside observer's point of view, entering after the shell enters, is waiting for eternity.

I've asked you at least four times to look at the Penrose diagrams you yourself brought up - only look at the ones showing the formation of a black hole by collapse - and you have not responded. Those diagrams show clearly what I keep telling you, and until you look at them, there is no point in discussing this further.I tried following the link to your diagram to see what kind of diagram it was. It points to someone's blog, but I am unable to find which blog entry of his refers to that diagram. I did find another blog on his site here (http://golem.ph.utexas.edu/~distler/blog/archives/000256.html), which discuss paradoxes regarding the interaction of infalling matter and hawking radiation. Apparently in terms of quantum gravity, it's not adding up to the unitary probability field we all like to see.

jmercer

3rd January 2008, 02:42 AM

DrBaltar, I think some of the confusion regarding the topic at hand stems from the following statements you've made:

sol invictus,

The part I don't get is if it takes an infinite amount of time for the star to completely collapse, then the region outside the black hole is not a vacuum. So no vacuum fluctuations and no virtual particles. So how does Hawking radiation happen? I am skeptical on this, especially since it has never been observed. I thought this theory applied to a 'classic' black hole in which the collapse has already happened and the region outside the event horizon was a vacuum.

You haven't addressed my point though. If it takes an infinite amount of time for a star to completely collapse, as seen from an outside observer, then the region above the event horizon is occupied by star matter. That is what I mean when I say there's no vacuum outside the event horizon. And I believe the virtual particles must appear just outside the event horizon. One particle goes in with negative energy, and the other escapes with positive energy.

So basically I can't imagine there would be much empty space in matter that dense, and even if there were, I certainly cannot imagine any of the virtual particles flying through matter that dense that is hurdling towards the event horizon at relativistic speeds. Any virtual particles speeding away from the black hole would easily get knocked back towards the event horizon with the infalling star matter.

jmercer

1) if the virtual particles only 'exist lest than a plank' then Hawking radiation couldn't happen. It is because one falls into the black hole, and the other escapes the region around the black hole that it can last.

2) See my previous post about matter falling into the black hole.

3) If a spherical star is collapsing into a black hole, then yes, you would have extremely dense star matter falling toward every point of the horizon simultaneously and constantly, for eternity.

4) Even neutrinos can interact with ordinary matter. It's rare, because atoms are mostly empty space, but it can happen. As I said the star matter of a star collapsing into a black hole is much denser than ordinary matter and it would certainly react.

sol is not correct. He is completely ignoring the fact that time slows down in a gravity well, and stops at the event horizon. Again see my previous posting.

You don't disagree with what exactly? What you quoted from me talked about the position of the outer shell of stellar matter as a function of time. The equation after what you quoted, and what the quote was about shows that matter accumulates forever (as seen from the outside) on the horizon.

I ran the numbers, and admittedly, the shell becomes very thin pretty quickly. With a 10 solar mass star, with an initial diameter of 1AU, assuming the outer 1/2 of the star free falls to the event horizon, after 10 days it would become a 1mm thick shell around the event horizon as measured from the outside. It continues to shrink rapidly but never completely goes away.

Virtual particles can form outside this shell, and some pairs of particles will have one fall in behind the shell, with the other particle leaving the black hole. The particle that falls in behind the shell will follow the shell into the event horizon (see I said it - I don't want any comments saying that in the particle's frame it passes into the event horizon without any problem - I know this). The problem is, that from an outside observer's point of view, entering after the shell enters, is waiting for eternity.

I tried following the link to your diagram to see what kind of diagram it was. It points to someone's blog, but I am unable to find which blog entry of his refers to that diagram. I did find another blog on his site here (http://golem.ph.utexas.edu/%7Edistler/blog/archives/000256.html), which discuss paradoxes regarding the interaction of infalling matter and hawking radiation. Apparently in terms of quantum gravity, it's not adding up to the unitary probability field we all like to see.

The issue is that you went from a eternal shell of matter so dense as to prevent particles from falling into the horizon... to one that thins eternally but never disappears, and doesn't prevent particles from falling in and Hawking radiation from forming.

How do you reconcile the series of above statements?

sol invictus

3rd January 2008, 05:46 AM

What jmercer said.

The diagram was in one of my posts, as well as linked to. Here it is again:

http://golem.ph.utexas.edu/~distler/blog/svg/bhformation.gif

For the fifth time, look at the part of the spacetime above and to the right of the blue shaded strip (which is the infalling matter/radiation which creates the hole). That region is part of the vacuum Schwarzschild black hole solution, with a horizon, singularity, asymptotic infinity, and no matter.

You don't disagree with what exactly? What you quoted from me talked about the position of the outer shell of stellar matter as a function of time. The equation after what you quoted, and what the quote was about shows that matter accumulates forever (as seen from the outside) on the horizon.

That equation shows that in a particular time coordinate matter never crosses the horizon. As I keep trying to explain to you, I can pick a system of coordinates in flat empty space for which any surface I like acts as a horizon - in fact the metric in those coordinates is locally identical to a BH horizon, and that exact same equation gives the distance from freely falling matter to that surface as a function of time.

It means nothing to observe that there exists a time coordinate in which it takes infinite time to cross some surface - you must be more careful.

It continues to shrink rapidly but never completely goes away.

In that time coordinate, yes. So what?

Virtual particles can form outside this shell, and some pairs of particles will have one fall in behind the shell, with the other particle leaving the black hole. The particle that falls in behind the shell will follow the shell into the event horizon (see I said it - I don't want any comments saying that in the particle's frame it passes into the event horizon without any problem - I know this). The problem is, that from an outside observer's point of view, entering after the shell enters, is waiting for eternity.

As you have noticed, the shell get extremely thin and close to the horizon in your coordinates. But Hawking particles form some distance outside the horizon - a distance inversely proportional to their energy. In fact it is absolutely meaningless to even discuss the position of the horizon to an accuracy better than one Planck length. So even in your coordinates, once the shell is that close, it is impossible to tell whether it is outside it or not, and it does not affect Hawking's calculation.

DrBaltar

3rd January 2008, 05:54 AM

The issue is that you went from a eternal shell of matter so dense as to prevent particles from falling into the horizon... to one that thins eternally but never disappears, and doesn't prevent particles from falling in and Hawking radiation from forming.

How do you reconcile the series of above statements?Because when I started on this topic I was going by intuition. I knew that, according to GR, time stops at the event horizon, as seen from the outside, and that collapsing star matter would have to stop short of entering the event horizon. I knew, but may not have mentioned it, that the star matter around the event horizon wouldn't be all that thick. I did not know specifically how fast it would become so thin.

As the topic went on, you guys understandably asked for the math. As you may have noticed I wasn't posting as frequently after that - I was reading up on it, and trying to follow the math. Then I found that the surface of the star would collapse according to:

R(t) = Rs + (R0 - Rs)*exp(-t/Rs)

where Rs is the Schwarzschild radius, and R0 is the radius of the shell before collapse.

For all t, R(t) > Rs.

After plugging in the numbers last night, I found that it collapses very quickly, i.e. down to 1mm in 10 days. So yes, the details may have changed, but the overall principles stayed the same, which are:
- As seen from the outside, there is a shell of collapsed star matter eternally wrapped around the event horizon.
- During the collapse, while there is significant thickness of star matter (which I recently found was on the order of days rather than millenia) assuming any virtual particles could form within that collapsing star matter, they would be dragged into the event horizon along with the star matter.
- The region inside the event horizon is not completely causally connected to the rest of the universe.

Before last night, I didn't see any reason to consider virtual particles forming outside the surface of the collapsing star because I didn't realize the shell would become so thin so fast, but it does. After I thought about it, it's pretty obvious that the particles outside the surface of the star would not overtake the shell of matter as they enter the event horizon. And if they could overtake the star matter, I still maintain that the shell is too dense to allow them to pass through. And if they could not overtake the infalling shell of matter, then since it takes an infinite amount of time for the surface of the star to completely collapse, as seen from the outside (ASFTO - time to turn that into an acronym), then the virtual particles must also take an infinite amount of time to enter the event horizon.

jmercer

3rd January 2008, 10:04 AM

Thank you for clarifying - the process you were undergoing to reach your current position wasn't visible to the rest of us. :)

I now return you all to your regularly scheduled debate. :D

sol invictus

3rd January 2008, 07:30 PM

I now return you all to your regularly scheduled debate. :D

Well, I for one have nothing further to add at this point. The physics is really quite clear.

DrBaltar

3rd January 2008, 09:10 PM

Sweet! Bartender, a round for everyone!
:alc::alc:

Agent : Orange

4th January 2008, 02:28 AM

Whooohoooo! Beer!

sol invictus

4th January 2008, 03:04 AM

Sweet! Bartender, a round for everyone!
:alc::alc:

Yeah, getting drunk might be the only reasonable option. Did you look at the diagram this time, or are you still pretending it doesn't exist?

jmercer

4th January 2008, 03:22 AM

I've noticed that alcohol - when taken in sufficient quantities - has a sort of a red-shift effect. ;)

DrBaltar

4th January 2008, 07:12 AM

Sorry Sol, I didn't see the post with the diagram you posted yesterday since you sumbmitted it at the same time I was writing my post.

The diagram was in one of my posts, as well as linked to. Here it is again:

<<image>>

For the fifth time, look at the part of the spacetime above and to the right of the blue shaded strip (which is the infalling matter/radiation which creates the hole). That region is part of the vacuum Schwarzschild black hole solution, with a horizon, singularity, asymptotic infinity, and no matter.I understand what the diagram seems to show. For the second time, I'm trying to find out what type of diagram that is, and would like to know the source of it.

That equation shows that in a particular time coordinate matter never crosses the horizon.Correct. And the particular time coordinate it shows is that of an outside observer, which is what is being discussed.

It means nothing to observe that there exists a time coordinate in which it takes infinite time to cross some surface - you must be more careful.Ok, you tell me, when the question is 'does an asymptotic observer ever see a star collapse into a black hole', what frame is the relevant frame?

In that time coordinate, yes. So what?So since that is the time coordinate of an asymptotic observer, that is what the asymptotic observer sees.

As you have noticed, the shell get extremely thin and close to the horizon in your coordinates. But Hawking particles form some distance outside the horizon - a distance inversely proportional to their energy. In fact it is absolutely meaningless to even discuss the position of the horizon to an accuracy better than one Planck length. So even in your coordinates, once the shell is that close, it is impossible to tell whether it is outside it or not, and it does not affect Hawking's calculation.That is interesting, can you cite an equation for the distance Hawking particles form outside the horizon?

sol invictus

4th January 2008, 07:24 AM

I understand what the diagram seems to show. For the second time, I'm trying to find out what type of diagram that is, and would like to know the source of it.

It's a conformal diagram (sometimes called Penrose or Carter-Penrose diagram) of the metric describing the formation of a black hole by the collapse of spherically symmetric radiation. It can be found in many papers and many sources on the web; I'm not going to dig them up for you, but I'm sure you can find them with google or another tool. You can also just solve Einstein's equations yourself and find the metric - it's actually totally trivial to do so using Birkhoff's theorem, especially if you take the collapsing shell to be thin.

You can also take the diagram for an eternal black hole you posted earlier and ask yourself what some radiation (or matter) falling into it will do. You need to solve for the geodesics and then plot them on the diagram. For radiation (massless particles), that's completely trivial - the whole point of a conformal diagram is that radial null geodesics are straight lines at a 45 degree angle, so an ingoing photon simply goes straight into the hole (just like the blue region in the collapse diagram above). For massive particles the path is curved and everywhere more steep than 45 degrees, but still qualitatively similar - it crosses the horizon at some finite spacetime point, and then later hits the singularity. In the future of that spacetime horizon-crossing point there is no longer any matter outside the horizon. While you can check this directly by solving for the geodesics, it is also proven by the fact (which you say you agree with) that the proper time for the infalling particle is finite.

Correct. And the particular time coordinate it shows is that of an outside observer, which is what is being discussed.

That depends on which outside observer. Probably you want to consider a observer very, very far away? In that case the difference between freely falling and fixed radial difference to the hole is negligible. Or you could consider an outside observer that dives in close to the horizon and then pulls out, in order to test whether there is any matter there. Or you could consider an observer outside that accelerates faster than necessary to stay out of the hole, in which case they will see a horizon outside the black hole horizon, and will see matter slow down and stop, and time come to a standstill, at a surface outside the hole's horizon. All of those see different things and have different coordinate systems in which they are at rest.

Ok, you tell me, when the question is 'does an asymptotic observer ever see a star collapse into a black hole', what frame is the relevant frame?

As far as I can tell, there has never been any disagreement on that question. Outside observers never see radiation cross the horizon - by definition. Just as outside listeners never hear anything cross an acoustic horizon.

That is interesting, can you cite an equation for the distance Hawking particles form outside the horizon?

Sure: d = 1/E (d=distance, E=energy, all in natural units).

DrBaltar

5th January 2008, 11:22 PM

It's a conformal diagram (sometimes called Penrose or Carter-Penrose diagram) of the metric describing the formation of a black hole by the collapse of spherically symmetric radiation. It can be found in many papers and many sources on the web; I'm not going to dig them up for you, but I'm sure you can find them with google or another tool.
Oh yeah, I found it on that website we looked at earlier. We already talked about this one, and I already debunked it on the last half of my post here (http://forums.randi.org/showthread.php?p=3279634#post3279634). Anyways it's a diagram showing an incoming sphere of radiation at the speed of light. That is different from the surface of a collapsing star which has rest-mass and is not falling at the speed of light.

You can also take the diagram for an eternal black hole you posted earlier and ask yourself what some radiation (or matter) falling into it will do. You need to solve for the geodesics and then plot them on the diagram. For radiation (massless particles), that's completely trivial - the whole point of a conformal diagram is that radial null geodesics are straight lines at a 45 degree angle, so an ingoing photon simply goes straight into the hole (just like the blue region in the collapse diagram above). For massive particles the path is curved and everywhere more steep than 45 degrees, but still qualitatively similar - it crosses the horizon at some finite spacetime point, and then later hits the singularity. In the future of that spacetime horizon-crossing point there is no longer any matter outside the horizon. While you can check this directly by solving for the geodesics, it is also proven by the fact (which you say you agree with) that the proper time for the infalling particle is finite.
Ok, I plotted the surface of the star according to the radius function as a function of time:
R(t) = Rs + (R0 - Rs)*exp(-t/Rs)
http://www.geocities.com/quarkburger/pics/KS_collapse.gif
In this Kruskal-Szekeres diagram, the surface of the star is in red. It has crossed the r=2.1M line and continues in. The curve of the red line shows that it asymptotically closes in on the diagonal line going to the upper right (the event horizon) where time goes to infinity. Nothing new here.

That depends on which outside observer. Probably you want to consider a observer very, very far away? In that case the difference between freely falling and fixed radial difference to the hole is negligible. Or you could consider an outside observer that dives in close to the horizon and then pulls out, in order to test whether there is any matter there. Or you could consider an observer outside that accelerates faster than necessary to stay out of the hole, in which case they will see a horizon outside the black hole horizon, and will see matter slow down and stop, and time come to a standstill, at a surface outside the hole's horizon. All of those see different things and have different coordinate systems in which they are at rest.In all these cases, the observer will see the outer shell of the star stop just short of the event horizon, unless he's going to free fall in with it.

Ok, you tell me, when the question is 'does an asymptotic observer ever see a star collapse into a black hole', what frame is the relevant frame?As far as I can tell, there has never been any disagreement on that question. Outside observers never see radiation cross the horizon - by definition. Just as outside listeners never hear anything cross an acoustic horizon.You didn't answer the question. What is the relevant frame?

Sure: d = 1/E (d=distance, E=energy, all in natural units).
I can't think of any combination of G or c that will put 1/E in units of meters??

sol invictus

6th January 2008, 07:17 AM

Oh yeah, I found it on that website we looked at earlier. We already talked about this one, and I already debunked it on the last half of my post here (http://forums.randi.org/showthread.php?p=3279634#post3279634). Anyways it's a diagram showing an incoming sphere of radiation at the speed of light. That is different from the surface of a collapsing star which has rest-mass and is not falling at the speed of light.

You didn't "debunk" anything. And yes, as I told you several times, that diagram is for infalling radiation - but the diagram describing the collapse of matter is not much different. The difference is that the blue region would be curved up a bit. That's it.

Ok, I plotted the surface of the star according to the radius function as a function of time:

You're using coordinates that are singular at the surface you're interested in, which is a bad idea. YOU introduced conformal diagrams, which was a good idea, because they show this kind of thing very clearly. In fact as I now notice, the diagram you posted actually shows precisely what I've been trying to explain to you.

Here's the diagram YOU posted earlier:

http://online.itp.ucsb.edu/online/colloq/hamilton1/oh/penrose_Schwpar.gif

Now, consider some infalling matter. If it doesn't accelerate, it follows a geodesic. Do you know what geodesics look like on this graph?

I'll tell you - a massless particle like a photon will follow a straight line at a 45 degree angle. If it's infalling, it will (say) come from the bottom right moving up and left, enter the horizon at some finite spacetime point, and hit the singularity. If it's massive, it will follow a line bounded by that one, but curved up somewhat. It will NOT suddenly make a right turn and avoid the horizon - not unless it has a rocket engine attached to it.

See that blue line with the arrow? That's the trajectory of a freely falling observer. See how it crosses the horizon?

Notice that there is no matter outside the hole above and to the right of the point where it crosses.

In all these cases, the observer will see the outer shell of the star stop just short of the event horizon, unless he's going to free fall in with it.

Yes, we all agree on that.

You didn't answer the question. What is the relevant frame?

I answered your question - there is no single relevant frame because there are many possible outside observers. Each sees something different, in some cases totally different. For example some see horizons which are nowhere near the hole.

I can't think of any combination of G or c that will put 1/E in units of meters??

You forgot about Planck's constant. This is a quantum effect - there is no Hawking radiation in classical physics. And G can't enter - Hawking (or more properly Unruh) radiation exists for accelerated observers even in flat space.

sol invictus

8th January 2008, 04:02 AM

This is kind of nice. What the say about the spaceship falling in is relevant to the discussion above.

http://www.thinktechnologies.com/portfolio/demos/Blackhole.swf

jmercer

8th January 2008, 07:42 AM

Very nice illustration - thanks, Sol. :)

sol invictus

8th January 2008, 12:11 PM

Very nice illustration - thanks, Sol.

You're welcome.

Ok, I plotted the surface of the star according to the radius function as a function of time:
R(t) = Rs + (R0 - Rs)*exp(-t/Rs)
http://www.geocities.com/quarkburger/pics/KS_collapse.gif
In this Kruskal-Szekeres diagram, the surface of the star is in red. It has crossed the r=2.1M line and continues in. The curve of the red line shows that it asymptotically closes in on the diagonal line going to the upper right (the event horizon) where time goes to infinity. Nothing new here.

By the way, what you said here is incorrect. It's not true that the red line "asymptotically" approaches the horizon. The only thing your equation tells you is that the red line reaches a point where t=infinity and R=Rs. But that entire black diagonal line is t=infinity, R=Rs. (Normally one wouldn't have multiple points described by a single coordinate pair, but you're using coordinates which are singular on the horizon - that's why it happens.)

So the equation does not tell you where the red line crosses the horizon, and in fact it doesn't happen infinitely far up and to the right. If you were careful, you'd discover that it crosses at a finite distance from the origin on that plot (at a point which is determined by R0). Actually, you should be able to see that if you extend your plot a little.

DrBaltar

9th January 2008, 08:00 AM

By the way, what you said here is incorrect. It's not true that the red line "asymptotically" approaches the horizon. The only thing your equation tells you is that the red line reaches a point where t=infinity and R=Rs. But that entire black diagonal line is t=infinity, R=Rs. (Normally one wouldn't have multiple points described by a single coordinate pair, but you're using coordinates which are singular on the horizon - that's why it happens.)

So the equation does not tell you where the red line crosses the horizon, and in fact it doesn't happen infinitely far up and to the right.Time in this diagram is represented by a hyperbolic angle, so it only reaches the 45 degree line at t=infinity. The radial distances are represented by hyperbolas. While it is true that the entire upper right black diagonal line is t=infinity, R=Rs, the limit of R as it approaches Rs from the outside is a hyperbola which only intersects the black line at a point infinitely far up and to the right.

If you were careful, you'd discover that it crosses at a finite distance from the origin on that plot (at a point which is determined by R0). Actually, you should be able to see that if you extend your plot a little.
If I were careful, I would not conclude that the curve representing the radius of the star would cross a line at a single point, when the entire upper right part of the line represents t=infinity, R=Rs.

sol invictus

9th January 2008, 08:37 AM

Time in this diagram is represented by a hyperbolic angle, so it only reaches the 45 degree line at t=infinity. The radial distances are represented by hyperbolas. While it is true that the entire upper right black diagonal line is t=infinity, R=Rs, the limit of R as it approaches Rs from the outside is a hyperbola which only intersects the black line at a point infinitely far up and to the right.

Nope.

As you have just agreed, that entire line is R=Rs, t=infinity. Hence where you end up on that line in the limit R->Rs, t->infinity depends on how you take the limit - in particular, it depends on R0 in your formula.

Let's see explicitly how that works. You've labeled the axes u and v (which is non-standard - those are usually null coordinates - but we'll stick with your notation). The horizon is the line u=v. Now, we need to know how u and v are related to t and R. The answer is

$v = \sqrt{R/R_s-1}~e^{R/2R_s}\sinh(t/2 R_s)$
$u = \sqrt{R/R_s-1}~e^{R/2R_s}\cosh(t/2 R_s)$.

Your formula was R(t) = Rs + (R0 - Rs)*exp(-t/Rs). Taking the limit of large t and plugging your formula into the above gives

$v \approx u \approx e^{1/2}\sqrt{R_0 - R_s}$,

valid in the limit t->infinity. So that proves that your red line crosses the horizon at a finite point on the diagram, NOT infinitely up and far to the right.

Of course you can also see this from the conformal diagram you posted with the blue line, or the one I posted with the blue shaded region, or simply by extending your plot a little.

DrBaltar

9th January 2008, 03:48 PM

sol invictus

9th January 2008, 04:53 PM

I got v=u=1/2 * Exp[1/2] * Sqrt((R0-Rs)/Rs), but yeah, pretty close. Interesting, because if I plug in large values of T and solve it, it's way beyond what 1/2 * Exp[1/2] * Sqrt((R0-Rs)/Rs) evaluates to.

I forgot about the 1/2 - you're correct. As for the rest of your comment, I'm not sure what you're trying to say.

In any case, do you now agree that your red line crosses the horizon at a finite point? If so, we're almost home...

Acleron

9th January 2008, 05:00 PM

DrBaltar

9th January 2008, 08:49 PM

I forgot about the 1/2 - you're correct. As for the rest of your comment, I'm not sure what you're trying to say.

In any case, do you now agree that your red line crosses the horizon at a finite point? If so, we're almost home...

I'm saying that yes, the math seems to show that it intersects the horizon at a finite point, BUT I can compute a radius R(t) using a large t, such as 10^10 and R(t) will be extremely close to Rs. When I plug in R(t) and t into the equations for u & v, u & v are WAY beyond the finite intersection point. This doesn't seem right. Why would any part of the curve R(t) outside of Rs be further to the upper right than the intersection point? Do you see the problem?

R(t) is only valid for all R(t)>Rs, but not R(t)=Rs. Perhaps this is causing the incontinuity.

DrBaltar

9th January 2008, 08:52 PM

Please don't agree :(
It will spoil the fun :)

LOL Sorry but I must follow the math. I still don't think it changes the fact that an outside observer will see the star collapse indefinitely.

ETA: Sol says he agrees with the equation R(t) = Rs + (R0 - Rs)*exp(-t/Rs). This equation marks position of the outter radius of the star as a function of t, not just where it appears to be.

sol invictus

9th January 2008, 10:11 PM

LOL Sorry but I must follow the math. I still don't think it changes the fact that an outside observer will see the star collapse indefinitely.

No one in this thread has ever disagreed with that. The disagreement is over what really happens, not what the outside observer sees - specifically whether or not there is a region of spacetime in which there is no matter outside the horizon.

ETA: Sol says he agrees with the equation R(t) = Rs + (R0 - Rs)*exp(-t/Rs). This equation marks position of the outter radius of the star as a function of t, not just where it appears to be.

Yes, but again - those coordinates are singular on the horizon. The coordinates u and v are better, because they can differentiate between different points on the horizon.

As for plugging in the equation and getting large u and v, you're apparently making a mistake somewhere. The square root term cancels the exponential, leaving you with the finite constant we both found.

69dodge

10th January 2008, 04:18 PM

sol invictus

11th January 2008, 06:56 AM

Can someone observing from afar ever say to himself, "I haven't yet seen the crossing but it has definitely already happened"?

Yes, absolutely.

First, consider a classical black hole (no Hawking radiation). Take a look at the diagram for the formation of a hole by collapse I posted above, and consider the region of the spacetime above the top of the shaded blue region. The metric in that region (in the ideal case with exact spherical symmetry at least) is precisely equal to the vacuum Schwarzschild black hole solution. No experiment an observer in that region can perform will ever indicate the existence of matter stuck to the horizon - because there isn't any. The situation is almost exactly analogous to a sonic horizon as observed by someone will access only to sub-sonic signals.

Turn on quantum mechanics. That region now contains Hawking radiation, but unless the observer can make exponentially (in the entropy of the black hole) accurate measurements, she will not be able to distinguish measurements made there (where it formed a moment ago from spherically symmetric radiation) from those she would have made in an eternal BH spacetime, or one that was formed 1,000,000 years in the past by infalling pink elephants. The hole will eventually evaporate entirely and disappear (in a specific way characteristic of black holes), leaving no doubt about whether it existed, but little clue when or how it formed.

As I keep saying, there is a consistent point of view (usually referred to as black hole complementarity) in which one can regard the quantum horizon as a Planck temperature impermeable membrane at the bottom of a Planckian gravitational well. When matter accretes onto it it burns up and is eventually radiated back out. If you wait a long time with nothing accreting, the radiation from the horizon causes it to shrink and eventually disappear. In this view, nothing crosses the horizon - not because time stops, but because the horizon is incredibly hot and dense, and nothing can penetrate it (anything coming close just melts, burns up, and is radiated out again).

DrBaltar

11th January 2008, 07:02 AM

Not so much what really happens, but when it really happens.

Someone who falls into a black hole experiences himself crossing the event horizon, which he wouldn't, of course, if the crossing didn't happen. So that's not the problem. The crossing happens.

But when?

(From the point of view of someone outside the black hole, that is.)Right, that is my question as well.

Can someone observing from afar ever say to himself, "I haven't yet seen the crossing but it has definitely already happened"?Someone might say that, but is there any way they could tell that it has already happened? I don't see how since all information that it has happened must travel at the speed of light. So if it has happened, and there's no way to see that it has, then it doesn't matter as far as this universe outside the event horizon is concerned.

As for plugging in the equation and getting large u and v, you're apparently making a mistake somewhere. The square root term cancels the exponential, leaving you with the finite constant we both found.
Possibly... I'll have to double check it again when I get some time.

sol invictus

11th January 2008, 07:36 AM

Someone might say that, but is there any way they could tell that it has already happened? I don't see how since all information that it has happened must travel at the speed of light. So if it has happened, and there's no way to see that it has, then it doesn't matter as far as this universe outside the event horizon is concerned.

See my post just above. One can make precisely the same arguments for aquatic creatures observing a sonic horizon. While they may never hear something cross the horizon, they can combine their theories with their observations to deduce that it actually has.

The case of classical black holes is really not any more subtle than that. The best way to describe the situation is to use diagrams as above - they make it clear precisely in what sense the infalling matter has crossed the horizon, and they also make it clear that your argument above (that Hawking's calculation will be affected by infalling matter) is incorrect.

If we include quantum mechanics the situation becomes less clear (c.f. the black hole information paradox), but I summarized the state of the art above.

Possibly... I'll have to double check it again when I get some time.

Well obviously you've made a mistake, since you've derived two contradictory results.

DrBaltar

11th January 2008, 07:47 AM

One can make precisely the same arguments for aquatic creatures observing a sonic horizon. While they may never hear something cross the horizon, they can combine their theories with their observations to deduce that it actually has.

This is the danger with analogies. It is not precisely the same, and it can lead you to make incorrect assumptions. The aquatic creatures may not hear the sonic horizon, but with the right equipment they could see it. In the case of the black hole, if the surface of the star has crossed the event horizon in space-like coordinates, but not in null-like or time-like coordinates, then if you can't see it, there is no other way to detect that it has happened. This is the basic problem of simultaneity in relativity. An observers reality is what he sees.

DrBaltar

11th January 2008, 08:29 AM

I think this debate boils down to what do we consider real.

Look at this (you may have to open this in a separate browser and make sure it's shown at full size) picture (http://www.geocities.com/quarkburger/pics/lorentz.jpg).

This is the classic 'make the long thing fit in the short box by making it approach the speed of light' trick. At the top of the diagram are stereo views of a space-time diagram (you can cross your eyes and turn it into a 3D image if you're so inclined).

There are two objects in this space-time diagram. The gray column is a square projected up into the time axis. Its x-y grid is the gray grid shown and would be 2 dimensions of our usual 3. Up and down are future and past.

The second object is the moving object and is shown in magenta. It is slanted because it is moving, and is moving at .85c as shown at the bottom. The magenta object's x-y grid is the distorted magenta grid which is actually at a hyperbolic angle relative to the gray stationary object's grid. The magenta object's time line is also angled at a hyperbolic angle corresponding to its speed.

Shown in the diagram is also the past and future light cone. The future light cone shows the spreading circle of light emanating from the observer at the origin. The past light cone shows what the observer can see.

The 'Simultaneous View' part of the diagram shows the magenta object length contracted and fitting inside the box. In the magenta frame, the magenta object's length is about 2.7. In the gray frame it turns out to be about 1.2 and is able to fit inside the gray box. This is what is usually illustrated in relativity classes to show how length contraction works.

Here's the spin on it though... The 'Observed View' shows what the stationary observer at the origin actually sees. What he sees is what is mapped to the past light-cone as seen in the 3-d diagram on top.

It is my opinion that the observed view is for all intents and purposes Reality for the observer. If he looks at the moving object this is what he sees. It would not appear to be rectangular. He can make the most accurate measurements he is able to make, and no matter how well he measures it, the observed view is as good as he can get. This is his reality.

sol invictus

11th January 2008, 09:06 AM

This is the danger with analogies. It is not precisely the same, and it can lead you to make incorrect assumptions. The aquatic creatures may not hear the sonic horizon, but with the right equipment they could see it.

As I mentioned above, to make the analogy work one simply needs to assume that these creatures do not have access to any signal that propagates faster than sound. (And in fact in the discussion we're having we've been making the analogous assumption - that no fields present propagate faster than light.)

In the case of the black hole, if the surface of the star has crossed the event horizon in space-like coordinates, but not in null-like or time-like coordinates, then if you can't see it, there is no other way to detect that it has happened. This is the basic problem of simultaneity in relativity. An observers reality is what he sees.

See my post above. After the hole evaporates, all observers are within the future lightcone of the crossing. Furthermore before it evaporates outside observers can perform a series of experiments, for example probing arbitrarily close to the horizon, all of which will indicate that the shell has crossed.

Finally, the observers outside may be familiar with general relativity, in which case they can solve for the metric both inside and outside the horizon and note that the shell has crossed.

sol invictus

11th January 2008, 09:17 AM

I think this debate boils down to what do we consider real.

I don't agree. The laws of physics never make inconsistent predictions between observers when the results of physical measurements are compared. On the contrary, the fundamental assumption of general relativity is that the laws of physics make the same predictions no matter what coordinates are used to write them down - otherwise there would be a major inconsistency, as Einstein realized. If the long thing actually fits in the box without breaking it, all observers will agree on that after the fact. If it doesn't fit, it doesn't fit. The only possible disagreement might be over when the object hits the wall, but that's because relative time is not a physical quantity in relativity.

There is a very specific, physical claim that you made earlier, which is the only one I've been disputing - that because the shell can never be seen by an observer that remains outside to cross the horizon, there is always matter close to the horizon and therefore Hawking's calculation was wrong.

That claim is factually incorrect, as I have demonstrated. There is a large (actually infinite) region of the spacetime in which no matter is present and the metric is vacuum Schwarzschild. In that region (which includes all of future infinity, part of past infinity, and all of the future of the horizon past a finite point) there is no matter present and Hawking's calculation can be performed without modification. The existence of the region is a physical fact, and therefore all observers should agree on it (including you :)).

Do you agree with that now?

DrBaltar

11th January 2008, 10:32 AM

I think this debate boils down to what do we consider real.I don't agree.

Well you keep saying you agree that an outside observer will forever see the star collapse:

In all these cases, the observer will see the outer shell of the star stop just short of the event horizon, unless he's going to free fall in with it.
Yes, we all agree on that.

It appears to me that you believe that regardless of what an outside observer sees, the space-like (have you heard this term before?) position, in the frame of an outside observer, of the outer shell has crossed the event horizon in finite time. That's why I think we're disagreeing on what is considered reality.

If the long thing actually fits in the box without breaking it, all observers will agree on that after the fact. If it doesn't fit, it doesn't fit. The only possible disagreement might be over when the object hits the wall, but that's because relative time is not a physical quantity in relativity.
No, this is basic stuff. In these two diagrams, the blue car is zipping through the gray garage at .5c. In the garage's frame, the length contracted car fits in the garage. In the car's frame the length contracted garage is too small to park in.
http://www.geocities.com/quarkburger/pics/length_contraction2.gif
In the blue car's frame it doesn't fit. In the garage's frame it does.
L' = L * sqrt(1-v^2) if we let c=1
This will get you the lengths in the simultaneous frame, not the observed frame as I talked about in my last posting, but I hope you get the picture.

There is a very specific, physical claim that you made earlier, which is the only one I've been disputing - that because the shell can never be seen by an observer that remains outside to cross the horizon, there is always matter close to the horizon and therefore Hawking's calculation was wrong.

That claim is factually incorrect, as I have demonstrated. There is a large (actually infinite) region of the spacetime in which no matter is present and the metric is vacuum Schwarzschild. In that region (which includes all of future infinity, part of past infinity, and all of the future of the horizon past a finite point) there is no matter present and Hawking's calculation can be performed without modification. The existence of the region is a physical fact, and therefore all observers should agree on it (including you :)).

Do you agree with that now?

No:
I still don't think it changes the fact that an outside observer will see the star collapse indefinitely.No one in this thread has ever disagreed with that.

jmercer

11th January 2008, 10:46 AM

sol invictus

11th January 2008, 10:55 AM

Well you keep saying you agree that an outside observer will forever see the star collapse:

Yes, that is what they will see (assuming the matter emits lots of intense radiation as it falls; otherwise they will see nothing at all).

It appears to me that you believe that regardless of what an outside observer sees, the space-like (have you heard this term before?) position, in the frame of an outside observer, of the outer shell has crossed the event horizon in finite time. That's why I think we're disagreeing on what is considered reality.

Again, you make the same elementary mistake. You're confusing coordinates with physics. The time coordinate in which the crossing takes infinite time is singular at the horizon. For the umpteenth time, I can choose a time coordinate which diverges at any null surface I choose. Any observer that accelerates so as to remain always outside that surface will never see anything cross it. But there is absolutely nothing special about that surface.

No, this is basic stuff. In these two diagrams, the blue car is zipping through the gray garage at .5c. In the garage's frame, the length contracted car fits in the garage. In the car's frame the length contracted garage is too small to park in.

Don't be silly. Either the car hits the back wall of the garage - because it's too long to fit, in the rest frame of both - or it doesn't, and every observer agrees (anyone that didn't would be hard-pressed to understand how it got all smashed up, and special relativity would be utter nonsense). The only disagreement can be whether the front fender hits the back wall before or after the rear fender has entered the garage.

No:

You don't agree that there is an infinite region of the spacetime of a collapsing BH in which there is no matter outside? You realize the computation you agreed with above proves mathematically that there is? You're contradicting yourself, and this latest statement also contradicts every diagram either of us has posted in this thread.

69dodge

11th January 2008, 10:57 AM

As I mentioned above, to make the analogy work one simply needs to assume that these creatures do not have access to any signal that propagates faster than sound. (And in fact in the discussion we're having we've been making the analogous assumption - that no fields present propagate faster than light.)

I'm still not getting it, I think.

The creatures may be assumed not to have actual access to faster-than-sound signals, but for them to say something like "I haven't yet heard event x, but it has already happened", there needs to be the possibility in principle of faster-than-sound signals, in order for the idea of simultaneity at a distance to make sense in the first place. Otherwise, "it has already happened" is just meaningless.

After the hole evaporates, all observers are within the future lightcone of the crossing. Furthermore before it evaporates outside observers can perform a series of experiments, for example probing arbitrarily close to the horizon, all of which will indicate that the shell has crossed.

Before the hole evaporates, outside observers are not within the future light cone of the crossing, yet they can perform experiments to demonstrate that it happened? Can that be right?

A single experiment, or any finite number of them, can't get "arbitrarily" close to the horizon.

It seems to me that the results of any particular experiment will be consistent with the shell having crossed the horizon but will also be consistent with the shell having gotten no closer that some particular nonzero distance from it.

sol invictus

11th January 2008, 11:13 AM

Before the hole evaporates, outside observers are not within the future light cone of the crossing, yet they can perform experiments to demonstrate that it happened? Can that be right?

Yes, in a certain (completely precise) sense.

A single experiment, or any finite number of them, can't get "arbitrarily" close to the horizon.

It seems to me that the results of any particular experiment will be consistent with the shell having crossed the horizon but will also be consistent with the shell having gotten no closer that some particular nonzero distance from it.

You can get in a spaceship and fly as close to the horizon as you want, as often as you want. You have an infinite amount of time to do these experiments (classically), so you can do an infinite number and get arbitrarily close. You will never detect any matter in any of those experiments. If the shell had stopped a non-zero distance away, there's an experiment you could do that would detect it (just get that close), but that's not the case.

Look - there's no point in arguing over semantics. We KNOW what the solution is, and we know from the solution that there is an infinite region of the spacetime with no matter in it, and we know that Hawking particles can appear there and be observed eventually. There shouldn't be any debate about that, because I, DrBaltar, and many others before us have proved it mathematically. The existence of that region of spacetime is a physical fact, and therefore all observers agree on it, just as all observers agree that the car hit the back wall of the garage.

The only debate is the pointless one over simultaneity and time-ordering. If you insist on saying it takes the shell infinite time to collapse, you are justified in doing so - but only because that is not a physical (coordinate independent) statement.

I am equally justified in saying it takes infinite time to travel from Boston to New York, because I can choose a time coordinate that diverges somewhere in between, and if I accelerate away from Boston fast enough, no one will ever catch up to me with the news that they went there!

DrBaltar

11th January 2008, 12:55 PM

On that last quote, DrBaltar; if a black hole is subject to evaporation, and eventually re-enters the "normal" universe (probably with a bang!)... then there will clearly be a point where an outside observer will no longer see the star collapse at all.

Do you mean if a black hole can evaporate even with the shell of matter around it? If that's what you mean, I have seen a paper that concludes basically what you say. It says that a black hole would evaporate faster than an asymptotic observer would see the star collapse.

jmercer

11th January 2008, 01:51 PM

Yep, that's what I was driving at. :)

sol invictus

11th January 2008, 02:07 PM

Yet another diagram showing matter collapsing to form a black hole, and crossing the horizon at a finite spacetime point. This one isn't conformal and might be easier to understand. The green wedge in the upper right might be a pair of Hawking particles, forming in a vacuum region long after the matter has crossed the horizon.

http://www.astro.ucla.edu/~wright/bh-st2.gif

The space-time diagram at right shows matter in blue collapsing to form a singularity [the solid black line], while the green curves are the future lightcones from events where light can escape to infinity, while the red curves are future lightcones from events where light cannot escape. The boundary between these two regions in space-time is the magenta curve. This boundary of the trapped region is the event horizon of the black hole.

(From http://www.astro.ucla.edu/~wright/bh-st.html.)

BillyJoe

12th January 2008, 04:26 PM

Credentials: Interested Layman.
(so I'm hoping this post makes sense)

In simple terms, is it correct to say the following:
(I hope I am using the word "observation" correctly in the scientific sense)

What is actually happening is what is happening from the point of view of someone at the point in space where that thing is...um...actually happening. Everyone else sees something different happening but they can agree observationally (by doing the math) with what is actually happening at that point in space.

It seems to me that everyone except Baltar is arguing that the (mathematical) observation is the reality, whereas Baltar is arguing that what he actually sees is what is real.

Is it also correct to say that Hawking radiation is the clincher as to who is correct. If Baltar is correct, there should be no Hawking radiation, but if everyone else is correct, there should be. And Hawking radiation has been confirmed hasn't it?

Thanks for any replies,
(It's been a hard slog ploughing through this thread.)
BillyJoe

sol invictus

12th January 2008, 05:00 PM

What is actually happening is what is happening from the point of view of someone at the point in space where that thing is...um...actually happening. Everyone else sees something different happening but they can agree observationally (by doing the math) with what is actually happening at that point in space.

Yeah, that's a good way to say it. In this case we know what the global spacetime looks like, both inside and outside the black hole, by solving Einstein's equations. One of the characteristics of that solution is that there is a large region of the spacetime which contains the horizon after all the matter has crossed and fallen in.

Is it also correct to say that Hawking radiation is the clincher as to who is correct. If Baltar is correct, there should be no Hawking radiation, but if everyone else is correct, there should be. And Hawking radiation has been confirmed hasn't it?

Not from black holes. However, perturbations during inflation are generated by a process almost identical to Hawking radiation. During inflation there is a horizon, and everything said about black hole horizons in this thread is equally valid there too. If DrBaltar were correct it would mean there would be no stars, planets, or galaxies in the universe (because it is "Hawking" radiation during inflation that seeds the density perturbations which collapse to form stars).

BillyJoe

12th January 2008, 05:29 PM

Thanks sol.

I just wanted to be sure that I got something out of reading this thread apart form just the enjoyment of reading this very interesting thread. It is often difficult to pick them out from amongst the mish mash.

This one started with nothing (:D), and ended with everything (:)), destroying free will along the way (;)).
What more can you ask for?

jmercer

13th January 2008, 06:31 AM

A Theory of Everything? :D

hammegk

13th January 2008, 06:54 AM

69dodge

13th January 2008, 06:22 PM

Sorry if I'm being a pain. Just trying to understand this stuff.

Which I don't, yet.

I bet you can agree with that, at least. :D

Yes, in a certain (completely precise) sense.

Ok. I still don't know what that sense is, though.

You can get in a spaceship and fly as close to the horizon as you want, as often as you want. You have an infinite amount of time to do these experiments (classically), so you can do an infinite number and get arbitrarily close. You will never detect any matter in any of those experiments. If the shell had stopped a non-zero distance away, there's an experiment you could do that would detect it (just get that close), but that's not the case.

Sure. The shell couldn't have stopped. But it could be continually approaching the horizon, though more and more slowly, so that, by the time you do any particular experiment, it is closer to the horizon than is the location of that experiment.

Look - there's no point in arguing over semantics. We KNOW what the solution is, and we know from the solution that there is an infinite region of the spacetime with no matter in it, and we know that Hawking particles can appear there and be observed eventually. There shouldn't be any debate about that, because I, DrBaltar, and many others before us have proved it mathematically. The existence of that region of spacetime is a physical fact, and therefore all observers agree on it, just as all observers agree that the car hit the back wall of the garage.

The only debate is the pointless one over simultaneity and time-ordering. If you insist on saying it takes the shell infinite time to collapse, you are justified in doing so - but only because that is not a physical (coordinate independent) statement.

I agree that there is an infinite region of spacetime with no matter in it. I just don't see any absolute sense in which, in that region or in a larger region that does contain matter, an event outside the horizon is "later" than an event on the horizon. The two event are, at best, spacelike.

I don't insist on saying that the shell takes infinite time to collapse, and in fact I complained (http://forums.randi.org/showpost.php?p=3289390&postcount=313) when DrBalter said that it does. I object to the insistence that it doesn't, because that is also a coordinate dependent statement.

I am equally justified in saying it takes infinite time to travel from Boston to New York, because I can choose a time coordinate that diverges somewhere in between, and if I accelerate away from Boston fast enough, no one will ever catch up to me with the news that they went there!

There is some difference. You feel yourself accelerating away from Boston. An observer far from a black hole doesn't feel himself accelerating away from the horizon, but he still can't see it.

sol invictus

13th January 2008, 06:44 PM

Sorry if I'm being a pain. Just trying to understand this stuff.

No problem - you're not being a pain at all. You're asking good questions.

Sure. The shell couldn't have stopped. But it could be continually approaching the horizon, though more and more slowly, so that, by the time you do any particular experiment, it is closer to the horizon than is the location of that experiment.

Except there is no experiment that any observer can do, once the shell has passed them, which will find it outside the horizon (let's assume the shell is made of null radiation for simplicity).

We can say this a little more mathematically. It is true that the shell crossing the horizon will never be in the past lightcone of an outside observer after finite time. However the shell crossing any surface outside the horizon, no matter how close, will be in the past lightcone of every outside observer eventually. Not only that, for any given shell outside the horizon, the time you have to wait before its crossing is in your past lightcone can be made arbitrarily small by diving in after the collapsing radiation and pulling out at the very last minute.

I agree that there is an infinite region of spacetime with no matter in it.

That's really the only thing I was attempting to argue for. The rest is mostly semantics.

I just don't see any absolute sense in which, in that region or in a larger region that does contain matter, an event outside the horizon is "later" than an event on the horizon. The two event are, at best, spacelike.

Well, see above. Strictly speaking you're correct.

There is some difference. You feel yourself accelerating away from Boston. An observer far from a black hole doesn't feel himself accelerating away from the horizon, but he still can't see it.

That's not a difference. If you're very far from Boston you only have to accelerate a very little bit for it to be behind a horizon, just like with the black hole.

69dodge

13th January 2008, 08:29 PM

That's not a difference. If you're very far from Boston you only have to accelerate a very little bit for it to be behind a horizon, just like with the black hole.

Isn't it? I don't have to accelerate away from the black hole at all. I can move inertially away from it, faster than escape velocity.

A more or less unrelated question: Consider three events along the worldline of a photon. Flat spacetime, say. The interval between any pair is the same, namely zero. But the situation is nevertheless not symmetric. The events can be ordered from earliest to latest, no? I'm not sure exactly what my question is, actually. But something seems wrong. Isn't the interval supposed to contain all of this kind of information? Maybe it's necessary to consider also various nonzero intervals between each of the three events and other nearby events?

Metrics that are not positive-definite are definitely weird!

sol invictus

14th January 2008, 05:46 AM

Isn't it? I don't have to accelerate away from the black hole at all. I can move inertially away from it, faster than escape velocity.

If you start off moving away, yes. But if you start off at rest in the standard coordinates and want to stay that way, you always have to accelerate a little no matter how far away you are. Making the hole bigger makes the acceleration required to remain outside it smaller and smaller and the situation closer and closer to that of an accelerating observer in flat space.

A more or less unrelated question: Consider three events along the worldline of a photon. Flat spacetime, say. The interval between any pair is the same, namely zero. But the situation is nevertheless not symmetric. The events can be ordered from earliest to latest, no?

Yes, they can be ordered. That's why it makes sense to talk about the horizon of the hole after the shell has crossed.

Isn't the interval supposed to contain all of this kind of information? Maybe it's necessary to consider also various nonzero intervals between each of the three events and other nearby events?

Metrics that are not positive-definite are definitely weird!

I certainly agree with the last statement... and yes, it's necessary to consider other intervals. I suppose one way to say it is to note that if the events are ordered somehow in one coordinate system, every other coordinate system related to the first one by a proper coordinate transformation (if you don't know what that means, I'll explain it) will preserve that same ordering.

More physically, imagine a big bunch of actual real observers equipped with clocks, mirrors, laser pointers, etc. Whichever observer happens to find herself at one of your three points sends out a pulse with her laser pointer at that moment. Then every observer who receives all three pulses will receive them in the same order in time, no matter what trajectory they are following.

If this wasn't true causality would fail.

DrBaltar

14th January 2008, 09:32 AM

I don't insist on saying that the shell takes infinite time to collapse, and in fact I complained (http://forums.randi.org/showpost.php?p=3289390&postcount=313) when DrBalter said that it does. I object to the insistence that it doesn't, because that is also a coordinate dependent statement.Not just a coordinate dependent statement. It's an observer dependent statement. If you are an observer outside the event horizon, you use Schwarzschild coordinates. If you are crossing the event horizon, then a Kruskal or Penrose diagram will describe what you see.

69dodge

14th January 2008, 06:54 PM

a proper coordinate transformation (if you don't know what that means, I'll explain it)

Yes, please. I don't.

More physically, imagine a big bunch of actual real observers equipped with clocks, mirrors, laser pointers, etc. Whichever observer happens to find herself at one of your three points sends out a pulse with her laser pointer at that moment. Then every observer who receives all three pulses will receive them in the same order in time, no matter what trajectory they are following.

If this wasn't true causality would fail.

Interesting.

Some observers will receive them all simultaneously, I think---those on the path of the original photon, farther along than the last of my three events.

Which makes sense, I guess. Lightlike events are just on the edge of causality. A tiny bit further, and some observers will begin to disagree about their order in time. But the very edge still belongs to the causal side, you're saying?

69dodge

14th January 2008, 07:15 PM

Not just a coordinate dependent statement. It's an observer dependent statement. If you are an observer outside the event horizon, you use Schwarzschild coordinates. If you are crossing the event horizon, then a Kruskal or Penrose diagram will describe what you see.

What do you mean? Coordinates don't determine what an observer sees. Any coordinate system can be used to calculate that, and they all give the same answer.

from http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html:On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon. That doesn't correspond to anyone's proper time, though; it's just a coordinate called t. In fact, inside the event horizon, t is actually a spatial direction, and the future corresponds instead to decreasing r. It's only outside the black hole that t even points in a direction of increasing time. In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.

At large distances t does approach the proper time of someone who is at rest with respect to the black hole. But there isn't any non-arbitrary sense in which you can call t at smaller r values "the proper time of a distant observer," since in general relativity there is no coordinate-independent way to say that two distant events are happening "at the same time." The proper time of any observer is only defined locally.

DrBaltar

14th January 2008, 08:53 PM

What do you mean? Coordinates don't determine what an observer sees. Any coordinate system can be used to calculate that, and they all give the same answer.

Because, as you quoted:
At large distances t does approach the proper time of someone who is at rest with respect to the black hole.

We all agree that the time it takes to cross the event horizon and get to the singularity in an observer's proper time is finite. But for an observer far from the black hole, they're going to see something different. They will see it take an infinite amount of time for an object to cross the event horizon.

I'm not saying coordinates determine what an observer sees, I'm saying that depending on where the observer is, you should use a coordinate system that is better suited for describing what that observer sees. See the difference?

sol invictus

15th January 2008, 05:09 AM

Yes, please. I don't.

Well, let's see... imagine rotating the coordinate axes through a very small angle. That's a coordinate transformation, and it's close to the identity transformation (which is just not doing anything: rotating by 0 degrees) in a sense I hope is obvious. But we can build up any rotation, even a large one, from a series of these small ones. So we can say all rotations are connected to the identity.

For an example of something that's not connected, imagine reflecting one coordinate: x->-x, for example. There is no way to go from that transformation to the identity transformation smoothly (via a series of infinitesimal transformations, that is). And unlike the case of a rotation, if you actually did that to an object, you'd really change it physically - in this case into its mirror image. So the group of coordinate transformations splits up into (at least) two disconnected components, and "proper" transformations are those in the part connected to the identity.

If you know a little math, the best way to characterize the difference is with the Jacobian of the transformation (the determinant of the matrix of partials) - whether it's positive or negative.

Some observers will receive them all simultaneously, I think---those on the path of the original photon, farther along than the last of my three events.

Well, there can't really be any actual observer that gets them all simultaneously, because to do so would require moving at the speed of light. But you can get arbitrarily close.

Which makes sense, I guess. Lightlike events are just on the edge of causality. A tiny bit further, and some observers will begin to disagree about their order in time. But the very edge still belongs to the causal side, you're saying?

Yeah, that's right - and it had better, because even if you yourself can't move at exactly the speed of light, you can send signals that do.

jmercer

15th January 2008, 10:35 AM

There is some difference. You feel yourself accelerating away from Boston. An observer far from a black hole doesn't feel himself accelerating away from the horizon, but he still can't see it.

That's not quite true. If you were station-keeping close enough to the black hole to be affected by its gravity to the point where you (as an individual) are physically aware of it, then you will indeed feel the acceleration. You'd feel it as weight from the gravitational pull of the black hole. (If you were falling into the black hole, you'd feel no acceleration.)

69dodge

15th January 2008, 03:02 PM

So the group of coordinate transformations splits up into (at least) two disconnected components, and "proper" transformations are those in the part connected to the identity.

Got it. Thanks.

Well, there can't really be any actual observer that gets them all simultaneously, because to do so would require moving at the speed of light. But you can get arbitrarily close.

Maybe I didn't understand your scenario, then.

Here's what I thought you meant:One red photon is travelling through space. Three observers, far from each other, each sees it as it passes them. Question: Who saw it first, and who last? Answer: There is a fourth observer, towards whom each of the original three fires their own blue photon when the red one passes them. Then, the order in which the three observers saw the red photon is the same as the order in which the fourth observer saw their blue photons.

If the fourth observer lies along the path of the red photon, past the other observers, the three blue photons will travel alongside the red one, and he'll see all four simultaneously.

sol invictus

15th January 2008, 03:14 PM

If the fourth observer lies along the path of the red photon, past the other observers, the three blue photons will travel alongside the red one, and he'll see all four simultaneously.

Well, yes - if they fire their blue photons in exactly the same direction as the red photon, all of them just move along together in a bunch. But they can emit blue photons in other directions, as a spherical pulse for example, and then everywhere except precisely along the direction of the red photon there will be non-zero intervals between the pulse arrival times (and the order will be the same everywhere).

69dodge

15th January 2008, 03:28 PM

Because, as you quoted:At large distances t does approach the proper time of someone who is at rest with respect to the black hole.

Yes, but you're considering t at small distances too to be when things "really" happen. What is the justification for that?

We all agree that the time it takes to cross the event horizon and get to the singularity in an observer's proper time is finite. But for an observer far from the black hole, they're going to see something different. They will see it take an infinite amount of time for an object to cross the event horizon.

I'm not saying coordinates determine what an observer sees, I'm saying that depending on where the observer is, you should use a coordinate system that is better suited for describing what that observer sees. See the difference?

Not really.

I agree that they won't ever see an object cross the horizon. But in any case they wouldn't see it cross exactly when it does cross, because they're far away. When they see it cross depends on (1) when it crosses and (2) how much time light takes to get to them from the crossing.

So they could say "I haven't yet seen it cross, because it hasn't yet crossed", but they could equally well say "I haven't yet seen it cross even though it has already crossed, because light from the crossing hasn't yet reached me".

Both of these accurately describe what they see. Which is "better suited" for that purpose?

69dodge

15th January 2008, 03:39 PM

If you were station-keeping close enough to the black hole to be affected by its gravity to the point where you (as an individual) are physically aware of it, then you will indeed feel the acceleration. You'd feel it as weight from the gravitational pull of the black hole. (If you were falling into the black hole, you'd feel no acceleration.)

I don't disagree. I did say "far from a black hole". And in a later post I pointed out that you could be falling away from the black hole, in which case you'd feel no acceleration but the horizon would still exist for you.

Ron_Tomkins

15th January 2008, 05:45 PM

Well, ten pages on nothing. That's gotta be something!

DrBaltar

15th January 2008, 09:56 PM

Yes, but you're considering t at small distances too to be when things "really" happen. What is the justification for that?I believe that I have always talked about an asymptotic observer, except when commenting on other people's discussions about observers near the event horizon.

So they could say "I haven't yet seen it cross, because it hasn't yet crossed", but they could equally well say "I haven't yet seen it cross even though it has already crossed, because light from the crossing hasn't yet reached me".

Both of these accurately describe what they see. Which is "better suited" for that purpose?If we go with the later, then why do astronomy? Everything that reaches the telescopes are from the past and only distracting us from the truth. We should instead compute the current state of the universe and render computer graphic images of what we would see if we didn't have to wait eons for light to finally reach us. Only then would we see reality. ;)

69dodge

16th January 2008, 02:50 AM

I believe that I have always talked about an asymptotic observer, except when commenting on other people's discussions about observers near the event horizon.

The observer is far from the black hole, but the event he's observing, whose time of occurrence is in question, is near to it. That event's t-coordinate may be very large, but it is the t-coordinate of an event near the black hole. So, what is its significance, exactly? How can we meaningfully compare it to the t-coordinate of a far-away event at the location of the observer, which he reads off of his watch, say?

If we go with the later, then why do astronomy? Everything that reaches the telescopes are from the past and only distracting us from the truth. We should instead compute the current state of the universe and render computer graphic images of what we would see if we didn't have to wait eons for light to finally reach us. Only then would we see reality. ;)

Yes. Don't we do just that?

For example, in 1675, Roemer estimated the speed of light by assuming that eclipses of a moon of Jupiter did not happen when he saw them happen, but rather some time earlier, where the light propagation delay depended on the distance from Jupiter to Earth at the time of the eclipse.

Actually, in relativity, "the current state of the universe" is not very well-defined, but I'm sure no astronomer assumes, even as a rough approximation, that the speed of light is infinite.

DrBaltar

16th January 2008, 05:07 AM

Actually, in relativity, "the current state of the universe" is not very well-defined, but I'm sure no astronomer assumes, even as a rough approximation, that the speed of light is infinite.Of course not. And even in my tongue in cheek statement above I was not suggesting they would. You and Sol seem to be suggesting that the mathematically calculated state of the black hole (but in no way verifiable) is more important than the observed state.

sol invictus

16th January 2008, 05:13 AM

I believe that I have always talked about an asymptotic observer, except when commenting on other people's discussions about observers near the event horizon.

The claim I found objectionable was that since matter always remained outside the horizon, the calculation of Hawking radiation would be strongly affected by it. That claim makes no reference to any specific observer - all observers that spend any time outside the horizon have the opportunity to observer Hawking particles, and their observations had better be consistent if the theory is sensible. (Note - "consistent" does NOT mean "the same".)

If we go with the later, then why do astronomy? Everything that reaches the telescopes are from the past and only distracting us from the truth. We should instead compute the current state of the universe and render computer graphic images of what we would see if we didn't have to wait eons for light to finally reach us.

That is precisely what astronomers do. They have a theory, and in that theory the convenient time coordinate is NOT the one you like - it's very different. It's what is usually called Friedmann-Robertson-Walker time, and it does not go to infinity at cosmological horizons. Surfaces of constant FRW time extend into regions we will never see, and from which photons will never arrive (because they are behind horizons).

When astronomers make observations, they interpret what they see in the context of that theory. In other words they observe rather than simply see.

sol invictus

16th January 2008, 05:19 AM

Of course not. And even in my tongue in cheek statement above I was not suggesting they would. You and Sol seem to be suggesting that the mathematically calculated state of the black hole (but in no way verifiable) is more important than the observed state.

The discussion was over the region outside the horizon, and the correct answers are perfectly verifiable to an observer capable of accessing that region and performing experiments.

Incidentally, you stopped responding on that topic around when you agreed with my calculation demonstrating that the shell of infalling matter crosses the horizon at a finite spacetime point. Do you now agree that there is an infinite region of matter-free spacetime outside the horizon after the shell has fallen in, or have you decided your (and my) calculation was wrong?

DrBaltar

16th January 2008, 06:32 AM

The claim I found objectionable was that since matter always remained outside the horizon, the calculation of Hawking radiation would be strongly affected by it. That claim makes no reference to any specific observer - all observers that spend any time outside the horizon have the opportunity to observer Hawking particles, and their observations had better be consistent if the theory is sensible. (Note - "consistent" does NOT mean "the same".)Yes, outside observers would see hawking virtual particles fall into the black hole. The mass of the black hole would decrease as seen from an outside observer. When would it decrease? After virtual particles described by Hawking's theory fall into the black hole. When do they fall into the black hole as seen from an asymptotic observer? After the star collapses. When does the star appear to be completely collapsed to an asymptotic observer? After an infinite amount of time has passed.

Incidentally, you stopped responding on that topic around when you agreed with my calculation demonstrating that the shell of infalling matter crosses the horizon at a finite spacetime point. Do you now agree that there is an infinite region of matter-free spacetime outside the horizon after the shell has fallen in, or have you decided your (and my) calculation was wrong?I agreed that it crossed the horizon on a finite point in Kruskal coordinates. With Kruskal coordinates you can track an object all the way to just short of the singularity. That is not however what an asymptotic observer sees. The timespan in his frame is still infinite.

I still think the core point of the debate is what is a valid reality to an asymptotic observer - what is known to the observer via information traveling at the speed of light to the observer, or what can be deduced mathematically. It seems that we all agree on what happens in each frame both mathematically and visually.

sol invictus

16th January 2008, 07:03 AM

Yes, outside observers would see hawking virtual particles fall into the black hole. The mass of the black hole would decrease as seen from an outside observer. When would it decrease? After virtual particles described by Hawking's theory fall into the black hole. When do they fall into the black hole as seen from an asymptotic observer? After the star collapses. When does the star appear to be completely collapsed to an asymptotic observer? After an infinite amount of time has passed.

This is wrong on several counts. First, no one sees Hawking particles form precisely at the horizon - they see them form some distance away. We already discussed this. Second, the infinity you're so concerned about does not exist except in some fictitious world without quantum mechanics. Hawking radiation changes the geometry radically - it makes the black hole disappear. So you can't use the classical geometry, which does not describe the physics at late times, to argue that there is no Hawking radiation. It's circular reasoning.

What really happens is that the radiation is always there. It's even there before the shell crosses the horizon in any coordinates, actually. (That might seem strange, but not once you recall that Hawking radiation and Unruh radiation are almost the same thing.) It makes the hole disappear, and it means that distant observers DO see photons from the horizon after finite time. I posted the conformal diagram for that some time ago.

I still think the core point of the debate is what is a valid reality to an asymptotic observer - what is known to the observer via information traveling at the speed of light to the observer, or what can be deduced mathematically. It seems that we all agree on what happens in each frame both mathematically and visually.

Not at all. You're arguing there is no Hawking radiation from black holes. That is false, and I (and just about everyone else) disagree.

You're also still confused by the term "asymptotic". If you really mean infinitely far away the debate is meaningless since that observer could never see anything in finite time, regardless whether or not there's a horizon. If you don't really mean infinite, then the coordinates you're using are an approximation. A distant observer can send her friend to probe arbitrarily close to the horizon. It will take a long time for the friend to come back, but not forever.

jmercer

16th January 2008, 12:03 PM

I don't disagree. I did say "far from a black hole". And in a later post I pointed out that you could be falling away from the black hole, in which case you'd feel no acceleration but the horizon would still exist for you.

Ooops! So you did. :blush:

My apologies.

DrBaltar

16th January 2008, 08:51 PM

This is wrong on several counts. First, no one sees Hawking particles form precisely at the horizon - they see them form some distance away. We already discussed this.That's nice. Not really relevant though since I didn't say they did form precisely at the horizon.

sol invictus

16th January 2008, 08:59 PM

That's nice. Not really relevant though since I didn't say they did form precisely at the horizon.

Then your comment above was pointless? What you were saying, as far as I could tell, was that you think infalling matter takes infinite time to cross the horizon. But you agree it gets very close very quickly, and if Hawking particles form some distance away rather than right at the horizon, what's the problem?

Not to mention the other errors I pointed out...

Matteo Martini

16th January 2008, 09:13 PM

Two questions with no answer:
- where is "nothing"?
- where is the Universe?

DrBaltar

16th January 2008, 09:35 PM

Then your comment above was pointless? What you were saying, as far as I could tell, was that you think infalling matter takes infinite time to cross the horizon. But you agree it gets very close very quickly, and if Hawking particles form some distance away rather than right at the horizon, what's the problem?Yes, outside observers would see hawking virtual particles fall into the black hole from some distance away from the event horizon. The mass of the black hole would decrease as seen from an outside observer. When would it decrease? After virtual particles described by Hawking's theory fall into the black hole. When do they fall into the black hole as seen from an asymptotic observer? After the star collapses. When does the star appear to be completely collapsed to an asymptotic observer? After an infinite amount of time has passed.

sol invictus

16th January 2008, 09:45 PM

jmercer

17th January 2008, 03:11 AM

Two questions with no answer:

Wrong.

- where is "nothing"?

"Nothing" is an illusion. There is no "nothing", so asking it's location has no meaning.

- where is the Universe?

Here. If it were elsewhere, it wouldn't be our universe.

jmercer

17th January 2008, 03:23 AM

Yes, outside observers would see hawking virtual particles fall into the black hole from some distance away from the event horizon. The mass of the black hole would decrease as seen from an outside observer. When would it decrease? After virtual particles described by Hawking's theory fall into the black hole. When do they fall into the black hole as seen from an asymptotic observer? After the star collapses. When does the star appear to be completely collapsed to an asymptotic observer? After an infinite amount of time has passed.

DrBaltar, you're equating visual observation with reality. It's not the same thing, or magicians would be pulling gold bars out of their hats and destroying the economy by doing so - it's an illusion of the star remaining that they see, leftover photons that slowly make their way out of the zone surrounding the star. The reason it takes so long for them to escape is due to time dilation from the gravity gradient, as you well know.

We know it's an illusion because the gravity well surrounding the hole is NOT the same as a gravity well of a collapsing star would be. Regardless of the photonic illusion, the gravity gradient of a fully-formed black hole would be (and is) different than one of a star in the process of collapse.

(Granted, I don't have the math to prove this myself, but I've read this in physics texts.)

Additionally, at some point the remaining photons from the original moment of the formation of the horizon eventually escape, so the images fades to the point where the asymptotic observer would see nothing - giving the (now correct) impression that the hole had finished forming and become invisible.

To paraphrase Galileo's possibly apocryphal statement... "Never the less, it collapses!"

BillyJoe

17th January 2008, 05:18 AM

Matteo,

Two questions with no answer:
- where is "nothing"?
- where is the Universe?

The meaning of "nothing" includes no existence and no time and no space.
This means "nothing" has no existence at no time in no place.

On the other hand, the universe is here, now.
So enjoy!

regards,
BillyJoe

The Painter

17th January 2008, 05:34 AM

Two questions with no answer:
- where is "nothing"?
- where is the Universe?

where is "nothing"? It’s right over there. >>>

where is the Universe? Don't move, It's right here.

Is nothing sacred?

DrBaltar

17th January 2008, 07:23 AM

OK, whatever you say. You're simply inventing things now.

Do you disagree with the order of events as I described?
1. Star collapse
2. Virtual particles fall into the black hole
3. The black hole's mass decreases as Hawking Radiation is emitted

Just to be clear, you do realize that according to every authority on the subject you're wrong? You must find it a little odd that all those (presumably smart and well-informed) physicists wasted 30 years thinking about Hawking radiation and the end state of black hole evaporation, when it actually never happens. If only they had consulted you...

In light of Hawking's great intelligence I am fully aware that I could be wrong on this, but not for the reasons you have been saying.

hammegk

17th January 2008, 07:34 AM

DrBaltar, you're equating visual observation with reality. ....

Or is the argument more about what constitutes an "observation"? I don't think visual is the key.

DrBaltar

17th January 2008, 07:43 AM

DrBaltar, you're equating visual observation with reality. It's not the same thing, or magicians would be pulling gold bars out of their hats and destroying the economy by doing so
The space-like representation of what an audience sees in a magic show would be extremely close to the null-like representation or visual representation of the show.

it's an illusion of the star remaining that they see, leftover photons that slowly make their way out of the zone surrounding the star. The reason it takes so long for them to escape is due to time dilation from the gravity gradient, as you well know.

We know it's an illusion because the gravity well surrounding the hole is NOT the same as a gravity well of a collapsing star would be. Regardless of the photonic illusion, [I]the gravity gradient of a fully-formed black hole would be (and is) different than one of a star in the process of collapse.If the star has evaporated away, why is there still a gravity well causing the time dilation? Answer: because changes in the gravitational field of the black hole also propagate away from the black hole at the speed of light.

sol invictus

17th January 2008, 08:30 AM

Do you disagree with the order of events as I described?

As several posters have been trying to explain to you here, temporal ordering of events is a slippery concept even in special relativity, let along non-trivial curved spacetimes such as the one we're discussing. The best way by far to understand the causal structure of these spacetimes is to use conformal diagrams, but you seem to be either unwilling or unable to understand what those diagrams show.

1. Star collapse
2. Virtual particles fall into the black hole
3. The black hole's mass decreases as Hawking Radiation is emitted

No, I don't agree with this. First of all, the horizon forms BEFORE the matter crosses its own Schwarzschild radius - or to be absolutely precise, it appears at a point precisely on the edge of the past lightcone of that event.

Secondly, star collapse isn't a single event - it's a process, even if we model the star by a thin shell. So I don't know how you assign a time to it, even aside from these other issues.

Third, I don't know how you think you're going to measure the mass of the hole. Different measures may give different answers at different times. The mass can only definitively be said to decrease once some Hawking radiation has passed outside the radial position of the observer. When that happens depends on the observer. If we use the coordinates you insist on, it happens BEFORE the shell crosses the horizon - so in your coordinates you have 1 and 3 in the wrong order (at least if by star collapse you mean shell crossing the horizon). The correct order (in your coordinates, assuming the star is a thin shell and that collapse means horizon crossing) is 3,1,2, or possible 3,2,1 for some of the Hawking particles.

In light of Hawking's great intelligence I am fully aware that I could be wrong on this, but not for the reasons you have been saying.

It's not just Hawking - it's hundreds of other physicists. I've simply been trying to explain to you what all of them know very well.

DrBaltar

17th January 2008, 08:50 AM

1. Star collapse
2. Virtual particles fall into the black hole
3. The black hole's mass decreases as Hawking Radiation is emitted
No, I don't agree with this. First of all, the horizon forms BEFORE the matter crosses its own Schwarzschild radius - or to be absolutely precise, it appears at a point precisely on the edge of the past lightcone of that event.Ok, feel free to correct where I placed "formation of horizon". I can't see where I placed it in the list, but perhaps your mathematically derived universe has revealed it to you.

Secondly, star collapse isn't a single event - it's a process, even if we model the star by a thin shell. So I don't know how you assign a time to it, even aside from these other issues.I'm not assigning a time. I'm just trying to see if we can agree on the order of these processes. Are you saying that virtual particles can cross the event horizon before the star has finished collapsing? For this to be true they would either form within the star matter as it's collapsing or overtake the star matter falling into the horizon.

Third, I don't know how you think you're going to measure the mass of the hole. Different measures may give different answers at different times. The mass can only definitively be said to decrease once some Hawking radiation has passed outside the radial position of the observer. When that happens depends on the observer. If we use the coordinates you insist on, it happens BEFORE the shell crosses the horizon - so in your coordinates you have 1 and 3 in the wrong order (at least if by star collapse you mean shell crossing the horizon).Yes I mean when the shell crosses the horizon. I'm not sure yet why you think 1 and 3 should be reversed, but your answer to my question above about how virtual particles cross the event horizon before the outer shell of the star does would clarify that.

sol invictus

17th January 2008, 09:10 AM

Ok, feel free to correct where I placed "formation of horizon". I can't see where I placed it in the list, but perhaps your mathematically derived universe has revealed it to you.

It's relevant to the ordering of 1 and 2, because once there's a horizon it will produce Hawking radiation. That can happen before the shell gets anywhere near its horizon.

Are you saying that virtual particles can cross the event horizon before the star has finished collapsing? For this to be true they would either form within the star matter as it's collapsing or overtake the star matter falling into the horizon.

Talking about when a virtual particle does something is going to get you into even more trouble... but yes, roughly speaking that can happen, although if we're talking about a star (rather than a thin collapsing shell) it would be awfully hard to tell, and for those particles Hawking's calculation would indeed be modified. So let's talk about Hawking particles produced outside the shell instead.

Hawking radiation can be thought of as the nucleation of pairs of particles, one with negative energy and one with positive (this is not completely correct, but good enough for now). Those pairs appear near the horizon of the hole, and their initial appearance conserves energy and does not change the mass of anything.

Next, the positive energy particle flies out, and the negative energy particle falls in. If there's an observer trying to determine the mass of the hole, the simplest thing she can measure is the total mass within a sphere defined by the radial distance she's at. That total mass doesn't change at all until the positive energy Hawking particle flies past her - so it's only at that time that she can determine that the mass of the hole has changed. But the particle flies past her at a finite time in your favorite coordinates, and the shell crossing of the star matter happens at infinite time - so the mass of the hole decreases by Hawking radiation BEFORE the star has collapsed (using your coordinates).

Yes I mean when the shell crosses the horizon. I'm not sure yet why you think 1 and 3 should be reversed, but your answer to my question above about how virtual particles cross the event horizon before the outer shell of the star does would clarify that.

Make sense now?

jmercer

17th January 2008, 07:24 PM

If the star has evaporated away, why is there still a gravity well causing the time dilation? Answer: because changes in the gravitational field of the black hole also propagate away from the black hole at the speed of light.

It would appear that the speed of gravity's propagation being limited to the speed of light is somewhat in debate, as you can see from this site (http://metaresearch.org/cosmology/speed_of_gravity.asp) and this one (http://www.space.com/scienceastronomy/gravity_speed_030116.html). Many edu sites are in agreement with your statement, however.

It's an interesting thing :)

Matteo Martini

17th January 2008, 11:13 PM

Matteo,

The meaning of "nothing" includes no existence and no time and no space.
This means "nothing" has no existence at no time in no place.

On the other hand, the universe is here, now.
So enjoy!

regards,
BillyJoe

What is something that does not exist?
Where is "no place"?
When is "no time"?

BillyJoe

18th January 2008, 12:53 AM

What is something that does not exist?
Where is "no place"?
When is "no time"?

My answers are respectively...

Nothing.
Nowhere.
Nowhen.

The last word is made up, but I think you understand, no?

DrBaltar

18th January 2008, 04:36 AM

So let's talk about Hawking particles produced outside the shell instead.

Hawking radiation can be thought of as the nucleation of pairs of particles, one with negative energy and one with positive (this is not completely correct, but good enough for now). Those pairs appear near the horizon of the hole, and their initial appearance conserves energy and does not change the mass of anything.

Next, the positive energy particle flies out, and the negative energy particle falls in. If there's an observer trying to determine the mass of the hole, the simplest thing she can measure is the total mass within a sphere defined by the radial distance she's at. That total mass doesn't change at all until the positive energy Hawking particle flies past her - so it's only at that time that she can determine that the mass of the hole has changed. But the particle flies past her at a finite time in your favorite coordinates, and the shell crossing of the star matter happens at infinite time - so the mass of the hole decreases by Hawking radiation BEFORE the star has collapsed (using your coordinates).
I'm still missing the part where the negative virtual particles that are formed outside the shell while the star is collapsing overtakes the shell and falls into the event horizon. Unless you're talking about the virtual particles that form after the star has collapsed?

DrBaltar

18th January 2008, 04:44 AM

It would appear that the speed of gravity's propagation being limited to the speed of light is somewhat in debate, as you can see from this site (http://metaresearch.org/cosmology/speed_of_gravity.asp) and this one (http://www.space.com/scienceastronomy/gravity_speed_030116.html). Many edu sites are in agreement with your statement, however.

It's an interesting thing :)

This is why I'm so hung up on what the observer sees and I'm thinking it's the most valid reality. It's not just a pretty picture. ALL information about what is happening at the black hole and about the properties of the black hole radiate away at the speed of light (or slower in the case of relativistic particles emanated from the BH).

sol invictus

18th January 2008, 06:27 AM

I'm still missing the part where the negative virtual particles that are formed outside the shell while the star is collapsing overtakes the shell and falls into the event horizon. Unless you're talking about the virtual particles that form after the star has collapsed?

Please re-read what I wrote. I never said anything about the "negative virtual particle" overtaking the shell (actually I don't think the question is even meaningful).

Instead, I pointed out that the only reasonable way for an outside observer to measure the mass of the hole is for her to measure the total mass inside a spherical shell of some radius centered on the hole (and larger than the horizon, obviously). That total mass will change only when some energy passes out through it - it is not affected by the motion of the collapsing shell or "negative virtual particles" inside. Positive energy Hawking radiation starts to pass out through any such shell at a finite time, the shell collapses at infinite time, so (in your coordinates) the black hole loses mass before the shell collapses.

If you don't agree, why don't you carefully specify how your outside observer will decide when the black hole has decreased in mass.

DrBaltar

18th January 2008, 07:34 AM

Instead, I pointed out that the only reasonable way for an outside observer to measure the mass of the hole is for her to measure the total mass inside a spherical shell of some radius centered on the hole (and larger than the horizon, obviously). That total mass will change only when some energy passes out through it - it is not affected by the motion of the collapsing shell or "negative virtual particles" inside. Positive energy Hawking radiation starts to pass out through any such shell at a finite time, the shell collapses at infinite time, so (in your coordinates) the black hole loses mass before the shell collapses.

Oh I see what you're saying. The positive virtual particle that escapes is from a point further outside the event horizon than the shell, so it is not time dilated as much and reaches the observer before the shell is seen to cross the event horizon. This is a distinction I should have made. #3 on the list should have been:
3. The black hole's mass decreases as negative virtual particles fall in.

Because the decrease in mass happens not when the Hawking radiation escapes, but when the negative virtual particles fall into the black hole.

If you don't agree, why don't you carefully specify how your outside observer will decide when the black hole has decreased in mass.

Mass can be measured by observing an accretion disk outside the black hole, or from the gradient of the red-shifting of infalling matter (assuming of course that the observer was capable of making these measurements). I take it you would measure the mass of the black hole by observing the energy of the Hawking radiation? At a given point outside the event horizon Hawking radiation Rh, the amount of radiation depends on the mass of the black hole. When does the gravitational potential at Rh change as antiparticles are falling into the black hole?

sol invictus

18th January 2008, 07:55 AM

Because the decrease in mass happens not when the Hawking radiation escapes, but when the negative virtual particles fall into the black hole.

OK, then you're simply inventing arbitrary definitions of "black hole mass", "when", etc. These definitions have nothing to do with physics, or even observations made by an asymptotic observer.

When does the gravitational potential at Rh change as antiparticles are falling into the black hole?

It changes when positive energy outgoing Hawking radiation passes through Rh - not a moment before and not a moment after (and therefore it changes before the shell collapses, in these silly coordinates). It has absolutely nothing to do with the positions of "negative virtual particles" at radii less than Rh. If you don't understand that, look up Gauss' law, or Birkhoff's theorem - the metric and gravitational potential at some point in a spherically symmetric configuration are affected ONLY by the mass inside the shell defined by that point, and that mass can only change when something passes through that shell. You should know that from elementary E&M if not from GR.

Of course this whole sub-topic is utterly pointless - it just serves to illustrate what we've all been trying to explain to you all along, which is that time ordering of events in some arbitrary coordinate system is a really poor way to characterize the physics of a curved space like this. The correct way to do it is to draw a Penrose diagram - that clearly and simply shows you the causal structure and immediately gives the answers to these kinds of questions.

DrBaltar

20th January 2008, 10:25 PM

Alright. So then Hawking radiation exists outside the collapsing shell. The event horizon still isn't going to shrink until the negative particles cross it, and the collapsing shell of star matter still will take an infinite amount of time to cross the event horizon as seen from an asymptotic observer since the size of the event horizon depends on the mass within it, as does the red shifting and time dilation of the collapsing star matter. The incoming negative particles from Hawking radiation cannot cross the event horizon before the star shell collapses, so an asymptotic observer will see the evaporation (from Hawking radiation) of the collapsing star end before he sees the star completely collapse, since that still takes an infinite amount of time.

sol invictus

21st January 2008, 04:36 AM

Alright. So then Hawking radiation exists outside the collapsing shell.

Great! Now that you've admited that, it's over.

The event horizon still isn't going to shrink until the negative particles cross it, and the collapsing shell of star matter still will take an infinite amount of time to cross the event horizon as seen from an asymptotic observer since the size of the event horizon depends on the mass within it, as does the red shifting and time dilation of the collapsing star matter.

You're still making the same mistakes... talking about when things happen even though you don't understand the causal structure. Once the effects of Hawking radiation are taken into account it is no longer true that it takes infinite time to cross the horizon - the observer will actually see that event at finite time (as part of the burst of radiation that's emitted at the last stage of the evaporation).

The incoming negative particles from Hawking radiation cannot cross the event horizon before the star shell collapses, so an asymptotic observer will see the evaporation (from Hawking radiation) of the collapsing star end before he sees the star completely collapse, since that still takes an infinite amount of time.

Impossible, once you've admitted there is Hawking radiation. You're making arguments based on the classical geometry, but the classical geometry is wrong at late times. We can easily compute the amount of radiation emitted per unit asymptotic-observer-time, and it's such that as the hole shrinks, in evaporates faster and faster. The total evaporation time is proportional to the mass of the hole cubed - long for a big hole, but always finite.

DrBaltar

21st January 2008, 06:35 AM

Impossible, once you've admitted there is Hawking radiation. You're making arguments based on the classical geometry, but the classical geometry is wrong at late times. We can easily compute the amount of radiation emitted per unit asymptotic-observer-time, and it's such that as the hole shrinks, in evaporates faster and faster. The total evaporation time is proportional to the mass of the hole cubed - long for a big hole, but always finite.You're not thinking this through. Yes the radiation associated with evaporation happens faster and faster. The rate increases, not because the negative particles have actually fallen into the event horizon, but because of, as you said, Gauss' Law. The negative particles occupy a sphere of smaller radius than the positions where virtual particles are being created. As seen by Asymptotic Bob, these negative particles will never cross the event horizon just as the outer shell of the star matter will never cross the event horizon. The collapse will still take an infinite amount of time despite the fact that Hawking radiation does not.

sol invictus

21st January 2008, 09:02 AM

You're not thinking this through. Yes the radiation associated with evaporation happens faster and faster. The rate increases, not because the negative particles have actually fallen into the event horizon, but because of, as you said, Gauss' Law. The negative particles occupy a sphere of smaller radius than the positions where virtual particles are being created. As seen by Asymptotic Bob, these negative particles will never cross the event horizon just as the outer shell of the star matter will never cross the event horizon. The collapse will still take an infinite amount of time despite the fact that Hawking radiation does not.

Suppose a star of mass M collapses to form a black hole. You agree it will start to radiate Hawking particles, at a faster and faster rate as time passes and it loses mass. Hence after some finite amount of time it will have emitted an amount of energy equal to M through the spherical shell defined by the observer's position. At that time (by conservation of energy) the total energy inside that sphere is zero.

Tell me - what do you think the spacetime is inside the shell at that time (or at any later time)?

(Hint - use Gauss' law.)

DrBaltar

21st January 2008, 04:05 PM

Suppose a star of mass M collapses to form a black hole. You agree it will start to radiate Hawking particles, at a faster and faster rate as time passes and it loses mass. Hence after some finite amount of time it will have emitted an amount of energy equal to M through the spherical shell defined by the observer's position. At that time (by conservation of energy) the total energy inside that sphere is zero.

Tell me - what do you think the spacetime is inside the shell at that time (or at any later time)?

(Hint - use Gauss' law.)I won't misuse Gauss' Law. The total energy inside the sphere (not the event horizon, but at the positions where virtual particles form) is zero. The energy inside the shell of infalling star matter is equal to mass M. Outside the star matter and inside where the virtual particles form is a flux of negative particles heading towards the event horizon.

Read this article here (http://arstechnica.com/news.ars/post/20070622-apotential-solution-to-the-black-hole-information-loss-paradox.html).
And the associated paper here (http://arxiv.org/PS_cache/gr-qc/pdf/0609/0609024v3.pdf).

The Case Western Team approached the problem from a relatively uncommon perspective. Instead of taking the view of an observer falling into a black hole, they took the more physically relevant viewpoint of the asymptotic observer: one who is so far away from all the action that s/he is not affected by it and cannot affect the process.

...

The physical equivalent problem that the researchers address is that of a shell of collapsing matter, or more technically a vacuum domain wall. Applying the equations of general relativity to this problem in the Schwarzschild metric, the team produced an expected result: the formation of the event horizon takes an infinite Schwarzschild time. Given that this result is well known, the researchers expanded their field of inquiry to include quantum effects. To see if this result still holds true in the quantum realm, the team made some assumptions and used what is known as the functional Schroedinger equation. They found that even when accounting for quantum effects, the black hole still takes an infinite amount of (Schwarzschild) time to form. Taking their results a step further, the physicists find that the collapsing shell will radiate away its energy in a finite amount of time. They found that the amount time needed to radiate away all of the shell's energy is shorter than the time needed for an object to fall through the event horizon. This leads to one of their major conclusions: an observer sufficiently far away "will see the evaporation of the collapsing shell before he can see any objects disappear".

So we conclude that the quantum domain wall [the outer surface of the collapsing star] does not collapse to RS in a finite time, as far as the asymptotic observer is concerned, so that quantum effects do not alter the classical result that an asymptotic observer does not observe the formation of an event horizon.

...

We therefore conjecture that the backreaction due to particle production will cause the collapsing domain wall spacetime to completely evaporate in a finite time.

sol invictus

21st January 2008, 04:53 PM

I won't misuse Gauss' Law. The total energy inside the sphere (not the event horizon, but at the positions where virtual particles form) is zero. The energy inside the shell of infalling star matter is equal to mass M. Outside the star matter and inside where the virtual particles form is a flux of negative particles heading towards the event horizon.

OK, let's review. So you agree that there is a sphere, larger than the event horizon, inside of which there is zero energy, but you don't think the metric inside is flat space. That violates several mathematical theorems (look up positive energy theorem). If you think it's true anyway, you should form a company selling power, because you will be able to extract unlimited energy from the vacuum.

This stuff about "negative energy particles" is just a cartoon to help people picture what's going on during black hole evaporation. It's not what the calculation actually shows. Hawking radiation is a perfectly physical process in which some of the energy in the curvature around by the hole is converted into outgoing particles. That reduces the curvature, leaving it a little smaller. When all the energy has evaporated, there is nothing left behind and the space is flat. That's the only possible endpoint consistent with physical law.

Read this article here (http://arstechnica.com/news.ars/post/20070622-apotential-solution-to-the-black-hole-information-loss-paradox.html).
And the associated paper here (http://arxiv.org/PS_cache/gr-qc/pdf/0609/0609024v3.pdf).

Aha!! I've thought for a while that paper might be part of this... I was wondering when you would mention it.

I read it when it came out. In fact I attended a seminar by one of the authors while the work was in progress (and he was unable to answer several basic questions asked at the end). The result is wrong, and it's known to be wrong by just about everyone else working in the field. I'd be happy to discuss why - although the reasons are not very different from what I've been telling you all along. Or if you'd prefer an argument from authority, I can quote hundreds of papers that contradict that one.

sol invictus

22nd January 2008, 04:13 AM

By the way, while I don't agree with the results of that paper, I want to point out that their results disagree completely with what you have been saying as well. For example they argue that the black hole evaporates in finite time by Hawking radiation, and they don't believe the spectrum of the radiation is strongly modified. They also don't believe (unlike both you and me) that an infalling observer will pass through the horizon (which is, frankly, dumb).

As I said, I'd be happy to explain in detail what's wrong with their analysis. The formalism they use is invalid from the very beginning, and I can demonstrate explicitly why.

jmercer

22nd January 2008, 07:50 AM

DrBaltar

22nd January 2008, 08:18 AM

As I said, I'd be happy to explain in detail what's wrong with their analysis. The formalism they use is invalid from the very beginning, and I can demonstrate explicitly why.

Please do.

sol invictus

22nd January 2008, 08:18 AM

Incidentally, I've been under the impression that at some point a black hole experiencing evaporation would lose enough mass to fall below the Chandrasekhar limit of density and mass required to form a horizon. If so, what happens then?

The Chandrasekhar limit applies to stars. The more massive a star is the stronger its self-gravity will be. Past a certain point the gravitational pull will exceed the pressures that support the star and it will collapse - either into a state with greater internal pressures, or into a black hole.

Once it's a black hole, it has already collapsed and there's no where to go - so there's no such limit. Black holes can be anywhere from the mass of a few protons to the mass of a galaxy.

sol invictus

22nd January 2008, 08:28 AM

Please do.

The biggest problem technically is their use of the mini-superspace approximation, which simply means you throw away all non s-wave degrees of freedom. That reduces the problem to a simple enough form that you can do some computations, but it's invalid in this case. It's not a good approximation to anything, and it invalidates their technical results.

One simple way to do that would be to apply their analysis to a real-world configuration rather than one with a scalar field. Take for example a collapsing shell of photons. Since there is no propagating s-wave mode of either the EM field or gravity, there would be precisely zero radiation according to their method, which is incorrect and also goes completely against their central claim, which is that Hawking radiation strongly modifies the spacetime. The method is just wrong, period.

Qualitatively, their results are wrong for the following reasons:

1) perturbative corrections of the type they claim to compute cannot resolve the information paradox. While it is true that the spectrum will be modified away from thermal by such corrections, a non-thermal spectrum doesn't necessarily contain any more information than a thermal one. Furthermore we now know the resolution of the info paradox for black holes in anti-de Sitter space, and it arises through non-perturbative corrections (as we know must be the case from scaling arguments).

2) their claim that an inertial infalling observer will never see themselves cross the horizon, and will instead see lots of Hawking radiation and a strongly modified spacetime, is utterly absurd. It explicitly violates the equivalence principle, and one can easily arrange a situation in which the backreaction of Hawking radiation is vanishingly small in the region of the horizon at the time of the collapse.

3) their arguments against the standard picture (as summarized on pages 10-11) are similar to and nearly as naive as the ones you've been making here (hence my suspicion that you would bring this paper up), and are incorrect for many of the reasons I've been explaining.

I could go on, but that's enough.