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  Why does gravity forbid local observables?

+ 2 like - 0 dislike

I heard in a conference that gravity forbids to construct local gauge invariants like $\mathrm{Tr}\left\{−\frac{1}{4} F_{μν}^{a}F_{a}^{μν}\right\}$ and only allows non-local gauge invariant quantities like Wilson Loops: $\mathrm{Tr}\mathcal{P}\exp\left[\oint_{\gamma} A^{a}dx_{a}\right]$. Could someone explain me where does it come from?

This post imported from StackExchange Physics at 2014-05-01 12:18 (UCT), posted by SE-user toot
asked Sep 11, 2011 in Theoretical Physics by toot (445 points) [ no revision ]

1 Answer

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General coordinate invariance lets you arbitrarily set the values of the metric and it's first derivative at any one point--http://en.wikipedia.org/wiki/Fermi_coordinates . Since you can do this, constructions like the maxwell term you describe above will be necessarily coordinate-dependent, and thus, not local observables.

This post imported from StackExchange Physics at 2014-05-01 12:18 (UCT), posted by SE-user Jerry Schirmer
answered Sep 11, 2011 by Jerry Schirmer (130 points) [ no revision ]
I am sorry but it still isn't clear for me, if I build a theory containing general coordinate invariance and U(1) local gauge symmetry, so a "world" containing gravity and electromagnetism, and I minimally couple the gauge boson to gravity, ie his kinetic term is $-\frac14 F^{\mu\nu}F_{\mu\nu}$ with $F_{\mu\nu} = D_{\mu}A_{\nu} - D_{\nu}A_{\mu}$ where $D$ is the covariant derivative associated with coordinate-diffeomorphisms, so this term is a scalar regarding both gauge and coordinate transformations; thus a good candidate for a local observable.

This post imported from StackExchange Physics at 2014-05-01 12:18 (UCT), posted by SE-user toot
ok my bad, I just read Lubos's answer (physics.stackexchange.com/questions/4359/…) and it now clear for me. Thanks Jerry (and Lubos). ;)

This post imported from StackExchange Physics at 2014-05-01 12:18 (UCT), posted by SE-user toot

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