> 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.
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?
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.
>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.
>> 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...
> 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...
> 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!
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!?
> 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.
> 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.
Don't we know about a bunch of black holes best measured in AU? Won't you have a good chunk of time inside those? Does time dilation work severely against you?
I vaguely recall doing this calculation on a problem set. For a black hole roughly the size of the one at the center of our galaxy, IIRC you have well under a microsecond. I could be remembering wrong.
Even worse: the way to maximize how long you have before you hit the singularity, you should do nothing. Firing your rocket in any direction gets you to the singularity in an even smaller amount of proper time: the singularity isn’t in front of you in space — it’s ahead of you in time.
Keep in mind that all of this is for the Schwarzchild metric, which is a nice solution to Einstein’s equations in the sense that you can derive it on a blackboard. It can’t describe what we think of as a real black hole for plenty of reasons, including the major one that a Schwartzchild black hole has existed forever and therefore could not have formed in a supernova. You need a different solution for a black hole that has only existed for a finite time.
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?
> "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)
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.
> 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.
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.
Ah, I gotcha, thanks for explaining. Yeah the 'accretion disk' are what this is normally called. Lots of matter is getting smashed right outside the EH, creating heat energy, and like you said it's able to blast out photons.
> 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
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.
> 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 :)
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.
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!
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.
It is absolutely untrue that GR predicts that you would be knocked unconscious crossing the horizon. In fact one of the most fundamental aspects of GR (equivalence) predicts the exact opposite - there is no local experiment you can do as a freely falling observer to detect the horizon.
You can do plenty of experiments to see if you are falling, e.g. hitting the surface of a planet you are falling towards. The event horizon is a surface like any other with a location in space and you can definitely see when you hit it (it's the bit where no light is coming out). And once you've crossed it, literally no EM radiation can move further from the singularity
I agree that if you are freely falling and then you are suddenly not freely falling because you hit the surface of a planet and experienced a huge acceleration, you will notice. That doesn't have anything to do with anything I said, but it is undeniably true.
An event horizon is not like the surface of a planet - you will not be accelerated as you pass through it.
It is, once again, irrelevant that light cannot propagate outward once you're behind the horizon because, again, you are falling towards the center, and in particular you are falling through the future light cone of your feet. Please look at some spacetime diagrams if you do not believe me, preferably ones in Kruskal-Szekeres coordinates.
In GR spacetime is locally flat and for an inertial observer special relativity applies, up to tidal corrections which can be made arbitrarily small at the horizon by considering a suitably large black hole. This is a deep and important fact about GR. The idea that falling through the horizon causes you to suddenly not be able to see your feet anymore appears to obviously violate this basic principle, so if you think your assertion is true you should be able to explain why either this principle of GR is actually not true, or why your assertion does not actually violate this principle.
Well, I disagree. Light literally can't move in a direction that makes it further from the singularity once inside the event horizon. I don't see what space being flat or not locally has to do with that. Check https://en.wikipedia.org/wiki/Event_horizon#/media/File:BH-n... for an example.
If your head is further from the singularity than your feet then you can't see them.
It doesn't have to move in such a direction! Look at a spacetime diagram and think about the trajectory of your head and feet! Read a book on GR! Do literally anything except have strong opinions about GR when you don't know any GR!
So when will we be able to just run general relativity numerical simulations on our desktop machines? So that you could set up Observer A at some point, and Observer B at some other point and and mass distribution, etc, then just crunch the numbers to see what each observer could see/measure as time evolves for each observer. Seems like the differential equations are straight forward enough(?). Is the possibility of singularities at the center of a black hole the hard part? What if you just simulated something that was 99.99% of the density needed to get a black hole? I suppose that you'd need a 4 dimensional matrix to hold the simulation (three space coordinates plus a time coordinate)? Is it that we just don't have enough RAM and storage yet in consumer machines? If your simulation did a 1,000 points in ever dimension, that would be 1e12 points. If there are 10 components of the tensor at each point and we are using 64-bit doubles per parameter, that means our simulation takes up ~80 TB. Or is it a that we don't have enough processing speed? Or are there still some philosophical issues that need to be decided when you program up the simulator? How many lines of code is a numerical general relativity solver using something like Euler's method? Is the core of a naive version less than, say 500 lines of C? I can see an optimized CUDA version being significantly larger of course.
The diagram on the Wikipedia page for Kruskal-Szekeres coordinates[1] does the job. There you see the trajectory of some infalling observer along with some future light cones[2] of points along that trajectory and the event horizon marked as the dashed line. The usual Schwarzschild r and t coordinates are also shown as the pale hyperbolas.
Say the trajectory that's drawn on the diagram is the trajectory of your feet. Now consider a second trajectory which begins slightly displaced "outwards" (that is, rightwards at t=0 on the diagram) from this first one - that's your head. Hopefully you agree that the head-trajectory would have to do something pretty strange to avoid crossing through the future lightcone of your feet, even behind the horizon. This doesn't require signals from your feet to travel "outward" - it's just that your head is travelling "inward".
K-S coordinates make it pretty clear that nothing drastic happens to the structure of spacetime at the event horizon - everything is perfectly regular. It's just that once you cross the horizon, the singularity (the thick hyperbola at the top of the diagram) is inevitably in your future: there is no trajectory within any future lightcone behind the horizon that doesn't run into the singularity.
You're doomed to run into it in finite time, and all your future lightcones lie entirely behind the horizon.
Thanks! Hmm, I think we're talking about slightly different things but it's been too long since I studied it to put it in the right words :)
I completely agree that spacetime can be "flatish" for a large block hole, but the event horizon does still represent a boundary right?
Consider the edge case of crossing the event horizon itself at some speed <<c (because you've got magic thrusters fighting the pull). At some point your feet will be through the event horizon and your head won't be. Do you agree that at that point you won't be able to see your feet?
I agree that your head will pass through the future light cone of your feet, and so could do somethign to affect your head (by emitting something falling slower than your head), but I'm not sure any light rays could follow that path.
Ok I drilled down a bit and looks like you are right, although I'm still not sure I've built up a clear understanding! (https://physics.stackexchange.com/questions/187917/thought-e...). In fact that question (series of onions) is exactly how I visualised it...
I'm not quite sure from that discussion why an event horizon is equivalent to a body moving outwards at the speed of light but it does make some sense. GR is always fun!
I still don't have a good idea of the "slow moving crossing the event horizon" case" but I'll read around it some more
Maybe the difference is between "free fall through an event horizon" vs "hover" (as much as is possible) at an event horizon
I was considering a stationary observer inside the event horizon, but that’s not possible. ldunn is correct that a free-falling observer will catch up with the photons reflected from their feet.
Space being flat locally is important because if the gravitational gradient is too high (i.e. you get too close to the singularity) your feet will be accelerated much faster than your head.
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.
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.
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.
> 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.
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