Black holes and time dilation around the event horizon

  • #1

Main Question or Discussion Point

As I currently understand it from the point of view of an observer falling into a black hole it takes a finite time to cross the event horizon and reach the singularity. From the point of view of a far away observer the person falling into the black hole never actually crosses the event horizon because time is dilated infinitely at the event horizon.

I am also told that because of Hawking radiation all black holes will evaporate fully in a finite (maybe very long) time. Therefore, from the point of view of a distant observer the black hole will actually disappear before the in-falling person crosses over the event horizon.

How is it possible that from one point of view a person can fall into the singularity being destroyed forever, while from another point of a view after a finite amount of time the in-falling person never even crosses into the black hole and can even rejoin the far away observer.
 

Answers and Replies

  • #2
1,538
0
As I currently understand it from the point of view of an observer falling into a black hole it takes a finite time to cross the event horizon and reach the singularity. From the point of view of a far away observer the person falling into the black hole never actually crosses the event horizon because time is dilated infinitely at the event horizon.
I would say no.

Indeed an observer falling into a black hole will reach the singularity in finite time. However a far away observer will never see that simply because light slows down the closer we get to the event horizon. So for instance the light coming from an object near the event horizon will take almost forever to reach a far away observer. When light is at the event horizon it will never reach an outside observer.

But just because an outside observer does not see him passing the event horizon does not mean he does not actually pass.
 
  • #3
I would say no.

But just because an outside observer does not see him passing the event horizon does not mean he does not actually pass.
But shouldn't both reference frames be equally valid?
 
  • #4
PAllen
Science Advisor
2019 Award
7,967
1,260
As I currently understand it from the point of view of an observer falling into a black hole it takes a finite time to cross the event horizon and reach the singularity. From the point of view of a far away observer the person falling into the black hole never actually crosses the event horizon because time is dilated infinitely at the event horizon.

I am also told that because of Hawking radiation all black holes will evaporate fully in a finite (maybe very long) time. Therefore, from the point of view of a distant observer the black hole will actually disappear before the in-falling person crosses over the event horizon.

How is it possible that from one point of view a person can fall into the singularity being destroyed forever, while from another point of a view after a finite amount of time the in-falling person never even crosses into the black hole and can even rejoin the far away observer.
You raise many deep questions, which ultimately have no answers until there is some complete quantum theory of black holes. However, ignoring Hawking radiation for a moment, a partial answer is:

Either the infalling person turned around before going through the horizon or they didn't. If they did, the distant observer will eventually see this. If they didn't, the distant observer will never see this.

As for Hawking radiation, I would characterize its derivations as heuristic. Attempting to use partial quantum theories of gravitation, all of the following have been derived by different researchers claiming expertise:

1) With quantum corrections, nothing ever passes the horizon.
2) With quantum corrections, everything reaches the horizon in finite time, even from an external view (the horizon still being a real phenomenon).
3) With quantum corrections, horizons don't ever form.

(1) and (2) I've seen published in peer reviewed journals. (3) I've only encountered recently, and the only proponent mentioned has published the papers in a single obscure journal.
 
  • #5
Thanks PAllen, I assumed this was the case. So basically I mixed a quantum effect with a purely classical model of black holes. I guess I will have to wait for a complete quantum theory of gravity.
 
  • #6
pervect
Staff Emeritus
Science Advisor
Insights Author
9,675
912
There's a couple of FAQ's on this, as I recall. See for instance http://cosmology.berkeley.edu/Education/BHfaq.html#q9

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

Wrong. When we said that Penelope would see it take forever for you to cross the horizon, we were imagining a non-evaporating black hole. If the black hole is evaporating, that changes things. Your friend will see you cross the horizon at the exact same moment she sees the black hole evaporate. Let me try to describe why this is true.
That answers part of the question, about why things appear differently. They don't appear differently if one is careful.

To untangle cause and effect, it's useful to consider a black hole whose evaporation is controlled by an external event. This is possible because a black hole that's big enough doesn't have to evaporate.

In that case, one can show that there is a "last event" that the infalling observer can observe, by tracing the light rays. Cause and effect only travel inwards, there's no way for causes to travel outwards, but there are still causal relationships that apply.

If the event that controls whether or not the black hole is allowed to evaporate occurs after this last event, it won't affect the fate of the infalling observer one whit, it will be "too late".

One can almost always imagine that an external event that can keep the black hole can happen, it just involves dumping enough mass down it.

So, basically, unless the black hole that one falls into is really, really small, the fate of the infalling observer can't depend on whether it is eventually allowed to evaporate or not.

(I know I didn't come up with the above argument, but I don't recall anymore which of the various FAQ's mention it.)
 
  • #7
1,538
0
But shouldn't both reference frames be equally valid?
Here is a plot (where the Schwarzschild radius is 1) that clearly shows what is going on, why a free falling observer increases his local velocity (wrt a static observer) the velocity observed from a remote location initially increases but closer to the event horizon decreases while eventually coming to a halt at the event horizon.

[PLAIN]http://img254.imageshack.us/img254/5750/velocitycompare.png [Broken]

What is the factor that causes the discrepancy?

It is the speed of light.
 
Last edited by a moderator:
  • #8
pervect
Staff Emeritus
Science Advisor
Insights Author
9,675
912
Some general remarks:

1) I think some sort of "toy metric" for an evaporating black hole could be useful in answering this question rigorously, but I've never seen such a metric written out. I suppose one could try a Schwarzschild metric with a time-dependent and very slowly varying total mass, I'm not sure if there are any hidden "gotcha's" in the approach, which would be an approximate solution to Einstein's equations rather than an exact one.

Answering questions rigorously about what someone sees when you don't have a metric is sort of difficult. Though one can make an educated guess, the results will be rather non-detailed.

2) The question is rather advanced, and I suspect a lot of people are worrying about the hard questions prematurely, in the sense that they'd learn more if they asked the easier questions where they could understand the answers fully. A quick pithy summary would be "walk before you try to run".

3) A large chunk of the problem is that people interpret "time slowing down" near the event horizon in terms of absolute time. People who have learned SR know that absolute time doesn't exist. Unfortunately, people who haven't learned SR not only don't know that absolute time doesn't exist, but don't even realize that you are talking about them when you talk about people who believe in absolute time.

This gets into the "walk before you can run" aspect I was making earlier, point #2. If the reader has some clue about why absolute time doesn't exist, trying to untangle the one-way causal relationships in a black hole is going to be a lot easier. I'm not sure it's even possible to do correctly if you're making false fundamental philsophical assumptions to start with.
 
  • #9
PAllen
Science Advisor
2019 Award
7,967
1,260
There's also the fact (mentioned in other threads, but not here so far) that except for hypothetical tiny primordial black holes, black holes are growing, not shrinking. The influx of energy from CMB is orders of magnitude greater than Hawking radiation. Only near the heat death of the universe will this likely reverse.
 
  • #10
1,538
0
I am surprised everybody is looking for explanations using quantum theory or Hawking radiation for something that is quite simple: light simply takes more time to get to the remote observer the closer it is to the event horizon and at the event horizon it never gets out. So there is really not much of a surprise a remote observer cannot see the object fall through the event horizon.
 
  • #11
(I know I didn't come up with the above argument, but I don't recall anymore which of the various FAQ's mention it.)


They used complex numbers to come up with the theory of black holes. The reality of black holes must be like the reality of the square root of negative unity, i=√(-1).
 
  • #12
DrGreg
Science Advisor
Gold Member
2,274
772
They used complex numbers to come up with the theory of black holes. The reality of black holes must be like the reality of the square root of negative unity, i=√(-1).
It's quite possible (and, I think, usual) to describe all aspects of black holes without the use of complex numbers.

(By the way, electrical and electronic engineers frequently describe alternating current and electronic circuits using complex numbers. I suppose that casts doubts on the reality of electronics?)
 
  • #13
My Theory would be that as the black hole evaporated you would see the person getting closer to it at an equal rate that it shrunk until it was just an immensly high gravity area that you see squash your friend. It most likely could not evaporate so fast as to vanish and your friend pass right on by, it would have to pass between being a blackhole and not being there.
 
  • #14
Drakkith
Staff Emeritus
Science Advisor
20,746
4,452
I am surprised everybody is looking for explanations using quantum theory or Hawking radiation for something that is quite simple: light simply takes more time to get to the remote observer the closer it is to the event horizon and at the event horizon it never gets out. So there is really not much of a surprise a remote observer cannot see the object fall through the event horizon.
How so? Wouldn't it be red shifted instead of taking longer to arrive since the speed is the same in all frames?
 
  • #15
1,538
0
How so? Wouldn't it be red shifted instead of taking longer to arrive since the speed is the same in all frames?
The speed of light is c locally. That does NOT mean that the speed of light between two objects A and B removed a distance from each other is c, this speed varies in a gravitational field.
 
  • #16
PAllen
Science Advisor
2019 Award
7,967
1,260
How so? Wouldn't it be red shifted instead of taking longer to arrive since the speed is the same in all frames?
No, passionflower is right. To a distant observer, light itself moves ever slower close to an event horizon. For example, this could be seen in 'light clock' reflecting light back and forth between two mirrors, oriented tangentially to the horizon and near it. To the distant observer, this clock (like any other) would run very slow - thus the light is moving very slow between the mirrors.

To the near horizon observer, everything is normal.

There is no sense in which one observer is right and the other wrong - they are just different frames of reference.
 
  • #17
Drakkith
Staff Emeritus
Science Advisor
20,746
4,452
I see. I'll have to go re-read my relativity book!
 
  • #18
1,538
0
Two formulas you want to keep in mind if you want to find the speed of light as observed by a far away observer:

For radial motion it is
[tex]\Large 1-{\frac {{\it r_s}}{r}}[/tex]
For other motion we can use an impact parameter b:
[tex]\Large \left( 1-{\frac {{\it r_s}}{r}} \right) \sqrt {1+{\frac {{\it r_s}\,{b}
^{2}}{{r}^{3}}}}[/tex]
Based on the value of this impact parameter we can also determine if an incoming light pulse plunges, goes into an (unstable) orbit or escapes and generally we can get an idea about gravitational lensing.
 
Last edited:
  • #19
Rs Being the Swartszchild radius? In that case then as the blackhole, and thefore the radius decayed, would the slow moving light reach us in what we would consider less time, giving the appearance of the person slowly moving forward with the radius?

What would we expect to see in a blackhole as its radius shrunk to just within the volume of the mass supporting it? I would guess the pressure would cause it to be a very hot plasma or something? Certainly not clear empty space that one could just float on through as it seems to me is assumed in the OP's question

How is it possible that from one point of view a person can fall into the singularity being destroyed forever, while from another point of a view after a finite amount of time the in-falling person never even crosses into the black hole and can even rejoin the far away observer.
 
  • #20
1,115
3
No, passionflower is right. To a distant observer, light itself moves ever slower close to an event horizon. For example, this could be seen in 'light clock' reflecting light back and forth between two mirrors, oriented tangentially to the horizon and near it. To the distant observer, this clock (like any other) would run very slow - thus the light is moving very slow between the mirrors...
I had taken as a given that such could be a well defined gedanken experiment - one could use the SC's to properly define transverse distance and light speed. But according to your reasoning here: https://www.physicsforums.com/showpost.php?p=3573328&postcount=79 what handle can one get on 'distance' between the mirrors and thence 'c' of light bouncing between? You say light moves ever slower closer to the EH. Fine, except that if one accepts a fuzzy/ambiguous notion of length in GR, how precisely and unambiguousy can 'slower light speed' be defined?
 
  • #21
pervect
Staff Emeritus
Science Advisor
Insights Author
9,675
912
Speeds and distances are both well defined and unambiguous in the local tangent space. Only when you start to look at larger distances over which the curvature of space-time is significant do you start to run into ambiguities.
 
  • #22
1,115
3
Speeds and distances are both well defined and unambiguous in the local tangent space...
Correct me if wrong - but is not 'local tangent space' here the situation where c is just the local, invariant, proper value? Which would be of no help when it comes to comparing 'here' with 'over there' - precisely the issue in question.
Only when you start to look at larger distances over which the curvature of space-time is significant do you start to run into ambiguities.
OK, but the static source, static observer problem I posed in the link specified what one would think a case where distance, on a relational basis, should have an unambiguous definition. Was left with the impression even for this simplest case, there is ambiguity. And yet astrophysicists have it seems precisely pinned down many parameters testing GR from the orbital mechanics of the Hulse-Taylor binary pulsar. Hard to believe that is possible without having a precise handle on length scale - but I could be wrong.
 
  • #23
PAllen
Science Advisor
2019 Award
7,967
1,260
I had taken as a given that such could be a well defined gedanken experiment - one could use the SC's to properly define transverse distance and light speed. But according to your reasoning here: https://www.physicsforums.com/showpost.php?p=3573328&postcount=79 what handle can one get on 'distance' between the mirrors and thence 'c' of light bouncing between? You say light moves ever slower closer to the EH. Fine, except that if one accepts a fuzzy/ambiguous notion of length in GR, how precisely and unambiguousy can 'slower light speed' be defined?
Absence of unique choice does not mean no choice can be made; it means more than one can be made, with some difference in the outcome. Pick most any plausible way of defining long distance and remote distance, and the distant observer will see lightspeed slow near an event horizon. The exact numbers will differ based on choice.

However, in some cases, the nature of a phenomenon is changed by such choices (which are really coordinate system choices). An example I like is the Shapiro effect. If I require that planets travel on near perfect ellipses, and measure orbital parameters of a planet at several points away from the sun, and then measure radar travel times to the planet as it passes behind the sun, I see Shapiro time delay. On the other hand, if I define distance by round trip light time times constant c, I see a different effect altogether: the Shapiro orbital bump. Both are completely valid, usable choices in GR. The former is more conventional, and simpler to use for astronomy.
 
  • #24
1,115
3
...An example I like is the Shapiro effect. If I require that planets travel on near perfect ellipses, and measure orbital parameters of a planet at several points away from the sun, and then measure radar travel times to the planet as it passes behind the sun, I see Shapiro time delay. On the other hand, if I define distance by round trip light time times constant c, I see a different effect altogether: the Shapiro orbital bump. Both are completely valid, usable choices in GR. The former is more conventional, and simpler to use for astronomy.
Surprised that both cases can be equally valid. Does not the assumption of a constant c in the second case make it suspect - the orbital 'bump' (apparent orbital speed-up/speed-down?) being an artifact of that assumption?

Getting back to the example given of light bouncing between mirrors. Let's forget the complications of being near an EH (and presumably required thus to be in free-fall). Pick a quiet, non-rotating perfectly spherical planet. Mirrors and bouncing light are on the surface, and an observer perched in a tall tower observes from above. We all agree that the observer detects a redshift of the light clock frequency, relative to the locally measured value at the surface. Can we or can we not derive this in an unambiguous manner via a gedanken experiment which assumes that light speed and/or distance between mirrors is gravitationally affected (sans the usual extraneous mechanical/optical effects of 'g') in a manner derivable from SC's? Or is the best we can do is refer to a 'quotient' of f = c'/(2*d') (f the redshifted clock frequency, c', d' the 'redshifted' values of light speed, mirror displacement, respectively), with neither c' or d' individually definable? Hard to see such a situation is mandated by GR.
[It is understood that by some authorities at least, in standard SC's c' is less in the radial direction than tangential, but not in ISC's. So just pick one, and work within it. Hopefully either system would agree as to net effects.]
 
Last edited:
  • #25
1,538
0
On the other hand, if I define distance by round trip light time times constant c, I see a different effect altogether
You can define whatever you want but that does not make it so. For instance how would you explain that the distance between A to B and B to A is different.

Can you provide some credible references where distance in GR is defined as the roundtrip time of light?
 
Last edited:

Related Threads on Black holes and time dilation around the event horizon

Replies
16
Views
1K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
11
Views
6K
Replies
3
Views
643
Replies
142
Views
22K
Replies
6
Views
1K
Replies
99
Views
20K
  • Last Post
Replies
5
Views
523
Top