I've exchanged a dozen of e-mails with Joe Polchinski, the most well-known physicist in the original team that proposed the firewalls. We haven't converged and Joe ultimately decided he didn't have time to continue and recommended me to write a paper instead (which he wouldn't read, I guess). However, he started to listen to what my resolution actually is and I could see his actual objection to it which seems flawed to me, as I discuss below.

Recall that \(\heartsuit\) represents the (near) maximum entanglement and the firewall folks demonstrate that because \(R\heartsuit R'\), the following things hold:\[

A\heartsuit B, \quad R_B \heartsuit B.

\] The degrees of freedom \({\mathcal O}(r_s)\) outside the black hole at \(t=0\) when the black hole gets old are maximally entangled with some part \(R_B\) of the early Hawking radiation (because \(R\heartsuit R'\)) as well as with the degrees of freedom inside the black hole \(A\) which are "mirror symmetrically" located in the other Rindler wedge from (infalling) Alice's viewpoint.

But because a system can't be maximally entangled with two other systems, there is a paradox and one of the assumptions has to be invalid. AMPS continue by saying that what has to fail is the "emptiness of the black hole" assumption. They make another step and say that all field modes inside the old black hole are hugely excited so an infalling observer gets burned once she crosses the event horizon.

My answer is that the resolution is that \(A\) and \(R_B\) aren't really "two other systems"; \(A\) is a heavily transformed subset of degrees of freedom in \(R_B\) so \(B\) is only near-maximally entangled with one system, not two, and everything is fine. I believe that this has been the very point of the black hole complementarity from the beginning.

*Yup, I took this picture in Santa Barbara, not far from the KITP.*

Now, Joe's objection is the following:

Even Alice must have a state, a pure state or a mixed state, ready to make predictions. Up to \(t=0\), she describes the early Hawking radiation in the same way as Bob (who stays outside). For example, this wave function may imply that \(N_b=5\) for an occupation number measured outside the black hole; the state may be an \(N_b=5\) eigenstate.This is Joe's paradox readjusted to "my" resolution. Is this paradox real?

It also means that the state is an eigenstate of a (complicated) observable at a later time which evolved from \(N_b\). On the other hand, this observable doesn't commute with \(N_a\), the occupation number for a field mode moderately inside the black hole, in the \(A\) region. The state can't be an \(N_b\) eigenstate and an \(N_a\) eigenstate at the same moment because they're generically non-commuting operators, and consequently, it can't be true that \(N_a=0\) which is needed for the emptiness of the old black hole interior from the viewpoint of an infalling observer.

I don't think so. What is true and what is not about Joe's statements above?

First, let us ask: Will her "later" state be an eigenstate of \(N_a\) or \(N_b\)? Here, the answer is clear. We assumed the state to be an \(N_b=5\) eigenstate so by the dynamical equations, the state will remain an eigenstate of an observable that evolved from \(N_b\) via Heisenberg's equations. For this quantity (probably involving an undoable measurement in practice), the measured value is sharply determined.

To predict the value of another quantity such as \(N_a\), we need to decompose the state into \(N_a\) eigenstates and the squared absolute values of the probability amplitudes determine the probabilities of different results. Joe is right that in principle, because the operators \(N_a\), \(N_b\), and their commutator are "generic", there will inevitably be a nonzero probability for \(N_a\neq 0\).

However, Joe isn't right when he suggests that this means a problem. Indeed, the probability of \(N_a\neq 0\) will be nonzero. Nevertheless, this probability may still be tiny. In other words, the probability that a particular mode will be seen as \(N_a=0\) may still approach 100 percent so in the classical or semiclassical approximation, quantum gravity will continue to respect the equivalence principle which implies that an observer falling into an old black hole sees no radiation (not even the Unruh one which he would see if he were not freely falling).

And I think that this is what happens. The probability of \(N_a\neq 0\) is tiny but nonzero. We may try to be somewhat more quantitative.

Choose a basis of the \(\exp(S)\)-dimensional space of the black hole microstates so that the basis vectors are \(N_a\) eigenstates. Now, my point is that the number of \(N_a=0\) basis vectors can be and almost certainly is greater (and probably much greater) than the number of \(N_a=1\) or higher eigenvalue eigenstates. We know how it would work if \(N_a\) were counting the occupation number from a non-freely-falling, "static" observer's viewpoint.

In that case, the probability of having a greater number of particles (by one) would be suppressed by a factor similar to the Boltzmann factor \(\exp(-\beta E_n)\) where \(E_n\) is the energy of the mode and \(\beta\) is the inverse black hole temperature, comparable to the Schwarzschild radius. In that case, we could prove that the probability of having a higher number of particles (by one) in some spherical harmonic \(Y_{LM}\) would be suppressed by something like \(e^L\); I am a bit sketchy here. This is just a description of the Unruh radiation that a "static" observer would experience right outside the black hole.

Things are harder for the freely infalling observer. Classically, she shouldn't see any radiation – because of the equivalence principle – so the higher values of \(N_a\) should be even more suppressed. The suppression should become "total" for macroscopic black holes.

At the same time, however, the working of the low-energy effective field theory means that in the relevant Hilbert space, the creation operator increasing \(N_a\) must relate states pretty much in a one-to-one fashion. So how it could be true that the number of \(N_a=1\) eigenstates in the basis is (much) lower than the number of the \(N_a=0\) eigenstates?

Before you conclude that my scenario is shown mathematically impossible, don't forget about one thing. If you create a quantum in a quantum field (in the black hole interior) and increase \(N_a\), you also change the total mass/energy of the black hole from \(M\) to \(M+E_a\). So the \(N_a=0\) states of a lighter black hole are in one-to-one correspondence with the \(N_a=1\) states of a heavier black hole.

In other words (or using an annihilation operator), the \(N_a=1\) states of the black hole of a given mass are in one-to-one correspondence with the \(N_a=0\) states of a lighter black hole whose mass is \(M-E_a\). But a smaller black hole has a smaller entropy, and therefore a smaller total number of all microstates. The ratio of the number of states is approximately equal to\[

\frac{ \exp(S_{\rm larger})} {\exp(S_{\rm smaller})} \sim \frac{\exp(M^2)}{\exp[(M-E_a)^a]}\sim\exp(2ME_a G)

\] where I restored Newton's constant in the final result. All purely numerical factors are ignored. This result is still the quasi-Boltzmannian \(\exp(C\beta E_a)\) with some unknown numerical constant \(C\). Well, the calculation was really more appropriate for a static observer but even for a freely infalling one, it should still be true that the action of a creation operator creates a larger and heavier black hole. In other words, the annihilation operators produce a lighter black hole with fewer states, and therefore the excited states are in one-to-one correspondence with the smaller number of states of a lighter black hole.

Even if the calculation above is wrong despite the tolerance for errors in the numerical factors (if the parameteric dependence is different, and I hope it is), I think it's true that a fixed-mass black hole has fewer eigenstates with \(N_a=1\) than those with \(N_a=0\), so it's more likely that we will see empty modes. This likelihood is becoming overwhelming for modes that are sufficiently localized on the event horizon (those proportional to \(Y_{LM}\) with a larger \(L\)). It means that if you pick a generic state such as the \(N_b=5\) eigenstate and decompose it to the \(N_a\) eigenstate basis, most of the terms will still correspond to \(N_a=0\) which means that there will be a near-certainty that you will measure \(N_a=0\) even though \(N_a,N_b\) refuse to commute.

The fact that \(N_a\neq 0\) may happen shouldn't be shocking. Look at a younger black hole and you will see that the interior can't be quite empty. It takes \({\mathcal O}(r_s)\) of proper time to suck "most of the material" of the star from which the black hole was created but because some of the material recoils etc., there's a nonzero amount of material inside the black hole at later times, too.

Joe neglects the fact that \(N_a=0\) is only an approximately valid statement and uses the strict \(N_a=0\) to derive a paradox. I think that \(N_a=0\) is just approximate and in fact, it's an interesting challenge to use the laws of quantum gravity – or specific laws in a formulation of string theory – to derive the percentage of states that have \(N_a=1\), for example. Classically, almost all microstates must correspond to an empty interior (as seen by an infalling observer) because the highest-entropy, dominant microstates are those that (because of the second law of thermodynamics) appear "later" once the black hole is sufficiently stabilized, almost perfectly spherical, and after it has consumed the star material and its echoes. The reason behind \(N_a\approx 0\) is therefore "entropic".

I don't know what the exact parametric dependence is so most of the formulae above were just "proofs of a concept" but I do think it shows that Polchinski et al. have overlooked a loophole that is arguably more plausible than all the loopholes they have discussed. The loophole says that the emptiness of the black hole interior simply isn't perfect but it is very good for large black holes and localized modes (certainly no deadly firewalls!). The equivalence principle at long distance scales, unitarity, and other assumptions of quantum mechanics and low-energy effective field theory may be preserved when complementarity is allowed to do its job and declare the information inside and outside the black hole as "not quite independent information".

And that's the memo.

I am working on an argument to hopefully complement yours. My chief belief is that the evolution of Alice after she falls into the black hole ends for Bob. There are simply no classical communication channels available to Alice by which she can communicate with Bob about the outcomes of her experiments. Equivalence principle also requires her local physics are always consistent.

ReplyDeletePolchinski et al seem to understand this but want a physical mechanism to destroy but want to destroy Alice's ability to perform local experiments. This seems counter to the equivalence principle.

Classical Alice will simply never emerge from the black hole to tell us anything, nor will any results of observations requiring "decoherence", at best all we can hope to know is what she was doing at the time she fell in. She simply has no ability to "collapse" a state relevant to Bob, since she is no longer describable as a classical object once she crosses the horizon. However she isn't destroyed locally.

I believe the proof of the counterargument to firewalls lies in the definition and boundary between free and entangled systems, since the direct product of

I totally agree with your comment, too (sorry if it was cut because you tried to write maths but it was interpreted as HTML tags), although a similar thing was the focus of a previous blog entry of mine.

ReplyDeleteWrite the paper.

ReplyDeleteLubos, I think many people would be very interested in reading an arXiv submission from you on this topic. It seems that you have a very clear understanding of the difference between AMPS and complementarity and I think the direct mapping is important. Even Bousso doesn't sound happy with firewalls. He makes the point that complementarity is still crucial for some applications though seemingly not enough for this paradox, and it seems like "overkill" to have both complementarity and firewalls.

ReplyDeleteI think that if you can show how these are really the same thing, many people would find it of value.

Thanks again!

I look at the holographic principle and I see the n+1 dimensions on the supersymmetric side with m=0 dimensional particles. I then see on the n dimensional conventional side strings with m=1 dimensions. Taking a metric shared by strings and space or particles and space we have ds^2= dx1^2 + dx2^2 + ...+ dx(n+1)^2 on one side and ds^2-dxn^2 = dx1^2 + dx2^2 + ... dxn^2 on the string side, not to mention the dualities between branes and such. Is anyone trying to develop a Quantum Mechanics that has a metric (ds^2) which is dimensionally invariant? I suppose someone is, it looks obvious to me, but maybe this is the solution to black hole physics. Lumo, do you have any input.

ReplyDeleteThanks

Forgive the math error on the left side of the 2nd equation, n should be (n+1).

ReplyDeletehi lumo,

ReplyDeletei have a very naive question : how the black hole can be rigorously empty (N_a=0) if there is an observer inside it ? it seems to me, your assumption that N_a=0 is approximate is automatically validate by the presence of an observer, even a skinny one. it feels to me that she "perturbes the measurement". couldn't she have (and see) her own contribution to the hawking radiation?

So if I understand your argument correctly, this is more a question of magnitude rather than effect. Namely the firewall argument is correct, but parametrically small and hence deviations from the equivalence principle are also unobservable?

ReplyDeleteThis is something I didn't quite understand from the original papers. How big is this effect calculated to be in the first place?

in fact, i can't even imagine how could a black hole evaporate if it is empty... and that explains the possible (approximate) vacuity of my previous question.

ReplyDeleteIt is counterintuitive but the black hole does evaporate even though it's empty - except for the very beginning when it's created out of a star etc. The matter is really "nonlocally tunneling" away from the black hole. Only quantum field theory at a curved background is the description needed to understand why this process is really going on.

ReplyDeleteRight. I think that in principle, it's true that there's "something" inside the black hole but AMPS don't really care much how *much* matter, energy, radiation - quantitatively - there is, and the right answer is that the number is tiny and going to zero in the classical limit of large black holes and localized field modes so the the equivalence principle survives exactly at the accuracy where one demands it should.

ReplyDeleteRight, Alice is of course an exception making the interior non-empty. But one may imagine it's a localized "exception to the rule" - Alice is falling in an otherwise empty environment - and one may study the modes of fields that are "outside Alice's body" where one expects N_a = 0.

ReplyDeleteThanks for the recommendation, Pig, but I don't consider most of this original research - many people are probably making similar arguments in front of their blackboards in their departments, and I have already been contacted by an example - and I am kind of happy that I don't have to face various hassles connected to the publishing of papers.

ReplyDeletea quantum tunnel directly from the singularity... do you have some references where such QFT in curved background calculation are done. i'm curious.

ReplyDeleteWell, the most appropriate paper is the paper that pioneered this calculation and discovered the radiation. It's the most famous paper written by the physicist. And you may have heard of the name of the physicist, too. It's Stephen Hawking. Over 6000 citations:

ReplyDeletehttp://www.itp.uni-hannover.de/~giulini/papers/BlackHoleSeminar/Hawking_CMP_1975.pdf

Could you comment on what the outside observer will see? Will Bob, standing far outside the black hole, actually see Alive get destroyed by Hawking radiation? Or what will Bob see going on? Will Bob see much at all, because the radiation is so weak?

ReplyDeleteAnonymous who penetrated into the defunct Blogger comments in some way asked:

ReplyDeleteCould you comment on what the outside observer will see? Will Bob, standing far outside the black hole, actually see Alive get destroyed by Hawking radiation? Or what will Bob see going on? Will Bob see much at all, because the radiation is so weak?

Answer: No. I am pretty sure that everyone agrees that no one outside can ever see what's happening inside the horizon, and not even what's happening at the horizon. It's causally impossible and causality can surely be trusted, at least approximately and at least outside the black hole.

The radiation is weak and redshifted when observed from a long distance from the black hole. But the corresponding radiation at various frames near the event horizon may be very strong. But the firewall is something that is supposed to go beyond any Hawking radiation or "transformed" radiation.

Hi Lubos,

ReplyDeletethanks. I asked the previous question.

I agree that Bob cannot see inside the black hole; I was thinking about seeing effects just outside the horizon. I ask this because Susskind used to say something like: `consider an elephant falling towards a black hole; to an outside observer (Bob) the elephant will get slowed down (time-dilated) then thermalized by the Hawking radiation, and eventually emitted, and so the elephant will never make it into the black hole according to Bob.' Does this make sense? Would Bob describe things this way? I thought that even though Alice will just see the elephant fall in with her [assuming firewalls are wrong], Bob will think of it as the elephant being destroyed, and that is the complimentarity picture.

Yes, it makes complete sense. Because the black hole interior is something that can't influence Bob, it's only a "final dumping ground", and Bob may reduce his world to the observable one that is only outside. So infalling Alice looks like a redder and redder slowing down lady, the signal evaporates, her experiences inside the hole are completely unobservable, and Bob literally sees - and may think - that Alice gets dissolved around the horizon as she becomes indistinguishable from the Hawking radiation.

ReplyDeleteBob will see that Alice and elephant are thermalized - their information becomes impenetrably encoded in "thermal mess" and "thermal mess" is usually identified with the death. However, Bob can't strictly conclude that Alice will feel like dying. And she won't.

thanks!

ReplyDeleteJust to clarify: is my summary of the "complimentarily picture" correct? (the very different description of the elephant between Bob and Alice) or am I missing something?

Could an event horizon be, ultimately, the result of a failure of mutual consistency between a pair of observers, where we define an observer in terms of an Abelian von Neumann subalgebra of observables?

ReplyDeleteDear Lumo,

ReplyDeletecould none of your colleagues who publishes papers more regularly help you avoid, circumvent, or ease the hassles a bit...? I suspect the hassle would less consist in writing the paper but rather in annoying "bureaucratic" obstacles in submitting ? It is not fair that you, who have such great insights into the topic at hand, should not be able to contribute to the arxiv discussion. Only if you'd like to of course ... ;-)

Anyway, I like this continution of the previous article and the arguments explaining that a large black hole appearse almost completely empty to an infalling observer (cool picture, ha ha!) :-)

Cheers

My question is probably very stupid and off-topic, but anyway:

ReplyDeleteHas anyone tried considering the event of the observation and the event of acquisition of the information as separate events? Can we postulate that the observer has limited "computational power", so recovering information (whatever that means) from Hawking radiation requires quite some time? Will that have any implications?

Dear Dilaton, I suspect that Lubos would have zero (0) trouble getting a paper of his accepted to arXiv and that he was in fact referring to getting a paper published in one of the established journals, which at a minimum would involve taking and incorporating feedback from reviewers and possibly also taking into account "political" considerations, a process which would take months, and he may well have decided that his time is better spent writing TRF articles... and working on his secret "projects".

ReplyDeleteOne of these projects, my crystal ball tells me, has to do with using neutrinos (controlled flavor changes and detection of same) for straightline broadband signaling through the earth and when completed will make him the world's first $ 10^11 person, with a Nobel thrown in for good measure ;)

Dear Eugene, no one who was recently enough sending papers to the arXiv can have a problem to send a paper to the arXiv. it's a non-peer-reviewed server for the community.

ReplyDeleteDear Ghandi, may I ask you to try to learn how to spell "complementarity" with at most one typo - you have two - before you ask someone to test whether you have actually understood the content of the principle? ;-) Thanks.

ReplyDeleteYour comment described one perspective of a process that is relevant for the complementarity, but not really the complementarity itself.

Yup, definitely, such limitations are important and have been appreciated but they don't affect too many things because the observer outside has infinite time to evaluate qubits from the Hawking radiation - the Universe outside lasts forever. While the observer inside sees no Hawking radiation. ;-)

ReplyDeleteDoes it help, for more qualitative descriptions, to consider a classical black hole, with a constant mass density ?

ReplyDeleteThis density will be of magnitude 1/(G (Rs)²) or 1/((G^3) M²). So maybe we could use this density, to be a qualitative indicator, of the percentage of excited modes inside the black hole.

Dear Trimok, note that your formulae contain no hbar, so they're classical results. But that's bullshit because classically, the black hole is empty, so the density is zero, except for the singularity where the density is divergent.

ReplyDeleteThe actual density must be much lower. It has also a certain dependence on the mode frequency etc.

Apologies, in a rush yesterday. What I wanted to say is that the direct product of free spaces is definitely not the direct product of the entangled basis states. in any reading of t he amps scenario the start with the collapse of a non entangled pure state. they then propose that the observers pop up afterword. So we have to assume in the AMPS construct that observers and the black hole were separable free states initially and entanglement entropy was zero. As soon as we say they interact we have to expand the applicable Hilbert space in the representative basis bases of the black hole and observers. So the schmidt expansion should be:

ReplyDelete|psi_s> = Sum sqrt(lamb_j) |j_hawkhole>|jhat_alicebob>

I am still working through my thoughts on the maximal entanglement and degeneracy (which leads to correlated observations). However, I think this is the right approach to refutation.

also Polchinski says "So this is effectively illegal cloning, but unlike earlier thought experiments a single observer can see both bits, measuring the early radiation and then jumping in and seeing the copy behind the horizon."

ReplyDeleteI don't think this possible because of relativistic considerations and the conservation of energy and momentum. Although Hawking radiation is a quantum phenomenon that does not violate relativity, from a classical view point the emission of radiation from the black hole could be viewed as an object from the interior overcoming the escape velocity. From a strictly classical perspective, and from conservation principles, the momentum and energy of the particle that had an inward direction would have to be viewed as having velocities of c. An infalling observer, having measured the outgoing radiation, would necessarily have a gap they would have to overcome in order to make a measurement of the infalling particle (from a classical perspective). So it simply is not true that a classically described observer like Alice could possibly "catch up" and perform a measurement on the early infalling radiation, not without a drastic violation of local physics.

An interesting perspective. Good luck with your paper that I will read if it appears and has less than 50 pages.;-)

ReplyDelete