Black Hole Event Horizon and the Observable Universe

[nicksauce]

I was watching a NOVA program, and they claimed that if you were falling into a black hole, when you crossed the event horizon you would be vaporized by the intense light that's built up at the surface that hasn't been able to escape the BH. This seems plausible, I suppose, but I have never heard such a thing before; in fact everything I've read said that you wouldn't notice anything while crossing the horizon. So is there any truth to this claim?

 

[atyy]

Maybe the blue sheet?

Are Physical Objects Necessarily Burnt Up By the Blue Sheet Inside a Black Hole? Lior M. Burko, Amos Ori
http://arxiv.org/abs/gr-qc/9501003
http://citeseer.ist.psu.edu/old/401984.html

 

[George Jones]

(Non-extremal) Rotating black holes have two horizons. The outer horizon, the boundary of the black hole, is an event horizon, and nothing too special happens as an observer crosses this horizon. The inner horizon is a Cauchy horizon, and some calculations indicate that measured energy density diverges at the Cauchy horizon. The Burki Ori paper to which atyy linked is about the Cauchy horizon (of Reissner-Nordstrom black holoes).

 

[wolram]

(Non-extremal) Rotating black holes have two horizons. The outer horizon, the boundary of the black hole, is an event horizon, and nothing too special happens as an observer crosses this horizon. The inner horizon is a Cauchy horizon, and some calculations indicate that measured energy density diverges at the Cauchy horizon. The Burki Ori paper to which atyy linked is about the Cauchy horizon (of Reissner-Nordstrom black holoes).

Please forgive this intrusion, but why is it that we know so much about black holes, i do not want to be antagonistic but what do we really know about them?

 

[nicksauce]

(Non-extremal) Rotating black holes have two horizons. The outer horizon, the boundary of the black hole, is an event horizon, and nothing too special happens as an observer crosses this horizon. The inner horizon is a Cauchy horizon, and some calculations indicate that measured energy density diverges at the Cauchy horizon. The Burki Ori paper to which atyy linked is about the Cauchy horizon (of Reissner-Nordstrom black holoes).

Ok thanks, that makes sense. I'm a little confused about the two horizons for rotating black holes. Is the outer horizon the same thing as the ergosphere, or is that something different all together?

 

[George Jones]

Ok thanks, that makes sense. I'm a little confused about the two horizons for rotating black holes. Is the outer horizon the same thing as the ergosphere, or is that something different all together?

No, the boundary for the ergosphere is outside the event horizon, except at the poles, where it touches the event horizon. The event horizon is a boundary of no return, but, using rockets, observers can cross and re-cross the boundary of the ergosphere as many times as they like.

 

[MuggsMcGinnis]

Do you know if anyone has published a solution to General Relativity that defines a frame of reference from which it would be possible to observe something to pass beyond an event horizon? That is, see it disappear?

Perhaps naively, I presume that, if it can happen, there must be some possibility of observing it. Or, at least some context from which the observation could be defined.

I'm referring, specifically, to objects with non-zero rest mass. Can an observer, in any reference frame, observe an object with non-zero rest mass, fall (disappear) through a Schwarzschild event horizon?

Thanks

 

[Chronos]

Reaching the event horizon would not be pleasant. Photons from the outside universe would be severely blue shifted.

 

[MuggsMcGinnis]

I can't disagree about how pleasant the trip would be... I'm just trying to understand if it is (observably) finite in duration.

By my reasoning, as an object falls closer to the event horizon, time (for it) slows.
Since the event horizon has a non-zero temperature, it's radiating a certain wattage of power per unit of surface area.
As time slows, the quantity of energy per second will increase.
Therefore, as the object falls closer to the event horizon, it will cook.

 

[Chronos]

It will not cook from the inside out, rather from the outside in. The horizon the observer infalls toward is redshifted relative to the observer. It is the external universe that is time accelerated and blue shifted relative to the observer.

 

[George Jones]

I can't disagree about how pleasant the trip would be... I'm just trying to understand if it is (observably) finite in duration.

By my reasoning, as an object falls closer to the event horizon, time (for it) slows.
Since the event horizon has a non-zero temperature, it's radiating a certain wattage of power per unit of surface area.
As time slows, the quantity of energy per second will increase.
Therefore, as the object falls closer to the event horizon, it will cook.

Since you refer to a black hole's temperature, I assume you're talking about Hawking radiation. Here what Birrell and Davies, pages 268-269, has to say.

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd to0 conclude that it acts as physical barrier to the falling observer.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuumm polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ..

It will not cook from the inside out, rather from the outside in. The horizon the observer infalls toward is redshifted relative to the observer. It is the external universe that is time accelerated and blue shifted relative to the observer.

For a freely falling observer, this cooking occurs not at the event horizon, but as the observer approaches the singularity. For a hoverig observer, cooking occurs "at" the event horizon.

 

[MuggsMcGinnis]

Regarding my previous posting... anybody have an answer?

I have yet to find equations for calculating how long it will take for an object with non-zero rest mass to be seen to fall through an event horizon. I thought that the Kruskal extensions might provide a means of calculating this, but I don't seem to be clever enough to figure it out.

Anybody?

Do you know if anyone has published a solution to General Relativity that defines a frame of reference from which it would be possible to observe something to pass beyond an event horizon? That is, see it disappear?

Perhaps naively, I presume that, if it can happen, there must be some possibility of observing it. Or, at least some context from which the observation could be defined.

I'm referring, specifically, to objects with non-zero rest mass. Can an observer, in any reference frame, observe an object with non-zero rest mass, fall (disappear) through a Schwarzschild event horizon?

Thanks

 

[Chronos]

An external observer will never see the infalling party cross the event horizon. A time dilation thing.

 

[MuggsMcGinnis]

Chronos

That's what I thought.

What bothers me is that the presumption that something can fall to (much less, through) an event horizon in finite time depends upon there being a `true reality' that contradicts the observable reality.

Imagine an observer external to the event horizon throws an object toward the center of mass of the `black hole'. No matter how long the observer waits, the in-falling object will always appear to not have reached the event horizon.

In fact, the observer can interact, causally, with the object, by bouncing light off of it. The object will approach the event horizon at an ever-slowing rate... a rate asymptotically approaching zero as the distance above the Schwarzschild radius approaches zero and as time approaches infinity.

If it's possible to interact with the in-falling object, no matter how long one waits, how can we presume that the object actually falls through the event horizon in finite time?

It seems that such a presumption is contrary to direct observation.

 

[xantox]

Regarding my previous posting... anybody have an answer? I have yet to find equations for calculating how long it will take for an object with non-zero rest mass to be seen to fall through an event horizon.

It will take exactly the same time as the black hole takes to disappear completely by Hawking evaporation.

 

[MuggsMcGinnis]

Yes, if one includes Hawking radiation. However, I was just considering the idealized case of a static Schwarzschild event horizon.

Regarding some previous posts in this (somewhat divergent) thread: It's easy to see how Hawking radiation would cook any in-falling object. As the black hole shrinks, its temperature increases and it radiates more energy per second. The in-falling object is getting closer to this surface that's getting hotter. Also, time is slowing for the in-falling object so the quantity of energy it sees radiating from the event horizon is even greater, per unit of time. All of these factors combine so that (I'm guessing) the apparent temperature that the in-falling object is subjected to approaches infinite as the remaining lifetime of the event horizon approaches zero. It would end with one infinitesimal instant of infinite temperature.

 

[xantox]

If it's possible to interact with the in-falling object, no matter how long one waits, how can we presume that the object actually falls through the event horizon in finite time?

By building this spacetime in general relativity and then measuring the proper time on the infalling body worldline.

It's easy to see how Hawking radiation would cook any in-falling object. As the black hole shrinks, its temperature increases and it radiates more energy per second. The in-falling object is getting closer to this surface that's getting hotter. Also, time is slowing for the in-falling object so the quantity of energy it sees radiating from the event horizon is even greater, per unit of time. All of these factors combine so that (I'm guessing) the apparent temperature that the in-falling object is subjected to approaches infinite as the remaining lifetime of the event horizon approaches zero. It would end with one infinitesimal instant of infinite temperature.

No, the infalling (free-falling) body will cross the event horizon right away, when the black hole is big, and the temperature of the horizon is low. Distant observers will see in the same distant future the last photons emitted in the distant past by the infalling body mixed with the photons from the black hole final explosion by evaporation.

 

[MuggsMcGinnis]

And the external observer will see the in-falling object disappear through the event horizon?

 

[xantox]

And the external observer will see the in-falling object disappear through the event horizon?

Yes, in the sense that he will receive at that moment the very last theoretical photons which may be possibly emitted by the object before entering the black hole.

 

[skeptic2]

Xantox, I wonder if you could explain your answer from post #17 a little. It is clear that to a distant observer an infalling object will appear to slow down, turn red and dim as it approaches the event horizon. However, apart from this optical effect, from the inertial frame of the distant observer, isn't time increasingly dilated and space radially contracted close to the horizon? It would seem that the object would still have to cross an infinite expanse of contracted space during increasingly dilated time. The infalling object wouldn't just appear to slow down and stop, to the distant observer it in fact, would. Could you please explain how the free falling object is able to cross the event horizon right away? I recognize of course that in its own inertial frame the infalling object would fall to the center of the BH rapidly as the event horizon evaporates beneath it.

 

[xantox]

The infalling object is following the strong curvature of the black hole spacetime – this is what makes the difference in respect to a distant observer in flat spacetime. The freefall path into the hole is finite (the proper time to cross the horizon is finite), it is only the "shadow" of this path projected on the coordinates of the distant observer which actually goes to infinity because of the curved geometry.

 

[MuggsMcGinnis]

There is no meaningful distinction between what "appears" to happen and what "really" happens. If there were, Einstein's relativity wouldn't work; it's entirely based on "appearance". Are you suggesting that there are two categories of "reality"? One that's how the universe "appears" to be and another that's how the universe "actually" is?

If the actual nature of reality isn't observable then empirical science has no basis.

Don't get me wrong... I'm a theoretician. But, I understand the value of empiricists. We do, occasionally, need someone to test even the most elegant theory. :c)

Any analysis that yields two logically incompatible results cannot be correct.

Here's a fact: There are no reference frames from which an infalling object can be observed to cross an event horizon... None.

However, all reference frames can observe that an infalling object is outside of the event horizon, forever. After 1 day. After 1 billion years. Only at the instant that the event horizon finally evaporates can any infalling object reach the event horizon. And then, only the instant it disappears.

If you can define a frame of reference from which an infalling object can be *observed* to disappear across the event horizon, I'll accept that it's a possibility. Have you read Kruskal?

For all objects of non-zero rest mass, the distance through spacetime to any event horizon is infinite.

 

[MuggsMcGinnis]

By the way, xantox, thank you for your reply. I appreciate you taking the time to argue with me. :c)

 

[skeptic2]

Thank you MuggsMcGinnis. I was hoping to get past this issue and on to the next which is, if nothing can cross the event horizon, how can the black hole exist? Although a case could be made for matter being trapped inside when a star collapses, that doesn’t explain 100 million solar mass black holes.

The obvious answer is that each particle of matter sees its own event horizon determined by the amount of matter that has fallen in ahead of it. This would imply that a BH is built up in layers like an onion. The matter at each radius remains where it is due to its inability to pass through its own event horizon. The only problem with this explanation is that it suggests the BH is solid instead of empty with a singularity at the center.

My understanding is that the existence of the singularity is required by the lack of any known degeneracy pressure sufficient to halt the collapse from gravity. Doesn’t the event horizon satisfy that requirement?

 

[MuggsMcGinnis]

skeptic2:

Here are my guesses as to what happens.

"if nothing can cross the event horizon, how can the black hole exist? Although a case could be made for matter being trapped inside when a star collapses, that doesn’t explain 100 million solar mass black holes."​

First off, I don't think that we should start from an assumption that 'black holes', as commonly described, do exist. Starting from that assumption makes it difficult to explore any other possibilities.

I prefer to think in terms of event horizons, without assuming that there's anything inside of an event horizon.

Let's assume that it is possible to pile so much matter together, in sufficient density, that no physical force can withstand the gravitational field and the stuff collapses under its own gravity.

I think it's worthwhile to consider the process. There will be a time, before the collapse, when repulsive forces can just barely resist gravity. Then, something happens and the collapse begins. This event won't occur simultaneously everywhere throughout the object. There will be one or more 'nucleation' sites, points at which collapse begins.

From the nucleation site(s), the collapse will spread outward to engulf the entire mass.

The question is, what does the nucleation and spread look like?

Initially, each site would be an infinitesimal event horizon; the nature of which is the point of this discussion.

Suppose that nothing can fall to an event horizon in finite time.


These nucleation sites would actually be 'empty'. That is, they would be empty of spacetime. They would be destinations unreachable. Things would fall toward them, but never reach them. Like the Oakland where Gertrude Stein spent her childhood, "There is no there there." A black hole's event horizon defines a 'region' from which spacetime has been excluded.

As the collapse continued, more matter became captured, falling toward one or more expanding event horizons.

As event horizons merged, this matter would be swept outward (from the perspective of an outside observer). It would still be falling toward the event horizon (EH) but, like any matter falling toward an EH, it would never reach the EH.

Like anything near an EH, it would appear almost frozen in time. I think, as two event horizons merge, any particle of matter about one of these event horizons would (discounting the distance of its fall during this process) retain a position of constant field. Gravitational field.

As you say, "This would imply that a BH is built up in layers like an onion. The matter at each radius remains where it is due to its inability to pass through its own event horizon."

However, this doesn't imply that the black hole is "solid". Think of a spherical, hollow onion. The outer part of the onion is formed of concentric shells. However, all of these shells have a radius > R_S. This R_S would correspond to the Schwarzschild radius of our event horizon.

Thus, all of the matter that has contributed to an event horizon (black hole) would remain outside of the event horizon, for any finite duration one might choose to observe.

The gravitational effects would look the same. A spherical ball with all of its mass located at its center has identical gravitational characteristics as a spherical ball with all of its mass uniformly distributed about its surface.

The only differences would be that this model doesn't require violating rules that I, personally, have grown quite fond of. E.g. Universality and Causality.

For anything to actually fall through an event horizon requires physics to work differently for observers in different reference frames and for information to be lost from the universe.

 

[xantox]

If you can define a frame of reference from which an infalling object can be *observed* to disappear across the event horizon, I'll accept that it's a possibility.

As the name suggests, external observers cannot indeed see events occurring at the event horizon.

So you wonder how it is possible for an infalling body to actually reach the horizon in finite time, which is the case, while distant observers never see it crossing. I'll attempt a handwaving explanation, while of course the actual proof is mathematical.

This is possible because in general relativity, not only time and space are relative but all which does matter is the geometry, eg the spacetime curvature, and the infalling body is following a path in extremely curved spacetime while the distant observer is not.

This leads to a huge asymmetry between the two observers. What is time for the first may cease to be time for the second, due to the tilting of light cones. It's sort of like being in a rollercoaster where "up" and "down" is not necessarily the same thing as for the guys watching.

In particular, approaching the horizon means traveling near what distant observers would consider the edge of their light cones, and reaching the horizon means traveling on what distant observers would consider a null interval (roughly, the one which light is able to cross in zero time). Such "eternal door" made of spacetime can thus close in a snap for the infalling body in the limit of the horizon.

 

[MuggsMcGinnis]

xantox:

Please consider this experiment and tell me what's wrong with it or my conclusions.

An observer located at a large, but finite distance (1,000 AU) from a Schwarzschild black hole (mass = 1 solar mass; 3 km radius) hurls a tiny highly-reflective sphere directly toward the gravitational center of the black hole. The sphere starts off moving pretty fast... maybe .1% of the speed of light.

The observer uses a laser range-finder to measure the distance to the in-falling sphere.

A black hole of one solar mass would be expected to evaporate in about 10^67 years, due to Hawking radiation.

At what point in time (less than 10^67 years), would the observer fail to receive a reflected signal from the sphere?

Results & Conclusions:

1. If the observer loses contact with the sphere before the black hole evaporates, then the sphere has been observed to fall through the event horizon.
2. If the observer can always, until the black hole evaporates, receive a reflected signal from the in-falling sphere, then one of the following two statements must be true:
   • The in-falling sphere falls through the event horizon but the observer can continue to bounce light off of some artifact of the sphere, that isn't really the sphere, because the sphere has disappeared below the EH.
   • The sphere does not reach the event horizon.

As you acknowledge, result (1) won't be observed.

We're left with result (2).

I'm arguing that what happens is (2B).

Am I correct that you feel result (2A) is possible?

Have I misstated your position?

 

[xantox]

Concerning 1., I agree with you, but to be exact we should add that the observer cannot say for sure -when- he loses contact with the body. This is because the number of photons emitted by the body before crossing the horizon is finite, and that it happens that the very last photon received by the distant observer is most likely to arrive extremely fast. But the observer still cannot say for sure when he received the -last- photon. Maybe the photon he received after 1 year was already the last one. Or maybe there is just one more left, but it will arrive 10^30 years later. So we should say more correctly that the last -possible- theoretical photon (the one which would be emitted on the limit of the horizon) received by the shell observer would arrive when the black hole completes its evaporation.

Concerning 2., 2B is not correct within general relativity, as the computation on the infalling body wordline gives a precise and finite duration to reach the horizon in terms of the body own clock. 2A is not correct too, as follows: the photons emitted by the shell observer will bounce on the sphere until it crosses the horizon (so very soon), then no more photons will bounce and complete blackness will follow. Most of those bounced photons, which are in finite number as said in 1., will come back soon to the shell observer, so blackness will be observed very soon. However there could be a few remaining photons which can be greatly slowed down and arrive only million years later, or at worst when the black hole disappears by evaporation, so that we're only sure at that point that there are no more left. Those late arriving photons are still from the set which bounced on the sphere zillions years earlier, just before it crossed the horizon.

 

[MuggsMcGinnis]

xantox:

RE: "Concerning 1"

1. Each pulse sent to the in-falling sphere travels, for all observers, at the speed of light. The return from each pulse will be received in less than 5.8 days.
2. If there's any question about the identity of each pulse, the signals can be modulated.
3. As an object falls closer to a Schwarzschild event horizon, it spreads horizontally, approaching in shape a spherical shell about the event horizon. Its cross-section should remain adequate for reflecting this signal.

Let's just presume that the laser is powerful enough, the reflectivity high enough, and the detector sensitive enough for the pulse returns to be reliably detected.

RE: "Concerning 2"

1. The transformation that you refer to, that, "within general relativity, as the computation on the infalling body wordline gives a precise and finite duration to reach the horizon" is the issue at question in this discussion. I am arguing that it is not a valid transformation because it yields results that are logically incompatible with the original, untransformed reference frame. The transformation yields results that, because they are incompatible, violate the assumption of universality, that reality works the same for all frames of reference. Universality is a premise of relativity and I argue that any deduction that contradicts a premise of the argument is an invalid deduction.
2. Can you please expand on your statement, "photons which can be greatly slowed down"? Under what circumstances does General Relativity predict that light will be slowed down?

Thank you for your replies.

 

[skeptic2]

Concerning 2B. When the computation was done of the time it takes for an infalling object to reach the event horizon, was the radius of the EH assumed to be fixed for the duration of that time? Unlike two observers traveling at relativistic velocities relative to each other and who both observe the others clock is running slow, an observer in curved spacetime and one in flat spacetime both agree whose clock is running slow, the one in curved spacetime. I think it is reasonable to assume that given the extreme time dilation very close to the EH, that the BH may evaporate as fast as the object falls toward it. How long would the time to reach the EH be if recalculated for the reference frame of flat spacetime?

Concerning 2A. I agree that some photons may take a very long time to return but instead of saying they were greatly slowed down can we say that they traveled a much greater distance due to the contraction of spacetime near the horizon?

 

[MuggsMcGinnis]

1) It's clear that, for the external observer, the event horizon evaporates before anything falls into it.

2) For the external observer, the distance to the event horizon is always less than 5.8 light-days. The only way for the laser pulse to require more than 11.6 days, round-trip, is for light to travel slower than the speed of light. By the way, I'm not willing to accept the possibility that, for any observer, light travels through vacuum at any speed other than c. No valid interpretation of general relativity can require this.

Also, GR requires universality: Observations made by any observer must be compatible with those of any other observer.

The problem here is, how can one reconcile these two statements?

1. The infalling object quickly (less than 580 days, in this example) falls below the event horizon.
2. The observer can interact (bouncing light off of the object & detecting the return) with the in-falling object for more than 10⁶⁶ years.

 

[xantox]

Each pulse sent to the in-falling sphere travels, for all observers, at the speed of light. The return from each pulse will be received in less than 5.8 days. [..] I am arguing that it is not a valid transformation because it yields results that are logically incompatible with the original, untransformed reference frame. [..] The problem here is, how can one reconcile these two statements? [..]

The photon emitted at the limit of the horizon will be received in infinite time measured by the distant observer, ie. never (supposing the purely classical case without evaporation). You're apparently considering this problem within special relativity, but that only allow to talk about flat spacetime with no gravity, where black holes do not exist indeed. I friendly recommend a reading of the beautiful and accessible introductory text by James Hartle, "Gravity: An Introduction to Einstein’s General Relativity", Addison Wesley (2002) (available on Amazon).

Can you please expand on your statement, "photons which can be greatly slowed down"? Under what circumstances does General Relativity predict that light will be slowed down?

I was supposed to mean that in terms of the distant observer clock, the photons which he would expect to receive much sooner if the spacetime was flat, will be received much later. This is again a manifestation of spacetime curvature, not a modification of properties of light which always move locally at the speed of light c. That the travel time of light increases within a strong gravitational field is one of the classic tests of general relativity which has been confirmed experimentally with a precision of about 0.1%.

When the computation was done of the time it takes for an infalling object to reach the event horizon, was the radius of the EH assumed to be fixed for the duration of that time?

Yes, since on that short timescale compared to the one of evaporation it may be considered fixed. But we may simplify the problem by ignoring evaporation, so the radius will never shrink. Also, another computation can be done of the time it takes for a star to collapse into a black hole, which gives again a finite (and short) result.

Unlike two observers traveling at relativistic velocities relative to each other and who both observe the others clock is running slow, an observer in curved spacetime and one in flat spacetime both agree whose clock is running slow, the one in curved spacetime. I think it is reasonable to assume that given the extreme time dilation very close to the EH, that the BH may evaporate as fast as the object falls toward it.

The event horizon may be considered a surface of infinite time dilation relative to the distant observer in flat spacetime. This does not mean that the infalling clock is stopping when it reaches the horizon, or that inward photons become freezed there. Inward photons continue to move at c at the horizon, infalling observers clocks do not stop at the horizon.

I agree that some photons may take a very long time to return but instead of saying they were greatly slowed down can we say that they traveled a much greater distance due to the contraction of spacetime near the horizon?

Perhaps you mean space contraction? but if space would contract then the distance would be smaller..

 

[skeptic2]

"This does not mean that the infalling clock is stopping when it reaches the horizon, or that inward photons become freezed there. Inward photons continue to move at c at the horizon, infalling observers clocks do not stop at the horizon."


True, clocks in their own reference frame do not stop but we are not talking about clocks in their own reference frame nor the time it takes an object to fall to the EH in its own reference frame. We are and we have been talking about observing an infalling object from the reference frame of flat space. So yes, for an object falling into a BH from its own reference frame it happens quickly in finite time. But for us in flatspace we see the object being affected by dilated time and contracted space. These effects become infinite at the EH. It would take an eternity to cross the EH and as Woody Allen once said, "Eternity is a long time, especially towards the end."

Are you suggesting that the reference frame of the infalling object is somehow the 'real' reference frame and what we in flat space see is only an illusion caused by gravitational distortion?

We have been talking about something happening quickly in one frame of reference and not happening at all in another. Perhaps it doesn't really happen in either frame of reference. In both frames of reference the BH evaporates before the object can cross except that in the infalling object's frame the BH evoporates very rapidly because of the extreme dilation of time. This is the problem with calculating the time to fall to the EH in the object's own reference frame. It ignores the fact that the BH is slowly evaporating in nondilated time and very much more rapidly in the dilated time of the object.

 

[Naty1]

What bothers me is that the presumption that something can fall to (much less, through) an event horizon in finite time depends upon there being a `true reality' that contradicts the observable reality.

If you accept in general that observers in different reference frames don't always agree on observed time,distance,mass,etc, elsewhere seems like this one should also also be accepted.

 

[skeptic2]

Naty1

I raised the question of how an object is able to cross the event horizon because I don't understand how it happens and what seems logical to me is at odds with the widely accepted interpretation of black hole geometries. Certainly I accept that observers in different reference frames don't always agree on observed time,distance,mass,etc. but I also believe that events in one reference frame can be mathematically transformed into any other reference frame to explain what those observers see. To suggest that simply because observers in different reference frames disagree about time, distance or mass, is sufficient reason to accept an ad hoc instance of differing observations without providing some sort of transformation between the reference frames is less than scientific.

I also raised the question hoping that someone here could point out the errors in my logic. The references I've seen, like xantox's posts, don't address the issue of the evaporation of the black hole during the extremely dilated time an object experiences as it falls towards the event horizon. Briefly put, is time infinitely dilated at the event horizon and if so, how does an object cross that event horizon in finite time? If it crosses in finite time in one frame of reference but not in another, what is the transformation between those reference frames that permits that?

 

[Chalnoth]

Briefly put, is time infinitely dilated at the event horizon and if so, how does an object cross that event horizon in finite time? If it crosses in finite time in one frame of reference but not in another, what is the transformation between those reference frames that permits that?

It doesn't. Basically in the case of a static black hole without any Hawking radiation, this means that it is impossible to do a transformation between the observer that has already passed the event horizon and an observer outside the event horizon. This is, in fact, what is meant by an event horizon in the first place: observers on different sides of an event horizon are causally disconnected, and it is therefore no longer possible to translate between their reference frames.

However, I'm beginning to think that with Hawking radiation, an infalling observer won't actually ever observe the interior of the black hole, but will instead just see the black hole evaporate to nothing, until the observer itself exits the black hole as Hawking radiation.

 

[skeptic2]

Basically in the case of a static black hole without any Hawking radiation, this means that it is impossible to do a transformation between the observer that has already passed the event horizon and an observer outside the event horizon.

Of course not. I guess I should have been clearer. I meant a transformation between an observer very close to the EH where time is extremely dilated and flat space.

However, I'm beginning to think that with Hawking radiation, an infalling observer won't actually ever observe the interior of the black hole, but will instead just see the black hole evaporate to nothing, until the observer itself exits the black hole as Hawking radiation.

Exactly, and this is the crux of my problem. If nothing can pass through the horizon, how can there be large black holes? (Small black holes may still be possible.) If the weight of all the matter can be supported by the EH, why is there required to be a singularity at the center?

 

[xantox]

Briefly put, is time infinitely dilated at the event horizon and if so, how does an object cross that event horizon in finite time? If it crosses in finite time in one frame of reference but not in another, what is the transformation between those reference frames that permits that?

The infalling body falls only in one place, not two. His path in that one place is finite, as it may be calculated by integrating the proper time on his worldline. So, the infalling body reaches the horizon, and enters the black hole, and fast. However, when this same path is measured by a distant observer, and since he is not in the same curved spacetime, he will use as a result coordinates having a radically different meaning. So he may well obtain an infinite result since his clock is not measuring the same thing that the infalling body calls "time".

Maybe this metaphor can help get a very rough and partial image: if you can only measure your shadow, then when the sun approaches noon exactly above you, the shadow will approach zero length, while at sunset it will become longer and longer, until going ideally to infinity. This does not mean your height is infinite, but just that it has been projected in some way onto different coordinates.

 

[Chalnoth]

Exactly, and this is the crux of my problem. If nothing can pass through the horizon, how can there be large black holes? (Small black holes may still be possible.) If the weight of all the matter can be supported by the EH, why is there required to be a singularity at the center?

Well, ultimately, I think that describing what a black hole actually is would require an understanding of quantum gravity. And we know that there isn't going to be a real singularity at the center: that's a feature of General Relativity, and a signature that General Relativity is wrong.

 

[Chronos]

GR may not be complete, but, nobody has proven it wrong. That hits my hot button. Show me the math before prophetizing.

 

[Chalnoth]

GR may not be complete, but, nobody has proven it wrong. That hits my hot button. Show me the math before prophetizing.

Let me be clear with what I mean. We know that GR must break down at some point, because it provides nonsensical predictions (singularities). But we don't yet know where it breaks down, because so far all experiments and observations are exactly in line with the theory's predictions.

 

[atyy]

Let me be clear with what I mean. We know that GR must break down at some point, because it provides nonsensical predictions (singularities). But we don't yet know where it breaks down, because so far all experiments and observations are exactly in line with the theory's predictions.

What exactly is so bad about a singularity, since it doesn't seem to be preventing predictions?

 

[Chalnoth]

What exactly is so bad about a singularity, since it doesn't seem to be preventing predictions?

Beyond the difficulties of having an actual infinity in the theory, General Relativity has a fundamental energy scale (the Planck scale). Singularities necessarily go far beyond that energy scale. And the reason why it's a problem for General Relativity is because GR predicts such singularities under very general conditions.

To put it another way, even with this fundamental energy scale sitting in the theory, if there was no reason to believe that the energy density could ever get high enough to contest this energy scale we might well think that the theory could potentially be absolutely correct.

And, of course, there are the incompatibilities with quantum mechanics to consider.

 

[Phrak]

I don't know what you all are talking about, but the horizon can't be crossed in finite time in the refererence frame of the exterior universe. So is it crossed, or is the question a nonsequiter?

Are we talking physics here, or UFOs?

 

[Chalnoth]

I don't know what you all are talking about, but the horizon can't be crossed in finite time in the refererence frame of the exterior universe. So is it crossed, or is the question a nonsequiter?

In basic General Relativity with no Hawking radiation, the answer is a definitive yes, because the proper reference frame to consider is not the reference frame of an external observer, but rather the observer falling into the black hole.

I think that the definitive answer to how this works in reality may potentially require an understanding of quantum gravity, which we don't yet have.

 

[xantox]

In basic General Relativity with no Hawking radiation, the answer is a definitive yes, because the proper reference frame to consider is not the reference frame of an external observer, but rather the observer falling into the black hole.

The answer is a definitive yes -also- with Hawking radiation for any macroscopic black hole. For a free falling body starting at rest in the distant flat spacetime, the time it takes on the body's own stopwatch to go from say 3x(horizon radius) to the horizon is around 100 microseconds – rather small compared to the evaporation timescale into a surrounding vacuum of such black hole of around 10^68 years, so that such effect may be ignored.

 

[stevebd1]

These lecture notes by Kim Griest might be of interest-
http://physics.ucsd.edu/students/courses/winter2007/physics161/Lectures/p161.8feb07.pdf

In all cases, r_s=2Gm/c^2'..Suppose we take the case where someone starts from rest at ∞ and falls into the hole.. as viewed from far away the person’s time slows down and then stops as it enters the Schwarzschild radius. The calculation is done starting from radial and time equations: dr/d\tau=\pm\sqrt{2Gm/r}=\pm\sqrt{r_s/r}, and dt/d\tau=\left(1-r_s/r\right)^{-1}, where we used conserved energy E=m which is valid starting at rest from infinity. Dividing these equation we find the relation between r and t, that is the speed as seen from from away:

v_{far\ away}=\frac{dr}{dt}=-\left(1-\frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}

We see again that as r→r_s, v→0. The far away observer never sees the person fall in..'The following is based on dr_{shell}=dr\left(1-r_s/r\right)^{-1/2} and dt_{shell}=dt\left(1-r_s/r\right)^{1/2}'..We can also find the speed measured in the shell frame:

v_{shell}=\frac{dr_{shell}}{dt_{shell}}=\left(1-\frac{r_s}{r}\right)^{-1}\frac{dr}{dt}

For the person falling in from far away, we put in the result for dr/dt above to find:

v_{shell}=\frac{dr_{shell}}{dt_{shell}}=-\sqrt{\frac{r_s}{r}}

This gives the result that to a shell observer, sitting at r=r_s, the falling objects goes by at v_shell=1, the speed of light! Isn’t it strange that the same object doing the same thing can be moving at 0 speed or c from different vantage points..'

Steve

 

[Chalnoth]

The answer is a definitive yes -also- with Hawking radiation for any macroscopic black hole. For a free falling body starting at rest in the distant flat spacetime, the time it takes on the body's own stopwatch to go from say 3x(horizon radius) to the horizon is around 100 microseconds – rather small compared to the evaporation timescale into a surrounding vacuum of such black hole of around 10^68 years, so that such effect may be ignored.

Remember that the important metric for determining whether or not the observer sees itself passing the event horizon is the lifetime of the black hole in the infalling observer's frame, as opposed to an outside observer's frame. Because any reference frame is a valid reference frame for computing the results, provided that you're not attempting to talk about behavior beyond an event horizon from your reference frame, and because from the outside observer's point of view an object never falls beyond a black hole's event horizon, it's beginning to look to me that the black hole will always evaporate before the observer passes the event horizon, when considered by an outside observer.

Now, I'm not certain on this. The idea just occurred to me in reading this thread, and I haven't heard any GR experts' take on it. But it seems to make sense to me. Perhaps I'll send an e-mail to my old GR professor...

 

[xantox]

because from the outside observer's point of view an object never falls beyond a black hole's event horizon, it's beginning to look to me that the black hole will always evaporate before the observer passes the event horizon, when considered by an outside observer

I fail to understand how you can consider that while without Hawking radiation the observer of the previous example can cross the final gap to the horizon in 100 microseconds, with Hawking radiation he would need to wait 10^68 more years. Or maybe you think the black hole will evaporate for him in 100 microseconds?

 

[Chalnoth]

I fail to understand how you can consider that while without Hawking radiation the observer of the previous example can cross the final gap to the horizon in 100 microseconds, with Hawking radiation he would need to wait 10^68 more years. Or maybe you think the black hole will evaporate for him in 100 microseconds?

That's precisely it. Remember that those 100 microseconds to the infalling observer are beyond positive infinity for the outside observer. So yes, I'm suggesting that perhaps the time dilation really is that extreme.