75.2k views
1 vote
I have seen different explanations to understand why there are no local gauge invariant observables in gravity.

Some of them explain that diffeomorphisms are a gauge symmetry of the theory and thus any observable evaluated at a spacetime point will be gauge dependent and therefore not an observable. This line of reasoning then argues for Wilson loops, or asymptotic charges as good (non-local) observables in gravity. This explanation, in my opinion, is purely classical, it doesn't rely on the uncertainty principle, or commutation relations, etc.

However, other explanations give the argument that any device trying to measure a local observable will have a finite size, and therefore a finite accuracy. If the device wants to probe local physics, it should be smaller. However, the uncertainty principle forces the device to collapse into a black hole before allowing the experiment to give us local information. Alternatively, it is also explained that the commutator of two (quantum) operators has to vanish for spacelike separations but that a metric that is dynamical and fluctuates will mess with the causality of the theory, therefore making the operators not observable. These arguments seem absolutely quantum mechanical.

I have a couple of very similar questions:

Is the statement no local gauge invariant observables in gravity true in classical GR, in quantum gravity or both?

If it is true in both, why do people treat the statement no local gauge invariant observables in quantum gravity as something special?

Do statements about observables in classical and quantum gravity mean different things?The arguments given to explain each one are pretty different and seem to involve different physics. The first one relies heavily on diffeomorhism invariance while the second one relies on holographic-flavoured arguments about how much information you can concentrate in a given volume before you form a black hole.

1 Answer

2 votes

Final answer:

In classical General Relativity (GR), there are no local gauge invariant observables due to diffeomorphism gauge symmetry. In quantum gravity, the uncertainty principle and the commutation relations make local observables impossible to measure.

Step-by-step explanation:

In classical General Relativity (GR), there are no local gauge invariant observables because diffeomorphisms are a gauge symmetry of the theory. This means that any observable evaluated at a spacetime point will be gauge dependent and therefore not a true observable.

However, in quantum gravity, the situation is more complex. Some argue that the uncertainty principle and the commutation relations make local observables impossible to measure due to the finite size of measurement devices and the effects of a fluctuating metric. Both arguments involve different physics and are not mutually exclusive.

User Veikko
by
7.7k points