Is the gravitational time dilatation a real effect?

[Kamil Szot]

Some people explain that gravitational time dilatation is just photons redshifting due to work they are doing against gravitational field while escaping from it, but I heard that the delay of the clock on the probe after passing near Jupiter was observed.

 

[D H]

Much closer to home: Both special and general relativistic time dilation are observed in the clocks on the GPS satellites. GPS would not work (or would not work very well) if both of these effects were not taken into account.

Much, much closer to home: http://www.wired.com/science/discoveries/news/2007/12/time_hackers?currentPage=all

 

[Dale]

Gravitational time dilation can even be observed in a tall building with sensitive equipment. Google the Pound-Rebka experiment.

 

[D H]

The Pound-Rebka experiment tested for gravitational redshift rather than time dilation. While the effects are certainly related, Kamil asked about time dilation in the OP.

To Kamil: Gravitational redshift (or blueshift) and gravitational time dilation are different effects/different ways of explaining the same phenomenon. You do not get one without the other.

 

[Dale]

As you say, they are the same phenomenon.

 

[Kamil Szot]

Ok. So I understand that this is a real effect. It does affect actual clocks, probably regardless of the principle on which they are operating. I doesn't have to involve redshifting photons.

Second question:

Difference of paces of time between two points is higher if difference of gravitational potential between them is higher, right?

 

[Dale]

Difference of paces of time between two points is higher if difference of gravitational potential between them is higher, right?

Yes.

 

[Kamil Szot]

Ok, thank you for confirming this. I read about this in books that I considered trustworthy.

Now few questions about some assumptions I'm much more unsure about.

At some distance from the center of mass of a sufficiently massive body there is a place with such a property that if you wanted to escape from this place (to some other spot, located, let's say 10 mld ly away) you would need to have nearly infinite energy, right?

Can I call this place as having gravitational potential equal nearly minus infinity?

Does it have nearly infinitely slower time pace than we do?

By saying that 'something is nearly equal infinity somewhere' I mean that it's very high, and if it's not high enough for purposes of my thinking I can move just a small bit to make it sufficiently higher.

I'll cut the chase to the problem that is truly bothering me.

If the answer to three questions above is 'yes' then how can anything in less than 14 mld years pass this zone of very low potential that has nearly infinitely slower time pace then we have?

If 14 mld years (by our count) from the moment of getting into slow zone something decided to use it's energy in effort to escape (this moment would be only fraction of a second after, from its point of view), we could see it out in some time in the future.

How can we know that from where we stand anything ever passed point of no return?

Can you explain to me what point of my reasoning is wrong?

 

[Dale]

(to some other spot, located, let's say 10 mld ly away)

What does "mld ly" mean? I assume ly is light-year, but I don't know what mld is.

you would need to have nearly infinite energy, right?

Can I call this place as having gravitational potential equal nearly minus infinity?

Does it have nearly infinitely slower time pace than we do?

Yes to all 3.

If the answer to three questions above is 'yes' then how can anything in less than 14 mld years pass this zone of very low potential that has nearly infinitely slower time pace then we have?

What is the significance of the change from 10 to 14 mld years?

How can we know that from where we stand anything ever passed point of no return?

The point of no return is called the event horizon. If we are outside the event horizion then we cannot know about anything past the horizon.

 

[Kamil Szot]

What does "mld ly" mean? I assume ly is light-year, but I don't know what mld is.

Ouch. Sorry. I'm from long scale country (http://en.wikipedia.org/wiki/Long_and_short_scales ). It was supposed to be shorthand from 10⁹, a billion in short scale. From now on I will use short scale.

Yes to all 3.

Great!

What is the significance of the change from 10 to 14 mld years?

Sorry again. 10 billion light years was arbitrarily chosen to be point far enough not to feel any gravitational influence of massive body. I should probably write: "escape to infinity" but I felt that I had already too many near infinities and wanted to avoid confusion. Seems that I introduced one instead.

I chose 14 billion years as time to convey suggestion that no matter how long (from our point of view) something is moving towards point of no return, we can't be sure if it won't turn back and escape to show up again some day.

The point of no return is called the event horizon. If we are outside the event horizion then we cannot know about anything past the horizon.

I used term point of no return because it's unambiguous (I hope). Wikipedia entry about event horizon mentions absolute horizons, apparent horizons and other notions of horizons. Again, I wanted to avoid confusion.

I'm not saying anything about anything behind or at event horizon. My question is about stuff near event horizon but outside.

 

[George Jones]

How can we know that from where we stand anything ever passed point of no return?

We can't know with certainty, but similar situations exist even in the flat spacetime of general relativity. See

https://www.physicsforums.com/showthread.php?p=2559940#post2559940

https://www.physicsforums.com/showthread.php?p=2215073#post2215073.

 

[Kamil Szot]

How can we know that from where we stand anything ever passed point of no return?

We can't know with certainty, but similar situations exist even in the flat spacetime of general relativity.

Let me get back to my original question because it better phrases my problem:

Can anything in less than our 13 bln years pass this zone of very low gravitational potential that has nearly infinitely slower time pace then we have?

By passing I mean entering it (from outside) and leaving it (towards surface of no return). I'm not even asking how anything passed event horizon.

I'm asking how can anything (for example neutrino) pass the zone between 1 meter from event horizon and 10^-15 meters from event horizon in our measly 13.7 bln of years. Please substitute 15 with sufficiently high value. I didn't really do the math on this.

See

https://www.physicsforums.com/showthread.php?p=2559940#post2559940

https://www.physicsforums.com/showthread.php?p=2215073#post2215073.

Thank you for the links but I unfortunately did not find answer to my question there. Also I'm not interested in black hole formation at the time but rather with black hole growth and merging.

 

[Dale]

no matter how long (from our point of view) something is moving towards point of no return, we can't be sure if it won't turn back and escape to show up again some day.

Essentially correct.

 

[Kamil Szot]

So if nothing ever has even got close to event horizon (yet) then nothing ever passed event horizon (irregardless of whether it is possible or not) and no black hole has since it's creation gained even one pound no matter how many stars it obliterated. Also no two black holes ever merged. Right?

If you agree on this with me then perhaps I'm discussing that matter with wrong person. ;-)

 

[Dale]

All of what you say is correct in Schwarzschild coordinates or in Rindler coordinates in flat spacetime. However, in both cases the singularity at the event horizon is only a coordinate singularity which can be removed by a suitable coordinate transformation. In both cases an object free falling crosses the event horizon in a finite amount of proper time even though it takes an infinite amount of coordinate time.

 

[Kamil Szot]

Are calculations done with Schwarzschild coordinates wrong outside of event horizon?

Don't calculations done in "coordinates after suitable transformation" indicate that there is difference in time paces just outside event horizon and in remote areas exactly same as calculations done in Schwarzschild coordinates do indicate?

Gravitational time dilatation is not some elusive mathematical glitch. It's a part of the reality we are living in and its existence should not depend on chosen system of coordinates.

I'm not even touching event horizon. I'm not trying to deal with discontinuity that occurs at that exact surface.

I just can't see how anything could get close enough to this discontinuity in mere 13.7 bln years of our time.


Let me make up a story illustrating why I think that no object falling at black hole has passed event horizon yet.

Let's say that shortly after big bang someone wanted to have a bit fun and tossed at some black hole device that contained equal amounts of matter and antimatter and a trigger mechanism. Trigger mechanism was designed to mix matter and antimatter when whole device gets as close to event horizon as 10^-6 meters. 13 bln years of outside time have past and the free falling device (after few hours of proper time) finally triggered itself and part of energy created by annihilation began ascending back from gravity well. We are going to see in in another few bln years or so.


P.S.

I really should calculate how close to event horizon can free falling object get in 13.7 bln years of remote time. Making up numbers like 10^-6 probably isn't doing me any good.

 

[Dale]

Are calculations done with Schwarzschild coordinates wrong outside of event horizon?

No, they are not wrong but they are also not unique.

Let me make up a story illustrating why I think that no object falling at black hole has passed event horizon yet.

The key word here is "yet". It implies a choice of some simultaneity convention and therefore the answer is coordinate dependent.

 

[Kamil Szot]

No, they are not wrong but they are also not unique. The key word here is "yet". It implies a choice of some simultaneity convention and therefore the answer is coordinate dependent.

You can describe given thing in different coordinates and get different speeds, accelerations even precedence. But you cannot turn reversible event into irreversible event by changing coordinates that you use to make calculations about physical reality.

But we are drifting away from my question.

Why do you dismiss result of reasoning based on Schwarzshild coordinates if I do not go outside its domain of applicability?

Can you give me an example of some simultaneity convention by which this exact spot of spacetime at which I am at the moment is after the event of anything falling to place being 10^-15 meters outside of event horizon?

 

[Dale]

you cannot turn reversible event into irreversible event by changing coordinates

I have never said that you could. Per your scenario we are dealing strictly with events outside the horizon, so escape is possible (I assume that is what you mean by reversible).

Why do you dismiss result of reasoning based on Schwarzshild coordinates if I do not go outside its domain of applicability?

I don't dismiss the results. I just point out that they are coordinate dependent.

 

[Kamil Szot]

Per your scenario we are dealing strictly with events outside the horizon, so escape is possible (I assume that is what you mean by reversible).

Thank you. That was exactly what I meant.

I don't dismiss the results. I just point out that they are coordinate dependent.

I think I can't agree on that with you. I think if escape is possible in one set of coordinates then it also must be possible after transformation to any other sets of coordinates that are valid for given situation.

I'll ask directly. Do you think that anything of what fallen at black holes for the last 13 bln year already crossed any of the event horizons making escape impossible? And if not then when it will happen? By 'already' and 'when' I'm taking about simultaneity as seen from our frame of reference.

 

[Dale]

I think if escape is possible in one set of coordinates then it also must be possible after transformation to any other sets of coordinates that are valid for given situation.

Yes. An event which is outside the horizon in one coordinate system will be outside the horizon in all systems regardless of whether or not the horizon itself is a coordinate singularity in the system.

Do you think that anything of what fallen at black holes for the last 13 bln year already crossed any of the event horizons making escape impossible? And if not then when it will happen? By 'already' and 'when' I'm taking about simultaneity as seen from our frame of reference.

Could you be more explicit about "our frame of reference"? We are dealing with curved spacetime so the usual meaning from SR will not apply globally.

 

[Kamil Szot]

Could you be more explicit about "our frame of reference"? We are dealing with curved spacetime so the usual meaning from SR will not apply globally.

I meant frame of reference associated with us on earth, traveling fairly inertially, with fairly low speed though fairly flat space.

 

[Dale]

I meant frame of reference associated with us on earth, traveling fairly inertially, with fairly low speed though fairly flat space.

In such a frame there is no event horizon and little time dilation so the whole question is moot.

 

[Kamil Szot]

In such a frame there is no event horizon and little time dilation so the whole question is moot.

Perhaps I failed to convey what I'm interested in.

I want to know if from my point of view (in my frame of reference) using notion of simultaneity appropriate for my frame of reference any object that was heading towards any black hole passed the point of no return and so by the laws of physics I can be sure that I will never see any part of it again.

 

[Dale]

I want to know if from my point of view (in my frame of reference) using notion of simultaneity appropriate for my frame of reference any object that was heading towards any black hole passed the point of no return and so by the laws of physics I can be sure that I will never see any part of it again.

What I am trying to ask is what you mean by "notion of simultaneity appropriate for my frame of reference" in a curved spacetime?

If you mean "radar time" then the answer is "no".

 

[Kamil Szot]

What I am trying to ask is what you mean by "notion of simultaneity appropriate for my frame of reference" in a curved spacetime?

I'm thinking about simultaneity like in special relativity. Events lying on a the same line parallel to x-axis here http://en.wikipedia.org/wiki/File:Relativity_of_Simultaneity.svg are simultaneous.

I guess that general relativity has some extension of that concept to non-flat spacetimes and allows for calculating it for any frame of reference including the one associated with earth.

If you mean "radar time" then the answer is "no".

I'm not sure what's a "radar time", but I'm guessing that this is somehow associated with sending signal, bouncing it from object and receiving it and timing this operation. That's not what I meant.

To phase my question even simpler:

I toss object at region of space that is occupied by a black hole. How much time will I have to wait to be sure that no parts of it will come back out at some point in the future the future and why would anyone think that this is some finite amount of time?

 

[Dale]

I'm thinking about simultaneity like in special relativity. Events lying on a the same line parallel to x-axis here http://en.wikipedia.org/wiki/File:Relativity_of_Simultaneity.svg are simultaneous.

I guess that general relativity has some extension of that concept to non-flat spacetimes and allows for calculating it for any frame of reference including the one associated with earth.

The special relativity concept you are describing is coordinate dependent, even in flat spacetime.

I'm not sure what's a "radar time", but I'm guessing that this is somehow associated with sending signal, bouncing it from object and receiving it and timing this operation. That's not what I meant.

Then you will have to be completely explicit on your simultaneity convention, because I am at a loss.

I toss object at region of space that is occupied by a black hole. How much time will I have to wait to be sure that no parts of it will come back out at some point in the future the future and why would anyone think that this is some finite amount of time?

It is not a matter of phrasing your question, but defining your coordinates (or at least your synchronization convention). In Schwarzschild coordinates and radar coordinates the answer is given above. If you are not satisfied with those coordinates then define the coordinate system you wish to use (preferably as a transformation from Schwarzschild coordinates).

 

[Kamil Szot]

The special relativity concept you are describing is coordinate dependent, even in flat spacetime.

I am aware of that.

In Schwarzschild coordinates and radar coordinates the answer is given above. If you are not satisfied with those coordinates then define the coordinate system you wish to use (preferably as a transformation from Schwarzschild coordinates).

Did I understood right that answer to my question in Schwarzschild coordinates (also in radar coordinates) is "You can never be sure that something you tossed at black hole won't come back out, no matter how long you wait, because from your point of view, as well as from point of view of anyone located farther from event horizon then the object you tossed, fall of this object towards even horizon will take forever" ?

If that is the answer in one set of coordinates valid for given situation then this is sufficient for me because transformation to any other valid set of coordinates can't change physical reality of my situation.

 

[Ich]

If that is the answer in one set of coordinates valid for given situation then this is sufficient for me because transformation to any other valid set of coordinates can't change physical reality of my situation.

Yes, there's no "time of no return" classically and in principle. In reality, the dissappearing of such an object happes very quickly, and after a finite time you can be sure that no more signals will reach you. For example, if the object (or half of it) turned into a single photon headed outwards, and the wavelength of this photon would be redshifted to more than the size of the universe when it arrives, you know that the object is gone for good.
Interestingly, there (also classically and in principle) a time after which you can't reach the infalling object with a signal.

 

[Kamil Szot]

Yes, there's no "time of no return" classically and in principle.

Great! Does that also mean that no event horizon will ever increase its size?

In reality, the dissappearing of such an object happes very quickly, and after a finite time you can be sure that no more signals will reach you. For example, if the object (or half of it) turned into a single photon headed outwards, and the wavelength of this photon would be redshifted to more than the size of the universe when it arrives, you know that the object is gone for good.

I don't think that's the case. I don't think there is a cap on how high energy can a photon have so if you could emit a single photon with energy of whole star it might have pretty reasonable wavelength after redshifting. This of course might not be possible but if you shined light at the black hole it will be blue shifted as it goes towards event horizon as much as it will redshift when coming back. So if nothing bad happens along the way with the light that falls onto black hole it should shine back out at some point in the future at similar wavelength.

Another question:

Does infalling object have a way to know how much kinetic energy it gained? Can it observe change in its inertial mass, for example when trying to move sideways with a thruster?

Let's say we have half matter, half antimatter object and we let it fall into a black hole. As it goes down it gains speed. When it is close to event horizon it annihilates. All energy is turned into photons. As they go out they are redshifted losing energy.

I think that annihilation of faster moving, free falling particles should result in higher energy photons otherwise there would be missing energy in that scenario.

If what I'm guessing is true, could you determine your speed in freefall while being completely blind to outside world by annihilating particle with its antiparticle and measuring wavelength of created photons?

If you could do such measurement shouldn't it give identical results if you had been accelerated to given speed not by gravity but by some other means?

Interestingly, there (also classically and in principle) a time after which you can't reach the infalling object with a signal.

Few days ago I wanted to make an example that you could go after falling object, grab it and pull it back at anytime, but after thinking a while I saw that this is impossible as there is cap on speed that forbids this from the point of view of infalling object. It's not obvious from the outside point of view, but you can imagine how gravitational shortening and time dilatation might forbid this in finite time.

 

[Ich]

Great! Does that also mean that no event horizon will ever increase its size?

Classically and in principle, yes. In reality, however, we expect Hawking radiation.

I don't think there is a cap on how high energy can a photon have so if you could emit a single photon with energy of whole star it might have pretty reasonable wavelength after redshifting.

The point is, redshift is exponatial with time. You tell me the mass, say, mass of the observable universe, and I tell you the (finite, short,) time in seconds it takes until the photon's wavelength would be redshifted to the size of the observable universe.

Does infalling object have a way to know how much kinetic energy it gained?

No.

I think that annihilation of faster moving, free falling particles should result in higher energy photons otherwise there would be missing energy in that scenario.

If what I'm guessing is true, could you determine your speed in freefall while being completely blind to outside world by annihilating particle with its antiparticle and measuring wavelength of created photons?

You're missing that kinetic energy or photon energy are frame-dependent. Your first sentence is true, but if you change to the comoving frame, kinetic energy is zero and the photon energy is unchanged.

It's not obvious from the outside point of view, but you can imagine how gravitational shortening and time dilatation might forbid this in finite time.

Not so complicated. The object simply hits the singularity before the photon can catch up with it. Look at the http://casa.colorado.edu/~ajsh/schwp.html#freefall",

 

[Kamil Szot]

Classically and in principle, yes. In reality, however, we expect Hawking radiation.

Good. I'm not sure about Hawking radiation though. If pair production was used to explain Hawking radiation it would still take forever (in our time) for the particle created outside event horizon (no matter how close) to reach it. Particles would need to be produced exactly at event horizon, and since event horizon has zero volume I'm not sure how this could work.

The point is, redshift is exponatial with time. You tell me the mass, say, mass of the observable universe, and I tell you the (finite, short,) time in seconds it takes until the photon's wavelength would be redshifted to the size of the observable universe.

Ok. Great. How about photons shining at black hole that would due to blueshift in finite, short time get more energy then that of entire observable universe and after sliding near event horizon might come out with same wavelength as the had? Is that possible?

No.
You're missing that kinetic energy or photon energy are frame-dependent. Your first sentence is true, but if you change to the comoving frame, kinetic energy is zero and the photon energy is unchanged.

Thank you. That suits my common sense better. Because why would one freefalling frame of reference be discernible from some any other freefalling frame that just moves slower.

I think I know now what happens to energy in the case I presented. Photons produced by annihilation have normal wavelength in freefalling frame of reference. But if you observe them in stationary frame of reference you see that photons emitted towards black hole are heavily blueshifted and photons going back are heavily redshifted. This way even after annihilation almost all of the energy is still heading towards black hole. Photons that shined back are heavily redshifted even before they started their climb out of gravity well. Even if you happened to produce photons traveling in a plane exactly orthogonal to your speed you still have only the energy of your rest mass that is nothing compared to energy required to climb out of gravity well and you end up with very high wavelength.

So even if you got rid of your mass by converting it to energy traveling at exact speed of light you still couldn't get significant part of you out.

Not so complicated. The object simply hits the singularity before the photon can catch up with it. Look at the http://casa.colorado.edu/~ajsh/schwp.html#freefall",

Thank you for this. I especially like the Penrose diagram that brings all infinities into view.

This is indeed obvious by looking at this graph http://casa.colorado.edu/~ajsh/stff.gif and comparing green and yellow lines. Although drafting such graph is not trivial if you have never seen it. Standard graph http://casa.colorado.edu/~ajsh/st0big_gif.html does not allow to observe this effect.



I really don't like stating that "The Schwarzschild spacetime geometry appears ill-behaved at the horizon, the Schwarzschild radius. However, the pathology is an artefact of the Schwarzschild coordinate system. Spacetime itself is well-behaved at the Schwarzschild radius, as can be ascertained by computing the components of the Riemann curvature tensor, all of whose components remain finite at the Schwarzschild radius."


If you have simple problem like this: "You see remote object traveling with constant speed v1 and you yourself have speed v2 then at what direction you should point your speed vector to intercept this object?" You can formulate quadratic equation to solve this problem. If you get delta < 0 than right course is stating that this problem has no solutions not stating that this pathology is just result of using real numbers, the problem itself is well behaved and when doing calculations using complex numbers everything is still ok no matter how far you are from object that you want to intercept. It's good math but it's not the reality.

 

[nismaratwork]

Good. I'm not sure about Hawking radiation though. If pair production was used to explain Hawking radiation it would still take forever (in our time) for the particle created outside event horizon (no matter how close) to reach it. Particles would need to be produced exactly at event horizon, and since event horizon has zero volume I'm not sure how this could work.




Ok. Great. How about photons shining at black hole that would due to blueshift in finite, short time get more energy then that of entire observable universe and after sliding near event horizon might come out with same wavelength as the had? Is that possible?



Thank you. That suits my common sense better. Because why would one freefalling frame of reference be discernible from some any other freefalling frame that just moves slower.

I think I know now what happens to energy in the case I presented. Photons produced by annihilation have normal wavelength in freefalling frame of reference. But if you observe them in stationary frame of reference you see that photons emitted towards black hole are heavily blueshifted and photons going back are heavily redshifted. This way even after annihilation almost all of the energy is still heading towards black hole. Photons that shined back are heavily redshifted even before they started their climb out of gravity well. Even if you happened to produce photons traveling in a plane exactly orthogonal to your speed you still have only the energy of your rest mass that is nothing compared to energy required to climb out of gravity well and you end up with very high wavelength.

So even if you got rid of your mass by converting it to energy traveling at exact speed of light you still couldn't get significant part of you out.


Thank you for this. I especially like the Penrose diagram that brings all infinities into view.

This is indeed obvious by looking at this graph http://casa.colorado.edu/~ajsh/stff.gif and comparing green and yellow lines. Although drafting such graph is not trivial if you have never seen it. Standard graph http://casa.colorado.edu/~ajsh/st0big_gif.html does not allow to observe this effect.



I really don't like stating that "The Schwarzschild spacetime geometry appears ill-behaved at the horizon, the Schwarzschild radius. However, the pathology is an artefact of the Schwarzschild coordinate system. Spacetime itself is well-behaved at the Schwarzschild radius, as can be ascertained by computing the components of the Riemann curvature tensor, all of whose components remain finite at the Schwarzschild radius."


If you have simple problem like this: "You see remote object traveling with constant speed v1 and you yourself have speed v2 then at what direction you should point your speed vector to intercept this object?" You can formulate quadratic equation to solve this problem. If you get delta < 0 than right course is stating that this problem has no solutions not stating that this pathology is just result of using real numbers, the problem itself is well behaved and when doing calculations using complex numbers everything is still ok no matter how far you are from object that you want to intercept. It's good math but it's not the reality.

You keep saying "our time", which is frame dependent again. The event horizon does not have zero volume, that would be the singularity. The event horizon probably fluctuates as the BH gains mass from infalling matter, or (in some huge amount of time as the universe cools) loses mass to HR. The whole point of HR is that the pair-production DOES occur "exactly at the event horizon", wherever and whenever that is. I don't see a problem with that, although any non-mathematical treatment of HR is necessarily somewhat misleading.

It seems to me, from reading this thread that you believe, or are asking how we can never toss a ball into a black hole and observe it passing the event horizon. If we remain in our earthbound coordinate system, then you are left with what Ich is saying, and you can be confident that you've received your final signal from the ball, but in a classical sense you never see that finally .00000000000001...% fall in. if you managed to glue the ball to your hand, and then you went on the one-way journey with the ball, there would be no problem; you would fall in with the ball in, as Ich said, a very short interval.

You just cannot ask this question while thinking of two different frames of reference, unless you accept that it is a matter of your coordinates which defines whether or not you get to see the infalling matter complete its journey. So, event horizons do grow, and eventually shrink at some future time. I think DaleSpam gave you the best answers that had technical content that you are going to get.

 

[Kamil Szot]

You keep saying "our time", which is frame dependent again.

I know. When I say "our time" I mean time in frame of reference associated with Earth. I didn't expect that this would be that confusing.

The event horizon does not have zero volume, that would be the singularity.

Singularity is a point and as such has no volume. In 3d space also lines and surfaces have no volume (unless they are fractal). Since event horizon is very precisely defined smooth surface so I guessed it has no volume. And that implied for me that probability of anything occurring exactly at event horizon is zero.

The event horizon probably fluctuates as the BH gains mass from infalling matter,

I think that ICH agreed with me on the first line of his post that this does not happen.

or (in some huge amount of time as the universe cools) loses mass to HR.

I fail to see how shrinking of event horizon could lead to absorption of one of the particles created.

The whole point of HR is that the pair-production DOES occur "exactly at the event horizon", wherever and whenever that is. I don't see a problem with that, although any non-mathematical treatment of HR is necessarily somewhat misleading.

I see it like this. Virtual pair production is a random event. Each point of space has some nonzero probability density of such event. You may calculate probability of such event occurring in some volume by calculating integral over this volume of that probability of density function.

Integral over zero volume equals zero so the event cannot happen exactly at event horizon.

The only way I can imagine particle getting inside is that by uncertainty principle. Probability of finding this exact particle behind event horizon is nonzero. But because on the other side of event horizon there is nothing (except singularity far away from event horizon), I don't see what observation could cause wave function to collapse to actual particle there.

It seems to me, from reading this thread that you believe, or are asking how we can never toss a ball into a black hole and observe it passing the event horizon. If we remain in our earthbound coordinate system, then you are left with what Ich is saying, and you can be confident that you've received your final signal from the ball, but in a classical sense you never see that finally .00000000000001...% fall in.

I'm not asking about what I can see. I'm interested in what actually happens in there in sense of the diagrams that Ich directed me to.

And what I got from this thread so far is confirmation of my own concerns:
If you are far from event horizon no matter how long you wait no falling object ever passes any event horizon.

Don't you agree with the above statement?

if you managed to glue the ball to your hand, and then you went on the one-way journey with the ball, there would be no problem; you would fall in with the ball in, as Ich said, a very short interval.

If I'm hovering millimeter from the horizon it will still take forever for the ball to fall in (as ball can have arbitrarily slower time pace than my own because of being closer), but if I'm co-moving with the ball (even not exactly glued to it but some distance after it) this 'forever' changes to short finite time. This means that when two objects fall into a black hole difference of time paces of the frame of references associated with them must be always finite and for two objects near event horizon quite small. Right?

You just cannot ask this question while thinking of two different frames of reference, unless you accept that it is a matter of your coordinates which defines whether or not you get to see the infalling matter complete its journey. So, event horizons do grow, and eventually shrink at some future time.

If you toss a star at a black hole and start a clock when see it disappear how long in your opinion will take the event horizon to grow? Less than 10¹⁰⁰⁰ years? Would you be able to detect this growth by seeing that black-hole now occludes larger portion of your view?

I think DaleSpam gave you the best answers that had technical content that you are going to get.

I liked answers from Ich better better because he understood what I meant.

 

[nismaratwork]

OK... I don't think I teach well enough to help you here, and if you feel that Ich is doing a better job, I won't interfere.