Understanding the Fall into a Black Hole: A Scientific Analysis

[TobyC]

I have been reading about black holes and am intrigued by the fact that when you use schwarzschild coordinates to describe a particle free falling radially into a black hole, it takes an infinite amount of coordinate time for the particle to reach the event horizon. Since the coordinate time in schwarzschild coordinates is the proper time for a distant observer, it will also take an infinite amount of a distant observer's proper time before the particle crosses the horizon. This happens despite the fact that the particle crosses the event horizon and reaches the singularity in a finite amount of its own proper time.

This seems to raise lots of questions that I hope some of you might be able to answer, like how is it that we, as distant observers, can observe black holes to grow as they pull in matter? Shouldn't all this matter get stuck at the event horizon?

I found an interesting website which claimed that this phenomenon was really just an optical illusion, since particles only appear to get 'stuck' at the event horizon because the light they send out becomes infinitely red shifted. The site seemed to suggest that in reality objects fall into a black hole in a finite time. But there seems to be something more than just an optical illusion going on which I can't get my head around.

If you take a supermassive black hole so that the horizon can be reached and crossed without being torn up by tidal forces, and if a spaceship then launches itself toward the black hole it will appear, to distant observers, to get stuck at the event horizon. You could interpret this as being an optical illusion but if you then wait a hundred years or more and send in a second spaceship traveling faster, won't the second spaceship be able to actually meet up with the first and chat to the century old astronauts about what they've missed during their fall? This seems to be what should happen when you look at it with schwarzschild coordinates, but I'm unsure.

 

[Ben Niehoff]

Edit: Everything I say here is wrong! See the diagrams posted by Jesse below.

[strike]This is a common fallacy. It does NOT take infinite time (w.r.t. a distant observer) for matter to fall into a black hole.[/strike]

When the infalling matter gets close enough to the horizon, you have to take its mass into account! The extra mass will actually increase the Schwarzschild radius of the combined system. In essence, when the falling matter gets close enough to the horizon, the horizon actually expands and swallows it.

This is something you can't see from the Schwarzschild metric alone, because the Schwarzschild metric describes a stationary system. But matter falling into a black hole is NOT a stationary system; it is obviously time-dependent. Most of the time, we can use the Schwarzschild metric as an approximation to a dynamical situation; however, the approximation breaks down at some point.

 

[TobyC]

This is a common fallacy. It does NOT take infinite time (w.r.t. a distant observer) for matter to fall into a black hole.

When the infalling matter gets close enough to the horizon, you have to take its mass into account! The extra mass will actually increase the Schwarzschild radius of the combined system. In essence, when the falling matter gets close enough to the horizon, the horizon actually expands and swallows it.

This is something you can't see from the Schwarzschild metric alone, because the Schwarzschild metric describes a stationary system. But matter falling into a black hole is NOT a stationary system; it is obviously time-dependent. Most of the time, we can use the Schwarzschild metric as an approximation to a dynamical situation; however, the approximation breaks down at some point.

Great thanks for this I was wondering if something along those lines would happen. This explains it clearly.

 

[JesseM]

This is a common fallacy. It does NOT take infinite time (w.r.t. a distant observer) for matter to fall into a black hole.

When the infalling matter gets close enough to the horizon, you have to take its mass into account! The extra mass will actually increase the Schwarzschild radius of the combined system. In essence, when the falling matter gets close enough to the horizon, the horizon actually expands and swallows it.

Do you have a reference for this, or is it your own argument? If the latter I'm pretty sure you're wrong since I've seen a number of physicists say it would take an infinite time visually for an object to cross the horizon, as seen by outside observers. Also if you look at the Eddington-Finkelstein diagram for a collapsing star on this page, here we are dealing with a non-static event horizon since the diagram does show the horizon (pink line) expanding as the star collapses, but it's still true that the worldlines of light rays moving outward from near the horizon (orange lines) experience more and more delay reaching a significant distance from the horizon the closer they were to the horizon initially, so I think it's still be true that in the limit as the starting distance above the horizon at the bottom of the diagram approached zero, the time needed to escape to a given radius would approach infinity in these coordinates.

What's more, if you look at a Kruskal-Szekeres diagram for the same collapsing star, you can see in these coordinates the pink event horizon is always a line at 45 degrees, and in these coordinates light rays always have worldlines at 45 degrees, meaning that a light ray emitted from any point on the horizon (even when the horizon was much smaller in Schwarzschild coordinates) would stay on the horizon forever, never escaping to reach outside observers:

 

[Bill_K]

When the infalling matter gets close enough to the horizon, you have to take its mass into account! The extra mass will actually increase the Schwarzschild radius of the combined system. In essence, when the falling matter gets close enough to the horizon, the horizon actually expands and swallows it.

Yes, you are absolutely correct. This is an interesting point.

 

[JesseM]

Yes, you are absolutely correct. This is an interesting point.

Again, any reference for this? It's true the horizon expands, but I think it's just incorrect to claim that this means the outside observer can see an object reach the horizon in finite time.

 

[Ben Niehoff]

Hmm, after seeing Jesse's diagrams, I stand corrected. Thanks.

 

[skeptic2]

So why are different coordinate systems needed and why do you get seemingly different results depending on the coordinate system you use?

 

[JesseM]

So why are different coordinate systems needed and why do you get seemingly different results depending on the coordinate system you use?

Different results for what? All coordinate systems agree about coordinate-invariant local facts, like what a particular observer's clock will read at the moment a particular light ray strikes him.

 

[TobyC]

Wow Ok I came back to this thread with another question about your original explanation. I was going to say that isn't General Relativity invariant under time reversal? And wouldn't this mean that if a distant observer saw an object fall into the horizon in a finite time they could also see an object come back out of the horizon in a finite time?

But now it looks like that explanation wasn't actually correct anyway, so I guess that my original post is still unanswered. Do you have an answer for it JesseM?

 

[JesseM]

Wow Ok I came back to this thread with another question about your original explanation. I was going to say that isn't General Relativity invariant under time reversal? And wouldn't this mean that if a distant observer saw an object fall into the horizon in a finite time they could also see an object come back out of the horizon in a finite time?

For a black hole that forms at some finite time from collapsing matter like a star, the time-reverse would be a white hole that "explodes" (singularity and event horizon disappear) at some finite time. On the other hand, the ideal Schwarzschild black hole solution is eternal in both past and future from the perspective of outside observers, and the maximally extended Schwarzschild black hole contains both a black hole interior region that matter falls into, and a white hole interior region that matter comes out of, so to an outside observer it looks like matter can both enter and exit the same object (but they would see infalling objects approaching the horizon for an infinite time in the future, and objects coming out would have been seen climbing away from the horizon an infinite time in the past). The maximally extended solution is easiest to understand in Kruskal-Szekeres coordinates, which unlike Schwarzschild coordinates can cover the entire spacetime rather than just a portion of it.

But now it looks like that explanation wasn't actually correct anyway, so I guess that my original post is still unanswered. Do you have an answer for it JesseM?

OK, I assume you mean the question here:

If you take a supermassive black hole so that the horizon can be reached and crossed without being torn up by tidal forces, and if a spaceship then launches itself toward the black hole it will appear, to distant observers, to get stuck at the event horizon. You could interpret this as being an optical illusion but if you then wait a hundred years or more and send in a second spaceship traveling faster, won't the second spaceship be able to actually meet up with the first and chat to the century old astronauts about what they've missed during their fall? This seems to be what should happen when you look at it with schwarzschild coordinates, but I'm unsure.

If a second ship approaches the black hole much later you won't ever see it catch up with the first one, it can get appear to get arbitrarily close but never quite close the visual gap. Meanwhile the second ship sees the apparent horizon (the black boundary that obscures background stars) as being far below it even after it has in reality crossed the horizon (see http://casa.colorado.edu/~ajsh/singularity.html#r=1, along with the "through the horizon" section here which has a link to a Penrose diagram that helps explain why this is), so although it can also see the first ship just slightly above the horizon, the first ship looks far away to it and it can't catch up or even send a light signal which is seen to reach the first ship before the second ship gets crushed at the singularity.

 

[nickthrop101]

surely your light would not get out once you are past the event horizon

 

[skeptic2]

Different results for what? All coordinate systems agree about coordinate-invariant local facts, like what a particular observer's clock will read at the moment a particular light ray strikes him.

In "How can Block [sic] Holes form in finite time", post #2, you say, "They can't cross in finite coordinate time in the Schwarzschild coordinate system, but that's just a feature of that particular coordinate system, in the Kruskal-Szekeres coordinate system (and others) objects do cross the horizon in finite coordinate time"

One would think they can't both be correct. Why are different coordinate systems needed?

 

[TobyC]

For a black hole that forms at some finite time from collapsing matter like a star, the time-reverse would be a white hole that "explodes" (singularity and event horizon disappear) at some finite time. On the other hand, the ideal Schwarzschild black hole solution is eternal in both past and future from the perspective of outside observers, and the maximally extended Schwarzschild black hole contains both a black hole interior region that matter falls into, and a white hole interior region that matter comes out of, so to an outside observer it looks like matter can both enter and exit the same object (but they would see infalling objects approaching the horizon for an infinite time in the future, and objects coming out would have been seen climbing away from the horizon an infinite time in the past). The maximally extended solution is easiest to understand in Kruskal-Szekeres coordinates, which unlike Schwarzschild coordinates can cover the entire spacetime rather than just a portion of it.

OK, I assume you mean the question here:

If a second ship approaches the black hole much later you won't ever see it catch up with the first one, it can get appear to get arbitrarily close but never quite close the visual gap. Meanwhile the second ship sees the apparent horizon (the black boundary that obscures background stars) as being far below it even after it has in reality crossed the horizon (see http://casa.colorado.edu/~ajsh/singularity.html#r=1, along with the "through the horizon" section here which has a link to a Penrose diagram that helps explain why this is), so although it can also see the first ship just slightly above the horizon, the first ship looks far away to it and it can't catch up or even send a light signal which is seen to reach the first ship before the second ship gets crushed at the singularity.

Ok thanks, that makes sense I think. But when a distant observer sees an object take an infinite amount of time to fall in, isn't that still a bit more than an optical illusion, contrary to what a website I read said? Am I right in saying that until the object actually crosses the horizon it is possible for it to escape in theory if it accelerated away, with rockets for example? This should mean that if a distant observer is watching a spaceship fall into a black hole, at any time between up to infinity it would be possible theoretically for the spaceship to turn its rockets on and return? Never having actually crossed the horizon? If this is true then surely the spaceship really is 'stuck' just above the event horizon, it's not merely an optical illusion.

Assuming that there is more than an illusion going on and objects really do take an infinite amount of a distant observer's proper time to fall into a black hole, how is it that we as distant observers are able to see black holes sucking in matter and growing? Shouldn't all the matter get stuck at the horizon?

 

[skeptic2]

TobyC, how would our observations or measurements be any different if all the matter were stuck at the horizon instead of falling through?

 

[TobyC]

TobyC, how would our observations or measurements be any different if all the matter were stuck at the horizon instead of falling through?

Well wouldn't one effect of it be that a black hole would lose its spherical symmetry? Or at least there would be no guarantee that it would keep it. If it picked up more matter from one direction than another then since all the matter accumulates at the event horizon there would be more matter on one side than the other.

In the book I'm reading I'm pretty sure it said that the Schwarzschild or Kerr metrics will describe any black hole because once an event horizon has formed it radiates away gravitational waves, all the matter is sucked to the singularity, and it quickly becomes spherically symmetric and unchanging in time. I can't seem to reconcile this with the fact that when a black hole accumulates matter it gets stuck at the event horizon.

 

[JesseM]

Ok thanks, that makes sense I think. But when a distant observer sees an object take an infinite amount of time to fall in, isn't that still a bit more than an optical illusion, contrary to what a website I read said? Am I right in saying that until the object actually crosses the horizon it is possible for it to escape in theory if it accelerated away, with rockets for example? This should mean that if a distant observer is watching a spaceship fall into a black hole, at any time between up to infinity it would be possible theoretically for the spaceship to turn its rockets on and return? Never having actually crossed the horizon? If this is true then surely the spaceship really is 'stuck' just above the event horizon, it's not merely an optical illusion.

I don't think so, it just means the distant person watching can never be sure whether it crossed the horizon, not that there isn't a definite truth about it. It's similarly true that an object undergoing constant proper acceleration in SR has a "Rindler horizon", and that light from events on or beyond the Rindler horizon can never reach them as long as they keep up the same constant acceleration, so that they'll see objects that approach it slow down and never quite reach it. But as you can see from the diagram on this page, the Rindler horizon is just the worldline of a light ray (with the accelerating observer's worldline being a hyperbola that approaches it asymptotically), so an observer who wasn't accelerating in the same way could easily cross the Rindler horizon in finite time (finite proper time, and also finite time in an inertial frame like the one used to draw the diagram):

The infinite visual time for an object to cross the event horizon as seen by an observer who hovers outside is no less an "optical illusion" than the infinite visual time for an object to cross the Rindler horizon as seen by the observer undergoing constant proper acceleration.

Assuming that there is more than an illusion going on and objects really do take an infinite amount of a distant observer's proper time to fall into a black hole, how is it that we as distant observers are able to see black holes sucking in matter and growing? Shouldn't all the matter get stuck at the horizon?

Well for one thing this is just an idealization, you would only see matter at the horizon forever if light were emitted in a perfectly continuous way rather than in discrete photons, and if you could detect light at arbitrarily high redshift, neither of which is realistic in practice (see the discussion in this entry of the Usenet Physics FAQ). But even in the ideal case, I would guess that you could still see the visual horizon (black region blocking out background stars) growing or shrinking, you'd probably just see the images of things "stuck on the horizon" move with it.

 

[TobyC]

I don't think so, it just means the distant person watching can never be sure whether it crossed the horizon, not that there isn't a definite truth about it. It's similarly true that an object undergoing constant proper acceleration in SR has a "Rindler horizon", and that light from events on or beyond the Rindler horizon can never reach them as long as they keep up the same constant acceleration, so that they'll see objects that approach it slow down and never quite reach it. But as you can see from the diagram on this page, the Rindler horizon is just the worldline of a light ray (with the accelerating observer's worldline being a hyperbola that approaches it asymptotically), so an observer who wasn't accelerating in the same way could easily cross the Rindler horizon in finite time (finite proper time, and also finite time in an inertial frame like the one used to draw the diagram):

The infinite visual time for an object to cross the event horizon as seen by an observer who hovers outside is no less an "optical illusion" than the infinite visual time for an object to cross the Rindler horizon as seen by the observer undergoing constant proper acceleration.

Well for one thing this is just an idealization, you would only see matter at the horizon forever if light were emitted in a perfectly continuous way rather than in discrete photons, and if you could detect light at arbitrarily high redshift, neither of which is realistic in practice (see the discussion in this entry of the Usenet Physics FAQ). But even in the ideal case, I would guess that you could still see the visual horizon (black region blocking out background stars) growing or shrinking, you'd probably just see the images of things "stuck on the horizon" move with it.

Hmmm... Ok but how will the horizon grow or shrink if no infalling matter is crossing it?

 

[JesseM]

Hmmm... Ok but how will the horizon grow or shrink if no infalling matter is crossing it?

In some coordinate systems like Eddington-Finkelstein or Kruskal-Szekeres it is crossing at a finite coordinate time, even if the light never reaches us. So in these coordinate systems the horizon can grow and shrink. Calculating the size of the apparent visual horizon would probably be complicated, but all coordinate systems should agree on what is seen visually, and from descriptions of black hole evaporation I'm pretty sure the result of such calculations is that the size of the horizon can appear to change visually.

 

[skeptic2]

Well wouldn't one effect of it be that a black hole would lose its spherical symmetry? Or at least there would be no guarantee that it would keep it. If it picked up more matter from one direction than another then since all the matter accumulates at the event horizon there would be more matter on one side than the other.

In the book I'm reading I'm pretty sure it said that the Schwarzschild or Kerr metrics will describe any black hole because once an event horizon has formed it radiates away gravitational waves, all the matter is sucked to the singularity, and it quickly becomes spherically symmetric and unchanging in time. I can't seem to reconcile this with the fact that when a black hole accumulates matter it gets stuck at the event horizon.

This can be analyzed using the bowling ball and rubber sheet analogy. The bowling ball represents a black hole and the rubber sheet stretches according to the formula 1/√(1 – ν^2/c^2 ) where v represents the escape velocity√(2GM/r) and r is distance from the center of the black hole. If we are observing an object rolling down the rubber sheet towards the bowling ball from above, corresponding to what a distant observer would see, it would appear that the closer it gets to the horizon the less rapidly it approaches it (it slows down). The object isn’t really slowing down, in fact it is falling faster and faster. It’s just that it is falling at a steeper and steeper angle. To the falling object, it is covering more and more distance but to the observer above it appears to be covering less and less. This demonstrates how to a distant observer, distance in the radial direction from the black hole is contracted. But it’s not just space that is contracted, the object itself also becomes thinner and thinner in the radial direction as it approaches the event horizon. So if matter were to accumulate on one side of a black hole, not only would the center of mass of the black hole shift so that the black hole would still be nearly symmetrical but all the accumulated matter would be so thin it wouldn’t be visible as a bulge.

What happens when the object reaches the event horizon? It never does actually, not even in its own proper time. If we apply the formula for how the rubber sheet stretches, we see the bowling ball has stretched it infinitely far downwards and infalling objects fall forever, never reaching the event horizon.

This model avoids many of the problems encountered with other models of black holes in which objects may cross the event horizon.
It doesn’t have to deal with infinite densities of the singularities.
It doesn’t have to deal with imaginary coordinates for space and time inside the event horizon.
It doesn’t have to explain how the effect of a change in mass of the singularity travels backwards in time, to escape the event horizon in order to change the gravity of the black hole.
There is no information paradox.
This model satisfies all of the predicted, observable effects of black holes.

It is interesting to note that the coordinates Schwarzschild’s defined, begin at the event horizon and only are defined outside the event horizon. Was that because Schwarzschild, himself did not believe anything could cross the horizon?

 

[JesseM]

This can be analyzed using the bowling ball and rubber sheet analogy. The bowling ball represents a black hole and the rubber sheet stretches according to the formula 1/√(1 – ν^2/c^2 ) where v represents the escape velocity√(2GM/r) and r is distance from the center of the black hole. If we are observing an object rolling down the rubber sheet towards the bowling ball from above, corresponding to what a distant observer would see, it would appear that the closer it gets to the horizon the less rapidly it approaches it (it slows down). The object isn’t really slowing down, in fact it is falling faster and faster. It’s just that it is falling at a steeper and steeper angle.

I don't think that really works, what you're really talking about is an embedding diagram which shows the curvature of space for a surface of constant t-coordinate in Schwarzschild coordinates. Distances along paths on this embedding diagram accurately reflect the 'proper distance' along spacelike curves which are confined to a single surface of constant t-coordinate, found by integrating the metric along these curves. The timelike worldline of a falling object would not stay within a surface of constant-t coordinate, but since the Schwarzschild geometry outside the horizon is a "stationary" one (none of the metric coefficients depend on t) the embedding diagram doesn't change from one t-coordinate to another, so you can plot where the in space the falling object is as a function of time on this embedding diagram. But since the distance from any finite R to the horizon is finite in the embedding diagram, yet it takes an infinite coordinate time for anything to reach the horizon, it can't be true that the object is "falling faster and faster" in this diagram, in fact it should be falling slower and slower as it approaches the horizon in the embedding diagram.

One of the problems with the "rubber sheet" analogy is that people take it too literally, it's not as if there's a gravitational field causing things to "fall" into depressions on the sheet, the rubber sheet is just supposed to represent the way curvature distorts spatial distances on a surface of simultaneity in some coordinate system, you could just as easily flip it over so that gravity wells became gravity hills and it would represent the distances just as accurately.

What happens when the object reaches the event horizon? It never does actually, not even in its own proper time.

But that's provably wrong according to the mathematics of general relativity, and the rubber sheet analogy is just meant to be a visualization of one aspect of GR. Please note the IMPORTANT! Read before posting thread at the top of the forum, this forum is not meant for people to use as a platform for advertising personal theories which conflict with mainstream ones like GR.