I'm having great conceptual difficulties with gauge transformations in general relativity. Let me explain my problem by making a comparison to gauge transformations in electrodynamics:
In electrodynamics one has equations of motion that dictate the evolution of the vector potential $A^\mu$. If these equations are satisfied then transforming the field as:
$$A^\mu \rightarrow A^\mu + \partial^\mu \Lambda \tag{1}$$
Will leave the equations still statisfied.
Now let's say we have a theorist solving an Electrodynamics problem and an experimentalist testing his solution, they agree on a certain coordinate system even before the theorist sits down to solve the equations of motion:
The theorist chooses a certain gauge to solve his problem. He is free to define $\Lambda$ anyway he likes, he doesn't even have to tell the experimentalist which gauge he has used. He hands his solution over and the experiment agrees with his prediction.
Now look at linearised gravity:
In linearised gravity we consider a metric perturbation $h_{\mu\nu}$:
$$g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu} \tag{2}$$
Then we can perform an infinitesimal coordinate transformation:
$$x^\mu \rightarrow x^\mu + \xi^\mu\tag{3}$$
Which results in changes to quantities in the theory such as:
$$h_{\mu\nu} \rightarrow h_{\mu\nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu \tag{4}$$
And would clearly also change the expressions for any vector fields and derivative operators in the equations.
Now we have the theorist predict the behaviour of a system of particles in a perturbed spacetime.
If all physical observables were unaltered by only using (4), then I would consider linearised gravity a true gauge theory because the theorist would not have to warn the experimentalist about whatever gauge he has chosen to use.
It turns out that in linearised gravity the Ricci tensor $R_{ij}$ is indeed unaltered if we use (4) only.
However, clearly the spacetime interval:
$$ds^2 = g_{ij}dx^i dx^j \rightarrow \left( g_{ij} + \partial_i \xi_j + \partial_j \xi_i \right)dx^i dx^j \tag{5}$$
Will change, and so will the Christoffel symbols and thereby the covariant derivative , geodesic equation and particle trajectories.
For the spacetime interval and the Christoffel symbols to be left unchanged the theorist cannot use only (4) and will need to adjust all his derivatives and vector fields in accordance with the complete coordinate transformation (3).
This is fine, the theory is generally covariant. However it seems to me that this is very different from what we just considered in the case of Electrodynamics, for if here the theorist performs a change (3), he had better warn the experimentalist that his solution is no longer expressed in the coordinate system they had initially agreed upon.
This is because he has performed a coordinate change and not, it would seem to me, a gauge transformation.
Do you, the reader, agree that in the Lorentz and transverse-traceless gauge we are no longer in Minkowski coordinates and that this fact is very much relevant if we are ever doing anything other than solving the linearised Einstein field equations in vacuum:
$$R_{ij} = 0 \tag{6}$$
The reason being the above mentioned invariance of $R_{ij}$ under the mere redefinition of the metric perturbation (4).
If not, then clearly I am missing the point and I would love to read you insights.
Much appreciation for you time!
