5 ms·

This is inaccurate. The object can fall in. Where the argument fails is that it relies on the idea that an observer will see the object redshift forever. But th

This is inaccurate. The object can fall in. Where the argument fails is that it relies on the idea that an observer will see the object redshift forever. But that applies in the classical GR realm only. However, when combined with QFT (which is required for BH evaporation) it no longer holds. In the classical approach, the light that the object emits a second before crossing the horizon will take years to reach the observer, at a millisecond it'll take decades, at a microsecond it'll take centuries, etc. But it's always still coming, never stops completely. So in the classical analysis, an observer will always see the object, unboundedly redshifted, infinitely near the horizon, but not quite there, forever. However, in the presence of Hawking radiation, the black hole loses mass, and thus the horizon shrinks. When this happens, that last bit of light the object emitted before crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick. You're correct that the object will experience this time dilation by seeing more radiation as it gets closer to the horizon. It'll also start getting pelted from behind by infalling radiation. But nothing will blow up as it crosses the horizon, so it'll safely continue on to experience whatever quantum-graviational phenomena happens at/near the "singlularity".

af3d 2y ago
Somewhat tangentially, what really perplexes me about Hawking radiation (HR) is this: What happens to the individual particles within the black hole as it evaporates? Like say we start with X particles. The black hole emits HR and shrinks a bit. But how exactly does it "give up" the energy from the black hole without destroying something in return? Does one particle just disappear and now we are left with X-1 particles (or what)? I haven't found any good explanations for this.
- daxfohl 2y ago
  At steady state, a classical black hole is fully described just mass, charge, and angular momentum. So there are no individual particles. Which itself was disconcerting to physicists because in the quantum world, information is supposed to be conserved. But they were okay-ish with it being "trapped in there somewhere". Hawking radiation is what blew that up because now the black hole evaporates. So now, nobody really knows. Lots of ideas, the most prominent being holograms on the boundary, but the math is still far from complete, and obviously experiments are even further.
  - jiggawatts 2y ago
    > At steady state, a classical black hole is fully described just mass, charge, and angular momentum. That's just wrong, we know this not to be the case from QM information theory and from thermodynamic arguments! This is the main point Hawking was making. While we don't yet have a good microscopic theory of what's going on, macroscopically we know that the information (entropy) doesn't just vanish into three numbers.
    - daxfohl 2y ago
      Correct, that's why I explicitly said "classical" in that sentence. Classical relativity means our understanding from Einstein's field equations, prior to the information-theoretic approaches.
    - af3d 2y ago
      > While we don't yet have a good microscopic theory of what's going on, macroscopically we know that the information (entropy) doesn't just vanish into three numbers. What information could one possibly extract from a "packet" of Hawking radiation? If the black hole was originally formed from a huge mass consisting of 80% iron and 20% xenon, for example, could such a thing be deduced by inspecting the radiation emitted by it? I would suspect that the answer would be "no". (Of course, I am just being an arm-chair physicist here.)
      - rcxdude 2y ago
        That is the core of the issue: hawking radiation would seem to be completely random, and therefore have no relation to what went into the black hole. But basically the entirety of physics works in a time-reversible fashion: if you could flip the direction of all the particles in a system, it would evolve back to its previous state (including such situations as two fluids mixing: entropy is how the precise arrangement of that mixed state that 'unmixes' itself is staggeringly unlikely to be seen randomly, but according to most models of physics, it should exist). But this seems to break down when it comes to black holes (it also breaks down in the various magical collapse interpretations of quantum mechanics, but the quantum wavefunction itself is also time-reversable)
jiggawatts 2y ago
> When this happens, that last bit of light the object emitted >>before<< crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick. (>>highlight<< mine) You're not contradicting my argument: I'm saying that there is only a "before", and never an "after". Saying that light emitted before crossing a horizon is visible in a finite time is not incompatible with what I'm saying. Keep in mind that Hawking's model is known to be oversimplified as well! In his simple evaporation model, the information encoded in the infalling matter is lost, violating QM information conservation. If you simply assume that no infalling matter ever crosses any horizon, that it all just "blows up" very slowly from the perspective of an outside observer, then there is no QM information paradox, no inconsistent observations, etc... If you disagree, please cite a recent paper. Here's a nice thought experiment for you: What happens at the last moment of an evaporating black hole's life? How does the horizon "disappear"? Whatever you imagine happens, now update your mental model for a relativistic observer. What do they see? I.e.: Does the increased apparent mass delay the observed disappearance of the horizon!? If so, how can this be? How can two observers disagree on the presence or absence of this horizon? Now consider what would occur at this juncture if there was only smooth curvature, no horizon, and no singularity. Would this enable the last moments of a BH's life to be consistently modelled for all observers?
- daxfohl 2y ago
  I mean, the paper you linked below describes the situation fairly well. We know the current semiclassical understanding of gravity is incomplete, but near the event horizon of black holes, QFT+GR are sufficient to model the physics. It's only at the Planck level that the math stops working, due to UV divergence. And under QFT+GR, the math shows that black holes can form, that objects can pass the event horizon from their perspective, and the black hole will evaporate later. Not much has changed since the 70's there. Now, it's entirely possible that a complete theory of quantum gravity comes out and upends our understanding of what happens everywhere in space, including at event horizons, and perhaps it's the case that black holes are never formed. But to date, a fully consistent theory of quantum gravity has yet to be created. But just saying "assume black holes don't form, and all contradictions go away" by itself doesn't really help, because that just says get rid of GR and/or QFT, but doesn't say what to replace them with, beyond "something that doesn't produce black holes". Which, sure, would be great, but the devil is in the details. The more important thing to most people in the field is solving the UV divergence problem anyway, which will answer in detail questions about Planck level effects. Whatever comes out of that will (hopefully) unfold naturally into the answer to the information paradox too. As far as the final question, nobody knows. That said, care is needed when describing events from different perspectives. The principle of relativity isn't that all observers agree on all events. It's that physics is the same in all frames of reference. Which is nuanced: what does one mean by "the physics"? In essence, it's a layman's statement of Noether's Theorem, relating symmetries and conservation laws. So, depending on quantum gravity's symmetries, different observers could see wildly different sequences of events, but they would agree that the conserved properties of the theory were indeed conserved. But until such a theory exists, it's anybody's guess as to what those symmetries actually are.

af3d 2y ago

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

daxfohl 2y ago
At steady state, a classical black hole is fully described just mass, charge, and angular momentum. So there are no individual particles. Which itself was disconcerting to physicists because in the quantum world, information is supposed to be conserved. But they were okay-ish with it being "trapped in there somewhere". Hawking radiation is what blew that up because now the black hole evaporates. So now, nobody really knows. Lots of ideas, the most prominent being holograms on the boundary, but the math is still far from complete, and obviously experiments are even further.
- jiggawatts 2y ago
  > At steady state, a classical black hole is fully described just mass, charge, and angular momentum. That's just wrong, we know this not to be the case from QM information theory and from thermodynamic arguments! This is the main point Hawking was making. While we don't yet have a good microscopic theory of what's going on, macroscopically we know that the information (entropy) doesn't just vanish into three numbers.
  - daxfohl 2y ago
    Correct, that's why I explicitly said "classical" in that sentence. Classical relativity means our understanding from Einstein's field equations, prior to the information-theoretic approaches.
  - af3d 2y ago
    > While we don't yet have a good microscopic theory of what's going on, macroscopically we know that the information (entropy) doesn't just vanish into three numbers. What information could one possibly extract from a "packet" of Hawking radiation? If the black hole was originally formed from a huge mass consisting of 80% iron and 20% xenon, for example, could such a thing be deduced by inspecting the radiation emitted by it? I would suspect that the answer would be "no". (Of course, I am just being an arm-chair physicist here.)
    - rcxdude 2y ago
      That is the core of the issue: hawking radiation would seem to be completely random, and therefore have no relation to what went into the black hole. But basically the entirety of physics works in a time-reversible fashion: if you could flip the direction of all the particles in a system, it would evolve back to its previous state (including such situations as two fluids mixing: entropy is how the precise arrangement of that mixed state that 'unmixes' itself is staggeringly unlikely to be seen randomly, but according to most models of physics, it should exist). But this seems to break down when it comes to black holes (it also breaks down in the various magical collapse interpretations of quantum mechanics, but the quantum wavefunction itself is also time-reversable)

jiggawatts 2y ago

> When this happens, that last bit of light the object emitted >>before<< crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick. (>>highlight<< mine) You're not contradicting my argument: I'm saying that there is only a "before", and never an "after". Saying that light emitted before crossing a horizon is visible in a finite time is not incompatible with what I'm saying. Keep in mind that Hawking's model is known to be oversimplified as well! In his simple evaporation model, the information encoded in the infalling matter is lost, violating QM information conservation. If you simply assume that no infalling matter ever crosses any horizon, that it all just "blows up" very slowly from the perspective of an outside observer, then there is no QM information paradox, no inconsistent observations, etc... If you disagree, please cite a recent paper. Here's a nice thought experiment for you: What happens at the last moment of an evaporating black hole's life? How does the horizon "disappear"? Whatever you imagine happens, now update your mental model for a relativistic observer. What do they see? I.e.: Does the increased apparent mass delay the observed disappearance of the horizon!? If so, how can this be? How can two observers disagree on the presence or absence of this horizon? Now consider what would occur at this juncture if there was only smooth curvature, no horizon, and no singularity. Would this enable the last moments of a BH's life to be consistently modelled for all observers?

daxfohl 2y ago
I mean, the paper you linked below describes the situation fairly well. We know the current semiclassical understanding of gravity is incomplete, but near the event horizon of black holes, QFT+GR are sufficient to model the physics. It's only at the Planck level that the math stops working, due to UV divergence. And under QFT+GR, the math shows that black holes can form, that objects can pass the event horizon from their perspective, and the black hole will evaporate later. Not much has changed since the 70's there. Now, it's entirely possible that a complete theory of quantum gravity comes out and upends our understanding of what happens everywhere in space, including at event horizons, and perhaps it's the case that black holes are never formed. But to date, a fully consistent theory of quantum gravity has yet to be created. But just saying "assume black holes don't form, and all contradictions go away" by itself doesn't really help, because that just says get rid of GR and/or QFT, but doesn't say what to replace them with, beyond "something that doesn't produce black holes". Which, sure, would be great, but the devil is in the details. The more important thing to most people in the field is solving the UV divergence problem anyway, which will answer in detail questions about Planck level effects. Whatever comes out of that will (hopefully) unfold naturally into the answer to the information paradox too. As far as the final question, nobody knows. That said, care is needed when describing events from different perspectives. The principle of relativity isn't that all observers agree on all events. It's that physics is the same in all frames of reference. Which is nuanced: what does one mean by "the physics"? In essence, it's a layman's statement of Noether's Theorem, relating symmetries and conservation laws. So, depending on quantum gravity's symmetries, different observers could see wildly different sequences of events, but they would agree that the conserved properties of the theory were indeed conserved. But until such a theory exists, it's anybody's guess as to what those symmetries actually are.