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> The conservation of baryon number is not consistent with the physics of black hole evaporation via Hawking radiation. There are other black hole models that

> The conservation of baryon number is not consistent with the physics of black hole evaporation via Hawking radiation. There are other black hole models that can conserve these quantum numbers! Speaking of things that are so obviously true that you must be an idiot not to agree, there are statements so obviously false that you have to be an idiot to agree: People keep repeating the nonsense put out by Penrose, which require non-physical timelike infinities to work. The current "pop science" (nearly science fiction) statement is that it is possible to fall into a black hole and there is "nothing special" about the event horizon. Quite often, just one paragraph over, the statement is then made that an external observer will never observe the victim falling in. The two observers can't disagree on such matters! To say otherwise means that you'd have to believe that the Universe splits (when!?) such that there are two observers so that they can disagree. Or stop believing in logic, consistency, observers, and everything we hold dear as physicists. This is all patent nonsense by the same person that keeps insisting that brains are "quantum" despite being 309K and organic. If the external observer doesn't observe the victim falling in, then the victim never falls in, full stop. That's the objective reality. Penrose diagrams say otherwise because they include the time at infinity, which is non-physical. Even if the time at infinity was "reachable", which isn't even mathematically sound, let alone physically, Hawking radiation is a thing, so it doesn't matter anyway: Black holes have finite lifetimes! There is only one logically consistent and physically sound interpretation of black holes: nothing can ever fall in. Inbound victims slow down relative to the outside, which means that from their perspective as they approach the black hole they see its flow of time "speed up". Hence, they also see its Hawking evaporation speed up. To maintain consistency with outside observers, this evaporation must occur fast enough that the victim can never reach any surface. Instead, the black hole recedes from them, evaporating faster and faster. This model (and similar ones), can preserve all quantum numbers, because there is no firewall, no boundary, nothing to "reset" quantum fields. Everything is continuous, consistent, and quantum numbers are preserved. Outside observers see exactly what we currently expect, black holes look and work the same, they evaporate, etc...

jodrellblank 1y ago
> "To maintain consistency with outside observers, this evaporation must occur fast enough that the victim can never reach any surface. Instead, the black hole recedes from them, evaporating faster and faster." If this is radiating a star's mass worth Hawking radiation particles, is it like the Solar Wind, and if it's happening ever faster is there a point where it would start pushing the victim away from the black hole again? (the 'victim' can be a solar sail if that helps)
- jiggawatts 1y ago
  Yes, infalling victims will have a rather unpleasant time as they discover that black holes are secretly supernovas frozen in time. Outside observers see the victim's own black body radiation become extremely redshifted, asymptotically matching the black hole's black body radiation. If you mathematically "undo" this distortion for both, then what you are really observing from the outside is a star's worth of matter getting converted to pure energy and the infalling victims getting blasted in the face by that. The victims can't make it back out "whole and intact" in the same sense that you're not going to keep your atomic integrity if you're up close and personal to a supernova. Your quantum numbers however... those can be preserved nicely.
  - quantadev 1y ago
    How are they getting blasted in the face when such a blast would necessarily have to be moving faster than light?
    - hparadiz 1y ago
      Because time slows down relative the outside observed as you get closer to the event horizon any matter falling inwards starts to compress blocking the matter above it until eventually that matter is compressed to an extreme against the event horizon. The photons that get fired off from that interaction away from the black hole are able to escape and that's why we can see some black holes as being extremely bright. However matter that is spiraling inwards will be blasted by hundreds of years worth of photons from every direction while inside this matter and energy goop and all sorts of other particles in a matter of moments relative to how it is experiencing time.
amluto 1y ago
> The two observers can't disagree on such matters! Why not? If a spaceship fell toward a black hole and, as it approached the event horizon, one observer saw it turn into a horse and the other saw it turn into a cat, that would be very strange indeed, and one would suspect at least one of the observers of being wrong. But if one observer sees it fall through the event horizon and the other observer waits… and waits… and gets bored and starts doing some math and determines that they could spend literally forever and never actually observe the spacecraft falling through the event horizon, then what’s the inconsistency? You might say “well, the first observer could fire up their communication laser and tell the second observer that ‘yes, the spaceship fell in at such-and-such time’, and the second observer would now have an inconsistent view of the state of the universe”, but this isn’t actually correct: the first observer’s message will never reach the second observer!
- jiggawatts 1y ago
  > Why not? Because that's not how relativity works! Two observers can disagree only on the order and relative timing of events, not what the events are or the total number of events. There are far more restrictions than that, but those are sufficient for my point. The whole quantum information loss problem is just this, but dressed up in fancy terminology. It's the problem with black holes that the "number of things" (particles, events, whatever) is "lost" when matter falls into them. The modern -- accepted -- resolution to this problem is that this information is not lost, preserving quantum numbers, etc... How exactly this occurs is still being debated, but my point is that if you believe any variant of QM information preservation, then the only logically consistent view is that nothing can fall past an event horizon from any perspective, including the perspective of the infalling observers. If you disagree and believe the out-dated GR model that an astronaut can't even tell[1] that they've crossed the event horizon, ask yourself this simple question: When does the astronaut experience this "non-event"[1]? Don't start with the mathematics! Instead, start with this simple thought experiment: The non-victim partner far away from the black hole holds up a light that blinks on an off once a second. The victim is looking outward and is watching the blinking speed up. How many blinks do they count at the time they cross the horizon? Now think through the scenario again, but this time assume the spaceship turns the light off when they observe that the black hole has finished evaporating. When does the in-falling astronaut observe the blinking stop? Keep in mind that every "toy model" makes the simplification here that the blinking rate goes to infinity as the astronaut falls in! (I.e.: "They see the entire history of the universe play out." is a common quote) [1] Isn't that a strong enough hint for everybody that there is no horizon!?
feoren 1y ago
This claim is different from the overwhelmingly accepted scientific consensus, so it's on you to provide evidence. You say the two observers can't disagree on whether the victim falls in in finite time; tens of thousands of Ph.D. physicists say they can disagree. Where is literally any citation, any evidence at all of what you're claiming?
- jiggawatts 1y ago
  > overwhelmingly accepted scientific consensus There is no consensus, quite the opposite: it was very well known that neither classical GR nor quantum mechanics are able to model a black hole! People like to argue this as if it is settled science, right after saying two contradictory things about it, both from simplified, incomplete models.
nextaccountic 1y ago
> The current "pop science" (nearly science fiction) statement is that it is possible to fall into a black hole and there is "nothing special" about the event horizon. How is this not true? From the point of view of whoever is falling, and supposing the black hole is very large
- quantadev 1y ago
  Nobody knows what happens at the event horizon, but we do know from the perspective of an outside observer things about physics 'break'. It makes sense that there's a flip-side to that 'breakage' (on the inside of the surface, or even "only at" the surface) that isn't just normal space as if nothing happened. For example there's no mathematics at all that mankind has ever known where an asymptotic approach towards some limit doesn't have a mirror version (usually inverted) on the other side of the asymptote. If we see time stop, at the EH it seems wrong to assume there's nothing "stopped" similarly from the other side too. So this means the surface has to be very special. You don't just pass by it and not notice as you fall in, imo.
  - amluto 1y ago
    > For example there's no mathematics at all that mankind has ever known where an asymptotic approach towards some limit doesn't have a mirror version (usually inverted) on the other side of the asymptote. That’s a strong statement. 1/sqrt(x), over the reals, doesn’t have an inverted world for x<0. Maybe you could argue that it does exist, weirdly rotated, outside the reals? In any event, the Schwarzchild metric itself is an actual example of this. From the perspective of a doomed spaceship at the event horizon, the Schwarzchild metric is quite civilized. The stuff after the horizon is a different story, but that’s not immediately after crossing the event horizon — it might be whole nanoseconds later :) Go take a GR class. It’s fun and mind-bending.
    - quantadev 1y ago
      What I meant to say was "asymptotically approaches infinity" for 'f(x)' at some limiting value 'x' and thus a left/right mirroring of the function. I shouldn't assume people know I mean vertical just because I say asymptotic, so thanks for catching that imprecision in my wording. As you probably know, horizontal asymptotes are never what we think of as the 'problematic' parts of Relativity, because when something approaches a constant that's never something that breaks the math. The Schwarzchild metric, being a relationship of 6 different variables I think, has some relationships that go to infinity asymptotically at the EH radius and some things that approach a constant at that radius, so it's an example of the kind of asymptotic I was talking about _and_ one like your "horizontal" example.
machina_ex_deus 1y ago
First of all, kruskal coordinates show beyond doubt that the event horizon is just a regular null hypersurface that the observer wouldn't notice crossing locally. (Of course if you look around, at the moment of crossing into the event horizon you see everything else that was falling into it unfreeze and continue crossing). If you want to take into account the evaporation of the black hole, then you should look at something like the vaidya metric. The mass function is a function of the ingoing Eddington coordinate v, which takes on a specific value when you cross the event horizon, and so you observe the black hole at a specific mass as you cross the event horizon. Contradicting your layman understanding of time dilation for the observer relative to the black hole. Once you cross the horizon, the r coordinate becomes timelike, and so you are forced to move to decreasing r value just like a regular observer is forced to move to increasing t value. Your entire future, all your future light cone is within the black hole and it all terminates at the singularity. Minewhile, the t coordinate is space like which is what gives you space like separation from the mess that had happened in the original gravitational collapse. You wouldn't be blasted by a frozen supernova like you have said. You can kind of say the universe splits at the event horizon, the time like coordinate changes from t to r and the future of the black hole branch of the universe is permanently cut off from the rest of the universe. In rotating and charged black holes it is different, and you observe the evaporation of the black hole once you cross the Cauchy horizon. If the black hole is eternal (because someone kept feeding radiation to the black hole, maybe by reflecting the hawking radiation inwards), then you would in fact see timelike infinity as you reach the Cauchy horizon, so this time like infinity is quite physical. You would need to avoid being vaporized by blue shifted incoming radiation.
- jiggawatts 1y ago
  Take a closer look at a picture of Kruskal coordinates, e.g.: https://upload.wikimedia.org/wikipedia/commons/1/1c/Kruskal_diagram_of_Schwarzschild_chart.svg https://upload.wikimedia.org/wikipedia/commons/1/1c/Kruskal_... Those closer-and-closer line spacings are hiding a mathematical infinity, which isn't physical for finite-lifetime black holes. Conversely, look at: https://en.wikipedia.org/wiki/Eddington%E2%80%93Finkelstein_coordinates#Schwarzschild_metric https://en.wikipedia.org/wiki/Eddington%E2%80%93Finkelstein_... The ordinary Schwarzschild metric diagram in that article makes it crystal clear that in-falling observers asymptotically approach the horizon, but never cross it. Read the next section as well, which uses the "Tortoise coordinate"... which again uses the mathematical infinity to allow the horizon to be crossed. I really don't understand why people keep arguing about this! If you find yourself writing an infinity symbol, you've failed at physics. Stop, go back, rethink your mathematics.

jodrellblank 1y ago

> "To maintain consistency with outside observers, this evaporation must occur fast enough that the victim can never reach any surface. Instead, the black hole recedes from them, evaporating faster and faster." If this is radiating a star's mass worth Hawking radiation particles, is it like the Solar Wind, and if it's happening ever faster is there a point where it would start pushing the victim away from the black hole again? (the 'victim' can be a solar sail if that helps)

jiggawatts 1y ago
Yes, infalling victims will have a rather unpleasant time as they discover that black holes are secretly supernovas frozen in time. Outside observers see the victim's own black body radiation become extremely redshifted, asymptotically matching the black hole's black body radiation. If you mathematically "undo" this distortion for both, then what you are really observing from the outside is a star's worth of matter getting converted to pure energy and the infalling victims getting blasted in the face by that. The victims can't make it back out "whole and intact" in the same sense that you're not going to keep your atomic integrity if you're up close and personal to a supernova. Your quantum numbers however... those can be preserved nicely.
- quantadev 1y ago
  How are they getting blasted in the face when such a blast would necessarily have to be moving faster than light?
  - hparadiz 1y ago
    Because time slows down relative the outside observed as you get closer to the event horizon any matter falling inwards starts to compress blocking the matter above it until eventually that matter is compressed to an extreme against the event horizon. The photons that get fired off from that interaction away from the black hole are able to escape and that's why we can see some black holes as being extremely bright. However matter that is spiraling inwards will be blasted by hundreds of years worth of photons from every direction while inside this matter and energy goop and all sorts of other particles in a matter of moments relative to how it is experiencing time.

amluto 1y ago

> The two observers can't disagree on such matters! Why not? If a spaceship fell toward a black hole and, as it approached the event horizon, one observer saw it turn into a horse and the other saw it turn into a cat, that would be very strange indeed, and one would suspect at least one of the observers of being wrong. But if one observer sees it fall through the event horizon and the other observer waits… and waits… and gets bored and starts doing some math and determines that they could spend literally forever and never actually observe the spacecraft falling through the event horizon, then what’s the inconsistency? You might say “well, the first observer could fire up their communication laser and tell the second observer that ‘yes, the spaceship fell in at such-and-such time’, and the second observer would now have an inconsistent view of the state of the universe”, but this isn’t actually correct: the first observer’s message will never reach the second observer!

jiggawatts 1y ago
> Why not? Because that's not how relativity works! Two observers can disagree only on the order and relative timing of events, not what the events are or the total number of events. There are far more restrictions than that, but those are sufficient for my point. The whole quantum information loss problem is just this, but dressed up in fancy terminology. It's the problem with black holes that the "number of things" (particles, events, whatever) is "lost" when matter falls into them. The modern -- accepted -- resolution to this problem is that this information is not lost, preserving quantum numbers, etc... How exactly this occurs is still being debated, but my point is that if you believe any variant of QM information preservation, then the only logically consistent view is that nothing can fall past an event horizon from any perspective, including the perspective of the infalling observers. If you disagree and believe the out-dated GR model that an astronaut can't even tell[1] that they've crossed the event horizon, ask yourself this simple question: When does the astronaut experience this "non-event"[1]? Don't start with the mathematics! Instead, start with this simple thought experiment: The non-victim partner far away from the black hole holds up a light that blinks on an off once a second. The victim is looking outward and is watching the blinking speed up. How many blinks do they count at the time they cross the horizon? Now think through the scenario again, but this time assume the spaceship turns the light off when they observe that the black hole has finished evaporating. When does the in-falling astronaut observe the blinking stop? Keep in mind that every "toy model" makes the simplification here that the blinking rate goes to infinity as the astronaut falls in! (I.e.: "They see the entire history of the universe play out." is a common quote) [1] Isn't that a strong enough hint for everybody that there is no horizon!?

feoren 1y ago

This claim is different from the overwhelmingly accepted scientific consensus, so it's on you to provide evidence. You say the two observers can't disagree on whether the victim falls in in finite time; tens of thousands of Ph.D. physicists say they can disagree. Where is literally any citation, any evidence at all of what you're claiming?

jiggawatts 1y ago
> overwhelmingly accepted scientific consensus There is no consensus, quite the opposite: it was very well known that neither classical GR nor quantum mechanics are able to model a black hole! People like to argue this as if it is settled science, right after saying two contradictory things about it, both from simplified, incomplete models.

nextaccountic 1y ago

> The current "pop science" (nearly science fiction) statement is that it is possible to fall into a black hole and there is "nothing special" about the event horizon. How is this not true? From the point of view of whoever is falling, and supposing the black hole is very large

quantadev 1y ago
Nobody knows what happens at the event horizon, but we do know from the perspective of an outside observer things about physics 'break'. It makes sense that there's a flip-side to that 'breakage' (on the inside of the surface, or even "only at" the surface) that isn't just normal space as if nothing happened. For example there's no mathematics at all that mankind has ever known where an asymptotic approach towards some limit doesn't have a mirror version (usually inverted) on the other side of the asymptote. If we see time stop, at the EH it seems wrong to assume there's nothing "stopped" similarly from the other side too. So this means the surface has to be very special. You don't just pass by it and not notice as you fall in, imo.
- amluto 1y ago
  > For example there's no mathematics at all that mankind has ever known where an asymptotic approach towards some limit doesn't have a mirror version (usually inverted) on the other side of the asymptote. That’s a strong statement. 1/sqrt(x), over the reals, doesn’t have an inverted world for x<0. Maybe you could argue that it does exist, weirdly rotated, outside the reals? In any event, the Schwarzchild metric itself is an actual example of this. From the perspective of a doomed spaceship at the event horizon, the Schwarzchild metric is quite civilized. The stuff after the horizon is a different story, but that’s not immediately after crossing the event horizon — it might be whole nanoseconds later :) Go take a GR class. It’s fun and mind-bending.
  - quantadev 1y ago
    What I meant to say was "asymptotically approaches infinity" for 'f(x)' at some limiting value 'x' and thus a left/right mirroring of the function. I shouldn't assume people know I mean vertical just because I say asymptotic, so thanks for catching that imprecision in my wording. As you probably know, horizontal asymptotes are never what we think of as the 'problematic' parts of Relativity, because when something approaches a constant that's never something that breaks the math. The Schwarzchild metric, being a relationship of 6 different variables I think, has some relationships that go to infinity asymptotically at the EH radius and some things that approach a constant at that radius, so it's an example of the kind of asymptotic I was talking about _and_ one like your "horizontal" example.

machina_ex_deus 1y ago

First of all, kruskal coordinates show beyond doubt that the event horizon is just a regular null hypersurface that the observer wouldn't notice crossing locally. (Of course if you look around, at the moment of crossing into the event horizon you see everything else that was falling into it unfreeze and continue crossing). If you want to take into account the evaporation of the black hole, then you should look at something like the vaidya metric. The mass function is a function of the ingoing Eddington coordinate v, which takes on a specific value when you cross the event horizon, and so you observe the black hole at a specific mass as you cross the event horizon. Contradicting your layman understanding of time dilation for the observer relative to the black hole. Once you cross the horizon, the r coordinate becomes timelike, and so you are forced to move to decreasing r value just like a regular observer is forced to move to increasing t value. Your entire future, all your future light cone is within the black hole and it all terminates at the singularity. Minewhile, the t coordinate is space like which is what gives you space like separation from the mess that had happened in the original gravitational collapse. You wouldn't be blasted by a frozen supernova like you have said. You can kind of say the universe splits at the event horizon, the time like coordinate changes from t to r and the future of the black hole branch of the universe is permanently cut off from the rest of the universe. In rotating and charged black holes it is different, and you observe the evaporation of the black hole once you cross the Cauchy horizon. If the black hole is eternal (because someone kept feeding radiation to the black hole, maybe by reflecting the hawking radiation inwards), then you would in fact see timelike infinity as you reach the Cauchy horizon, so this time like infinity is quite physical. You would need to avoid being vaporized by blue shifted incoming radiation.

jiggawatts 1y ago
Take a closer look at a picture of Kruskal coordinates, e.g.: https://upload.wikimedia.org/wikipedia/commons/1/1c/Kruskal_diagram_of_Schwarzschild_chart.svg https://upload.wikimedia.org/wikipedia/commons/1/1c/Kruskal_... Those closer-and-closer line spacings are hiding a mathematical infinity, which isn't physical for finite-lifetime black holes. Conversely, look at: https://en.wikipedia.org/wiki/Eddington%E2%80%93Finkelstein_coordinates#Schwarzschild_metric https://en.wikipedia.org/wiki/Eddington%E2%80%93Finkelstein_... The ordinary Schwarzschild metric diagram in that article makes it crystal clear that in-falling observers asymptotically approach the horizon, but never cross it. Read the next section as well, which uses the "Tortoise coordinate"... which again uses the mathematical infinity to allow the horizon to be crossed. I really don't understand why people keep arguing about this! If you find yourself writing an infinity symbol, you've failed at physics. Stop, go back, rethink your mathematics.