Does the discreteness of spacetime in canonical approaches imply good bye to STR?

Question

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?

Answer 1

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=(V_1,V_2,V_3) $ 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 either $0 $, or $\hbar$, or $2\hbar$ ..., that it, precisely multiples of a fixed quantity.

Now supose you have measured that the $x$ component of the angular momentum was hbar. Rotate slowly 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 $\mathrm{cm}^2$, that is are areas.

Answer 2

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.

Answer 3

@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) ;)

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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"
Link
http://arxiv.org/abs/0908.3232

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.