5 ms·

> It would also mean that quantum field theory in curved spacetime can only be consistent if baryon number fails to be conserved! This would be utterly shockin

by cvoss 1y ago

> It would also mean that quantum field theory in curved spacetime can only be consistent if baryon number fails to be conserved! This would be utterly shocking. Is it really shocking (today)? I mean, isn't this a logical consequence of Hawking radiation for black holes? I thought we were shocked by this a long time ago, but now we're ok with it. The authors of the paper in question may very well be wrong in their calculations (I can't say), but this blog post doesn't smell good to me because of doubtful statements like these, passed off as so obviously true that you must be an idiot not to agree. That kind of emotional writing does not become someone whose profession should focus on scientific persuasion. From Wikipedia [0], itself citing Daniel Harlow, a quantum gravity physicist at MIT: > The conservation of baryon number is not consistent with the physics of black hole evaporation via Hawking radiation. [0] https://en.m.wikipedia.org/wiki/Baryon_number https://en.m.wikipedia.org/wiki/Baryon_number

jiggawatts 1y ago
> 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.
AlecBG 1y ago
I'm not sure what more you want from him, there are many papers and even a textbook linked? It's bloody John Baez, the man knows his stuff. On you actual point, it is shocking because its claimed that baryon number is not conserved without black holes getting involved
- lupire 1y ago
  Are you saying that when Baez referred to "curved spacetime" he was excluding black holes (because the paper was claiming that non--black-holes have Hawking radiation?) or are you saying something else?
  - AlecBG 1y ago
    well he certainly mentions a result where if there is an everywhere timelike Killing vector field (+ some other assumptions) you can prove that Hawking radiation doesn't occur and that does not include for example the Schwarzschild solution because the Killing vector field partial/partial t becomes non-timelike on the horizon. So for example if you take a dead star in a vacuum with nothing else in the universe (and make certain technical assumptions) then you can prove that the star does not emit Hawking radiation. That's quite a strong result, and certainly does make the result seem shocking.
- throwawaymaths 1y ago
  > shocking because its claimed that baryon number is not conserved without black holes getting involved Isn't it also speculated that there's hawking radiation caused by the event horizon at the edge of the visible universe in an accelerating frame?
molticrystal 1y ago
>That kind of emotional writing does not become someone whose profession should focus on scientific persuasion. What you'd probably prefer reading is one of the sources John Carlos Baez cites [0]: Comment on “Gravitational Pair Production and Black Hole Evaporation” Antonio Ferreiro1, José Navarro-Salas, and Silvia Pla Where they take the equation used in the paper, and outline how there is a better way than using that equation "... is obtained to the lowest order in a perturbative expansion, while the standard way to obtain the non-perturbative Schwinger effect using the weak field approximation is to perform a resummation of all terms" and how the one in the paper being critiqued can't handle situations arising from electromagnetic cases, much less the gravitational one properly. These are the statements Baez makes but the cited paper gives in a much more professional tone and method. https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.133.229001 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.13...
gus_massa 1y ago
>> if baryon number fails to be conserved! This would be utterly shocking. > Is it really shocking (today)? Moreover, there are a few experiments that try to measure the proton decay (that would break the baryon number conservation.) They are run on Earth, far away form any black hole. For now, all of them failed to find a decay, and the conclusion is that the half life of protons is at least 2.4E34 years. https://en.wikipedia.org/wiki/Proton_decay#Experimental_evidence https://en.wikipedia.org/wiki/Proton_decay#Experimental_evid... I found an old article by quantamagazine explaining one of the experiment. It's a huge pool of very pure water and a lot of detectors. No black hole required. https://www.quantamagazine.org/no-proton-decay-means-grand-unification-must-wait-20161215/ https://www.quantamagazine.org/no-proton-decay-means-grand-u... (HN discussion https://news.ycombinator.com/item?id=13201065 https://news.ycombinator.com/item?id=13201065 )
pfdietz 1y ago
Also, the Standard Model does allow nonconservation of baryon number, nonperturbatively.

by cvoss 1y ago

jiggawatts 1y ago

> 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.

AlecBG 1y ago

I'm not sure what more you want from him, there are many papers and even a textbook linked? It's bloody John Baez, the man knows his stuff. On you actual point, it is shocking because its claimed that baryon number is not conserved without black holes getting involved

lupire 1y ago
Are you saying that when Baez referred to "curved spacetime" he was excluding black holes (because the paper was claiming that non--black-holes have Hawking radiation?) or are you saying something else?
- AlecBG 1y ago
  well he certainly mentions a result where if there is an everywhere timelike Killing vector field (+ some other assumptions) you can prove that Hawking radiation doesn't occur and that does not include for example the Schwarzschild solution because the Killing vector field partial/partial t becomes non-timelike on the horizon. So for example if you take a dead star in a vacuum with nothing else in the universe (and make certain technical assumptions) then you can prove that the star does not emit Hawking radiation. That's quite a strong result, and certainly does make the result seem shocking.
throwawaymaths 1y ago
> shocking because its claimed that baryon number is not conserved without black holes getting involved Isn't it also speculated that there's hawking radiation caused by the event horizon at the edge of the visible universe in an accelerating frame?

molticrystal 1y ago

>That kind of emotional writing does not become someone whose profession should focus on scientific persuasion. What you'd probably prefer reading is one of the sources John Carlos Baez cites [0]: Comment on “Gravitational Pair Production and Black Hole Evaporation” Antonio Ferreiro1, José Navarro-Salas, and Silvia Pla Where they take the equation used in the paper, and outline how there is a better way than using that equation "... is obtained to the lowest order in a perturbative expansion, while the standard way to obtain the non-perturbative Schwinger effect using the weak field approximation is to perform a resummation of all terms" and how the one in the paper being critiqued can't handle situations arising from electromagnetic cases, much less the gravitational one properly. These are the statements Baez makes but the cited paper gives in a much more professional tone and method. https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.133.229001 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.13...

gus_massa 1y ago

>> if baryon number fails to be conserved! This would be utterly shocking. > Is it really shocking (today)? Moreover, there are a few experiments that try to measure the proton decay (that would break the baryon number conservation.) They are run on Earth, far away form any black hole. For now, all of them failed to find a decay, and the conclusion is that the half life of protons is at least 2.4E34 years. https://en.wikipedia.org/wiki/Proton_decay#Experimental_evidence https://en.wikipedia.org/wiki/Proton_decay#Experimental_evid... I found an old article by quantamagazine explaining one of the experiment. It's a huge pool of very pure water and a lot of detectors. No black hole required. https://www.quantamagazine.org/no-proton-decay-means-grand-unification-must-wait-20161215/ https://www.quantamagazine.org/no-proton-decay-means-grand-u... (HN discussion https://news.ycombinator.com/item?id=13201065 https://news.ycombinator.com/item?id=13201065 )

pfdietz 1y ago

Also, the Standard Model does allow nonconservation of baryon number, nonperturbatively.