How are frames of reference releated to each other in general realitivity?

Question

Suppose their are 3 objects(O1, O2 and M) in a region of space with large variations in the momentum stress tensor.

They all have 3 different trajectories through a very inhomogeneous spacetime manifold, given by O1($\lambda$), O2($\lambda$), M($\lambda$).

I understand that the manifold is defined by a set of definite points, so that in an abstract point of view, O1, O2 and M define objective trajectories across this set of definite points. But, if we want to describe them, we need to define an atlas and chose a chart (which specifies an origin). This will then specify a point in the range in the chart (4 numbers) to each object.

It makes sense to me that the chart you would choose will be based on the measurement apparatus of some observer. So O1 sets up an apparatus with 3 orthogonal space-like vectors (say rulers that are placed in what O1 perceives as orthogonal) and a clock, which define the spacial coordinates and a time coordinate. Measurements are then a set of 4 numbers made a different times: $$ (x_1,y_1,z_1,t_1)$$ $$ (x_2,y_2,z_2,t_2)$$ $$ (x_3,y_3,z_3,t_3)$$ $$ \dots $$ $$ (x_n,y_n,z_n,t_n)$$

My first question is: Are these set of measurement all in the same chart? I'm uncertain about this because at each point in time, O1 may have moved to a different point (objective definite point) in the space-time manifold and thus changed the origin of the chart.

I'm guessing that their is a way to define a chart such that all these measurements are in the same one, but I don't know how to do that.(?) Given that their is, O1, will have set of measurements for O2 and M as a set of points in the domain of O1's chart. O1 also has a set of measurements of various points(say the end points of the orthogonal rulers) of O2's apparatus(the same type of apparatus O1 is using). O1 can then construct 3 vectors (in his chart) which have some relation to the ones that define O2's chart.

My Second question how does does O1, predict the measurements O2 will make of M, given that they both use the same type of apparatus and clock. O1 also knows the space-time structure and can write down a metric and connection in O1's chart.

Edit To be more specific, O1 has two 4-vector that describes O2 and M's path and the metric in it's coordinate system (the domain of the chart that describes O1's reference frame): $$ \vec{x}_{O2} $$ $$ \vec{x}_{M} $$ $$ g_{i,j}(\vec{x}) $$

How does O1 calculate $ \vec{x}_{M}' $ in O2 coordinate system:

$$ \vec{x}_{M}'(\vec{x}_{O2},\vec{x}_{M},g_{i,j}(\vec{x}))$$

Answer 1

There is no way of describing an 'objective definite point', just like you @Shane, implied in your comment there is no 'absolute frame'. There is an infinite number of coordinate frames (atlas/charts) that can be defined, and they are related to each other by diffeomorphisms. The sequence of spacetime points you reference are in the $O_1$ reference frame, which includes wherever that observer moved to (if you wanted to define it so the origin was always where he started no problem, just transform it as he moves, including the clock's rest frame if he took it or left it behind. Just define it one way or the other.).

$O_1$ can predict what the others measure by transforming the coordinate system to theirs, of course measuring their position over time, i.e. their trajectory, and then transforming to them. The transformations are just what defines one in terms of the other. The metric then transforms as a two index tensor, mthe connection has its equation, and the curvature tensor or Ricci tensor transform as the appropriate rank Tensors. Position and velocity (wrt an affine parameter) transform as 4-vectors.

It is what defines a differentiable manifold, what GR uses for its spacetime. It is basic GR.

As to using the Hubble flow as a special coordinate frame in cosmology, indeed it is, and it is defined by the physics of that flow. We measure in what coordinate frame the CMB is isotropic and homogeneous, and that defines the frame for us. And another oberver, if comoving, will define the same frame, with the Big Bang at t and r equal zero. But in many other situations, say near a black hole, that's not a very useful coordinate system, and we are smarter to define it differently. Similarly in our solar system for the local gravitation.