How can a black hole absorb matter?

[Line_112]

TL;DR

According to the general theory of relativity, time does not flow in a black hole, which means that processes do not occur in it. As a result, logically, it cannot absorb matter.

According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a classical black hole, from which not even light comes out.
Or is there some other model of matter absorption that allows a black hole to form and continue to suck in matter?

 

[Ibix]

According to the General Theory of Relativity, time does not pass on a black hole

This is not correct.

"Time slows down as you approach a black hole" is an over-simplification. Things can fall into a black hole in finite time by their own clocks, and there is a moment after which an external observer cannot (even in principle) stop the infall, but the external observer can never observe the horizon crossing. That does not mean that it does not happen.

Even "on" a black hole is problematic - the event horizon is a null surface, which isn't really a place.

 

[willyengland]

Even "on" a black hole is problematic - the event horizon is a null surface, which isn't really a place.

What is a "null surface"?

 

[Ibix]

What is a "null surface"?

A surface with at least one null tangent vector everywhere on it.

Light cones are another example. In flat spacetime, at any instant the light cone of some chosen event is a sphere, but it's expanding/contracting at the speed of light. That means that one direction in the surface (the inward/outward and future/past pointing one) is null. So the surface isn't really a place because no frame says any point on it is at rest, although you could talk about where it is at a given instant.

In curved spacetime it's possible to have a null surface that stays where it is in some senses, which is what an event horizon is. If you fall through one, a better local description is that the horizon passed upwards through you at the speed of light, and once you are through it the reason you can't go through it again is that it is receding from you at the speed of light. Nevertheless, a distant observer outside the hole could argue (loosely) that the event horizon is stationary.

Curved spacetime is a bad place to make intuitive arguments, in short. It's really easy to make mistakes like the OP if you don't study the maths. (Of course, even if you do study the maths it's still very possible to say misleading or outright wrong things, but hopefully it's less likely.)

 

[Nugatory]

Or is there some other model of matter absorption that allows a black hole to form and continue to suck in matter?

There is, and the “similar threads” section below will take you to some of the older threads about this. An excerpt from one of them:

Consider a black hole of mass $M$, surrounded by a spherically symmetrical shell of dust with total mass $m$ at a great distance from the black hole and falling/collapsing into the black hole. Consider $R'$, the Schwarzschild radius of a hypotherical black hole of mass $M+m$ and $R$, the Schwarzschild radius of the black hole. Clearly $R'$ is greater than $R$ so is outside the event horizon; thus the infalling shell of dust will reach $R'$ in finite time according to you and other external observers. But once it gets there... we have a spherically symmetric mass distribution all inside its Schwarzschild radius $R'$, and that is a black hole with radius $R'$. So our initial state is a black hole of radius $R$ and our final state is a black hole of radius $R'$, and we get from one to the other in a finite external time. We just can't use the Schwarzschild solution (which describes a black hole of constant mass) to describe what's happening in between.​

 

[JimWhoKnew]

Clearly $R'$ is greater than $R$ so is outside the event horizon; thus the infalling shell of dust will reach $R'$ in finite time according to you and other external observers.

The event horizon is a global concept and can never be in the causal past of external observers.
(Ignoring the possibility of completely evaporating black holes)

Late edit:
See, for example, fig. 59 in Hawking & Ellis.

 

[Dale]

According to the General Theory of Relativity, time does not pass on a black hole,

This is too broad of a statement. The correct statement would be that in Schwarzschild coordinates the coordinate time asymptotically approach infinity as an in falling object approaches the horizon.

The Schwarzschild coordinates do not cover the horizon, so you cannot use them to make any claims about the horizon itself.

Or is there some other model of matter absorption that allows a black hole to form and continue to suck in matter?

Yes, there are many alternative coordinates in GR, such as Lemaitre. Also, if you want to speak of the formation of a black hole there are different spacetime manifolds you can use, such as Oppenheimer Snyder.

 

[Tomas Vencl]

Consider a black hole of mass $M$, surrounded by a spherically symmetrical shell of dust with total mass $m$ at a great distance from the black hole and falling/collapsing into the black hole. Consider $R'$, the Schwarzschild radius of a hypotherical black hole of mass $M+m$ and $R$, the Schwarzschild radius of the black hole. Clearly $R'$ is greater than $R$ so is outside the event horizon; thus the infalling shell of dust will reach $R'$ in finite time according to you and other external observers. But once it gets there... we have a spherically symmetric mass distribution all inside its Schwarzschild radius $R'$, and that is a black hole with radius $R'$. So our initial state is a black hole of radius $R$ and our final state is a black hole of radius $R'$, and we get from one to the other in a finite external time. We just can't use the Schwarzschild solution (which describes a black hole of constant mass) to describe what's happening in between.​

As JimWhoKnew has already noted: In other words, this scenario assumes that the photons emitted from the shell itself during the collapse are not affected by the mass of the shell, but only by the mass of the original black hole, which I think is quite an unphysical assumption.

 

[PeterDonis]

Clearly $R'$ is greater than $R$ so is outside the event horizon

Careful. The event horizon expands during this process. By the time the shell reaches $R'$, the event horizon is also at $R'$. That is a consequence of this:

once it gets there... we have a spherically symmetric mass distribution all inside its Schwarzschild radius $R'$, and that is a black hole with radius $R'$.

In other words, a radially outgoing light ray emitted from the shell at the instant it reaches $R'$ will remain at $R'$ forever--it won't escape. Which means this...

we get from one to the other in a finite external time.

...is also problematic, even apart from the fact that we can't define a single Schwarzschild chart covering the entire spacetime in the model described, so we don't actually have a well-defined "external time" to use.

 

[Ibix]

I think @Nugatory's post actually highlights the contradiction in the OP's thinking quite nicely.

We start with a black hole of mass and radius $M$ and $R$. Then we allow the spherical shell of mass $m$ to drop. By the OP's reasoning it's perfectly possible for the shell to reach radius $R'=(1+m/M)R$ in finite time - but now we've ended up with mass $M'=M+m$ inside radius $R'$, which is just a slightly more massive black hole. So the event horizon cannot not grow - if we try to argue that the matter isn't really crossing the horizon we end up admitting that it had to have crossed anyway, a self-contradiction.

A correct model of black hole growth would not be based on Schwarzschild spacetime (although it's a decent approximation away from the growth phase).

 

[Line_112]

"Time slows down as you approach a black hole" is an over-simplification. Things can fall into a black hole in finite time by their own clocks, and there is a moment after which an external observer cannot (even in principle) stop the infall, but the external observer can never observe the horizon crossing. That does not mean that it does not happen.

But the whole trick here is that we are the external observers, not the little people falling into the black hole. For us, time does not pass there. Then, for the formation of a classical black hole, an infinite timefrom our point of view would be required from the moment the primary gas-dust cloud begins to compress until the threshold of the black hole is reached. It should also follow from this that for supermassive black holes to appear on the boundary of the observable Universe, an infinite time in our frame of reference would be required, that is, from this point of view, the Universe should exist forever, and not 13.7 billion years.

 

[PeroK]

But the whole trick here is that we are the external observers, not the little people falling into the black hole. For us, time does not pass there.

This is a serious misconception, albeit one promoted by many popular-science sources.

 

[Ibix]

For us, time does not pass there

No. As I said, intuitive reasoning in curved spacetime is a mistake. Your objection turns out to be similar to claiming that you can't walk through the north pole because you'd fall off the edge of the map. Just because the coordinates you use misbehave there does not mean there's any real physical consequences.

Schwarzschild coordinate time (which is what you are using) isn't defined at the horizon. Basically this is because (on a radial line) its concept of simultaneity is radar simultaneity, and you can't get an echo back from the horizon. But there are plenty of other simultaneity conditions available where an external observer agrees that the object fell in. @Dale named several in #7.

 

[JimWhoKnew]

But there are plenty of other simultaneity conditions available where an external observer agrees that the object fell in. @Dale named several in #7.

Consider a case of a collapsing progenitor, say. Suppose we are given an arbitrary event on the world-tube of an eternally-external observer. We should be able, at least in principle, to construct a coordinate chart in which the metric is non-singular at the event horizon, and the simultaneity (constant $t$ spacelike hypersurface) that includes our given event, precedes the formation of the horizon.

My point is that all events on an EH are either future-like or spacelike relative to the world-tube of an eternally-external observer. Unlike coordinates and simultaneities, this is an invariant property of the spacetime in discussion. In that sense, the claim that EH hasn't formed yet from the perspective of an external observer (at any proper time on her wristwatch), is just as good (or bad) as the claim that it has.

 

[Dale]

But the whole trick here is that we are the external observers

That is only a small part of the trick, not even close to the whole trick. Other parts of the trick include

1) we can use any coordinates we choose, including ones like Lemaitre

2) the Schwarzschild coordinates do not cover the horizon itself, so no claim can be made at the horizon based on it

3) the Schwarzschild spacetime is static, so it does not apply for any question about forming or growing black holes

4) the Schwarzschild spacetime is a vacuum spacetime, so it does not apply for any discussion of black holes over cosmological scales

So no, that is not the “whole trick”. You are missing a lot. The stuff you are missing is important in this context. In fact, point 3 basically precludes Schwarzschild spacetime from being used in any discussion about a black hole absorbing matter.

 

[Ibix]

My point is that all events on an EH are either future-like or spacelike relative to the world-tube of an eternally-external observer. Unlike coordinates and simultaneities, this is an invariant property of the spacetime in discussion. In that sense, the claim that EH hasn't formed yet from the perspective of an external observer (at any proper time on her wristwatch), is just as good (or bad) as the claim that it has.

I would observe that there are two distinct periods in the history of a distant observer of a stellar collapse: the period where the entire event horizon formation is unambiguously in the observer's future, and the period where some of it is spacelike separated from the observer. Similar remarks apply to the horizon crossing of any matter falling in later.

So while I agree that you can always make the claim "the horizon hasn't formed yet" there is a time when doing so becomes a matter of choice. And it's perfectly ok to make your choice so that your coordinates don't cover the horizon forming. But picking coordinates that don't cover the horizon doesn't mean that the horizon doesn't form, any more than the fact that my road atlas only covers Britain doesn't mean America doesn't exist.

 

[PeterDonis]

For us, time does not pass there

Wrong. For someone falling into the black hole, time passes normally. Their clocks read one second per second, just as ours do. We see them slowing down and slowing down (and the light from them getting more and more redshifted) as they approach the horizon, because of the way the outgoing light from them to us is affected by the spacetime geometry. What we see is not what is actually happening to them.

for the formation of a classical black hole, an infinite timefrom our point of view would be required from the moment the primary gas-dust cloud begins to compress until the threshold of the black hole is reached.

Wrong. For someone falling in with the cloud, the time by their clock to the formation of the event horizon is finite. This has been known since Oppenheimer and Snyder published their groundbreaking paper in 1939 that described a simple model of spherically symmetric gravitational collapse.

 

[PeterDonis]

We should be able, at least in principle, to construct a coordinate chart in which the metric is non-singular at the event horizon

Sure, at least three such charts have been known for decades and are well covered in the literature: Painleve, Eddington-Finkelstein, and Kruskal.

and the simultaneity (constant $t$ spacelike hypersurface) that includes our given event, precedes the formation of the horizon.

"Our given event" where? If you mean an event on the worldline of a distant observer who stays at the same altitude above the horizon, no, your claim is not correct; in any of the charts I named above, you will reach a point on the distant observer's worldline where the constant time hypersurfaces now cross the horizon.

ll events on an EH are either future-like or spacelike relative to the world-tube of an eternally-external observer.

If by "future-like" you mean "within the future light cone", yes, this is true. However, it doesn't mean what you appear to think it means.

Unlike coordinates and simultaneities, this is an invariant property of the spacetime in discussion. In that sense, the claim that EH hasn't formed yet from the perspective of an external observer (at any proper time on her wristwatch), is just as good (or bad) as the claim that it has.

In the sense that both claims are meaningless, yes. The invariant property of the spacetime that you reference is valid, but again, it doesn't mean what you appear to think it means. And it certainly does not mean that this invariant property is a valid reason to claim that the black hole can't form at all.

Indeed, you used the phrase "formation of the horizon" (I quoted it above)--but that only makes sense in a model where the horizon, i.e., the black hole, does form. And such a model is self-consistent--nothing you have said implies otherwise, and, as I pointed out in response to the OP just now, such a model was first published in 1939, so it's not a new idea--and it has the invariant property you describe. In other words, a black hole can form and produce a spacetime with that invariant property.

 

[JimWhoKnew]

And it's perfectly ok to make your choice so that your coordinates don't cover the horizon forming. But picking coordinates that don't cover the horizon doesn't mean that the horizon doesn't form, any more than the fact that my road atlas only covers Britain doesn't mean America doesn't exist.

The coordinates that I was talking about should cover the horizon (and possibly the interior). How else can the the metric be non-singular on the EH? Of course the horizon forms. It is an invariant property of spacetime. The coordinates are just constructed in a way that the simultaneity of the specific single given event on the world-tube, precedes the EH formation (in this specific chart).

As for the existence of America, as well as the landing on the moon, I don't want to go there...

 

[Ibix]

Of course the horizon forms.

I agree. I don't think the OP does, though.

 

[JimWhoKnew]

Sure, at least three such charts have been known for decades and are well covered in the literature: Painleve, Eddington-Finkelstein, and Kruskal ... in any of the charts I named above, you will reach a point on the distant observer's worldline where the constant time hypersurfaces now cross the horizon.

Ever occurred to you that charts like Eddington-Finkelstein can be matched smoothly with other charts outside the horizon?

Indeed, you used the phrase "formation of the horizon" (I quoted it above)--but that only makes sense in a model where the horizon, i.e., the black hole, does form. And such a model is self-consistent--nothing you have said implies otherwise,

I explicitly opened with "Consider a case of a collapsing progenitor...".

"Our given event" where?

I understand you missed the "Suppose we are given an arbitrary event on the world-tube of an eternally-external observer". Regardless of its path, or whether the collapse has any symmetries.
(compactness is assumed...)

And it certainly does not mean that this invariant property is a valid reason to claim that the black hole can't form at all.

I claimed the simultaneity can precede formation. Big difference!

Please read posts more carefully before replying.

 

[Ibix]

Please read posts more carefully before replying.

I have to say I don't really understand the point you're making.

What you seem to be saying is that it is possible for an external observer to a collapse to construct a simultaneity criterion such that their "now" never includes the horizon. Am I understanding you correctly? Because if so, I don't think anybody is disputing this. What we are disputing is the OP's conclusion from such a coordinate system that the horizon never forms (or the black hole never grows). That's only true in the tautological sense that if you define "now" to exclude the horizon then it does not include the horizon.

 

[PeterDonis]

Ever occurred to you that charts like Eddington-Finkelstein can be matched smoothly with other charts outside the horizon?

You could do that, I suppose, but why would you need to? Eddington-Finkelstein covers all the way out to $r \to \infty$.

I explicitly opened with "Consider a case of a collapsing progenitor...".

Yes. What does that have to do with what I said?

I understand you missed the "Suppose we are given an arbitrary event on the world-tube of an eternally-external observer".

Yes, sorry, I missed that. Thanks for clarifying.

I claimed the simultaneity can precede formation.

You did, but not in the passage I quoted and responded to, that you were responding to. In that passage, I was responding to your statements about an invariant property of the spacetime--which has nothing to do with simultaneity conventions.

Big difference!

To whom? Maybe you didn't intend to claim that the invariant property of the spacetime you described implies that the black hole can't form at all. But you and I aren't the only ones posting in this thread. I wanted to make clear to the OP of the thread that nothing you're saying supports his claims. I wasn't saying that you intended to support his claims; I honestly can't figure out whether you did or not, but it doesn't matter either way to the point I was trying to make clear to the OP.

 

[PeterDonis]

Please read posts more carefully before replying.

Please be clearer about what point relevant to the thread topic you are actually trying to make. So far I'm still not sure what it is.

 

[martinbn]

The way I understood @JimWhoKnew point is that given an observer, who stays out of the hole, there are no events on the EH, which are in the observer's causal past. This is coordinate/chart independent and in daily language can be phrased as "the horizon never formed for that observer", because "formed" means that it has happened in the past.

 

[JimWhoKnew]

Please be clearer about what point relevant to the thread topic you are actually trying to make. So far I'm still not sure what it is.

Post #14 was written as a response to the quoted sentences:

But there are plenty of other simultaneity conditions available where an external observer agrees that the object fell in. @Dale named several in #7.

In #14 I devised a cunning plan to challenge this argument. Judging by the feedback, my explanation was not clear. Now I have something much simpler:

Suppose I hover over a Schwarzschild BH at constant $~(r,\theta,\phi)=(R,\pi/2,0)$ . $R/M~$ is a measurable quantity in principle. If I send a light pulse radially toward the hole, after what interval of proper time $\Delta\tau$ on my watch can I "agree" that the pulse must have crossed the horizon?

If I use the ingoing Eddington-Finkelstein coordinates, I get $~\Delta\tau=0$ . Makes sense?
If I'll use other charts, like Gullstrand-Painlevé, I expect a positive $~\Delta\tau$ . Considering all possible "ingoing" coordinate charts of this sort, for which the metric is non-singular at the horizon and matter/light can fall in in a finite coordinate time, is there an upper bound on $~\Delta\tau$ ?
(after which I can declare for certain that the pulse has crossed the horizon)

If there is no upper bound, the quoted argument is void.

Note: challenging arguments does not necessarily mean I support OP's claim. It means that I want the correcting arguments to be good and valid.

 

[Dale]

If there is no upper bound, the quoted argument is void.

Why? The existence of any finite $\Delta \tau$ is sufficient. @Ibix argument doesn’t rely on there being an upper bound.

 

[Ibix]

In #14 I devised a cunning plan to challenge this argument. Judging by the feedback, my explanation was not clear. Now I have something much simpler:

I don't think you were unclear. I just don't see that you are saying anything different from what I said. In a universe with a black hole you can pick a simultaneity condition where it exists or where it doesn't. The fact that you can pick one where the hole doesn't exist doesn't mean the hole doesn't exist at all, just that you've picked a bad map that excludes the hole.

This is a separate problem from verifying whether the spacetime you are in contains black holes or not. You can only confirm that it doesn't. It might contain compact objects, but if it contains true black holes you can't confirm it except by falling in.

 

[PeterDonis]

This is coordinate/chart independent

True.

and in daily language can be phrased as "the horizon never formed for that observer"

False. The proper phrasing in daily language is "the observer never sees light signals from the horizon formation". But it is a simple-minded logical error for that observer to conclude that because he didn't see something, it never happened.

 

[PeterDonis]

I want the correcting arguments to be good and valid.

Here's one simple good and valid argument: this...

all events on an EH are either future-like or spacelike relative to the world-tube of an eternally-external observer.

...already assumes that there is a black hole. A black hole is defined as a region of spacetime that is not in the causal past of future null infinity (and therefore is not in the causal past of an eternally-external observer). So if the statement of yours quoted just above is true, then a black hole exists by definition. Which means the statement of yours quoted above can't possibly show that a black hole doesn't exist.

More here:

https://www.physicsforums.com/insights/black-holes-really-exist/