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I don't understand why physicists are surprised that the information of a black hole is proportional to its boundary.

From a pure GR perspective, in my understanding, an observer outside a black hole will never actually see anything fall into the black hole. An object thrown in will just get closer and closer to the event horizon, but never cross.

Therefore to the outside observer, the black hole never actually really forms, but there is just a bunch of matter on the boundary of this "almost" black hole.

But this means all the information is smeared on the boundary of this thing, which is exactly what the holographic principle is saying.

What am I wrong about here?

Níckolas Alves
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puzzleshark
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  • It's because in weak gravity, like on Earth, that entropy is proportional to M, not M^2 like with a black hole. – shawn_halayka Dec 01 '22 at 00:12

  • "Therefore to the outside observer, the black hole never actually really forms" Black hole can hypothetically form inside a star, without matter outside falling in through the hole boundary.

     – Ján Lalinský Dec 01 '22 at 02:15
  • You might be interested: https://physics.stackexchange.com/q/739108/ – Allure Dec 01 '22 at 03:08
  • @JánLalinský “Black hole can hypothetically form inside a star” - Not in a finite time by an outside clock. Such solutions use coordinate transformations, of which people tend to forget. When you transform the solution back to the coordinates of an external observer, the star matter never crosses the horizon in the eternity of time. – safesphere Dec 03 '22 at 03:21
  • @safesphere I've read that a body falling towards an existing horizon never crosses it. But I can imagine that maybe the horizon can somehow appear at non-zero radius when density becomes high enough, without the matter inside ever falling through... – Ján Lalinský Dec 03 '22 at 23:43
  • @JánLalinský The horizon first forms at the point of the maximal gravitational potential (time dilation). In a collapsing star, it is at the center. In a collapsing hollow shell, it is at the shell while there is nothing inside. If there is a particle anywhere inside the hollow shell, then this particle adds to the curvature, so the horizon first forms at this particle and quickly expand to the shell. In any case, there is never anything inside. – safesphere Dec 04 '22 at 19:01
  • @safesphere > "In a collapsing star, it is at the center." Maybe there can be a star where it is not. A shock wave caused by some falling material or just contraction may hypothetically compress matter in some thin spherical layer of the star to critical density where horizon forms at each point of the layer. Then not a single point, but a whole sphere touching the layer may become the horizon. – Ján Lalinský Dec 04 '22 at 22:26
  • @JánLalinský This cannot happen for a number of reasons. Any matter adds to the curvature, so if matter is present, there always will be the point of the maximal time dilation. Also, as observed from outside, the horizon has a property of the null infinity, which matter cannot reach ever regardless of the configuration or dynamics. The sphere you are referring to has to be empty for the curvature inside to be constant and maximal. – safesphere Dec 05 '22 at 19:04
  • @safesphere What exactly do you mean by curvature? Why is curvature inside being constant and maximal important for the horizon to exist? – Ján Lalinský Dec 05 '22 at 22:53
  • @JánLalinský The part of the spacetime curvature responsible for gravity is the time dilation, slowing down of time as observed remotely. The horizon is where the time dilation diverges, meaning the horizon forms where time stops. If you want the horizon to form on a sphere, then time must stop at the sphere. If time at any point inside moves slower, then the horizon forms there instead. Time inside cannot move faster either (think of the gravitational potential, the shell or Birkhoff theorem). So the curvature (time dilation) must be constant and maximal on the sphere and everywhere inside. – safesphere Dec 06 '22 at 00:50

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Let's suppose Alice flies a rocket ship into a black hole. Poor Alice unknowingly crosses the event horizon. Too late, she realizes something is wrong as the tidal forces begin to grow, and she takes an evasive maneuver to try to get out of the gravitational field. Does she turn her ship left or right?

Note -- if you get bothered when free will gets brought into physics conversations, you can replace the above example with a robotic ship that will turn right or left with 50% probability depending on whether a radioactive atom decays or not, at some point after the ship crosses the horizon. Free will isn't a relevant issue here, Alice's spaceship is just meant to be a vivid example of information beyond the event horizon that is inaccessible to an outside observer in classical GR.

That information "exists" in spacetime, even though it's not directly apparent to an observer outside the black hole. Now, if I asked you "where" that information existed, you would probably say something like the information about whether Alice moved left or right "exists" on the trajectory of Alice's rocket -- in other words, the information exists in a region inside the event horizon. But that kind of intuition is precisely what leads to the degree of freedom counting where a black hole's entropy is proportional to its volume. If the holographic principle is correct, then that intuition must be wrong, and somehow the information about what Alice did after crossing the horizon gets encoded on the boundary of the black hole.

Now in classical GR, Alice's situation is strange, but not a problem. There's no real need to invoke the holographic principle in classical GR, so we might as well just assume the information about Alice's "right or left" choice is lost beyond the event horizon, inaccessible to outside observers who (like you said) never see the black hole form at all. Sure, there is some information in the spacetime that the outside observers will never be able to access, but, so what?

The issue arises when we include Hawking radiation. Then the black hole evaporates. So the outside observer will be able to access information about what happened inside the black hole, through the Hawking radiation. If unitary evolution of the quantum mechanical state is correct, then in principle it should be possible to evolve the quantum state back in time with unitary evolution and learn exactly what happened inside the black hole (however, this is completely impractical, since you would need to know a huge amount of information spread out in subtle ways over all of spacetime, and also you can never never learn the full state by measuring one copy of a quantum system). Various calculations of the entropy of the Hawking radiation show that it is proportional to the black hole's area. In order for the degrees of freedom of the black hole to be consistent with the entropy of the Hawking radiation, it must be the case that Alice's choice about whether to turn left or right did not just occur inside the black hole, but also must somehow have happened on the black hole horizon as well, even though Alice was nowhere near the horizon. That's certainly not something you would expect in classical GR, since an outside observer would never be able to learn anything that happened beyond the event horizon.

This is just a silly example, but maybe it helps elucidate why the area law is so weird.

Andrew
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  • "If quantum mechanics is correct, then it should be possible to observe the state of the Hawking radiation, and then evolve the quantum state back in time with unitary evolution and learn exactly what happened inside the black hole."

    But QM (the theory) is not meant to do this, and it was never demonstrated that it is possible to retrodict from measurements of radiation the past psi function of the radiating system. Not even for a single excited atom, because spontaneous emission is random in QM.

     – Ján Lalinský Dec 01 '22 at 02:32
  • @JánLalinský You're right, I'm speaking loosely. More technically I mean that if unitary evolution is correct, then knowledge of the full state of the Hawking radiation + other degrees of freedom after the black hole has evaporated should let you reconstruct the quantum state at earlier times, including the state of fields inside the horizon. – Andrew Dec 01 '22 at 02:37
  • That is the same argument though. It's completely unrealizable. There can't be knowledge of the full state (psi function, or quantum field). And we are not that sure that unitary evolution is always correct, when in all practical calculations we have to reject it at some point and use additional rules such as the Born probability rule or the golden rule. – Ján Lalinský Dec 01 '22 at 02:40
  • @JánLalinský Yes I agree, "unitary evolution isn't correct" is a possible resolution of the information paradox. I'll rephrase the part of my answer that says "if quantum mechanics is correct" to "if unitary evolution of the quantum state is correct," that's a more precise version of what I meant anyway. – Andrew Dec 01 '22 at 02:46
  • OK but then if deterministic evolution and omnipotent measuring powers are assumed, then there is no issue at all, because whatever we can measure on the black hole radiation was already determined by state of things in the past before Alice fell into the hole, including whether she turned left or right inside. How is entropy distributed in space seems irrelevant to this. – Ján Lalinský Dec 01 '22 at 02:57
  • @JánLalinský "whatever we can measure on the black hole radiation was already determined by state of things in the past before Alice fell into the hole" -- agreed. "How is entropy distributed in space seems irrelevant to this." -- But normal QFT doesn't behave like this, so that seems to suggest quantum gravity must not be a normal QFT. Anyway, I am not trying to take a stand one way or the other, just summarizing my understanding of "mainstream" views on the information loss paradox and black hole entropy. My personal feeling is that this is so speculative that it's basically philosophy. – Andrew Dec 01 '22 at 03:10
  • Thanks for clarification. > "But normal QFT doesn't behave like this, so that seems to suggest quantum gravity must not be a normal QFT." -- Could you please clarify this? Do you mean normal QFT does imply something about how entropy is distributed in space and it is inconsistent with concentration of entropy on the surface? – Ján Lalinský Dec 01 '22 at 03:15
  • @JánLalinský Imagine a free field. If you discretize space into a cubic lattice with $N$ grid points on each size, then the dimension of the Hilbert space will grow as $\sim N^3$. (Effectively you're quantizing $N^3$ coupled harmonic oscillators). So the number of degrees of freedom grows like the volume. If true, the holographic principle says that this counting must be wrong for black holes. I'll try to find a source that talks about it in more depth, but that's essentially how I've seen it presented before. – Andrew Dec 01 '22 at 03:27
  • Number of degrees of freedom is infinite in field theories, it is not "growing like the volume". In discrete theories we have finite numbers, and the number of degrees of freedom can be made proportional to volume, but then we don't have differential equations for EM or gravity fields. So this seems like connecting two very different ideas that are not compatible. – Ján Lalinský Dec 01 '22 at 17:23
  • In continuous theory, it does not matter whether information is stored in patterns in a volume or a surface; due to infinite number of degrees of freedom of these sets, even the smallest dimensional set is capable to carry the same information as 3D space region does. E.g. one can encode any function on 3D space as function on 2D or 1D space. – Ján Lalinský Dec 01 '22 at 17:31
  • For example, to store information contained in values of field in a 3D region into a 1D region, store each value at a point of 3D region with real coordinates $x,y,z$, at that point of 1D space that has the coordinate that is constructed as interleaving of digits of $x,y,z$. – Ján Lalinský Dec 01 '22 at 17:35
  • This mapping is not continuous, but it does preserve all the information about the original field in the 3D region. – Ján Lalinský Dec 01 '22 at 17:36
  • @JánLalinský I don't think this conversation is particularly productive in the comments -- at this point I'm going to point you to a review: https://arxiv.org/abs/0806.0402. If you don't like it that's fine -- I'm not sure I do, all of it is so speculative -- but these aren't just some random ideas I came up with last night, it's my best attempt to summarize what mainstream theoretical physicists in this field have said about why the area law is surprising. – Andrew Dec 01 '22 at 17:40
  • Alright! Thanks. – Ján Lalinský Dec 01 '22 at 17:50
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    From an outside observer's perspective this "choice" that Alice makes, happens infinitely far in the future. She never reaches the interior. We would only see her become part of the black hole exterior, and therefore additionally the information about which choice she would have made becomes a part of the exterior. There is no interaction between the choice she makes in the interior and the information on the exterior (since she never gets there). This is the way I would understand this thought experiment at least. – puzzleshark Dec 01 '22 at 17:54
  • I'd like to point out that this site is valuable mainly for mainstream physics education, so I think it is good to question and challenge unclear claims and point out their speculative nature when warranted. – Ján Lalinský Dec 01 '22 at 17:56
  • @puzzleshark indeed but then there is the idea that black hole will eventually evaporate in finite time. These (speculative) ideas (freezing at the horizon and evaporating in finite time) seem to go against each other. – Ján Lalinský Dec 01 '22 at 18:01
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    @JánLalinský That's a good point. I agree challenging speculative ideas is good. I enjoyed this discussion, I just think it's a little too complicated to have in a comment thread. To be fair, I think I tried to give the best possible answer I could to the stated question -- "why is the area law surprising." Anyway, thanks for the discussion! – Andrew Dec 01 '22 at 18:07
  • @Andrew I agree, thanks for the attempt. – Ján Lalinský Dec 01 '22 at 18:09
  • @puzzleshark Your understanding is correct and so this answer becomes moot from the second sentence claiming that Alice has crossed the horizon and thus can’t communicate with the outside. No black hole is needed in this example. Once we cross the moment of midnight, we no longer can communicate with yesterday either. – safesphere Dec 03 '22 at 03:11
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If you form a black hole from a star (or neutron star), there is already some matter at $r = 0$, and indeed, all radii $r < r_{s}$ contain some matter. Are you going to push it out to keep it on the boundary? If you consider Penrose diagrams and results from papers, the answer is no.

Even if not every black hole is formed this way, you can surely construct a black hole with matter on the interior. Should those black holes play by different rules?

The "frozen star" picture of a black hole leads to a lot of confusion. Do not assume that all the matter is on the surface.

Two major areas where general relativity is interesting are black holes and cosmology. Cosmology provides a great example of why "never observing" is different from simultaneous reality. Consider the cosmological event horizon. Light emitted past this radius now will never reach us--this is only 5 Gpc (16 light years) away. Since we are nothing special and the Universe is isotropic and homogenous on large scales, there is not some great wall of matter trapped at this horizon, and we are not trapped within some great wall of matter on the event horizon of a distant galaxy either, even though our photons will never reach them. We know that there are spacelike slices where galaxies are moving through the cosmological event horizon just fine, even if we cannot see it.

Alwin
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  • I'm not sure that this is true. A black hole still needs to form starting with a radius of 0. therefore all particles that become part of it, need to "fall into it" as I've described. At least this is how I would understand it – puzzleshark Dec 01 '22 at 17:56
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    I refer you to this excellent answer: https://physics.stackexchange.com/a/146852/307737 – Alwin Dec 01 '22 at 20:07
  • Hmm thanks for the reference! Clearly this is much more complicated then what I was thinking! – puzzleshark Dec 02 '22 at 15:11
  • Your answer is incorrect. “Are you going to push it out to keep it on the boundary?” - Yes. Imagine you observe a flying black hole hitting a rock. You’ll see the horizon pushing the rock and never see the rock crossing the horizon. This is called “linear frame dragging”. Another way of looking at this is that the spatial radius of the horizon is always zero $\sqrt{x^2+y^2+z^2}=0$ for any black hole, so the radius never increases despite the circumference growing. Geometry of strongly curved spaces is neither Euclidean nor intuitive. – safesphere Dec 03 '22 at 03:32
  • @safesphere thanks for your comments. What do you make of that reference shown above https://physics.stackexchange.com/a/146852/307737 ? It seems to claim not only can an outside observer “see” matter falling into a black hole but can even “see” the singularity. To me this doesn’t make too much sense – puzzleshark Dec 03 '22 at 05:34
  • That's not what the answer claims. In fact, the answer says: "When we say that a distant observer never sees matter hit the event horizon, the word "sees" implies receiving an optical signal. It's then obvious as a matter of definition that the observer never "sees" this happen, because the definition of a horizon is that it's the boundary of a region from which we can never see a signal." It just distinguishes seeing from being. – Alwin Dec 03 '22 at 06:29
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    @puzzleshark Ben is a smart guy, but his diagrams are flawed, the star collapses to a line, not to a point. His interpretation of them is also wrong, the vertical line is not an axis of rotation, because the singularity is a line, not a disk or a sphere. His most fundamental error is equating "simultaneity" (which he also claims doesn't exist) to timeslices (e.g. blue and red). This is absurd. The Schwarzschild spacetime is globally hyperbolic, so causality is preserved. This directly leads to the fact that nothing ever crosses the horizon, as far as the entire external universe is concerned. – safesphere Dec 03 '22 at 06:50
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Edit: What you are saying is correct, except for one thing. Just because an observer throwing something into a BH takes an infinite amount of time to see it fall into the BH does not mean the observer cannot detect the presence of a BH. The BH will be a dark region (volume) in space. This the observer can detect. However the fact that this already-existing region can be described using an area is the interesting part.

In virtually all physical systems that we know of, entropy is proportional to volume. In other words, in $n$ dimensions, entropy is described using an $n$-dimensional ‘‘hypersurface’’. The fact that the entropy of a black hole can be described using an $(n-1)$-dimensional surface is therefore a surprising result. Moreover, this result means that if you throw a physical system inside a black hole, the result of the sum of the entropy of this system (which we used to think of using a volume) plus that of the black hole is described using a surface not a volume.

Y2H
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    From OP's question, I think he's aware of that. So, it doesn't really address his question. – Raskolnikov Nov 30 '22 at 06:53
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    An observer outside a black hole sees just the light emitted by matter falling into it. However, due to time dilation this picture does not show what happens to that matter. If you fall into black hole "your information" will cross the event horizon but the outside observer will never see it. Why the information is "encoded" into event horizon is not a matter of course and therefor surprising. – JanG Nov 30 '22 at 16:08
  • As observed from outside, the volume of a black hole is zero. A black hole is not a ball in space. Roughly speaking, the inside is filled with time, not with space (the radius inside is a direction in time, not in space). Thus it would be much more surprising if entropy of a black hole were proportional to its volume and therefore zero. – safesphere Dec 03 '22 at 03:39
  • @safesphere that’s not true but I don’t have time to get into these basics. If you care to learn more about why this is not true maybe check Chapter 5 in Carroll. – Y2H Dec 03 '22 at 18:26
  • @Y2H It is unfortunate that your busy schedule leaves no room for a logical comment, but thank you for posting at least a patronizing opinion. It is not clear what exactly you think is not true, but just in case you refer to the zero volume, please see: https://arxiv.org/pdf/0801.1734.pdf - “There is zero volume inside the black hole in any Schwarzschild time slice of a Schwarzschild black hole spacetime.” - That is if you care to learn and have time to go into these basics. – safesphere Dec 04 '22 at 19:21