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Are there any versions of LQG that claim to not violate Lorentz symmetry?

+ 6 like - 0 dislike
173 views

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?

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Gordon
asked Jan 28, 2011 in Theoretical Physics by Gordon (400 points) [ no revision ]
Most voted comments show all comments
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.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user user566
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.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user user346
I agree, space_cadet---they are virtually the same and could be merged. I did not see the other question. Having read the answers though, I see that there is disagreement, as I suspected that there would be and this could devolve into string theory vs lqg--continuous vs discrete.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Gordon
@Gordon - Indeed. It is the devolution aspect of the debate that I'm worried about ;)

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user user346
Not sure what this has to do with string theory. This is a fairly concrete question that could (and should) be answered without any reference to string theory. I for one would be curious to read answers from people in the know.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user user566
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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.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user QGR
@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.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user user346

1 Answer

+ 1 like - 0 dislike

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: http://www.physicsforums.com/showthread.php?t=281951

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: http://arxiv.org/abs/1012.1739

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user genneth
answered Jan 28, 2011 by genneth (565 points) [ no revision ]
@Genneth you said it better than I ever could!

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user user346
Well, so you say. Saying it three times does not make it true.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Gordon
This is the correct answer. LQG is Lorentz invariant. To see details, see the two papers fr.arxiv.org/abs/1012.1739 (recent, shows the Lorentz covanraince of LGG explicitly) and fr.arxiv.org/abs/gr-qc/0205108 (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

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Carlo Rovelli
@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...

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user genneth
@Carlo-- Dittrich, Thiemann have a paper denying lqg discreteness and say that "detailed construction of gauge invariant versions of geometrical operators" must be proven- arxiv.org/abs/0708.1721 .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.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Gordon
I will accept this answer as a response from the lqg people. I do not think many (any) from the string community will agree, including me.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Gordon
Not having read the paper, I can just say from the conclusion that the paper doesn't actually prove that LQG is Lorentz invariant, since the paper appears to appeal to holography, which isn't satisfied by LQG.

This post imported from StackExchange Physics at 2014-05-14 20:46 (UCT), posted by SE-user Dimensio1n0

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