Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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

share|improve this question
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. –  QEntanglement Jun 20 '11 at 8:53
@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. –  Marek Jun 20 '11 at 9:14
I think the question should be restated to be more specific because "there is a lot to be said on this topic." –  QEntanglement Jun 20 '11 at 9:54
He did not say "no observables", he said "no localised observables" –  Philip Gibbs Jun 21 '11 at 6:36
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. –  Ron Maimon Sep 4 '11 at 0:07
show 2 more comments

2 Answers

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.

share|improve this answer
add comment

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

Strangely, Lubos Motl has just posted a new article about (a similar to tthis) idea..

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 qyuote 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.

Sam#e with things like Twi#stor theory (I think, correct me if I'm wrong.); the real space is the space of twistors.

In LQG , though, there is indeed a vierbin; though, ...... (But read Lubos Motl's objections to Loop Quantum Gravity)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.