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The event horizon of a black hole is where gravity is such that not even light can escape. This is also the point I understand that according to Einstein time dilation will be infinite for a far-away-observer.

If this is the case how can anything ever fall into a black hole. In my thought experiment I am in a spaceship with a powerful telescope that can detect light at a wide range of wavelengths. I have it focused on the black hole and watch as a large rock approaches the event horizon.

Am I correct in saying that from my far-away-position the rock would freeze outside the event horizon and would never pass it? If this is the case how can a black hole ever consume any material, let alone grow to millions of solar masses. If I was able to train the telescope onto the black hole for millions of years would I still see the rock at the edge of the event horizon?

I am getting ready for the response of the object would slowly fade. Why would it slowly fade and if it would how long would this fading take? If it is going to red shift at some point would the red shifting not slow down to a standstill? This question has been bugging me for years!

OK - just an edit based on responses so far. Again, please keep thinking from an observers point of view. If observers see objects slowly fade and slowly disappear as they approach the event horizon would that mean that over time the event horizon would be "lumpy" with objects invisible, but not passed through? We should be able to detect the "lumpiness" should we not through?

Waffle's Crazy Peanut
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Matt Luckham
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    Further thoughts on this. If a black hole exists the event horizon is in a different time reference to all other objects outside of the event horizon. Does that mean that nothing could ever cross a black hole event horizon as from our reference (and the rest of the universe) the object will always halt at the horizon? – Matt Luckham Feb 23 '12 at 09:53
  • Look at this answer: http://physics.stackexchange.com/a/18993/1186 – Anixx Feb 23 '12 at 10:24
  • I like the answer, but what I am saying is that objects cannot pass through an event horizon "in either direction" as everything outside of a black hole as a time reference that would result in the objects halting at the horizon. Everything I read is about "black holes swallowing up" stars. How can they if the accelerated reference frame for all objects outside of the black hole would mean that as they hit the horizon they would stop - forever! – Matt Luckham Feb 23 '12 at 10:39
  • You are right. Read the comments to that answer. Also look at this and comments: http://physics.stackexchange.com/a/9016/1186 – Anixx Feb 23 '12 at 10:43
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    It seems paradoxical that we think we have black holes with many million solar masses, but our current theories seem to indicate that a black hole would take an infinite amount of time to consume anything! Is the answer to my question we don't know? – Matt Luckham Feb 23 '12 at 10:50
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    If you're on the outside, an event horizon can never be in your past, more or less by definition of "event horizon". So it's a trivial truth that an outside observer can never observe the formation of a black hole. I strongly urge you to learn what a Penrose diagram is and then look at a Penrose diagram for a collapsing star. You'll then be able to answer this and many related questions for yourself. – Dan Piponi Feb 24 '12 at 20:13
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    @DanPiponi I have read a lot about this, but never had a clear explanation as to why in poplar science black holes exist and are active in "consuming matter", when at the same time we are taught through GR that from any reference point outside of the event horizon as an object approaches the horizon its time, as it appears to all objects outside of the horizon, wil stop. I see you can't explain it either... – Matt Luckham Feb 24 '12 at 23:36
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    @MattLuckham If I shut my eyes, I might miss events taking place near me. That doesn't mean they didn't take place. Photons can't reach you from anything that passes through an event horizon so you can't see them. This is, nonetheless, consistent with things falling through an event horizon. What exactly is the difficulty you're experiencing with this idea? – Dan Piponi Feb 25 '12 at 00:50
  • @danpiponi oh dear dan. Do you think that it is just a red shift issue. That the object continues on its way? What about gravitational time dilation? The objects time stops as from the observer. You not be able to see it anymore but gr clearly states the object will slow and will never cross the horizon. Read all the other posts on this page! You appear confused on this issue... – Matt Luckham Feb 25 '12 at 07:40
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    @MattLuckham "gr clearly states the object will slow and will never cross the horizon" Do you have a reference for this claim? "The objects time stops as from the observer." Time is what is measure by clocks. How are you measuring the "objects time" "from the observer"? Describe your procedure. – Dan Piponi Feb 26 '12 at 22:38
  • @daNpiponi read the original question. It is pretty clear from where we are measuring. – Matt Luckham Feb 27 '12 at 08:46
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    duplicate of http://physics.stackexchange.com/questions/5031/can-black-holes-form-in-a-finite-amount-of-time –  Nov 14 '14 at 22:05
  • I encourage everyone to have a look at the related question that @BenCrowell points out and especially his own very good answer. The important point there, that's so far missing in the answers here, is that for the notion of "has something fallen into a black hole by now" one has to consider the definition of "now". And the concept of simultaneity is not a fixed one in GR even for a given observer. – Emil Mar 22 '15 at 11:46
  • I don't quite know what you're trying to ask. Do you mean how can an object act as a source of gravity to make the black hole bigger after it crosses the event horizon? It's similar to how gravity escapes a black hole as described at https://physics.stackexchange.com/questions/937/how-does-gravity-escape-a-black-hole/256500#256500. The object with its own gravitational field changes the gravitational field of the black hole before it passes the event horizon then that gravitational field sustains itself. – Timothy Aug 09 '18 at 22:09

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It is true that, from an outside perspective, nothing can ever pass the event horizon. I will attempt to describe the situation as best I can, to the best of my knowledge.

First, let's imagine a classical black hole. By "classical" I mean a black-hole solution to Einstein's equations, which we imagine not to emit Hawking radiation (for now). Such an object would persist for ever. Let's imagine throwing a clock into it. We will stand a long way from the black hole and watch the clock fall in.

What we notice as the clock approaches the event horizon is that it slows down compared to our clock. In fact its hands will asymptotically approach a certain time, which we might as well call 12:00. The light from the clock will also slow down, becoming red-shifted quite rapidly towards the radio end of the spectrum. Because of this red shift, and because we can only ever see photons emitted by the clock before it struck twelve, it will rapidly become very hard to detect. Eventually it will get to the point where we'd have to wait billions of years in between photons. Nevertheless, as you say, it is always possible in principle to detect the clock, because it never passes the event horizon.

I had the opportunity to chat to a cosmologist about this subject a few months ago, and what he said was that this red-shifting towards undetectability happens very quickly. (I believe the "no hair theorem" provides the justification for this.) He also said that the black-hole-with-an-essentially-undetectable-object-just-outside-its-event-horizon is a very good approximation to a black hole of a slightly larger mass.

(At this point I want to note in passing that any "real" black hole will emit Hawking radiation until it eventually evaporates away to nothing. Since our clock will still not have passed the event horizon by the time this happens, it must eventually escape - although presumably the Hawking radiation interacts with it on the way out. Presumably, from the clock's perspective all those billions of years of radiation will appear in the split-second before 12:00, so it won't come out looking much like a clock any more. To my mind the resolution to the black hole information paradox lies along this line of reasoning and not in any specifics of string theory. But of course that's just my opinion.)

Now, this idea seems a bit weird (to me and I think to you as well) because if nothing ever passes the event horizon, how can there ever be a black hole in the first place? My friendly cosmologist's answer boiled down to this: the black hole itself is only ever an approximation. When a bunch of matter collapses in on itself it very rapidly converges towards something that looks like a black-hole solution to Einstein's equations, to the point where to all intents and purposes you can treat it as if the matter is inside the event horizon rather than outside it. But this is only ever an approximation because from our perspective none of the infalling matter can ever pass the event horizon.

N. Virgo
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  • Thanks, good answer. But there is a tiny question that arises. It is claimed by some that freely infalling observer will not see the Hawking radiation. I personally do not believe in that for a falling observer there is no horizon, and as he sees the horizon, he necessarily sees the radiation. But this may serve as a base for another question. – Anixx Feb 25 '12 at 02:08
  • Also thanks for noting that the cosmologists do not believe in actual black holes. – Anixx Feb 25 '12 at 02:10
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    I suspect that the infalling clock does see Hawking radiation, because although it doesn't see the event horizon in the same place as the outside observer (and hence notes nothing special at 12:00), there is still an event horizon ahead of it, which is closer to the singularity and hence (presumably) more tightly curved than the one we observe from outside. But the people who claim it sees no radiation (e.g. Susskind) are extremely smart, so to be honest I don't know. – N. Virgo Feb 26 '12 at 10:26
  • Indeed. This is also my own point of view - that any observer should see the horizon, possibly in different location. Otherwise one could just set himself in a free falling above a BH for short time to see no horizon. And if there is an apparent horizon, then there is also necessarily a Hawking radiation. – Anixx Feb 26 '12 at 10:42
  • By the way as we see, the falling observer will be approaching the horizon until the final BH evaporation. This means that he will touch the horizon at the time of the final explosion, which by his clock should be 12:00. I would not call this observation "nothing special". Apparently he will see the BH quickly evaporating with decreasing radius as he approaches, but exacly when he approaches the surface, the BH should shrink to a tiny point with Planck temperature and explode. – Anixx Feb 26 '12 at 10:52
  • Good point - but actually I think the explosion will happen at some time after 12:00, because the BH will have less mass by then, and hence a smaller event horizon. In the classical story for a non-radiating black hole you would pass the event horizon (noting nothing special) and then some time later hit the singularity. I think the time of witnessing the evaporation event for a real black hole must be somewhere between these two times. – N. Virgo Feb 26 '12 at 11:08
  • By the way, @Anixx, you might be interested on a blog post I wrote about black holes a while ago, which touches on some of these points. http://jellymatter.com/2011/02/26/falling-into-a-black-hole-part-1/ (I never did get around to writing part 2.) – N. Virgo Feb 26 '12 at 11:10
  • For an external observer the clock of the falling observer will freeze at 12:00. This will last until the BH evaporation. At any moment before the evaporation the falling observer can signal the external observer that his clock passed 12:00, but it will not pass. So I conclude that the falling observer touches the horizon, and at the same time, the final explosion point, exactly at 12:00 by his own clock. – Anixx Feb 26 '12 at 11:43
  • let us continue this discussion in chat – N. Virgo Feb 26 '12 at 14:36
  • Oh - aparrently we can't continue in chat because it won't let me log in.

    I think that we as outside observers will eventually see the clock pass 12:00 as the hole shrinks. Remember, 12:00 was the time at which the hands would freeze for a non-radiating black hole.

     – N. Virgo Feb 26 '12 at 14:38
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    @Nathaniel - are you saying that when astronomers search for evidence of black holes, they're actually searching for evidence of approximations to black holes (dark grey holes if you like)? – Dawood ibn Kareem Mar 09 '12 at 10:50
  • @DavidWallace yes :) – N. Virgo May 07 '12 at 12:42
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    Nathaniel, did you talk about with your cosmologist friend the idea mentioned in @Anixx 's answer that there could be "actual" black holes with an horizon that have been in existence since the big bang? Anyhow, excellent post, you are a most formidable technical writer and nowadays tell people at work to browse Physics SE posts of various authors (you included) for examples of clear technical writing. A new graduate was most surprised when I gave her the assignment of "browsing Physics SE" before she wrote a report that she needed to do at work! – Selene Routley Jan 07 '14 at 03:39
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    @WetSavannaAnimal we didn't discuss that possibility. To my mind the idea that some black holes are "actual" and some aren't seems a bit fishy - I would guess the primordial ones would turn out to be just as "approximate" as the ones formed from collapsed stars. But I haven't really thought much about primordial black holes and the processes that might form them. Many thanks for the compliments, they're much appreciated! – N. Virgo Jan 07 '14 at 06:40
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    This answer is along the lines of "my friend the cosmologist said..." or "I think what my friend said was..." Not very convincing. We have better answers to the question that this one duplicates: http://physics.stackexchange.com/questions/5031/can-black-holes-form-in-a-finite-amount-of-time –  Nov 16 '14 at 22:54
  • How can you guys believe in Hawking Radiation if there can't be anything inside the event horizon, thus, no mass inside the "real" black hole? – lvella Aug 27 '15 at 06:12
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    @Ivella It is a little bit more complex thing. Anyways, Hawking Radiation isn't experimentally verified so you are right in the sense that it is at most "believe" and not fact. I suggest to make a new question about this, it would be interesting. – peterh Jul 14 '16 at 23:28
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    Your answer is written as though you can derive a contradiction from the facts that you never see an object pass the event horizon and a black hole eventually evaporates and other simple assumptions about generally but they in fact don't contradict one another. https://physics.stackexchange.com/questions/411909/could-this-be-a-solution-to-the-black-hole-information-paradox shows why there's no paradox with an evaporating black hole if you don't make additional assumptions. – Timothy Jul 01 '18 at 03:29
  • John Rennie gives a great quantitative treatment in his answer to a related question. He explicitly shows that for a realistic black hole formed from matter collapse, a faraway observer never actually sees the event horizon form. – tparker Oct 07 '18 at 20:38
  • "from an outside perspective, nothing can ever pass the event horizon" – if this is true, how can we observe a merger of two black holes? – Vojtěch Sep 26 '21 at 18:58
  • @Vojtěch my understanding is that strictly speaking we can't. As distant observers we never actually see the event horizon form (see this answer that tparker linked above), so we only ever observe approximations to black holes, which may merge to form an approximation of a bigger black hole. These approximations are extremely good and the convergence time is extremely rapid, but they are approximations nonetheless. (Disclaimer: it's been years since I did any serious thinking or reading about this - I wouldn't count myself an expert.) – N. Virgo Sep 27 '21 at 03:27
  • (The above is not to say that black holes don't form or don't merge. It's just that the time dilation is so extreme that the light from those events can never quite reach us, so strictly speaking we can't observe them.) – N. Virgo Sep 27 '21 at 03:31
  • "As distant observers we never actually see the event horizon form" – but we can "hear" the gravitational waves. If the time dilation really stopped the time at the horizon, we should never be able to "see" the merger, not? – Vojtěch Sep 27 '21 at 19:45
  • @Vojtěch we also hear gravitational waves from neutron star collisions, and there are no event horizons there. The observable, "approximate" black holes should make gravitational waves indistinguishable from those made by "real" black holes. – N. Virgo Sep 28 '21 at 00:16
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    @Nathaniel "we only ever observe approximations to black holes, which may merge to form an approximation of a bigger black hole" - While this is correct +1, it may sound misleading to some. "Approximation" here applies to what merges (the "almost horizons"), but not to the fact that they merge. Two "almost horizons" do merge into one larger "almost hirizon" in seconds by the clock of a remote observer. Anyone thinking they "almost merge" would be incorrect. They do merge into one without approximation. Here is an animation based on numerical gravity: https://m.youtube.com/watch?v=Y1M-AbWIlVQ – safesphere Oct 10 '21 at 18:12
  • @safesphere Thanks for the comment. The phrase "by the clock of an external observer" is a bit tricky - it means a foliation of space-time into sets of events that are considered simultaneous. If you had two 'real' black holes their event horizons can merge in finite time according to that definition, but the light from any point on their horizons takes an infinite time to reach the observer, so neither the horizons nor their merging can truly be observed. I don't know the details of that simulation but I would guess the finite speed of light isn't accounted for in the rendering. – N. Virgo Oct 11 '21 at 04:45
  • @Nathaniel Agreed. These are theoretical predictions or expectations. Testing them experimentally is a different matter. – safesphere Oct 11 '21 at 10:19
  • What about horizon growing in radius after mass falls to it? Does it mean a distant observer will never measure the horizon grow? Won't it also cause the accretion disk to grow in size? – Nayeem1 Nov 17 '21 at 14:34
  • "(I believe the "no hair theorem" provides the justification for this.)"

    Is this true? I thought that the "No Hair Theorem" assumed the problem away by assuming that the collapse into singularity has already occurred. Maybe I am wrong?

     – sf_jeff May 23 '22 at 23:05
  • So a black hole is basically impenetrable, and matter basically bounces off the horizon as radiation? – the_prole Dec 10 '22 at 08:41
  • @the_prole no, not really. Objects don't bounce off the black hole, they keep on falling toward the black hole forever (or until the black hole evaporates at least), it's just that because time slows down near a black hole they never actually reach it, at least when seen from outside. From the perspective of someone falling toward a black hole they will just pass the event horizon and fall in. – N. Virgo Dec 11 '22 at 07:45
  • @user161096 I appreciate the edit, but the explanation you gave isn't the one I would have gave, I don't think. Substantial content like that should really belong in a separate answer on this post I think, or possibly in a new Q&A. I'm really sorry to reject it but it feels odd to put words in my mouth that I wouldn't have said. – N. Virgo May 14 '23 at 01:26
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Assume the object falling in is a blue laser that you launched directly (radially) towards the Schwarzchild (non-rotating) black hole that is aimed directly at you and that you are far from the black hole. The massive object is the laser itself, the light that you are watching is your way to "see" the object as it approaches the event horizon.

First of all just because the laser is moving away from you it will be slightly red-shifted just by the Doppler effect. As it approaches the black hole that slight red-shift will become more and more significant. The laser light will go from blue, to green, to yellow, to red, to infrared, to microwave and to longer and longer wavelength radio waves as it appears to approach the event horizon from your point of view. Also the number of photons it emits per second (as you detect them) will decrease with time as the horizon is approached. This is the dimming effect - as the wavelength increases, the number of photons per second will decrease. So you will have to wait longer and longer between times when you detect the longer and longer wavelength radio waves from the blue laser. This will not go on forever - there will be a last photon that you ever detect. To explain why, let's look at the observer falling in.

Your friend who is the observer riding on the laser does not even see anything happen when he crosses the event horizon (if he is freely falling). The point is that the event horizon is not at all like a surface that you hit or where anything unusual happens from the freely falling observers point of view. The reason why there will be a last photon you will ever detect is because there are only a finite number of photons emitted between the time the laser starts to fall and the time the laser crosses the event horizon. So that last photon emitted just before it goes over the event horizon will be the last photon you will ever see. That photon will be a very long wavelength photon and you may not see it until some time in the distant future - how far in the future will depend on the number of photons per second that the laser emits - but there will be a last photon and after that you will not see any more photons.

So, I claim that the laser does disappear from an outside observer's point of view. Note that trying to "illuminate" the object near the event horizon by shining a different laser on the object and looking for scattered photons will not work. (It will not work even if you throw the second laser in to try to illuminate the first laser.) From the point of view of the laser that fell in, these photons will only hit the laser after it has already crossed the event horizon and therefore the scattered light cannot escape from the black hole. (In fact, if you wait too long before you try to illuminate the object, the infalling laser will have already hit the singularity at the center of the black hole.) From the outside observer's "point of view" (but he cannot "see" this), the infalling laser and the photons that are trying to illuminate the laser will get "closer and closer" to each other as they get frozen on the event horizon - but they will never interact and there will never be a scattered photon that you might try to detect.

FrankH
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    I am talking about objects with mass falling into a black hole. My point is that for any observer the object will halt at the event horizon. Is that not what GR says will happen? – Matt Luckham Feb 23 '12 at 09:34
  • That is what I essentially said in my example - the laser is the object with mass falling. I only talked about the laser because you need to have some illumination on the "object" that is falling into the BH. So instead you can imagine that you are shining the laser on the object as it falls in, but then you get to the issue of the photons having to catch up with the object, instead you can just watch the laser light to see what happens to the object. As I said, the photons will get more and more redshifted and less and less frequent as it approaches the event horizon. – FrankH Feb 23 '12 at 17:51
  • ...and it will take forever and will never appear to you to pass the event horizon - you will just have to wait longer and longer between lower and lower energy photons - all coming from the laser that has not yet crossed the event horizon. – FrankH Feb 23 '12 at 17:52
  • @FrankH: does the recent work about black hole "firewalls" change the generally held view that nothing particular happens to an infalling observer at the horizon. – Richardbernstein Jan 22 '13 at 03:53
  • @Richardbernstein - I think the issue of black hole "firewalls" is not settled yet. In fact, I had a class with Prof Lenny Susskind tonight where he said the same thing - it is not settled and he thinks they may disappear... – FrankH Jan 22 '13 at 07:06
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    Why does this Answer have only one upvote? +1 for writing so clearly. – Eugene Seidel Jan 22 '13 at 12:02
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    Can you provide a reference/is it difficult to prove that you cannot illuminate infalling objects seen near the horizon? It goes strongly against my intuition. Let me explain. You may receive photons from the infalling observer arbitrarily far into the future (however redshifted), depending on his temporal precision of emission. If you simply time-reverse this received photon it would seem you could in fact send a photon to the infalling observer at arbitrary times after he has started falling. The problems of finite power and finite temporal precision seem like technicalities. – Real Sep 20 '18 at 12:42
  • @Real Did you read my answer? There WILL be a last photon from the infalling laser that you will ever see. We cannot predict when you will receive that last photon. In fact, the only way you will know you received the last photon is for you to continue to watch the BH forever, waiting for another photon. Similarly, the situation where an object is falling in while you illuminate it is exactly the same situation. Some of your photons may illuminate the infalling object after it has passed the event horizon, you will never see that photon bounce back to you. Clear? – FrankH Sep 20 '18 at 18:50
  • @Real Here is a link that says what I said: http://www1.phys.vt.edu/~jhs/faq/blackholes.html#q11 "A finite number of photons were emitted by A before A crossed the horizon, and a finite number of photons were emitted by B (and collected by A) before A crossed the horizon." – FrankH Sep 20 '18 at 18:57
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    This does not answer the following: - imagine a very heavy body (let's say, a moon) that is going on the trajectory chosen to fall into the black hole asap. While it moves, we can observe how the curvature of spacetime changes due to its mass. – Eugene Dudnyk Aug 10 '20 at 21:52
  • What happens at the moment when moon's center of mass crosses the event horizon? We can't see the moon moving into BH, but it actually should add up to the mass of the BH at some point. Will it NEVER add up to the mass of the BH? Or, if it will, what will be the trajectory of the curvature, following the moon's path from the event horizon to the singularity? – Eugene Dudnyk Aug 10 '20 at 21:52
  • @EugeneDudnyk, Please ask this as a question, not as a comment. – FrankH Aug 11 '20 at 00:10
  • @FrankH Yep, here: https://physics.stackexchange.com/q/572476/147418 – Eugene Dudnyk Aug 11 '20 at 00:13
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Everything you say in your question is true, and your comment "the event horizon is in a different time reference" is also true, though it needs to be stated more precisely.

If you've read much on relativity you've probably come across terms like "frame of reference" and "inertial frame". A "frame" is a coordinate system, i.e., a system of distances, angles and times used to measure the location of things. For example, the map grid references are a coordinate system used to measure locations of things on the Earth's surface.

General relativity (GR) gives us a way to describe the universe that is independent of any frame of reference. However, for us observers, to calculate what we see, we have to do the calculations in our frame of reference i.e. in meters and seconds that we can measure. The static black hole is described by the Schwarzschild metric, and it's not hard to use this to calculate things like how long it takes to fall onto the event horizon. One common coordinate system is co-moving coordinates, i.e., the observer falling into the black hole measures distances from himself (putting himself at the origin) and time on the stop watch he's carrying. If you do this calculation you find the observer falls through the event horizon in a finite time, and in fact hits the singularity at the centre of the black hole in a finite time.

But where things get odd is we calculate the time taken to reach the event horizon in our coordinate system as observers sitting outside the black hole. This is an easy calculation, that you'll find in any introductory book on GR, and the answer is that it takes an infinite time to reach the event horizon.

This isn't some accounting trick; it means we will never see an event horizon form. At this point someone will usually pop up and say that means black holes don't really exist. In a sense that's true in our coordinate system, but all that means is that our coordinate system does not provide a complete description of the universe. That's something we've been getting used to ever since Galileo pointed out that the Sun doesn't revolve around the Earth. In the coordinate system of the freely falling observer the event horizon does exist and can be reached in a finite time.

You ask:

If this is the case how can a black hole ever consume any material, let alone grow to millions of solar masses.

As long as you stay outside the event horizon a black hole is nothing special. It's just an aggregation of matter like a star. In the centre of our galaxy we have a compact region, Sagittarius A*, containing millions of star masses, and from the orbits of stars near Sagittarius A* it contains enough matter in a small enough space to make it a black hole. However, the orbits of those stars just depend on the mass they're orbiting and whether Sagittarius A* is actually a black hole or not is irrelevant.

Peter Mortensen
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John Rennie
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    Thanks John. I think you have put my view across far more succinctly. How can a black hole consume matter if everything outside of it is a in a reference frame which would mean that any matter approaching the black hole could never cross the event horizon. Is the answer to my question we just don't know? – Matt Luckham Feb 23 '12 at 10:48
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    John - my question is about black holes. I know about the centre of our galaxy, but that is off topic. If black holes exist can they consume matter? – Matt Luckham Feb 23 '12 at 10:53
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    Suppose I'm sitting here at my desk and I pick up my computer and hurl it at the Sun (and assume for the sake of this argument the Sun is a black hole). I will never see my PC reach the event horizon. The bugs inside my PC will see the PC reach the event horizon and indeed cross it. Who is right? Me, or the bugs in my PC? The answer is both! You cannot ask a question "can matter cross an event horizon?" without specifying what frame of reference you want the answer given in, and the answer will depend on the frame of reference. – John Rennie Feb 23 '12 at 10:56
  • I only mentioned the centre of the galaxy because we're fairly sure there's a black hole there. – John Rennie Feb 23 '12 at 10:57
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    John - everything outside of the event horizon, e.g. the universe, will have a frame of reference which will result in nothing ever reaching the event horizon? So, my conjecture is that nothing can ever fall into a black hole... – Matt Luckham Feb 23 '12 at 11:03
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    No, you're getting mixed up about what a frame of reference/co-ordinate system is. A frame of reference isn't a thing that locks in all the matter within it. There is no sense that we are all locked into a frame of reference that stops us crossing an event horizon. A frame of reference is just a measurement system. It's true that if you and I choose the most natural frame of reference we will never see matter cross the event horizon, but I can assure you that if I threw you into a black hole you would see matter crossing the event horizon. – John Rennie Feb 23 '12 at 11:09
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    John, I did not say the same frame of reference, I said all other frames of reference. Any reference point outside of the event horizon would result in the object being time dilated to a point of freezing as it reaches the event horizon hence I do not believe it possible for any object to ever cross the event horizon of a black hole. – Matt Luckham Feb 23 '12 at 23:46
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    Suppose you and I are on a satellite above a block hole. I stay on the satellite while you jump in. You will see yourself pass through the event horizon and shortly afterwards hit the singularity (where you will meet a messy but quick death :-). I will never see you reach the horizon but that doesn't mean you didn't, only that my co-ordinate system doesn't reach to the horizon. You say "I do not believe it possible for any object to ever cross the event horizon of a black hole" but that's only because you have never jumped into one! – John Rennie Feb 24 '12 at 09:02
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    See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html for a reasonably accessible description of the maths involved. – John Rennie Feb 24 '12 at 09:03
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    John - you are missing my point. If from all reference points, and I mean all, outside of the black hole the object will never fall into the black hole, regardless of whether from the object perspective it can fall into the black hole (and I believe it can), if for every reference point outside of the black hole objects cannot pass then how can a black hole consume anything? We and the rest of the universe live in reference points that mean that a black holes event horizon is a barrier for any matter pass. Do you agree? – Matt Luckham Feb 24 '12 at 09:51
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    I think we agree that observers outside the black hole will never see anything passthrough the event horizon. But this seems to me a trivial observation. It's like saying the Sun seems to go around the Earth, which is perfectly true from one perspective but is a poor description of reality. If you insist on this perspective you'll have an impoverished view of the universe. GR is a wonderfully, and I mean wonderfully rich theory, but you need to embrace it to appreciate it and that means giving up anthropocentrism. – John Rennie Feb 24 '12 at 10:06
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    John, I am afraid you are missing the point. By fixating on the reference point of the object falling into the black hole you are avoiding the question I asked. Are you able to answer it, succintly: - "How can anything ever fall into a black hole as seen from an outside observer?" – Matt Luckham Feb 24 '12 at 10:19
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    How about the Penrose experiment: http://en.wikipedia.org/wiki/Penrose_process. If we did the experiment and detected our test mass leaving with more energy than it arrived with wouldn't that prove (indirectly) that the other half of the test mass had passed through the event horizon? – John Rennie Feb 24 '12 at 11:15
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    Actually, the more I think about it the more I think the Penrose process is proof. For example, we don't "see" quarks, we believe they exist because QFT says the jets we observe prove they exist. Well assuming you believe GR the momuntum of the outgoing particle in the Penrose process proves that the event horizon exists and that the other half of our test mass crosses it. If you deny the black hole consumed our test mass you have to either disbelieve GR or disbelieve virtually all of modern physics :-) – John Rennie Feb 24 '12 at 12:01
  • It is because of GR that this paradox exists. I think it just underlines how much we don't understand. – Matt Luckham Feb 24 '12 at 13:18
  • The same process can be repeated with any rotating star. Even the split of the particle is not necessary. – Anixx Feb 24 '12 at 16:55
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    Why do you use the work "paradox"? The Schwartzchild metric describes a static black hole in a precise and internally consistent way. I don't see what is paradoxical. – John Rennie Feb 24 '12 at 18:12
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    If nothing can pass the event horizon from the outside point of view, this means, nothing can pass through it till the BH completely evaporates. So, everything falling on the BH will survive before reaching the event horizon and will see the end of the BH. – Anixx Jun 10 '17 at 09:40
  • @John It seems to me that the big point of misunderstanding here is of philosophical kind: what does it mean that something happens. For me it is something along the lines "there is a point in space-time and a reference frame (in our context: outside the black hole) that agrees with being simultaneous with the event of an object crossing the horizon". In this sense I agree with Matt, that a crossing will never happen. But if you have a different philosophy, this might not satisfy. – M. Winter Oct 14 '18 at 14:02
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    Also: if you are in the co-moving frame of an infalling observer, the moment of crossing with the frame will not be simultaneous with anything outside of the BH. So assuming that the black hole does not exist for eternity, how can this be? From what I understand, the moment of "crossing" is simultaneous to: the end of the universe, OR the end of the black hole (e.g. you get radiated away by the accumulated Hawkin radiation) – M. Winter Oct 14 '18 at 14:05
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Indeed, nothing can get under the horizon. The stuff close to the event horizon does move outwards as the black hole radius increases. Even more with any black hole deformations such as waves on its surface, the tidal deformations or the change of the rotation speed, all the objects close enough to the horizon remain "sticked" to it and follow all the changes of the black hole form. All objects close enough to a rotating black hole horizon, rotate with it at the same speed. If a black hole moves, so does everything close enough to its surface, including the things located on the side of the direction of the move. If anyone interested what mechanism make such sticking possible, it is called frame-dragging.

You may ask then, how a black hole can appear then and the horizon form. It is conjectured that they cannot, and the only possible black holes are the hypothetical primordial black holes that existed from the very beginning of the universe.

The objects that can be very similar to black holes are called collapsars. They are virtually indistinguishable from actual black holes after a very short time of the formation. They consist only of matter outside the radius of the event horizon of a black hole with the same mass. This matter is virtually frozen on the surface like with actual black hole, due to high gravity level.

Such collapsars possibly can become black holes for a short time due to quantum fluctuations and thus emit hawking radiation.

Astrophysicists do not separate such collapsars from actual black holes and call all them black holes due to practical reasons because of their actual indistinguishability.

Here is a quote from one paper that supports such point of view:

Our primary result, that no event horizon forms in gravitational collapse as seen by an asymptotic observer is suggestive of the possibility of using the number of local event horizons to classify and divide Hilbert space into superselection sectors, labeled by the number of local event horizons. Our result suggests that no operator could increase the number of event horizons, but the possibility of reducing the number of pre-existing primordial event horizons is not so clear and would require that Hawking radiation not cause any primordial black hole event horizons to evaporate completely.

Source

Peter Mortensen
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Anixx
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  • Thank you Anixx! I thought I was flying solo with my view on this! Completely agree with everything you have said. This makes far more sense when compared to what GR states. – Matt Luckham Feb 23 '12 at 11:06
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    @Matt: The "collapsar" is a silly idea--- it is just the exterior point of view for black holes. The modern ideas of black hole complementarity resolve the issue of black hole exterior and interior picture, and since I have explained these to Anixx, and yet he persists with these nonsense ideas, I will downvote. – Ron Maimon Feb 23 '12 at 16:37
  • Ron, if you can explain how an object can fall in a black hole when for all other reference points the object will stop at the event horizon I would love to hear it. – Matt Luckham Feb 23 '12 at 23:42
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    It should be noted that even Ron agrees that all information about the falling object will always remain outside the event horizon. That is even if something falls under the horizon (as a kind of abstraction), if falls without any information (hence no structure). At the same time anything unstructural in the center of mass of a massive body behaves exactly the same way as if it was spread over its surface. – Anixx Feb 23 '12 at 23:56
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    Thus even in best case we would not be able to distinguish whether something went under the horizon or just got spread over the surface (information about fallen things will always be available from the surface). – Anixx Feb 24 '12 at 00:00
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    @ Ron Maimon is not the question exactly about exterior picture (I don't think interior picture is relevant because it cannot be verified with scientific method)? – Anixx Feb 24 '12 at 00:11
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    @Anixx: The interior picture can be verified with scientific method, by throwing a camera in, taking pictures, and letting it come out again, in the case I believe, namely that for charged or rotating holes, it comes out again. If I am wrong, the same information can be extracted by watching the Hawking radiation very carefully. The statement that the infalling observer falls through from her point of view is well established science. In the exterior, the time only freezes until the observer is smeared over the entire horizon due to string-spreading. – Ron Maimon Feb 25 '12 at 01:21
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    @Anixx: The idea that the exterior picture is complete is correct, in that the interior picture can be extracted from the exterior picture, but it is incorrect because the infalling object is not frozen from its point of view--- the freezing world line ends at a finite proper time on a point which is not a special point, which contains extensions to the interior. The connection between interior and exterior picture is subtle, but it is understood. The answer is not a collapsar, but Susskind's black-hole complementarity. – Ron Maimon Feb 25 '12 at 01:27
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    @Matt Luckham: The object doesn't quite "stop" at the event horizon. It smears over the horizon, and thermalizes with other stuff on the horizon, all at a Planck-length skin near the horizon. This process is the neutral-version of string theory particle emission and absorption. – Ron Maimon Feb 25 '12 at 01:29
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    Well, Ron. You claim that the camera can come back from the black hole. But what lets you to believe that the camera was indeed under the horizon, if 1) information about this camera's state was always available for outside observer 2) as such the camera could not make any pictures of anything that could not be seen by the outside observer directly. If I say I go to a sex-shop for 5 minutes, but you always see me outside of a sex shop and I return with things which can be bought outside the sex shop, and do not bring things that only sold in sex shop, do you believe I actually was in sex shop? – Anixx Feb 25 '12 at 01:30
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    "the freezing world line ends at a finite proper time on a point which is not a special point, which contains extensions to the interior." - lol. This is only if the black hole is eternal and does not evaporate. Again GR is applied outside of its domain. Please learn that GR is not applicable to times exceeding or comparable to the supposed time of BH evaporation. – Anixx Feb 25 '12 at 01:38
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    @Anixx: your new answer is the non-controversial part, and Susskind complementarity really does make this answer incorrect. The reason one believes complementarity allows the camera to go inside (from its point of view) is because otherwise it's path ends on the horizon at a finite proper time from the point of view of the camera, which is physically preposterous. Complementarity is the statement that the interior of the sex-shop can be reconstructed from the exterior, so that there is no difference between saying that you went in and saying that you got spread out. – Ron Maimon Mar 24 '12 at 05:12
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    But the spread-out picture is suboptimal, because the camera doesn't break as it crosses the horizon (although from your point of view, it atomizes, and all the protons' strings blow up, it is completely destroyed, then reformed when it comes out, like a ripple rebounding over a pond and recollecting at the point it entered). The interior spacetime is a reconstruction using externally available boundary data, but all spacetime is a reconstruction in the same way--- this is the holographic principle. It's beautiful and subtle, and it really works in string theory. – Ron Maimon Mar 24 '12 at 05:14
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    This answer is completely wrong. There are several correct answers by people competent in relativity for the question that this one duplicates: http://physics.stackexchange.com/q/5031/ –  Nov 16 '14 at 22:52
  • "The stuff close to the event horizon does move outwards as the BH radius increases." I agree that it has to, but how is it happening? – BartekChom Jan 25 '15 at 14:21
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    Now I think that it does not move outwards. Nothing can fall into a black hole, so black hole radius cannot increase. Actually, black holes do not exist. But in practice there are things that for all practical purposes are black holes and if thing A starts to fall on one of these almost-BH, thing A will freeze on the almost-BH surface. If other things fall onto the almost-BH, thing A will be inside the almost BH. But there still be no true event horizon, so nothing will be under event horizon. – BartekChom Feb 05 '15 at 13:49
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    @Anixx: I like this answer and I'd like to upvote it, but there is one problem. I noted it on Jan 25 and now I've solved it: Matter has no possibility to move outwards, but collapsars have no event horizon, so it does not have to - it can be in a place that would be under event horizon if the collapsar were true black hole. I hope that you'll agree that it's obvious. If you can see my rejected edit, you should understand what I mean. Plus one spelling mistake: oblects->objects. – BartekChom Feb 06 '15 at 13:52
  • @BartekChom Since Anixx edited the question after your comment I don't know if you like the current edit, it still says things move out, when they don't. – Timaeus Sep 17 '15 at 00:38
  • @Timaeus If I see correctly, Anixx edited in February 2012 and I commented in February 2015. However, thanks to your comment I saw an edit with source about possibility of primordial black holes. I am not sure if they are really possible, especially because Hawking wrote in 2014 that black holes in the oryginal meaning do not exist and should be redefined. Besides, I do not see how the primordial horizon could grow in finite time. Anyway, now I am not sure that this part of the answer is wrong. I am still waiting for an explanation. – BartekChom Sep 17 '15 at 06:36
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    @ben Crowell that question you posted has all wrong answers every single answer there is completely wrong given the context of the question . It irritates me so much when a good question gets locked up with bad answers spreading false knowledge and locking the question up. Those answers failed to understand when the most basic context that op is referring to an outside observer not one inside blackhole or near it why is that sob hard to understand? – user1062760 Feb 18 '17 at 23:23
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    I think this information is wrong. According to one coordinate system that describes all of space, space is moving towards the singularity in an approximately Newtonian way and beyond the event horizon, space is moving so fast that light can't fight the current, and an object crosses the event horizon in finite time. – Timothy Aug 09 '18 at 22:04
  • Sometimes theories have consequences not particularly intended or expected by their writers (with the best-known example having been Einstein's addition of the cosmological constant to GR), and my answer to a PSE question ("How can black holes even exist?"), very similar to the one at hand, mentions one in connection with the Borde-Guth-Vilenkin Theorem and Poplawski's relativistic hypothesis for cosmogenesis within black holes. (The CC was superfluous when Einstein added it to GR after Hubble's 1929 discovery, although it has since become useful in some hypotheses re cosmic acceleration.) – Edouard Feb 03 '19 at 16:55
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    @BenCrowell From time to time I revisit this question (because I have my own questions on the subject), and this is not the first time I am direct by your comment to that other question, whose top 3 answers talk about what happens from the standpoint of an infalling object. If you feel this and the following answers are wrong, please link to the correct answer that actually approaches the problem from the standpoint of an outside observer, not an infalling object. – lvella Aug 29 '19 at 21:16
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    @BartekChom the matter gets sticked to the event horizon by means of frame-dragging. So, if the event horizon moves, grows, the BH moves as a whole or changes rotating speed, the matter near the event horizon follows its movements, including moving outwards in a sense (but it always moves towards the event horizon). Keyword: frame-dragging. – Anixx Jul 02 '20 at 13:35
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I would like to add a fact that, perhaps, is not controversial.

Namely, that all the information about any infalling object will be available for the outside observer at any time. The information cannot get lost under the horizon, otherwise we have the information loss paradox.

This means that it is theoretically possible for an outside observer to restore any object that went in the direction of the BH, because all of its information still kept.

This is true not only regarding objects that are falling after BH formation but also for those objects which were there at the time the star collapsed. So even if you were in the center of a star when it was collapsing, all information about you is still preserved, available outside the horizon and your body can be reconstructed.

Anixx
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    Sorry, definitely controversial. Intuitively, how could the information at the center possibly get out? And even if it does eventually get out, how can it possibly be available *at all times?* Nobody can currently answer this. – Peter Shor Jan 22 '13 at 03:16
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    @PeterShor As I understand it, the information never even gets to the center, so there is no need for it to be able to get out. All information remains on the event horizon and is therefore available at all times. – Wouter Jan 22 '13 at 09:53
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    @Wouter: What if the information starts at the center? Suppose after a star collapses into a black hole, you want to reconstruct the quantum state of the matter in the center of the star before its collapse. How does this information get out? As far as I can tell, nobody has an uncontroversial explanation. – Peter Shor Jan 25 '13 at 01:35
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    @Peter Shor, the horizon initially appears at one point and then raises. It is not that a volume suddenly becomes under the horizon. – Anixx Jan 31 '13 at 18:26
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    @Anixx: The "horizon" of a black hole is an entirely imaginary surface. If you have light rays coming in from all sides and forming a black hole, the "horizon" appears at a point *before* relativity says the information that a black hole is forming can arrive at that point. Is there non-local physics here? Does the matter at the center of a collapsing star suddenly turn into photons (or gravitons) for no locally discernible reason and start racing to get out before it's trapped by the black hole horizon? How can that not be controversial? – Peter Shor Jan 31 '13 at 18:35
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    @Peter Shor You said "If you have light rays coming in from all sides and forming a black hole, the "horizon" appears at a point before relativity says the information that a black hole is forming can arrive at that point." - incorrect, because 1)light rays cannot create a black hole 2)if some massive particles are coming together so to create a concentration at which a horizon appears, the information about this was available long before. Knowing the particles' gravitational fields it is always possible to predict will they form a horizon or not. – Anixx Jan 31 '13 at 19:35
  • there is no faster-than-light information transfer here. – Anixx Jan 31 '13 at 19:36
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    @Anixx: Maybe you should give a correct answer to this stackexchange question then. Especially if you can find an authoritative source saying that black holes cannot be formed by radiation, contrary to the accepted and upvoted answer. – Peter Shor Jan 31 '13 at 19:55
  • @Peter Shor thank you for the link. Anyway, this does not affect this question, even if the BH is formed by the radiation. We experience gravitational attraction of a photon's energy even if it still did not reach us, because radiation does not appear from nothing. – Anixx Feb 01 '13 at 06:01
  • @Anixx: But the photons are coming towards us, which is what makes the black hole form. The same gravitational attraction, if it stayed far away, would not help form a black hole. – Peter Shor Feb 01 '13 at 12:03
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    @PeterShor Sorry for reviving a decade old argument, but Anixx is correct, light cannot create a BH. The reason for a controversy is equating light with classical EM radiation. In GR light is Null Dust instead giving different solutions. Consider a collapsing shell of ultra high energy neutrinos. Their energy is enough to create a BH, but a kinetic energy of a relativistic object does not bend spacetime, so the neutrinos will just shoot through and fly away. Both neutrinos and photons are Null Dust in GR. Therefore a collapsing shell of light (unlike classical EM radiation) cannot create a BH. – safesphere Jun 10 '19 at 21:09
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Seems to me the faller is part of the black hold and so will itself be evaporating

If one throws a log in a fire is it the fire that burns the log, or is the log now part of the very fire. I see the faller as part of the event horizon, so rather than say the faller is destroyed by a fire-wall, maybe the faller itself evaporates.

Perhaps this is just quibbling over semantics.

Shiekh
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This is most easily seen from the mathematical point of view: the Schwarzschild metric reads: $$ds^2=-c^2(1-\frac{2GM}{c^2r})dt^2+(1-\frac{2GM}{c^2r})^{-1}dr^2+r^2d\theta^2+r^2sin^2\theta d\phi^2$$ for bodies crossing the horizon slowly, we may have $$\frac{dt}{d\tau}=\frac{1}{\sqrt g_{00}}=(1-\frac{2GM}{c^2r})^{-1}$$ The radius of a black hole is its Horizon, and this radius is the Schwarzschild Radius, which equals $r=\frac{2GM}{c^2}$. So at the horizon, the ratio of the coordinate time and the proper time is infinite, which means you would measure the bodies' time to slow down by a factor of infinite, which means you wll see them stop at the horizon.

PM 2Ring
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YuJuchong123
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There is a good answer by John Rennie, and I think the continued discussion in the comments rests on a misunderstanding by the OP, who asks in a comment to John:

"How can anything ever fall into a black hole as seen from an outside observer?" – Matt Luckham Feb 24 '12 at 10:19

The misunderstanding of the OP is in the definition of the "outside observer" and in assuming that all observers are "outside".

"Outside" means outside the gravitational influence/pull of the black hole, i.e., not attracted and falling into it. This of course does not define "all observers". There will be observers falling into the black hole because their trajectory falls towards the black hole, no matter how far they are. This includes all the matter falling towards the singularity, seen from whatever framework.

Peter Mortensen
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anna v
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I recommend reading the answers to some of the questions at right --> -->

Particularly this one.

I expect that this question will be closed as exact duplicate, but what you'll find in response to the other questions is that what someone falling into a black hole observes, and what someone outside who is watching them fall in, are not the same. The exact nature of the change of the image can be (and has been) worked out, but again, I recommend looking at the other questions here.

Community
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AdamRedwine
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    That was a different question you linked me to! I am talking about crossing the event horizon. I appreciate the object crossing the event horizon does detect any slow down, but I am talking about the "observer". If I am "observing" will I ever see an object go into the black hole? If not then I will never "observe" the hole consume anything? Is that correct? – Matt Luckham Feb 22 '12 at 13:40
  • I wasn't directing you to the question... I was directing you to the answers to the question... hence why I said "I reccomend reading the answers to some of the questions..." – AdamRedwine Feb 22 '12 at 16:23
  • Re "some of the questions at right": That is a too unstable reference. For example, the example question you linked to is not there (2021-10-08). It is on the linked page "See more linked questions" (there are currently 99 questions linked to this one). The "Related" part is even more unstable. – Peter Mortensen Oct 08 '21 at 12:39
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Third paragraph onwards are slightly speculative on my part; I'm not exactly sure of them. Comments appreciated

This is all due to the queerness of relativity. In your reference frame, the rock stops at the horizon. The rock senses no such stopping. The rock will see the stars condense (this 'condensation' is more apparent for massive black holes), due to gravitational lensing. It will see the horizon approaching, and will fall through it.

Remember, time and space are relative. This is a rather extreme case where time appears to flow infinitely faster in a different frame.

About the 'dimming', I'm not exactly sure what happens. IIRC, the claim that the rock 'freezes' is a half-truth. Theoretically, the rock is frozen in your reference frame, but a telescope can't see that. To make life easier, let's assume the rock to be covered in lamps (a normal rock would become invisible long before it reaches the horizon)., At the horizon, the light emitted by these lamps is $\infty$ redshifted, so it basically doesn by exist. This can also be looked at as the photon turning tail and being absorbed (actually the photon gets frozen at the horizon) So a rock at the horizon is invisible. A rock near the horizon is very dim, as almost all the light emitted is re-absorbed(also, there is redshifting of light, more redshift $\implies$ less energy. ). So, what we see is that the rock gradually dims as it reaches the horizon. It also appears to go slower. The rate of dimming and slowing converge at the horizon, where the rock is frozen, but completely invisible. IMHO, this happens at $t=\infty$ in your frame.

So a black hole stays black. You won't see any stars, gas, rocks, or ambitious researchers stuck to the horizon, though you may manage to see dim versions of them as they fall in near the horizon.

As to how the black hole grows, its due to the absolute horizon. The horizon of a black hole grows in 'anticipation' of infalling material.

Manishearth
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    Thanks for your reply Manishearth. So if you are saying that the rocks have dimmed to be invisible, but are still there (from the observers point of view) and if the observer was able to watch the black hole for a billion years would the rock still be there (invisible, but at the event horizon). So, from the observers perspective, would anything ever actually cross the event horizon? – Matt Luckham Feb 22 '12 at 13:52
  • Remember that the rock itself has a gravitational field. So I think that the rock extends the black hole just by being there. But I'm not sure. GR does fail after the event horizon, and it might even fail at the event horizon. I'm not an expert. – Manishearth Feb 22 '12 at 14:02
  • Don't say redshift and deflection are the same, because they are different: Color change vs. intensity change. By the way, intensity of an ideal laser beam will be decreased too. – kartsa Feb 22 '12 at 17:34
  • @kartsa Yes, but they give the same effect (I clarified that). IIRC, $\infty$ redshift of the photon is equivalent to it being stuck at the horizon. And color change $\implies$ intensity change (though it's not by deflection). Take a beam and redshift it. $E=h\nu,I=E/A$. If $\nu$ decreases, so does $E$, and consequently $I$. – Manishearth Feb 23 '12 at 03:02
  • I quote FrankH: "Also the number of photons it emits per second (as you detect them) will decrease with time as the horizon is approached" THAT's what I meant. Number of photons per second. Not intensity. Intensity is proportional to redshift factor squared. – kartsa Feb 23 '12 at 04:54
  • @kartsa Yes, so there are multiple factors here:redshift decreasing intensity, and gravity decreasing intensity. – Manishearth Feb 23 '12 at 04:59
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    Guys - whether you can see the object or not is irrelevant is it not? The question is can the object cross the event horizon and move on to to the singularity from any observer not in the same reference as the object? I think GR says no... – Matt Luckham Feb 23 '12 at 09:39
  • @MattLuckham Aah: what happens here is that the minute,it reaches the horizon, it extends it. The rock has a grav field as well. So it sort of gobbles itself up. The 'stuck at horizon' is only true if the body itself does not gravitate. – Manishearth Feb 23 '12 at 09:56
  • And the horizon won't become lumpy, due to the no-hair conjecture. – Manishearth Feb 23 '12 at 09:56
  • Does GR not say that not only time dilation becomes infinite, but also the objects mass becomes infinite? To be honest my brain is hurting. Need to find a black hole and test this stuff! – Matt Luckham Feb 23 '12 at 10:05
  • Why would the mass become infinite? Mass-energy is still conserved if you have a black hole.Black holes have finite mass; black hole+object=black hoe;thus any object that falls into a black hole has finite mass. – Manishearth Feb 23 '12 at 10:17
  • This is showing my ignorance. I know with SR that when an object approaches the speed of light its mass becomes infinite (as does time dilate) to any observer. My understanding was that GR says the same effect occurs within gravitational fields. So, when my rock of 100 kg hits the event horizon not only does it appear to freeze, but its mass is also infinite to any observer? Is that rubbish? – Matt Luckham Feb 23 '12 at 10:26
  • Who said its speed becomes infinite? SR time dilation is different from gravitational time dilation IIRC. The former deals with velocities, the latter sort of deals with acceleration. But not infinite acceleration. – Manishearth Feb 23 '12 at 10:35
  • I never said speed becomes infinite. Everything I have read states the SR time dilation and gravity time dilation is the same. I quote: "This is because gravitational time dilation is manifested in accelerated frames of reference or, by virtue of the equivalence principle, in the gravitational field of massive objects." – Matt Luckham Feb 23 '12 at 10:43
  • @MattLuckham Sorry, I meant speed becomes $c$. Yes, SR and GR time dilation are fundamentally the same thing, but you're mixing it up a bit too much. Your logic is as follows, right: $\infty \text{time dilation in SR}\implies v=c\implies\gamma=\infty\implies m=\infty;$

    $\therefore\infty \text{time dilation in GR}\implies m=\infty;$ $\because \text{time dilation in SR}\cong\text{time dilation in GR}$. Here, you're mixing up cause and effect. (continued in next comment)

     – Manishearth Feb 23 '12 at 11:21
  • $v=c\implies \text{time dilation}=\infty$. Not the other way around. In GR, we can have infinite time dilation without any velocity issues. – Manishearth Feb 23 '12 at 11:23
  • "It will see the horizon approaching, and will fall through it. " - the falling observer will never see himself crossing the horizon. The horizon will always be at the distance, even whe he is already inside the BH, because he will see a different horizon, not the BH horizon. – Anixx Dec 07 '18 at 19:22
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One version I heard is this : the radius of the event horizon can be defined using the mass it envelopes. Now although from an outside observer, an object never enters the event horizon, but in a finite time it will be very very close to it. Now if you include this object as part of the black hole and recalculate the event horizon, you'll find that the new event horizon already includes this object, therefore the object can be seen as inside the newly formed black hole.

seilgu
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You see objects freezing outside the event horizon. You see event horizon moving outwards when more stuff falls into the event horizon. Stuff in the event horizon does not move outwards, when event horizon moves outwards, therefore objects become engulfed by the event horizon.

kartsa
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  • Interesting? I have not thought of this one. Is that what happens when we have a 1 million solar mass black hole and a 100 kilo rock meets the event horizon? – Matt Luckham Feb 23 '12 at 09:36
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    This is not true. The stuff close to the event horizon does move outwards as the BH radius increases. Even more with any BH deformations such as waves on its surface, the tidal deformations or the change of the rotation speed, all the oblects close enough to the horizon remain "sticked" to it and follow all the changes of the BH form. All objects close enough to a rotating BH horizon, rotate with it at the same speed. – Anixx Feb 23 '12 at 10:30
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    @Anixx Well maybe stuff does follow any motion of horizon. I just repeated what some big boys said. – kartsa Feb 24 '12 at 00:22
  • @MattLuckham I don't know. But hey, does a rock have any energy besides kinetic energy at the horizon? If it has some energy, then there's an event horizon around the rock. Then the problem becomes a collision of event horizons problem. (Near event horizon just a little bit of extra curving of space-time will make an event horizon) – kartsa Feb 24 '12 at 00:40
  • @ kartsa, the downvote is not due to me, I know that some big boys claim such things. – Anixx Feb 25 '12 at 02:14
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The problem with testing if anything but the image/information remains floating above where the event horizon should be is that any signals an object emits get both slower and red-shifted to the point where a probe will become essentially unresponsive and invisible. One possible way to verify if objects are really still hovering there is if you equipped a bunch of test probes with mirrors that reflect short wavelengths really well (the reason to have a bunch of them is that eventually, even single photons will transfer a destructive amount of momentum). I'm guessing you will get reflections as long as you don't run out of probes (though you WILL have to wait exponentially longer for each additional measurement).

Edit: Based on a comment by Anixx, the probes won't be "swallowed" as their mass expands the black hole, but will simply be pushed outward slightly.

Dustin Soodak
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  • "eventually, even single photons will transfer a destructive amount of momentum" - This would be true for hovering observers, but it is technically impossible to hover there. However, for free falling observers the radial outside light is actually redshifted (asymptotically twice at the horizon). – safesphere Dec 23 '19 at 05:41
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If my knowledge of time dilation is correct, it goes both ways. Time slows down for the object in this situation. But, time does not slow down for an outside observer. Therefore, no, time would be infinite only for the object nearing the black hole. (I might be wrong, if I am, please tell me in the comments)

user37390
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At the event horizon, the person getting sucked in will see light twice as fast. Then as he falls in he will eventually see light 3 times as fast, then 4 times as fast, then 5 times as fast, until light seems to be infinitely fast and time starts to grow infinite to him, and probably he will seem to take a split-second to fall in, then he will be destroyed because there is only space for less than a cubic-planck (infinitesmall) while he's bigger than a cell. However outside, you'll never see something falling in the event horizon, and when the escape velocity needed is half the speed of light, you will see objects going half the speed, which isn't so bad, but after a while, when the speed needed becomes 299, 792, 457 meters per second, and pretend there's a clock, it will be 1/299, 792, 458 times as fast and therefore take around 9 years to go 1 second, which probably the clock already fell in. The person falling in, however feels nothing special when crossing the event horizon, and can even communicate with another person falling in, until one is destroyed by gravity singularity. There's no way to go into a black hole and tell the tale to everyone, but there's a way to identify every object that has fallen in the black hole since the big bang, but you'll need a super duper fast running program and much better eye-sight than normal. Since the faster you go the slower time is for you because the speed of light is only a little faster than you, you'll have to wait until you're almost the speed of light like 299, 792, 457.99[...]9 meters per second, then immediately without waiting 299, 792, 458/ the number I wrote -1 seconds, and move back at the speed of light all the way for a very, very long time, probably more than the time big bang took to form and tell the tale. And no, you age at the same speed, but time just SEEMS much slower because light is only going a bit faster than you.

Brandon Enright
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bacca2002
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    This is completely wrong. Observers will never observe any speed for light other than c but they will see red / blue shifting. – Brandon Enright Nov 27 '13 at 22:32
  • I didn't say that for observers. I said that for the person that falls in, which probably will get destroyed firsts unless the black hole has a lot of mass. And the second part was about before the object falls in the horizon, which time for the person will feel much slower, while the person falling in will feel the light coming about twice as fast, assuming it goes in a straight line which it should without any force(which is not true). And the object would rip apart right before falling if part of the object was through the horizon and part isn't, and it's going the speed of light. – bacca2002 Nov 27 '13 at 22:49
  • This is wrong: "while the person falling in will feel the light coming about twice as fast". So is this "the object would rip apart right before falling if part of the object was through the horizon and part isn't". This is impossible: "and it's going the speed of light". – Brandon Enright Nov 27 '13 at 23:49
  • This would be true except for the fact that the observer falling into the BH is also getting closer and closer to light speed so you have to also take SR effects into account. I believe the SR doppler effect ends up being the exact inverse of the GR time slowing effect plus SR time slowing effect so they should end up seeing both the same frequency and same duration if they are in free fall. Different story completely, of course, if they have rockets powerful enough to hover just above the event horizon. – Dustin Soodak Jan 02 '19 at 22:41