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wendisf

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- #1

wendisf

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Just because you can't see something cross the horizon doesn't mean it doesn't cross the horizon. Locally (at the EH) there is nothing unusual going on at all; stuff just crosses the EH without even noticing that there is such a thing.

- #3

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But locally at the EH, wouldn't time be dilated to where a hypothetical observer would see the end of the universe (as we know it) in a few seconds on their clock? It sounds like you're saying that even though co-moving observers will never see anything cross a BH's EH, they can still see the effect of material having done so via an increase in its mass?Just because you can't see something cross the horizon doesn't mean it doesn't cross the horizon. Locally (at the EH) there is nothing unusual going on at all; stuff just crosses the EH without even noticing that there is such a thing.

- #4

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The effect of infalling mass for a remote observer happens LONG before the actual infall, so the infall itself is irrelevant as far as that is concerned. You speak of time dilation in a confusing way. Locally there IS no time dilation; it's something only seen by a remote observer. Red shift kills effective observation long before the BH starts to evaporate from Hawking Radiation, to say nothing of long before the end of the universe.But locally at the EH, wouldn't time be dilated to where a hypothetical observer would see the end of the universe (as we know it) in a few seconds on their clock? It sounds like you're saying that even though co-moving observers will never see anything cross a BH's EH, they can still see the effect of material having done so via an increase in its mass?

If there WERE an infinitely-long-lived remote observer, he would see things redshifted out of his ability to see them and then after a staggeringly long time (10^80 years or so) he would see the BH start to get smaller due to the radiation

- #5

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I understand that time dilation is never experienced locally, just as time contraction isn't. But in the case of contraction (since gravitational time dilation isn't symmetric as with velocity) the EH observer is the remote observer. I also understand that it's impossible (given our current tool set) to observe the current known universe from that vantage. But hypothetically...The effect of infalling mass for a remote observer happens LONG before the actual infall, so the infall itself is irrelevant as far as that is concerned. You speak of time dilation in a confusing way. Locally there IS no time dilation; it's something only seen by a remote observer. Red shift kills effective observation long before the BH starts to evaporate from Hawking Radiation, to say nothing of long before the end of the universe.

If there WERE an infinitely-long-lived remote observer, he would see things redshifted out of his ability to see them and then after a staggeringly long time (10^80 years or so) he would see the BH start to get smaller due to the radiation

- #6

George Jones

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http://www.physics.uoguelph.ca/poisson/research/agr.pdf;

see Firgure 5.7 on page 134 (pdf page 150).

Radiation falls into a black hole from v1 to v2, but the left diagram of Figure 5.7 shows that the event horizon (EH) starts to grow before the first radiation crosses the event horizon.

- #7

wendisf

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That is not known. The formulation of supermassive BH's is still a bit of a mystery.Interesting discussion, but I'm still confused. Let me re-phrase my conundrum. I will assume that the articles I've read are correct... the super-massive BH in the center of the Milky Way started out at a typical stellar mass, and acquired its "super massiveness" by "swallowing" everything in its neighborhood.

What paradox? Again, the fact that you can't observe something crossing the EH is utterly irrelevant to what's actually going on. Also, for a feeding BH, you would observe the EH grow.So, had I been alive a few billion years ago, I would have observed a stellar mass BH in the center of the galaxy. If I were also immortal, I would have had the privilege of watching this BH for the past few billion years, all the way to the present. Hence, I would have observed this BH growing to its current multi-million stellar mass size. Yet, over all of those billions of years, I would never observe any mass actually crossing the EH. To me, that suggests that the BH should be the same mass today as it was billions of years ago. But, obviously, it is not. Can someone explain this paradox?

- #9

Nugatory

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No. The infalling observer reaches the singularity and is destroyed very quickly. He sees only the light that crosses the horizon almost immediately after he does and that catches up to him before he reaches the singularity.But locally at the EH, wouldn't time be dilated to where a hypothetical observer would see the end of the universe (as we know it) in a few seconds on their clock?

By "co-moving" do you mean distant hovering observers? If so, the answer is yes. Do remember that the Schwarzschild solution describes a static black hole, one whose mass does not change - and that's not the case when you drop an object of non-negligible mass into the black hole. At a very hand-waving level, you cannot trust the Schwarzschild solution with its infinite time dilation arbitrarily close to the horizon when the infalling object is at or below Schwarzschild ##r## coordinate ##R'## where ##R'## is the Schwarzchild radius of a mass equal to the original mass of the black hole plus the infalling mass.It sounds like you're saying that even though co-moving observers will never see anything cross a BH's EH, they can still see the effect of material having done so via an increase in its mass?

- #10

russ_watters

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I think the OP's confusion is different from what you are addressing: it seems the "never see the infaller cross the event horizon" thing is for a black hole of fixed size. The question is, if the black hole grows, does the infaller disappear?Just because you can't see something cross the horizon doesn't mean it doesn't cross the horizon. Locally (at the EH) there is nothing unusual going on at all; stuff just crosses the EH without even noticing that there is such a thing.

- #11

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Oh? It didn't see that way to me. I think he was just asking about the mass falling into the BH. @wendisf , what say you?I think the OP's confusion is different from what you are addressing: it seems the "never see the infaller cross the event horizon" thing is for a black hole of fixed size. The question is, if the black hole grows, does the infaller disappear?

- #12

russ_watters

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Given that black hole growth is in the title, I think it is critical to his question...Oh? It didn't see that way to me. I think he was just asking about the mass falling into the BH. @wendisf , what say you?

....actually, I think it is both sides of the same coin; how do they grow and how can an insfaller not be observed to cross if it grows.

- #13

George Jones

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I thinkhewas just asking about the mass falling into the BH.

hmm

- #14

wendisf

- #15

Nugatory

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I very highly recommend this paper: https://arxiv.org/abs/0804.3619But locally at the EH, wouldn't time be dilated to where a hypothetical observer would see the end of the universe (as we know it) in a few seconds on their clock?

Even if you find the math of the coordinate transformation to be a bit overwhelming, you will be able to get a qualitative understanding of Kruskal cordinates. You cannot use Schwarzschild coordinates at and around the event horizon because those coordinates are singular at the horizon.

- #16

Nugatory

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This apparent paradox is a result of misapplying the Schwarzschild solution, which is only valid if the infalling mass is negligible compared to the mass of the black hole so that the size of the horizon doesn't change (and note that even in that situation, Schwarzschild coordinates don't work at the horizon - you have to use some other coordinates that are not singular at the horizon to describe the Schwarzschild spacetime there).

Back up for a moment and 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 to describe what's happening in between.

- #17

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The manifold has 4 dimensions, so it can be regarded as a 4-dimensional geometric object. We can and do make a general claim that three of these four dimensions of the manifold representing space, and one of them represents times.

The problem, I think, consists of applying this mathematical 4d model in understandable terms. When put into popular language, especially, there are some false pictures that block a proper understanding, most specifically the tendency to attempt to carry over the Newtonian idea of absolute time into General relativity, when this notion of time won't even work in special relativity.

Trying to interpret this 4d mathematical model (which works fine) into a model with an absolute model of time and 3 dimensions of space just doesn't work, and will never work. But if the only model one has of time is absolute time, saying that this model doesn't work isn't necessarily very helpful. Unfortunately, it's not clear just what is more helpful, perhaps this really is as helpful as one can be. The only route I see for progress is going back to special relativity, and understanding the nature of time in SR and how and why time in SR is unified with space into an entity we call "space-time".

This tends to be too much of a digression, from the original question.

I think that talking about "the block universe" MAY be more helpful than talking about "4 dimensional manifolds", but I'm not really sure. The point is that a consistent model exists, but this consistent model may not fit into the mold that a reader wants it to fit into.

- #18

wendisf

- #19

Ibix

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I think @Nugatory's point is that the idea that you never see anything crossing the event horizon comes from the Schwarzschild spacetime. But the Schwarzschild spacetime is (strictly speaking) only a description of a black hole in an otherwise empty universe. It's a decent approximation for a black hole with negligible mass (like a rocket) outside it (*Edit: or if there are any large masses they're far enough away that we can ignore them *). But it's not an accurate description of a spacetime where there is any significant mass outside the hole. So it doesn't describe a growing black hole.

Nugatory's example is showing you the contradiction in trying to imagine significant mass in a Schwarzschild spacetime. It "ought" to be able to fall to R' in finite time, but it also "ought not" to. The solution is to use a more complicated spacetime that describes significant mass outside the black hole. Presumably matter can cross the horizon in finite time in such a spacetime, but I'm not sure if we have an analytical description of such a thing. If we do, I don't know it.

Nugatory's example is showing you the contradiction in trying to imagine significant mass in a Schwarzschild spacetime. It "ought" to be able to fall to R' in finite time, but it also "ought not" to. The solution is to use a more complicated spacetime that describes significant mass outside the black hole. Presumably matter can cross the horizon in finite time in such a spacetime, but I'm not sure if we have an analytical description of such a thing. If we do, I don't know it.

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- #20

PeterDonis

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Presumably matter can cross the horizon in finite time in such a spacetime, but I'm not sure if we have an analytical description of such a thing.

The ingoing Vaidya metric describes null dust (basically radially infalling light rays) falling into a growing black hole. It is an exact solution. See here:

https://en.wikipedia.org/wiki/Vaidya_metric#Ingoing_Vaidya_with_pure_absorbing_field

I don't know if there is a corresponding exact solution for, say, ordinary dust (i.e., matter moving on radially ingoing timelike worldlines).

As far as matter "crossing the horizon in a finite time", it's a bit more complicated than that. Schwarzschild spacetime has a timelike Killing vector field (more precisely, it's timelike outside the horizon), so there is a natural notion of "time" associated with that Killing vector field (which basically corresponds to "time according to an observer at infinity"); it's according to that notion of "time" that matter cannot cross the horizon in "finite time". But the Vaidya metric does not have a timelike Killing vector field (which is to be expected since the black hole is growing, so its mass is not constant), so there is no natural notion of "time" according to which we can assess the "time" it takes for infalling matter to cross the horizon (apart from the proper time along the infalling matter's worldline, which is always finite, in both the Schwarzschild and the Vaidya case). So I would not say that "matter can cross the horizon in finite time" in the Vaidya metric; I would say that the notion of "the time it takes for matter to cross the horizon" you are trying to use is not well-defined, since the spacetime is not static. I would expect similar remarks to apply to any spacetime describing a non-static black hole.

- #21

Ibix

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Does this apply to Nugatory's infalling thin shell, though? The Vaidya metric has all of space filled with radiation, I think. But outside Nugatory's shell (sorry that makes you sound like a mollusc, Nugatory) we have a Schwarzschild metric, so there's a timelike Killing vector field there. So it seems at least possible that an observer outside the hole will have a sensible definition of time to use if the matter falls in.I would expect similar remarks to apply to any spacetime describing a non-static black hole.

- #22

PeterDonis

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Does this apply to Nugatory's infalling thin shell, though?

Yes, because the timelike KVF on the past side of the shell is different from the timelike KVF on the future side. So the notions of "external time" are different in the two regions, and in the region actually occupied by the shell, there is no well-defined notion of "external time" at all. In short, there is no single notion of "external time" that you can use to analyze the infall of the shell.

outside Nugatory's shell (sorry that makes you sound like a mollusc, Nugatory) we have a Schwarzschild metric, so there's a timelike Killing vector field there. So it seems at least possible that an observer outside the hole will have a sensible definition of time to use if the matter falls in.

Yes, for matter (of negligible mass compared to the hole) that falls in

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Just because you can't see something cross the horizon doesn't mean it doesn't cross the horizon...

Well, it means that from point of view of some observers it does not cross the horizon ever... yet, you made the assertion "something cross the horizon" without the reference to the particular point of view it happens; it implies that such fact is not relative but "absolute" what is apparently a false assertion. I may rephrase your statement "Just because you can't see something cross the horizon doesn't mean that another observer may in fact witness it (the one who falls in BH)..."

I would like to elaborate the question the author of this thread has asked - the measurement of BH mass happens not by one who crosses the horizon but by observer who has no chance to observe such event so BH mass for such outside observer should remain zero... Well, the solution is apparent - the outside observer measures BH mass by its gravitational effect on the stars around what does not require that falling matter into BH must cross the horizon.... thus, the mass of BH for outside observer is a composition of all masses falling into BH....

- #24

PeterDonis

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it means that from point of view of some observers it does not cross the horizon ever

No, it doesn't. It means that from the point of view of some observers, the event of the object crossing the horizon is not visible. That's not the same as saying "it does not cross the horizon ever", because there is no guarantee that the entire spacetime will be visible to a given observer.

you made the assertion "something cross the horizon" without the reference to the particular point of view it happens; it implies that such fact is not relative but "absolute" what is apparently a false assertion

No, it isn't. The worldline of the object and the horizon itself are both invariant, well-defined geometric objects--a curve and a surface--and whether they intersect is also an invariant geometric fact. So it is absolute, not relative. The distant observer simply cannot receive light signals from the event of the intersection.

- #25

PeterDonis

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the mass of BH for outside observer is a composition of all masses falling into BH....

More precisely, the "mass" measured by a given observer, heuristically, includes all objects that are closer to the "source of gravity" (in this case the BH) than the observer is. (Note that this only works in an asymptotically flat spacetime, where the notion of "closer to the source of gravity" has a well-defined meaning.)

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- #27

PeterDonis

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such notion as "entire spacetime" is what I may not comprehend

You're going to have a really tough time with GR (and arguably should not be posting in an "I" level thread on it) if you can't comprehend that notion. Do you understand what "spacetime" means in the simpler context of SR?

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And measurements relative to your remote observer of events up to a point just outside the EH of a BH are as far as he can go. The rest of spacetime inside that sphere are not visible to him.

- #29

PAllen

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Yes, but all of physics involves putting together many observations and extrapolations into a coherent whole. We can model what someone on mars would have seen in the sky a thousand years ago. Of course, we can never verify this, but we routinely assume "something similar to that" is part of the past of the universe. Similarly, all observers observations (modeled as needed) relating to BH can be put into one coherent whole. There are no alternate versions of what happens to infalling matter. It is just that an infaller can continue getting signals from me at and beyond horizon crossing, while their replies never reach me. It is no harder a concept than that the past can pass information to the future but not vice versa (assuming no wormhole, or the like).

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You're going to have a really tough time with GR (and arguably should not be posting in an "I" level thread on it) if you can't comprehend that notion. Do you understand what "spacetime" means in the simpler context of SR?

It was a rhetorical figure to make the point that "entire spacetime" (not a "spacetime") is the way to imply that even if the information about "event" can not be acquired by any means still may belong to realm of observer (in this case you ;o)...

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