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  Does spacetime really exist in quantum gravity?

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If there are no localized observables in quantum gravity, does spacetime really exist, or might spacetime really be an illusion?

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user Jim
asked Jun 20, 2011 in Phenomenology by Jim (50 points) [ no revision ]
If there are no observables in quantum gravity, then it means that the theory of quantum gravity is useless. It does not imply that space time does not exist or is an illusion. I vote to close this question because the question doesn't make sense.

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user QEntanglement
@QEntanglement: partially true but I feel like the question could make sense. More or less everyone agrees that the classical notion of space-time should break down at Planck scale and get replaced by some non-commutative version. There is lot to be said on this topic.

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user Marek
I think the question should be restated to be more specific because "there is a lot to be said on this topic."

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user QEntanglement
I think questions on "really exist" are border metaphysics, or at least philosophy. I could answer on those lines: "do you really exist or are you a construct of my mind". After all for each of us the only "existence" of the "other" comes through convolutions upon convolutions of electromagnetic pulses that end up in our brain with the illusion "there is something out there" :) . I would also vote to close, unless it changes to "what is the experimental evidence that ...." , because we start with a framework where existence and reality are not in question.

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user anna v
He did not say "no observables", he said "no localised observables"

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user Philip Gibbs
This question should be reformulated as: how do you reconstruct a space-time from the observables admitted in quantum gravity, namely the S-matrix in flat space, or the AdS boundary theory in asymptotically AdS spaces? This is a topic of very active research.

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user Ron Maimon
@Jim: I agree the question could be reformatted for clarity and will probably recieve more attention if it were ask in this way Postulate: If there are no localized observables in quantum gravity, does spacetime really exist, or might spacetime really be an illusion. Scientifically Relevant Question: How do you reconstruct a space-time from the observables admitted in quantum gravity, namely the S-matrix in flat space, or the AdS boundary theory in asymptotically AdS spaces

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user Argus

2 Answers

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Andrew Strominger thinks spacetime is an illusion. It's all a Cosmic Hologram at the future boundary of spacetime at the end of time. It projects out the illusion we see around us.

This post imported from StackExchange Physics at 2014-03-07 14:34 (UCT), posted by SE-user Maya Lila Brahman
answered Feb 2, 2013 by Maya Lila Brahman (10 points) [ no revision ]
+ 1 like - 0 dislike

I think this question is a bit too broad, as it depends on the theory you're talking about.

Lubos Motl has just posted a new article about (a similar to this) idea on TRF.

Well, you do have spacetime in string theory, for example. But as Lubos Motl mentions in the article linked, the metric tensor isn't exactly that well-defined at the stringy scales. To quote the article:

However, quantum gravity doesn't allow you things like that. The metric tensor is only good and well-defined in an effective description of quantum gravity. At shorter distances, it just ceases to be a good observable. Well-defined observables in quantum gravity are different; the gauge fields in the $\mathcal N=4$ Yang-Mills theory involved in the most famous example of the AdS/CFT correspondence are an example. The matrices $X,P,Θ$ in Matrix Theory are another example.

Same with things like Twistor theory, where the real space is the space of twistors.

In LQG, though, there is indeed a vierbin; though.

answered Aug 31, 2013 by dimension10 (1,985 points) [ revision history ]
edited Jan 31, 2015 by dimension10

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