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Observables have to be gauge invariant, but clearly the spacetime interval:

$$ds = g_{ij}dx^idx^j = (\eta_{ij}+h_{ij})dx^idx^j\tag{1}$$

is not invariant under a transformation:

$$ h_{ij} \rightarrow h_{ij}+\partial_i\xi_j+\partial_j\xi_i\tag{2}$$

Since:

$$g_{ij}dx^idx^j \rightarrow (g_{ij}+\partial_i\xi_j+\partial_j\xi_i)dx^idx^j\neq g_{ij}dx^idx^j\tag{3}$$

So how can (2) be a valid gauge transformation?

Quanta
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  • $ds$ and the proper time $\int ds$ is an obserable, assuming one can read a clock hahah – Quanta Jun 07 '23 at 19:45
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    Who claims is this a "valid gauge transformation" and what do they mean by that? See also https://physics.stackexchange.com/q/46324/50583 and its linked questions. – ACuriousMind Jun 07 '23 at 19:58
  • it is valid gauge transformation because it leaves the linearized EOM: $R_{ij}=0$ invariant.gauge transformations, like the third answer in you linked post states: "preserve the physics we care about". But it doesn't do that for all physics like the spacetime interval. unless we change the coordinate differential but then were just performing a cordinate transformation. please read my response to the answer to my above question below. how can we say the the transverse-traceless gauge is in minkowski coordinates?? – Quanta Jun 07 '23 at 20:21

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The transformation you're talking about is generated by the infinitesimal coordinate transformation $$ x^{\mu} \rightarrow \hat{x}^{\mu} = x^{\mu} + \xi^{\mu}(x) $$ where we assume $\xi^{\mu}$ is small (i.e., working up to first order in $\xi$). Applying the appropriate transformations on the coordinate basis one forms $dx^{\mu}$ shows the interval $ds^2$ is indeed invariant.

Eletie
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  • so if we write down the equation of motion for Linearized gravity in minkowski coordinates and perform a gauge transformation, we are now in a new coordinate system? yet the gravitational waves in the transverse-traceless gauge are always advertised as being the solutions in minkowski coordinates... how can this be? – Quanta Jun 07 '23 at 20:13
  • I was playing a bit fast and loose here and treating $ds^2$ as a scalar, but that is what it is: the distance between two points is a scalar. If you work in different coordinates then you must make sure you're $\xi$'s are adapted to those coordinates too. (Think about Killing vectors: we can work in any set of coordinates, but clearly for Minkowski space in Cartesian coordinates they take a simpler form). But I did not say that the gauge transformations are coordinate transformation! Only that scalars are necessarily gauge invariant. – Eletie Jun 07 '23 at 20:24
  • the $dx$'s are only dependent on the coordinates so if you transform them you are necessarily performing a coordinate transformation – Quanta Jun 07 '23 at 20:56
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    This is the issue with talking about gauge transformations in GR, which isn't strictly a gauge theory. But when people talk about fixing a gauge in GR they really mean a coordinate choice. The point is that the transformation you wrote is exactly the infinitesimal diffeomorphism I wrote, and of course $ds$ is invariant under those. – Eletie Jun 08 '23 at 07:22
  • @Eletie Gauge choice in GR are not (always) the same thing as a coordinate choice. Even in a completely coordinate free formulation of perturbation theory, you would still have a gauge ambiguity related the identification of the physical manifold to a background spacetime. Calling it a coordinate choice leads to confusion over how a scalar can be both invariant under coordinate transformations, but not gauge invariant. Linearized GR around a background is a gauge theory in most senses of the term. – TimRias Jun 12 '23 at 07:30