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Posts posted by md65536


13 hours ago, Markus Hanke said:
No, we have to take this locally. It’s Schwarzschild coordinate time, so this is what a faraway stationary clock measures locally in its own frame of reference. It is not what physically happens anywhere else. In GR, time is always a purely local concept.
Sure, but rjbeery is talking about multiple clocks (infalling A, far away B, etc). If the different observers remain able to communicate (A before reaching the event horizon), they must be able to relate their times to each other. Does "time is purely local" mean that GR just doesn't say anything about how the clocks relate, and doesn't depend on them relating? In reality, even using GR to model spacetime, observers still can relate their clocks in ways that GR doesn't care about.
13 hours ago, Markus Hanke said:Again, this is not possible. Time is a purely local concept in GR; there is, in general, no notion of simultaneity across extended regions of curved spacetime, and you can’t map notions of space and time local to some faraway observer into anything that happens anywhere else. In particular not to test particles in free fall, which aren’t stationary. You can define static hypersurfaces of simultaneity based on the coordinate system you have chosen (in Schwarzschild, these will be nested spheres), but that is not the same thing.
But you can do that, even if not in all cases. If "Pick a method of determining simultaneity" is understood to mean that you're making a choice of what you mean by simultaneity and how you define it, then for example a clock hovering above a black hole, at rest relative to a distant clock, can use "radar time" to define simultaneity of events at the two clocks' locations. In this example, they can agree on simultaneity.
Eg. if the hovering clock is gravitationally time dilated so that its clock is ticking at half the rate of distant clock, the clocks can be set so that every tick of the hovering clock happens "at the same time" (by their choice of simultaneity definition) as every second tick of the distant clock, and both observers can agree, and the choice of simultaneity can remain consistent and useful indefinitely.
In the case described (A is infalling), each observer can have their own notion of simultaneity, but they won't agree with each other. I don't see how this is a problem in this thread, it's not like any claimed physical effects are based on simultaneity??
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2 hours ago, rjbeery said:
Parse this paragraph carefully. It says that it's assumed that an event horizon can form with Hawking radiation...but we have no model for it. Therefore we "turn off" Hawking radiation to allow the event horizon to form, and then "turn it on" to produce the graph that I used to show that a contradiction resides in this logic.
That paragraph is talking about the apparent horizon, and you're talking about the event horizon. See https://en.wikipedia.org/wiki/Apparent_horizon :
QuoteIn the simple picture of stellar collapse leading to formation of a black hole, an event horizon forms before an apparent horizon.^{[2]} As the black hole settles down, the two horizons approach each other, and asymptotically become the same surface. If the AH exists, it is necessarily inside of the EH.
Apparent horizons depend on the "slicing" of a spacetime. That is, the location and even existence of an apparent horizon depends on the way spacetime is divided into space and time. For example, it is possible to slice the Schwarzschild geometry in such a way that there is no apparent horizon, ever, despite the fact that there is certainly an event horizon.^{[3]}
I don't know enough to see any problems here. We *need* to make assumptions to model the interior of a black hole (inside the event horizon) because we can't make any observations to test our models. But that's exactly why there are no real problems; if you say some model or assumption "logically" implies some physical phenomena or paradox, but it has no observable consequences, how can you claim it's a real problem?
It's like the movie Interstellar, which made up a paradoxical imagination of the interior of a black hole. Yet Kip Thorne says something like that the movie doesn't break any scientific laws, but that's because there are no laws that say what is observed inside a black hole. The event horizon and evaporation are things that have physical significance outside of the black hole, including effects that can be observed. The apparent horizon can't be observed from outside.
2 hours ago, rjbeery said:This is close, but regarding #1  the interior of a black hole cannot be described by the remote observer, and does not need to be. The infinite future is represented for a coordinate time observer B before mass crosses the event horizon. We have to take this literally. Pick a method of determining simultaneity, and then map those events of object A approaching the event horizon to the remote observer B; there will be events A that match to events B for any and all times/events for B from "now" until "eternity".
Now, let the black hole evaporate, such that the black hole and A are simply gone and replaced by a new observer, C. Now, B and C can match events using our method of simultaneity, forever. In other words, If B were to describe what was happening in the region of the black hole at some certain time, he would claim that both A is asymptotically approaching it, and C is calmly residing there with no black hole in sight.
Oh, I see a little clearer the problem that you're describing. But it's easily resolved.
As Markus pointed out, a Schwarzschild BH doesn't evaporate. An infalling object A gets stuck on the event horizon "forever" (in B's coordinates), but the event horizon continues to exist forever. If on the other hand the BH evaporates in finite time, the event horizon no longer exists when the BH has evaporated away. Observer B doesn't have events occurring at the event horizon at times when the event horizon doesn't exist. Either A falls in and the black hole evaporates and A's world line ends with a finite coordinate time (in B's coords), or the black hole doesn't evaporate and the event horizon lasts forever with A on it (in B's coords), but not both.
If you describe A and B in terms of causal connections, or events involving the other that they can observe, they're going to agree, no matter what realistic thing you have them do. In terms of simultaneity alone, they don't need to agree, and there's nothing paradoxical about that. But yes, object A can't both be trapped on a static event horizon forever, and let the event horizon evaporate in finite time. The event horizon can't be both static and nonstatic in a given reference frame.
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5 hours ago, rjbeery said:
If you could summarize MY position on this, how would you do so?
This is what I got from your posts:
1) A black hole and all the events in its interior can be described in the coordinates of an observer at infinity.
2) A Penrose diagram of an evaporating black hole shows that the formation and disappearance of a black hole have the same time coordinate.
3) If an event A has a coordinate time that is less than the coordinate time of an event B, then A happened before B (maybe even in B's past?).
Problems with this: (1) The interior events do not have meaningful time coordinates for this observer.
(2) If that's what the diagram really shows, then the coordinates used in that diagram can't be the same as for the observer in (1).
(3) You're comparing coordinate times of events that have no causal connections, and their ordering is irrelevant, but you see "logic problems" by treating it as something physical.
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49 minutes ago, rjbeery said:
I'm referring to coordinate time (i.e. the infinite observer) when I make the statements below.
In my experience there can be a tendency in forums to obfuscate through complexity, or at least unintentionally overcomplicate a problem. Looking at the Schwarzschild solution for a black hole of radius zero should make it obvious that any sort of evaporation leads to a contradiction.
I'm not seeing any problem, except maybe mixing of different time coordinates.
In my understanding, the point of having the observer at infinity is that it is "shielded" from any effects of spacetime curvature. In its coordinates, you could say eg. the black hole formed very far away and at coordinate time (ie. observer's local time) t=0, remained at rest, had a lifetime of 100 units of coordinate time, and finished evaporating at t=100, then sometime later at t>100 some other event happened at the location of the black hole.
The same could be said if instead of a black hole, you're talking about a snowball with negligible mass. Neither has any effect on the coordinate time of the observer at infinity. There is nothing contradictory in the coordinate times of this observer on its own. So clearly we're comparing the times of different observers???, but it's not clear to me what other times you're speaking about here.
Also, it should be possible to choose foliations of spacetime such that any events in the interior of the black hole are assigned meaningless coordinate times anywhere in [0, 100]. However, they would have no physical significance to the observer, and there'd be no way to break causality or create a contradiction through your choice.
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Earlier you wrote:
On 6/17/2020 at 4:41 PM, rjbeery said:To see that this is a problem, consider the coordinate of the black hole completing its evaporation at t=100 in coordinate time. The point (r=0, t=100) is represented both at the event horizon formation and also after the black hole is gone.
Can you remind us of the situation you're describing (I've lost track)? Whose coordinate time are you talking about here? It sounds elsewhere like you're talking about a Schwarzschild black hole, at rest (and evaporating) in the coordinates of an observer at infinity. However it also sounds like the Penrose diagram doesn't show those time coordinates, and the statement above doesn't match those coordinates either??? When you speak of time, if you could mention in each case whose coordinates you're referring to, that might make it clearer.
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6 hours ago, Markus Hanke said:
But the answer to your question will be “no” regardless, because in both cases we are in a curved spacetime, so you can’t naively use the transformation rules of SR.
You're speaking of a general case, but if you compared two observers in flat spacetime (eg. at infinity, or in the location of a single event), wouldn't it have to be the case? I didn't see it specified what observers are being compared (eg. infalling vs. one at infinity), but if you're given the choice of which observers to compare, it should be possible?
8 hours ago, Strange said:Also, an observer in free fall does not see Hawking radiation, and so does not see the black hole evaporate. Does this mean that, from the point of view of an external observer, someone in free fall towards a black hole must always reach the event horizon before it disappears, however fast the black hole is evaporating?
I think this is related to https://en.wikipedia.org/wiki/Black_hole_complementarity
I don't think the issue is settled in accepted science. However, seeing the infalling object reach the event horizon doesn't make sense (unless the BH has zero size). Maybe the object evaporates just like the BH does. Maybe the object simply fades out of existence as it's infinitely redshifted to nothing.
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2 hours ago, Mordred said:
Unfortunately this is not true in the case of a coordinate singularity such as the EH.
A coordinate singularity is not invariant under coordinate change. The r_s=2GM is an artifact of the Schwartzchild metric.
Doesn't the null surface correspond with the event horizon, which has the same physical significance (re. lightlike intervals, causality, etc.) regardless of coordinates? Whether or not the event horizon at the Schwarzschild radius is a singularity depends on choice of coordinates, but that doesn't determine its existence or behaviour. It's still an event horizon.
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15 hours ago, Mordred said:
Agreed but let's ask a question. Where is the null surface located ? Under one coordinate choice its r_s= 2GM. However this isn't true for the Kruskal.
Now ask yourself is it the spacelike or the time like that are invariant under the Lorentz transforms ?
This question becomes important to understand the region's of the Penrose diagrams.
Here is an examination of different causal connections in different coordinate systems the article is specifically dealing with Penrose diagrams.
[...]
The null surface is the event horizon? Isn't it located at the same place in different coordinates, just with different numerical representations in the different coordinate systems?
What are you asking is invariant? There are aspects of spacelike and timelike things that are relative, and other aspects that are invariant, right?
Mordred, can you please answer my previous question so that I know I'm not just wasting my time here? Is rapidity some kind of acceleration?
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22 minutes ago, Mordred said:
Now given the last paper can you state the two observers experience identical causal connections ?
Can you state an observer in the interior of the EH is causally disconnected from the universe outside the EH ? I will supply the coordinate system that describes this region.
https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates
Exploring the causal connections with black holes seems related enough to the topic.
I'd say that causal connections are invariant. I think light cones are invariant. A light signal from one event either reaches (causes) another event, or it doesn't. That doesn't change depending on observer. I don't know what you mean by "experience" but of course different observers will see things appear differently, observe different parts of spacetime etc.
Obviously if a particle can survive (as expected) falling into a BH, there can be causal connections from outside to in. As far as I know, it is not known exactly what an observer inside the black hole would observe, as that requires some speculative extrapolation of testable physics. It is only causal connections from inside to out that are prohibited.
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32 minutes ago, Halc said:
In the coordinate space defined by such an observer, the black hole doesn't exist and never (yet) existed.
I don't think that's correct. It did exist, and it evaporated in the observer's past. It's only the interior events that aren't causally connected to the observer in the moment described.
2 hours ago, Mordred said:Sigh your still not getting it.
Here there is a clear example of the acceleration aspects and rapidity under the twin paradox
How is this relevant to the thread? Your link confirms that rapidity is not acceleration. Quote: "rapidity η (v) ≡ tanh^−1 (v)". Constant rapidity implies constant velocity. Do you understand that rapidity is not acceleration? See https://en.wikipedia.org/wiki/Rapidity :
QuoteIn relativity, rapidity is commonly used as a measure for relativistic velocity. Mathematically, rapidity can be defined as the hyperbolic angle that differentiates two frames of reference in relative motion, each frame being associated with distance and time coordinates.
For onedimensional motion, rapidities are additive whereas velocities must be combined by Einstein's velocityaddition formula. For low speeds, rapidity and velocity are proportional, but for higher velocities, rapidity takes a larger value, the rapidity of light being infinite.
If you still think rapidity is acceleration or requires acceleration or whatever, can you please provide a definition that you're using?
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23 minutes ago, Mordred said:
Look at the formulas do you require trigonometric functions for constant velocity ?
Sure, if you're using hyperbolic functions with the Lorentz transformation. Not relevant here.
32 minutes ago, Mordred said:The example they give here is a rotation in spacetime which is a form of acceleration.
It isn't. Have you actually read your link? You should.
None of these terms are relevant to this thread. They're worse than irrelevant when used incorrectly.
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8 minutes ago, Mordred said:
No rapidity is a term used to describe the hyperbolic rotations due to Lorentz boosts. It is not involved under constant velocity.
??? From https://en.wikipedia.org/wiki/Lorentz_transformation : "Transformations describing relative motion with constant (uniform) velocity and without rotation of the space coordinate axes are called boosts, and the relative velocity between the frames is the parameter of the transformation." Do you think a boost is an acceleration? Can you please provide references to the definitions you're using (including rapidity)? I don't trust what you're saying because it's not making any sense to me with respect to anything else I read.
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2 hours ago, Mordred said:
An infalling observer is undergoing rapidity (acceleration)
You keep referring to rapidity as acceleration. The only definitions I've seen are that it's a measure of relativistic velocity. What definition are you using? Your statement makes as little sense as saying an infalling observer is undergoing velocity.
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3 hours ago, Mordred said:
So how one defines the light one also involves the coordinate choice.
Are you saying that you get different light cones with different coordinate choices? But if you have light from one event reaching another event on its future light cone, that doesn't change depending on coordinates, so I don't know what you mean.
4 hours ago, Mordred said:Also there is a significant difference from what an observer at rest will note and an infalling observer. They will have significantly different lightcones.
This makes no sense to me. Isn't it the events that have light cones, not observers? For example, if you have an event where an infalling observer contacts a hovering observer as it passes, that event has a light cone. Are you saying it has different light cones for the different observers?
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2 hours ago, rjbeery said:
Being in my past is exactly the problem. It's directly contradicts the descriptive definition of event horizons and black holes.
Is it fair to say your argument is basically, "if a black hole's entire existence is within an event's past light cone, then any interior events of the black hole are also within that event's past light cone"?
I think the argument is false. The interior events are not in your past light cone. For one thing, a light cone is based on the paths of light from one event to another, and a black hole doesn't have such paths from interior events to exterior.
I think what you're doing is using a mathematical definition of a light cone in "Your" flat spacetime. Then with a black hole placed in that past light cone, you're effectively assigning flat spacetime coordinates to events within the black hole, like you might do if the black hole wasn't there at all and the spacetime remained flat. You're effectively giving physically meaningless flatspacetime coordinates to events in a curved spacetime, which I think is okay, but then you're drawing conclusions about those events based on physics that applies in flat spacetimes, which is not correct.
In the curved spacetime, the black hole's interior events are geometrically outside of your past light cone, I think... or, I have no idea.Maybe another way to put it is that black holes tilt light cones, and you're not accounting for that.
As an amateur, I think I'm missing the maths and vocab that would make this clear and precise.
edit: Thinking more about tilted light cones... Say the event of the BH evaporating is in your past light cone. You can say there's a causal connection between you and it because its future light cone intersects your past light cone. But for an event within the BH's horizon, the event's future light cone is tilted more than 45 degrees such that its future light cone does not intersect your past light cone, and there is no causal connection. I suspect there's more to it than that.
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8 hours ago, rjbeery said:
We come back the next morning and have equipment that recorded the MBH's existence. We can also verify that the MBH no longer exists. This clearly puts the entire history of this MBH in our causal past, and CERTAINLY in the causal past of future null infinity. That region in spacetime of our lab containing this theoretical MBH cannot have included an event horizon.
The definition you quoted, "The black hole region, B, of such a spacetime is defined to be the points of M not contained in the causal past of future null infinity." If there was a black hole then there were such points, which means there must be an event horizon. There's no contradiction there.
Can you give an example of an event from inside the theoretical MBH's horizon that causes an effect outside, such that it has some recorded effect "in the lab the next morning"? If so, then you're on to something. If not, you're making extraordinary claims without evidence.
8 hours ago, rjbeery said:We can always travel to the region in space where the black hole "used to exist" and declare that region to be completely within our past light cones, contradicting the notion that an event horizon ever existed there in the first place.
It's events that are in past light cones, not spatial regions. Assuming you know that, you mean something like that all the events at that location in a given Euclidean coordinate system, but with earlier times, are in our event's past light cone. However, the events within the past black hole's event horizon are not part of that Euclidean coordinate system, I think. If you were claiming that black hole event horizons can't exist in a Euclidean spacetime, I think you'd be right.
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2 hours ago, studiot said:
It means that it does not have one value but many depending upon which direction you look in.
The (Riemann) curvature has one tensor value for each location+time, and that one tensor value is made up of "other stuff" that gives you the different scalar curvature values in different directions?
The "other stuff"'s beyond me and probably not important for the conversation but wikipedia says it's "the Christoffel symbols and their first partial derivatives"... with values corresponding to the 4 dimensions? And there's a scalar curvature in the direction of time as well as any other direction?
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4 hours ago, Markus Hanke said:
Spacetime curvature is a tensorial quantity (the Riemann curvature tensor) and is thus covariant, i.e. not frame dependent. In GR it is calculated directly from the connection coefficients and its derivatives, which in turn are functions of the metric and its derivatives.
Edit: I'm trying to wrap my head around this. Does this mean that the curvature has a specific value for every event in 4D spacetime (ie. a field), and that at each event, it's the same for all observers? Conversely, something like the magnetic field also has a value for every event in 4D spacetime, but the value at a given event can differ depending on inertial frame.
Then 'globally' would refer to all of 4D spacetime, not 'spatially everywhere at a particular moment' like I've been thinking of it? If so then then my original reply below might not make sense.

Is the curvature locally frame independent, or globally? It is timedependent, right? OP's experiment involves changing curvature?
If you set up two of OP's experiment, say some light years apart, then the relative timing of the experiments depends on inertial frame. Can the global curvature (or the tensor field?) be described without that mattering? Does the concept of a "global curvature at a given time (a Cauchy surface?)" even make sense? Or does the tensor field necessarily extend through time or something? Now I'm confused.
My guess would be that it's all local, and trying to apply it globally to causally disconnected 4D regions would require some arbitrary choice of ... how you want to connect the regions into a global thing.
Basically, OP's experiment describes an event at the midpoint observer's location, and all observers everywhere in the universe agree on the spacetime curvature near that event?
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I have more questions about my understanding of this than I have answers... hopefully someone can correct me.
3 hours ago, geordief said:two massive bodies [...] separated by a distance of 100 light seconds [...] approach each other directly with a speed of c/1000
You're specifying a moment in time here. The curvature of a static spacetime should be frameindependent, but if it's changing over time in different locations, the timing of those changes will be framedependent (right?), and you won't have a single description of the spacetime at a given moment (eg. the moment of some event), because there are no frameindependent individual moments that span the space.
The frame is still needed to define the moment. You're specifying the moment, and you have the choice of specifying whose frame of reference you're using. Since you didn't specify any other frame, the midpoint observer's frame (ie. the frame where the two black holes are symmetric) is the only sensible frame of reference that can be assumed by your description. (However, your description is missing some info, like whether you mean the black holes are traveling relative to each other at c/1000, or that's their closing speed according to the midpoint observer, and neither seems like your obvious intention.)
Changes in spacetime curvature are propagated as gravitational waves, and they propagate when there is a change in acceleration of a mass (right?). When you have two masses approaching each other headon in freefall, it is symmetric and no gravitational waves are emitted (right?). I'm guessing that means that there's no framedependence here, and all observers (all frames) would agree on the mathematical description of the spacetime at the moment that the two black holes are 100 light seconds apart according to the midpoint observer.
Also, I suspect that if what you were describing wasn't symmetric, it would matter that the black holes are approaching each other (rather than eg. moving apart after a flyby). In your example, without any gravitational waves being propagated, I suspect that you'd have the same curvature with the masses at the given separation, regardless of their speed.
(Right?)
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On 6/2/2020 at 5:42 PM, geordief said:
Is there something in the equations that only allows them to ,as it were be attracted rather than repelled by the sources of massenergy?
Quote from https://en.wikipedia.org/wiki/Antigravity:
QuoteUnder general relativity, gravity is the result of following spatial geometry (change in the normal shape of space) caused by local massenergy. This theory holds that it is the altered shape of space, deformed by massive objects, that causes gravity, which is actually a property of deformed space rather than being a true force. Although the equations cannot normally produce a "negative geometry", it is possible to do so by using "negative mass". The same equations do not, of themselves, rule out the existence of negative mass.
[...]
Bondi pointed out that a negative mass will fall toward (and not away from) "normal" matter, since although the gravitational force is repulsive, the negative mass (according to Newton's law, F=ma) responds by accelerating in the opposite of the direction of the force. Normal mass, on the other hand, will fall away from the negative matter. [...]
The Standard Model of particle physics, which describes all currently known forms of matter, does not include negative mass.
Also, you're getting a lot of answers to "why" things fall toward each other. If people didn't have answers, you'd probably see something like "Science doesn't deal with 'why', it deals with 'what'", and that applies here too. The answers aren't the root cause of attraction or anything like that, they're just other measurements that correspond. Masses correspond with a certain geometry of spacetime, geodesics of that geometry have certain configurations. If 'what' makes sense enough, you tend to stop thinking about 'why' (eg. one wouldn't wonder "why can't a circle curve outward instead of in on itself?" if one knows what a circle is).
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On 5/17/2020 at 4:08 AM, Strange said:
I may have misread the question as to how you could do it from a diagram like the one in the post.
To do it from a diagram like that, you can rotate the image 180 degrees, then overlap the ellipses. Here the two suns are the foci. Obviously the accuracy will be limited by image quality but for example you can see that the foci of Mercury's ellipse are farther apart than Earth's, and that Mercury is on the order of 3/2 as far from the sun at aphelion as it is at perihelion.
Mercury:
Earth:
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On 5/11/2020 at 4:47 PM, Flavio Manna said:
Here how can i find the other focus of Mercury's orbit?
On 5/11/2020 at 5:36 PM, Strange said:Not easy, because the eccentricity the orbit is so small.
Couldn't you just find the distance from the first focus to Mercury at perihelion, and the distance from the first focus to Mercury at aphelion, then take a vector from the first focus toward the point of aphelion, with a length that's the difference of the two distances? Even if those values are determined by an accurateenough simulation, that should be easy.
According to wikipedia, the difference is 23.8 million km, so that's approximately how far apart the foci should be (neglecting the sun's wobble). That's more than half the distance between Mercury and the sun at perihelion.
Mercury's orbit eccentricity is huge (0.2056) compared to Earth's (0.0167).
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On 5/11/2020 at 3:17 PM, Flavio Manna said:
i'm tying to do a solar system simulation, i alredy did the gravity simulator, the first problem now comes with the planets' orbit.
So my question is, is there a value, a data or something else that indicates the orentation (or direction) of an orbit? Problably i'm searching for the rotation around the yaxis.
If you're simulating Newtonian gravity, then all you need to get an elliptical orbit is to give the smaller mass a velocity. Then calculate updated positions and velocities, using gravity for acceleration. The data would be mass of the sun, location of the planet relative to center of sun, and velocity at that location. Or if accurate relative positions aren't important, just distance from the center of the sun at apogee or perigee, and speed at that point (oriented 90 degrees from the sun) will do. Simply updating position, velocity, and acceleration iteratively ("Euler method") can give a good approximation for simple visualization.
I'd be curious to see if you can find eg. the max deviation in distance between Mercury and the sun, based only on those data. You would know you have a small enough step size, when making it a lot smaller (eg. half) doesn't change the results much.
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21 hours ago, GaryV said:
If you used many different synchronized wave emitters to create a point of constant high energy in front of the ship, as per Relativity, the amount of energy at that point would have the equivalent effect as the corresponding amount of matter. That means that you would warp space as much as that amount of matter would.
As MigL suggested, your idea is to warp space with a system of energy that has much more mass than the ship? And it pulls you along as you fall toward it? But to travel anywhere, wouldn't you have to accelerate it (using much more energy than accelerating the ship, since it's much more massive)?
Are you imagining something with a lot of gravitational mass but little inertial mass? I don't think anything known or predicted is like that.
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Black holes and evaporation
in Relativity
Posted
rjbeery's example didn't require the two observers (infalling A and distant B) to agree on simultaneity. Though, B and C (observer at location of black hole in B's coordinates, after it has evaporated) agree. It was brought up to illustrate the idea of A existing "forever" in B's coordinates, never passing the Schwarzschild BH event horizon.
I agree the topic really has nothing to do with simultaneity, and that's why I'm commenting on it. You've said Schwarzschild BHs don't evaporate, and it's not possible to determine simultaneity across extended regions of spacetime. If someone's not following the details of the thread, they might think those are equally problematic and that "GR says the example's not possible." But, not being able to unambiguously define simultaneity resolves nothing of rjbeery's paradox, while Schwarzschild BHs not evaporating completely destroys it. If anyone else is struggling to see and understand the resolution of the "paradox", simultaneity's not a problem, but evaporation is. (Because, rjbeery has both the event horizon disappearing, and existing forever for the infalling object to be caught above it, both in the distant observer's coordinates.)