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I have been studying Birkhoff's theorem and, separately, co-moving coordinates.

One of the final steps in Birkhoff's theorem (from Weinberg's GR pg. 337) is to redefine the time coordinate to absorb the $f(t)$ that arises, as $$t' = \int^{t}f(t)^{1/2}dt$$ Where $t$ is the time coordinate (which has incidentally been redefined already once through the derivation, but clearing this up will clear up the previous changes as well).
Why is this $f(t)$ able to be redefined away like this?

Co-moving coordinates, like used in FRW cosmology, are defined such that the cosmic fluid is in free-fall, and thus, $d\tau = dt$. If we did not define the FRW metric as co-moving, it might mean the $g_{00}$ term is a function of coordinate time. This would in turn mean that an object at a constant spatial point would have a proper time that diverges from the coordinate time.

What would this mean physically? (To have the proper time diverge from coordinate time as a function of time)

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Physically this makes no difference. Without this redefinition you just have an arbitrary lapse function $N(t)$ in the $g_{00}$ component, but it is not dynamical. See my answer here Why the FLRW metric was chosen with constant $g_{00}$ time component?

Essentially this is just another acceptable coordinate transformation stemming from the reparametrisation invariance of GR, and it's of course convenient to pick coordinates that are comoving w.r.t some background flow.

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  • $\begingroup$ I read your other answer as well. Very helpful, thanks. What exactly do you mean non dynamical? Because it seems g00 as a function of t will dynamically change the metric $\endgroup$
    – perchlorious
    Commented Jun 6, 2022 at 13:13
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    $\begingroup$ They are non-dnamical in the sense that they act purely as Lagrange multipliers, these answers expand on this a little (and links if you want more detail) physics.stackexchange.com/questions/171720/… physics.stackexchange.com/questions/554028/… $\endgroup$
    – Eletie
    Commented Jun 6, 2022 at 13:47
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    $\begingroup$ A different choice of lapse leads to a different 3-geometry on your spatial hypersurface of constant $t$, but the same 4-geometry (regardless of coordinate choice), which is where our physics comes from. We can therefore set $g_{00}$ to one for simplicity. There's actually a nice short discussion of this in MTW in the context of the ADM formulation. $\endgroup$
    – Eletie
    Commented Jun 6, 2022 at 13:50

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