How do Black Holes Grow? A Far-Away Observer's Perspective

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SUMMARY

This discussion centers on the paradox of black hole growth from the perspective of a far-away observer, particularly in relation to General Relativity. It establishes that while objects falling into black holes appear to never cross the event horizon due to time dilation and redshift, they do indeed fall in from their own perspective. The conversation highlights the need for a deeper understanding of black hole formation and growth, suggesting that astrophysicists may refer to massive dense objects as "black holes" even before they reach the event horizon. Key models mentioned include the Oppenheimer-Snyder model and the Vaidya metric, which describe black hole dynamics in the presence of matter.

PREREQUISITES
  • Understanding of General Relativity principles
  • Familiarity with black hole terminology, including event horizon and Schwarzschild radius
  • Knowledge of the Oppenheimer-Snyder model of gravitational collapse
  • Basic concepts of time dilation and redshift in astrophysics
NEXT STEPS
  • Research the Oppenheimer-Snyder model for insights on black hole formation
  • Study the Vaidya metric for understanding black hole absorption of radiation
  • Explore the implications of time dilation and redshift on observational astrophysics
  • Investigate numerical solutions for growing black holes in various spacetime models
USEFUL FOR

Astronomers, astrophysicists, and students of physics interested in black hole dynamics, particularly those exploring the implications of General Relativity on black hole growth and observation.

  • #31
Rene Dekker said:
I need to mull that over. Isn't that just another way of saying the same thing? At a moment you'd call "now", the horizon is infinitely far away?
No, the value of the integral is finite and constant at all times. It's just that no time coordinate can be assigned to any event on the horizon in this coordinate system. Trying to do it is like asking what's east of the north pole on Earth. The coordinate system just doesn't work in a way that lets the question make sense.
 
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  • #32
Rene Dekker said:
Maybe they don't define reality, but they should at least describe reality, right?
Sure, but the end of a description does not imply the end of the thing described. Similarly, a description that goes forever does not imply the thing goes forever.

Rene Dekker said:
Coordinates are a generalisation of the measurements that we can make about reality.
No. Coordinates are just an assignment of a set of numbers to events such that the mapping is smooth and invertible. Coordinates are not a generalization of measurements. In particular, measurements are invariant and coordinates are not.

Rene Dekker said:
If the real measurements don't correspond to the coordinate system that we have chosen, then we need to change our coordinate system
No. That is not a requirement for coordinates. Coordinates are only required to be smooth and invertible. Using any such coordinates the measurements can be expressed as invariants.

Rene Dekker said:
Measurements are our only way to observe and describe reality. If something cannot be measured, then it is not really part of reality.
Yes.

Rene Dekker said:
Hence if it is outside our coordinate system (carefully chosen such that all possible measurements can be described with them), it is not really part of our reality.
No. The correct “hence” is: “Hence coordinates are not part of reality”.

This should not be a surprising conclusion since we began by saying that coordinates just describe reality. So the idea that a mere description of reality is not reality, should not be objectionable.

Rene Dekker said:
Are there physical situations where it is not possible at all to assign any coordinate system that reflects the measurements? That is, where it is not possible to define a coordinate system where spacetime distances between events are identical to the distance in the coordinate system?
Any time there is spacetime curvature (tidal gravity) it is impossible.
 
  • #33
Dale said:
Any time there is spacetime curvature (tidal gravity) it is impossible.
A trip to Google finds this post from Physics Forums back in 2009 by @George Jones in #5.
https://www.physicsforums.com/threads/embedding-curved-spacetime-in-higher-d-flat-spacetime.290098/ said:
Chris Clarke* showed that every 4-dimensional spacetime can be embedded isometically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

* Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
manifolds," Proc. Roy. Soc. A314 (1970) 417-428
This would suggest that it is possible to have a flat space time (with a lot of dimensions) with an ordinary pseudo-Riemanian metric so that an embedded four dimensonal, intrinsically curved subspace can share the same metric.

Likely the required embedding is so convoluted that the conclusion is rendered meaningless for all practical purposes.
 
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  • #34
jbriggs444 said:
This would suggest that it is possible to have a flat space time (with a lot of dimensions) with an ordinary pseudo-Riemanian metric so that an embedded four dimensonal, intrinsically curved subspace can share the same metric.
"The same metric" on the much higher dimensional space, yes. But "distance" in the higher dimensional space is not the same as "distance" restricted to the 4-D subspace. So the fact that the metric on the higher dimensional space is flat is useless, since it doesn't give you the "spacetime distance" that you are looking for.
 
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  • #35
PeterDonis said:
But "distance" in the higher dimensional space is not the same as "distance" restricted to the 4-D subspace.
Well, “distance” between two separate events is not the same. But “distance” along a path is the same in the embedding space as in the curved space.
 
  • #36
Dale said:
“distance” along a path is the same in the embedding space as in the curved space.
Yes, but the flat metric "distance" that the OP is interested in (what the OP calls "coordinate distance") is the distance along a geodesic of the flat metric, not along a curve lying in the embedded curved space.
 
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  • #37
phinds said:
You are missing the point. If appearance and reality don't match, there's something wrong with the appearance. The object falls in. Period. The fact that it does not appear to do so to a remote observer is irrelevant.
The problem is that one cannot say when the object falls in, in terms of time that makes sense to the outside observer.
 
  • #38
Demystifier said:
The problem is that one cannot say when the object falls in, in terms of time that makes sense to the outside observer.
I'm not sure I agree. For example, imagine an external observer watching an electric charge fall toward a neutral Schwarzschild black hole. When the charge is at any finite height above the hole, the observer sees the charge's electric field as a complicated pattern of superposed multipole moments arising from the distorted spacetime around the hole. But as the charge approaches the horizon, the field smoothly transitions to the electric monopole field of a charged Reissner-Nordstrom black hole. So even though the observer never sees the charge actually pass the horizon, by monitoring the distribution of the E-field outside the hole could she not reasonably declare the time of in-fall to be the moment the external field is sufficiently close to a coulomb field?
 
  • #39
Demystifier said:
The problem is that one cannot say when the object falls in, in terms of time that makes sense to the outside observer.
I don't entirely agree with this. You can't say when it crosses the horizon in Schwarzschild coordinates, I agree, because it left the region covered by Schwarzschild coordinates. But that isn't the only definition of "now" available to an external observer - merely one of the simplest, as it corresponds to radar time.

There is a time for any external observer when the event "infalling object crosses the horizon" is no longer in their future light cone. Up until this time they might do something that stops the infalling object from entering the hole, but after this time it is too late. Even if all they need do is send a light pulse to the object asking it to turn on its arbitrarily powerful rocket motor, they are too late - the object crossing the horizon is no longer in their causal future. At any later event on the external observer's worldline there is an acausal hyperplane that passes through that event and the horizon crossing event. So you can call any time after the horizon crossing leaves your future light cone "when" the object crossed the horizon if you want.
 
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  • #40
Ibix said:
I don't entirely agree with this. You can't say when it crosses the horizon in Schwarzschild coordinates, I agree, because it left the region covered by Schwarzschild coordinates. But that isn't the only definition of "now" available to an external observer - merely one of the simplest, as it corresponds to radar time.

There is a time for any external observer when the event "infalling object crosses the horizon" is no longer in their future light cone. Up until this time they might do something that stops the infalling object from entering the hole. But after this time it is too late. Even if all they need do is send a light pulse to the object asking it to turn on its arbitrarily powerful rocket motor, they are too late - the object crossing the horizon is no longer in their causal future.
There's a calculation of this here:

https://www.physicsforums.com/threa...into-a-black-hole.1012103/page-3#post-6599762
 
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  • #41
renormalize said:
I'm not sure I agree…. could she not reasonably declare the time of in-fall to be the moment the external field is sufficiently close to a coulomb field?
Ibix said:
I don't entirely agree with this…. There is a time for any external observer when the event "infalling object crosses the horizon" is no longer in their future light cone.
That’s two different but equally reasonable definitions, which I think demonstrates @Demystifier’s point.
 
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  • #42
Nugatory said:
That’s two different but equally reasonable definitions, which I think demonstrates @Demystifier’s point.
I'd say it illustrates the opposite - there are many definitions that make sense.
 
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  • #43
The outside observer can reason as follows. I define time as the thing measured by my own local clock. With this definition of time, it takes an infinite time that a falling object reaches the horizon, so the falling object can never (where "never" means not after a finite time) enter the black hole interior. Moreover, according to classical general relativity, there is no experiment that I (the outside observer) can perform that would prove me wrong.
 
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  • #44
Demystifier said:
The outside observer can reason as follows. I define time as the thing measured by my own local clock. With this definition of time, it takes an infinite time that a falling object reaches the horizon, so the falling object can never (where "never" means not after a finite time) enter the black hole interior. Moreover, according to classical general relativity, there is no experiment that I (the outside observer) can perform that would prove me wrong.
That is not clear. He can define time by his watch only along his world-line. For these questions he needs time for much more than his world line. And he can define a time function, whose level surfaces will constitute his convention of nows and at the events of his world line will have values equal to his watch readings, and it can cover parts of the space-time inside the black hole. This way he can say when the object entered the black hole.
 
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  • #45
Ibix said:
No, the horizon is a finite distance away (or, at least, ##\int\sqrt{ g_{rr}}dr## is finite). It's just that the horizon is never at a time you'd call "now" using those coordinates.
The earlier question was about light pulses that are sent from an outside observer, reflect back from the object falling towards the black hole, and then return back at the observer again. They apparently would take longer and longer time to return back, the closer the object comes to the event horizon, until they take infinitely long.
If the distance to the horizon is finite in Schwarzschild coordinates, then how do they describe these increasing return times? Is the coordinate speed of light close to the black hole less than c?
 
  • #46
Demystifier said:
The outside observer can reason as follows. I define time as the thing measured by my own local clock. With this definition of time, it takes an infinite time that a falling object reaches the horizon
This reasoning does not follow. That definition of time only assigns time to events on your worldline. To extend it anywhere else you must also adopt a simultaneity convention.
 
  • #47
Dale said:
This reasoning does not follow. That definition of time only assigns time to events on your worldline. To extend it anywhere else you must also adopt a simultaneity convention.
Fair enough. Is there a reasonable simultaneity convention which I, as an outside observer, can apply to the black hole interior?
 
  • #48
Demystifier said:
Fair enough. Is there a reasonable simultaneity convention which I, as an outside observer, can apply to the black hole interior?
Depends on what you mean by reasonable.
 
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  • #49
Demystifier said:
Fair enough. Is there a reasonable simultaneity convention which I, as an outside observer, can apply to the black hole interior?
Sure. You can use the simultaneity convention of any of the following:

https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

https://en.m.wikipedia.org/wiki/Lemaître_coordinates

https://en.m.wikipedia.org/wiki/Eddington–Finkelstein_coordinates

https://en.m.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

Or any other simultaneity convention you like. Your simultaneity convention is arbitrary.
 
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  • #50
Rene Dekker said:
Is the coordinate speed of light close to the black hole less than c?
Not only is the coordinate speed of light less than c, it is also anisotropic in Schwarzschild coordinates.
 
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  • #51
Rene Dekker said:
Is the coordinate speed of light close to the black hole less than c?
For a radial null path the Schwarzschild metric tells us that ##0=\left(1-\frac{R_S}r\right)dt^2-\left(1-\frac{R_S}r\right)^{-1}dr^2##, which means that in these coordinates the coordinate speed of light is ##\frac{dr}{dt}=\pm\left(1-\frac{R_S}r\right)## (minus sign for inward-going rays). This tends to zero as ##r## approaches ##R_S##.
 
  • #52
Dale said:
Not only is the coordinate speed of light less than c, it is also anisotropic in Schwarzschild coordinates.
Yes, I see it myself now. It is ## (1 - \frac{r_s}{r}) c## in the radial direction, correct?
I would have thought that it would be more practical to always use a coordinate system where the coordinate speed of light is c everywhere. But I guess in such a system it is not possible to describe things like a black hole.
 
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  • #54
Rene Dekker said:
I would have thought that it would be more practical to always use a coordinate system where the coordinate speed of light is c everywhere. But I guess in such a system it is not possible to describe things like a black hole.
In Kruskal–Szekeres coordinates the radial speed of light is ##c## everywhere, i.e. inside, outside and on the event horizon (though not the tangential speed).
 
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  • #55
Rene Dekker said:
Yes, I see it myself now. It is ## (1 - \frac{r_s}{r}) c## in the radial direction, correct?
I would have thought that it would be more practical to always use a coordinate system where the coordinate speed of light is c everywhere. But I guess in such a system it is not possible to describe things like a black hole.
You are still fundamentally confused by the relationship between a geometry, which is the same geometry in all coordinate systems; and coordinates.

A black hole is a black hole regardless of your coordinate system.

Your other mistake is to elevate coordinate dependent things to the statis of universal laws that must apply in all coordinate systems.
 
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  • #56
Rene Dekker said:
Yes, I see it myself now. It is ## (1 - \frac{r_s}{r}) c## in the radial direction, correct?
Yes, that is correct.

Rene Dekker said:
I would have thought that it would be more practical to always use a coordinate system where the coordinate speed of light is c everywhere. But I guess in such a system it is not possible to describe things like a black hole.
There are fundamental limitations to the types of spacetimes for which such a coordinate system is possible. And even in well-behaved spacetimes often such coordinates cannot cover the entire spacetime.
 
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  • #57
PeroK said:
You are still fundamentally confused by the relationship between a geometry, which is the same geometry in all coordinate systems; and coordinates.
What I am definitely confused by, is the habit of changing coordinate system if they don't suit your needs. In my mind, coordinates are related to measurements, and you can only (and must) change a coordinate system, if it does not correspond to the measurements you are making.

I can fully understand that different coordinate systems can describe the same reality in different ways, like cartesian and polar coordinates. But the measurements (distances, proper time, etc) that you calculate with those coordinates should be the same. The should reflect reality.

Like in this discussion, when one says: "there is no event on the event horizon that corresponds with 'now' for the remote observer". And then somebody else says: "But we can just choose another coordinate system, and then there will be a 'now' on the event horizon". That is something I don't yet grasp. The event horizon is either part of reality or it isn't. If it is part of reality, there should be events that happen at the same time as events on Earth. I cannot understand that you can simply whip that into or out of existence by changing coordinate systems.
And yes, I understand that simultaneity is a much more fluid concept in GR than it is in Newtonian physics and Special Relativity. But my mind cannot follow why you can just change that by choosing another coordinate system, without making any physical changes in reality, or taking another observer's viewpoint.

But those are just my personal struggles with the theory, and I love to learn more about it. It was a great discussion, and I learned a lot. I thank everybody for the insightful comments.
 
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  • #58
Rene Dekker said:
I can fully understand that different coordinate systems can describe the same reality in different ways, like cartesian and polar coordinates. But the measurements (distances, proper time, etc) that you calculate with those coordinates should be the same. The should reflect reality.
Yes, but this is modern physics (relativity) and we have not separate time and space, but a single spacetime continuum! Spacetime distances are invariant. But, unlike classical physics, spatial distances and time intervals are not invariant. They are coordinate dependent.

This is why it's important to learn the basics before trying to study black holes!
 
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  • #59
Rene Dekker said:
Like in this discussion, when one says: "there is no event on the event horizon that corresponds with 'now' for the remote observer". And then somebody else says: "But we can just choose another coordinate system, and then there will be a 'now' on the event horizon". That is something I don't yet grasp.
The point is that "now, over there" doesn't have any possible assumption-free definition. You need to decide what you mean by "now, over there", which you can do in several different ways. And once you've decided what you mean you can set up a measurement process that reflects that definition - or simply define a mathematical transform between convenient physical measurements and convenient coordinates.

The teaching of SR based on inertial frames obscures this somewhat, because people come away with the notion that there's "my frame's time" and "your frame's time" and there's a Right Way to define these things. There isn't. Draw a Minkowski diagram and draw an arbitrary curve on it, with the only restriction being that the gradient must be everywhere strictly less than one. That's a valid definition of "now" for any event on that line, albeit a pointlessly complicated one. Note that this process allows you to draw the straight line simultaneity plane of a moving Einstein frame and claim it for your own. That's fine - nastier maths, but fine.

GR, with its curved spacetime, gives you a reason to use more complicated definitions than Einstein's simple "bounce a light pulse off it guys" approach. And sometimes you absolutely have to do it because "bounce a light pulse off it" can't work to define "now" at an event horizon because the pulse never comes back.
 
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  • #60
Rene Dekker said:
In my mind, coordinates are related to measurements
And that is something you will have to unlearn in GR, because it is simply not possible to always find coordinate charts in a curved spacetime that work the way your intuitions say they should work.

Rene Dekker said:
The event horizon is either part of reality or it isn't.
It is.

Rene Dekker said:
If it is part of reality, there should be events that happen at the same time as events on Earth.
This is the part you will need to unlearn. I understand that your intuitions are telling you this should be true. That means you need to retrain your intuitions.

One avenue towards retraining is to substitute the phrase "spacelike separated" for "at the same time". The advantage of this is that "spacelike separated" is an invariant--whether or not two events are spacelike separated from each other does not depend on your choice of coordinates. So if we rephrase your statement here to "there are events on the horizon that are spacelike separated from events on Earth", that statement is true without qualification.

Rene Dekker said:
I cannot understand that you can simply whip that into or out of existence by changing coordinate systems.
That's not what you're doing. The horizon is there regardless of what coordinates you choose; it's part of reality. But "simultaneity" is not part of reality. It's a human convention.

Rene Dekker said:
And yes, I understand that simultaneity is a much more fluid concept in GR than it is in Newtonian physics
Yes.

Rene Dekker said:
and Special Relativity.
No. SR has the same "fluidity" about simultaneity that GR does. Unfortunately many treatments of SR do not properly emphasize this fact.

Rene Dekker said:
my mind cannot follow why you can just change that by choosing another coordinate system, without making any physical changes in reality, or taking another observer's viewpoint.
Because simultaneity is not a "real thing"; it's a human convention. Changing coordinates just means changing the convention.
 
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