The gravitational pull of a black hole depends on its mass. At a fixed distance from the center, it isn’t any stronger or weaker than that of a star with the same mass. The difference is that, since a black hole doesn’t have a surface, the gravitational pull can continue to increase as you approach the center.

The gravitational pull itself isn’t the problem, the problem is the change in the pull, the tidal force. It will stretch any extended object in a process with technical name “spaghettification.” That’s what will eventually kill you. Whether this happens before or after you cross the horizon depends, again, on the mass of the black hole. The larger the mass, the smaller the space-time curvature at the horizon, and the smaller the tidal force.Leaving aside lots of hot gas and swirling particles, you have good chances to survive crossing the horizon of a supermassive black hole, like that in the center of our galaxy. You would, however, probably be torn apart before crossing the horizon of a solar-mass black hole.

It takes you a finite time to reach the horizon of a black hole. For an outside observer however, you seem to be moving slower and slower and will never quite reach the black hole, due to the (technically infinitely large) gravitational redshift. If you take into account that black holes evaporate, it doesn’t quite take forever, and your friends will eventually see you vanishing. It might just take a few hundred billion years.

In an article that recently appeared on “Quick And Dirty Tips” (featured by SciAm), Everyday Einstein Sabrina Stierwalt explains:

“As you approach a black hole, you do not notice a change in time as you experience it, but from an outsider’s perspective, time appears to slow down and eventually crawl to a stop for you [...] So who is right? This discrepancy, and whose reality is ultimately correct, is a highly contested area of current physics research.”No, it isn’t. The two observers have different descriptions of the process of falling into a black hole because they both use different time coordinates. There is no contradiction between the conclusions they draw. The outside observer’s story is an infinitely stretched version of the infalling observer’s story, covering only the part before horizon crossing. Nobody contests this.

I suspect this confusion was caused by the idea of black hole complementarity. Which is indeed a highly contest area of current physics research. According to black hole complementarity the information that falls into a black hole both goes in and comes out. This is in contradiction with quantum mechanics which forbids making exact copies of a state. The idea of black hole complementarity is that nobody can ever make a measurement to document the forbidden copying and hence, it isn’t a real inconsistency. Making such measurements is typically impossible because the infalling observer only has a limited amount of time before hitting the singularity.

Black hole complementarity is actually a pretty philosophical idea.

Now, the black hole firewall issue points out that black hole complementarity is inconsistent. Even if you can’t measure that a copy has been made, pushing the infalling information in the outgoing radiation changes the vacuum state in the horizon vicinity to a state which is no longer empty: that’s the firewall.

Be that as it may, even in black hole complementarity the infalling observer still falls in, and crosses the horizon at a finite time.

The real question that drives much current research is how the information comes out of the black hole before it has completely evaporated. It’s a topic which has been discussed for more than 40 years now, and there is little sign that theorists will agree on a solution. And why would they? Leaving aside fluid analogies, there is no experimental evidence for what happens with black hole information, and there is hence no reason for theorists to converge on any one option.

The theory assessment in this research area is purely non-empirical, to use an expression by philosopher Richard Dawid. It’s why I think if we ever want to see progress on the foundations of physics we have to think very carefully about the non-empirical criteria that we use.

Anyway, the lesson here is: Everyday Einstein’s

*Quick and Dirty Tips*is not a recommended travel guide for black holes.

While I can kind of understand the effects of falling into one, I'm just curious as to exactly what is inside a blackhole. If the information it can contain is a function of its surface area as opposed to its volume then does that suggest that there is nothing inside or am I missing something?

ReplyDeleteCharles,

ReplyDeleteThat's an excellent question... Nobody knows what's inside a black hole. According to General Relativity there is vacuum and then in the center there's the singularity. However, this theory breaks down close by the singularity and quantum gravity should become important. So, to know what's at the center of a black hole, you need to know quantum gravity.

As long as you're inside the horizon but away from the center, it should be empty space. But even that is questioned by some. It might be that black holes actually do have long-range quantum effects which causes the classical (general relativistic) approximation to break down much earlier.

since more than 40 years ---> for more than 40 years

ReplyDeleteThis is a tricky one, since (different meaning of "since" here) "since" is usually "seit", but not always. :-|

While I'm at it, only slightly on-topic but very interesting: what do you think of

this recent paper? Like you, Paddy is interested in various aspects of gravitation and pursues a research programme someone off of the beaten path. He also knows ahugeamount about physics.Dear Bee,

ReplyDeleteI have a question on how the time your in-fall past the event horizon of a black hole takes from the perspective of an outside observer. Last September LIGO detected a signal from two merging black holes - a merger that we didnt observe as taking an infinite amount of time obviously. I am sure this has something to do with the relative masses involved and different scenarios for "test masses" that don't bend space time much themselves, and larger masses that do, but could you explain a little more? It would be fascinating to see some sort of plot of how the time dilation effect varies with the mass falling into the black hole.

Great post, thanks. Regarding the discrepancy -or not- between the predictions of the infalling and distand observers: you say there is none. But isn't the crossing of the horizon a discrepancy itself? For the distant observer you never cross it, while for the infalling you always do. They give different physical predictions and this seems a paradox to me. How can we verify -as being distant, not infalling observers- that particles do indeed cross the horizon without crossing it ourselves? And how do black holes increase their mass due to the infalling gas when the gas cannot enter the horizon in the first place for Earth's reference frame? Thank you for any feedback

ReplyDeleteBee,

ReplyDeleteOne thing (among others) I do not understand, is what happens to the atoms (or subatomic particles) of damned scientist falling towards BH, when the BH is small enough to evaporate and pop out of existence before the scientist's remnants reach the BH horizon (as seen from the outside observer).

Looking from outside, the falling object newer reaches horizon (if I understood correctly), and if the BH ceases to exist in finite time, the falling object, seen form outside, should be outside of horizon at the time the external viewer sees the BH to die.

Can you elaborate this a bit?

BR, -Topi

Phillip,

ReplyDeleteThanks, I fixed that...

Padmanabhan's papers seem forever to remain on my "to read" pile. I read some of his earlier work (on the zero point length and thermodynamics of gravity) but not this one. So really I can't say anything about it, I'll just add it to my list of good intentions ;)

Waterbergs,

ReplyDeleteThe signal that LIGO detects doesn't come directly from the horizon, but from the surrounding space-time. It is, basically, a deformation that continues to travel outwards. It carries the redshift (time-dilation) effects in it, but these are finite and not infinite.

Giannis,

ReplyDeleteNo, crossing the horizon is not a discrepancy. You write

"For the distant observer you never cross it, while for the infalling you always do. They give different physical predictions and this seems a paradox to me. "This is only paradoxical if you think that both observers have the same notion of time, but they don't. There are, in General Relativity, an infinite number of different times, and each observer has its own "internal" proper time. This is a different time for the observer who falls in an the one who is far away. You can convert one time into the other by a change of coordinates and it turns out that what is an infinite time for the distant observer is only a finite amount of time for the infalling observer. They both *do* describe the same situation, if you know how to properly compare their measurements. There is no internal disagreement in this.

The outside observer though lacks part of the story because he can't look inside the horizon. Best,

B.

Topi,

ReplyDeleteThere are two different aspects to your question, one is what happens to the particle (what's its worldline), the other question is what does the outside observer see.

As to the first question, the particle's worldline crosses the horizon, then inevitably hits whatever replaces the singularity if quantum gravitational effects are taken into account.

In general the answer to the second questions what the outside observer sees depends on how the black hole evaporates. If you have a case with a shrinking (apparent) horizon, the outside observer will eventually see the particle vanish and then later reappear. If the horizon remains light-like, the observer will probably at some point just see the horizon turn into a quantum gravitational thing, see the particle enter that, and then who-knows-what. Really it's a good question, but one that one can't answer without knowing the solution to the black hole information problem. Best,

B.

Hi Sabine,

ReplyDeleteThank you for your explanation. What you describe is exactly what happens in special relativity as well; a phenomenon takes place at different times for different observers. However, it *does take place*, eventually. In the black hole scenario, on the other hand, if we define the phenomenon to be the crossing of the event horizon, then i disagree with you and argue that we do have a discrepancy; namely, for different observers (distant or infalling) the "crossing of the horizon" may or may not happen no matter whether you look at a finite or infinite time.

I would have agreed with you if both observers can see the crossing of the horizon but at different times; then, yes, we would have the well-known relativity of when an event takes place, just like in special relativity. But here, the event simply does not take place at all for some observers, even for infinite times.

For the outside observer, you take an infinite amount of time to cross the horizon, but the black hole evaporates in a finite time. Can you clarify my mixup here?

ReplyDeleteGiannis,

ReplyDeleteThe event does happen, in the only meaningful definition of the term: it is a location in space-time that is not a boundary and the worldline of the observer does not end there.

See, you can always take space-time and introduce strange time-coordinates that go to infinity somewhere and hence cover only parts of the complete space. This is possible, but physically entirely meaningless. You can indeed do this already in flat space-time, for example by using the Rindler coordinate system. This coordinate system does not cover all of space-time and has a coordinate that goes to infinity for what is, in "normal" (Minkowski) coordinates a perfectly ordinary time.

Or to give you a simple example, suppose you fancy the idea of introducing a new time coordinate which is tan(t), where t is my normal time. General Relativity rests on the freedom to chose your coordinate system and there is nothing that can prevent you from doing this. You'd then come to conclude that it takes an infinite amount of time for me to reach any age older than Pi/2. Does this have any physical relevance? No, because all you have done is using a coordinate that covers only part of the full space.

Let me assure you that this is entirely uncontroversial. There is nothing whatsoever paradoxical about coordinates covering only part of space-time, it frequently happens in General Relativity. Schwarzschild coordinates for example also only cover part of the static black hole's metric. That any value of a coordinate goes to infinity does not mean that space-time "ends" there.

Best,

B.

Bee,

ReplyDeleteIf an elementary particle falls towards horizon, and can not reach it as seen from the external viewer (when the BH is not evaporating, or expanding). From the external viewer it looks like the BH does not increase in mass/size, as all the infalling matter nearly freezes before reaching horizon.

Could it be seen from external point of view by the following way?

When the particle has reached distance from horizon, where the combined mass of original BH plus the mass of the particle justifies to "new" BH with horizon radius such that the particle is already inside this new horizon. Thus two smaller junks of matter (whether previously BH or not) combine and form a (new) BH with radius to cover both junks.

BR, -Topi

CIP,

ReplyDeleteIf the black hole evaporates it does not take an infinite amount of time, it's as simple as that. The black hole is gone at a finite time, and by that time, at the very latest, everyone has been seen falling in.

There is one case where this isn't so, which is the case in which the black hole connects two disconnected asymptotic regions (future infinity) of space-time (in the other region you have a white hole). In this case however, it also takes an infinite amount of time for the black hole to evaporate. This is admittedly a somewhat contrived scenario, and I don't actually know anyone who works with it. Best,

B.

Topi,

ReplyDeleteRoughly speaking, yes, I think you can think of it this way. This isn't really well understood, how a black hole "eats up" small chunks of matter. It's an interesting question, but I don't know of a good reference for this. The reason it's difficult is that this scenario deviates from the spherical (or at least axial) symmetry that is normally assumed. I think the picture is that the horizon somewhat dents outwards at the point where the mass approaches it, then eats up the mass, rings down, and becomes spherical again at a slightly bigger radius. Best,

B.

Hi Sabine,

ReplyDeleteI guess all this was written before WW2 (not the parts on BH evaporation or information paradox), I also read (maybe 15-20 years ago) that an equivalent theory exist where c is a relative constant. Meaning that it is locally observed as constant. Is that right? If so, what would a BH be?

Thanks

J.

Thanks Sabine,

ReplyDeleteYour reply is very explanatory. I like your intuitive examples, but they make feel as the problem is just a matter of re-labeling, hence, just a mathematical artifact, like your Pi/2 example. As you say,

"Does this have any physical relevance? No, because all you have done is using a coordinate that covers only part of the full space."

But the situation where observer is sitting far away from a black hole and never seeing an object crossing the horizon is something physical and it has nothing to do with what coordinates i am going to describe it with.

The fact that the external observer cannot give a coordinate description of the object inside a black hole is very strange and counter-intuitive to me. It seems to imply that "what can be observed" by an observer in space-time is in one-to-one mathematical correspondence to whether the coordinate system that is used can be extended in the region the observer is interested in. E.g., the external observer cannot assign a coordinate system that covers the internal of the black hole, as you said, and this mathematical impossibility of coordinate-assignment coincides exactly with the fact that the internal is fundamentally unaccesible by all physical means.

Notice that this could not be necessarily the case. What i would intuitively expect from a theory would be to give coordinates for all space-time, no matter whether some regions can be physically observed from the standpoint of an observer or not.

Let me give you an example; in the case of the external observer standing outside of the black hole, i would `reasonably' expect that the observer can mathematically assign a coordinate system to the whole space -including the inside and the outside of the black hole- and therefore she could assign a trajectory of the in-falling particle until it reaches the singularity. Now, does it mean that just because a continuous trajectory can be assigned (from the outside to the inside of the black hole) therefore the internal trajectory could be observable by the external observer? No! Just because you can assign a trajectory, it does not necessarily mean that you can observe some parts of it by using physical systems. For example, for the inside part of the trajectory, since photons cannot escape the black hole then the in-falling system, that emits them, would be unobservable. Also, since the photon wavelength approaches infinity when the object approaches the horizon we could argue that even the horizon-crossing would be unobservable. *But*, mathematically, even though we cannot physically observe the system, we *know* where in space-time the system is located because we have its trajectory written down.

Sorry for the long post. My point is that different observers see a fundamentally different physical universe that is independent of any mathematical artifacts of the theory and how they label their coordinates. For the external observer the object never crosses the horizon *at all times* finite or (infinitely) infinite. The impossibility of assigning a single coordinate system in GR seems to have some deep physical meaning which i don't understand at the moment.

@Phillip Helbig

ReplyDeleteI stumbled upon Thanu Padmanabhan's work some years ago when I found him quoted in Erik Verlinde's paper on gravity. Since then I've been reading up on his work regularly. Hugely interesting work.He is very productive and progresses steadily with increments. It seems it would need combining with other current angles of approach.He's also very correct in his papers about what he has and has not answered yet. A true professional.

This blog is the best blog I have ever came across. I stumbled upon your site via google searching for topics about reality & computer simulation. This site is incredibly informative and inspiring! I love your writing style and your knowledge on various subjects. I really appreciate your book reviews as well. Keep up the great work and please don't stop. :-)

ReplyDeleteGiannis,

ReplyDeleteThere *are* coordinate systems that cover the inside of the black hole. In fact there are infinitely many of them. It's just that the coordinate system in which the infalling time is infinite isn't one of them.

That there are regions of space-time which the observer in the infinite distance cannot see is exactly what it means for space-time to have an event horizon. I understand that this may be unintuitive. But there is nothing logically inconsistent about it. Best,

B.

Thank you Sabine.

ReplyDeleteI am confused. In the original post you said that in the external frame it takes an infinite time to fall into a non-evaporating BH but in your second reply to Topi it seems that u argue that it does only take a finite time, after all (even without evaporation). Can u clear this up, please?

ReplyDeleteMaurice,

ReplyDeleteNot sure what you mean. It takes an infinite time to fall in in the coordinate-time of the asymptotic observer if the black hole does not evaporate. It takes a finite time in the proper time of the infalling observer. It takes a finite time in both frames if the black hole does evaporate. This is the test-particle approximation (Geodesics, no backreaction). What happens if you take into account that the mass of the particle changes the background I don't know, and I'm not sure it is known. Best,

B.

Hi Bee,

ReplyDeleteExcellent summary.But all these scenarios are for static Schwarzschild BH. I understand, most of the BH are rotating, Kerr type where there are multiple horizons, branch sheets etc. Do they change the scenarios of a falling observer in an essential way? Would you discuss these?

Dear Sabine,

ReplyDeleteThank you very much for this blog.

I've been teaching General Relativity at my university this semester and it's been a huge adventure (at least for me, can't say the same for my students).

Excellent post.

best wishes

Marcelo

kashyap,

ReplyDeleteThe evaporating case isn't static, but yes, I only referred to the non-rotating case. In the rotating space-time there are more cases to distinguish because of the ergosphere and the different causal structure. But the main conclusions still remain the same: there's no paradox in using different time coordinates, you'll die when you fall in, nobody knows what happens at the singularity, and the black hole information loss problem is still unsolved. Best,

B.

> The black hole is gone at a finite time, and by that time, at >the very latest, everyone has been seen falling in.

ReplyDeleteWhy is that true in the test-particle limit?

> What happens if you take into account that the mass of the >particle changes the background I don't know, and I'm not >sure it is known.

The LIGO theorists have surely calculated what happens when a neutron stars falls into a BH of similar mass up to the time the NS crossed the horizon in the external frame, don't u think?

If observer A lives forever and is outside the black hole, while observer B takes an infinite amount of time to fall in the black hole from observer A's perspective, how can observer A also see the black hole evaporate, or do they? From observer A's perspective it seems like there are two different futures to view. Which one is right?

ReplyDeleteJust because it's a neat idea: I recently noticed this paper on arXiv talking about how tiny, little black holes could be much, much bigger on the inside.

ReplyDeletehttp://arxiv.org/abs/1604.07222

Maurice,

ReplyDeleteA geodesic can't just end at some point in space-time that's not a boundary. If it doesn't end at some point, the proper time to reach that point is finite. Consequently if the black hole evaporates, it can only take a finite proper time to fall in.

Yes there are almost certainly numerical studies of the case where both masses are comparable, but I'm not familiar with the literature on numerical GR. Best,

B.

Bee,

ReplyDeleteIf the outside observer pings (send photons to be reflected back, with a radar for example) the falling object, how does the observer see the measured distance and doppler shift?

Br, -Topi

ReplyDelete"I stumbled upon Thanu Padmanabhan's work some years ago when I found him quoted in Erik Verlinde's paper on gravity. Since then I've been reading up on his work regularly. Hugely interesting work.He is very productive and progresses steadily with increments. It seems it would need combining with other current angles of approach.He's also very correct in his papers about what he has and has not answered yet. A true professional."Like John Barrow and Carlo Rovelli, he manages to combine productive research with writing (semi-)popular books and being an all-around nice guy. I recommend one of his semi-popular book to everyone interested in physics, but keep in mind that it is not easy going (but still worth it), as I point out in my review of

Sleeping Beauties in Theoretical Physics. (The link here is to a page which contains a link to the proof of the review. It contains one slip-up which was corrected in the published version. I'll leave it to readers here to spot it.)> If [a geodesic] doesn't end at some point, the proper time to >reach that point is finite.

ReplyDeleteSorry, this sentence makes no sense to me. There is no endpoint but the time to reach it is finite?? I am really no expert on this but really, if it takes an infinite time for a massless test-particle to fall into a non-evaporating BH of arbitrary mass, why should it take a finite time to fall into a BH whose mass is decreasing? Can u give us a reference for a detailed discussion?

Maurice,

ReplyDelete"Sorry, this sentence makes no sense to me. There is no endpoint but the time to reach it is finite??"If the geodesic doesn't end at point X, the proper time to reach point X is finite. Does that make more sense?

" but really, if it takes an infinite time for a massless test-particle to fall into a non-evaporating BH of arbitrary mass"This statement is correct only in a particular coordinate system.

"why should it take a finite time to fall into a BH whose mass is decreasing? Can u give us a reference for a detailed discussion?In the proper time of the infalling observer it takes a finite time either way. If the black hole is evaporating in a finite time, it takes a finite time to fall in because by the time the black hole has evaporated you must surely have fallen in. No, I can't give you a reference for this, sorry, but look at the causal diagram and it will become apparent, you can eg use the ones in this paper.

The asymptotic time goes to infinty at the upper end of the boundary labeled with I^+. If you look at Fig one, which shows the non-evaporating case, that's directly at the horizon. Hence, for the asymptotic observer it looks like it takes an infinite amount of time to fall in.

Now the scenarios with complete evaporation, Fig 2 and 3. Whatever time it is at which the black hole is gone is (by assumption) finite. Consequently any geodesic that hits the horizon (or apparent horizon) must have hit it at a finite (asymptotic time). Best,

B.

Once I have crossed the event horizon, nothing (including light) can increase its distance to the singularity anymore, right?

ReplyDeleteWould I then see the wall of my spaceship in the direction of the singularity?

If yes, how can this light reach my eyes without increasing its distance from the singularity?

If no, how can I still live on (for a short while)? I'd imagine, for example, that signal transfer in nerve cells relies in part on processes involving photons. So signals from my lower brain couldn't reach my upper brain anymore?

Hello,

ReplyDeleteAwesome Blog and your knowledge on this is fantastic.

However, the time coordinate mapping problem, leading to discrepancies in black hole event horizon between the far away observer vs the falling in observer problem... I understand it is NOT controversial in the study of General Relativity because all we can see and measure (with) is based on the speed of light.

Just a crazy question though. It's probably impossible to actually do this... but what if there is quantum entanglement mixed into the equation? Of course, there's the tricky problem that you can never send meaningful information through this, but it doesn't dispute that suddenly we have a means and mapping system, that is

faster than lightand could in theory reveal some controversies that some of the commentators are trying to say.Essentially, with such a "link" between the falling in and far away observer -- the definition of "now" changes. In fact, if the theories are to be believed, it "crosses time". It might be an absolutist "now" rather than the relative time-flow based "now" we experience. What I mean is, if time is a flowing river and we define "now" as any particular wave of water moving through. Then a more absolute definition of "now" would be observed from "outside" the river, and freeze frame the whole thing, like taking an instantaneous photo. That is much more absolute, in comparison to the river's point of reference, because it can see entire stretch and define a more concrete "instant" than otherwise. Essentially, with the tangled particles, there is now a more "faster" definition of instantly, which could definitely create issues trying to describe the black hole scenario.

I'm sorry if I'm not making sense and spewing garbage. I'm not from this field and I'm truly just an enthusiast and amateur.

'Messier and messier' said Alice

ReplyDeleteI am confused by the BH evaporation process itself. I can understand matter going in and making the BH bigger but I don't understand the BH getting smaller if no matter is able to exit.

ReplyDeletePut another way, while the outside observer sees an evaporating BH what does would an inside observer see? I am expecting an inside observer would not see any change other than the outside universe fading away such that when the outside observers see the BH finally disappear, the inside observer would see the outside universe completely disappear, but the inside of the BH would still continue to exist as it had before.

It is expected [York, 1983] that the quantum gravitational fluctuations can lead to the quantum “trembling” of the event horizon. Then the formally infinite expression for the duration of the fall of a body into a black hole must be replaced with a finite quantity.

ReplyDeleteIvan,

ReplyDeleteYes, one shouldn't take the exact position of the horizon too seriously.

Michael,

ReplyDeleteThe infalling observer sees thermal particles. If you fall into a black hole you can still receive signals (or images) from the outside.

Hon Chong,

ReplyDeleteThe particles in the Hawking radiation *are* entangled across the horizon. This is the origin of the black hole information loss problem. As you say, this doesn't allow one to submit information though. (Except for one bit). Best,

B.

Matthias,

ReplyDeleteThe redshift, or the relative slowing down of time, is a global effect (ie over long distances), you don't notice it locally (over short distances). You wouldn't notice anything unusual until the tidal forces become large. The easiest way to see this is that it would violate the equivalence principle, which can't happen in GR by construction. Best,

B.

So I've thought about this for a couple days now and read the comments but this is still a puzzle to me: If my "friends will eventually see [me] vanishing" only upon evaporation of the black hole, wouldn't they see all other matter that ever crossed the horizon (from its own perspective) as well? Or does "see" in that context mean "we know it's there, frozen in time for the outside observer because of the infinite gravitational red-shift but we can't actually see it with our eyes" (because it's red-shifted so much that the hole appears black)? Or what is it I'm not getting here?

ReplyDeleteHerr Weh,

ReplyDeleteYes, that's right, your friends will not only see you but everything else that ever got close to the horizon. (Depending on the direction.) And, yes, "seeing" here doesn't actually mean they might see it with their own eyes, they would need to use very sensitive instruments. In practice, you will only "see" anything approaching the horizon up to a finite distance which is where the sensitivity limits of your IR detector are. Best,

B.

Thanks for your answer! So if in my spaceship inside the event horizon light goes in all directions, it can (locally) increase its distance from the singularity?

ReplyDeleteMatthias,

ReplyDeleteNo, the light cannot increase the distance to the singularity. All the light rays are eventually bent around and end in the singularity. It's just that locally you wouldn't be able to tell.

Hi again Matthias,

ReplyDeleteYou might find this image useful from this website. You always move within the cones.

> If the geodesic doesn't end at point X, the proper time to reach point X is finite. Does that make more sense?

ReplyDeleteAt least it makes logical sense. But where is "point X"? But I now realized that your claim:

>If you take into account that black holes evaporate, it doesn’t quite take forever, and your friends will eventually see you vanishing.

is certainly wrong in this generality: you can always find

a start position from which even massive objects will not reach

the horizon before the BH completely evaporates. Take

a proton a meter from a microscopic BH that evaporates

within a nsec. So there seem to be two errors in your post:

1. the claim that massive objects ("you") take forever to reach the horizon in the external frame (true only for massless test particles).

2. the claim that in general objects do reach the horizon in finite time if the BH evaporates.

Hi Sabine,

ReplyDeletethanks a _lot_ for taking the time to find this link for me! I get it now. Great blog - and even greater to get such answers :-)

In the case of Schwarzschild black holes, the situation with coordinate time and proper time is simple and straightforward; however, I am wondering what happens in other types of spacetimes, for example Vaidya-Bonnet. Is the coordinate in-fall time here still infinite ? Are there ANY spacetimes where the coordinate time is finite and well defined ? After all, I wouldn't expect to actually find Schwarzschild-type black holes in the real universe.

ReplyDeleteMaurice,

ReplyDeleteI have throughout this thread only referred to observers, hence not to massless particles. If you are trying to say that not all timelike geodesics reach the horizon to begin with, that is true: Luckily it is indeed possible to forever stay outside the black hole. Needless to say, I was only referring to curves that cross the horizon to begin with.

Wandering Wolf,

ReplyDeleteWhile it is difficult to actually compute the time in general backgrounds, the question whether the infalling time is finite or not depends only on the causal structure and hence carries over to collapse scenarios with asymptotically flat regions, provided the collapse ends at some point. (Infinitely expanded collapsing matter distributions can do all kind of weird things.)

No one talks much about what the infalling observer sees outwardly. In that infalling signals or photons would be seen or detected by the infalling observer would there be the impression that things are speeding up outside the influence of the black hole? An example might be seeing the planets of a solar system moving in their orbits. That is if an observer or probe survived a while before being shredded or spaghettified, probably way before it hit the so called singularity. Since the observer or probe has moved radially inward would things ahead of him/her be red shifted also such that the event horizon is in a sense a future ocurrence which is always going to be smaller and smaller and therefore hotter and hotter?

ReplyDeleteHi Bee,

ReplyDeleteabout the "before crossing the horizon of a solar-mass black hole". We don't really have any of those, of course, as you appear to need at least 2.5-3.5 solar masses to make a black hole. Otherwise, you become a neutron star (or worse). Unless, of course, you are willing to wait for a long time :-)

Chris,

ReplyDeleteI'm a theorist, I don't bother with factors of order one.

Sabine, Sorry for the OT comment. But once you get a chance would be interested in your review of Janna Levin's latest book.

ReplyDeleteThinking whether to order it or not.

Dear Bee,

ReplyDeletethank you for making that clear to me. Turns out that reading 'Black Holes and Time Warps' (twice!) did after all leave me with at least some comprehension. And following your blog of course, which is really remarkable and certainly one of the best out there.

-w.

Shantanu,

ReplyDeleteMy husband read the book and tells me it's very good. Alas, I will not have time to read it any time soon, sorry. (I'll have a review of Sean Carroll's new book at some point though.)

Hi,

ReplyDeleteI'm very late to this party, but perhaps you'll still see my comment.

You write:

"It takes you a finite time to reach the horizon of a black hole. For an outside observer however, you seem to be moving slower and slower and will never quite reach the black hole, due to the (technically infinitely large) gravitational redshift. If you take into account that black holes evaporate, it doesn’t quite take forever, and your friends will eventually see you vanishing. "

I'm a bit confused by this: If the BH has a finite life time and a remote Schwarzschild observer sees you still frozen in at the end of this lifetime (albeit incredibly red-shifted), how do you ever cross the event horizon? Wouldn't you just fall towards it, but even in your time, before you reach it, the event horizon is gone because the BHs lifetime ends?

If the black hole evaporates, it doesn't have an event horizon. It has an apparent horizon. And yes, you will actually cross it.

ReplyDeleteThanks, but I'm afraid I'm still puzzled: What does the external observer see? She sees me freezing at the (apparent) event horizon, never crossing it (and taking infinitely long for that), whereas I actually do cross it, but this can never be seen from the outside? If I were to send out light signals while falling, what would their worldlines look like so that they never reach the outside? I think I understand hoe it works for an eternal BH, but for an evaporating BH, I don't really grasp it.

ReplyDeleteIs there a spacetime/Penrose diagram or something you could point me to that shows how different observers see this?

If the black hole entirely evaporates, the outside observer will eventually see you crossing (just before she sees you coming out again). The causal diagram is Figure 3 of this paper.

ReplyDeleteThanks a lot; I'll need to brush up my Penrose-diagram reading skills a bit to fully understand the figure...

ReplyDeletePS: I really like your blog, BTW