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In all the canonical approaches to the problem of quantum gravity, (eg. loop variable) spacetime is thought to have a discrete structure. One question immediately comes naively to an outsider of this approach is whether it picks a privileged frame of reference and thereby violating the key principle of the special relativity in ultra small scale. But if this violation is tolerated doesn't that imply some amount of viscosity within spacetime itself? Or am I writing complete nonsense here? Can anybody with some background in these approaches clear these issues?

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The quantization of angular momentum does not imply that there's a preferred direction in space. Similarly, there's no reason to believe that the quantization of spacetime implies that the laws of physics are not Lorentz invariant. It's not clear that anybody has come up with a Lorentz-invariant theory of loop quantum gravity (if they have, it certainly hasn't been proven Lorentz invariant so far). –  Peter Shor Jan 23 '11 at 15:28
Isn't there a difference between quantization of angular momentum of a particle and the quantization of the very arena (spacetime) at which this quantization takes place? Doesn't the story become much more complicated when the very structure of spacetime itself subject to the uncertainty principle? –  user1355 Jan 23 '11 at 15:35
There is a difference; quantizing spacetime is much harder, and I don't believe that anybody has done it successfully yet. But that doesn't mean that it can't be done in principle. –  Peter Shor Jan 23 '11 at 16:04
I know, it is much harder. But that was not my point. I wanted to express my disagreement with the analogy expressed in your first two sentences of your initial comment. –  user1355 Jan 23 '11 at 16:13
I still like the analogy. I've heard a lot of arguments that you can't come up with a Lorentz-invariant space-time, and I don't see why you can't use most of them to show there's also no isotropic quantization of angular momentum (which we know is incorrect). –  Peter Shor Jan 23 '11 at 19:29
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7 Answers

up vote 9 down vote accepted

Here is as I see the problem. The best way to understand it is to think what happens with rotations and quantum theory. Suppose that a certain vector quantity V=(V1,V2,V3) is such that its components are always multiples of a fixed quantity d. Then one is tempted to say that obviously rotational invariance is broken because if I take the vector V=(d,0,0) and rotate it a bit, I get V=(cos(phi) d, sin(phi) d,0), and cos(phi) d is smaller than d. Therefore, either rotational invariance is broken, or the vector components can be smaller. Right? No, wrong. Why? Because of quantum theory. Suppose now that the quantity V is the angular momentum of an atom. Then, since the atom is quantum mechanical, you cannot measure all the 3 components together. If you measure one, you can get say eithe 0, or hbar, or 2hbar ..., that it, precisely multiples of a fixed quantity. Now supose you have measured that the x component of the angular momentum was hbar. Rotate slwly the atom. Do you measure them something a bit smaller that hbar? No! You measure again either zero, or hbar ... what changes continuously is not the eigenvalue, namely the quantity that you measure, but rather the probabilities of measuring one or the other of those eigenvalues. Same with the Planck area in LQG. If you measure an area, (and if LQG is correct, which all to be seen, of course!) you get a certain discrete number. If you boost the system, you do not meqsure the Lorentz contracted eigenvalues of the area: you measure one or the other of the same eigenvalues, with continuously changing probabilities. And, by the way, of course areas are observables. For instance any CERN experiment that measures a total scattering amplitude amplitude is measuring an area. Scattering amplitudes are in cm2, that is are areas.

carlo rovelli

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Hi Carlo, just wanted to say welcome, I hope you stick around to answer questions, not just make corrections. I’d admit for example that I don’t understand the discreteness of geometrical operators, how can those be diif invariant, and how can the planck scale discreteness be resolved with any measuring device which does not back-react on the geometry. As for your last sentence, I think we can both agree we don’t have to worry about this set of issues for the LHC... –  user566 Jan 27 '11 at 0:17
Thanks! This is time consuming, and there is so much else to do... But I'll try to stick around for what I can. I am disturbed by many wrong things I read, and would like to help bringing a bit of clarity, for the little I understand. And of course I am disturbed by all the nonsense Lubos is shouting around... –  Carlo Rovelli Jan 27 '11 at 7:54
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Yes, sb1, all your statements are totally correct.

A discrete spacetime immediately implies lots of bulk degrees of freedom that remember the detailed arrangement of the discrete blocks - unless it is regular and unique. Consequently, the "vacuum" carries a gigantic entropy density - the Planck entropy density if the concept is applied at the Planck scale.

Such an entropy density $(huge,0,0,0)$ immediately violates the Lorentz symmetry because in different reference frames, it will have nonzero spatial components.

Also, the privileged reference frame will mean that any moving object will instantly - within a Planck time or so - dissipate all of its kinetic energy to the extra degrees of freedom in the discrete structure. So the "vacuum" will actually behave as a superdense liquid that immediately stops any "swimmer": inertia becomes totally impossible. One could continue with other flagrant contradictions.

All these things dramatically violate basic observations of the real world. In fact, the Fermi satellite has verified that even at the Planck scale, the Lorentz symmetry works much better than up to O(100%) errors. Also, one can see that any non-stringy, non-QFT unification attempt for quantum gravity has considered the discreteness in the very sense we discuss, which is why it is instantly excluded.

One can use special dictionary and special proofs for each of them. For example, spin foam models define the proper area of surface $\Sigma$ more or less as the number of its intersections with the spin foam. However, it's clear that this number can't go to zero even if the shape of $\Sigma$ in spacetime is chosen to be light-like or near-light-like.

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It's very much reminiscent of the 19th century aether Lubos!! I hope that gloomy history is not repeating itself. –  user1355 Jan 23 '11 at 8:13
Exactly! Did you get to all these conclusions independently? Cool... I've been pointing the analogy - it's more than analogy, it's almost an equivalence - with the luminiferous aether for years... People like to "make things out of pieces" - and the electromagnetic and gravitational vacua are no exceptions. Moreover, there is this statement that "distances smaller than Planck length don't exist" that confuses many people. They heard it and misunderstood it, thinking it applies to coordinate differences. It can only apply to Lorentz-invariant "proper" distances, otherwise relativity is broken. –  Luboš Motl Jan 23 '11 at 8:19
you say - discrete spacetime immediately implies lots of bulk degrees of freedom that remember the detailed arrangement of the discrete blocks - unless it is regular and unique. Consequently, the "vacuum" carries a gigantic entropy density - the Planck entropy density if the concept is applied at the Planck scale. This is true only if you don't believe what the holographic principle or the covariant entropy bound has to say about the limits of observability of these bulk degrees of freedom. Also you claim the Fermi satellite has verified that even at the Planck scale. AFAIK that is –  user346 Jan 23 '11 at 10:50
simply untrue. I'm certain that there is no experiment in orbit or on earth which has ever probed any such effects anywhere close to the Plank scale. If there is please give me some references so I can share the sad news with my peers :) –  user346 Jan 23 '11 at 10:53
@space_cadet: I already gave you one reference on bounds on Lorentz violation in response to your question about graphene. One of the Fermi observations Lubos refers to is nature.com/nature/journal/v462/n7271/edsumm/e091119-06.html, although I'm not sure why anyone would have expected an effect there given the already strong bounds.... –  Matt Reece Jan 23 '11 at 15:31
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It's worth making an analogy here. LQG does not impose an fixed lattice structure. The only discreteness existing is that area and volume operators have a discrete spectrum, specifically meaning that there is a "smallest" area and volume possible before zero. This should be compared to the familiar spin algebra --- there is rotational invariance, but the states are not rotationally invariant (individually)! The situation is almost exactly analogous.

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Welcome to Physics.SE @genneth. Good point +1 –  user346 Jan 24 '11 at 12:34
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Loop Quantum Gravity is an attempt to quantise four dimensional general relativity directly by canonical quantisation methods. If it succeeded it would have to be able to recover the original curved spacetime and gravitational dynamics as a classical limit, but in its current form it does not quite do that. In this context it does not even make sense to discuss whether STR is lost due to discreteness. (STR is taken to mean Special Theory of Relativity. More precisely in this context it means that we recover Lorentz symmetry for local reference frames) Some physicists such as Carlo Rovelli have not given up trying to fix the problems of LQG.

However, the mathematics of knot theory and spin networks that arises naturally in the LQG approach is itself very interesting and this may be telling us something about how a more successful theory of quantum gravity should work. There is too much ground to cover in discussing such possibilities so I'll mention just one feature of spin networks that gives a clue about how a discrete spacetime may not be in contradiction with STR.

As a preliminary point, canonical quantisation methods treat time and space differently which is why the question asks whether a preferred reference frame arises. The best answer to this is that canonical quantisation can be used in relativistic quantum field theory (for example in QED). It works and is equivalent to Lagrangian path integral methods that explicitly preserve Lorentz invariance, so no preferred reference frame arises in this case. However, since the symmetry is not explicitly preserved in canonical methods it has to be demonstrated that the final solution preserves it. As I've said already, this is the case in QFT but in LQG we don't get a final solution where this can even be tested.

However, The interesting thing to note is that you can use spin networks in three dimensional space to define a theory of 3D quantum gravity which is discrete but yet preserves diffeomorphism invariance (and therefore also local Lorentz invariance) due to algebraic relations that allow the spin networks to be modified without affecting the overall path integral sum. See e.g. arxiv:hep-th/9202074 This result has inspired all the work on spin foams and group field theory which attempts to get a similar result for 4D quantum gravity. So far a fully working solution has not been found, but the message is that discrete spacetime does not necessarily have to violate Lorentz invariance and one day we may learn how to see that for a real theory of quantum gravity.

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Dear Phil, I agree that you don't even get smooth space, so you don't have to ask whether it's Lorentz-symmetric. However, the violation of the vital local (Poincare) symmetries is a major part of the reason why you don't get the smooth space, and even if you got it, you would still face the problems with the symmetries. –  Luboš Motl Jan 23 '11 at 9:09
One more comment, about "canonical methods". The problem is with physics, not with methods. If you have Lorentz-invariant theories, they're Lorentz-invariant even in the canonical language. The canonical language makes the Lorentz symmetry harder to prove, but if the symmetry is there, it's there independently of the "methods". I think that this whole focus on "methods" is just fundamentally physically flawed. Physics is about physics, the phenomena and their properties, not about the "formalism". –  Luboš Motl Jan 23 '11 at 9:10
Concerning 3D gravity, is has no physical bulk degrees of freedom, so one may argue that it's the "same theory" as another theory that has no physical bulk degrees of freedom. However, when you study the Chern-Simons theory non-perturbatively, it is actually different from 3D gravity; I may give you a reference to a paper by Witten if you want and I find the time. So even the 3D gravity which has no nontrivial dynamics fails to work. But of course, in the absence of the bulk degrees of freedom, one can't prove the Lorentz violation by the "viscosity" method above. –  Luboš Motl Jan 23 '11 at 9:13
@Lubos, I don't think we are in great disagreement here but it is interesting that you distinguish between the formalism/methods and the physics. It is a subtle metaphysical question whether a formalism that fails to produce a real model actually made contact with some incorrect physics, or whether it was just failed to make contact with some correct physics. –  Philip Gibbs Jan 23 '11 at 9:30
@Lubos, I agree that the 3D example is too unphysical to be a strong indication that a similar system could work for realistic quantum gravity. People have tried for a long time and failed. However, with Witten working on the relationship between Khovanov homology and quantum theories there is always hope that something new is round the corner. –  Philip Gibbs Jan 23 '11 at 9:41
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@sb1 you've opened up a can of worms here. This is a good question but it is subject to the standard misunderstandings about the nature of LQG and quantum geometry.

There is no problem with the question of Lorentz Invariance (LI) in a discrete spacetime. If your spacetime becomes discrete then your notion of LI must change accordingly. Keep in mind that LI is nothing more than a statement that physics should be causal ref1. For a continuous $3+1$ manifold this requirement is expressed in terms of invariance of the space-time interval $ds^2 = -dt^2 + dx^2$. The corresponding symmetry group is the continuous Lie group $SO(3,1)$ or $SL(2,\mathbb{C})$.

In a discrete spacetime $ds^2$ will have to be replaced by its discrete generalization $ \hat{d} s^2 = -\hat{d}t^2 + \hat{d}x^2 $ with $\hat{d}$ being the discrete ("quantum" or "q") differential. The invariance group of this interval is $SL(2,\mathbb{Z})$.

There is a more physical route involving considerations of the transformation of punctures of black hole horizons w.r.t an external observer which leads us to see how $SL(2,\mathbb{C})$ must reduce to $SL(2,\mathbb{Z})$ in the event that the numbers of these punctures is small.

Also your intuition regarding the dissipative effects of discreteness, is correct:

But if this violation is tolerated doesn't that imply some amount of viscosity within space-time itself?

It absolutely does and this is a result that we have know from a completely different directions - primarily from AdS/CFT and the fluid-gravity correspondence tells us that there is a lower bound for the viscosity to entropy ratio of horizons - those of black holes or those experienced by accelerated observers in an otherwise flat spacetime. Eling - Hydrodynamics of spacetime and vacuum viscosity, Son and Starinets - Viscosity, Black Holes, and Quantum Field Theory. This leads to the question of information loss. Having a dissipative horizon seems to suggest that information is lost in quantum gravity. However, this is only as seen by a "local" observer - who only has access to a portion of the spacetime. For a universal observer who has access to all the regions of the spacetime, this problem will not occur.

The debate around this topic is reminiscent of that surrounding Einstein's introduction of the notion that a Lorentzian manifold, rather than a Galilean one, was the right tapestry for events in spacetime. Then people struggled mightily to adapt all the physical inconsistencies of Newtonian's theory without having to give up the cherished Galilean property of absolute space and time. Now we see a similar reluctance to abandon the safe confines of continuous manifolds for a more general framework which can also describe discrete geometries. If one is willing to take this intellectual leap then there is no problem in becoming comfortable with a notion of LI which is adapted to the discrete setting.

I expect a strong response from the usual suspect(s) ;)

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"Keep in mind that LI is nothing more than a statement that physics should be causal" -> uh, what? Causality, locality and LI are connected but are by no means the same thing. I hope you'll clarify this somehow. –  Marek Jan 23 '11 at 11:08
The viscosity effects you're talking about involve particles interacting with a thermal background. The original question is asking about viscosity in empty space. –  Matt Reece Jan 23 '11 at 15:39
Usual suspects are tired of arguing. Some leaders of LQG go to say that discreteness and the resulting violation of LI generates falsifiable predictions for the theory. Are they mistaken? –  user566 Jan 23 '11 at 16:19
@Moshe in my answer I have tried to explain why discreteness and LI are not incompatible if one adapts the notion of LI to the discrete case. If you insist on applying LI, as it is, to the discrete case, then of course, it will be "violated". As for "some leaders" its hard to reply without any names or references. I'm surprised that LQG gets so much criticism for generating falsifiable predictions. Isn't that a plus point for any physical theory? In contrast what falsifiable (or otherwise) predictions has string theory generated? –  user346 Jan 23 '11 at 16:31
I'm sympathetic to the view that LQG can be LI, and it is a good project to prove that it is. But, then you lose the ability to gloat about Lorentz violations as your Popperian trump card. Make your choice. –  user566 Jan 23 '11 at 16:47
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As Lubos points out, LQG requires the imposition of a vast amount of data or degrees of freedom on spacetime. The amount of data for separable states is larger than when those states are in entanglements. The holographic principle also indicates how the degrees of freedom in an $N$ dimensional spacetime (say by counting vertices in a tessellation of the spacetime) may be reduced to that on an $N-1$ dimensional boundary or horizon. This reduces the huge combinatorics the LQG work requires (which makes some of their papers nearly unreadable), and further this also yields some interesting results on entanglement entropy and correlations between fields in a space and its boundary.

There is a cut off in the scale of spacetime given by the Planck length. However, this does not mean spacetime is sliced and diced up, but rather indicates some limit on the information we can extract from spacetime. This connects in some ways to entanglement entropy of a black hole, where the elemental unit of a black hole has a Planck mass. This discrete system is then a limit on the information content we may observe about quantum gravity or spacetime, but it does not necessarily mean that spacetime has some sliced and diced characteristic that leads to actual violations of Lorentz symmetry.

I think the numbers of degrees of freedom may be reduced far further.

The Fermi spacecraft observed widely different wavelengths of photons from a very distant quasar. These photons were emitted in a pulse. The time of arrival was nearly simultaneous, which means the photons traversed a multi billion light year distance with no dispersion. This dispersion is predicted by violations of the equivalence principle and Lorentz symmetry breaking by LQG was not observed. These photons should couple to the quantum foam and exhibit a slight dispersion over this vast distance. I think this puts the kibosh on a lot of LQG.

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"which makes some of their papers nearly unreadable" - I completely agree with you on that point. Some folks, including me, are trying to remedy that situation. Again as for the Fermi measurements I have to see what bounds people obtained from LQG methods. The problem is people keep arguing as if LQG and ST are two independent, free-standing structures. They are not. A complete theory will contain essential elements of both. As for "kibosh" well LQG at least has a prediction. What phenomenologically testable predictions have come out of String theory? –  user346 Jan 23 '11 at 16:17
The correspondence between gravitons and gauge particles or AdS/QCD does make predictions, and the low viscosity fluid behavior of quark-gluon plasmas is in line with so called “soft black holes” predicted. What intrigues me is whether the over counting of degrees of freedom in LQG might be adjusted by string theory. If that could be done the LQG might merge into string theory in some way. I don’t completely dismiss the LQG program, but as it stands there seem to be deep problems. –  Lawrence B. Crowell Jan 23 '11 at 18:40
No, not at all. The degrees of freedom of LQG are less, and not more that in strings (see my detailed answer to this in the "Can Loop Quantum Gravity connect in any way with string theory?". I havent leqrnerd how to link directly from here). There first combinatorial calculations were heavy, but now there are very simple and straighforward way to do the counting, and very elegant as well. See for instance the last paper by Eugenio Bianchi on this arXiv.org/abs/arXiv:1011.5628, where the counting is done elegant statistical techniques from polimer physics. –  Carlo Rovelli Jan 27 '11 at 8:01
And the observations supporting Lorentz invariance at hight energy are not in contradictions with LQG. See again the reply mentioned in the previous comment. –  Carlo Rovelli Jan 27 '11 at 8:03
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There are great answers above. One of the features of STR is the Klein-Gordon equation. I'd like to point out that the Klein-Gordon equation models the vibrations in an elastic solid. This is well known among engineers dealing with the subject. An elegant derivation is:
R A Close, Advances in Applied Cliord Algebras, "Exact Description of Rotational Waves in an Elastic Solid"

The point is that despite the discrete substructure that might contribute to an elastic solid, the final result does satisfy the Klein Gordon equation quite well.

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+1. There are plenty of examples such as this one, but as they "you can't teach a man how something works, if his livelihood depends on not knowing how"! –  user346 Jan 24 '11 at 8:57
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