My problem with the relativity representation on gravity.

WannabeNewton

The quote by Pervect never made any claim about being at absolute rest in space. Of course that is nonsensical. It just said that given a space - time possessing a certain one parameter isometry group, we can utilize the orbits of the group action of said isometry group to define a notion of rest with respect to the orbits. In the above example of defining locally non rotating observers in stationary axisymmetric space - times notice how I said an observer is 'at rest' with respect to the spatial hypersurfaces ##t = \text{const.}## if his 4 - velocity ##u^{a} = \alpha \triangledown^{a} t## (the proportionality scalar field ##\alpha## is just the normalization); this isn't alien from galilean relativity wherein we describe observers in collinear uniform motion with respect to one another. This example was meant to show that we can deduce physical properties of the observer with precisely that definition of locally non rotating e.g. the fact that the observer's angular momentum vanishes, as shown above.

I don't recall me nor Pervect claiming absolute rest. Do you agree with the above however?

Passionflower

Certainly things can be at rest wrt to other things in curved spacetime except of course when a spacetime is non-stationary. I also see no issues with considering things at rest wrt certain coordinate values, for instance a Schwarzschild radius or a shell with a given r-value.

WannabeNewton

We are in agreement then; I'm not sure anymore what this little quibble was about lol. What exactly is the issue in conclusion?