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LQG formulations have a minimum length/area. Since say, a Planck area can always be boosted, any minimum area in space can be shrunk. Do LQG proponents worry about local Lorentz invariance violation, and if not, why not? In LQG, does considering length to be a quantum operator really get rid of the boost problem?

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My impression as an outsider is that the majority of that community believes (and always believed, despite some isolated claims to the contrary) that LQG ought to be Lorentz invariant, partially because they recognize what a disaster it would be if it's not. But, I'd be interested to hear from the insiders, especially about current ideas how this feat can be achieved. – user566 Jan 28 '11 at 2:50
This question seems identical to a previous one. Perhaps @Gordon is asking for something different from what is being asked in that question. But if not, then I'd suggest this question be merged with the previous one. – user346 Jan 28 '11 at 4:58
Moshe: Lorentz violation is surely a terrible catastrophe. However, it's equally obvious that LQG - at least all LQG models and descriptions I've ever seen - maximally violate Lorentz symmetry at the Planck scale. Well, try to describe microstates of space as an irregular lattice with a Planckian density of vertices and edges (or spin form - connected pieces of areas in Minkowski spacetime), and now try to Lorentz transform it. What will you get? What's your guess? :-) This violation is such an obvious thing that I can't imagine a physicist questioning it. – Luboš Motl Jan 28 '11 at 6:37
If there's no Lorentz violation, we should be able to write down an explicit boost generator operator, and show it is a symmetry of the theory. – QGR Jan 28 '11 at 9:30
@QGR I strongly disagree with the practice of editing questions or answers to add or remove content which was not originally present, unless the OP expresses such a wish. This violates a principle of neutrality than any editor should respect. Consequently I am rolling back your edit, consisting of the second paragraph: "If LQG turns out to be ..." This also happens to be identical to a comment you left earlier. – user346 Jan 29 '11 at 8:07

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This has been asked and answered before: see Does the discreteness of spacetime in canonical approaches imply good bye to STR?

Also, this question has popped up many times on other sites such as physicsforums:

The answer is roughly that LQG does not in fact violate Lorentz invariance. The discretisation of area and volume operators does not imply a broken symmetry, any more than discretisation of angular momentum states imply breaking of rotational symmetry --- symmetries in quantum theories are equations of the operator algebra, not of the states!

See also:

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@Genneth you said it better than I ever could! – user346 Jan 28 '11 at 12:45
Well, so you say. Saying it three times does not make it true. – Gordon Jan 28 '11 at 13:50
This is the correct answer. LQG is Lorentz invariant. To see details, see the two papers (recent, shows the Lorentz covanraince of LGG explicitly) and (explains in detail why the argument about shrinking of the minimal area is wrong. that is, gives the details behind the point made by gennetg.) carlo rovelli – Carlo Rovelli Jan 28 '11 at 14:32
@Gordon Wilson: if you look at the abstract of the 2nd paper that Carlo linked to, you will see the argument developed a little more: all observers see the same spectrum, and it's the probability distribution which gets Lorentz boosted. I'm not sure how to state things any clearer than that... – genneth Jan 28 '11 at 14:37
@Carlo-- Dittrich, Thiemann have a paper denying lqg discreteness and say that "detailed construction of gauge invariant versions of geometrical operators" must be proven- .This paper is from 2007. I will try to read your recent paper, but I am no expert in LQG. Thanks for your response and links. – Gordon Jan 28 '11 at 17:35

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