20k 2 days ago

What this thread is really missing is a simulation. I can't promise its the enterprise (as the performance constraints are crippling), but here's a bunch of cubes instead:

https://www.youtube.com/watch?v=iTw0pJvTkGw

It seems that what you see is the object flattened on the shadow in front of you, and it remains flattened. Apparently past me didn't implement redshift on objects, but its likely extremely redshifted

Edit:

Here's a second better clip, showing this more clearly

https://www.youtube.com/watch?v=npC6lCwYUN0

_petronius 2 days ago

There's a good episode of PBS Spacetime about Penrose diagrams: https://www.youtube.com/watch?v=T4oYvSH6jJ8

It gets a little far into the weeds in a very "the map is not the territory" kind of way, but is fun none the less.

I think (it is not addressed to my satisfaction in the video) that it is implied that from inside the black hole, assuming it was formed from a collapsing star, you would see all around you the event horizon: behind you, the unverise at the time you crossed the event horizin, in front of you, the surface of the star at the moment of its collapse. The singularity, as the post covers, exists only in your future, so you could not see it, even though you will always end up there.

cvoss 2 days ago

It is argued that Bob sees light from Alice's crossing of the horizon at the same instant Bob himself crosses. Isn't this true of all matter that enters? When Bob enters, he sees everything that ever fell into the black hole "before" him, at all once? Is it blinding? Does it fry and scramble Bob? Or is it so redshifted that Bob survives?

  • daxfohl 2 days ago

    The article is missing on a couple points. The analysis assumes all Alice's light is emitted radially exactly outward from the center of the black hole. In reality, light is emitted in all directions, and anything emitted at even slightly different angles would get sucked into the black hole. But, Bob might still see it, because he can catch up with it. So when the article says Bob sees Alice cross the horizon when Bob crosses the horizon, it really means that Bob won't see any photons they emitted from inside the BH until Bob crosses into the BH. But similarly, one second prior Bob will be encountering photons Alice emitted roughly 0.99 seconds prior to crossing the horizon, and so on, because those are all getting redshifted too.

    It's similar to if you and the car in front of you are accelerating at the same rate, they have a small head start, and they've got someone throwing fastballs at you at 100mph. When you get to the point where you are going 100mph relative to the ground, you'll hit the ball that was thrown from exactly that spot relative to the ground. So, sure, crossing the 100mph barrier implies something interesting mathematically, but it's not something that the observer would particularly notice. The math for GR isn't exactly the same (baseballs won't redshift), and in particular there isn't even a piece of dirt to compare the photon's motion to, as the horizon is just a mathematically-defined "place", but to a first-order approximation, it's the same thing going on with photons and spacetime distortion. It's a continuous function.

    There's a bit more to it than can be explained in a comment, but the main thing to know is that (as far as we know) nothing special happens at the horizon if you're falling in.

  • kobalsky 2 days ago

    this is my uninformed guess.

    why would bob see anything? I understood that the event horizon is a threshold, not a shell that you cross and suddenly can see inside.

    to see something photons have to bounce on something and reach our eyes, we stop seeing stuff inside the horizon because those photons don't bounce back and they are pulled into the singularity.

    my logic said that if light can't escape the horizon, then it can't escape alice to reach bob, even if he's inside the horizon, photons can't suddenly go backwards from alice until bob, they are being pulled further inside.

    • cyberax 2 days ago

      The objects can emit photons by themselves.

      The problem is that (classically) when you cross the event horizon, the photons that you emit at just that moment will _stay_ _in_ _place_ forever.

      • raattgift 2 days ago

        > (classically) ... photons

        Uhhh... one of those words should go.

        Let's keep it fully classical and drop "photon": we're interested in gravitational effects rather than quantum ones (uncertainty, fluctuations, tunnelling, details about scattering and more). Really what we want is something to illuminate (pardon the pun) interesting null geodesics, so a thin collimated beam -- a pencil of light -- will do.

        The relevant surface here is the apparent horizon, which can be measured by infalling apparatuses, and not the event horizon, the location of which is determined by the configuration of the entire spacetime. (See Visser PRD 2014 <https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.12...> or the corresponding arxiv version <https://arxiv.org/abs/1407.7295>).

        > stay in place forever

        Note the region inside the shell and the downward-pointing wedge in Fig 1. of Visser 2014 is flat Minkowski spacetime. Everything in that region will work like Special Relativity, as one would expect from the shell theorem.

        In particular, a pencil of light directed outwards through the apparent horizon in Fig. 1. will ride the AH down to the singularity, but a receiver intercepting the pencil of light just inside the shell would notice nothing unusual: spacetime is flat there.

        Eventually the collapsing shell collides with observers floating weightlessly inside it, and they have a bad time. But they can direct a pencil of light inwards just before the shell hits them.

    • Sharlin 2 days ago

      No, we're talking about images of things, photons emitted by everything that has fallen in before, before they crossed the horizon.

  • cyberax 2 days ago

    > Isn't this true of all matter that enters?

    Not quite. He will see the light emitted by _all_ of the matter that has fallen in before him, but only in an infinitely small area.

    • mtdewcmu 2 days ago

      A single photon can't be seen multiple times, right? So, if photon A goes into Alica's retina, then Bob can't see photon A. If a big, opaque object passes through the event horizon right in front of you, it would absorb or scatter the photons in its path, and you would not see them.

      • cyberax 2 days ago

        Yep, I explained it a bit more here: https://news.ycombinator.com/item?id=42299891

        • mtdewcmu 2 days ago

          So you don't see _everything_ that went in before you, mostly just the _last_ thing.

          • cyberax 2 days ago

            You'll see kind of a "cone" where the light emitted from all the objects just ahead of you can be seen, but as you look further away, you can see only more and more recent objects.

  • mtdewcmu 2 days ago

    I'm going to guess. From Bob's perspective, Alice's ship would still be able to block light. So he wouldn't be able to see what was ahead of him through the back of Alice's ship; Alice's ship would occlude his view.

itishappy 2 days ago

Fun! I think there's an interesting hidden puzzle in the first sentence:

> 101 starship captains, bored with life in the Federation, decide to arrange their starships in a line, equally spaced, and let them fall straight into an enormous spherically symmetrical black hole—one right after the other.

Does this problem have a globally consistent solution? In the curved spacetime around the blackhole, can everyone agree on what equal spacing means?

  • codethief 2 days ago

    No, they can't. One could very well come up with some other coordinate system whose time & space coordinates are (non-linear) combinations of the usual Schwarzschild time & space coordinates. In that coordinate system, spatial slices of equal time would be different, so spatial distances would be measured differently.

    TL;DR Just pretend the author wrote "[…] equally spaced with respect to some observer's frame of reference" (e.g. a stationary observer at infinity that uses the usual Schwarzschild coordinates).

readyplayernull 2 days ago

If I got BH theory right, each particle of the ship entering the event horizon will almost fully stop, while the next particles entering will still slightly move, compressing the whole ship into a thin shell or crust that, due to atomic mechanics, won't function as normal matter, so the crew won't know what happened to them. The entering ship will redshift until it dissapears. Their particles will "spacetime-travel" far far away into the future until the compressed crust bounces back to space as radiation. Did I get it right?

  • pierrec 2 days ago

    Ignoring the "atomic mechanics" and "spacetime travel" parts, what you're describing is roughly what might be seen by an external observer, far from the black hole. From the ship's point of view, passing the event horizon happens (and in the case of a large-enough black hole, could be quite uneventful). This puzzle is all about reconciling those two points of view, and exploring intermediate points of view. If you're trying to describe things from a single authoritative point of view, of course, there is no such thing (although here's the disclaimer: I am also not an astrophysicist)

    • readyplayernull 2 days ago

      Granted. Thinking twice, BHs move in space really fast, for example when they orbit other BHs. Meaning their mass is not frozen in time, otherwise these will elongate or rip apart. So the time dilation wouldn't be have such big effect, and atoms might not be compacted too much?

      • raattgift a day ago

        BHs move in space pretty slowly!

        Singleton black holes from collapsed stars move in their galaxy about the same as uncollapsed star: on the order of 200 km/s or so (faster towards the middle but still in the disc where motion is roughly circular around the galaxy's centre, slower in the bulge where motion is randomly around the galaxy's centre). Really unusually high-velocity stars move about 65-100 km/s faster than these, which is still not that fast, and "hypervelocity stars" (we've seen some twenty, compared to the 100 billion or so stars in our galaxy) move about only five to ten times faster still, but we're still at only about a thousandth of the speed of light.

        Black hole binaries can be arbitrarily wide, even up to many thousands of light-years, taking hundreds of thousands of years or more to orbit each other, possibly at speeds comparable to those of stars in galaxies. MEERKAT just announced its pulsar timing array results which focuses on orbital periods of some tens of years ("nanohertz gravitational waves"), which means not moving very fast. Here's a nice cartoon: https://physics.aps.org/articles/v16/116 LIGO is sensitive to much higher frequencies, once a black hole binary's mutual orbit has shrunk considerably - at the final chirp, they are moving at double-digit percentages of the speed of light and the orbital periods are in the milliseconds. In that regime black hole horizons do have bumps raised on them: https://www.youtube.com/watch?v=Y1M-AbWIlVQ

        The black holes can't rip apart though: everything inside stays inside.

        I'm afraid I can't figure out what you're talking about in terms of frozen in time, time dilation, or atoms compacting.

  • mtdewcmu 2 days ago

    Let's see if I can state this properly. The atoms of the ship will pass right through the event horizon like nothing. To see the ship, though, photons have to travel from the ship to your eyes. As the ship goes deeper into the black hole's gravity, the photons will "appear to" be getting slowed down by the black hole's gravity well, each photon more than the last. So, an outside observer would have to wait longer and longer to get the next photon. In fact, he'd have to wait an infinitely long time to get all of them. It's like an optical illusion, except that a real physicist would say it's not an optical illusion; it's time dilation, etc.

cyberax 2 days ago

> However, when any of the starships behind him crosses the horizon, the captain of that starship will see Bob in front of them, also crossing the horizon!

This is a tricky point. The 51-st captain will see the ships in front of him getting slower and slower, the distance between them growing less and less. At the horizon, they'll be separated by zero distance.

But how can that be? Imagine that you're looking at a line of cars in front of you, while sitting in a bus. You can see far ahead, and you can project the distances between cars onto your windshield. Now you go out of the bus and look at the cars ahead of you again from your regular height. Now you get down on your knees, and look again. And then lie down on the road and look ahead again.

That's exactly how it would feel for that captain, any lateral distances between ships will keep getting smaller and smaller, until they completely disappear at the horizon. All the ships in front of you will be squished into a line, and the ship in front of you would block the view.

coolThingsFirst 2 days ago

Pure existential dread. There is a force out there that can eat our solar system as if though it’s a cracker.

This universe is bizarre af, it shouldn’t even exist and us too but somehow it does.

jiggawatts 2 days ago

Any such blog / article / video that features a Penrose diagram is just wrong, because it's using mathematics that doesn't apply to the physical universe.

Penrose diagrams draw black holes as if they have existed forever, and will last forever -- that's what the "future infinity" line means.

Obviously black holes form at some finite time, and Stephen Hawking showed that they evaporate in a finite time.

This matters. A lot!

Fundamentally, relativity does not allow observers to disagree on observations of what, only when. If an outside observer never observes someone falling into a black hole then they cannot fall in. It's just that simple! Again, for the slow people at the back of the class: Observers cannot disagree on this. If there is a paradox in your model, then your model is broken, end of story.

There is this weird aspect of modern physics of holding on to the almost-mystical "woo" aspects far more than is justified in either the mathematics or observations. Quantum Mechanics and cosmology are especially rife with popularisations of what amounts to science fiction story telling. It's fun to think about wormholes, white holes, alternate universe accessible trough black holes, etc... I've read these novels, they were fun! Not physical though, natch.

Much more realistically: An infalling observer is slowed down relative to the outside universe and the stellar remnant that formed the black hole. Effectively, the hapless infalling victim sees the black hole and the universe both "speed up".

The error most people make here is that they assume that the black hole is already fully formed in the infinite past and is "completed", making it a mathematically perfect sphere in a sense.

No!

It hasn't finished forming because as its gravitational field increases its time distortion increases. Its formation is "frozen in time" (actually just very very slow), it is never fully formed.

Infalling observers see this slowdown speeding back up, so what they observe is the light of the black hole evaporation getting blue-shifted and brighter until right before mathematically they would have "fallen past the event horizon" what they're actually seeing is akin to a supernova exploding in their face. They're blasted into subatomic particles, joining the radiation of the black hole evaporation many trillions of years into the future... but not infinitely into the future.

This model solves every paradox of black holes, making them boring and not sci-fi exciting again, which is why you haven't heard about it! Uncool stories don't get repeated on blogs for ad-clicks.

Just remember: there are no infinities in the universe, and anyone using one in their theory is almost certainly just making stuff up to sound cool.

PS: Penrose is also behind the kook notion that brains are quantum despite all evidence.

  • daxfohl 2 days ago

    This is inaccurate. The object can fall in. Where the argument fails is that it relies on the idea that an observer will see the object redshift forever. But that applies in the classical GR realm only. However, when combined with QFT (which is required for BH evaporation) it no longer holds.

    In the classical approach, the light that the object emits a second before crossing the horizon will take years to reach the observer, at a millisecond it'll take decades, at a microsecond it'll take centuries, etc. But it's always still coming, never stops completely. So in the classical analysis, an observer will always see the object, unboundedly redshifted, infinitely near the horizon, but not quite there, forever.

    However, in the presence of Hawking radiation, the black hole loses mass, and thus the horizon shrinks. When this happens, that last bit of light the object emitted before crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick.

    You're correct that the object will experience this time dilation by seeing more radiation as it gets closer to the horizon. It'll also start getting pelted from behind by infalling radiation. But nothing will blow up as it crosses the horizon, so it'll safely continue on to experience whatever quantum-graviational phenomena happens at/near the "singlularity".

    • af3d 2 days ago

      Somewhat tangentially, what really perplexes me about Hawking radiation (HR) is this: What happens to the individual particles within the black hole as it evaporates? Like say we start with X particles. The black hole emits HR and shrinks a bit. But how exactly does it "give up" the energy from the black hole without destroying something in return? Does one particle just disappear and now we are left with X-1 particles (or what)? I haven't found any good explanations for this.

      • daxfohl 2 days ago

        At steady state, a classical black hole is fully described just mass, charge, and angular momentum. So there are no individual particles. Which itself was disconcerting to physicists because in the quantum world, information is supposed to be conserved. But they were okay-ish with it being "trapped in there somewhere". Hawking radiation is what blew that up because now the black hole evaporates. So now, nobody really knows. Lots of ideas, the most prominent being holograms on the boundary, but the math is still far from complete, and obviously experiments are even further.

        • jiggawatts 2 days ago

          > At steady state, a classical black hole is fully described just mass, charge, and angular momentum.

          That's just wrong, we know this not to be the case from QM information theory and from thermodynamic arguments! This is the main point Hawking was making. While we don't yet have a good microscopic theory of what's going on, macroscopically we know that the information (entropy) doesn't just vanish into three numbers.

          • af3d 2 days ago

            > While we don't yet have a good microscopic theory of what's going on, macroscopically we know that the information (entropy) doesn't just vanish into three numbers.

            What information could one possibly extract from a "packet" of Hawking radiation? If the black hole was originally formed from a huge mass consisting of 80% iron and 20% xenon, for example, could such a thing be deduced by inspecting the radiation emitted by it? I would suspect that the answer would be "no". (Of course, I am just being an arm-chair physicist here.)

            • rcxdude a day ago

              That is the core of the issue: hawking radiation would seem to be completely random, and therefore have no relation to what went into the black hole. But basically the entirety of physics works in a time-reversible fashion: if you could flip the direction of all the particles in a system, it would evolve back to its previous state (including such situations as two fluids mixing: entropy is how the precise arrangement of that mixed state that 'unmixes' itself is staggeringly unlikely to be seen randomly, but according to most models of physics, it should exist). But this seems to break down when it comes to black holes (it also breaks down in the various magical collapse interpretations of quantum mechanics, but the quantum wavefunction itself is also time-reversable)

              • af3d a day ago

                > But basically the entirety of physics works in a time-reversible fashion: if you could flip the direction of all the particles in a system, it would evolve back to its previous state

                What does that even mean though? Certain systems may indeed time-reversible, but I would argue that most are not (practically speaking). Imagine for example a meteorite which has fallen to Earth. In order to "reverse the process", not only would it have to "reassemble itself" from the innumerable pieces embedded in the ground, it would also have to be flung back passed the escape velocity of our planet!

                > But this seems to break down when it comes to black holes (it also breaks down in the various magical collapse interpretations of quantum mechanics, but the quantum wavefunction itself is also time-reversable)

                I still don't understand the issue. Entropy is essentially just a measure of how close to a system is to the "average value". A high-entropy system being very close to it (and hence, "highly disordered"), while a low-entropy system might be two or three standard-deviations from the mean. A black hole with little angular momentum, charge, and/or mass would necessarily have a lower entropy than otherwise, but in any case we can calculate that without knowing a thing about what is going on inside of it. Moreover we can deduce that such a black hole would indeed be "easier to time-reverse" than one with a higher-entropy, but what does that even tell us? As far as I can tell, not a whole lot.

          • daxfohl 2 days ago

            Correct, that's why I explicitly said "classical" in that sentence. Classical relativity means our understanding from Einstein's field equations, prior to the information-theoretic approaches.

    • jiggawatts 2 days ago

      > When this happens, that last bit of light the object emitted >>before<< crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick.

      (>>highlight<< mine)

      You're not contradicting my argument: I'm saying that there is only a "before", and never an "after". Saying that light emitted before crossing a horizon is visible in a finite time is not incompatible with what I'm saying.

      Keep in mind that Hawking's model is known to be oversimplified as well! In his simple evaporation model, the information encoded in the infalling matter is lost, violating QM information conservation.

      If you simply assume that no infalling matter ever crosses any horizon, that it all just "blows up" very slowly from the perspective of an outside observer, then there is no QM information paradox, no inconsistent observations, etc...

      If you disagree, please cite a recent paper.

      Here's a nice thought experiment for you: What happens at the last moment of an evaporating black hole's life? How does the horizon "disappear"? Whatever you imagine happens, now update your mental model for a relativistic observer. What do they see? I.e.: Does the increased apparent mass delay the observed disappearance of the horizon!? If so, how can this be? How can two observers disagree on the presence or absence of this horizon? Now consider what would occur at this juncture if there was only smooth curvature, no horizon, and no singularity. Would this enable the last moments of a BH's life to be consistently modelled for all observers?

      • daxfohl 2 days ago

        I mean, the paper you linked below describes the situation fairly well. We know the current semiclassical understanding of gravity is incomplete, but near the event horizon of black holes, QFT+GR are sufficient to model the physics. It's only at the Planck level that the math stops working, due to UV divergence. And under QFT+GR, the math shows that black holes can form, that objects can pass the event horizon from their perspective, and the black hole will evaporate later. Not much has changed since the 70's there.

        Now, it's entirely possible that a complete theory of quantum gravity comes out and upends our understanding of what happens everywhere in space, including at event horizons, and perhaps it's the case that black holes are never formed. But to date, a fully consistent theory of quantum gravity has yet to be created. But just saying "assume black holes don't form, and all contradictions go away" by itself doesn't really help, because that just says get rid of GR and/or QFT, but doesn't say what to replace them with, beyond "something that doesn't produce black holes". Which, sure, would be great, but the devil is in the details. The more important thing to most people in the field is solving the UV divergence problem anyway, which will answer in detail questions about Planck level effects. Whatever comes out of that will (hopefully) unfold naturally into the answer to the information paradox too.

        As far as the final question, nobody knows. That said, care is needed when describing events from different perspectives. The principle of relativity isn't that all observers agree on all events. It's that physics is the same in all frames of reference. Which is nuanced: what does one mean by "the physics"? In essence, it's a layman's statement of Noether's Theorem, relating symmetries and conservation laws. So, depending on quantum gravity's symmetries, different observers could see wildly different sequences of events, but they would agree that the conserved properties of the theory were indeed conserved. But until such a theory exists, it's anybody's guess as to what those symmetries actually are.

  • while_true_ 2 days ago

    It irks me so many physicists/cosmologists jump from the mathematical GR singularity at the center of a BH to "matter there has infinite density." That's highly unlikely, it's probably quark plasma.

    • af3d 2 days ago

      From what I understand of it the singularity is analogous to what happens in the 2D Cartesian plane with functions such as f(x) = 1/x. When x equals 0, the function itself "breaks down" because the y-coordinate extends to infinity (ie f(0) is "undefined"). In the context of a black hole that means that neither time nor space (hence neither does matter) "exist" at this singularity. In other words upon arrival the falling object has essentially "reached the end of time".

    • jiggawatts 2 days ago

      My simple model of it is to just think about spatial surfaces and do "accounting" of the total flux through them.

      If you draw a sphere around a star collapsing into a black hole, you can treat it as a closed system. If the black hole evaporates, then all of the mass-energy of its progenitor original star needs to leave through these concentric surfaces. This is on the same order of magnitude as a supernova, as it is equal to the collapsed core of the star being converted into pure radiated energy!

      An infalling observer accounting of the mass-energy flows must match this external view point. As they cross smaller and smaller bounding spheres, they must see this energy flowing out through those boundaries, adding up to the same total. (There is nowhere else for the energy to go; it has to be blasting you in the face as you fall in!)

      Oversimplified models of black holes concentrate the mass-energy to a point, leaving spacetime around it an empty vacuum. So an infalling observer will see zero, zero, zero, zero... infinite energy density for an infinitesimal time. This is non-physical nonsense, and isn't even mathematically sound!

      From Hawking we know that infalling (and distant) observers see some finite energy flux, and from Einstein's GR we know that infalling observers will see this blue-shifted and time-accelerated on the way in.

      The logical conclusion is that the flux is observed to increase smoothly by infalling observers until the entire amount is accounted for in a finite time. This is a staggering total amount of radiated energy, equivalent to an matter/anti-matter explosion of matter the density of a neutron star core! There is no way anything could "fall through" this while jotting down their observations. It's not a survivable journey.

      There are no wormholes, reachable parallel universes, and there are no separate white holes "elsewhere" in the universe. A black hole is the white hole, smeared out trillions of years into the future so that the enormous total energy is radiated out so slowly from an outside perspective that they just look black. Infalling observers see the "true" white nature of these explosions frozen in time.

      • codethief 2 days ago

        > If you draw a sphere around a star collapsing into a black hole, you can treat it as a closed system. If the black hole evaporates, then all of the mass-energy of its progenitor original star needs to leave through these concentric surfaces.

        That would be great if energy were a conserved quantity in General Relativity, which it isn't. Heck, we don't even know how to write down the total energy/momentum of a given spacetime volume.

        • jiggawatts 2 days ago

          > energy were a conserved quantity in General Relativity, which it isn't

          It is conserved, except at cosmological scales. Locally GR conserves energy the same as any other self-consistent physical theory.

          • codethief a day ago

            > It is conserved, except at cosmological scales

            That's incorrect. Energy conservation can be violated at a much smaller scale, e.g. when gravitational waves are involved, or redshift phenomena (e.g. in Schwarzschild).

            Yes, we usually talk about the fact that gravitational waves can carry energy but what exactly is their energy content? And what exactly is the conservation equation here?

            > Locally GR conserves energy the same as any other self-consistent physical theory.

            Locally, you can write down a divergence equation for the energy momentum tensor, yes. However, locally we're in Minkowski space anyway, so that is not really surprising.

            The point is that the local divergence equation doesn't take into account the energy carried by the gravitational field itself. To give another example beyond gravitational waves: A Schwarzschild black hole carries mass, even though it is a vacuum solution to the Einstein field equations.

            See also https://en.m.wikipedia.org/wiki/Mass_in_general_relativity and in particular the section on quasi-local mass.

  • tofof 2 days ago

    Is there a source you would recommend for further elaboration and exploration of this model?

    • jiggawatts 2 days ago

      One good paper is this one: https://arxiv.org/pdf/1907.04879

      Generally you can search for finite black holes and read things written in the last half a decade or so.

      • raattgift 2 days ago

        From the very top of the second page of that "One good paper":

           Can a BH form in a finite time as viewed by a
           distant observer? (Answer: Yes.)
        
        From your comment you appear to believe that preprint supports (emphasis yours):

           It hasn't finished forming because as its gravitational
           field increases its time distortion increases. Its 
           formation is "frozen in time" (actually just very very 
           slow), *it is never fully formed.*
        
        and you also write:

           If there is a paradox in your model, then your model
           is broken, end of story.
        
        Finally you write "relativity does not allow observers to disagree on observations of what" and also "Hawking showed that they evaporate in a finite time."

        Static and non-static observers in general curved spacetimes immersed in relativistic QFTs generically disagree on particle counts. The Unruh, Hartle-Hawking, Rindler, Minkowski and Boulware vacuums seem especially apposite search terms for someone who isn't among "the slow people at the back of the class".

        (Ginzburg & Frolov 1987 is a good starting point: https://iopscience.iop.org/article/10.1070/PU1987v030n12ABEH... (one can also stick "sci-hub.se/" in front of the URL). It is cited a lot <https://scholar.google.co.uk/scholar?cites=12518697499517381...>).

        • jiggawatts 2 days ago

          Keep reading to the end:

          "Moreover, even without invoking nonsingular models, it’s not clear that rotating or charged BHs form an event horizon at all when evaporation is taken into account."

          Even this paper using modern (2024-era) numerical methods struggles to model the complexities of physically realistic black holes, and ends with a "... more research needed" at the end.

          > Ginzburg & Frolov 1987

          That was 37 years ago! We may have figured out one or two things about black holes since then.

          • raattgift a day ago

            > Keep reading to the end

            I did. In particular I did not miss the whole preceding apparent horizon vs event horizon context of §VIII or the second half of the second column of p. 17, all of which conflicts with your:

              It hasn't finished forming because as its gravitational 
              field increases its time distortion increases. Its 
              formation is "frozen in time" (actually just very very 
              slow), it is *never fully formed*.
            
            How do you square the emphasized part with, "Once the trapped region is formed, continued collapse is inevitable"? You should also note that your picture in your paragraph starting "Infalling observers..." is very different than theirs.

            And then this choice statement and your previous paragraph in mixed order:

            > That was 37 years ago! We may have figured out one or two things about black holes since then

            "Tell all physicists you don't read physics papers without saying you don't read physics papers..." [1]

            I mean, did you even look at the range of dates in your One Good Paper's bibliography?

            For starters, Ginzburg & Frolov was just a useful foundational paper (which I suspected you had never heard of) that shows that generically in curved spacetimes, different observers count different numbers of particles, and in particular one observer's vacuum can be another observer's cloud of electrons, positrons, and photons. This was a nice way of saying that your "relativity does not allow observers to disagree on observations of what" is just wrong.

            I gave you a google scholar link to the 37 year old paper paper so that it would be clear anyone with a slightly different academic background (and I hoped you) that it's a foundational theory paper. Frolov is the well-known author of two standard textbooks on the physics of black holes (with Novokov and with Zelnikov), both of which deal with Frolov's 37-year-old paper in the context of evaporation and of different observers counting different particle numbers and how that an acceleration between past and future observers accounts for Hawking quanta. Both Frolov textbooks appear early on the first page of google scholar results.

            (The 37-year-old concern is especially funny. Frolov's 21st century textbook, like practically all textbooks on gravitational phenomena and theory, even reference papers which are now more than a hundred years old, oh no! Choosing a recent GR textbook -- Carroll 2014 -- the author lists under Advanced General Relativity: Hawking & Ellis 1973, de Felice & Clarke 1990, Sachs & Wu 1977; and in the Graduate section: Wald 1984, MTW 1973, Weinberg 1972, ...)

            > modern (2024-era) numerical methods

            which are built to be compatible with ... what? Analytical and/or perturbation theory, right?

            (BTW, Stark & Piran published their computer results in 1985: <https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.55...>. Again, can be found by placing sci-hub.se/ before that URL. That's 39 years ago! "We have performed an extensive series of tests [including] known perturbation solutions.")

            One more kick at the ridiculous dead horse: at the bottom of p. 17 one finds, "We are not the first to propose a picture like the one presented throughout this section. As far back as theoriginal discovery of the Hawking effect, similar ideas were invoked in [4], and by various comments of [2, 3]". The dates of those are respectively 1974, 1975, and 1976. Oh no!

            Could it be that among the "one or two things about black holes since then" that "We" may have "figured out" is that early theory papers (the 80s are not early) turn out to have good support in things like astrophysics, magnetohydrodynamics, gravitational wave observations, Hulse-Taylor / PSR J0737−3039 etc. etc.?

            - --

            [1] https://journals.aps.org/125years but uh oh that was 2018 and surely the list would be totally different now!