My understanding is that an observer outside a black hole would see an object falling into the hole as taking forever to cross the event horizon. But black holes evaporate due to quantum effects, so they will not last forever. So if you could wait around the 10^65 years for a stellar mass black hole to evaporate, what would you observe?
I just wanted to say, damn good question. I knew, or thought I knew, both these attributes about black holes but never noticed the apparent contradiction. Sorry I can’t offer an answer.
Um.
UM.
At first, I thought I had an easy answer for you. But now I wonder…
A person falling into a BH would appear to an outside observer to take essentially forever to fall in. Not exactly forever, though; the person’s time would appear to slow down more and more as they approached the event horizon, the place where the escape velocity equals the speed of light. That’s the so-called “edge” of the BH.
But if BHs do evaporate, that adds a fun twist. Let’s say a BH emits a particle once every 1 minute (I’m just making that number up). To us, we see it spurt out a particle every minute, and it loses a tiny fraction of its mass. Over zillions of years, it goes away entirely.
Zillions of our years. Now, I don’t claim to be an expert on relativity, but it seems to me that to the person falling in, time appears to flow greatly faster than to someone in flat space. They don’t see the BH emitting every minute, they see it pouring out at a much higher rate. If they could somehow orbit the BH just above the event horizon, they would see the BH glowing as it emitted radiation much faster than we see it, from our vantage point far away.
By the time they actually get to the event horizon, enough time will have passed in flat space (by definition!) that the BH will have evaporated.
Yikes. Can this be right? Someone falling in would see the BH getting brighter and brighter until it exploded underneath them. Only minutes or even seconds will have passed for them, while vast amounts of time will have passed outside.
I need to run this past some friends of mine. I wonder if I missed something here?
From the Sci.physics FAQ
http://www.phy.hr/~kkumer/BHfaq.html#q9
Won’t the black hole have evaporated out from under me before I reach it?
We’ve observed that, from the point of view of your friend Penelope who remains safely outside of the black hole, it takes you an infinite amount of time to cross the horizon. We’ve also observed that black holes evaporate via Hawking radiation in a finite
amount of time. So by the time you reach the horizon, the black hole will be gone, right?
Wrong. When we said that Penelope would see it take forever for you to cross the horizon, we were imagining a non-evaporating black hole. If the black hole is evaporating, that changes things. Your friend will see you cross the horizon at
the exact same moment she sees the black hole evaporate. Let me try to describe why this is true.
Remember what we said before: Penelope is the victim of an optical illusion. The light that you emit when you’re very near the horizon (but still on the outside) takes a very long time to climb out and reach her. If the black hole lasts forever, then the light
may take arbitrarily long to get out, and that’s why she doesn’t see you cross the horizon for a very long (even an infinite) time. But once the black hole has evaporated, there’s nothing to stop the light that carries the news that you’re about to cross the
horizon from reaching her. In fact, it reaches her at the same moment as that last burst of Hawking radiation. Of course, none of that will matter to you: you’ve long since crossed the horizon and been crushed at the singularity. Sorry about that, but you
should have thought about it before you jumped in.
Actually, I have a problem with his explanation. Thje light you emit will not slow down as you get closer to the event horizon. It redshifts infinitely, but the velocity is still the same. I need to think about this some more. As I said, the details of relativity are not my forte.
I think this link covers what you’re talking about
http://www.public.iastate.edu/~physics/sci.physics/faq/fall_in.html
basically stuff still disappears to outside viewers when that stuff approaches and goes into the event horizon.
I also have a couple of problems with the usual explanations of why a BH isn’t a frozen star. The singularity at the event horizon is dismissed as only a coordinate singularity, but the coordinate system is WRT to a faraway stationary observer. Isn’t that us? Doesn’t this mean that for us, here on Earth, time does come to stop at the EH?
The frozen star syndrome is also explained away by invoking the Finkelstein frame of reference, but this FoR doesn’t correspond to anyone’s proper time, so what good is it?
That BH would most likely be gaining mass via light and mass falling into it faster then HR depleting it (since it was so close you could orbit it - I am assuming that there must be something near it)
Also if the BH is rotating you get a ring ‘singularity’ (as opposed to a point) and most are assumed to rotate due to conservation of angular momentum. - doesn’t really matter for the op though.
In the present Universe, a stellar-mass black hole will gain mass from the cosmic microwave background far faster than it’ll lose it due to Hawking radiation (the CMB has a temperature of 2.7 K; the hole has a temperature of about 10[sup]-6[/sup] K). But that’s not really relevant. If we can wait around for the hole to evaporate in the first place, then we can certainly wait long enough for the Universe to cool down some more, and if we like, we can suppose that our hole is in the middle of one of the supervoids, so as to eliminate accretion from anything else (except our intrepid explorer, of course).
That said, though, I don’t know either. It’s not just a matter of the light escaping; there really is a difference in elapsed time according to the two observers. For instance, if our suicid-- er, intrepid, explorer were to stop short just outside the horizon and return to a safe distance, the rest of the Universe would have aged far more than him. I would presume that the answer is that the Schwartzchild solution (which leads to the prediction of infinite time before infall) is only exact for a steady-state hole, and that a hole of changing mass can swallow its food in a finite time, but that’s just a SWAG.
First, someone falling into a black hole takes a finite amount of time to do so in their frame of reference. In fact, if you head directly into it, it’s a very short period of time. Thus, if you’re one a trajectory that will cross the event horizon, you’ll never see the black hole evaporate. In any event, the event horizon, where Hawking radiation is occuring, is always deeper in the gravity well than any observer, including one in the process of falling into the black hole. Thus, while you’ll see the radiation less red-shifted than an observer in flat space, you’ll never see it blue-shifted. Translation: falling into a black hole won’t dramitically increase the amount of Hawking radiation you observe.
Second, an observer in flat space never observes an object falling into a black hole cross the event horizon even if the black hole evaporates. Rather, the light reaching the observer from the object get more and more red-shifted. The light is infinitely redshifted just at the instant the black hole completely evaporates.
But how can this be? you cry. If some burst of Hawking radiation occurs after the object falls into the hole, how can light from the object, even very redshifted light, be reaching a distant observer after the Hawking radiation does?
It is, as they say, a game of inches, or in this case, Angstroms, or even Planck lengths. Hawking radiation occurs when a virtual particle pair forms. The particles are created some very small distance apart. One of the pair is sucked into the event horizon and disappears. The other, on a slightly different trajectory, eventually escapes into flat space and becomes a real particle. The reason that light from an object that has already fallen into a black hole reaches you after Hawking radiation that actually occurred “later” is that the “late” light you receive from the object was emitted closer to the event horizon than the point were the virtual particle that eventually escapes to free space was created.
Let’s say that, sometime after an object has fallen into a black hole and crossed its event horizon, a virtual photon pair are created one Angstrom apart. (I’m making up these distances for illustration. The actual “allowable distance” between virtual particles depends on their energy.) One of the pair is created right next to the event horizon – or even inside the event horizon – and disappears. The other heads off for free space and is seen by a distant observer as Hawking radiation. Light from the object will still reach the distant observer later if it was emitted by the object closer to the event horizon, say at a distance of half an Angstrom.
By the way, someone mentioned a "singularity at the event horizon. The singularity is always inside the event horizon in a black hole.
Chronos
I had exactly this thought once. I think the solution is that time dilation is meaningless unless we bring the two observers back into the same frame of reference. Since the person falling into the black hole disappears from the universe, we can’t do that. FTR, IIRC, most black holes are at least 2.4 stellar masses.
Yes, that was me, but what I said was that there was a coordinate singularity at the event horizon and there most certainly is. In fact this is the definition of an event horizon. In other words the Schwartzschild radius defines a mathematical surface which is a coordinate singularity and which defines the event horizon.
You rock Lumpy. You get the prize as the first person that I’ve seen stump both ** The Bad Astronomer ** and ** Chronos**. Thease guys are some of the smartist people I’ve ever met (well virtualy met that is).
What do you mean there is no prize!:mad:
There has to be a prize!
We demand a prize!
By the way TBA congrats on your book I’m going to pick up a copy the next time I’m at B&N.
Heh. You need to get out more. 
I know more about black hole creation and the care and feeding of them in active galactic nuclei (well, a bit about that). The mechanics of relativity still amaze me. I’ll have to read these posts again several times and think on them.
Thanks. Luckily, I didn’t do a black hole chapter!
Here’s a lil notion of the Universe (can’t claim credit for it myself, forgot which friend told it):
Mass falls into a singularity (black hole).
As it does so the singularity increases in size.
This means that the diameter of the event horizon also increases.
Okay so far?
Now, suppose a whole bunch of mass is falling in.
The event horizon boundary increases in size, and as it increases more mass is within the boundary, so it increases faster and faster.
Could the event horizon end up increasing at light speed?
Creepy, no? So what happens then?
Well, suppose this singularity suddenly encompasses a humble Milky Way Galaxy.
Suddenly, a bunch of galaxies and stuff that we couldn’t see before (because they were inside the event horizon) are visible where there was nothing before.
Now we are inside a singularity. But do we really notice much difference? Isn’t a great big singularity a heck of a lot like a universe in itself?
–kooky, I know. I forgot all the details but there’s the main idea. Supposed to explain the appearance of quasars too.
Any takers on confronting it? I’d like a good counterargument.
I don’t think this is right. An observer “hovering” (accelerating to maintain position outside of the event horizon) would see this, but an infalling inertial observer won’t.
Even though an external observer (“Alice”) would see the infalling observer (“Bob”–poor Bob) taking forever to fall in, that doesn’t mean that Alice sees all of Bob’s future. In fact Alice only sees Bob’s future up to the point where he passes the event horizon. The portion of Bob’s future occuring after he passes the horizon is inaccessible to Alice.
This isn’t hard to see if you believe Penrose diagrams. The Penrose diagram for an evaporating, astrophysical black hole (figure on p.4 of the PDF available here ; I can’t find a nice HTML picture) shows that if Bob passes the event horizon he will end up at the singularity. Alice will see Bob slow down as he approaches the event horizon; his more-and-more-redshifted photons reach Alice until the black hole evaporates, but Alice will never see the part of Bob’s history that lies within the horizon (except maybe through information encoded in the final remnant or explosion; no one really knows what happens at this point).
No. The singularity is a point. It does not increase. the diameter of the event horizon does increase.
Acutally, the event horizon is already at light speed. An outgoing ray of light just at the horizon is suspended there. If you increase the mass and hence the size of the horizon, it moves faster than the light suspended there. This is not a violation of relativity, because the horizon is not a physical thing, but a feature of the geometry of space.
Crossing the event horizon does not make everything inside suddenly become visible to you. You would see nothing to the left or right of you and nothing ahead. Or rather, what appeared to be to you left and right would actually be in a narrow V with you at the point. The horizon (if it were visible) would appear to be a sphere with you outside and the universe inside.
As noted above you are always outside the singularity. (Always is a very short time inside the horizon.)
How am I doing so far? 
For what it’s worth, I asked my GR professor about this today, and he confirmed my guess: The Schwartzschild metric is an approximation (albeit a very good one) based on the assumption that the black hole is unchanging in time. If your black hole is changing, then it won’t quite take an infinite amount of time, in the external reference frame, for the infalling particle to cross the horizon, so the particle can still cross the horizon before the hole evaporates.
As for an expanding horizon encountering more matter, if the matter were that close, the horizon would already have formed around it.
Using the standard assumptions and all:
The larger a black hole’s mass, the smaller its density. For a galaxy-mass black hole, the density is significantly less than that of water. For a universe-mass black hole, the density is extremely small indeed. In fact, it’s comparable to the density of the universe itself. Therefore, the universe may be a black hole. (!) This sort of follows the lines of Smart Ask Mother For Car’s notion of the universe. So, we could be inside a black hole and not realize it. This idea of a black hole is very different from the idea we have of a stellar-mass black hole - all the mass concentrated at the center. That’s when all the stuff that DrMatrix said (eg, you can’t see what’s around you inside the event horizon) comes into play.
Now, I do not have an advanced understanding of GR, so there could be some monkey wrench in this idea, about which I don’t know.
If one ignores the Cosmological Constant, then the condition for the Universe to be a black hole is exactly the same as the condition for the Universe to be closed. The singularity at the “center” is the Big Crunch singularity. Assuming that the Schwartschild metric is valid inside the event horizon (and we’ve nothing better to assume, without being able to see inside) the time and radius coordinates would switch roles inside a hole, so the center is always in the future (and can’t be seen), but you can move through “time” as a spatial dimension.
If one does not ignore the Cosmological Constant, then I’ven’t a friggin’ clue what a cosmic hole would be like, or even if the correspondence still holds. All of the black holes known or theorized are far to small for [symbol]L[/symbol] to be significant, and I don’t think that anyone’s ever calculated it.