________________________________________________________________________________
"Trick though it is, it has real physics in it. The gravitational path integral doesn’t distinguish replicas from a real black hole. It takes them literally. This activates some of the latent topologies that the gravitational path integral includes. The result is a new saddle point containing multiple black holes linked by space-time wormholes. It competes for influence with the regular geometry of a single black hole surrounded by a mist of Hawking radiation."
Ah yes, the complex of wormhole interlinked blackholes surrounded by mists of Hawking radiation...
This style of writing, without the actual math, seems to more so obscure true understanding than reveal it for such a complex construct as english words are very likely insufficient for actually capturing the true meaning.
Adding the (partial?) equations of the path integral and/or the matrices would allow for an actual understanding without the confining nature of human words.
In fact just thinking and writing this comment made me more confused, just a few paragraphs prior the path integral could not be fully calculated, but now here it is certain that "the gravitational path integral doesn’t distinguish replicas from a real black hole". Just how powerful is this 'mathematical trick'?
>Adding the (partial?) equations of the path integral and/or the matrices would allow for an actual understanding without the confining nature of human words.
Go read the paper if you want that. Quanta isn't a journal its a popular science mag.
"But in terms of making sense of black holes, this is at most the end of the beginning. Theorists still haven’t mapped the step-by-step process whereby information gets out. “We now can compute the Page curve, and I don’t know why,” said Raphael Bousso at Berkeley."
So it does not seem that the physicists fully understand their calculation results either.
Edit: here's the arxiv link:
https://arxiv.org/abs/1911.11977
Thanks for the link. I find it very frustrating when pop science magazines refuse to link to the actual paper. I don't want to read their attempt to put what the paper says into layman's language; their attempts are usually way off. I want to read the actual paper.
Edit: I was finally able to find links to actual papers in the article:
https://arxiv.org/abs/1810.02055
https://arxiv.org/abs/1905.08255
https://arxiv.org/abs/1905.08762
https://arxiv.org/abs/1908.10996
https://arxiv.org/abs/2002.08950
https://arxiv.org/abs/2004.05863
These appear to be the papers giving the various calculations the article discusses. There are also links to some previous work that was built on:
https://arxiv.org/abs/0804.0055
https://arxiv.org/pdf/1304.4926
https://arxiv.org/abs/1408.3203
For what it's worth I find these criticisms have little value. If your brain is so big that you find reading Quanta beneath you then why are you reading Quanta at all? Just read Arxiv or Nature or Physical Review D and shut up.
_> If your brain is so big that you find reading Quanta beneath you then why are you reading Quanta at all?_
I'm not. I only found out about this through HN, so all I had to start with was the Quanta article since that's all that HN linked to.
_> Just read Arxiv or Nature or Physical Review D and shut up_
Please save your insults for someone who has asked for them.
> Please save your insults for someone who has asked for them.
frankly, you did - other than links, your post consists of a unprompted, charged, polemic insisting that you refuse to read the article being discussed. This is especially uncouth given you effectively stepped into a room that says on the door "Group Assembled To Comment on Article"
I find the criticism had a lot of value because they pointed out all the proper arxiv resources related to the article. I think it should be a top rated comment directly under the article posting.
I'm a layperson. Does this mean there's no firewall but they don't know how?
To someone like me, who believes that the universe is, _must be_, fundamentally a computer, the actual academic paper is hardly any better. Whatever the paper is doing is not describing any kind of computational process. It's a mille-feuille of semi-rigorous mathematical abstraction that sounds far more like a kind of liturgical incantation than a description of the software our universe is running.
That's not to say it is definitely not useful! It could be that, for some bizarre reason, reading the mathematical tea leaves of our current incomplete (and incoherent) theories might well yield insight into post-quantum physics. I do doubt it though.
_To someone like me, who believes that the universe is, must be, fundamentally a computer..._
Why do you believe that?
In defence of the idea, if all physics is found to have a logical and mathematical basis, then doesn't all physical expression and interaction become computation?
"Computable" is a bit more strict than "logical and mathematical". We can define many perfectly consistent processes with perfectly consistent "observable values", which are inherently uncomputable (as in, there is no algorithm, even an inefficient one, that runs on Turing machines, capable of calculating the observable).
Here is my favorite example:
https://en.wikipedia.org/wiki/Chaitin%27s_constant
Couldn't it be the reverse? Couldn't computation be the closest approximation to pure expression and interaction? Seems to me that this is more natural.
Also, I will bet 500$ that no one will be able to prove that all physics has a logical and mathematical basis within my lifetime :)
It'd be hard for it not to be yet remain consistent with observations and conservation laws.
Depends what you mean though. Define "computer".
See this sibling comment for context
https://news.ycombinator.com/item?id=24948188
The definition you are looking for is very well established and intensely studied in physics and theoretical computer science. This is a good place to start a wikipedia exploration of the topic:
https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis
Because I don’t think infinite amounts of information, and uncomputable numbers, exist. They are fantasies, and relatively recent ones at that. Unfortunately those fantasies came to dominate 20th century mathematics, and in the guise of string theory obtained a death grip on fundamental physics. Fortunately a new generation of smart people, with intuitions honed on computers, is well placed to capitalize on this arbitrage opportunity.
Heartening to see that unusual perspectives and opinions are rewarded with downvotes.
I suspect the downvotes are due to (1) the lack of intellectual humility and (2) misrepresentation in your comments. Plenty of physicists agree that uncomputable numbers are unphysical and the idea that infinite information density can not exist is in no way new or controversial (it is pretty much the starting point for the holographic principle, one of our main paths to a theory of everything). They are beautiful ideas worth discussing, but you used your knowledge of them to disparage people and fields which it seems you only have passing familiarity with.
"To astronauts who ask whether they can get out of a black hole, physicists can answer, “Sure!” But if the astronauts ask how to do it, the disquieting reply will be: “No clue.”"
My intuition is that these astronauts would exit the black hole the way a candle exits a flame, i.e., at the atomic or subatomic level, correct?
Information, they now say with confidence, does escape a black hole. If you jump into one, you will not be gone for good. Particle by particle, the information needed to reconstitute your body will reemerge. Most physicists have long assumed it would; that was the upshot of string theory, their leading candidate for a unified theory of nature.
Does this mean string theory produced a prediction? I thought it had no predictive power and that's why people didn't like it.
Maybe not. From the article: "But the new calculations, though inspired by string theory, stand on their own, with nary a string in sight. Information gets out through the workings of gravity itself — just ordinary gravity with a single layer of quantum effects...Over the past two years, physicists have shown that the entanglement entropy of black holes really does follow the Page curve, indicating that information gets out. They did the analysis in stages. First, they showed how it would work using insights from string theory. Then, in papers published last fall, researchers cut the tether to string theory altogether."
Sounds like there is no dependency on string theory.
I took that to mean that they don't use string theory to _prove_ anything per se. But it sounds like they got "insights" from string theory, which seems like it was useful for once? IANAP.
I am a mathematician with some physics background and have been skimming this article: they took _mathematical_ inspiration from string theory in terms of how they formulated the path integrals, but the physical concept of a string doesn’t actually appear. So string theory provided analytic tools but didn’t actually provide physics.
Not in a useful sense, it just matched with another theory. The problem with string theory remains that we can't actually measure any of its predictions.
The title is incorrectly copied. The original says "Nears Its End"
The title probably changed. The URL slug still has "comes-to-an-end".
It seems like every few weeks there are these huge, complicated and profound discoveries in physics. I feel like i missed the boat on this, as if I chose the wrong profession or am missing out on something. I should have pursued math further but I probably would not have been smart enough anyway. Wikipedia has no mention of the page curve, so this is pretty cutting-edge and theoretical even by the standards of theoretical physics.
If it makes you feel better, I studied physics (although only at the undergraduate level) and I still can’t really follow articles like this. If you really want to understand this stuff you basically have to dedicate your life to it, genius or not.
Discussed on Sean Carroll's podcast, Mindscape:
https://www.preposterousuniverse.com/podcast/2020/09/21/115-...
He seems a little non-committal if not skeptical.
One telling exchange near the end regarding the gravitational path integral they used:
"1:18:41 SC: And there is this trick that you can introduce, ’cause what you’re supposed to do is say, well, integrate up all of the spacetimes that match on to this particular wave function you’re looking at. But the trick is, instead of integrating all the four-dimensional spacetimes that match on to this condition you’re looking at, you can just say, well, I’m going to integrate over all four dimensional spaces, so I’m going to forget about spacetime. I’m just going to do what we call the Euclidean path integral because Euclid just talked about space, not time. And…
1:19:13 NE: Oh, you went there. [laughter]
1:19:15 SC: I did, I did. This is where I’m going. And so it was sort of like you could justify… It’s a trick. It’s a mathematical trick. And it’s very rigorously justifiable in certain simple cases in quantum mechanics, and it certainly has the smell of being correct in certain more subtle cases in quantum field theory. In quantum gravity, what they were doing with it, it just seemed to be a trick so they could get a finite answer at the end of the day, and it was very unclear why it had anything to do with the real world, but they suggested it did. Maybe they were right. And since then, I think we’ve become a little more comfortable with the idea that we can use this trick of calculating quantum gravity wave functions by integrating over the Euclidean path integral, the set of all the spaces that end up looking like what we want, instead of all the spacetimes that look like what we want.
1:20:05 NE: Yes.
1:20:05 SC: And that’s what you’re doing, isn’t it? That’s the kind of wormholes that you’re invoking.
1:20:09 NE: Yes, right. That’s what I was trying to sweep under the rug.
1:20:11 SC: I know. [laughter] And you were right to do so, but I just like to live dangerously here.
[chuckle]
1:20:18 SC: So Lenny and Juan have wormholes that are literally good old in spacetime wormholes, and you have wormholes that are in these fake Euclidean spaces that you used to calculate the entropy.
1:20:29 NE: That’s exactly right. Yeah, that’s exactly right. And these fake Euclidean spacetimes have more boundaries. There are more edges than our original spacetime, which means that these wormholes are connecting these… More edges than we have in our original spacetime, and therefore, it’s difficult to make sense of them in terms of the original spacetime that we’ve started with."
So if enough matter feel into a black hole, we'd get an entangled parallel universe?
Big question then it's what is the entanglement "key" and its properties. Or how to even brute force it.
Blind leading blind:
My impression was that parallel universes do not feature.
Rather, it was that, contrary to appearances, Hawking radiation contains all the same information that went into the black hole.
However, this information appears only in the joint distribution over those photons rather than in any of the marginal distributions.
Broken metaphor: You throw a coin into a black hole. Was it heads or tails? Two photons escape the black hole, both spin-up* with 50% probability -- they each appear to carry no information about the coin. But suppose that if they are the /same/ spin, then the coin was heads, and if they are /different/ spin then the coin was tails. In this way the coin's information is not destroyed, but is also not carried by any individual photon.
* I have no idea if spin is actually the property of the radiation that carries the information.
I also get the impression that it is later photons that must somehow be entangled with earlier ones. That -- referring to that V-shaped curve -- the black hole is storing up information about what went in up to about the halfway point in time, which it later radiates away during the second half.
Waiting for an actual physisict to chime in, as, disclaimer, I don't actually know anything about this.
I read it as "every photon emitted carries information about everything that fell into the hole up until that point, but you don't get complete information about everything until the very last photon is emitted".
>You throw a coin into a black hole. Was it heads or tails?
being heads or tails is a property of landing with all of one side of the coin facing down. This does not seem possible in the situation described.
Of course, but it doesn't matter. Any bit of information can do the trick here.
I’m glad Quanta changed the title (it should be changed here too). The research is interesting, but the paradox is in the same place it’s been in for years, ie “locality probably doesn’t survive the paradox”.
"Muted at first, these effects come to dominate when the black hole gets to be extremely old. The hole transforms from a hermit kingdom to a vigorously open system. Not only does information spill out, anything new that falls in is regurgitated almost immediately."
> extremely old
Completely OT, but this gives an indication of 'old' in this context:
https://m.youtube.com/watch?v=uD4izuDMUQA
At least to me, it's pretty obvious that the shadow follows the form. Gravity is like an echo for mass. It seems that we are now realizing it was wrong to assume that the echo was homogenous. The electron movement process in atoms is emitting gravitational waves. The movement of any type of energy causes gravitational waves. Gravity won't be homogenous if the pattern of mass-energy emitting it, isn't homogenous. We know most mass comes from cyclic processes, ie. the inner processes of hadrons. So it stands to reason that those processes should have a unique and animated gravitational field.
one hydrogen atom doesn't have enough mass to produce a graviton.
General Relativity has no quantization, so all minor changes are infinitely perturbated. Now if you choose to go with some kind of quantum gravity, the energy of a graviton wouldn't be fixed, it would vary depending on the frequency of it.
https://en.wikipedia.org/wiki/Graviton#Energy_and_wavelength
It would make more sense to me if black holes had no internals, so the matter exists on the surface of the horizon. No infinities, and the matter can slowly escape over time. Maybe something in the nature of gravity which prevents more than a certain density to exist in any place and time.
Right, except (my lay understanding is) that would require something special to happen to infalling matter at the event horizon, though we don't have a plausible theory for what that thing is--the prevailing interpretation is that passing through the event horizon is a non-event for the observer.
On the other hand, the holographic principle says that the event horizon surface is big enough to contain all the information, so in a way of speaking we seem to be free to model black hole information as though it is "on the event horizon".
To an external observer, new material takes an infinite amount of time to cross the event horizon. So your intuition is kind of right there. But there is matter that starts out inside the horizon, and there needs to be something going on in there because otherwise the gravity would keep pulling it tighter and tighter to infinite density in an infinitely small spot (the original meaning of "singularity").
_> To an external observer, new material takes an infinite amount of time to cross the event horizon._
While this is a common pop science statement, it is not correct. The correct statement is that the external observer never _sees_ the material cross the event horizon, because, heuristically, it would take an infinite time for light emitted exactly on the event horizon to get out to the external observer. However, this does _not_ equate to the external observer being able to assign an "infinite time" to the material crossing the horizon. The notion of "time" the external observer would have to use to do that is mathematically undefined on the horizon.
_> there is matter that starts out inside the horizon_
No, there isn't. When an ordinary object like a star collapses to a black hole, there is no horizon at all to start with, and there is no matter starting out inside any horizon.
>It would make more sense to me if black holes had no internals, so the matter exists on the surface of the horizon.
That may still be the case, but for that we need a theory of quantum gravity.
In my mental model the internal structure of the black hole, composed of all the material that fell into it, is always preserved from some frame of reference.
As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. Thats true of every particle or photon it traps. Of course they get twisted and distorted by the extreme gravitational forces, but in principle there is always structure. I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy.
I'm sure I'm wildly off the mark, but it's the best I can do.
_> As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it._
No, this is not correct. To the extent that the horizon is something that can be "seen" at all, you see it when you cross it. It doesn't shrink away below you.
_> I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy._
This is a common heuristic picture, but it is very limited. No real particles fall into the hole as a result of Hawking radiation. An observer already inside the horizon might be able to deduce that Hawking radiation was being emitted from local observations on, say, the tidal forces in his vicinity, but I'm not sure even that would be true: the models I am familiar with for evaporating black holes do not require any causal connection at all between the emission of Hawking radiation and any particular observer inside the horizon (in more technical language, the two remain spacelike separated for the entire trajectory of the observer inside the horizon).
>As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it.
You have it backwards. From the POV of an external observer you never cross the event horizon, because that's a function time dilation, and light that you emit being more and more red-shifted as you get closer to the event horizon. From the falling observer POV, nothing really changes - or rather, what happens close to the event horizon or just past the event horizon depends on what the theory of quantum gravity says.
I didn't address the POV of an external observer but I'm sure you're right. But what you describe for the POV of someone falling in seems consistent with what I suggested. I don't understand what you think is backwards about it.
You said that from the perspective of a falling observer, they never see themselves cross the event horizon. That's not right. The falling observer certainly will see themselves cross the event horizon. In their frame of reference, time flows normally. For smaller black holes, they may get spaghettified, but for larger ones, crossing event horizon will be an unremarkable occurrence.
There will always be an event horizon below them beyond which light cannot reach them, and they will always be able to observe the external universe. So how will they see themselves crossing the event horizon?
_> There will always be an event horizon below them beyond which light cannot reach them_
Wrong. See my other response upthread.
_> they will always be able to observe the external universe_
But the light coming to them from the external universe will be more and more redshifted, and they can calculate from the observed redshift when they have crossed the horizon.
Also, they will only see a limited portion of the future of the universe; anything that happens later than a fairly short time after they cross the horizon, they will never see, because they will hit the singularity inside the hole before light from such events can reach them.
The usual definition of event horizon is from the perspective of the external observer—the falling astronaut can calculate the threshold at which no amount of thrust can ever move them away from the singularity (in fact, this is what is meant that a time and space coordinate switch places—you can now only move 'down' in space, just like you can only move forward in time).
_> The usual definition of event horizon is from the perspective of the external observer_
Strictly speaking, no, the usual definition of event horizon is completely observer independent: the event horizon is the boundary of the region of spacetime that cannot send light signals to infinity. That region, and its boundary, are invariant geometric properties of the spacetime, independent of any observer.
_> time and space coordinate switch places_
This is a common pop science statement, but it can be very misleading. It would be better to say that, inside the horizon, moving "down in space" (meaning, decreasing the surface area of the 2-spheres you pass through, since the radial coordinate that you are using is defined in terms of those areas) _is_ moving forward in time: the "forward in time" direction in spacetime is _also_ the "decreasing areal radius" direction in spacetime.
Even that, however, depends on a particular choice of coordinates. There are other choices of coordinates where it is not true (at least not in the simple form in which I have just stated it).
> no amount of thrust
Are you sure that's right? Thrust implies acceleration, not just velocity.
Suppose you're just inside the event horizon, from the perspective of an external observer. You're traveling outward at a velocity of .95c. That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. But if you were close to the event horizon you might cross back over it before you fall back in, right? At which point the external observer could see you again.
And once you're outside the event horizon, e.g. to the place where escape velocity is .9c, you might by then have slowed to .8c, but if you still have the capacity to accelerate back above .9c again, you could still get out, couldn't you?
> Are you sure that's right? Thrust implies acceleration, not just velocity.
Yes. The simpler answer is that even if you shoot a laser straight up the light can't get out, and the corollary is that you can't get out yourself even with arbitrarily high thrust.
No matter how much thrust you have to fight against gravity, light will always beat you, because light cannot be slowed down by gravity. If light is trapped then you are trapped.
> You're traveling outward at a velocity of .95c.
Let's make it simpler, and calculate what it takes for the outside observer sees you moving away from the black hole _at all_, even 1mph.
The closer you get to the black hole, the higher your local speed has to be. If you're 10 meters outside the event horizon, you have to get up to 99.999etc. percent of light speed (locally measured) just to slowly inch away from it (as measured by the outside observer), or even just to stay still. Once you fall inside the event horizon, you would have to go _faster_ then light locally for the outside observer to measure you as moving upward.
_> Are you sure that's right?_
It's right.
_> Suppose you're just inside the event horizon, from the perspective of an external observer._
Whether or not you are inside the event horizon is not a matter of perspective. All observers will agree on it.
_> You're traveling outward at a velocity of .95c._
Relative to what?
Outside the horizon, velocities, such as the escape velocity you mention, are usually meant as velocities relative to static observers--observers who are maintaining the same altitude above the horizon.
But below the horizon, _there are no static observers_. It is _impossible_ to maintain a constant "altitude" (which here just means the surface area of the 2-sphere at your location, centered on the hole) even for an instant. So "velocity" relative to a static observer has no meaning inside the horizon.
_> That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer._
No, that's not what happens. Relativity is not Newtonian physics.
The event horizon is not a place in space. It is an outgoing null surface: a surface formed by radially outgoing light rays. Spacetime around the black hole is curved in such a way that this particular outgoing null surface has a constant surface area: in other words, even though there is a full 2-sphere's worth of radially outgoing light rays at the horizon, the surface area of the 2-sphere formed by those light rays is _constant_--the 2-sphere does not expand, as you would expect a radially outgoing spherical wave front to do, because of the curvature of spacetime.
If you are just a little bit inside the horizon, you are on a slightly smaller 2-sphere, and on that 2-sphere, even radially outgoing light rays cannot stay in the same place, much less expand; they _contract_--that spherical wave front, even though it is radially outgoing, _decreases_ in area with time, i.e., it is falling. So you, having to move slower than light, will fall even more. You can't move outward at all; you can't even stop moving inward. (The fact that there are no static observers inside the horizon is a consequence of this.) That is true no matter how hard you accelerate, since no amount of acceleration will let you move faster than light.
I suppose they could calculate the theoretical event horizon from infinity, but they won't observe it, however they will observe one below them.
_> they will observe one below them_
No, they won't. This is simply wrong. See my other responses upthread.
_> From the POV of an external observer you never cross the event horizon_
More precisely, the external observer never _sees_ you cross the horizon. But he can still _deduce_ that you did from other data. So the external observer does not say "you never cross the horizon", period. He just can't _see_ it happen.
This still isn't quite right. In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time.
In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. There is a coordinate singularity around this switch, which is where the infinite coordinate time comes from.
An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon, which they would find to be finite.
_> In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time_
No: the path of the free-falling observer at and beneath the horizon _is not covered at all_ by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make _any_ claims about what happens.
_> In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike._
No: there is a _separate_ coordinate patch that covers the region inside the horizon (but _not_ the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense.
_> There is a coordinate singularity around this switch_
Yes.
_> which is where the infinite coordinate time comes from._
No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all.
_> An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon_
Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will _not_ be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, _all four_ coordinates are spacelike below the horizon.)
I'm not sure what you mean by coordinate "patch" in a format sense; although I think I informally get what you are going for. Since we seem to agree that an external observer sees a coordinate singularity at the horizon [0], it is not clear to me that there is an objective answer to what "their" coordinate system looks like on the interior. I tend to use a hyperbolic coordinate system. In the region on the outside of the horizon, it matches the natural coordinate system for an observer maintaining constant acceleration away from the black hole. From a purely geometric perspective, it is also defined on the region inside the horizon.
By "time coordinate" I mean the hyperbolic angle. This corresponds to the proper time of our constantly accelerating observer. This remains a well defined coordinate on the interior of the black hole. As you identify, in the language of relativity, this is a spatial coordinate on the interior of the horizon. I specifically refer to this as "coordinate time" because it is the coordinate that our observer is using for time; not because it is actually a time coordinate in all regions of space-time.
> No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all.
You do not need to consider the horizon itself. Consider just portion of the path that occurs before the horizon. This path crosses infinite coordinate time. Put another way, the singularity in this coordinate system is precisely that coordinate time diverges towards infinity as you approach the horizon. Put another way, if you pick any finite amount of coordinate time, you can find a portion of free-falling path from before the horizon that crosses that amount of coordinate time .
[0] Although I think we need to be a bit more careful about what we mean by external observer for this to be formally true. A free falling observer is "external" until they cross the horizon, but they don't see a coordinate singularity there.
_> I'm not sure what you mean by coordinate "patch"_
I mean that there are two separate, disjoint regions of spacetime (one outside the horizon and one inside the horizon), and each one is covered by a _separate_ coordinate chart. Informally people often talk as though those two charts are "the same", but they're not. They're two separate coordinate charts, that just happen to be described using the same symbols.
_> we seem to agree that an external observer sees a coordinate singularity at the horizon_
No, we don't agree about that. The coordinate singularity is only there for the particular coordinates we are talking about (which cannot cover the horizon). There are plenty of other possible choices of coordinates that do not have a coordinate singularity at the horizon, so a single coordinate patch covers the entire region of spacetime of interest.
There is a certain sense in which the coordinates we have been using (Schwarzschild coordinates) are the "natural" ones for an observer who is at rest relative to the hole. But one has to be very careful not to put too much weight on that; it can easily lead to errors if taken too far.
_> I tend to use a hyperbolic coordinate system._
Can you give a reference for this? What you are describing in your next paragraph (about "hyperbolic angle" and so forth) does not seem like any of the coordinate charts I am aware of for Schwarzschild spacetime (Schwarzschild, Painleve, Eddington-Finkelstein, Kruskal-Szekeres, or Penrose).
_> In the region on the outside of the horizon, it matches the natural coordinate system for an observer maintaining constant acceleration away from the black hole._
This is just Schwarzschild coordinates; any observer who is at rest relative to the hole (a static observer) has to maintain constant outward proper acceleration to stay that way. Close enough to the horizon, these coordinates can be approximated by Rindler coordinates on flat spacetime, which are a commonly used coordinate chart in special relativity for treating observers with constant proper acceleration.
However, if Schwarzschild coordinates are what you have in mind, first, your next paragraph doesn't seem to be describing them at all, and second, the key property you give here does _not_ hold on the patch inside the horizon, since there are no static observers there.
_> Consider just portion of the path that occurs before the horizon. This path crosses infinite coordinate time_
No, it doesn't. Every event on the path that is outside the horizon has a finite value of Schwarzschild coordinate time. You even say so later in this paragraph, so you appear to be contradicting yourself.
It is true that, for some cases, you can take limits as Schwarzschild coordinate time increases without bound to compute quantities that are finite on the horizon. In fact, the _proper_ time experienced by an infalling observer from some altitude above the horizon, to the horizon itself, is one of them. But you cannot use this limiting process to justify saying that the Schwarzschild coordinate time itself "is infinite" at the horizon.
_> A free falling observer is "external" until they cross the horizon, but they don't see a coordinate singularity there._
Any observer can choose any coordinates, regardless of their state of motion. Whether any observer "sees a coordinate singularity" at the horizon is purely a matter of which coordinates they choose, and has no physical meaning.
https://www.youtube.com/watch?v=JDNZBT_GeqU&t
I think you might be a bit reversed; an outside observer will never see you fall past the event horizon (because it would be impossible for light from you to reach them), but unless you happen to fall into a black hole with a "fire wall", you won't experience anything special when passing through the event horizon, aside from the already high tidal forces.
>already high tidal forces.
Tidal forces are low for supermassive black holes, right?
https://en.wikipedia.org/wiki/Supermassive_black_hole#Descri...