Exploring the Contradictions of Black Holes

In summary, there are three statements about black holes that the conversation attempts to reconcile: 1. The experience of crossing the event horizon for a massive black hole will be normal and finite, 2. An observer far away from the black hole will never see an object cross the event horizon, and 3. An observer far away from the black hole will see it explode in finite time due to Hawking radiation. It is suggested that the third statement may be incorrect, and that the observer falling into the black hole will experience crossing the horizon before the black hole evaporates. However, this assumes that the black hole is not shrinking to nothing from evaporation. In that case, the observer would still not see the object cross the event horizon
  • #1
Ich
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I´m sure that I am not the first to ask this question, but I did not find the answer in recent threads-

There are three statements about black holes that I can´t bring together:
1. I will experience nothing special when I cross the event horizon of a massive black hole, and I will cross it in finite proper time.
2. For an observer in flat space far away from the black hole I will never reach the event horizon - the procedure takes infinite time.
3. An observer in flat space far away from the black hole will see it explode in finite time due to hawking radiation.

So will I suddenly get killed crossing the event horizon? Or will I reach he singularity?

Could someone please point out what I got wrong?
 
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  • #2
Ich said:
I´m sure that I am not the first to ask this question, but I did not find the answer in recent threads-

There are three statements about black holes that I can´t bring together:
1. I will experience nothing special when I cross the event horizon of a massive black hole, and I will cross it in finite proper time.
2. For an observer in flat space far away from the black hole I will never reach the event horizon - the procedure takes infinite time.
3. An observer in flat space far away from the black hole will see it explode in finite time due to hawking radiation.

So will I suddenly get killed crossing the event horizon? Or will I reach he singularity?

Could someone please point out what I got wrong?
All seems correct accept #3. As I recall, since there is a 3k radiation background the black hole will absorb energy at a rate which is greater than it radiates energy. What you're speaking about is micro-black holes in the absence of 3k radiation.

Pete
 
  • #3
pmb_phy said:
All seems correct accept #3. As I recall, since there is a 3k radiation background the black hole will absorb energy at a rate which is greater than it radiates energy. What you're speaking about is micro-black holes in the absence of 3k radiation.

Pete
As I understood the temperature of a black Hole depends inversely on it´s mass.
So in 10^xyz years, when the background radiation is 10^-xy K the black hole will start to lose energy and finally explode.
And then the problem returns.
 
  • #4
Ich said:
I´m sure that I am not the first to ask this question, but I did not find the answer in recent threads-

There are three statements about black holes that I can´t bring together:
1. I will experience nothing special when I cross the event horizon of a massive black hole, and I will cross it in finite proper time.
2. For an observer in flat space far away from the black hole I will never reach the event horizon - the procedure takes infinite time.
3. An observer in flat space far away from the black hole will see it explode in finite time due to hawking radiation.

So will I suddenly get killed crossing the event horizon? Or will I reach he singularity?

Could someone please point out what I got wrong?
The last section of this page seems to say that the observer falling in will observe himself crossing the horizon long before the black hole evaporates, but an observer outside would see him reach 0 distance from the horizon at the exact moment the black hole evaporated to nothing:
What about Hawking radiation? Won't the black hole evaporate before you get there?

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

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

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

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

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

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

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

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

(Due to considerations I won't go into here, some physicists think that the black hole won't disappear completely, that a remnant hole will be left behind. Current physics can't really decide the question, any more than it can decide what really happens at the singularity. If someone ever figures out quantum gravity, maybe that will provide an answer.)
 
  • #5
The commonly accepted answer is that you will fall through the horizon, whatever an external observer sees. However, I find that hard to accept too. See www.chronon.org/articles/blackholes.html

From the relativity faq in JesseM's post:
I won't experience that cataclysm myself, though; I'll be through the horizon, leaving only my light behind. As far as I'm concerned, my grisly fate is unaffected by the evaporation.
Note that this has a hidden assumption of an absolute time - not allowed!
 
  • #6
thanks JesseM, that gave me something to think about.
As I understood, the external observer will never see me crossing the horizon, but will see me dissolved together with the horizon.
And I will experience falling in the black hole the same way I expected to, i.e. the black hole has roughly the properties it had back then when I started falling in - I will not see the explosion at t~10^xyz.

There still remains something that bothers me: Does the external observer really merely see my image frozen at the event horizon while I have already been crushed?
Or, to avoid the simultaneity issue, is it physically possible to rescue me after, say, 1000 years flat time? After 10^xy years? When am I out of reach for the universe?
If I could be reached until the black hole vanishes there would really be a paradox.
 
  • #7
chronon said:
From the relativity faq in JesseM's post:
...
Note that this has a hidden assumption of an absolute time - not allowed!
Could you explain that?
 
  • #8
Ich said:
Note that this has a hidden assumption of an absolute time - not allowed!
Could you explain that?
The verb tenses used in the passage quoted imply that while the external observer goes on seeing your image, you have already fallen through the horizon. However, this implies a comparison of the times of two separated events, and so is not meaningful - it is only meaningful to say event A is before event B if B lies in the future light cone of A.

Ich said:
Or, to avoid the simultaneity issue, is it physically possible to rescue me after, say, 1000 years flat time? After 10^xy years? When am I out of reach for the universe?
No after a certain time light from the external observer will not be able to catch you up. Since the external observer can't travel faster than light it will be impossible to rescue you. (Of course if you believe my arguments that the black hole will evaporate away before you reach it then that complicates things - but then you won't need rescuing).
 
  • #9
chronon said:
The verb tenses used in the passage quoted imply that while the external observer goes on seeing your image, you have already fallen through the horizon. However, this implies a comparison of the times of two separated events, and so is not meaningful - it is only meaningful to say event A is before event B if B lies in the future light cone of A.
Ok, that´s the "simultaneity issue" I brought up. I read the original statement different: It sais that
1. I will fall through and
2. The observer will not see me fall through.
It doesn´t explicitly compare times. But still there´s the other point-->
chronon said:
No after a certain time light from the external observer will not be able to catch you up. Since the external observer can't travel faster than light it will be impossible to rescue you. (Of course if you believe my arguments that the black hole will evaporate away before you reach it then that complicates things - but then you won't need rescuing).
So after a certain time nobody outside can interact with me. That´s sufficient for me to state that it´s not me there at the horizon but merely my image.
I´m trying to resolve the problem like that:
At the time I lost contact with the universe, I´m as good as through the event horizon - I can´t return (can I?) and nothing from the outside can bring me back. So I will be killed by singularity.
What remains after that time is an image which will fade away and finally be destroyed in the explosion. It´s merely an artifact of curved space time (something like trapped photons) and has nothing to do with me.

Does this make sense?
And how does one calculate that time?
 
  • #10
It doesn't really make sense to say that what an external observer sees is just an image rather than the real thing (unless you say the same thing about everything you see). In particular, the external observer may see you and your clock (running slower and slower) falling into the black hole. Suppose that he eventually gets tired of watching, when he sees your clock at some time T, thinking that you must have in reality fallen through the event horizon. However, you have a super-powerful rocket which you start when your clock says time T, (when you haven't yet crossed the event horizon) and so you escape from the black hole, much to his surprise. Hence at no time can the external observer claim that you have actually passed through the event horizon.
 
  • #11
For what it's worth, here are my own weird ideas about what happens to objects falling into a black hole. See http://www.rfjvanlinden171.freeler.nl/idea/index.htm at sections 2 and 3. The assumptions behind it are: a black hole (and gravity) is 5D; everything always moves at c (also in 5D). The clue is that the velocity vector of the falling object rotates towards [tex]x_5[/tex] during the fall.
So the effect of the frozen image at the event horizon is merely the linear projection to 3D space of the moving image in [tex]x_5[/tex].
I'm not completely alone in my assumption that black holes may be 5D objects. Just Google with http://www.google.com/search?hl=en&q=5+dimensional+black+holes to see that.
As already stated in the beginning of my page, don't take it too serious.

BTW have a look at Andrew Hamilton's homepage where you can watch yourself falling into a black hole! http://casa.colorado.edu/~ajsh/home.html
 
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  • #12
This spacetime diagram might help (adapted from Geroch's General Relativity from A to B).
 

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  • #13
Ich said:
Ok, that´s the "simultaneity issue" I brought up. I read the original statement different: It sais that
1. I will fall through and
2. The observer will not see me fall through.
It doesn´t explicitly compare times. But still there´s the other point-->

So after a certain time nobody outside can interact with me. That´s sufficient for me to state that it´s not me there at the horizon but merely my image.

I think you may need to think about this some more.

The person who is falling into the black hole can still receive signals from the outside world. But, once he passes the event horizon, he cannot send signals back.

The question arises - if you have an observer sitting at infinity, sending out pulses once a second, how many of them does the infalling observer see? He can see these pulses even after he crosses the event horizon.

It takes math to really calculate the answer - the only really interesting part of the answer is that he sees a finite number of pulses from the external observer, not an infinite number, before he reaches the black hole.

There is a fairly simple argument to show that the infalling observer can't see an infinite number of pulses, without a lot of math. You have to believe that photons are redshifted when they go "uphill", and blueshifted when they go "downhill". You also have to believe that the tidal forces near a black hole at a given point don't act weirdly as a function of velocity. You also have to believe that the observer hits the singularity in a finite amount of his own proper time.

Given all these assumptions, you can say that from the infalling observers POV, the stretching tidal forces mean that the photons from infinity have to climb "uphill" to reach him. This means that the photons are redshifted.

The point of this is that your mental model must give the following results to be correct:

Photons emitted by the infalling observer don't make it back out
Infalling photons tracing a radial path (ones that fall directly into the black hole) are redshifted when they reach the infalling observer.
The observer reaches both the event horizon (and even the singularity!) in a finite amount of proper time (umm, we're doing a non-rotating black hole here to keep life simple).

Because the infalling photons are redshifted, the observer sees only a finite number of pulses from the outside world before he reaches the event horizon (or before he hits the singularity for that matter). (Not only photons are redshifted, but any time interval is lengthend. If you send pulses out at once a second with a redshift factor of 2, pulses willl arive once every two seconds).

A model that the infalling ovserver never reaches the event horizon usually fails on this point (the finite number of pulses) - because if you think of him never reaching the point, you think that he must receive an unlimited number of pulses. This is wrong.

At the time I lost contact with the universe, I´m as good as through the event horizon - I can´t return (can I?) and nothing from the outside can bring me back. So I will be killed by singularity.
What remains after that time is an image which will fade away and finally be destroyed in the explosion. It´s merely an artifact of curved space time (something like trapped photons) and has nothing to do with me.

Does this make sense?
And how does one calculate that time?

To actually calculate the paths of light and objects, one needs to use a well behaved coordinate system like the Finklestein ingoing coordinates, or Kruskal coordiantes. Of the two, Finklestein coordinates are the easiest to understand. The Finklestein coordinate 'v' is basically just the number of pulses an observer has seen while falling from infinity. Another way of saying this - by consturction, radially infalling light rays have a constant Finklestein coordinate 'v'. This coordinate 'v' is neither a space coordinate, nor a time coordinate, but a 'null' coordinate, becuase it desccribes a light-like path with a constant number.

There's a graph at http://casa.colorado.edu/~ajsh/collapse.html#kruskal which may be of some help if you study it enough of an infalling black hole in Finklestein coordinates (and in Kruskal coordinates, too).

Given the right coordinates, the job of calculating the trajectories is basically a matter of satisfying the geodesic deviation equations, which are a system of second order partial differential equations.

For specificity we say that the particle follows some parameterized path

[tex]x^i(\tau)[/tex], i.e. t(tau), x(tau), y(tau), z(tau), in the usual cartesian coordinates. In Finklestein coordinates it's actually r(tau), v(tau), theta(tau), phi(tau), where theta and phi are two angles (zero for someone falling straight in, so we can deal only with r(tau) and v(tau) for an observer with no angular momentum.

Then we write down the geodesic deviation equations

[tex]\frac{d^2 x^c}{d \tau^2} + \Gamma^a{}_{bc} \frac {d x^b}{d
\tau} \frac {d x^c} {d \tau} = 0[/tex]

Then we solve these equations.

The [tex]\Gamma[/tex] in the above expression are the Christoffel symbols for the given metric - they can be calculated by hand (or much more easily by programs like GRTensor II) from the metric coefficients. They are basically partial derivatives of the metric coefficients.

Working the problem out in Finklestein coordinates, I get the following equations for what it's worth (I've represented differentiaton with respect to tau by a prime mark. You can see these are second order equations in r'', r', r, v'', v', v, where r and v are the Finklestein coordinates. (r is equivalent to the Schwarzschild radial coordinates).

r'' - 2m/r^2 r'v' +(1/r^2-2m^2/r^3) v' v' = 0
v'' + 1/r^2 v' v' = 0

Oh yeah - you need the boundary conditions to solve the equations - specifically, what are r, r', and v (the radial position, radial velocity, and you can safely set v=0 at tau=0, but you need to specify that that's what you're doing).
 
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  • #14
chronon said:
However, you have a super-powerful rocket which you start when your clock says time T, (when you haven't yet crossed the event horizon) and so you escape from the black hole, much to his surprise. Hence at no time can the external observer claim that you have actually passed through the event horizon.
:cry:
There is no time T when I am beyond any hope to return? I really have to cross the event horizon? Isn´t there some calculation that shows that it has to be decided in a finite time whether I will cross the horizon or not?
My problem with it is the following:
An observer will already have collected part of my remains as Hawking radiation at the time when he sees me cross the border. How could I come back then and say "April Fool! I never went down there!".
How would one resolve this?
 
  • #15
Ich said:
Or, to avoid the simultaneity issue, is it physically possible to rescue me after, say, 1000 years flat time? After 10^xy years? When am I out of reach for the universe?
If I am maintaining a fixed distance from the horizon, I think there should be a point beyond which I can't rescue you. The reason I think this is because I know that, despite the fact that an observer on the outside sees you take an infinite amount of time to cross the horizon, you do not see the entire infinite future history of the universe pass before your eyes as you cross it, you only see a finite amount of time pass on my clock (pervect discussed this above); this must mean that there is a time beyond which a light signal emitted from me will not have time to catch up with you before you cross the horizon. So beyond this point, nothing I do should be able to have any causal effect on you before you cross it.

Of course, this doesn't mean that you may not independently choose to fire your rockets right before you reach the horizon, so perhaps there is no time beyond which I can't be sure you won't avoid crossing the horizon, even if there is a point beyond which it's out of my hands to save you.
 
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  • #16
JesseM,

so my argument was flawed. I could still escape the black hole. But then, if I decided so just before entering the horizon, how could I return undamaged to the observer when he collected already a considerable amount of me as Hawking radiation?
 
  • #17
Ich said:
JesseM,

so my argument was flawed. I could still escape the black hole. But then, if I decided so just before entering the horizon, how could I return undamaged to the observer when he collected already a considerable amount of me as Hawking radiation?
I don't know. The problem is even worse if you assume that the Hawking radiation somehow encodes information about what objects fell into the black hole, as a few of the proposed resolutions to the black hole information loss paradox suggest. If I was able to decode the information in the Hawking radiation, could I determine that you did indeed cross the event horizon at a time when I still see you almost frozen outside of it? If this happened, would this allow me to predict with 100% certainty that I would not later see you decide to fire your rockets and move away from the event horizon? I guess as long as I didn't receive this Hawking radiation until after the point where it's no longer in my power to interfere with what happens to you, then this wouldn't lead to any time-travel-type paradoxes, but it still seems a bit weird. Maybe there's some way to avoid this conclusion, but I have no idea what it would be.
 
  • #18
Thanks pervect. I really appreciate the time you spent answering, and I will think about it further.
But right now, I don´t think I understand how the puzzle is resolved.
OK, I only see a finite portion of the universe during my fall. I interpret this fact similar to our universe: as it´s expansion is accelerated, the visible universe shrinks.
When I cross the event horizon, my past light cone does NOT extend to infinity.
Still, if I decided to accelerate and escape, is there a last flat time to do so? Is there a time when it´s safe to say "ok, all information about me is lost to the universe, now I can come back as Hawking radiation"?
 
  • #19
Ich said:
Thanks pervect. I really appreciate the time you spent answering, and I will think about it further.
But right now, I don´t think I understand how the puzzle is resolved.
OK, I only see a finite portion of the universe during my fall. I interpret this fact similar to our universe: as it´s expansion is accelerated, the visible universe shrinks.
When I cross the event horizon, my past light cone does NOT extend to infinity.
Still, if I decided to accelerate and escape, is there a last flat time to do so? Is there a time when it´s safe to say "ok, all information about me is lost to the universe, now I can come back as Hawking radiation"?

There is a last time to do so in the infalling observers coordinate system. This is the time when he (the infalling observer) reaches the event horizon. Once he / you passes the event horizon, there is no return.

However, from the outside perspective, light being emitted close to the event horizon will be delayed, and as you get closer and closer to the horizon, the delay approaches infinity. So the outside observer will never actually see you reach the point of no return. He can see you get arbitrarily close, and if he knows the capabilities of your rocket, he can predict when that particular rocket doesn't have the capability to escape, but the outside observer will never actually see you pass through the critical point where you actually reach the event horizon, no matter how long he waits.
 
  • #20
pervect said:
There is a last time to do so in the infalling observers coordinate system. This is the time when he (the infalling observer) reaches the event horizon. Once he / you passes the event horizon, there is no return.

However, from the outside perspective, light being emitted close to the event horizon will be delayed, and as you get closer and closer to the horizon, the delay approaches infinity. So the outside observer will never actually see you reach the point of no return. He can see you get arbitrarily close, and if he knows the capabilities of your rocket, he can predict when that particular rocket doesn't have the capability to escape, but the outside observer will never actually see you pass through the critical point where you actually reach the event horizon, no matter how long he waits.
So no matter how long you wait, isn't it still possible that just before the astronaut crossed the horizon he fired his rockets and escaped the black hole, but you just haven't seen it happen yet because the light from the event of his beginning to fire the rocket hasn't reached you yet?
 
  • #21
Ich said:
:cry:
There is no time T when I am beyond any hope to return? I really have to cross the event horizon?

Yes, you have to cross to not be able to come back in principle.

My problem with it is the following:
An observer will already have collected part of my remains as Hawking radiation at the time when he sees me cross the border. How could I come back then and say "April Fool! I never went down there!".
How would one resolve this?

If he sees the hole fade away (or whatever) due to Hawking radiation, then the image of you approaching the horizon will fade away with it and presumably you crossed the horizon. Or he may see you turnaround and come back before the hole fades away. If he has collected any of your remains, that’s a piece of you that you left behind.
 
  • #22
One more thing I probably should mention is the similarity between the event horizon of a black hole and that of an accelerating observer.

When an observer accelerates, stationary objects "behind" him seem to pass through an "event horizon" that looks very similar to the event horiozon of a black hole.

However, the stationary observer (it could be you or I) obviously doesn't see anything peculiar happen when he passes the horizon.

The major between the event horizon due to acceleration and that of a black hole is that if the accelerating observer stops accelerating, the event horizon disappears.

Probably the main lesson here is that different observers have different ideas of simultaneity. When horizons enter the picture, the problems with assuming that simultaneity means anything becomes even more accute. What one needs to do is to focus on causal chains - what events can cause other events, because light can reach them from those events, and not worry so much about what happens "at the same time", which doesn't have any operational defintion that's independent of some arbitrary cooridnate system.

A secondary point is that the striking similarity between the acceleration horizon (aka the Rindler horizon) and the event horizonof a black hole suggests that it's not very fruitful to think of one crossing the horizon as "never happening". Clearly, when an accelerating observer accelerates so that we disappear from his view, we are actually still there, and we have actually experienced the events that the accelerating observer can't see. It's just that he can't see them, not that they didn't happen.
 
  • #23
pervect said:
What one needs to do is to focus on causal chains - what events can cause other events, because light can reach them from those events, and not worry so much about what happens "at the same time", which doesn't have any operational defintion that's independent of some arbitrary cooridnate system.
I gave this some more thought. The paradox I tried to create is the following:
There´s a time just before the explosion of the black hole where the observer has collected enough hawking radiation to be sure that I have been destroyed.
However, at the same time (same place, too), the observer still did not see me cross the horizon (according to the FAQ JesseM quoted), and so there would be still the possibility that I could manage to escape.

Just now I think that I misunderstood the statement
If the black hole is mortal, you'll instead see those events happen closer and closer to the time the black hole evaporates. Extrapolating, you would calculate my time of passage through the event horizon as the exact moment the hole disappears! (Of course, even if you could see me, the image would be drowned out by all the radiation from the evaporating hole.)
somewhat: The observer can not see me hanging on the event horizon until the black hole explodes as a matter of principle. The horizon can hold only a certain amount of information (if I recall correctly), and at the time I mentioned there would not be enough information left to hold my picture there.

But that means that there is a time before the explosion where an outside observer can be sure that I will not return, even if he did not see me cross the horizon. That contradicts some of the posts in this thread - maybe because there´s QM involved.
 
  • #24
I see that you were talking about the case where the black hole evaporates all along - due to some combination of inatentaviness on my part, and "lag" where I didn't see all the discussion at the time I composed my response, I missed this point.

With an evaporating black hole, there is some finite time at which the observer will see you cross the event horizon - it's not infinite. (If the black hole didn't shrink, the time would be inifniite). I'm not quite sure how to calculate it exactly, but I think the time at which you are seen crossing the event horizon will be earlier than the time the black hole explodes.

If we consider a hypothetical light wave emitted exactly at the event horizon of a non-shrinking BH, it will "hang" there indefinitely. The only observers that will ever see it (in the non-shrinking case) are those observers who pass through the event horizon at the same point.

Now, if the black hole shrinks, the light that was originally trapped in the event horizon will be able to escape. Even a small shrinkage will let it escape, but this process will take a very long time for a small shrinkage. So it's not very clear operationally how to calculate the time at which an observer will see the photons that were emitted when you cross the event horizon and the BH later shrinks. I suspect a rather detailed calculation would be required.
 
  • #25
pervect said:
With an evaporating black hole, there is some finite time at which the observer will see you cross the event horizon - it's not infinite. (If the black hole didn't shrink, the time would be inifniite). I'm not quite sure how to calculate it exactly, but I think the time at which you are seen crossing the event horizon will be earlier than the time the black hole explodes.
The FAQ answer I linked to in my first post said that an outside observer would see you crossing the event horizon at the exact moment the black hole evaporated to nothing. It's from John Baez's page so I'd be inclined to trust it, although it wasn't written by him.
 
  • #26
JesseM said:
The FAQ answer I linked to in my first post said that an outside observer would see you crossing the event horizon at the exact moment the black hole evaporated to nothing. It's from John Baez's page so I'd be inclined to trust it, although it wasn't written by him.

Be careful, that was what I thought, too. But when I re-read it, I found out that he talks about the extrapolated time - not the actual observation.
Maybe I interpret more into it than he means, but, as pointed out in my previous post, there might be some QM reasons that the image is principally invisible at that time.
 
  • #27
Ich said:
Be careful, that was what I thought, too. But when I re-read it, I found out that he talks about the extrapolated time - not the actual observation.
You're talking about this paragraph:
If the black hole is mortal, you'll instead see those events happen closer and closer to the time the black hole evaporates. Extrapolating, you would calculate my time of passage through the event horizon as the exact moment the hole disappears! (Of course, even if you could see me, the image would be drowned out by all the radiation from the evaporating hole.)
When he says "extrapolating", I took him to mean that the observed distance from the horizon would approach zero as the time approaches the moment of final evaporation, although of course at the exact moment of final evaporation there will be no event horizon left to cross, and the image of the ship would presumably shrink to nothing at that moment too. Was your interpretation different?
 
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  • #28
JesseM said:
Was your interpretation different?
No, it was the same. But now I´m convinced that the image has to fade well before the black hole explodes.
 
  • #29
Ich said:
No, it was the same. But now I´m convinced that the image has to fade well before the black hole explodes.
Why? And are you talking about it fading to the point where it's no longer visible in practice, or are you talking about the idealized case where light isn't quantized and you can detect it at arbitrarily large wavelengths?
 
  • #30
I remember some statement (wasn´t it hawking?) that the event horizon contains but a finite amount of information, proportional to radius (mass) or surface, I don´t know.
That means that my image is principally not more contained in it after some time.
 
  • #31
I've been reading these postings, just trying to get a grip on black holes. I am pretty new to this. Something that I have been wondering about... If space traveller A moves directly toward a black hole, into an ever-increasing spacetime compression (while spacetime seeming to decompress as traveller A compresses), would the perceived remaining distance to the black hole decrease at an ever slowing rate, making the trip to the black hole seem endless? From an outside observer, would traveller A seem to assume a spiral path to the black hole, due to spacetime twisting, even though traveller A is moving directly toward it? Also, could it be possible that black holes are actually extremely large (in size, possibly larger than entire galaxies) but distort our perception of themselves by pulling in the space around themselves to a "virtual" singularity. Sorry, don't mean to interrupt a previous discussion, but it seemed liked you guys might have some input. Thanks!
 

Related to Exploring the Contradictions of Black Holes

1. What is a black hole?

A black hole is a region in space where the gravitational pull is so strong that nothing, not even light, can escape from it. This is due to the fact that the black hole has a singularity, which is a point of infinite density and zero volume.

2. How are black holes formed?

Black holes are formed when a massive star dies and collapses in on itself, creating a singularity. This can also happen when two or more smaller black holes merge together.

3. What are the different types of black holes?

There are three main types of black holes: stellar, intermediate, and supermassive. Stellar black holes are formed from the collapse of a massive star, intermediate black holes are larger than stellar black holes but smaller than supermassive black holes, and supermassive black holes are found at the center of most galaxies.

4. Can anything escape from a black hole?

No, once something crosses the event horizon of a black hole, it cannot escape. This is because the gravitational pull is so strong that not even light can escape.

5. How do scientists study black holes?

Scientists study black holes using various methods, such as observing the effects of a black hole's gravity on its surroundings, studying the radiation emitted from a black hole, and using computer simulations to model and understand their behavior. They also use instruments like telescopes and gravitational wave detectors to gather data on black holes.

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