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Time discreteness and lorentz invariance

  1. Aug 1, 2009 #1
    I understand that some do not accept LQG in particular, but any discrete spacial geometry in general, because that would be a violation of lorentz symmetry. It was explained to me as meaning that If a 2D plane were discretized into a grid or lattice, a vector would not have a continuous rotational symmetry group, there would be certain rotations that could not be performed. Therefore, as part of lorentz symmetry, the physics would be lorentz variant, something which we experimentally believe to be always true.

    I was thinking about what it would mean for space to be continuous, but time to be discrete, and wondering if such ideas have been explored, or even how the math on such a vector space would work. In my mind, this would make the physics rotationally invariant still, since time as a single dimension can only undergo one rotation that will always map onto a previous time on an evenly spaced metric. Or is it that this discontinuity in the translational symmetry for time would violate the invariance too?

    Anyhow, I'm very new to the mathematics used to talk about all this stuff, is there a place to get a good crash-course with applications along the way to demonstrate it?
     
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  3. Aug 1, 2009 #2

    marcus

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    Whoever "explained" that to you evidently didn't know what they were talking about. LQG does not represent space as a grid or lattice. Space and spacetime are each modeled by a continuum in LQG---one by a 3D differential manifold and the other by a 4D manifold.

    There is no violation of lorentz symmetry in ordinary 4D LQG as far as we know.

    And there is definitely no violation of rotational symmetry!

    If you would like a good introduction to contemporary LQG there is a presentation on it that Carlo Rovelli (probably the top authority on it) was invited to give at the main Strings conference last year---Strings 2008 at Cern. I will get a link for you.

    Interestingly, LQG predicts that when areas and volumes of physical objects are measured they will come out in quantized levels, like the energy levels of a hydrogen atom. The area and volume operators have discrete spectrum. This is not put in by defining the theory on a lattice, it comes out as an advanced result, because the geometric operators corresponding to measurements are part of a quantum theory. Discrete spectra of operators is a kind of discreteness, and it turns out to be entirely compatible with Lorentz symmetry! Rovelli had a paper about that in 2002 or 2003. It's actually kind of neat how it works out. I will get the link for that in case you want to pursue it further.
    http://arxiv.org/abs/gr-qc/0205108
    Reconcile Planck-scale discreteness and the Lorentz-Fitzgerald contraction

    ================================

    Nonstring quantum gravity is essentially the quantization of space(time) geometry, and if you are new to it the best place to start is with a Scientific American article (the "signallake" link in my signature down at the bottom of this post, in small print. There are several different allied approaches to quantizing the geometry of general relativity and LQG is not the easiest to begin with. By far the easiest is the random building block approach of Jan Ambjorn and Renate Loll. They generate small quantum/random universes in the computer and study them individually and make averages of their properties---the path integral approach. They call their method "causal dynamical triangulations" (CDT) but that is not very descriptive.

    They get a heap of little triangle-like building blocks to self-assemble in the computer into a quantum spacetime that obeys a quantized version of General Rel. (higher dimensional triangles called "simplices" or "simplexes").
    BTW just because they use discrete building blocks as an approximation, in their theory, does not mean they think space or spacetime is "made" of little blocks! :biggrin: People use discrete lattices and blocks all the time and then they let the size go to zero. Like you do in calculus. You take limits of discrete finite approximations, limits as the grid gets finer and finer. Different approaches do this differently, but you get the idea.

    CDT is a good introduction to the whole field of quantum gravity/geometry. I would advise you to read that Scientific American article by Jan Ambjorn and Renate Loll. Only later try some article by Rovelli about LQG.
     
    Last edited: Aug 1, 2009
  4. Aug 1, 2009 #3

    tom.stoer

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    There are too very short arguments against "discretized spacetime violates Lorentz-invariance":
    - the operator algebra respects the Lorentz-algebra
    - discrete eigenvalues do not signal any such violation

    Compare the situation with standard rotational symmetry: the angular momentum has a discrete spectrum and leads to a discrete basis of eigenvectors; nevertheless the rotational symmetry is not violated.
     
  5. Aug 1, 2009 #4
    Marcus posts:
    When I posted similar thoughts in other threads I was promptly attacked for essentially "crazy" posts,and it was in a quantum physics thread(!!) even after quoting four of five well known physicsts, like Brian Greene and Lee Smolin; I know relativity is smooth and I guess continuous, but why would there be such myopic views just because discrete spacetime is not in the standard model?? It seems like arguing "spacetime" cannot be a physical entity or arguing on the other hand that it must be.

    And Tom posted,
    Could you post a few underlying explanatory comments or provide a reference where I might read a bit more....for example how might such eigenvalues signal a violation if one exists??

    I have Penrose THE ROAD TO REALITY and likely he discusses this somewhere...although frankly he loses me at times...but I know enough to recognize that with several chapters devoted to discrete spacetime and his own research and papers on twister theory for many,many eyars, HE sure has not ruled out a discrete spacetime...and in Chapter 33 he discusses "more Radical Perspectives.." related to discretness, even referencing Ahmavaara who proposed replacing the real number system used in conventional physics!
     
  6. Aug 2, 2009 #5

    tom.stoer

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    If the first violation occures it would be called an anomaly. That means that the classical equation

    [Qa,Qb] = ifabcQc

    for a Lie algebra with generators Q receives Quantum corrections like

    [Qa,Qb] = ifabcQc + h Aab

    where A is no the anomalous term.

    I have no idea how the second argument can be turned round. A trivial example would be if the spectrum does not correspond to the spectrum of the angular momentum operator.

    My argument is slightly different. My claim is that the sentence "as the spacetime model of LQG is discrete, LQG violates Lorentz invariance" is simply wrong. I provide too reasons how this can be understood: there is no operator anomaly and therefore the symmetry is respected on the level of the operators; there are discrete eigenvalues - but that is true for the angular momentum as well, and nobody would say that "the hydrogen atom breaks rotational symmetry because the spectrum is descrete".

    The problem in LQG is that you don't see the generators of the symmetry in the calculations after you have fixed the symmetries (I am talking about the canonical approach). You can compare that with QCD or QED: After fixing the gauge, the Gauss law goes away - it has been solved. So you need no longer check if it's respected. The q.m. gauge fixing does not break the gauge symmetry, it only removes the unphysical dgrees of freedom. The same happens in LQG.
     
  7. Aug 3, 2009 #6
    great,simple, example...thanks!!
     
    Last edited: Aug 3, 2009
  8. Jan 5, 2011 #7
    I'm sorry but this analogy with the angular momentum sounds a bit fishy to me. Let us look at some details. First of all, SO(3) rotational invariance is already broken down to SO(2) (rotations around the direction of vector L) as soon as one turns on a non-zero angular momentum, i.e. a preferred direction in space. This breaking of the SO(3) rotational symmetry has nothing to do with quantization.

    One may further inquire if the SO(2) rotational symmetry, which changes the components of the angular momentum contunuosly, is further broken by the quantization of the angular momentum since the eigenvalues of Lz and L2 are discrete. Upon further examination we see that while the SO(2) rotations change Lx and Ly components (these do not have a discrete spectum), the Lz component as well as the square of the magnitude L2, which do have a discrete spectrum, stay invariant under SO(2). Therefore, there is no conflict between SO(2) symmetry remaining a good symmetry and the quantization of the the angular momentum.

    On the other hand, unlike the case of the angular momentum where the Lz and L2 remain invariant under SO(2) rotations, an area is NOT invariant under Lorentz boosts and therefore the analogy is completely misleading.
     
  9. Jan 5, 2011 #8
    Actually, even the SO(2) symmetry does become broken for the higher, e.g. d-orbitals, in the hydrogen atom to some discrete subgroup. You can easily see this if you examine the shapes of the hydrogen wavefunctions.
     
  10. Jan 5, 2011 #9

    marcus

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    Lorentz covariance of loop quantum gravity

    http://arxiv.org/abs/1012.1739
    Lorentz covariance of loop quantum gravity
    Carlo Rovelli, Simone Speziale
    6 pages, 1 figure
    (Submitted on 8 Dec 2010)
    "The kinematics of loop gravity can be given a manifestly Lorentz-covariant formulation: the conventional SU(2)-spin-network Hilbert space can be mapped to a space K of SL(2,C) functions, where Lorentz covariance is manifest. K can be described in terms of a certain subset of the 'projected' spin networks studied by Livine, Alexandrov and Dupuis. It is formed by SL(2,C) functions completely determined by their restriction on SU(2). These are square-integrable in the SU(2) scalar product, but not in the SL(2,C) one. Thus, SU(2)-spin-network states can be represented by Lorentz-covariant SL(2,C) functions, as two-component photons can be described in the Lorentz-covariant Gupta-Bleuler formalism. As shown by Wolfgang Wieland in a related paper, this manifestly Lorentz-covariant formulation can also be directly obtained from canonical quantization. We show that the spinfoam dynamics of loop quantum gravity is locally SL(2,C)-invariant in the bulk, and yields states that are preciseley in K on the boundary. This clarifies how the SL(2,C) spinfoam formalism yields an SU(2) theory on the boundary. These structures define a tidy Lorentz-covariant formalism for loop gravity."

    A few years back it was conjectured by some people that LQG might not be Lorentz invariant---a few tried without success to give a rigorous mathematical proof that the theory deviated from invariance. They gave up. That was back in 2006-2007 as I recall.
     
    Last edited: Jan 5, 2011
  11. Jan 5, 2011 #10
    Re: Lorentz covariance of loop quantum gravity

    In this case is the area, which is the dual variable to the gauge field, still discrete even though the monodromies of the gauge field live in a non-compact space? My understanding was that the quantization of the area simply follows from the compactness of SU(2).
     
  12. Jan 6, 2011 #11

    atyy

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    Although LQG may be Lorentz covariant, there is no indication that this is Lorentz covariance of the classical spacetime. All present indications are that the classical spacetime is Regge, which should fail at some level the most common tests of "Lorentz covariance". Actually, that is too generous, since there is no indication yet of an emergent classical spacetime (DT started from Regge and showed that was insufficient), and the theory is divergent, so it's not even a theory.
     
  13. Jan 6, 2011 #12

    tom.stoer

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    Of course each single state |lm> breaks the invariance, but the set of all states |lm> for fixed m generate a faceful rep. of SO(3); that also "cures" singling out z. In that sense the hydrogen atom IS SO(3) symmetric.
     
  14. Jan 6, 2011 #13
    Ok, I'm not sure I'm following here. Fixing m implies that you are fixing the Lz component in which case you are not doing a general SO(3) rotation, but instead an SO(2) rotation that preserves Lz. Maybe you meant to take a superposition state by summing up over all m for some fixed l ?
     
  15. Jan 7, 2011 #14

    tom.stoer

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    Fixing m = Lz is arbitrary. You could fix m = Le according an arbitrary axis e.

    If you have |nlm> = |100> the state is explicitly SO(3) symmetric. If you now use a photon to create the state |211> you have chosen a specific photon with specific angular momentum which breaks SO(3) invariace. But if you only can say that the atom is in |21.> the symmetry is still preserved as you do not know about the photon. That's why you have to consider all three |21.> states and these three states together form the full 3-dim. rep. of SO(3). In that sense SO(3) is preserved.

    But that's not the point. The point is that every 2l+1 dim. rep. of SO(3) is spanned by a discrete set of eigenvalues, that there are discrete values for l and m, but nevertheless the Hamiltonian is explicitly SO(3) symmetric. If you look at the hydrogen atom you get discrete states, but still you have full rotational invariance. If YOU now pick a specific state then YOU break the SO(3) symmetry but the math of the problem doesn't break the symmetry only because states and spectra are discrete. That was my point.

    (this argument is not always correct, e.g. the discrete p-states for a particle in a box are due to the broken translational invariance; I only want to make clear that not every discrete structure, spectrum or set of states indicates a broken symmetry)
     
    Last edited: Jan 7, 2011
  16. Jan 7, 2011 #15
    Ok, I'm going to say something trivial that you already know but I still want to make my point. Let's say you have a lagrangian that's invariant under a particular symmetry. Then you solve the equations of motion and find stationary points where the symmetry is broken and some points where the symmetry is unbroken. I say that the statement of whether the symmetry is broken or not depends on the specific solution (suppose there are dynamical transitions,e.g. tunneling, so the choice of the solution is not adhoc) although the lagrangian is still symmetric. In other words, while the theory is still symmetric, the corresponding symmetry may or may not be spontaneously broken depending on the solution.

    Suppose I have a Lorentz covariant lagrangian and after solving the EOMs I obtain a solution where some vector field gets a non-zero vev. This would imply that the Lorentz symmetry is spontaneously broken in the vacuum, which is probably not so good.
     
    Last edited: Jan 7, 2011
  17. Jan 7, 2011 #16

    tom.stoer

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    AOne specific solution sillof course break the symmetry. Suppose you have a Lagrangian of a free particle in R3. Every single solution x(t) breaks rotational and translational invariance.

    But again, this is not my point My point is that introducing discrete structures like |lm> for SO(3) do not necessarily break continuous symmetries.
     
  18. Jan 8, 2011 #17
    I agree with that statement but only because in quantum mechanics systems with a finite number of degrees of freedom always have a symmetric ground state. So, when the system is in the ground state, the symmetry is unbroken. However, in the above context the analogy with area quantization vs Lorentz invariance looks completely misplaced.
     
  19. Jan 9, 2011 #18

    tom.stoer

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    It has nothing to do with the ground state and with a finite number of degrees of freedom. Spin networks need not contain infinitly many vertices (for finite volume there will be finitely many vertices), nevertheless local Lorentz invariance is guarantueed.

    The analogy is quite clear: you have a certain basis of the Hilbert space; the set of all states can be labelled according to representations of a certain symmetries. For the hydrogen atom it's SO(3) (or SO(4)~SU(2)*SU(2) if you take the enlarged symmetry of 1/r into account). For spin networks it's SU(2) acting on each individual spin. In both cases the Hilbert space is labelled according to representations of the symmetry.

    Then there are certain operators in the Hilbert space with discrete spectrum. In the hydrogen atom it's the angular momentum operator, in LQG it's the area operator.
    In both cases the relevant operators have the correct operator algebra.

    There is no reason why a discrete spectrum and a countable basis of the Hilbert space must break a certain symmetry, neither in the hydrogen case, nor in the LQG case.

    Then there is another misconception: in LQG you have LOCAL Lorentz invariance which means that Lorentz invariance is a gauge symmetry. This symmetry has nothing to do with a certain state (a solution). For a solution you would always talk about GLOBAL Lorentz invariance which we do not need in the context of GR; of course there are many solutions that DO break global Lorentz invariance (an FRW universe, for example)
     
  20. Jan 9, 2011 #19

    atyy

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    If the classical spacetime is Regge-like as may be in the EPRL/FK models, how can the path of a light ray be frequency independent? Won't the edges in space put some sort of wavelength dependent boundary condition on the light wave?
     
  21. Jan 9, 2011 #20

    tom.stoer

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    I agree that it may be a problem in Regge-like / CDT discretizations if there is no appropriate "continuum limit". But that is not the point; if you simply start with a spin network Hilbert space on which all operators either act in a covariant way or are reduced to the identity (via "gauge fixing"), then no symmetries are violated.

    Of course there are "good" and "bad" discretizations, of course CDT breaks Lorentz invariance explicitly; but SU(2) spin networks don't - just in the same way as the spherical harmonics do not break rotational invariance in the hydrogen atom
     
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