May 01 2008
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Can black holes ever really form?
On my live video chat the other day, I was asked a very cool question: if time slows down as you approach a black hole, how can they ever really form? Won’t they slow down to zero before they can actually be created?
I recorded my answer and put it on YouTube. Sorry about the jumpiness, but this was recorded off the live stream and the connection was laggy.
I love answering questions, so drop by the next time I do a live chat, and maybe yours will make it into the video!
Very well explained!
If the following paper is correct, black holes may not actually form due to other reasons…
http://arxiv.org/abs/astro-ph/0405353
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Great vid.!
To explain such a complex matter in a few minutes requires some talent.
I’ve shown it to my kids and they loved it.
Keep on with the great job.
I don’t know why I was thinking about this yesterday, but I wondered if we would expect light rays/particles to be in orbit around a black hole and would that occur right at or just outside the event horizon? I’m guessing the black hole would have to be one that isn’t consuming lots of matter at the time. If there were lots of light orbiting a black hole and you passed through the event horizon (with your eyes intact somehow) would you pass through a blinding flash of light?
I’m not a scientist and I didn’t stay at a Holiday Inn Express last night, so this may be a stupid question.
Ah, I got all laggy and with connection errors when you answered this question. Now I can finally hear it good!
I recommend that when you post a video, you add diagrams because somehow your hand puppetry doesn’t quite convey the forces involved, yuk yuk yuk.
You keep switching frames. Of course, if you are the viewpoint of falling through the event horizon your time marches on normally.
But to an outside observer, the time slows to zero at the event horizon. The perception of motion is time dependent, so as you watch something fall into a black hole from outside, your measurement of the velocity of the falling object also slows. When the object reaches the event horizon, your perception of the speed of it falls to zero. Stopped.
Or to put it another way, if you see something a centimeter outside the event horizon, it might take 100M external years to fall the next millemeter at it’s current perceived velocity, so even if it’s not actually stopped, it looks so slow as to LOOK stopped.
So one would think that a black hole would have a “picture” of all the stuff it’s swallowed, forever sitting on it’s event horizon.
hi Phil! quick question, why does the escape velocity depend on density rather than just the total amount of mass? if you’re outside of a mass, doesn’t all of the gravity mathematics still work if you assume all of the mass is at the point of the center of mass for the object? i guess i’m just unclear as to why the volume of the mass matters rather than just the mass itself.
Didn’t you visit this idea b4 (star trek reference)? If I remember correctly, there was some conjecture that the formation of a black hole can’t be witnessed due to the escape velocity > c problem.
Also, you are wrong in everything you say ’cause goddidit! loljk.
ok, think i figured out my own quesiton. in order to get close enough to the total mass to get to the event horizon, the volume of that mass must be really small. otherwise, i’d be inside the volume before i got to the event horizon and as i moved inwards, i’d have less and less mass between me and the center. is that it?
Very well explained… so, the friend being kicked out the airlock and falling in, would look back at you in the spaceship, and see the universe end, because relative to him, the rest of it (outside the gravity well) has sped up to infinity.
Which begs the next question: when you fall in and this universe ends… then where/when exactly are you going?
I seem to recall a statement on one of those astronomy shows that there seems to be some disagreement on the idea that the normal rules of space-time may break down in such circumstances. If that’s true, then I guess all bets are off as to what actually may happen insode one of these stellar Cheshire Cats.
Very nice explanation to a darn good question.
Now, on to the “angels on top of a pin” question.
Thanks for the site, and the blog!
I’m glad everyone’s enjoying this one. And you know, folks, there is a Digg button on this post.
Phil, great answer to that old chestnut. We get letters all the time announcing that astronomers (and Astronomy magazine) are wrong, wrong, wrong, and black holes CANNOT exist because of the “information loss” paradox. - Dan P.
i’d have less and less mass between me and the center. is that it?
Bingo, Steve. Nice job.
Doesn’t the same thing, time slowing down to nothing (infinite number of seconds between ticks) happen at the speed of light? Therefore, if one could build a spaceship to accelerate to the speed of light, wouldn’t they be able to explore the entire universe? However, when they returned to earth, it would have ceased to exist?
Errr, how do people who believe that black holes can’t form explain what appear to be observable effects surrounding black holes in the real universe?
there is a Digg button on this post.
“But I hate shoveling.” — Mike Nelson, MST3K
You say that the person falling through the event horizon will be fine. Won’t he reach some point where the gravitational field-tangent gets so big that his feet (given that he’s falling feet first) will feel a much larger gravitational force than his head causing him to be stretched thin?
I realize that you are only talking about time here, but just feel like knitpicking
@ Yssi
When BA states that the person would be “fine”, I don’t think he meant in terms of survivng the event. I think he meant “fine” in terms of his perceived passage of time. To him, were he somehow able to overcome the myriad of issues that would kill him instantly in such an event, time would tick away just “fine” at it’s normal pace.
And, IIRC, I think the “stretched out” appearance would be to an observer looking in at the falling person… but someone please correct me if I’m wrong on that…
Great post Phil…
causing him to be stretched thin?
Yep, though interestingly that won’t necessarily happen at the event horizon itself for sufficiently massive black holes.
I think the “stretched out” appearance would be to an observer looking in at the falling person
No, that’s a tidal effect because of the difference in the “tug” of gravity between your head and feet.
http://casa.colorado.edu/~ajsh/singularity.html
Wouldn’t the matter forming the black hole itself appear to asymptotically approach its event horizon but never actually cross it? Of course, any light coming off its surface would be so incredibly red-shifted that it would appear black anyway.
I have heard this same question many times, and the same answer. But one thing occurs to me. Sure the guy outside the black hole sees the watch the guy falling in has as going slower and slower. But as we have said, at the event horizon, his time stops so from the external reference frame, he never crosses the black hole. So, even though he is always accelerating as he falls, at some point the time dilation effect increases faster than his velocity, so that from the outside he will appear to be decelerating.
So, first, is there anything particularly interesting about the point with the greatest apparent velocity? And second, since to the guy falling, time is proceeding normally, if he looks out, does he see the complete life of the universe as he crosses the event horizon?
Isn’t the current theory that black holes will eventually evaporate due to Hawking radiation? And this occurs from the point of view of outside the black hole, right? If my understanding is correct, this means we could potentially observe black holes evaporate given long enough time scales.
Does this mean that, as your unlucky friend is falling towards a black hole, as we see their watch slow to a crawl, and they watch millions, billions, trillions of years pass in the outside universe, wont the black hole evaporate before they actually cross the event horizon?
Relativity hurts my brain. -_-
Adam: You are exactly right about the black hole dissolving due to Hawking Radiation before you cross the event horizon. Basically if you were to fall into a black hole, you’d see the event horizon recede below you as you fell, and you would *never* cross the event horizon.
However, it would not be a pleasant experience. You’d basically be bathed in Hawking radiation the whole time, and since to you this is happening on a scale measure in seconds, the power (energy/time) incident on your spacesuit would be tremendous. So if you managed to survive the tidal forces ripping your body apart, you could look forward to being bathed in ~10^30-40 Joules of radiation over a few seconds.
Thanks for the clarification, naked bunny…
Oh man, this reminds me of way back in highschool, the ubernerdy discussions of weird theoretical physics. I had concluded way back then that theoretically, a black hole is just an incredibly dense thin shell right at the event horizon. Thanks Phil!
Essentially, if you fall into a black hole, the universe ends in heat death before you cross the EH. Interestingly, I believe that would also happen if you could travel at light speed (even just close to it will work). Before you’re able to hit the brake, the universe dies a quiet death.
Jolley Bloger: Yeah, I once read a so-so novel where the agent of an immensely powerful alien accelerated a stellar system to near light speed as a diversionary tactic in an ongoing war. By the time that agent decided to slow the system back down (it had never been instructed what to do after initiating the acceleration), several centuries had passed for the people living there, but the rest of the universe had entered heat death and all other stars were long extinguished.
Here is an enhancement to the “stuck at the edge of the event horizon” argument. It is not original, of course.
As an observer sees an object’s time slow to infinity, the “temperature” of the object approaches 0 K, so it’s DeBroglie wavelength expands and covers the event horizon, so everything just gets all blurred together, kinda spread out all across the black hole’s surface.
Make sense?
Here’s something I don’t understand about black holes… Why don’t we see a bright ring or light around them?
Consider this…
From a long distance away, as you look closer and closer to the event horizon, the light is being bent more and more. At some distance away, the light is, for the same of argument, basically “straight”. As it bends more and more as you get closer, a given “width” in your view covers more and more space behind the black hole, condensing all the light in that larger swath into narrower and narrower views.
For example, at a far distance from the black hole, one micro-arcsecond of your vision gets the light from a smidgen more than one micro-arcsecond of space behind the black hole. As you get closer, at some distance, your 1 micro-arcsecond view covers 2 micro-arcseconds of space behind the black hole. As you get closer and closer, at some point, your 1 micro-arcsecond view covers 1 arcsecond, 1 arcminute, 1 degree, 90 degrees of space behind the black hole. And so on.
So, as I said, why don’t we see a bright ring of light around a black hole?
Just curious.
That should, of course, be “bright ring of light around them”.
Interesting video. Black holes show how elastic spacetime is.
I doubt your friend would feel fine though. He would be microwaved and shred to spaghetti way before the event horizon was reached. Hmm, did I say Spaghetti?
@# Ken Bon 01 May 2008 at 1:30 pm
Apart from the fact this actually happens, it would be way too tiny to see it from this distance. Black holes are relatively small.
We do know about the gravitational lensing effect though.
As your frenemy, let’s call him George, drops into the black hole, you as the outside observer would also see a tremendous red shift as George gets apparently closer to the event horizon. George’s light would never go out, but simply fade, as the light is “stretched” out to longer and longer wavelengths.
Ok, so, if matter and light falling into a black hole appears to stop in time (as seen from the outside) then wouldn’t the black hole appear to have a solid surface that never moves? That “surface” would consist of all the mattery-wattery-stuff the black hole was gobbling up, right?
And would that surface appear to be infinitely bright, as even the light from all the stars it consumed would also appear to freeze? Which brings up a brain-mooshing question…
What would frozen light look like?
Would also then this surface keep an infinite record of all the “stuff” the black hole had consumed?
Frozen light doesn’t look like anything… it’s not hitting your eye, you wouldn’t see it. We see things either because we bounce light off them and it hits our eyes, or because they themselves are light hitting our eyes.
This seems to be a common misconception in the way we mentally perceive the universe. It’s somewhat related to the uncertainty principle (http://en.wikipedia.org/wiki/Uncertainty_principle), wherein we cannot observe something (i.e. bounce light off it) without changing it.
inverse says, “Frozen light doesnâ??t look like anythingâ?¦ itâ??s not hitting your eye, you wouldnâ??t see it. We see things either because we bounce light off them and it hits our eyes, or because they themselves are light hitting our eyes.”
Ahh, so I can’t shine a flashlight at a black hole and see the things stuck in time just before the event horizon, or almost-but-not-quite-stuck? I’m not talking about the quantum level, but in a larger human-sized scale.
According to that wiki-link, “Physically, the uncertainty principle requires that when the position of an atom is measured with a photon, the reflected photon will change the momentum of the atom by an uncertain amount inversely proportional to the accuracy of the position measurement.”
Since the momentum of the object (and its atoms) stuck at the the “surface” of the BH is zero, then does this uncertainty principle still apply? I mean, we know the object isn’t going anywhere anytime soon, so we SHOULD be able to see it, right?
When light is slowed down, does it just get dimmer, does it red-shift, or does it disappear completely?
I think the core of the problem is, any light you shine in is going to experience the same sort of difficulties getting back out to your eye that whatever you’re trying to observe is having.
And if I recall correctly (I’m far from an expert, and it was some time ago that I read it), Hawking radiation occurs precisely because of the uncertainty principle - since the momentum of a particle is known, its position is uncertain and may manage to fall outside the event horizon. If anyone would like to clarify that, it would be appreciated…
I have a nit to pick here. Gravity does not “shoot up”, or increase in any way. Black holes do not generate more gravity than a normal star of the same mass.
Thanny,
It isn’t that the gravity shoots up, per se, but that the density of the object concentrates the effects of gravity to a point (or very nearly a point). It is this density of mass that causes the escape velocity to be so high - at the surface of the mass.
I would imagine that at 6100 km or so from the surface of a Earth mass black hole that the escape velocity from that distance is still 11 km/s. But at the surface, the velocity would be impossibly high. And at the event horizon, it’s the speed of light. An Earth mass black hole would have an event horizon about 1 cm from the center of that black hole.
Because the Earth isn’t concentrated into a black hole space, and is, in fact, spread throughout a rough sphere 12000 km or so across, you never reach that kind of intense gravity - as the effect of gravity decreases by your distance from it. Think of it this way, in the case of the black hole, all of the mass is inside that 18 mm shell (event horizon) and no part of you would be more than a couple meters away if the closest was touching. You would feel the full effect of all the gravity of all that mass at an average distance of 1 meter (and you’d be shredded by tidal forces at the same time).
On the Earth, you might be an average of one meter from the surface you’re standing on, but you are also 12000km from the furthest part of the planet, and sorta kinda 6000 km on average from the mass.
Now, not being a rocket scientist, I’m sure I’ve screwed something up, and if I’m corrected, we’ll both learn something!
JBS
absolutely cannot wait for your book Dr Plait, fan from Barnsley, Yorkshire, England!
Nice explaination there - there are a few extra things you need to add
1) The falling person does not see the universe end before they fall in - there is a finite “outside time” from which they can receive a photon before they hit the singularity.
2) Even though the outside observer does not see the infalling person cross the horizon, there is a finite time in which they could rescue the infalling person (i.e. send a message inwards telling the person to fire a rocket pack and escape.
It should be noted that there’s more to a black hole’s inescapability than the classical escape velocity, c, at the event horizon. If that’s all there was to it, then it’d still be possible to escape. Think of it this way: escape velocity at Earth’s surface is about 11,000 meters per second. This doesn’t prevent me from leaping skyward at much less than that velocity, rising briefly above the surface and ultimately falling back to it. If classical escape velocity were the only principle at work, then an astronaut could dip below the horizon (black, because massless light would indeed be unable to escape) then give out a burst of thrust on its rocket pack and rise above, to be snatched up by an orbiting rescue ship. From above the horizon it’s possible for massive matter to achieve escape velocity because it’s less than c.
In general relativity, the reason that nothing can escape is because the space & time axes trade roles at the horizon. Inside there are three timelike axes and one spacelike. Escaping back out the event horizon becomes an exercise in traveling backward through time. Just being passively inside means falling endlessly toward the singularity, because the timelike dimensions all approach that point asymptotically.
I admit I find black holes mind-boggling and I don’t really understand this, but I read a few months back (I think it was on a BBC website, but cannot yet refind the article) that a black hole may never completely get to exist as such because a singularity would never manage to fully form as time would slow to infinity as the supernova remnant shrank, so the black hole would evapourate due to Hawking radiation before it could achieve a proper singularity? Does that make sense? I’ve just come back from a lecture on pulsars by their discoverer, Jocelyn Bell Burnell, and they’re bad enough to try and get your head around! She said an object approaching within 1 metre of a 10 kilometre wide neutron star would be accelerated to impact the surface at 60% the speed of light! Ouch!
Thanks guys. I wasn’t confused before, you sure cured me of that :p
Still puzzled. Black holes exist from out point of view. They Hawking radiate energy from our point of view, which energy is from the matter within them. But if, from our point of view, nothing has ever fallen into the black hole (including the matter that forms the black hole itself?), the energy must come from the infinite future from our point of view.
The engineer in me says that what we’re describing is simply the point where the approximation we call physics breaks down, and another level of explanation is necessary.
Still can’t understand how you could add mass to a black hole in finite (external) time. Say a black hole existed 10 billion y ago. Something started falling into it at that time. At this moment, because to me as an external observer it is slowing down, it still hasn’t gone through the event horizon, it is still falling and is not a part of the black hole.
And this will go on forever.
As far as I understood that was the question, and it isn’t answered.
Hi, Phil,
I couldn’t watch the whole video — it got all jumpy and started resetting a few minutes in, but I wanted to say that I have come to really dislike the “escape velocity goes to c” approach to explaining the event horizon. Perhaps you clarified this later in the video. In my experience, the classical idea of escape velocity leads to misconceptions, because if you think about the definition of escape velocity, it’s the velocity needed to leave the object and not come back. A body going slightly slower than that will go a long way before having to fall back. So you could imagine someone inside the event horizon throwing something out to someone just outside the event horizon — the thrown object doesn’t *have* to go at the escape speed, because it doesn’t have to get out to infinity. So the “can’t go faster than light” constraint doesn’t seem to lead to “can’t get back once you’ve crossed the event horizon” if you think of the event horizon as “the place where the escape velocity goes to c”. If you have the time to explain it, I find tipping light cones do a much better job of helping people visualize what’s going on at the event horizon. I’ll be slightly embarrassed if you do explain this later in the video. I’m off to go grade final exams!
Yours,
Don Smith
Hey AstroProf-
If you don’t like the “escape velocity goes to c” approach, how about “the sphere upon which a photon could maintain a stable orbit around the BH” approach?
I’ve always been slightly amazed that nobody has taken umbrage at the term “black hole”. I say this because there have been lawsuits and bylaws passed over the computer terms “master” and “slave” as they apply to hard disks and directories. One might think that the type of person who would somehow equate a master X.500 directory sending data to a slave directory would get a bug up their own hole about the term “black hole”.
In any case, I’m delighted to see that at least in astromomy, saner heads prevail than prevail in IT.
Or better yet, the event horizon is that sphere where Major Tom, falling, falling, reaches C. Once you do that, it’s sure to mess up your orbital mechanics, in such a way that it’s pretty impossible to escape.
Still, in an armchair physics sort of way, I wonder how much of a rocket you’d need to hover a meter above the event horizon. What happens if you lower down a guy on a rope?
DaveS,
Major Tom will approach c as he approaches the horizon if he starts from a radius of infinity and zero velocity. Let’s say the black hole has built about it a rigid cage with finite radius greater than the event horizon. Major Tom could jump off that cage and would only acquire velocity less than c as he reached the horizon.
And it’s interesting to point out that the value of g can be a very comfortable size even at the horizon, if the hole is massive enough. A supermassive hole will have an enormous radius for its event horizon. That places anything just above the horizon very far from the hole’s center of mass. Even approximating the circumstances with Newton’s law, F = GMm/r^2, we can see that F will be small at the horizon for very massive holes. For a massive enough black hole, Major Tom might hover outside the event horizon with a small rocket pack.
But according to GR, if Tom lowers himself below the horizon on a rope, it’s goodbye.
I noticed a few mistakes in your video.
It is known, that the speed of light is constant.
So escape velocity greater than the speed of light can’t prevent light from escaping.
So the black hole can’t be formed because of escape velocity becoming equal to speed of light.
I hope this helps
Some years ago, Neil deGrasse Tyson wrote a column in Natural History about horizons. He made some comments about the event horizons around black holes, so I emailed him with a question that had bothered me for a long time. He was gracious enough to repl;y and not make me feel like an idiot, and I think his response is relevant here.
What I asked was, since escape velocity applies to ballistic objects, IOW objects that are essentially thrown and left to follow their own inertia, that a non-ballistic object should be able to escape a black hole, since it never has to reach escape velocity. IOW, you can escape Earth without reaching escape velocity by climbing a very long ladder.
So, my question went, if Superman were to fall into a black hole, and he had an infinitely strong ladder, couldn’t he climb out?
His answer was to point out that I had used a classical definition of a black hole. That the relativistic version defines the event horizon asn where space-time folds in on itself, so there is no path out of the black hole.
As to poor George, as Zippy called him, he would be ripped apart by tidal forces well before getting to anything really interesting, as I understand it.
Oh, and Naked Bunny, your alias is causing me to have thoughts that a married schoolteacher shouldn’t admit to.
niin,
The speed of light is invariant for all inertial frames of reference. But immersed within a gravitational field, the only such frames are infinitesimally small, localized regions. General relativity supersedes special relativity (that’s why it’s called general relativity; it models a whole spectrum of circumstances which are outside the scope of the more restricted special relativity). Given a gravitational field radiating from a massive object, there’s a ratio between clock rates in two different reference frames at two different altitudes in the field. The speed of light is indeed different for those two frames. Observers within each frame will, however, arrive at the same speed for light when measured locally. This is due to the uniform scaling of time within a properly localized frame of reference.
Hi Mark.
Major Tom will approach c as he approaches the horizon if he starts from a radius of infinity and zero velocity. Let’s say the black hole has built about it a rigid cage with finite radius greater than the event horizon. Major Tom could jump off that cage and would only acquire velocity less than c as he reached the horizon.
I think Poor Ol’ Major Tom would still acquire velocity=c relative to the horizon even if he jumps from rest at radius r_0 > r_horizon. For a stationary observer at radius r (r_horizon <= r r_horizon, v->1 (i.e., c). Of course, this is an asymptotic result, since no stationary observer can be located at the horizon. Then again, the same is true if the object falls from rest at infinity.
Mark Martin
hi mark, thanks for trying to help.
if you say that light do slow down because of time issues, then you would have the situation that for the lights frame of reference, the light would still be able to get out. Just like phil says that stuff do fall into the black hole in his video. So I don’t think this can be the cause of a black hole forming. Or what?
I don’t think you can explain lights failure to escape with escape velocity and that was why i pointed out phils mistake.
The biggest problem here is anyone speaking in terms of escape velocities. They simply don’t exist in special relativity, it’s a non-applicable Newtonian idea. A simplification for things not at all like black holes.
Spacetime at the event horizon is moving toward the singularity at the speed of light relative to any observer who is maintaining hight above it. Spacetime inside that horizon doesn’t exist in the normal sense for anyone who’s accelerating to maintain altitude.
Mike’s example about building a structure above the horizon from which to hang a rope should mention that the total gravitational acceleration of any object hanging from such a structure down to the horizon places an infinate strain on the rope. You can’t maintain hight at the horizon by any means.
The strains on any such structure would also be immense anywhere near the horizon.
Now, an observer that freefalls toward the singularity sees things a little differently, and (other than the extremely distorted viewpoint from all the light bending) it’s all normal enough in terms of how you’d expect to fall and what the outside universe would look like as you fell. Regardless, any attempt to stop falling requires them transferring into the inertial frame of those things stationary at an infinate distance, which requires infinate acceleration at the event horizon, and pretty extreme amounts further out too, depending on the local spacetime curvature.
The appearance of stopping at the horizon is simply that below there none of your light emissions escape, the last light seen above was from just before you crossed, which takes a very long time to escape to any significant distance.
Theories about the potential for non-collapse of the original matter are quite complex, and involve pressures being generated slightly beyond where the horizon would form due to complete mass to energy conversion through various relativistic electromagnetic effects, keeping the total density below the critical point.
I am very sorry but I must say that as someone who’s been asking himself this question for a long time, I am not at all satisfied with the answer
It seems to me that the only observer for whom the black hole would ever form in finite time would be the observer falling into the black hole.
The explanation saying that the black hole just “is actually there”, behind the event horizon, is extremely frustrating (sorry…), since unless you are suggesting that gravitational information can travel faster than light, what’s behind the event horizon (assuming it has formed in the first place) should have no effect at all on what’s outside!
The whole point of the question is that for an outside observer, an event horizon would never actually form! Assuming information cannot travel faster than the speed of light, saying “time only APPEARS to slow down” is an error, and throws what most of general relativity is about out the window
If you still want your classical black holes with an event horizon and a singularity to exist in our universe as something else than a limit as time goes to infinity… Then you must believe that gravity can travel faster than light, because you are talking about receiving gravitational information from a singularity which has formed later than when the light you receive was emitted. I have some trouble accepting that this can be true…
I hope someone can explain this to me :(…
Bob,
Gravity doesn’t need to “travel” faster than the speed of light. Gravity doesn’t escape from a black hole. In GR gravity is a function of local conditions. I fall toward Earth because of the differential properties of space right here where I am, not way over there where Earth is. The field equations represent a continuously differentiable curve from the singularity all the way to infinity. There’s no hiccup at the event horizon. The field just above or below the horizon knows nothing of the horizon itself; it only knows how it’s being tugged upon by the field just infinitesimally off to one side.
Thank you very much for your answer! I thought I understood how gravity is only dependent on local conditions… but…
Just to make sure: If the sun suddenly magically stopped existing (along with its whole mass), the earth would still be subjected to the exact same gravitational field until the moment when it stops receiving photons from the sun, right? This is what I meant when I talked about gravity “travelling” at the speed of light…
I might be wrong and I probably didn’t give enough thought to this but… It appears to me that anything else would cause all sorts of problems with GR…
The thing about the black hole never forming is maybe just a question of semantics? The singularity “exists” in the sense that things that fall into it do it in finite time from their own point of view… But for any outside observer, the singularity and event horizon would never be “observed” to form in finite time.
Does that make sense? For an event horizon to be “observable” from outside in the sense that there is this “visible” radius from which no light escapes, there has to be a singularity beyond it, right? Let’s say the event horizon starts with no radius and grows as more matter falls into it, but for an outside observer, nothing ever falls into it!
So there is no “obsevable” radius that we can call the event horizon of a black hole, just some very dense matter that’s going slower and slower and denser and denser… Never reaching the infinite density required for us to “observe” a black hole because that would take infinite time! Or is that wrong?
I guess the question I’d like to have answered is: what would the formation of a regular black hole look like for an outisde observer looking at the star collapsing? Because according to my own mental differential equation solver it would look like just what I described above (no visible event horizon, ever)
I hope I’m not just confusing myself even more….