How can anything ever fall into a black hole as seen from an outside observer?

Question

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?

Answer 1

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.

source

Answer 2

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.

Answer 3

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 there is no point at which the laser disappears, you will just have to wait longer and longer between times when you detect the longer and longer wavelength radio waves from the blue laser. This will go on forever - you will never see the laser pass the event horizon.

On the other hand, your friend who is riding on the laser does not even see anything unusual 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. 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?

Answer 4

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.

Answer 5

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.

Answer 6

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:

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 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.

Answer 7

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.

Answer 8

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.