Bandersnatch said:
Peter, can you provide an estimate for what is the influence of the gravitational well of the Milky Way w/r to the comoving observer? Is it not more significant than Earth's gravity?
It's more significant than Earth's, yes, but that's a very low bar. It's not significant in any sense that matters for this discussion.
The general formula we're interested in is the deviation of the metric coefficient ##g_{tt}## from 1. A good estimate of that for objects with weak fields (we'll come back to how we know which fields are weak in a moment) is simply ##GM / c^2 R##, where ##M## is the mass of the object and ##R## is its radius. We have ##G / c^2 \approx 10^{-27}## meters per kilogram, so we can do some quick order of magnitude estimates:
Earth: ##M \approx 6 \times 10^{24}##, ##R \approx 6 \times 10^6##, so ##GM / c^2 R \approx 10^{-9}##.
Sun: ##M \approx 10^{30}##, ##R \approx 10^9##, so ##GM / c^2 R \approx 10^{-6}##.
Sun's field at Earth's orbit: ##M \approx 10^{30}##, ##R \approx 10^{11}##, so ##GM / c^2 R \approx 10^{-8}##.
Milky Way: ##M \approx 10^{42}##, ##R \approx 10^{22}##, so ##GM / c^2 R \approx 10^{-7}##.
All of these are obviously weak fields, since the correction is much less than 1; a strong field would be one in which this correction, calculated as above, was significant compared to 1. (In the strong field regime the formula I gave above, which is an approximation, breaks down, and we need to use the exact metric coefficient formula.) As you can see, the correction due to the Milky Way's field, at Earth's position, is the largest of the three; but at the surface of the Sun, the correction due to the Sun's field is larger than that due to the Milky Way's field. At Earth's orbit, the correction due to the Milky Way makes a difference of about 1000 years in the observed age of the universe (i.e., we here on Earth see a universe with an apparent age about 1000 years younger than a comoving observer far out in intergalactic space).