User:Spinningspark/Black hole anomaly

Black Holes!

snipped from answers on the reference desk science page

What would happen if you had a space ship out in intergalactic space, and they tied a rope around an astronaut and dropped him thousands of light-years into a black hole (the space ship is out of danger etc etc, and the astronaut doesn't starve). He falls past the event horizon.

• Does the rope continue to play out? How far does it play out? Does it go 1 event-horizon radius and then go slack (or at least not be pulled anymore by the astronaut.. to say nothing of gravitational effects on the rope itself), or is it "bigger inside"?
• How much force is put on the rope? Would it pull the space ship toward the black hole?
• If the astronaut, after falling past the event horizon, pulled on the rope with all his strength, would his crewmates playing out the rope feel the tug? Could he communicate in morse code to them or something?

--ffroth 01:56, 31 December 2007 (UTC)

The ship's crew would see the astronaut slow down (and become more and more redshifted) so that he approaches the event horizon but never crosses it. From their point of view, neither the astronaut nor the rope ever touches the event horizon, so the experiment gives them no new information about the inside of the black hole (as required by the definition of "event horizon").

The experience of the astronaut himself depends on what the ship's crew eventually does. If we assume they can't keep holding the rope forever, then they must either pull him back up, or let go of the rope (or the rope breaks somehow, which amounts to the same thing). If they pull him back up, then nothing particularly unusual happens (except that the astronaut will have aged less than the crew). If they let go of the rope, then the astronaut falls in and gets to see the inside of the black hole, but he can't tell the crew about his experience, not even by tugging Morse code on the rope, because it's no longer connected (P.S. also because the crew has already died of old age).

The one important point is that unless the ship's crew allows the rope to continue paying out forever (by their watches), the astronaut never gets across the event horizon. In any event, the crew never observes anything cross the event horizon. That's why the word horizon is so appropriate, because you can never observe anything cross it. —Keenan Pepper 04:34, 31 December 2007 (UTC)

So how did "stuff" get inside the event horizon in the first place in order to create the black hole? I've always failed to grasp the theory of black hole creation by gravitational collapse for this very reason. Spinningspark (talk) 17:21, 1 January 2008 (UTC)

The original star that became the black hole had "stuff" - otherwise it wouldn't have stayed over the Chandrasekhar limit (I'm not sure I got that right) and become a black hole. Effectively the event horizon means that things get inside, but nobody sees them get inside because their time moves much, much, much faster than ours. I remember the thought experiments about "what if the moon fell into a black hole, astronauts included", and apparently the astronauts would get to see the universe age with their own eyes at superspeed as they fell in, but after that, who would know ('cause it's a singularity). Altho there is a very small chance that would happen. ~user:orngjce223 how am I typing? 19:48, 1 January 2008 (UTC)

Yes, the original star had "stuff". But from our inertial frame it will take forever to collapse beyond its own event horizon. In its own inertial frame (or an astronaut falling in) it happens in a finite normal time, but to an outside observer it does not. So given that a black hole exists and postulating that it was formed by the collapse of a super heavy star, it must have happened forever ago in our frame. Which leads to the conclusion that black holes existed before the creation of the universe. This is a contradiction, or if it is not, no ione has ever succeeded in satisfactorily explaining it to me. Spinningspark (talk) 23:42, 1 January 2008 (UTC)

So, you could just pull someone/something out from beyond the event horizon of a black hole using a rope? --Kurt Shaped Box (talk) 22:43, 1 January 2008 (UTC)

I don't want to put words in anyones mouth but I think the previous posters were saying just the opposite. Spinningspark (talk) 23:45, 1 January 2008 (UTC)

Yeah, I think I misread the response slightly. Still, how could they continue playing out the rope, yet not have the astronaut in question eventually cross the EH? Surely it's just a question of falling a certain distance in a certain period of time? I mean, does the astronaut actually fall any slower in reality, just because the effects of the black hole are (to the observer) screwing with human time perception? EDIT: Yes, I realize that I may be moving the discussion along to a complex debate on the existence of reality outside of human perception.... ;) --Kurt Shaped Box (talk) 01:52, 2 January 2008 (UTC)

No, the spaceship NEVER observes the astronaut cross the EH. The descent of the astronaut becomes slower and slower from their perspective as the EH is approached. This is down to time dilation, the astronaut also appears to age very slowly. The astronaut, from his perspective, on the other hand, is falling at a perfectly normal rate but the spaceship and everone on it is ageing very fast. This scenario is similar to the twins paradox only gravity is the cause of the time dilation in the case of black holes. Spinningspark (talk) 07:14, 2 January 2008 (UTC)

Careful, there's been some misinformation in this thread. I'm not sure whether the astronaut is being slowly lowered or whether he's bungee-jumping. I'll assume bungee-jumping. In that case, as others have said, the ship sees the moments just before the astronaut crosses the event horizon stretched out and redshifted into the indefinite future. (Classically, anyway. In reality the astronaut emits only finitely many photons before falling through, and the last one will reach the ship after a finite, and typically very short, time.) This is the Doppler effect, not time dilation. The astronaut does not see a reciprocal blueshift. He sees only a part of the ship's history before falling in, and more of its history after. (You can see out of a black hole from inside.) The ship does not have an unlimited time to pull the astronaut back out. The time in the ship's history that the astronaut sees when crossing the event horizon is also the moment at which it becomes impossible for the ship to do anything to save the astronaut. After that, although they still see light from before the astronaut falls through, it's causally impossible for them to save him; in effect he has "already fallen through". I hope these ugly ASCII-art diagrams will make things clearer.

            |      /.
            |     /.
            |    /.
            |   /.
            |  / .
            | / .
            |/  .
            /  .
           /|  .
          / | .
         /  | .
        /   | .
       /    |.
      /     |.
     /      |.
    /       .

This is a portion of a Kruskal-Szekeres diagram of a black hole (or something like that, anyway). The diagonal line is the event horizon (a null line). The vertical line is the astronaut (this needn't be straight, it's just hard to draw otherwise). The dotted line is the ship; its worldline is a hyperbola with the event horizon as one asymptote. Light from the astronaut to the ship travels like this:

            |      /.
            |     /.
            |    /.
            |   /.
            |  //.
            | //.
            |///.
            ///.
           /|//.
          / |/.
         /  |/.
        /   |/.
       /    |.
      /     |.
     /      |.
    /       .

It's hard to see in this picture, but the hyperbolic shape of the ship's worldline means that light emitted arbitrarily close to the event horizon reaches the ship arbitrarily far in the future (unbounded redshift) and light from behind the event horizon never reaches the ship at all. Light from the ship to the astronaut [error corrected —KP] travels like this:

            |      /.
            |\    /.
            |\\  /.
            |\\\/.
            |\\/\.
            |\/\.
            |/\\.
            /\\.
           /|\\.
          / |\*
         /  |\.
        /   |\.
       /    |.
      /     |.
     /      |.
    /       .

The * is the point in the ship's history that the astronaut sees when crossing the event horizon, and also the first point at which the ship can no longer take action to save him (since the effect of any action is limited to its future light cone). In practice the point of no return occurs far earlier, since real materials only have so much tensile strength and the force you need to save the astronaut goes to infinity as you approach the critical time.

The hole will continue to pull the rope in indefinitely; it's just like throwing anything else into a black hole. Suppose the ship decides to stop releasing the rope some time after it's too late to save the astronaut. Then, assuming they can stop the rope at all, it will exert a continuous pull on the ship for as long as they continue to hold it. This pull has a familiar name: weight. The weight of the rope is ∫ a dm = ∫ ρ(r) a(r) dr. The acceleration a goes to infinity as r approaches the event horizon, but the rope does not extend that far; it has necessarily broken somewhere above the horizon where its tensile strength was exceeded. Since the rope is broken, the ship can't use it to signal the astronaut, but it can signal him by other means (like a light beam). The astronaut can't signal the ship by any means. -- BenRG (talk) 09:11, 2 January 2008 (UTC)

I was wondering about the same thing, found inconsistent answers on this page, and eventually found a great reliable summary here: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

How can you ever fall in a black hole?

Another attempt to get an answer from the reference desk

Since time slows per the reference frame of an outside observer as an object approaches the event horizon, and 'freezes' at the horizon (thus the Russian term for black hole: 'frozen star'), how can a black hole ever increase in mass? This was once considered a flaw in the concept of black holes, but evidently was somehow resolved. However, I've never seen an account of how it was resolved. (And yes, I do understand that for an observer falling past the event horizon, supposedly nothing unusual would appear to happen - except that you'd think s/he'd witness the end of the universe.) — kwami (talk) 07:03, 10 March 2008 (UTC)

This question has been previously answered here [1] and probably many times before. However, I never felt satisfied with the answer given then, for the same reason given by Kwamikagami. If someone can elaborate, please do. SpinningSpark 08:15, 10 March 2008 (UTC)

I’m no physicist. But as the mass of the black hole increases, the event horizon will increase in radius and objects that previously were hovering on the edge will be enveloped. And objects don’t need to actually be at the center of the singularity to contribute to the black hole’s mass/gravitational effect. (This is just my speculation.) — Knowledge Seeker দ 08:33, 10 March 2008 (UTC)

But the BH can't increase in mass unless something crosses the event horizon in finite time in the external reference frame. — kwami (talk) 08:46, 10 March 2008 (UTC)

I’m not sure I agree with that statement. — Knowledge Seeker দ 08:51, 10 March 2008 (UTC)

How would it increase in mass, without mass being added to it? — kwami (talk) 08:54, 10 March 2008 (UTC)

Perhaps I should be more clear. I contend that a mass placed just outside the event horizon will increase the gravitational field/curvature of space (as it would anywhere) and that therefore a small volume of space which previously had almost, but not quite enough gravity to have an escape velocity greater than c (or equivalent formulation) now will; in effect, the combined event horizon around the two objects will be larger. (Again, speculation.) — Knowledge Seeker দ 08:57, 10 March 2008 (UTC)

A group of massive objects, act as if all their gravity were concentrated at thier center of mass. So from a distant observer's POV anything frozen at the edge of an event horizon would still contribute to the total mass of the black hole system. Does that resolve the problem? Theresa Knott | The otter sank 12:28, 10 March 2008 (UTC)

Does this mean that an unfortunate astronaut just outside the event horizon might be observed by a "local" observer (say, another astronaut also falling into the black hole) as not having quite reached the horizon, but that both might be "observed" by a distant observer to be within the event horizon (i.e. part of the black hole)? dbfirs 12:54, 10 March 2008 (UTC)

Hi. On a Discover magazine there's this guy named João Magueijo that has a theory called Varying Speed of Light (VSL). When he introduced this theory, one physicist said it really stood for "Very SiLly". However, according to this theory, time and the speed of light slow near a black hole and never allow anything to escape nor go in (after all, black holes are infinitely dense). Hope this helps. Thanks. ~AH1^(TCU) 14:22, 10 March 2008 (UTC)

As far as I know Magueijo's VSL theory is supposed to be an alternative to cosmic inflation, so it only matters in the very early universe and doesn't have any bearing on (present-era) black holes. -- BenRG (talk) 21:33, 10 March 2008 (UTC)

Objects can and do fall in. They only appear to stop at the event horizon. 64.236.121.129 (talk) 16:43, 10 March 2008 (UTC)

The objects may see themselves falling in. An outside observer sees the objects dim and disappear as they approach the Schwartzchild radius. An alternate theory Gravastar suggests that matter falling into a compact stellar object will appear to cool and form a Bose-Einstein condensate. One consequence of general relativity is that perception is affected by perspective. Jehochman ^Talk 16:51, 10 March 2008 (UTC)

64 and Jehochman, it’s not that Kwami’s really disputing whether they fall in or not (the phrasing of the heading, unfortunately, is misleading). His question is (as I understand it), essentially: If, from the reference frame of an outside observer, an object never passes the event horizon [due to time dilation], how/when will he observe an increase in the black hole’s mass? — Knowledge Seeker দ 18:11, 10 March 2008 (UTC)

If matter has fallen inside the event horizon, how can you get back any information about mass inside? If light can't get out, neither can any other sort of information. If an outside observer can never see anything cross the event horizon, then presumably from their point of view, nothing ever has. A sphere or shell of mass around a point will be indistinguishable to an outside observer from all that mass being located at the center point. There would be no observable difference on the force of gravity affecting an outside observer. There are a lot of contradictions if we assume the existence of black holes. This may mean that our black hole model is wrong. Jehochman ^Talk 18:16, 10 March 2008 (UTC)

I don't know that much about black hole dynamics, but I think Knowledge Seeker is correct. Certainly it's a mistake to think that infalling matter has to fall through the event horizon to add to the mass, because nothing that happens inside the event horizon can be relevant to physics outside. Whatever it means for "the black hole to gain mass", it has to involve only the physics outside the event horizon.

The event horizon is acausal; it will happily expand faster than light to engulf infalling matter. Here's a special relativistic analogy. I'll use 2+1 dimensional Minkowski space because it's easier to visualize than 3+1 dimensions and 1+1 isn't enough for this example. At time t = 0 pick some region of the xy plane and magically destroy everything there. There's then a region of spacetime (a collection of events) with t < 0 which is unobservable to anyone who survives past t = 0, because any signal that might have originated from those events was destroyed. For example, say you destroy everything in the circular region x² + y² < 1 light year². Then the unobservable region is a cone in Minkowski space whose apex is x = 0, y = 0, t = −1 year and whose base is the destroyed disc. If you destroy a square (|x| < 1 ly, |y| < 1 ly) then the unobservable region is a square pyramid. The boundary of the unobservable region is the event horizon. The event horizon is always a null surface because of how it was defined, but it may have "creases" which are not null. For example, in the square pyramid case you have corners which go from x = y = 0 at time t = −1 year to x = ±1 ly, y = ±1 ly at time t = 0; they're effectively moving at c√2.

Let S be the destroyed set. Then the observable region at time −t is $\bigcup _{x\in S^{c}}B_{ct}(x)$, where S^c is the complement of S and B_r(x) is the ball of radius r centered at x. Imagine S being eroded away from all sides at the speed of light as you go backward in time. Now say S is a sort of peanut shape, or an overlapping union of two circles. As you go back in time it will become a thinner peanut; then it will break into two teardrop shapes, with the pointy bits pointing toward one another; then the teardrops will shrink and become more circular until they disappear. Now run this forward in time: two expanding teardrops appear out of nowhere, extend toward each other faster than light, and merge. That's kind of what a black hole merger is like. The event horizon is not at all like a physical object obeying dynamical laws; it's a global property of the whole spacetime and it "knows about the future". In fact it's a theorem that the event horizons of two black holes that will eventually merge have the teardrop point on them from the beginning. That's for merging black holes, i.e. tossing one black hole into another, but I think tossing ordinary matter into a black hole works similarly. Whether any of this answers the original question I'm not entirely sure. This is a tricky subject. -- BenRG (talk) 21:33, 10 March 2008 (UTC)

I wonder if Knowledge Seeker might be on the right track: Once an object approaches a black hole close enough that their center of mass lies within the EH, does the EH then expand correspondingly? But this would happen while a small mass is still quite far from the EH, so it wouldn't be engulfed. (If you were to drop a sandwich into a Solar-mass black hole, it would hardly affect the EH at all, so we're left with my original question.)

It's not true that we can't know what's within the EH, only that no signal can be sent out. Fields extend beyond the EH, but any modulations of those fields would be red-shifted to a frequency of zero. However, with those fields we can still measure the total electronic charge, mass, and momentum (linear and angular). — kwami (talk) 22:23, 10 March 2008 (UTC)

Also, pace another comment, it isn't the case that the object just appears to slow down, due to some kind of optical illusion. It isn't light that slows down, but time itself. In the external reference frame, it really doesn't reach the EH. Thus the problem. — kwami (talk) 22:50, 10 March 2008 (UTC)