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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?

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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 9:53
Look at this answer: physics.stackexchange.com/a/18993/1186 –  Anixx Feb 23 '12 at 10:24
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
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
@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

12 Answers 12

up vote 9 down vote accepted

Indeed, nothing can get under the horizon. 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.

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 BH with the same mass. This matter is virtually frozen on the surface like with actual BH, due to high gravity level.

Such collapsars possibly can become BHs 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 BHs 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.


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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 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 0:11
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 1:30
"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 1:38
This answer is completely wrong. There are several correct answers by people competent in relativity for the question that this one duplicates: physics.stackexchange.com/questions/5031/… –  Ben Crowell Nov 16 '14 at 22:52

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.

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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 2:08
Also thanks for noting that the cosmologists do not believe in actual black holes. –  Anixx Feb 25 '12 at 2:10
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. –  Nathaniel 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
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! –  WetSavannaAnimal aka Rod Vance Jan 7 '14 at 3:39

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 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.

See these questions and answers for more details: How does the star that has collapsed to form a Schwarschild black hole appear to an observer falling into the black hole? and How does an object falling into a plain Schwarschild black hole appear from near the black hole?

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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 9:34
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 3:53
@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 7:06

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 co-ordinate 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 co-ordinate system used to measure locations of things on the Earth's surface.

General Relativity gives us a way to describe the universe that is independant 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 Schwartzchild 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 co-ordinate system is co-moving co-ordinates 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 co-ordinate 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 co-ordinate system, but all that means is that our co-ordinate 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 co-ordinate system of the freely falling observer the event horizon does exist and can be reached in a finite time.

You ask:


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 Sgr 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 Sgr A* is actually a black hole or not is irrelevant.

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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
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
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
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
See math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… for a reasonably accessible description of the maths involved. –  John Rennie Feb 24 '12 at 9:03

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.

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+1: indeed, not controversial. –  Ron Maimon Mar 24 '12 at 5:09
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 3:16
@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 9:53
@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 1:35
@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

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.

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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.

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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

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.

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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 9:36
@Anixx Well maybe stuff does follow any motion of horizon. I just repeated what some big boys said. –  kartsa Feb 24 '12 at 0:22
@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 0:40
@ kartsa, the downvote is not due to me, I know that some big boys claim such things. –  Anixx Feb 25 '12 at 2:14

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.

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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 3:02
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 9:39

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.

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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)

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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.

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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

protected by Qmechanic Feb 18 '14 at 22:44


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