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The FRW metric can be written in conformal coordinates to give: $$ds^2=a^2(\eta)(-d\eta^2+d\mathbf{\Sigma^2}),$$ where $\eta$ is the conformal time and $\mathbf{\Sigma}$ ranges over 3-dimensional space of uniform curvature.

Using the above metric one can state that an interval of proper time $d\tau$ is given by: $$d\tau = a(\eta)\ d\eta.$$ Additionally an interval of proper distance $ds$ is given by: $$ds=a(\eta)\ d\mathbf{\Sigma}.$$

Now one normally interprets the FRW metric in standard coordinates $(t,\mathbf{\Sigma})$ as describing a space that expands with respect to cosmological time $t$.

But the choice of coordinates is arbitrary.

By using conformal coordinates $(\eta,\mathbf{\Sigma})$ is one equally justified in asserting that the FRW metric describes a spacetime whose scale $a(\eta)$ changes with respect to conformal time $\eta$? In other words the units of length and time change with respect to conformal time $\eta$.

Now the Einstein Field Equations can be written in natural units ($\hbar=c=1$): $$G_{\mu\nu}=l_P^2\ T_{\mu\nu},$$ where $l_P$ is the reduced Planck length and the cosmological constant is assumed to be the vacuum energy component of $T_{\mu\nu}$.

If I decide to use the FRW metric expressed in conformal spacetime coordinates then I think this implies that the units of length and time scale with the universal scale factor $a(\eta)$.

As the proper reduced Planck length $l_P$ must always be constant as my units of length expand then perhaps Einstein's Field Equations should be modified to give:

$$G_{\mu\nu}=\frac{l_P^2}{a^2(\eta)}T_{\mu\nu}.$$

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    $\begingroup$ why do you say "co-ordinates", they are called "coordinates" without the dash $\endgroup$
    – Yukterez
    Commented Mar 24, 2020 at 2:43
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    $\begingroup$ Well, did you calculate $G_{\mu\nu}$ for your metric? $\endgroup$
    – Toffomat
    Commented Mar 26, 2021 at 10:45

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It's true that FLRW metric can be written in a conformal form. But there is still an absolute sense to the notion of space expansion, so it is not just about units and coordinates. I am not quite sure if this is what you are asking, but if it is then here is an argument.

If you take an ordinary physical object such as a steel ball bearing, then you can only fit a finite number of them into a space having the form of a 3-sphere (I take this case for convenience in order to make the point). If that space then expands or contracts, then the number of ball bearings that will fit into it will change. This is an absolute statement: it does not depend on coordinates (except that I take for granted the standard foliation into spacelike surfaces). The statement relies on the equivalence principle, or local flatness, in order to argue that these ball bearings will all have the same size each in their own local inertial rest frame (assuming they are all made from the same number of atoms etc.).

Applied to cosmology, instead of a ball-bearing you can use a galaxy cluster of some standard size. And if the average curvature is zero or negative, then the argument can either be framed for a finite space with a non-standard topology, or one can simply talk about density of ball bearings (galaxy-clusters).

This type of argument is appealing to materials science in order to define units of length. One could equally use a clock and a fixed speed of light. In the question it appears that you might be wanting to bring in some other way to define units of length. One can always announce "starting from the year 2100, the definition of 1 metre will change annually, in such a way that the number of metres around the circumference of Earth will grow by 1 each year." Such an announcement is, it appears to me, what the question is proposing, but if I am mistaken then perhaps the question could be clarified.

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This is right with respect to the FLRW metric being a conformal coodinate system. The metric $$ ds^2 = \Omega^2(u)(du^2 - d\Sigma^{(3)}) $$ is with the change of variables $du^2 = \Omega(u)^{-2}dt^2$

lets us identify the above metric as the conformal time dependent metric $$ ds^2 = dt^2 - \Omega^2(u)d\Sigma^{(3)} $$ with the expansion factor $a(t) = \Omega^2(u)$.

I am not sure about $G^{\mu\nu} = \ell_p^2T^{\mu\nu}$. How did $\hbar$ creep into Einstein's field equations that are purely classical? I suppose one could write the Einstein field equations $$ G^{\mu\nu} = 8\pi\frac{\ell_p^2}{\hbar c}T^{\mu\nu} $$ The inclusion of $a(\eta)$ in the Einstein field equation is complicated, for it depends upon how $T^{\mu\nu}$ behaves.

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