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Author Topic: What would you see of someone falling through an event horizon?  (Read 6812 times)

Offline Bill S

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I know this is a subject that has received a lot of attention, but I am still trying to tie up a few ends in my thoughts.

The following is an extract from the notes I wrote when I was trying to sort out my ideas about falling through an event horizon.

“To an outside observer any object approaching the Schwarzschild radius appears to take an infinite time to penetrate the event horizon.  If that object were an astronaut armed with a clock, her observation would be that she passed through the event horizon in a finite period of time, nor would she be aware of any abnormalities of time as she did so.  The reason for this is that her clock would deviate from that of an observer outside, and at a constant distance from, the event horizon, so grossly that the same event that, observed from the outside, took an eternity, would occur within a finite time in the frame of reference of the free-falling astronaut.  Obviously, the two observers are seeing the same event – a unique and immutable spacetime event – yet their interpretations of the time taken for that event to play out could hardly be more different.  For the outside observer the other figure is “frozen” for ever at the point of passing through the event horizon; while the falling observer records her uninterrupted passage through the event horizon, without a suspicion of a brief delay, much less an infinite hold-up.
 
A spacetime event must be unique and immutable; so, how can something that is unique and immutable appear to be so different in different frames of reference? If this interpretation of the effects of spacetime distortion is correct, it must mean that any body, any fragment of cosmic debris, that has ever (in its own frame of reference) fallen through an event horizon into a black hole, must, in the frame of reference of any outside observer who stops to look, be still stuck at the event horizon.  The entire accretion history of a black hole since the formation of its event horizon should be visible to any observer whose technology allows him or her to manoeuvre into the right position.  Strange as this might seem, it is even stranger to realise that, outside the frame of reference of the observer, all this material is not there, because it has long since plunged down the ever steepening  gravity well into the depths of the black hole. 

There is an alternative way of looking at the problem of the seemingly everlasting accretion sphere of the black hole.  Perhaps what popular science books tend to present as a problem is, in reality, nothing more than a recurrence of Zeno’s paradox.  If we consider the situation from the point of view of the outside observer as an example of asymptotic decay, in which the infalling object is not simply stuck for ever in the same state, but is gradually vanishing, with its progress being characterised by an asymptotic curve, then, in theory, it would never actually vanish, but in reality, like Zeno’s arrow, it would come to a conclusion.  In other words, it would vanish.  This seems to be the simplest explanation, and the simplest may well be the best.”         

On reflection, I find myself wondering about a couple of things.

1. I have recently read that the speed of the infalling astronaut, relative to the outside observer, would approach “c” as she neared the event horizon.  Would this necessarily be the case?

2.  As the infalling astronaut neared the event horizon, would the light from her, as seen by the outside observer, be red shifted out of the visible spectrum, thus causing the disappearance of the astronaut? 

I would really appreciate comments on my notes, as well as, hopefully, answers to the questions.


 

Offline JohnDuffield

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“To an outside observer any object approaching the Schwarzschild radius appears to take an infinite time to penetrate the event horizon.
IMHO this is right.

If that object were an astronaut armed with a clock, her observation would be that she passed through the event horizon in a finite period of time, nor would she be aware of any abnormalities of time as she did so.
IMHO this is wrong. Even if she started falling a billion years ago, the infalling observer hasn't seen herself pass through the event horizon yet, and she never ever will. So it isn't the case that she doesn't see anything unusual, it's more like she doesn't see anything.

The reason for this is that her clock would deviate from that of an observer outside, and at a constant distance from, the event horizon, so grossly that the same event that, observed from the outside, took an eternity, would occur within a finite time in the frame of reference of the free-falling astronaut.
That's what they say. But that finite time take infinite time to happen, so it hasn't happened yet. 

Obviously, the two observers are seeing the same event – a unique and immutable spacetime event – yet their interpretations of the time taken for that event to play out could hardly be more different.
Yep. And they can't both be right. Something's got to give.

For the outside observer the other figure is “frozen” for ever at the point of passing through the event horizon; while the falling observer records her uninterrupted passage through the event horizon, without a suspicion of a brief delay, much less an infinite hold-up.
I think the clincher for this is to flip it around and imagine you're at the event horizon with a laser pointing straight up, and then you ask this: why doesn't the light get out?   
 
A spacetime event must be unique and immutable; so, how can something that is unique and immutable appear to be so different in different frames of reference?
Because the notion that the infalling observer sees everything happening as normal is a fairy tale.

If this interpretation of the effects of spacetime distortion is correct, it must mean that any body, any fragment of cosmic debris, that has ever (in its own frame of reference) fallen through an event horizon into a black hole, must, in the frame of reference of any outside observer who stops to look, be still stuck at the event horizon.
Yep. But the black hole grows like a hailstone. You're a water molecule. You can't pass through the surface, But you can alight upon the surface, and then you can be buried by other water molecules. So the surface passes through you.

The entire accretion history of a black hole since the formation of its event horizon should be visible to any observer whose technology allows him or her to manoeuvre into the right position.  Strange as this might seem, it is even stranger to realise that, outside the frame of reference of the observer, all this material is not there, because it has long since plunged down the ever steepening  gravity well into the depths of the black hole.
No it hasn't. The force of gravity at some location depends on the local gradient in the speed of light. At the event horizon the "coordinate" speed of light is zero. It can't get any lower than that. There is no more gravity.   

There is an alternative way of looking at the problem of the seemingly everlasting accretion sphere of the black hole. Perhaps what popular science books tend to present as a problem is, in reality, nothing more than a recurrence of Zeno’s paradox.  If we consider the situation from the point of view of the outside observer as an example of asymptotic decay, in which the infalling object is not simply stuck for ever in the same state, but is gradually vanishing, with its progress being characterised by an asymptotic curve, then, in theory, it would never actually vanish, but in reality, like Zeno’s arrow, it would come to a conclusion.  In other words, it would vanish.  This seems to be the simplest explanation, and the simplest may well be the best.
I think there's another simplest explanation, which is that the original frozen-star black hole interpretation is correct. See The Formation and Growth of Black Holes by Kevin brown for a mention of it. He doesn't favour it, but I think it's just got to be right.         


On reflection, I find myself wondering about a couple of things.

1. I have recently read that the speed of the infalling astronaut, relative to the outside observer, would approach “c” as she neared the event horizon. Would this necessarily be the case?
That's what they say. But gravity is only there because the coordinate speed of light is reducing with altitude, and the infalling astronaut is falling faster and faster. For myself I think Freidwardt Winterberg's GRB firewall has to be correct. Matter is annihilated on the way in. 

2.  As the infalling astronaut neared the event horizon, would the light from her, as seen by the outside observer, be red shifted out of the visible spectrum, thus causing the disappearance of the astronaut?
Leaving firewalls out of it, again that's what they say. And yes, the infaller would cease to be visible. But note that light is not redshifted when it climbs out of a gravitational field. That's a myth. You can work this out for yourself by imagining that you send a 511keV photon down into a black hole. The mass increase is 511keV/c². The photon wasn't blueshifted at all. In similar vein if that photon was coming up instead of going down, it isn't redshifted at all. Conservation of energy applies.
 

Offline Bill S

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Thanks John.  Your comments are interestingly at variance with the "authorized version".  Hopefully they will spark a lively response. 
 

Offline JohnDuffield

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The thing is Bill, is that "my" version is the original version. It's the frozen-star black hole, and it dates back to Oppenheimer's time. The people who advocate the "authorized version" don't want you to know that it's just one interpretation. Again, take a look at the Kevin Brown article. See this bit:

"Historically the two most common conceptual models for general relativity have been the "geometric interpretation" (as originially conceived by Einstein) and the "field interpretation" (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural "limit" of this process is that the clock asymptotically approaches "full stop" (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's "frozen" due to the strength of the gravitational field. Within this conceptual framework there's nothing more to be said about the clock's existence. In contrast, according to the geometric interpretation, all clocks run at the same rate, measuring out real distances along worldlines in curved spacetime. This leads us to think that, rather than slowing down as it approaches the event horizon, the clock is following a shorter and shorter path to the future time coordinates. In fact, the path gets shorter at such a rate that it actually reaches the future infinity of Schwarzschild coordinate time in finite proper time."

See that future infinity in finite proper time? That finite proper time takes forever. It hasn't happened yet. And it never ever will.
 

Offline yor_on

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You're correct Bill. In the eyes of the far observer it would take 'forever'. Don't get stuck on that one though, locally defined your clock will tick, and it will take you a finite time. Just exchange 'time' for 'c'.
 

Offline jeffreyH

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This is a very interesting region of 'spacetime' and is constantly argued over. I am developing my own theories on this which are currently proving a challenge so I have little to say at the moment. It is not necessarily the way the mainstream opinion views it. My view may shift on that one. I am jumping to no easy conclusions.
 

Offline Bill S

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Having established that the perception of the rate of descent is different in different Fs of R, could we establish the rate of descent in the F of R of the descending astronaut, at the EH?
 

Offline PmbPhy

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Quote from: Bill S
“To an outside observer any object approaching the Schwarzschild radius appears to take an infinite time to penetrate the event horizon.  If that object were an astronaut armed with a clock, her observation would be that she passed through the event horizon in a finite period of time, nor would she be aware of any abnormalities of time as she did so.
For the most part that's correct. There are finite tidal forces acting on an object which is falling into a black hole so, depending on the mass of the black hole, the object would experience those tidal forces.

Quote from: Bill S
The reason for this is that her clock would deviate from that of an observer outside, and at a constant distance from, the event horizon, so grossly that the same event that, observed from the outside, took an eternity, would occur within a finite time in the frame of reference of the free-falling astronaut.
That is correct.

Quote from: Bill S
Obviously, the two observers are seeing the same event – a unique and immutable spacetime event – yet their interpretations of the time taken for that event to play out could hardly be more different.  For the outside observer the other figure is “frozen” for ever at the point of passing through the event horizon; while the falling observer records her uninterrupted passage through the event horizon, without a suspicion of a brief delay, much less an infinite hold-up.
Correct. But you need to keep in mind that outside observers will eventually loose contact, in every sense of the term, with the in-falling astronaut.

Quote from: Bill S

A spacetime event must be unique and immutable; so, how can something that is unique and immutable appear to be so different in different frames of reference?
The event has a specific set of spacetime coordinates, that's true. However its relationship between other events depends on the observer. That's what relativity is all about.

Quote from: Bill S
If this interpretation of the effects of spacetime distortion is correct, it must mean that any body, any fragment of cosmic debris, that has ever (in its own frame of reference) fallen through an event horizon into a black hole, must, in the frame of reference of any outside observer who stops to look, be still stuck at the event horizon.
As observed from outside observers, yes. That's correct.

Quote from: Bill S
In other words, it would vanish.  This seems to be the simplest explanation, and the simplest may well be the best.”         
If you mean that outside observers would eventually loose all contact with it then yes, that's correct.

On reflection, I find myself wondering about a couple of things.

Quote from: Bill S
1. I have recently read that the speed of the infalling astronaut, relative to the outside observer, would approach “c” as she neared the event horizon.  Would this necessarily be the case?
No. Not as observed from outside observers. From the astronauts point of view she'd see objects held at rest near the horizon pass by here near the speed of light.

Quote from: Bill S
2.  As the infalling astronaut neared the event horizon, would the light from her, as seen by the outside observer, be red shifted out of the visible spectrum, thus causing the disappearance of the astronaut? 
Yes.

Quote from: Bill S
I would really appreciate comments on my notes, as well as, hopefully, answers to the questions.
You're welcome as always my friend. :)
 

Offline Ilinca Sergiu

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What would you see of someone falling through an event horizon?
 
I think you can see the same as when the total atomic disintegration. A more accurate image can be obtained by placing an astronaut inside a hydrogen bomb and detonate it.
 

Offline JohnDuffield

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An H-bomb is only about 1% efficient. Check out Friedwardt Winterberg's firewall. Here's the abstract:

http://www.znaturforsch.com/aa/v56a/56a0889.pdf

"In the dynamic interpretation of relatively by Lorentz and Poincaré, Lorentz invariance results from real physical contractions of measuring rods and slower going clocks in absolute motion against an ether. As it was shown by Thirring, this different interpretation of special relativity can be extended to general relativity, replacing the non-Euclidean with a Euclidean geometry, but where rods are contracted and clocks slowed down. In this dynamic interpretation of the special, (and by implication of the general) theory of relativity, there is a balance of forces which might be destroyed near the Planck energy, reached in approaching the event horizon. In gravitational collapse, the event horizon appears first at the center of the collapsing body, thereafter moving radially outward. If the balance of forces holding together elementary particles is destroyed near the event horizon, all matter would be converted into zero rest mass particles which could explain the large energy release of gamma ray bursters".

What he's saying is that there's a 100% conversion of matter to energy.
 

Offline jeffreyH

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The term "near the event horizon" is crucial. The statement was not "at the event horizon". You need to ask why it was phrased this way. How near is near?
 

Offline Ilinca Sergiu

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-Absolutely agree less than-1%. H-bomb is the most optimistic picture. However, only she, today is suitable for demonstration the approximate processes for supporters traveling through black holes.
-Near the event horizon? Then let the astronaut near-bombs. Or look what makes a black hole with the stars who approached her.
 

Offline evan_au

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I tried doing a simplistic numerical simulation of what you would see of an astronaut dropping into a black hole from the distance of Earth's orbit - ignoring the immense energies needed to bring an astronaut to rest this close to a black hole!

I tried it with a black hole of 10 solar masses (the low-end of stellar black holes), and one similar in size to the one in the center of our galaxy (about 3 million solar masses). To make it simpler, I assumed that it is non-rotating and has no electric charge -and there is no other matter falling in.

Unfortunately, even with adaptive time-steps, the simulation blows up when you skip past the event horizon. And the astronaut is traveling so fast that even small time-steps are far too crude.

I think that one of the interesting details is how long it takes the light from the astronaut to disappear from visibility.
For a stellar-mass black hole, I estimate that the astronaut takes about 21 days to reach the event horizon (as seen by a distant observer), and is traveling about 16,000 km/sec when he reaches the event horizon, which has a Schwarzchild radius of about 25km. This speed is not too close to the speed of light, so you can almost ignore the  Lorentz factor. The astronaut spends so little time in a zone with significant gravitational time dilation that he would appear to disappear almost instantly.

For a galactic-mas black hole, the astronaut takes something like an hour to reach the event horizon, and is traveling at almost the speed of light; the Doppler shift alone would make it impossible to see the astronaut. Even at his starting point there is significant gravitational time dilation, which means you need a solution based on the General theory of relativity (doing this with any precision is beyond my abilities!).

If the astronaut were illuminated by 1kW of violet LEDs (around 400nm), that would give the best chance of seeing him for the longest time before the light is red-shifted out of the visible spectrum (>750nm).

Is there anyone out there who has a better handle on the maths, and can solve the equations for what you would see of an astronaut dropping into a black hole?
« Last Edit: 24/11/2014 15:50:45 by evan_au »
 

Offline JohnDuffield

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IMHO there's big issues with this, evan. When you dig down, you find that the given explanation just doesn't add up. Have a look at the Baez website where Don Koks says this:

"Einstein talked about the speed of light changing in his new theory.  In the English translation of his 1920 book "Relativity: the special and general theory" he wrote: "according to the general theory of relativity, the law of the constancy of the velocity [Einstein clearly means speed here, since velocity (a vector) is not in keeping with the rest of his sentence] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity.  A curvature of rays of light can only take place when the velocity [speed]  of propagation of light varies with position."  This difference in speeds is precisely that referred to above by ceiling and floor observers."

The coordinate speed of light at the ceiling is greater than the coordinate speed of light at the floor. When you drop a brick, the speed of the brick when it hits the floor relates to the difference in the two coordinate speeds of light. When you drop a brick into a black hole from an "infinite" distance, it is said to be travelling at the speed of light when it crosses the event horizon. That sounds reasonable, in that the coordinate speed of light where you are is 299,792,458 m/s, and the coordinate speed of light at the event horizon is zero. But there's a problem. How can the remote observer measure the speed of the falling brick to be greater than his measurement of the coordinate speed of light at that location? Something has got to give.
« Last Edit: 24/11/2014 18:33:05 by JohnDuffield »
 

Offline jeffreyH

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IMHO there's big issues with this, evan. When you dig down, you find that the given explanation just doesn't add up. Have a look at the Baez website where Don Koks says this:

"Einstein talked about the speed of light changing in his new theory.  In the English translation of his 1920 book "Relativity: the special and general theory" he wrote: "according to the general theory of relativity, the law of the constancy of the velocity [Einstein clearly means speed here, since velocity (a vector) is not in keeping with the rest of his sentence] of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [...] cannot claim any unlimited validity.  A curvature of rays of light can only take place when the velocity [speed]  of propagation of light varies with position."  This difference in speeds is precisely that referred to above by ceiling and floor observers."

The coordinate speed of light at the ceiling is greater than the coordinate speed of light at the floor. When you drop a brick, the speed of the brick when it hits the floor relates to the difference in the two coordinate speeds of light. When you drop a brick into a black hole from an "infinite" distance, it is said to be travelling at the speed of light when it crosses the event horizon. That sounds reasonable, in that the coordinate speed of light where you are is 299,792,458 m/s, and the coordinate speed of light at the event horizon is zero. But there's a problem. How can the remote observer measure the speed of the falling brick to be greater than his measurement of the coordinate speed of light at that location? Something has got to give.

The black hole doesn't trap light at the event horizon. That happens at some distance outside the horizon. OK so everybody will disagree. Do I care? No. I've worked it out.
 

Offline jeffreyH

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I tried doing a simplistic numerical simulation of what you would see of an astronaut dropping into a black hole from the distance of Earth's orbit - ignoring the immense energies needed to bring an astronaut to rest this close to a black hole!

I tried it with a black hole of 10 solar masses (the low-end of stellar black holes), and one similar in size to the one in the center of our galaxy (about 3 million solar masses). To make it simpler, I assumed that it is non-rotating and has no electric charge -and there is no other matter falling in.

Unfortunately, even with adaptive time-steps, the simulation blows up when you skip past the event horizon. And the astronaut is traveling so fast that even small time-steps are far too crude.

I think that one of the interesting details is how long it takes the light from the astronaut to disappear from visibility.
For a stellar-mass black hole, I estimate that the astronaut takes about 21 days to reach the event horizon (as seen by a distant observer), and is traveling about 16,000 km/sec when he reaches the event horizon, which has a Schwarzchild radius of about 25km. This speed is not too close to the speed of light, so you can almost ignore the  Lorentz factor. The astronaut spends so little time in a zone with significant gravitational time dilation that he would appear to disappear almost instantly.

For a galactic-mas black hole, the astronaut takes something like an hour to reach the event horizon, and is traveling at almost the speed of light; the Doppler shift alone would make it impossible to see the astronaut. Even at his starting point there is significant gravitational time dilation, which means you need a solution based on the General theory of relativity (doing this with any precision is beyond my abilities!).

If the astronaut were illuminated by 1kW of violet LEDs (around 400nm), that would give the best chance of seeing him for the longest time before the light is red-shifted out of the visible spectrum (>750nm).

Is there anyone out there who has a better handle on the maths, and can solve the equations for what you would see of an astronaut dropping into a black hole?

I skimmed over this post originally. Although I doubt I would be able to shed any light on the problem, I just don't know enough to do that, I would be interested to see what you did to create the model.
 

Offline Traid1942

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Your post is very informative for me. I have read the whole postings and comments and want to say special thanks.
 

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