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This question is kind of inspired in this one:

Diff(M) as a gauge group and local observables in theories with gravity

The conundrum i'm trying to understand is how is derived the (quite) extraordinary statement that in GR there are no local observables. I just want to stress that this is indeed an extraordinarily counter-intuitive assertion (with extraordinarily dramatic consequences for any compatible theory of quantum reality), and it deserves an extraordinarily robust explanation.

Among the statements i see are mostly involved in this argument are:

  • It is usually argued that Diff(M) is a gauge transformation. This point i don't have any sort of issue; the atlas of reference frames we use to describe space-time are indeed a human convention and physical laws should not depend on such conventions

  • A physical observable should be invariant under any gauge transformation. whaaa? i mean, this is clearly preposterous non-sense; actually i think this is more dangerous than simple non-sense, is just circular self-justification. This is actually saying that observables should be scalars?

    I can provide you a proof that this is non-sense by saying that the momentum of a particle is an observable in my reference-frame, and in other Diff(M) gauge (that is, another reference frame) i will see a different momentum of the same observable. All eigenvalues and eigenvectors transform according to the vector representation of the Poincare group. You are now of course free of dismiss such 'proof' as non-sense because my assumption that the momentum of a local particle is an observable is non-sense, but then, how are you sustaining that part of the argument, without actually saying again that Diff(M) is a gauge transformation? how do you avoid the circularity in this argument?

    U(1) gauge is a bad example because this is an internal symmetry; Diff(M) are space-time symmetries and i don't think that what is true in there (A eletromagnetic vector potential being unobservable, but B and E being observable, hence observables are invariant under U(1) )

  • It is usually argued that observables in GR formally exists in the asymptotic boundary of the space-time. Is there an argument for this that is unrelated/independent to the previous point?

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I didn't understand how this question differs from the previous one, so I voted it's an exact duplicate. –  Luboš Motl Feb 23 '11 at 7:25
    
Let me just answer the 3 questions here. The first one is not a question; it says that lurscher agrees that Diff is a gauge group. The second question: diffeomorphisms that don't become trivial at infinity are not included in the gauge group, so the global Lorentz symmetry generators don't have to annihilate physical states. But if they had to, the gauge-invariant quantities would have to be not only scalars but scalars unrelated to any point in spacetime - universal scalar integrals over spacetime! –  Luboš Motl Feb 23 '11 at 7:55
    
But in reality, as I have mentioned and as is relevant for the third question, diffeomorphisms that transform points even at infinity by a finite or infinite amount don't have to keep the physical states invariant. That means that observables may be tensors in the Minkowski background. However, they can't be fields at points in the middle of the bulk - because those points are given by coordinates which are not Diff-invariant. However, because the Diff as a gauge group turns off at the asymptotic boundary, fields over there are gauge-invariant and physical. –  Luboš Motl Feb 23 '11 at 7:57
    
@Lubos, 'diffeomorphisms that don't become trivial at infinity are not included in the gauge group' so you are not including, among many things, trivial Lorentz transformations. You need lots of covers of an atlas to define a global diffeomorphism, but if you start with a local lorentz boost locally, you need to add covers far away (i.e: far galaxies) that blue-shift/red-shift appropiately relative to your 'trivial map' at infinity. Why those transformations don't belong to Diff(M)? these are also gauge transformations! –  lurscher Feb 27 '11 at 14:17
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2 Answers 2

Instead of re-inventing the wheel, here is a bunch of references:

Here is something I wrote back when I was blogging, the discussion was kind of fun.

There is also a pretty good explanation in this paper by Giddings- Hartle-Marolf on how this statement doesn't contradict all kind of semi-intuitive ideas people have (and started commenting on here already) about local observables.

Finally, this paper by Arkani-Hamed and collaborators has very nice introduction explaining why dynamical gravity, and not just Diff invariance, is what prevents the existence of local observables.

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Note the above papers are referring to general considerations about quantum gravity and the initial question was it seemed to me more classical. Instead I think the OP's problem lies more with the formulation of GR as a gauge theory where there is a bit of a semantic game that is confusing if you haven't seen it before. Note that what it means to be an 'observable' has a sharp meaning, and only properly has even a chance of making sense if the asymptotics are fixed (eg the SMatrix). –  Columbia Feb 23 '11 at 3:42
    
I just read the blog posting you wrote (which was nice and intuitive). However, if the comment section doesn't confuse people, nothing will! The irony is that there are imo multiple correct answers that are apparently mutually contradictory. I think ultimately the problem is that there are different mathematical definitions floating around for all these loaded words. –  Columbia Feb 23 '11 at 4:32
    
I only said it was fun for me, your mileage may vary...But, yes, it is subtle business to pin down the correct statement in this case, though not impossible. –  user566 Feb 23 '11 at 4:40
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Who says that there are no local observables? There are two local degrees of freedom in General Relativity, related to the two polarizaiton modes of local gravitational waves. And they are locally observable--hence the LIGO experiment.

Now, general relativity in 2+1 dimensions turns out to be topological--all the degrees of freedom of the theory are determined by the boundary conditions and the matter equation of state. But that's certainly not true of the 3+1 dimensional world that we inhabit.

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yes, but in the quantum version of the theory they seem to dissapear, and i just need to know why –  lurscher Feb 23 '11 at 3:00
    
@lurscher: where does anyone make that claim? If they disappear, there are no gravitons, amongst other things. –  Jerry Schirmer Feb 23 '11 at 3:02
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Jerry, you are very confused about the meaning of the term "observable". It is not "anything that is observable". An observable is - classically - a quantity that parameterizes the configuration space. Quantum mechanically, it becomes a Hermitean operator that evolves in time via Heisenberg equations, if we adopt the Heisenberg picture. Polarizations of the graviton are not observables; they're states. There are no local gauge-invariant observables in GR. Moreover, even the wave is not really "local" in the technical sense - you have to be distorting the meaning of "local" as well. –  Luboš Motl Feb 23 '11 at 7:27
    
@Lubos: My definition of local is a field that satisfies a first or second order PDE that depends only on terms inside of its light cone. Linearize Einstein's equation, and such a field shows up clear as rain. –  Jerry Schirmer Feb 23 '11 at 13:34
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