Is the growth of black holes paradoxical?

[BitWiz]

Much of what people say about the vicinity of black holes doesn't seem to make sense.

For instance, it seems to be impossible for a black hole to grow by "ingestion" by scooping up matter around it or in its path, at least in the traditional sense. Gravitational time dilation takes care of that -- no particle having mass will ever reach the event horizon, much less travel through it, and because of the asymptotic partitioning of space-time at the horizon, I don't think that even a photon can penetrate a black hole as it would have to raise itself to an infinite frequency. So an event horizon seems to be impenetrable -- from either direction.

However, it seems that a black hole can ingest matter by growing. If a massive object approaches a black hole, and comes close enough such that the two combined masses (or portions of a mass) now fit within their paired Schwarzschild radius, a new shell-like event horizon will form behind the intruding mass, and in the process, any other matter around the original black hole is now engulfed within the new expanded radius.

For instance, if a neutron star of about two solar masses approaches a black hole containing about 60 million solar masses such that its entirety is within about 6 kilometers of the event horizon, a new event horizon will form behind it, and in the process engulf enough space to contain the volume of the Sun (if my math is correct).

Thus the structure of black holes could be a series of horizon shells around the original dense core, each one partitioning its contents out of the accessible universe, but also partitioning themselves from each other. The internal structure of each shell and its contents would have the same properties as it did before the shell was formed, but would be inaccessible except by its own contents.

In an another example, two black holes could be rotating around each other, but their combined mass would cause a new event horizon to form some distance away, appearing externally as a single entity, but internally, there would still be two black holes and the matter orbiting their center(s) of gravity.

In the collapse of a super nova core or a neutron star > 3 solar masses, the resulting black hole does not have to form all at once, but could be a cascade of event horizons, each outer partition seeing any inner partition(s) as a black hole, and any outer partitions as infinite space-time(?). This could solve the infinite mass paradox -- "time" would take the place of the Pauli exclusion principal.

As far as the oft told story that passengers in a rocket approching and entering an event horizon would never notice that they had sailed past the end of time -- in years, an infinite number raised to the infinite power an infinite number of times -- pretty darned unlikely I would think. Instead, one of two things might happen: 1) they could eventually be engulfed by a new event horizon and seem to appear in a new universe that they don't recognize except for nearby objects; and/or 2) if Mr. Hawking is correct, the black hole, having run out of material with which to grow, and over a vast amount of time (for a large object), might evaporate as the rocket approached it; to the passengers, it would be as if the black hole became smaller as they were about to touch it, disappearing entirely as they passed through its center -- several quadrillion years in the future. Good luck finding your way home after that.

If Mr. Hawking is correct, and the shell hypothesis is also correct, then we would have Hawking radiation leaking from shell to shell, greatly increasing the time it takes for a black hole to evaporate, but also providing a way for inner shells to become visible once again, either externally or internally. I think this would solve the "missing data" conundrum as well.

Comments please?

Chris

 

[PeterDonis]

Much of what people say about the vicinity of black holes doesn't seem to make sense.

The main issues you are raising here have been discussed many times here. A typical recent thread:

https://www.physicsforums.com/showthread.php?t=656805

A good quick answer to your basic objection is here:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

I would recommend reading that Usenet Physics FAQ entry before posting further. Many threads on this topic end up getting locked because people get told the correct answer repeatedly, but it's counterintuitive so they refuse to accept it. I'll respond to the main points you raise below, but you shouldn't depend just on my responses. The issues you raise, and the responses to them, have been well understood by physicists for decades.

no particle having mass will ever reach the event horizon, much less travel through it

This is wrong. You haven't said why you believe this, but I suspect it's because you have seen that the Schwarzschild time coordinate goes to infinity at the horizon. That's true, but it doesn't mean anything physically. In GR, coordinates are just numbers that label events; they have no intrinsic physical meaning. The Schwarzschild time coordinate has a direct physical meaning only for observers that are at rest very, very far from the horizon. For things that happen close to the horizon, Schwarzschild coordinates are highly distorted (kind of like Mercator coordinates for the Earth near the poles) and don't directly reflect anything physical.

To investigate whether an object with mass can fall through the horizon, we have to look at the actual physics, not coordinates. The easiest physical quantity to look at is the proper time along the infalling object's worldline; this can be calculated, and it directly reflects the time registered on a clock following the worldline. When you calculate it, you find that it's finite: that is, an infalling clock registers a finite time elapsed to the horizon. This shows that your claim, that "no particle having mass will ever reach the event horizon", is false; such a particle *will* reach the horizon in a finite time by its own clock.

and because of the asymptotic partitioning of space-time at the horizon

Can you be more specific about what you mean by this? I think I know, but this is an unusual way of stating it, so I'd like explicit confirmation of where you are getting this from.

I don't think that even a photon can penetrate a black hole as it would have to raise itself to an infinite frequency.

Same thing here; I suspect that you are attributing a meaning to the Schwarzschild time coordinate that it doesn't have. Photons can fall through the horizon just fine, and they don't get infinitely blueshifted when they do. You can calculate these things using similar math to the math you use to calculate the finite proper time for an infalling particle with mass.

Thus the structure of black holes could be a series of horizon shells around the original dense core, each one partitioning its contents out of the accessible universe, but also partitioning themselves from each other. The internal structure of each shell and its contents would have the same properties as it did before the shell was formed, but would be inaccessible except by its own contents.

This is not correct; a black hole has a single event horizon. When two black holes merge (which can happen), their horizons merge also, resulting in a single horizon. Similar remarks apply to your other scenarios.

This could solve the infinite mass paradox

What is the infinite mass paradox?

As far as the oft told story that passengers in a rocket approching and entering an event horizon would never notice that they had sailed past the end of time

They won't notice that they have crossed the horizon (at least, not by any observations they make locally), but crossing the horizon does not equate to "sailing past the end of time". See my comments on the Schwarzschild time coordinate not having any direct physical meaning, above.

 

[jk22]

I read that the schwarzschild singularity was just of coordinates but not of space-time. However there seem to be a singularity at the origin.

 

[haushofer]

Yes. The singularity at the horizon r=2M is due to bad coordinates, just as the point r=0 in spherical coordinates; the Jacobian vanishes there. A simple way to find this is to calculate curvature scalars; these are independent of the coordinates choses.

These curvature scalars will diverge for "physical singularities", indicating that EVERY observer will measure that the curvature diverges if you approach such a point.

 

[PAllen]

I read that the schwarzschild singularity was just of coordinates but not of space-time. However there seem to be a singularity at the origin.

The singularity at the origin is an invariant space-time singularity. The horizon is only singular in certain coordinate systems.

 

[Dale]

Much of what people say about the vicinity of black holes doesn't seem to make sense.

It certainly is confusing and counter-intuitive to learn, but it is logically self-consistent as well as being consistent with current evidence.

For instance, it seems to be impossible for a black hole to grow by "ingestion" by scooping up matter around it or in its path, at least in the traditional sense. Gravitational time dilation takes care of that -- no particle having mass will ever reach the event horizon, much less travel through it, and because of the asymptotic partitioning of space-time at the horizon, I don't think that even a photon can penetrate a black hole as it would have to raise itself to an infinite frequency. So an event horizon seems to be impenetrable -- from either direction.

However, it seems that a black hole can ingest matter by growing. If a massive object approaches a black hole, and comes close enough such that the two combined masses (or portions of a mass) now fit within their paired Schwarzschild radius, a new shell-like event horizon will form behind the intruding mass, and in the process, any other matter around the original black hole is now engulfed within the new expanded radius.

You may be interested in the Oppenheimer-Snyder metric which describes the formation of an EH from a dust cloud.

http://grwiki.physics.ncsu.edu/wiki/Oppenheimer-Snyder_Collapse

Thus the structure of black holes could be a series of horizon shells around the original dense core, each one partitioning its contents out of the accessible universe, but also partitioning themselves from each other. The internal structure of each shell and its contents would have the same properties as it did before the shell was formed, but would be inaccessible except by its own contents.

This is pretty speculative. It doesn't seem to be in keeping with mainstream GR.

In an another example, two black holes could be rotating around each other, but their combined mass would cause a new event horizon to form some distance away, appearing externally as a single entity, but internally, there would still be two black holes and the matter orbiting their center(s) of gravity.

There are no stable orbits inside the EH. In fact, there are no stable orbits within the photon sphere, which is well outside of the EH.

http://en.wikipedia.org/wiki/Photon_sphere

Comments please?

Two comments: avoid speculation and be brief.

Re: speculation. It is against the forum rules to be overly speculative or to promote personal theories. This forum is for learning mainstream physics only. Please learn GR as it is, this is not the place to try to fix it. Also, take advantage of the fact that many of the experts have faced and overcome the same mental hurdles you are facing.

Re: brevity. Spend some time to consider your confusion or question and find the root misunderstanding you are facing. Once you have figured that out, ask it as clearly and succinctly as possible. If you embellish or go off on tangents then you run the risk of having people who could answer your main question becoming distracted and wasting time and effort addressing your tangents and embellishments instead.

 

[BitWiz]

Hi, Peter,

Thanks for your time and your response.

The main issues you are raising here have been discussed many times here. A typical recent thread:

https://www.physicsforums.com/showthread.php?t=656805

I've gone through about 200 posts so far, and many seem to be on target. Thank you. I'm making an interim response now, and will follow up later.

A good quick answer to your basic objection is here:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

This is an excellent link, thank you.

I would recommend reading that Usenet Physics FAQ entry before posting further. Many threads on this topic end up getting locked because people get told the correct answer repeatedly, but it's counterintuitive so they refuse to accept it.

I'm not looking to get myself banned. ;-) My goal is to natively understand GR, not to alter it to fit my preferences.

The Schwarzschild time coordinate has a direct physical meaning only for observers that are at rest very, very far from the horizon. For things that happen close to the horizon, Schwarzschild coordinates are highly distorted (kind of like Mercator coordinates for the Earth near the poles) and don't directly reflect anything physical.

The problem for me is the "infinite" word. If I as an observer at a finite distance see that an object takes an infinite amount of time to approach a horizon, then it will take an infinite amount of time at any observer distance.

Thus the structure of black holes could be a series of horizon shells around the original dense core, each one partitioning its contents out of the accessible universe, but also partitioning themselves from each other. The internal structure of each shell and its contents would have the same properties as it did before the shell was formed, but would be inaccessible except by its own contents. ​

This is not correct; a black hole has a single event horizon. When two black holes merge (which can happen), their horizons merge also, resulting in a single horizon. Similar remarks apply to your other scenarios.

Even two BHs orbiting each other?

What is the infinite mass paradox?

I misspoke. It's the density paradox. A point mass with undefined density. The paradox is that, under GR if the density of an object is high enough, it becomes undefined (gravity shrinks its volume to zero).

Again, thanks for your time.

Chris

 

[PeterDonis]

there are no stable orbits within the photon sphere, which is well outside of the EH.

A small technical point: actually there are no stable orbits inside r = 6M. The photon sphere is at r = 3M (and the horizon is at r = 2M); photons can orbit the hole at the photon sphere, but such orbits are unstable; small perturbations will cause an orbiting photon to either spiral into the hole or escape to infinity. The same goes for orbits of timelike objects between r = 3M and r = 6M.

 

[PAllen]

The problem for me is the "infinite" word. If I as an observer at a finite distance see that an object takes an infinite amount of time to approach a horizon, then it will take an infinite amount of time at any observer distance.

This is false, and is shown in pure special relativity. A uniformly accelerating rocket sees a horizon form behind it, and objects fail to reach it in infinite time, as observed by the rocket (as long as it continues to uniformly accelerate). Meanwhile, the non-accelerating objects know nothing of this horizon and 'fall' through it in finite time. Look up Rindler Horizon.

There are many parallels between the Rindler horizon and BH horizons; in particular, external static observers must be always accelerating (however slightly) to stay static. Thus external observers are analogous to Rindler observers in SR.

Also, note that static observers near a BH horizon are increasingly implausible - the force needed to maintain the static position approaches infinite. As soon as such an observer stops experiencing near infinite g-force from the acceleration, they rapidly (per their watch) cross the horizon.

 

[PeterDonis]

I'm not looking to get myself banned. ;-) My goal is to natively understand GR, not to alter it to fit my preferences.

Ok, good. For future reference, the tone of your first post in this thread did not convey that impression. I realize that it's hard to know people's expectations when you're new to a forum.

The problem for me is the "infinite" word. If I as an observer at a finite distance see that an object takes an infinite amount of time to approach a horizon

But you don't see that. You see that there is a *coordinate* that goes to infinity at the horizon, but that's not the same as the object actually taking an infinite amount of time to reach the horizon. The actual time experienced by the object is finite, as I said in my last post.

The key point here is that, as I said in my last post, coordinates are just numbers that label events; they don't have any intrinsic physical meaning. In particular, a coordinate labeling can be highly distorted. An example from ordinary experience is the Mercator coordinate chart, which is used to map the Earth's surface. In this chart, the coordinates of the North and South poles are infinite, but that doesn't mean the actual physical distance to the poles is infinite. The chart just becomes more and more distorted as you approach the poles, to the point of "infinite distortion" *at* the poles.

Similarly, Schwarzschild coordinates become more and more distorted as you approach the horizon, to the point of "infinite distortion" *at* the horizon. So curves of finite length can "look infinite" in these coordinates at the horizon, but that doesn't mean they really are infinite.

Even two BHs orbiting each other?

Two BHs orbiting each other are still separate BHs, so they have separate event horizons; but if they merge, they merge into a single BH with a single event horizon.

The paradox is that, under GR if the density of an object is high enough, it becomes undefined (gravity shrinks its volume to zero).

Ok, that clears things up a bit. There are really two issues here:

(1) Some people think it's a "paradox" that you can have black holes with different masses but singularities of "the same size" (zero size) at the center. This isn't actually a paradox, because the spacetime curvature of the hole, which is what we measure that leads us to attribute "mass" to it, is not actually coming from the singularity at the center. It's coming from the past, from the object that originally collapsed to form the hole. The Usenet Physics FAQ has a good, if brief, entry on this:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

So black holes with different masses have different spacetime curvatures, because they were formed from different collapse processes with different amounts of matter; that's what makes them different. The singularities at the center don't have to be different.

(2) There is, however, an issue (at least many people, including me, think it's an issue) with the fact that the singularity at the center of a black hole has infinite density. That means it also has infinite spacetime curvature, and *that* means that mathematical quantities that tell us about the physical characteristics of spacetime become singular there. The mainstream view in GR, as I understand it, is that this tells us that GR as a theory breaks down at the singularity.

Of course the big question then is, if GR breaks down at the singularity, what replaces it? This is a major reason why physicists talk about searching for a theory of quantum gravity: such a theory would be the most promising candidate to take over from GR in situations like this. Basically, the idea is that when spacetime curvature gets strong enough (the usual definition of "strong enough" is that the radius of curvature is of the order of the Planck length, 10^-35 meters), quantum effects become significant, and the behavior of spacetime changes--possibly to the point that "spacetime" is no longer even a good description of physics at this scale.

None of this, however, affects the physics far enough away from the singularity; and for any black hole of practical interest (which means holes of stellar mass or larger), "far enough away" from the singularity is still well inside the horizon. So the issue of what the correct physics is at the singularity, while it is a genuine issue, doesn't affect any of the points under discussion in this thread.

 

[BitWiz]

Hi, Peter,

Two BHs orbiting each other are still separate BHs, so they have separate event horizons; but if they merge, they merge into a single BH with a single event horizon.

Let's worst-case this. Let's say I have two black holes with a combined mass of 1000 solar masses orbiting (really fast) at 3500 km separation, and for the sake of simplicity, that their event horizons remain roughly spherical. I thus have two separate BHs and two horizons.

Now add some infalling mass that goes into orbit around either the BHs or their combined center of gravity (at their LaGrange points). At some time, the orbiting mass + all (or portions of) the orbiting BHs reaches critical density around the common center of gravity. At first, this new even horizon is likely to pierce the horizons of the original BHs -- would you consider this a merger? -- but after more mass is accumulated, in this case, just a few hundred solar masses since the new entitity "borrows" mass from the original pair -- the "central" horizon moves beyond the original BH horizons. Now what do we have?

Thanks,
Chris

 

[BitWiz]

Ok, good. For future reference, the tone of your first post in this thread did not convey that impression. I realize that it's hard to know people's expectations when you're new to a forum.

Thanks, Peter. I'm more engineer than scientist, more logician than math, so I'm probably more like a foreigner than "new," but I will learn to get along. My specialty is systems, and I want to understand GR as a system. That seems to be an uncommon approach; or perhaps I'm in the wrong forum.

Thanks for being direct.

Chris

 

[PeterDonis]

Let's say I have two black holes with a combined mass of 1000 solar masses orbiting (really fast) at 3500 km separation, and for the sake of simplicity, that their event horizons remain roughly spherical. I thus have two separate BHs and two horizons.

I see your intent here, but it won't work as you've described it. Your intent appears to be to have the horizons separated by a very small distance, but there is no way to have such a configuration be stable. As I posted earlier in this thread, there are no stable orbits inside r = 6M, which is three times the horizon radius; so even a very small object can't orbit the BH just outside its horizon.

Furthermore, another BH of the same mass is not "a very small object", and you can't treat it as one. You simply can't construct a stable scenario with two BHs orbiting each other this way. Gravity in GR is nonlinear, so you can't just superpose two individual BH solutions and get another solution. That's not to say that it's impossible for two BHs to orbit each other, just that it's not as simple as just having them orbit each other like two billiard balls.

In what follows, I'm going to pretend for the sake of argument that we *can* construct a stable system with two BHs orbiting each other fairly closely (but not as close as you've said). "Stable" here means the BHs stay in their mutual orbit for a long enough time compared to whatever experiments we are going to run; but it's important to note that such a system, even if it can be constructed, will *not* stay stable indefinitely. The two BHs will gradually spiral into each other because the system as a whole will be emitting gravitational waves and therefore losing energy, just as a binary pulsar system does (this has been confirmed by observation):

http://en.wikipedia.org/wiki/Binary_pulsar

In view of the above, please bear in mind that everything I'm saying is only heuristic; I am not working from an actual known solution of the GR equations. So this is really just handwaving--educated handwaving, I hope, but still handwaving. The strict answer would simply be that the scenario you have tried to construct is not valid; but I know that's not very satisfying, so I'm trying to do more than that, with caveats as above.

Now add some infalling mass that goes into orbit around either the BHs or their combined center of gravity (at their LaGrange points). At some time, the orbiting mass + all (or portions of) the orbiting BHs reaches critical density around the common center of gravity.

Yes, but that doesn't mean an event horizon instantaneously forms there. An event horizon is a globally defined surface: it's the boundary of a region of spacetime (a "black hole") that can't send light signals to infinity. If two black holes are merging, then there is really only one event horizon; it just has two "branches" in the past instead of a single one. But since the final configuration is a single black hole, there is only one region of spacetime as a whole that can't send light signals to infinity; again, that region just has two "branches" instead of one, so if you drew a spacetime diagram, for instance, with time vertical and spatial dimensions horizontal, the black hole region would look like a pair of trousers, so to speak, instead of a cylinder.

Also, the event horizon doesn't "jump" from one radius to another; it moves smoothly between them. Consider a simpler case for a bit: a single black hole that gains mass from a spherically symmetric, thin shell of infalling matter. The mass of the BH plus the shell is larger than the mass of the BH by itself, so what happens when the shell reaches the new, larger horizon radius due to the combined mass (which is slightly larger than the original horizon radius)? When the shell reaches that point, the new event horizon with a larger radius must be formed, right?

Yes, it is, but now consider a light ray that is moving outward, just outside the original horizon radius, in such a way that it just happens to hit the infalling shell at exactly the instant that the shell reaches the new (larger) horizon radius. That light ray will be trapped: it will stay at the new horizon radius forever (because that's what the horizon *is*, locally--it's a surface where outgoing light rays are trapped at the same radius forever). But that also means that an event just inside the path of that light ray, even though it is outside the original horizon radius, can't send light signals to infinity, so it must be part of the global black hole region.

In other words, globally, the event horizon expands smoothly from the original radius to the new radius as the infalling shell approaches the new radius; at the instant the shell hits the new radius, the event horizon has just reached that new radius as well. That means that we can't know exactly where the event horizon is without knowing the entire future history of the spacetime--for example, if we ourselves were hovering just outside the original BH, before the infalling shell of matter came in, we could find ourselves stuck inside the new BH without realizing it, if we didn't know the shell was falling in, and if we were inside the new horizon radius, because the boundary of the global region that can send light signals to infinity could pass by us, moving outward, *before* we saw the infalling shell. There is no way to tell, locally, that you can no longer send light signals to infinity from your current location.

This is kind of long-winded, but the point is that the event horizon is not a "thing" that you can keep track of just by looking at local phenomena. It's a globally defined boundary, and you can be misled if you try to think of it as a local thing.

At first, this new even horizon is likely to pierce the horizons of the original BHs

No. What will happen is that the event horizons of the two original BHs will start expanding, *before* the new matter has accumulated; they will probably merge with each other even before all the new matter has fallen in, and then the single combined EH will continue to expand until all of the accumulating matter has fallen inside the new horizon radius due to the final total mass present. After everything is all done, there will be a single BH, and a single event horizon.

I realize this is not easy to visualize, and there are a lot more complications that I haven't even gone into: the infalling matter is likely to emit X rays, and as the horizons merge, gravitational waves will be emitted. There are lots of efforts ongoing to numerically simulate black hole mergers to learn more details.

If you want to try to get a handle on how black holes gain mass, I would back away from the complicated scenario you've proposed, and start with the simpler case I gave above: a single, non-rotating, spherically symmetric BH that gains mass from a thin, spherically symmetric shell of infalling matter. Understanding that scenario will give a good baseline to go on to more complicated ones.

 

[PeterDonis]

I'm more engineer than scientist, more logician than math, so I'm probably more like a foreigner than "new," but I will learn to get along. My specialty is systems, and I want to understand GR as a system. That seems to be an uncommon approach; or perhaps I'm in the wrong forum.

The questions you're asking are appropriate for this forum, and they're good questions.

 

[PeterDonis]

If you want to try to get a handle on how black holes gain mass, I would back away from the complicated scenario you've proposed, and start with the simpler case I gave above: a single, non-rotating, spherically symmetric BH that gains mass from a thin, spherically symmetric shell of infalling matter.

Btw, a very good, readable popular book on GR that discusses this type of scenario is Kip Thorne's Black Holes and Time Warps. I would highly recommend it if you want a reasonably non-technical description of the kinds of things I've been talking about here.

 

[Lino]

Peter, Your *simple* analogy is great and I understand the basics (although I will need to read a lot more) of what you are saying. Can I ask what the distant observer would see in the scenario that you outlined? (Assuming the light ray is from a constant signal and that the shell of infalling matter is transparant) I assume that the distant oberver will first see the signal, then as the event horizon expands, see the frequency change (redden), and finally disappear (as it is swallowed by the expanding horizon). Is this correct? It seems at odds with the standard *frozen at the event horizon* description that one reads.

Regards,

Noel.

 

[PeterDonis]

Can I ask what the distant observer would see in the scenario that you outlined? (Assuming the light ray is from a constant signal and that the shell of infalling matter is transparant) I assume that the distant oberver will first see the signal, then as the event horizon expands, see the frequency change (redden), and finally disappear (as it is swallowed by the expanding horizon). Is this correct?

More or less; I assume that by "the signal" you actually mean a series of light signals emitted by an astronaut that is hovering outside the original horizon radius but inside the new horizon radius (after the shell falls in). Those signals will already be redshifted when they reach the distant observer, even a long time before the shell falls in, when the horizon is at its original radius. As the shell gets closer to the new horizon radius, the signals will get more and more redshifted until they disappear, when the expanding horizon overtakes the hovering astronaut.

This will be *before* the shell reaches the new horizon radius, so if the distant observer is also watching light emitted by the shell, he will still be able to see that light when he sees the hovering astronaut disappear (though that light, from the shell, will also have been getting more and more redshifted); then, some time after that (probably very soon, but the time lag depends on the mass of the hole), he will see the shell disappear.

It seems at odds with the standard *frozen at the event horizon* description that one reads.

That description mistakenly equates the increasing redshift of the light received from objects as they get closer and closer to the horizon, with an actual "slowing down of time" for those objects. The objects themselves don't see their time "slow down"; to them (for example, to an observer riding on the shell as it falls in), they reach the horizon in a finite amount of time.

 

[Lino]

Peter, Thanks for the reply. I have been reading around the subject to try to understand the principles ... but it is taking a lot longer than I thought it would! Oh well!

I appreciate what you are saying from the astronauts perspective (finite time ... growth of the event horizon ... etc.) and uderstand the basics, but I'm trying to understand the experience from the distant observers perspective. I appreciate that the astronaut, and his / her clock, will see time pass at the *normal* rate, but from the distant observers perspective I was under the impression that the rate at which the at which the astronaut is seen moving toward the event horizon (and the red shift of his signal) will approach infinity and so never actually (be seen to) cross the event horizon.

But my reading of the previous posts is that from the perspective of the distant observer, irrespective of the astronaut falling-in or the event horizon growing, the astronaut / signals will drop behind the event horizon and disappear. But ... I'm still not certain so on with my reading!

Regards,

Noel.

 

[PeterDonis]

I'm trying to understand the experience from the distant observers perspective.

The first thing to understand is that the distant observer does not experience what is happening to the astronaut. He only experiences the light signals coming from the astronaut; and those light signals have to pass through the intervening spacetime. See further comments below.

I was under the impression that the rate at which the at which the astronaut is seen moving toward the event horizon (and the red shift of his signal) will approach infinity and so never actually (be seen to) cross the event horizon.

A quick correction: the redshift approaches infinity, but this corresponds to the apparent rate of the astronaut's fall approaching zero. Note, however, that this is only the *apparent* rate: the infinite redshift means that the spacetime in between the astronaut and the distant observer is distorting what the distant observer sees--the distortion grows larger and larger as the astronaut gets closer to the horizon, until it becomes an infinite distortion when the astronaut is *at* the horizon.

But my reading of the previous posts is that from the perspective of the distant observer, irrespective of the astronaut falling-in or the event horizon growing, the astronaut / signals will drop behind the event horizon and disappear.

This is saying the same thing as the above, just in different words. The astronaut "dropping behind the horizon and disappearing" is equivalent to the astronaut's light signals redshifting to infinity.

 

[Lino]

Thanks Peter. The Usenet Physics FAQ was great, & it references MTW's Gravitation that I'm trying to work through at the moment - so good to know that my reading is on the right track!
The really interesting (& obvious when you think about it) thing that I've read so faris that the astronaut gets dimmer and fades to nothing as he approaches the event horizon, so I never actually see him *freeze*. I hadn't connected those dots!

Regards,

Noel.

 

[Lino]

I'm afraid that I need to come back to ask a further question. I seem to be reading "around" the subject without actually getting to the nub of the matter!

In relation to my previous scenario (the difference between an astronaut / signal free falling toward an event horizon versus the event horizon expanding to encompass astronaut / signal, from the perspective of a distant observer), what I envisaged was that for the free falling astronaut would appear to slowdown / fade but never stop or disappear, but for the expanding event horizon the astronaut would disappear completely.

I can appreciate that the speed of the expanding event horizon might slow as it approaches its new resting place, inline with the approach of the in falling shell, and thus if the astronaut were at the location of the new horizon the effect (infalling astronaut or expanding horizon) would be the same. But if the astronaut is inside the new horizon, then he will be completely emcompassed and disappear. Thus the perspective of the distant observer would be different for each situation, and that doesn't seem correct! Am I mis-representing something?

Regards,

Noel.

 

[PeterDonis]

what I envisaged was that for the free falling astronaut would appear to slowdown / fade but never stop or disappear, but for the expanding event horizon the astronaut would disappear completely.

Not quite. As the expanding horizon gets closer and closer to the radius where the astronaut is, light signals emitted to the distant observer from that astronaut will get more and more redshifted, and the time between them will get longer and longer. At the instant when the expanding horizon engulfs the astronaut, his outgoing light signals, from the standpoint of the distant observer, become infinitely redshifted and take an infinite time to get out. So there's not really a difference.

 

[Lino]

Much appreciated again Peter.

I'm still stuck comparing infinite redshift and "froozen" time. When I transpose this analogy into "fading to invisibility" it makes sense for the free falling astronaut, but if the rate at which the horizon expands slows and "freezes" (as it approaches the astronaut) how does it catch up with the smooth trajectory to the enlarged horizon?

Regards,

Noel.

 

[PeterDonis]

if the rate at which the horizon expands slows and "freezes" (as it approaches the astronaut)

It doesn't. Light signals emitted outward by the astronaut get more and more redshifted and take longer and longer to get out to the distant observer, but that's not the same as saying the expansion of the horizon itself slows and freezes. The astronaut will find himself at the new horizon in a finite length of time by his own clock; it just takes an infinite amount of time by the distant observer's clock for that information to get out to him.

 

[Lino]

It doesn't. Light signals emitted outward by the astronaut get more and more redshifted and take longer and longer to get out to the distant observer, but that's not the same as saying the expansion of the horizon itself slows and freezes. The astronaut will find himself at the new horizon in a finite length of time by his own clock; it just takes an infinite amount of time by the distant observer's clock for that information to get out to him.

I agree with what you are saying here and (mostly) understand it.

Now, this led early on to an image of a black hole as a strange sort of suspended-animation object, a "frozen star" with immobilized falling debris and gedankenexperiment astronauts hanging above it in eternally slowing precipitation. This is, however, not what you'd see. The reason is that as things get closer to the event horizon, they also get dimmer. Light from them is redshifted and dimmed, and if one considers that light is actually made up of discrete photons, the time of escape of the last photon is actually finite, and not very large. So things would wink out as they got close, including the dying star, and the name "black hole" is justified.

This paragraph is describing what the distant observer would see as the astronaut free falls toward the event horizon, and I understand the logic of the astronaut dimming but not why it would “wink out” (I don’t think that he “runs-out” of photons). However, if you consider the scenario where the event horizon expands (ahead of a shell of infalling material) from A to B – it does seem to make sense (to me).

Consider we have super sensitive equipment and a line of astronauts stretching for A to B. In order for the event horizon to move from A to B in a smooth / timely fashion, it will encompass the individual astronauts in turn. As it does (from the perspective of the distant observer), the signal / view of the initial astronaut will greatly redshift and then (as the FAQ says) wink out. But this means that the perspective of the distant observer is different if the astronaut is free infalling or the event horizon is expanding – which does not make sense (to me).

Which brings me back to the FAQ quote, maybe both do “wink out”, but why?

Regards,

Noel.

 

[harrylin]

Much of what people say about the vicinity of black holes doesn't seem to make sense.

For instance, it seems to be impossible for a black hole to grow by "ingestion" by scooping up matter around it or in its path, at least in the traditional sense. Gravitational time dilation takes care of that -- no particle having mass will ever reach the event horizon, much less travel through it, and because of the asymptotic partitioning of space-time at the horizon, I don't think that even a photon can penetrate a black hole as it would have to raise itself to an infinite frequency. So an event horizon seems to be impenetrable -- from either direction.

However, it seems that a black hole can ingest matter by growing. If a massive object approaches a black hole, and comes close enough such that the two combined masses (or portions of a mass) now fit within their paired Schwarzschild radius, a new shell-like event horizon will form behind the intruding mass, and in the process, any other matter around the original black hole is now engulfed within the new expanded radius.

For instance, if a neutron star of about two solar masses approaches a black hole containing about 60 million solar masses such that its entirety is within about 6 kilometers of the event horizon, a new event horizon will form behind it, and in the process engulf enough space to contain the volume of the Sun (if my math is correct).

[..]

Comments please?

Chris

The black hole growth paradox has been discussed somewhat in earlier threads. In particular with reference to mathpages which also brings it up (click on the links for more):
https://www.physicsforums.com/showthread.php?p=4195617

 

[PeterDonis]

I understand the logic of the astronaut dimming but not why it would “wink out” (I don’t think that he “runs-out” of photons).

He doesn't run out of photons, but once he is at or inside the horizon, the photons he emits can't escape back out to the distant observer. This is true whether he falls to the horizon or the horizon expands to him.

The "wink out" description assumes that there is a finite limit to how low a frequency (or how long a wavelength) the distant observer can detect; when the astronaut's photons are redshifted beyond that limit, he "winks out". This happens *before* the astronaut actually reaches the horizon (or the horizon reaches him). If the distant observer could detect photons of any finite frequency, however low, he would continue to detect them (at lower and lower frequency, and coming further and further apart) forever by his clock; but all the photons he detects would have been emitted by the astronaut from some point above the horizon.

Which brings me back to the FAQ quote, maybe both do “wink out”, but why?

Yes, they both do. See above.

 

[someGorilla]

I can appreciate that the speed of the expanding event horizon might slow as it approaches its new resting place

Notice that whether the horizon "slows" depends on your choice of coordinates. In a local inertial frame, the horizon expands at c! Even a "static" horizon expands at c from the point of view on an observer who is crossing it.
If you are not familiar with the Rindler horizon, I suggest you have a look at it. It's a simple case, involving no gravity and no spacetime curvature, of an event horizon with many features of a black hole's horizon, including redshift, objects falling past the horizon but never appearing to do so to a distant observer, causally disconnected regions of spacetime etc. It's like the horizon of a very large black hole, with negligible tidal forces. See for example this.

 

[Lino]

Notice that whether the horizon "slows" depends on your choice of coordinates. In a local inertial frame, the horizon expands at c! Even a "static" horizon expands at c from the point of view on an observer who is crossing it.
If you are not familiar with the Rindler horizon, I suggest you have a look at it. It's a simple case, involving no gravity and no spacetime curvature, of an event horizon with many features of a black hole's horizon, including redshift, objects falling past the horizon but never appearing to do so to a distant observer, causally disconnected regions of spacetime etc. It's like the horizon of a very large black hole, with negligible tidal forces. See for example this.

Thanks someGorilla. I do (mostly) appreciate this but will be checking out the referenced link. Thanks again.

Regards,

Noel.

 

[Lino]

Peter, First off, thanks for the time you have spent responding on this topic, I really appreciate it. I also recognise that I might just have a "mental block" and need to read (alot) more on / around the subject. If I am frustrating you, please consurve your patience and feel free to ignore this post.

I am mostly in general agreement with what you are saying:

... all the photons he detects would have been emitted by the astronaut from some point above the horizon. ...

Accepted. I understand how this comment is consistent in both scenarios.

... If the distant observer could detect photons of any finite frequency, however low, he would continue to detect them (at lower and lower frequency, and coming further and further apart) forever by his clock ...

I understand how this comment will be correct for the scenario of the infalling astronaut. I also understand that for the expanding horizon scenario the distant observer could detect photons at lower and lower frequency ... but I would have assumed for a finite period of time.

(I hope that this doesn't add confusion, but it strikes me like a "radioactive half life" problem: in that (for the distant observer) the time for the horizon to move to the new location appears infinite, but the time taken for the horizon to reach half way (i.e. the location of the astronaut) can be measured specifically.)

Regards,

Noel.

 

[PeterDonis]

I also understand that for the expanding horizon scenario the distant observer could detect photons at lower and lower frequency ... but I would have assumed for a finite period of time.

(I hope that this doesn't add confusion, but it strikes me like a "radioactive half life" problem: in that (for the distant observer) the time for the horizon to move to the new location appears infinite, but the time taken for the horizon to reach half way (i.e. the location of the astronaut) can be measured specifically.)

Not really. The distant observer can't see the horizon at all; light emitted outward at the horizon stays at the horizon. So the distant observer can't see the horizon "move" either.

Here's a way to illustrate what's going on. Suppose there are *three* astronauts. One, astronaut A, free-falls into the hole when it's at its original mass (with a smaller horizon). The second, astronaut B, is hovering at a radius halfway between the old (smaller) and the new (larger) horizon radius; the third, astronaut C, free-falls into the hole after it's reached its new mass (with a larger horizon). Then the distant observer will see all three astronauts' light signals get more and more redshifted and take longer and longer to get to him. But he will see this happen first to astronaut A, then to astronaut B, and finally to astronaut C.

 

[Lino]

Much appreciated Peter.

Regards,

Noel.