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.

 

[Ich]

If pair production was used to explain Hawking radiation

It isn't. It's something to do with positive and negative frequencies going to future null infinity, the latter not cancelling the former as they should in the presence of an event horizon. Or something like that, I didn't understand the derivation.
HR can't be a process happening exactly at the EH, because the local spacetime at the EH is not different from the spacetime somewhere else.

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?

Of course, if you have a reflector ten times thes mass of the observable universe somehow hovering there, within <<1 attometer from the horizon.

I think I know now what happens to energy in the case I presented.
[...]
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.

Yes.

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

it's exactly the same with Schwarzschild coordinates. These are supposed to be static coordinates. You'd like to place static observers (with r=const.) everywhere, and the solution blows up and then gets imaginary for r<=2M. So it tells you that there's no solution to the problem, just like in your example.
Also like in your example, that doesn't mean that spacetime is broken. You just can't place a static observer there. If you allow non-static observers, you can place them everywhere (the singularity remains a problem of course). That's reflected in the presence or absence of coordinate failures in the coordinarte systems that are based on the respective observers.
If you allow for arbitrary v2 in your example, you'll get it fixed the same way.


From your answers to nismaratwork:

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.

I'm usually talking about infalling test particles, which do not affect the horizon. If you'd throw in a significant mass, the EH would expand. But it would do so before the mass actually reaches it, so it still wouldn't cross it.

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.

Contrary to what you can see, there are different diagrams showing different things "actually happening". There are some where there is no horizon at all*, and others, where the horizon cannot be crossed in finite time.

*: as far as the specific class of observers is concerned.
The interior of a black hole is defined in a coordinate independent way, so I think you should say that in these coordinate systems, particles "actually" cross the horizon in a finite time.

 

[Kamil Szot]

It isn't. It's something to do with positive and negative frequencies going to future null infinity, the latter not cancelling the former as they should in the presence of an event horizon. Or something like that, I didn't understand the derivation.
HR can't be a process happening exactly at the EH, because the local spacetime at the EH is not different from the spacetime somewhere else.

Ok. Let's leave HR for now. It's way too complicated subject for my current understanding.

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?

Of course, if you have a reflector ten times thes mass of the observable universe somehow hovering there, within <<1 attometer from the horizon.

That's not what I meant. I thought about hovering at a safe distance from event horizon, having normal reasonably powered reflector (or laser) and aiming it so the beam would slide as close to event horizon as possible, wait and observe if anything got out. I think that photons of the beam should blue shift as they go down the gravity well possibly gaining arbitrarily high energy (in my frame of reference) and then redshift as they are getting out returning to wavelength I sent them with. If my aim is precise I might have to wait billions of years for it to get out. If I have computing power high enough maybe I could even get to know what was near event horizon as light passed there.

Yes.

I hope you confirm not only my final conclusion but also the way I reached it. ;-)

[...], that doesn't mean that spacetime is broken. You just can't place a static observer there. If you allow non-static observers, you can place them everywhere

That's an interesting thought but I don't think that this is all that is to it. From the point of view of remote static observer something very peculiar happens with non-static objects that travel near event horizon. In the frame of reference of remote static observer not only static clocks at event horizon have infinitely slower time pace. Same thing goes for clocks freefalling there.

I'm usually talking about infalling test particles, which do not affect the horizon. If you'd throw in a significant mass, the EH would expand. But it would do so before the mass actually reaches it, so it still wouldn't cross it.

That's interesting. Could you please point me to some explanation how this works that I might understand?

Contrary to what you can see, there are different diagrams showing different things "actually happening". There are some where there is no horizon at all*, and others, where the horizon cannot be crossed in finite time.

This diagrams are distorted in clever way to allow you to see things that are not visible on plain diagram. They allow you to discern between different ways (speeds) of asymptotically going to infinity. It allows you to see that (even in remote static frame of reference) light won't catch up freefalling object if it was shined too late. But as these diagrams expose some aspects they hide others. It's much harder to see from such deformed diagram that freefalling to event horizon as well as travel of light chasing frefalling object will both take infinite time (in remote static frame of reference).

What's actually happening might depend on chosen frame of reference (to some degree) but it shouldn't depend on how you illustrate it with a diagram.

About the things that shouldn't depend on the frame of reference: I think whether it is possible for two objects to meet should not depend on frame of reference from which you observe them so it's very good that light won't catch up falling object if it was sent to late after it no matter what frame of reference you pick. Other thing I think should be independent of frame of reference is the difference between how much two clocks meeting clocks aged since their previous meet up.

 

[Dale]

About the things that shouldn't depend on the frame of reference: I think whether it is possible for two objects to meet should not depend on frame of reference from which you observe them ... Other thing I think should be independent of frame of reference is the difference between how much two clocks meeting clocks aged since their previous meet up.

Those are both invariant. What is not invariant above is your question about whether or not something has crossed "yet", which requires a synchronization convention and is therefore coordinate-dependent.

 

[Kamil Szot]

Those are both invariant. What is not invariant above is your question about whether or not something has crossed "yet", which requires a synchronization convention and is therefore coordinate-dependent.

I know that and that's why I specified that I'm interested in "yet" in frame of reference associated with Earth.

 

[Dale]

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" ?

Distance from the event horizon is not relevant, only that Schwarzschild coordinates are used to define the "point of view" (which is reasonable for any stationary observer).

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.

If this is your assertion then you cannot consider any coordinate dependent things to be "physical reality" (I agree). Whether or not something has crossed the horizon "yet" or "already" is therefore clearly not a question about "physical reality".

 

[Kamil Szot]

If this is your assertion then you cannot consider any coordinate dependent things to be "physical reality" (I agree). Whether or not something has crossed the horizon "yet" or "already" is therefore clearly not a question about "physical reality".

You say that there is a frame of reference in which something "already" crossed event horizon? And what time is on Earth then if you are looking from that frame of reference?

 

[Ich]

I thought about hovering at a safe distance from event horizon, having normal reasonably powered reflector (or laser) and aiming it so the beam would slide as close to event horizon as possible

As close as possible means the photosphere at r=3M. Any light ray entering it will end up in the singularity. Your scenario is not possible.

I hope you confirm not only my final conclusion but also the way I reached it. ;-)

Yes.

[...], that doesn't mean that spacetime is broken. You just can't place a static observer there. If you allow non-static observers, you can place them everywhere

That's an interesting thought but I don't think that this is all that is to it.

Yes, these non-stationary observers all will move towards the singularity inside the EH.

In the frame of reference of remote static observer not only static clocks at event horizon have infinitely slower time pace. Same thing goes for clocks freefalling there.

Tell me, what is going wrong with freefalling observers there except that remote static observers attribute some crazy numbers to them? These numbers refer to a class of observers that cannot possibly extend beyond the horizon, and it's this impossibility that you see in the infinities.

Could you please point me to some explanation how this works that I might understand?

I keep forgetting the source where I read about it, and it wouldn't be too helpful either. Simply put, the EH is defined as the boundary of the regio from which nothing can escape to infinity. If you bring another mass close to the one already existing, escaping to infinity becomes harder because you'd have to run from the combined potential of those masses. That why the EH becomes larger.

What's actually happening might depend on chosen frame of reference (to some degree) but it shouldn't depend on how you illustrate it with a diagram.

These diagrams are actually expressions of different frames of reference, i.e. coordinate systems. "distorted diagram" == different coordinate system. Are you aware that coordiates in GR can be defined quite arbitrarily?

I think whether it is possible for two objects to meet should not depend on frame of reference [...] Other thing I think should be independent of frame of reference is the difference between how much two clocks meeting clocks aged since their previous meet up.

Yes.
Whether something goes through the horizon in finite or infinite coordinate time is not something you can measure. It does not pertain to "what's actually happening". The other things do.

 

[Kamil Szot]

As close as possible means the photosphere at r=3M. Any light ray entering it will end up in the singularity. Your scenario is not possible.

Could you please elaborate bit on that? Is there area outside event horizon that light can't escape from? What r=3M mean? Does it exist for non-rotating uncharged black-holes?

Tell me, what is going wrong with freefalling observers there except that remote static observers attribute some crazy numbers to them? These numbers refer to a class of observers that cannot possibly extend beyond the horizon, and it's this impossibility that you see in the infinities.

It's not that you can't extend beyond event horizon, you can't even extend to it. I see infinities as I'm getting close to event horizon not when I've passed it.

As for what's wrong with free falling observers: Theoretically they can pass information back. They can pass back what their clock indicates so these crazy numbers are not an unobservable math trick to get calculations in line with reality. These crazy numbers show what's actually happening, at least as far as we in our frame of reference are concerned.

I keep forgetting the source where I read about it, and it wouldn't be too helpful either.

If you ever recall that please share it with me.

Simply put, the EH is defined as the boundary of the regio from which nothing can escape to infinity. If you bring another mass close to the one already existing, escaping to infinity becomes harder because you'd have to run from the combined potential of those masses. That why the EH becomes larger.

Potential at EH is infinite. Potential outside of EH is finite. As you bring new mass close you add finite potential to finite potential. There is no way that sum of two finite numbers gave you infinity.

I'm not saying it's not happening. Maybe you just put it too simple.

These diagrams are actually expressions of different frames of reference, i.e. coordinate systems. "distorted diagram" == different coordinate system.

Yes. I noticed that.

Are you aware that coordiates in GR can be defined quite arbitrarily?

If you mean that you can consider any physical situation from any frame of reference (also non-inertial) using not necessarily Cartesian set of coordinates but also any other set of coordinates you can transform them to (keeping some physical invariants like two things I gave example of), and draw right diagrams and write right equations to properly describe this situation in these coordinates, then yes I'm aware of that.

Whether something goes through the horizon in finite or infinite coordinate time is not something you can measure.

But you can discern between short finite time and waiting billions of years for something to happen.


Can you answer the question how long will I have to wait after feeding a star to a black-hole to observe it EH grow?

 

[Ich]

Could you please elaborate bit on that? Is there area outside event horizon that light can't escape from? What r=3M mean? Does it exist for non-rotating uncharged black-holes?

I always try to supply key words to look up in Wikipedia, for example. But here my own elaboration:
There's a minimum radius from which light can escape when it's moving tangentially - like an incoming beam grazing that region. That radius is 3M in geometric units, it's 1.5 times the Schwarzschild radius. Time dilation there is very finite. Worst thing that could happen is that you shoot the light in a long-lasting orbit, but that's arbitrarily unlikely.

Theoretically they can pass information back. They can pass back what their clock indicates so these crazy numbers are not an unobservable math trick to get calculations in line with reality.

What crazy numbers? 12:47:05 PM? There's nothing wrong with the numbers such a probe could relay. The problem is that you'll receive the updates less and less freequently, until they finally cease.

These crazy numbers show what's actually happening, at least as far as we in our frame of reference are concerned.

Yeah, that it was 12:47:05 PM when the probe sent the signal. And your clock shows that it was, say, 3:00 am the next day when you picked up the signal. So what?

Potential at EH is infinite. Potential outside of EH is finite. As you bring new mass close you add finite potential to finite potential. There is no way that sum of two finite numbers gave you infinity.

I'm not saying it's not happening. Maybe you just put it too simple.

Yes, I put it too simple. Just wanted to explain that the definition of an EH is highly nonlocal (you have to wait for an eternity to be sure which photon escaped and which not), and make plausible that it can grow before the matter actually falls in.
Doesn't matter, I remembered thathttps://www.physicsforums.com/showthread.php?p=2370131#post2370131". (The review paper).

But you can discern between short finite time and waiting billions of years for something to happen.

Yes. But what's going to happen to you when something crosses the horizon? (Hint: nothing.)

 

[Kamil Szot]

I always try to supply key words to look up in Wikipedia, for example. But here my own elaboration:
There's a minimum radius from which light can escape when it's moving tangentially - like an incoming beam grazing that region. That radius is 3M in geometric units, it's 1.5 times the Schwarzschild radius. Time dilation there is very finite. Worst thing that could happen is that you shoot the light in a long-lasting orbit, but that's arbitrarily unlikely.

Thank you. I think that's sufficient amount of keywords for me and also an interesting information that I was not aware of.

What crazy numbers? 12:47:05 PM? There's nothing wrong with the numbers such a probe could relay. The problem is that you'll receive the updates less and less freequently, until they finally cease.

Yeah, that it was 12:47:05 PM when the probe sent the signal. And your clock shows that it was, say, 3:00 am the next day when you picked up the signal. So what?

I mean that in probes message there is 12:47:05 PM year 13.7 bilionth since big bang while on Earth there is 12:01:03 PM year 18 billionth.

Being unable to detect signal because it has too huge wavelength doesn't mean that it is impossible to receive the message. Maybe in a billion year humanity will discover how to pump this photons back with energy and blue shift them enough to read the message?

Also you'll never cease to receive the updates. You'll just have to wait longer and longer for each one, but that might be actually good because this gives you time to upgrade your technology so you could capture next update with even higher wavelength.

Doesn't matter, I remembered thathttps://www.physicsforums.com/showthread.php?p=2370131#post2370131". (The review paper).

Thank you!

Yes. But what's going to happen to you when something crosses the horizon? (Hint: nothing.)

How about end of universe if it will have an end?

 

[Ich]

I mean that in probes message there is 12:47:05 PM year 13.7 bilionth since big bang while on Earth there is 12:01:03 PM year 18 billionth.
[...]
Also you'll never cease to receive the updates. You'll just have to wait longer and longer for each one, but that might be actually good because this gives you time to upgrade your technology so you could capture next update with even higher wavelength.

You're talking about "classically and in principle" again. There will still be a last signal, the one shortly before the probe enters the horizon. From then on, there will be no more updates.
If you like to imagine a continuous stream, you get the notion of an "frozen and redshifted" image of the probe. But usually it is not appreciated just hoz frozen and redshifted that image really is.
Rest assured that, even classically, after a day or so there is absolutely nothing to receive, no matter how advanced your technology is.

How about end of universe if it will have an end?

Maybe. I was rather thinking about something that makes the "crossing event" relevant one way or another. Like the obsever hearing the EH crack when the probe runs into it.

 

[Dale]

I know that and that's why I specified that I'm interested in "yet" in frame of reference associated with Earth.

We have discussed this over and over. Please see above.

You say that there is a frame of reference in which something "already" crossed event horizon? And what time is on Earth then if you are looking from that frame of reference?

That is completely arbitrary. You don't seem to understand that in GR the whole concept of simultaneity is up to the whim of whoever specifies the coordinate system. It is only a label attached to a set of events, with no significance other than to identify the events. You are free to re-label as desired.

 

[nismaratwork]

Is it wrong to just say that he doesn't understand GR or its underpinnings, and leave it at that?

 

[Kamil Szot]

You're talking about "classically and in principle" again. There will still be a last signal, the one shortly before the probe enters the horizon. From then on, there will be no more updates.

But won't this last signal arrive to Earth after infinite Earth time? - "classically and in principle" whatever you mean by that

Rest assured that, even classically, after a day or so there is absolutely nothing to receive, no matter how advanced your technology is.

So you are saying that in Earth frame of reference travel to event horizon of a black hole takes less than a day? That's completely opposite to what I understood from your previous answers.

I think you rather mean that signal will just be undetectable because of its wavelength and delay between each two consecutive photons. I don't know on what basis do you dismiss technical possibility of receiving such signal.

I can think of one way how detecting such signal might be possible. You might put precisely designed set of moving masses on the way of the signal and pump it up with energy, same way passing near Jupiter pumped energy into Voyager probe.

Maybe. I was rather thinking about something that makes the "crossing event" relevant one way or another. Like the obsever hearing the EH crack when the probe runs into it.

Now you are just pulling my leg. ;-)

What I wanted to say is that universe might end before you receive last signal no matter how long finite time of existence universe has ahead of itself. Maybe not if universe final fate is big crunch.


One thing again about reality of last signal: When I ask you if function y=1/x as you approach zero from the right has some last non-negative point, if you look at plot of this function in normal coordinates you may say 'no'. But if you never knew of normal coordinates and used coordinates labeled x and 1/y and plotted this function in those coordinates your answer might be 'yes'.

Of course that's not the same situation because the answer to the above is always 'no' and I just erroneously extended domain of function y=1/x with point x=0 while plotting it in hyperbolic coordinates.

In case of EH changing plots is not just simple coordinate transformation. It is changing the frame of reference, and I think that existence of 'last signal' might be frame dependent.

 

[Kamil Szot]

Is it wrong to just say that he doesn't understand GR or its underpinnings, and leave it at that?

If I understood it to the extent you do I wouldn't be asking questions about it here.

Purpose of my questions is broadening my understanding. And I already understand few things I have not understood before so from my point of view your responses are far from being pointless, and I'm very thankful to members of this forum that respond to this thread.