Comment by cvoss

>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.13...

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AlecBG a day 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

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throwawaymaths 21 hours 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?

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lupire a day 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?

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- AlecBG a day 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.
  
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gus_massa a day 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_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-u... (HN discussion https://news.ycombinator.com/item?id=13201065 )

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jiggawatts a day 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...

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amluto a day 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!

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- jiggawatts a day 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!?
  
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  amluto a day ago
  
  > 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.
  No, and this has nothing to do with quantum mechanics or the no-hair theorem or anything particularly fancy.
  As a toy example, suppose you have a frame with a (co-moving, but it doesn’t really matter) time coordinate t. A series of events happen at the origin (x=y=z=0 in this frame) at various times t.
  There’s another observer in a frame with a time coordinate t'. The frames are related by t' = t - 1/t for t<0. The t=-10 event happens at t'=-9.9. The t=-4 event happens at t’=-3.25. The t=-1 event happens at t’=0. t=-1/100 happens at t'=99.99. t=0 gets closer and closer to happening but never actually happens. t=1 doesn’t even come close.
  Critically, the t' observer does not observe t=0 or t=1 in some inconsistent manner. There is no disagreement between the observers as to what happens at t>=0. To the contrary, those events are simply not present in the t' observer’s coordinate system!
  Note that the transformation above isn’t about when light from an event gets to the t' observer — it’s the actual relativistic transformation between two frames.
  The Schwarzchild metric has a nastier transformation than this. If you toss a rock into an isolated black hole from far away, you will see the rock get progressively closer to the event horizon, and you will never see it fall in. But the rock is in trouble: its co-movimg coordinate system ends not long after it crosses the horizon. That latter phenomenon is called the “singularity”, it’s solidly inside the event horizon, and it’s not avoidable by coordinate system trickery. While general relativity does not explain what happens when one encounters the singularity, one might imagine that it’s fatal to the rock the reaches it.
  edit: FWIW, you also say:
  > 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.
  I’m not sure what you’re talking about. In a pure GR model of an isolated black hole, as you fall in, you will observe tidal forces. In a smaller black hole, the tidal forces will squash you long before you reach the event horizon. In a large enough black hole, they will not! Your view of the sky would certainly look very, very distorted and delightfully and possibly dangerously blue-shifted, but we’re talking about an isolated black hole. Nothing to see in the sky, and you may well survive your visit to the event horizon. Then, dramatically less than one second later for any credibly sized black hole, you will meet the singularity, and IIRC you should probably expect to be squashed by tidal forces before that. Source: I took the class and did the math. I assume this is what the “pop science” you’re talking about is saying, and it’s not wrong.
  P.S. I’ve never tried to calculate how lethal the blue-shifted sky would be. Naively considering just the time transformation, it should be infinitely lethal at the event horizon. But trying to apply intuition based on only part of a relativistic transformation is a great way to reach incorrect conclusions.
  
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jodrellblank a day 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)

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- pixl97 a day ago
  
  I don't think the hawking radiation occurs at the edge of the event horizon itself.
  Arvin Ash just did a video on this
  https://www.youtube.com/watch?v=UxVssUb0MsA
  It appears to occur outside the event horizon in a large area.
  
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- jiggawatts a day 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.
  
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  Ygg2 a day ago
  
  > black holes are secretly supernovas frozen in time.
  I don't think that's true. What kills you isn't radiation of the singularity, but cosmic microwave background (and other infalling radiation) turned to visible light, then x-rays, then gamma rays.
  
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  quantadev a day ago
  
  How are they getting blasted in the face when such a blast would necessarily have to be moving faster than light?
  
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feoren a day 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?

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- jiggawatts a day 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.
  
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nextaccountic a day 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

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- bencyoung a day ago
  
  Consider that every "surface" inside the event horizon is like a stronger event horizon so passing through you'd certainly notice things like not being able to see your feet any more as the light wouldn't be able to travel out to your eyes! There would be a lot of other stuff happening too so you may not notice exactly, but the event horizon is definitely noticeable!
  
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  ldunn 19 hours ago
  
  Why wouldn't you be able to see your feet? Your head is also falling through the horizon (hopefully - otherwise you are going to be very unhappy), so the light from your feet doesn't need to escape the horizon for you to see it.
  
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- quantadev a day 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.
  
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  amluto a day 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.
  
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machina_ex_deus a day 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.

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- mr_mitm 16 hours ago
  
  > 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).
  Is that so? Isn't that a continuous effect? Things falling into the black hole appear to be frozen at the event horizon only for an observer at infinity.
  
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- jiggawatts 19 hours ago
  
  Take a closer look at a picture of Kruskal coordinates, e.g.: 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_...
  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.
  
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  ubercow13 11 hours ago
  
  The article you linked says precisely that Kruskal–Szekeres coordinates are not singular at the event horizon. The event horizon is completely regular: https://en.wikipedia.org/wiki/Gravitational_singularity#Curv...
  You can choose stupid coordinates that introduce a singularity wherever you like, in GM or in classical mechanics just the same. The coordinates have no meaning.
  
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