Regulars from Cosmic Variance will be well acquainted with the idea of “firewalls” around black holes, from reading Joe Polchinski’s guest post on the subject. And then you heard more about them from John Preskill’s post at Quantum Frontiers. Or maybe George Musser’s post at Scientific American. Long story short: there is a believable claim on the market that, if you believe that information is preserved in the evaporation of black holes via Hawking radiation, an infalling observer should be incinerated by a wall of high-energy radiation when they cross the event horizon, in dramatic contradiction to everything classical general relativity would lead you to believe. Important stuff, if true. (“True” might mean “the argument is valid but one of the underlying assumptions is wrong, therefore teaching us something important about quantum gravity.)

Word is now finally leaking out into the more popular press, courtesy of my lovely wife Jennifer Ouellette’s article at Simons Science News, a new initiative from the Simons Foundation. It’s a great article, which I would say even if we were not notoriously nepotistic back-scratchers.

Here’s my attempt to squeeze the firewall argument down to its essence, for people who know a little quantum mechanics. If information escapes from a black hole, the radiation emitted at late times must share quantum entanglement with radiation that escaped at early times, in order to describe a pure quantum state (from which the black hole presumably formed). At the same time, to an observer near the event horizon, the local conditions are supposed to look almost like empty space — the quantum vacuum. But within that vacuum are virtual particles, some of which will eventually escape in the form of radiation and some of which will eventually fall into the black hole. In order for the state near the horizon to look like the vacuum, that outgoing radiation and the ingoing radiation must also be entangled. Therefore, it appears that the outgoing radiation is both entangled with the ingoing radiation, and with the radiation that escaped at earlier times. But that’s impossible; quantum mechanics won’t let degrees of freedom be separately (maximally) entangled with two different other sets of degrees of freedom. Entanglement is monogamous. A simple — but unpalatable — way out is to suggest that the state near the horizon is not a quiet state of maximal entanglement, but a noisy thermal state of high-energy radiation — a firewall.

It’s a slightly tricky business, as you expect it to be when we’re mixing up quantum mechanics with things happening in spacetime, in the absence of a once-and-for-all theory of quantum gravity. Probably most people who have thought about the issue don’t believe firewalls really exist (although some do), but in that either there’s a secret flaw in the argument, or one of our fundamental assumptions is out of whack. Maybe information is not conserved, or maybe it’s transferred faster than light, or maybe quantum mechanics doesn’t work quite the way we thought. The story should continue to be interesting no matter what happens.

I have to admit that I don’t understand this firewall stuff in detail, but I don’t understand your simplified explanation either. Going from a maximally entangled state to a nonmaximally entangled state is supposed to help get over the monogamy problem by allowing the infalling particle to be entangled with two different things, one inside and one outside the black hole. However, the partial trace of a maximally entangled state is always maximally mixed and there is no hotter state than a maximally mixed one, i.e. it is a thermal state with infinite temperature. So, in the original explanation where we did not worry about monogamy and had maximally entangled states, surely things were even worse. Why didn’t people worry about infinite temperature super firewalls before the problem was pointed out?

Matt, I’m not sure what the relevance of the partial trace is. Pre-firewall-controversy, I think most people would have said that infalling observers think the environment looks close to a pure state (the vacuum) as they fell through the horizon. Sure, tracing over some of the modes in the vacuum would leave you with a thermal state, but why would you do that?

Without access to empirical verification physics regresses to philosophy and it’s never ending arguments about irrelevant technicalties.

OK, I’m still confused. Pre-firewall did people not think that the Hawking radiation was entangled with things inside the black hole?

Yargh!

http://www.nature.com/nphys/journal/v9/n1/full/nphys2492.html

Not that I understand anything about this. But your simplified description makes it sound like the two contradictory “entanglements” are seen by two different observers, who are (almost) completely causally separated by the event horizon. (If the infalling observer was behind the horizon, there would be no problem, right? The two observers could never compare observations, so there would seem to be no actual contradiction.)

Matt– Yes, they did. And they thought that an infalling observer would locally see a maximally-entangled state. They hadn’t realized that it was difficult for both of these things to be true at the same time.

Charles R– That could be the kind of thing that is relevant. But in conventional QM, it doesn’t matter who “sees” the entanglement; a state is either entangled or it isn’t.

Right. I guess I’m imagining applying something like Rovelli’s relational interpretation of QM here, where a quantum state can only be defined in relation to a particular observer.

Does this apply to wormholes too? That will make life harder for science fiction writers.

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I didn’t know that the entanglement of a wavefunction is invariant under spacetime coordinate transformations (I guess locally Lorentzian transformations). Is there then an operator observable E of which its eigenfunctions would be those that are entangled? And E would then transform as a symmetry in spacetime?

I had imagined, probably falsely, that entanglement was like simultaneity in Special Relativity – where it is not invariant.

“there is no difference between free fall and inertial motion”

This is an assumption. Gravity is considered to be acting like a force, it has not been unified yet, so this is an assumption. An assumption that is perhaps incorrect…

@matt:

When thinking of giving up the equivalence principle, beware that this very same principle (or rather the “strong” formulation of it) was used to formulate the Standard Model as well. All four interactions (gravity, electroweak and strong) obey it, as we can see so far from all possible experiments. “All laws of physics locally look the same as in flat space” — where “space” is an appropriate fiber-bundle over the base 4-manifold. Curvature of this fiber-bundle gives rise to all known interactions (Higgs notwithstanding). This curvature never couples directly to matter fields, but only through “minimal substitution”, coming from symmetry localization. This is just the strong equivalence principle (though rarely explicitly named as such).

As for the firewall — in the absence of any experimental data, it boils down to one’s personal taste about which principles to keep and which to give up. Personally, I’d rather have gravity violate unitarity than the equivalence principle. 🙂

Best, 🙂

Marko

I just want to know that if Firewalls become a reality, will I be able to get a refund for The Black Hole War: My Battle with Stephen Hawking to Make the World Safe for Quantum Mechanics by Susskind. Will Preskill return his Encyclopedia to Hawkings? Finally, how would this affect the basic holographic principal relating to string theory, allowing 3d info on a 2d surface, outside of issue of black holes?

I’m of course no expert on the subject, but wouldn’t the existence of a singularity in the math indicate that the equivalence principle would be more likely to break down than the other 2 possibilities? If normality is an accurate thought, then I would assume that when physics laws break down at a singularity, then so would the equivalence principle. I’ll admit that this may be my ignorant assumption.

These arguments about “information” in black holes seem terribly ill-defined to me. Basically, is there a way to carry the solutions of the pde of some kind into, and then importantly, back out of the black hole?

In practice, can we construct a powerful enough laser/radio wave, such that the signal crosses the even horizon and then later we can detect (even in principle) an emitted signal from the black hole at a later time? Keep in mind that for entanglement purposes, our signal could be encoded in the polarisations/ sequential polarisations of the signal.

If the interior and exterior of the black hole are continuous, and you allow a signal to travel in some fashion though and out of it, then it seems to me that no matter how strong this firewall, you must allow the possibility of information transmission in theory.

However, if you cannot carry the solution of a pde through the black hole and out the other side then it seems you must entirely abandon the concept of information altogether, before you even consider the properties of black holes. If you cannot describe the path or the state of “information” after it enters the black hole then you can say nothing about what it should or should not do when or if it ever emerges. It could be destroyed, it could be randomised, it could be bifurcated and non-linearly interfered with in which case it’s practically gone away.

Anyway, since I don’t know much else about it, I’ll ask a gendanken-question: If you fired a laser/light/radio pulse at the black hole, with equal strength from all directions simultaneously(a collapsing spherical wave), would the black hole emit a corresponding pulse at some later date?

I just want to know that if Firewalls become a reality, … Will Preskill return his Encyclopedia to Hawkings?If there is a firewall, he should only return the ashes 🙂

Hi Sean —

This is a question relevant to BH horizon firewalls, and though doesn’t get into anything as deep as entanglement I thought this might be a natural place to mention it.

Consider an arbitrarily thin, spherical shell of matter infalling symmetrically onto/into a Schwarzschild (ie un-charged, non-rotating) black hole, and the set of null trajectories, ie “signal paths”, originating on the shell’s trajectory and heading out radially. In classical GR these signal paths take longer and longer to emerge up out of the black hole as seen by stationary observers at some safe distance away; in fact, the “last” signal line which originates just before the shell crosses the event horizon doesn’t reach the distant observer until an infinite amount of that observer’s proper time passes. Thus, even if some equipment on the shell were able to keep sending signals along those signal paths, the distant observer would never see the shell actually cross the horizon, but it would appear “frozen” on the horizon’s surface. Thus, I learned in Kip Thorne’s book, one term in early use for the classical Schwarzschild solution was “frozen star”.

So far, so classical. Now here’s the interesting question. If the BH is evaporating during the time of the shell’s infall, albeit slowly, due to Hawking radiation, then the stationary observers watching from up at a safe distance will — eventually, after a long but finite amount of their proper time — see all of that radiation, ie the entire BH’s worth of energy, emerge and cross their radius on its way out. Thus, the distant observers “see” a very odd time ordering: from their view, just looking at sheaves of signal paths, the infalling shell didn’t actually cross the horizon and enter the BH interior before the entire hole was evaporated! This has two (at least) very strange immediate consequences: (1) If the Hawking radiation did originate at the event horizon itself, then the entire BH’s worth of energy must have passed up through the shell on its way out — and all in a _very_ short amount of proper time as experienced by the shell itself. Talk about a fire wall! and (2) Since the shell didn’t enter the horizon before the evaporation was complete, it will, again after a short amount of its own proper time, find itself floating in empty space! once the BH is completely gone.

I _think_ one can escape these odd consequences along the following lines, but this is where you big-brained people will have to take over for a more solid answer: (1) If the Hawking radiation is actually formed at some finite distance “above” the horizon, then the shell will quite quickly (as measured both through its own proper time and the distant observers’ proper time) be “under,” ie closer to the horizon than, where the radiation is coming into being. So is the shell safe from immolation because the radiation is being formed “overhead” and going upward? or does some of it threaten to scatter back inward and menace the shell from above?

And (2) Generally if the mass of the BH is shrinking due to Hawking evaporation then the radius of its EH must also be decreasing proportionately — can you draw a Finkelstein diagram for this? — as measured by the distant observers. So, presumably the infalling shell will then follow the horizon as it shrinks inward, still frozen onto a deflating surface? If so, then eventually the shell will be compressed down to some very small radius, whatever is the smallest radius the hole reaches before its final moment of evaporation. Thus the shell _does_ eventually meet the unclothed singularity, even if it never crosses the horizon first!

So, it seems to me that there is already plenty of weirdness that we have to face once we allow for Hawking radiation, even before getting into such subtler questions as entanglement, unitarity and the preservation of information. So, can you enlighten me on this basic time-ordering puzzle? will the distant observers see a BH evaporate completely _before_ they see an infalling inertial trajectory cross the horizon?

[Posted just as we cross the horizon from 2012 to 2013 in my time zone; happy new year!]

Paul– These are complicated questions, and often depend on subtle assumptions. If a black hole evaporates completely away, there wasn’t *really* ever an event horizon at all. (There can be an apparent horizon: http://en.wikipedia.org/wiki/Apparent_horizon)

From the point of view of the infalling shell, there is a moment when they enter the apparent horizon. That’s long before the black hole has evaporated, so from their perspective they just fall right into the hole — they don’t see it grow smaller and smaller. Every signal they emit after that stays inside the black hole, or at least appears to. But the external observer will in principle see the effects of those signals, but it will all be mixed up with the Hawking radiation. (If information is conserved.)

If the four forms of the Bose-Einstein condensaat could be explained by interaction with both foton and electron,is this the way how to proove the real behaviour of the electron.

Einstein versus Bohr; Einstein said: God doesn’t play dice. Who is right?