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  How does the holographic principle imply nonlocality?

+ 6 like - 0 dislike
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For example in the discussions here and here there are comments by Ron Maimon:

Your complaint about locality would be more serious if holography didn't show the way--- the CFT in AdS/CFT produces local AdS physics, even though the description is completely and ridiculously nonlocal

and

Once you realize that gravity is defined far away on a holographic screen, the idea of hidden variables becomes more plausible, because the physics of gravity is nonlocal in a way that suggests it might fix quantum mechanics

How is gravity nonlocal? I thought GR was explicitly Lorentz Invariant? Or are these statements more philosophical (something I would not expect from Ron), ie, just a statement that the boundary is "far away" and isomorphic to the interior...


EDIT:

Ron gave an answer that is very difficult for me to parse. Can someone who is a bit more pedagogically inclined interpret what he says? I asked him to clarify various points in the comments, with little luck. I'm not even sure how he is defining 'locality':

The nonlocality of gravity doesn't mean that Lorentz invariance is broken, Lorentz invariance and locality are separate concepts. It just means that to define the state of the universe at a certain point, you need to know what is going on everywhere, the state space isn't decomposing into a basis of local operators.

I do not see how this does not violate Lorentz invariance. If your state at time t depends on parts of the universe outside your light cone, this is clearly a-causal.

"Locality" is a bit of an overloaded term, and for this discussion I will assume that it means there are bosonic operators at every point which commute at spacelike separation (Bosonic fields and bilinears in Fermi fields). This means that that the orthogonal basis states at one time are all possible values of the bosonic field states on a spacelike hypersurface, and over Fermi Grassman variables if you want to have fermions.

I do not understand this definition, and frankly it seems unnecessarily complicated and non-transparent. Is this a different definition of 'locality' compared to what is used, for example, in Bell's famous paper?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
asked Aug 18, 2012 in Theoretical Physics by user1247 (540 points) [ no revision ]
Related physics.stackexchange.com/questions/33954/…

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user drake
To clarify your doubts: Lorentz invariance just tells us that all inertial frames are equivalent. Locality/causality is a separate physical input into the framework. (Afaik) I could build a theory which is globally coupled but rotationally invariant, for eg: a system of N spins coupled to all other spins while respecting d-dimensional rotational invariance of the space in which they sit.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Siva
Operators commuting at spacelike separation is like saying that two points at spacelike separation are not correlated (since you calculate their correlation by taking a "vacuum" sandwich of those bosonic operators)

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Siva

1 Answer

+ 4 like - 0 dislike

The nonlocality of gravity doesn't mean that Lorentz invariance is broken, Lorentz invariance and locality are separate concepts. It just means that to define the state of the universe at a certain point, you need to know what is going on everywhere, the state space isn't decomposing into a basis of local operators.

"Locality" is a bit of an overloaded term, and for this discussion I will assume that it means there are bosonic operators at every point which commute at spacelike separation (Bosonic fields and bilinears in Fermi fields). This means that that the orthogonal basis states at one time are all possible values of the bosonic field states on a spacelike hypersurface, and over Fermi Grassman variables if you want to have fermions.

If you extend this idea to curved spacetime and to arbitrarily short distances, you get a completely ridiculous divergence in the number of black hole states. This was the major discovery of 'tHooft, which is the basis of the holographic principle.

To see this, consider the exterior Schwarzschild solution, The local t temperature is the periodicity of the imaginary time solution, and it diverges as 1/a where a is the distance to the horizon (this distance is measured by the metric, which is diverging in r coordinates, so it is not $r-2m$ for r near the horizon, but proportional to $\sqrt{r-2m}$. With this change of variables, the horizon is locally Rindler).

Assuming that the fields are local near the horizon, the thermal fluctuations of the fields consists of a sum over the entropy of independent thermal field fluctuations at the local temperature. You can estimate the entropy (per unit horizon area) in these fluctuations by integrating the entropy at any r with respect to r. The entropy density of a free field (say EM) at temperature T goes as $T^{3}$, so you get

$$ \int_{2m}^A {1\over (r-2m)^{1.5}} dr $$

The convergence at large A is spurious, the redshift factor asymptotes to a constant in the real solution, so you get a diverging entropy. This is sensible, it is just the bulk entropy of the gas of radiation in equilibrium with the black hole. But this integral is divergent near the horizon, so that the black hole Hawking vacuum in a local quantum field theory in curved spacetime is carrying an infinite entropy skin.

This divergent entropy is inconsistent with the picture of a black hole forming and evaporating in a unitary way, it is inconsistent with physical intuition to have such an enormous entropy in an arbitrary small black hole, it is just ridiculous. So any quantum theory of gravity with the proper number of degrees of freedom must be nonlocal near a black hole horizon, and by natural extension, everywhere.

The divergence is intuitive--- it is saying you can fit an infinite amount of information right near the horizon, because nothing actually falls in from the extrerior point of view. If the fields are really local, you can throw in a Gutenberg bible and extract all the text by careful local field measurements a hundred years later. This is nonsense-- the information should merge with the black hole and be reemitted in the Hawking radiation, but that's not what the semiclassiclal QFT in curved space says.

'tHooft first fixed this divergence with a brick wall, a cutoff on the integrals to make the entropy come out right. This cutoff was a heuristic for where locality breaks down. In order to fix information loss, around 1986, he considered what happens when a particle flies into a black hole, and how it could influence emissions. He realized that the only way the particle could influence the emissions was through the gravitational deformation the particle leaves on the horizon.

This deformation is nonlocal, in that the horizon shape is determined by which light rays make it to infinity. The backtracing showed that an infalling particle leaves a gravitational imprint on the horizon, like a tent-pole bump where it is going to enter. He could get a handle on the S-matrix by imagining that the bumps are doing all the physics, the horizon motion itself, and this bump-on-the-horizon description was clearly similar to the vertex operator formalism in string theory, but with crazy imaginary coupling, and all sorts of wrong behavior. This is now known to be because he was considering a thermal Schwartzschild black hole, rather than an extremal one. In extremal black holes, the natural analog to 'tHoofts construction is AdS/CFT.

String theory

In string theory, you have a nonlocality which was puzzling from the beginning--- the string scattering is only defined on-shell, and the only extension to an off-shell formalism requires you to take light-cone coordinates. This was considered an embarassment in string theory in the 1980s, because to define a space-time point, you need to know off-shell operators which you can Fourier transform to find point-to-point correlation functions.

In the 1990s, this S-matrix nonlocality was reevaluated. Susskind argued heuristically that a highly excited string state should be indistinguishable from a large thermal black hole. One of the arguments was that the strings at weak coupling at large excitation numbers are long and tangled, and should have the right energy-radius relationship.

Another of Susskind's arguments is that a string falling into a black hole should get highly thermally excited, and get longer, and it becomes as wide as the black hole at 'tHooft's brick wall, so that the brick wall is not an imaginary surface to cut off an integral, but the point where the strings in the string theory are no longer small compared to the black hole, and the description is no longer local.

Susskind argued that at large occupation numbers, it is thermodynamically preferrable to have one long string rather than two strings with half the excitation. This is essentially due to the exponential growth of states in string theory, to the Hagedorn behavior. But it means that the picture of a string falling into a black hole is better considered a string merging with the big string which is the black hole already.

The d-branes were also identified with black holes by Polchinsky, and the dualities between D-branes and F-strings made it clear that everything in string theory was really a black hole. This resolved the mystery of why the strings were described by a 2d theory which was so strangely reproducing higher dimensional physics--- it was just an example of 'tHooft's holographic descriptions.

All this stuff made a tremendous pressure to find a real mathematically precise realization of the holographic principle. This was first done by Banks Fischler Shenker and Susskind, but the best example is Maldacena's.

AdS/CFT

In AdS/CFT, you look near a stack of type IIB 3-branes to get the near-horizon geometry (which is now curved AdS, not flat Rindler, because the black holes are extremal), and you identify the dynamics of string theory near the horizon with the low-energy theory on the branes themselves, which consists of open strings stuck to the branes, or N=4 SUSY SU(N) gauge theory (the SU(N) gauge group comes from the Chan-Paton factors, the N=4 SUSY is the SUSY of the brane background, and the superconformal invariance is identified with the geometeric symmetry of AdS).

The correspondence maps the AdS translation group to involve a dilatation operator on the field theory, so that if you make an N=4 field state which is sort-of localized at some point in AdS, and you move in one of the AdS directions, it corresponds to making the blob bigger without changing its center. This means that there is absolutely no locality on the AdS side, only on the CFT side. Two widely separated points are represented by a CFT blobs of different scale, not by a CFT state of different position, so they cannot possibly be commuting, except in some approximation of low energy. The CFT is local, but this is boundary locality, analogous to light-cone locality, not a bulk locality. There is no bulk locality.

This nonlocality is so obvious that I don't know how to justify it any more than what I said. There aren't four dimensions of commuting bosonic operators in the N=4 theory, just 3 dimensions. There aren't five spacetime dimensions, just four. The remaining dimension is emergent by different scales in the CFT. So this example is airtight--- string theory is definitely nonlocal, and nonlocal in the right way suggested by 'tHooft and Susskind's arguments.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
answered Aug 19, 2012 by Ron Maimon (7,730 points) [ no revision ]
we discard infinities in gauge theories like its nothing and we don't even blink because as long as the observable quantities are finite, we are all good and dandy. Are we supposed to chicken out from some infinite entropy by asymptotically integrating near the horizon? but we cannot observe or measure anything arbitrarily close to the horizon either, that entropy is not observable. if we are "good" with infinities in QED non observable quantities we'll be fine with infinite entropies in the horizons as well. In any case this a topic i find exciting and your answer is very interesting Ron, +1

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user lurscher
@lurscher: the "infinities" in gauge theory are unphysical and not troublesome--- they are just mass corrections and charge corrections. This is a real honest to goodness physical divergence, it means the entropy is wrong in a local theory. It's not something you can fix by formal tricks, or by redefining parameters. Entropy is an absolute quantity in quantum mechanics, it is the log of a counting number.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
Hi @Ron Maimon. Thanks for the answer, but it's a bit over my head. I took field theory in grad school, so I have some background here, just not quite enough I guess. Below I have some questions to help clarify my understanding.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
"It just means that to define the state of the universe at a certain point, you need to know what is going on everywhere" -- So, for example, a measurement I make at point A may depend on what is happening at point B outside my light cone? In such a case wouldn't boosting mix cause and effect, and further wouldn't communication faster than light be naively possible?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
"This deformation is nonlocal, in that the horizon shape is determined by which light rays make it to infinity." -- But you go on to say that the deformation is caused by the infalling particle, not the outgoing particle making it to infinity. So I'm confused, and left not completely understanding how and where the nonlocality is manifested between the bumps on the horizon and the infalling and outgoing particles. Can you give a simple example involving one infalling particle, one bump, and one outgoing particle?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
"In string theory, you have a nonlocality which was puzzling from the beginning--- the string scattering is only defined on-shell, and the only extension to an off-shell formalism requires you to take light-cone coordinates. This was considered an embarassment in string theory in the 1980s, because to define a space-time point, you need to know off-shell operators which you can Fourier transform to find point-to-point correlation functions." -- Again an example might help, because here I don't understand how using light-cone coordinates means that the physics is non-local.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
"The correspondence maps the AdS translation group to involve a dilatation operator on the field theory, so that if you make an N=4 field state which is sort-of localized at some point in AdS, and you move in one of the AdS directions, it corresponds to making the blob bigger without changing its center. This means that there is absolutely no locality on the AdS side" -- OK, this is sounding like an example I can wrap my mind around, but as it stands I can't tell what you mean. What do you mean by "you move in one of the AdS directions, it corresponds to making the blob bigger" continued...

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
... a blob on the AdS or CFT side? And what you do mean by "you move"? An observer moves? Maybe if you make this example more clear and explicit and it give me a very clear example, that would be really helpful.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
@user1247: Forget "causality", this notion is redefined entirely in S-matrix theory in terms of analyticity. In AdS/CFT you have a different notion of "boundary causality", the boundary states lead to later boundary states in the usual way. If you took field theory, this is enough. Imagine that you have a field theory where translation in x,y are normal translations, but translation in z is implemented by a dilatation operator. The fields commute at different x,y, but moving in z can't possibly keep things local. This is a caricature of AdS/CFT (the translations are in a uniform curved space).

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
@Ron Maimon -- did you read my questions ("a measurement I make at point A may depend on what is happening at point B outside my light cone?"...)? I'm still struggling to understand what you mean by 'local' in your above response, so as it stands it doesn't provide me with any added insight. Also, I can't figure out what a 'dilatation operator' is.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
@user1247: A "dilatation operator" is one that moves x to $(1+a)x$, in other words it blows up the space by a rescaling factor. If interpret a bulk translation as a boundary dilatation, this is obviously a nonlocal map. You are right about the issue with light cones, this is why you have to be careful to say that the states are "in" scattering states, and causality is S-matrix causality. This is what people spent almost all the 1960s clarifying. The measurements are not made at "point A", you don't make local measurements. They are made asymptotically on scattering states.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
I added an edit to my question to reflect the fact that I'm hoping for someone else to come in and interpret what you are saying. I'm still not sure how you are even defining 'locality'.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
@user1247: I am defining "locality" as spacelike separated observables commute. For example, in light-cone, you have u-v coordinates, and only a codimension 2 collection of observables on the initial value null-sheet necessarily commute, so you don't have locality from light-cone locality.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
So... an example of something non-local would be space-like spin |z> measurements on an entangled pair of electrons. In that sense ordinary QM is non-local, so how are you saying anything non-trivial?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
This terminological confusion is what lead me to ask this question, which perhaps you would be interested in engaging: physics.stackexchange.com/questions/34650/…

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
@user1247: No, spacelike spin measurements are implemented by commuting observables: the spin operators on separated electrons commute. this implies the no-signalling theorem (and it is obvious in QFT). The property that observables don't have a localization, so that there are observables which do not commute at spacelike separation, is a statement that is not correct in ordinary QFT, it's what makes string theory nonlocal. Regarding your other question, I want to answer "causality", but I want to do it respectful of Mandelstam, and I am not sure how to do that well yet.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
I thought that if two observables commute, then a measurement of one does not constrain the eigenvalue spectrum of the other. But in the case of spacelike spin measurements, a measurement of spin UP on one electron constrains the measurement of the other electron to spin DOWN. Isn't this equivalent to a case of non-commutation of |Z(x1,t1)> with |Z(x2,t2)>? What am I missing?

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user user1247
@user1247: The state is what is special there--- the two observations commute, the reduced density matrix doesn't care if you measure the other electron, so long as you don't tell me the result. This is the difference between non-commuting observables and entangled states, it's a little confusing, I know, but I said it right above.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon
@Gugg: Great job, thanks, I always misspell the dude's name, it's embarassing.

This post imported from StackExchange Physics at 2015-04-11 10:31 (UTC), posted by SE-user Ron Maimon

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