Oh boy... falling into a black hole

[pavelcherepan]

Say we have an object falling into a black hole. As it approaches the even horizon external observers who are farther away from EH than the object in question will see it slow down and get progressively red-shifted the closer it gets to EH. Then finally, the from the perspective of the object it will cross the horizon, but all external observers will disagree, as they will see it forever stuck just above the EH.

The object will go more and more dim, but as I understand, with an infinitely sensitive devices, external observers should be able to observe it being stuck at the same spot indefinitely. Is that correct?

Now, the object, before venturing on this one-way trip would have had a mass and energy and would cause the local space-time to curve, however small that effect might have been. So the question is: if my previous statements were correct, we should be able to observe electromagnetic radiation of the object as being stuck above EH indefinitely, but would we be able to observe the curvature of space-time corresponding to that object in that same location?

[beecee]

24 minutes ago, pavelcherepan said:

Say we have an object falling into a black hole. As it approaches the even horizon external observers who are farther away from EH than the object in question will see it slow down and get progressively red-shifted the closer it gets to EH. Then finally, the from the perspective of the object it will cross the horizon, but all external observers will disagree, as they will see it forever stuck just above the EH.

The object will go more and more dim, but as I understand, with an infinitely sensitive devices, external observers should be able to observe it being stuck at the same spot indefinitely. Is that correct?

Now, the object, before venturing on this one-way trip would have had a mass and energy and would cause the local space-time to curve, however small that effect might have been. So the question is: if my previous statements were correct, we should be able to observe electromagnetic radiation of the object as being stuck above EH indefinitely, but would we be able to observe the curvature of space-time corresponding to that object in that same location?

I'm pretty sure the answer to that is yes, remembering all frames of references are valid.

[pavelcherepan]

6 minutes ago, beecee said:

I'm pretty sure the answer to that is yes, remembering all frames of references are valid.

Ok, but then an infalling object will have increased the mass of the BH. So in this case we'd have once "copy" of the object that became a part of the BH and another just hanging out at the event horizon? This doesn't happen with EM radiation, because we simply can't see the object inside the BH, but gravitationally it would then appear to have created duplicates. This is what confuses me the most.

[Strange]

That last point is part of the answer why the image won’t be “frozen” there forever. At some point, the event horizon expands to include the location of those final photons. 

Also, there can only be a finite number of photons emitted before the object falls through the event horizon so, again, it can only be visible for a finite time. 

[pavelcherepan]

28 minutes ago, Strange said:

Also, there can only be a finite number of photons emitted before the object falls through the event horizon so, again, it can only be visible for a finite time. 

But as far as I understand the photons emitted close to EH will get more and more red-shifted and if we use something like a Penrose diagram for illustrative purposes, then final light will arrive infinitely far in the future.

36 minutes ago, Strange said:

That last point is part of the answer why the image won’t be “frozen” there forever. At some point, the event horizon expands to include the location of those final photons. 

OK, this makes sense, but let's assume we're dealing with very lonesome BH that's not feeding on anything, except CMB and its event horizon is expanding extremely slowly.

[Strange]

1 hour ago, pavelcherepan said:

OK, this makes sense, but let's assume we're dealing with very lonesome BH that's not feeding on anything, except CMB and its event horizon is expanding extremely slowly.

Then there is nothing for you to see falling into it.

But as each photon falls in, the event horizon expands to meet it. 

BTW the Schwarzschild description of a black hole only applies to a black hole that has always existed and is unchanging (in an empty universe) so situations like these need to be modelled by numerical methods (I think).  

[pavelcherepan]

7 minutes ago, Strange said:

Then there is nothing for you to see falling into it.

But as each photon falls in, the event horizon expands to meet it. 

Visibility aside, I'm still more interested in gravitational effects on the curvature of space-time from the body. 

[Strange]

15 minutes ago, pavelcherepan said:

Visibility aside, I'm still more interested in gravitational effects on the curvature of space-time from the body. 

Well ... gravitational effects are the curvature of space-time so ...

[MarkE]

4 hours ago, Strange said:

Well ... gravitational effects are the curvature of space-time so ...

I'm still confused about gravity and gravitational waves. It is known that Theodor Kaluza's equations (with one added dimension) imply that gravity can be considered another dimension. But if this is actually true, then how come we can directly detect these gravitational waves? In Einstein's view gravity is not a force, rather a curvature of spacetime. Does anybody know what Einstein's view on Newtons constant 'G' was? Because, if gravity is not a force, then how come there can be a number/value attached to it?

To fully grasp the inner workings of a black hole, requires to fully understand gravity @pavelcherepan, because of the strong gravitational field within the event horizon of a black hole.

Edited February 11, 2018 by MarkE

[Strange]

3 hours ago, MarkE said:

I'm still confused about gravity and gravitational waves. It is known that Theodor Kaluza's equations (with one added dimension) imply that gravity can be considered another dimension.

It describes gravity in the same way as GR but the n from ve dimensions, what ch allows it to include gravity and electromagnetic interactions in the same equations. This reduces to either GR or Maxwells equations in 4D. 

3 hours ago, MarkE said:

But if this is actually true, then how come we can directly detect these gravitational waves? In Einstein's view gravity is not a force, rather a curvature of spacetime. 

And gravitational waves are cyclical changes in that curvature. 

3 hours ago, MarkE said:

Does anybody know what Einstein's view on Newtons constant 'G' was?

No. But I think it is only needed because of the arbitrary human units of measurement used. You can choose a system of units where G=1. 

[MarkE]

19 hours ago, Strange said:

This reduces to either GR or Maxwells equations in 4D. 

Has it been widely accepted by the scientific community that gravity is indeed another dimension? Why (not)?

19 hours ago, Strange said:

You can choose a system of units where G=1. 

In the International System of units 'c' has a value because the photon has an actual speed that can be measured. G has neither a speed nor a mass (because it has no particle that mediates it), so where does the measured value of G (6.6740831×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²) come from?

Edited February 12, 2018 by MarkE

[MigL]

Aside from MarkE's confusion about Kaluza-Klein theory and ( gravitational ? ) dimensions...

Gravitational change information ( or any information, for that matter ) is subject to the same dilation as light or EM radiation, to a far-away observer.

[Strange]

1 hour ago, MarkE said:

Has it been widely accepted by the scientific community that gravity is indeed another dimension? Why (not)?

No. Because it appears to be a misunderstanding on your part. The Kaluza-Klein model extends GR to 5 dimensions so it can include electromagnetism as well as gravity.

1 hour ago, MarkE said:

In the International System of units 'c' has a value because the photon has an actual speed that can be measured. G has neither a speed nor a mass (because it has no particle that mediates it), so where does the measured value of G (6.6740831×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²) come from?

In the natural units where G = 1, c is also equal to 1.

G is the measured constant of proportionality between force and mass, in Newtonian gravity, or (simplifying massively) between mass and the amount of curvature of space-time. I guess you could think of it as one of the things that defines how "stiff" space-time is.

 

[MarkE]

42 minutes ago, Strange said:

No.

Indeed it's not widely accepted by the scientific community, but isn't the statement "Whether or not the universe is five-dimensional is a topic of debate" suggesting that it's a 'Maybe' rather than a 'No'?

Edited February 12, 2018 by MarkE

[Strange]

1 hour ago, MarkE said:

Indeed it's not widely accepted by the scientific community, but isn't the statement "Whether or not the universe is five-dimensional is a topic of debate" suggesting that it's a 'Maybe' rather than a 'No'?

Sorry, I was responding to your “gravity is another dimension” statement. 

Whether there are 4, 5, 11 or 26 dimensions is indeed a matter of scientific research. 

[pavelcherepan]

Thanks everyone for your inputs. Now can we please look into the conundrum I'm looking at. I think I shouldn't have brought infinity into question, it will be easier without it:

1) Object falls into a black hole

2) After a finite time it reaches singularity and its mass is added to the mass of BH

3) In EM spectrum we should still be able to see it stuck just above EH. We can't see into BH, so there is only one copy of the object in the observable universe in EM spectrum.

4) If, like MigL commented, gravitational information is subject to the same time dilation as any other information, we might still be able to measure space-time curvature of the object outside EH

5) Then we have "two copies" of the object's mass-energy in the universe. One being a part of BH mass and the other causing curvature outside EH. How is that possible?

[Strange]

9 minutes ago, pavelcherepan said:

2) After a finite time it reaches singularity and its mass is added to the mass of BH

It adds to the mass of the black hole as soon as it passes the event horizon - which includes the horizon expanding to meet it when it passes the point of no return. 

12 minutes ago, pavelcherepan said:

5) Then we have "two copies" of the object's mass-energy in the universe. One being a part of BH mass and the other causing curvature outside EH. How is that possible?

It isn't possible which shows there must be something wrong in your reasoning. Note that the event horizon is invariant so the distant observer will see the size of the event horizon increase as the object falls through it. 

[pavelcherepan]

1 hour ago, Strange said:

It isn't possible which shows there must be something wrong in your reasoning.

That's the whole reason I started this discussion. There is obviously something wrong with my reasoning and I want to know what it is.

1 hour ago, Strange said:

Note that the event horizon is invariant so the distant observer will see the size of the event horizon increase as the object falls through it. 

Indeed, the Schwarzchild radius of the BH will increase somewhat. If we take a 2-solar-mass BH, the radius of EH will be 5932.(4) m, but if we throw in a 100 kg object, it will increase to 5932.44444444444444444444444459. Then, if the measurable extent of space-time curvature caused by this 100 kg object prior to falling into the BH is greater than the difference between these numbers, we should still be able to detect it as being outside EH, regardless of its expansion. And this would still result in "duplicating" the object's mass-energy.

I'm very confused.

Edited February 13, 2018 by pavelcherepan

[beecee]

All the distant observer would see is infalling mass redshifted to infinity before it ever crosses the EH, until fading from view: I don't believe the distant observer will ever see the EH of the BH increase in size for the same reasons....gravitational red shift. At least that's my take on it and going from a very fuzzy memory many years ago, when answered on another forum by an astronomer.

Edited February 13, 2018 by beecee

[J.C.MacSwell]

The only thing from it left at the event horizon is the radiation it gave off before it entered. That rapidly gets redshifted into oblivion...theoretically there forever but decreasing (quantum effects aside... in any event (pun intended), at some point it will be indistinguishable from the Hawking Radiation or if the black hole grows in size it (the "signature") is gone.

 

But in any case the object is no more at the event horizon than you are a light year from earth (in every direction) due to any unencumbered radiation you sent off last year.

[Strange]

6 hours ago, pavelcherepan said:

Then, if the measurable extent of space-time curvature caused by this 100 kg object prior to falling into the BH is greater than the difference between these numbers, we should still be able to detect it as being outside EH,

The event horizon will not remain spherical as something falls towards it; it will bulge out to meet the object. I don’t know if this by itself is enough to answer your question. Also, the object will be moving at the speed of light as it crosses the event horizon, which may be relevant. (My answers are a bit vague; I think you would need someone with a good understanding standing of the math to fully answer your question. Trying to apply “common sense”  or even Newtonian physics is almost bound to fail) 

[Philip K. Dick]

Well first of "Pavel", if that is indeed your actual real name. 

À "black hole" is supposedly caused by a star's gravitational collapse upon itself, Ha Ha Mr. Hawking! So basically, your ass is grass or another material if you even dare to get close, Ha Ha Mr. Hawking! 

And as for you "Strange", now I know that's your nickname, unless your Dad was Dr. Strange. Which I believe is very unlikely, because he is in the movies now. Read more fool! 

Sincerely,

Philip K. Dick

[beecee]

4 hours ago, Philip K. Dick said:

Well first of "Pavel", if that is indeed your actual real name. 

À "black hole" is supposedly caused by a star's gravitational collapse upon itself, Ha Ha Mr. Hawking! So basically, your ass is grass or another material if you even dare to get close, Ha Ha Mr. Hawking! 

And as for you "Strange", now I know that's your nickname, unless your Dad was Dr. Strange. Which I believe is very unlikely, because he is in the movies now. Read more fool! 

Sincerely,

Philip K. Dick

So obviously you do not accept the existence of BH's? Have you any other potential model to describe the effects that we see? Please if you do, take the time to start another thread in the speculative section, and put forward your hypothetical for the forum's professionals to judge....thank you, Otherwise of course, like the many other anti science and anti GR nuts that infest forums such as this, open to any Tom, Dick or Harry, you are just pissing into the wind.

Edited February 19, 2018 by beecee

[Strange]

5 hours ago, Philip K. Dick said:

Well first of "Pavel", if that is indeed your actual real name.

Says the guy posting under a pseudonym.

[swansont]

8 hours ago, Philip K. Dick said:

Well first of "Pavel", if that is indeed your actual real name. 

À "black hole" is supposedly caused by a star's gravitational collapse upon itself, Ha Ha Mr. Hawking! So basically, your ass is grass or another material if you even dare to get close, Ha Ha Mr. Hawking! 

And as for you "Strange", now I know that's your nickname, unless your Dad was Dr. Strange. Which I believe is very unlikely, because he is in the movies now. Read more fool! 

Sincerely,

Philip K. Dick

!

Moderator Note

This needs to stop, right now. Discuss the topic, and refrain from personal comments.