Physical interpretation of FRW normal coordinates

Question

The Friedmann-Robertson-Walker metric (I consider for notational simplicity the flat space case): $$\text d s^2 = \text d t^2 - a(t)^2\text d \boldsymbol{x}^2$$ can be brought to normal (Minkowski) form at the origin by a quadratic change of coordinates (see e.g. Eq. (10) of Ref.1): $$\boldsymbol x = \boldsymbol X -H_0\boldsymbol X T, \\t=T-\frac{1}{2}H_0\boldsymbol X^2, $$ where $H_0=\dot a (0)$ and I assume $a(0)=1$.

My question is: does the above coordinate transformation have any physical interpretation, for instance in terms of accelerations or Newtonian gravitational fields?

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The $\boldsymbol {x}$ transformation is telling me that conformal coordinates and locally Minkowski coordinates are related by a simple rescaling $X=a(t)x$, and moreover looks like a Lorentz boost with velocity $\boldsymbol V = H_0 \boldsymbol X$. Which suggests me to rewrite: $$\boldsymbol x = \boldsymbol X -\boldsymbol V T, \\t=T-\boldsymbol V\cdot \boldsymbol X + \frac{1}{2 H_0}\boldsymbol V^2. $$ However, provided I'm on the right track, I don't know how to make sense of the last $V^2$ term in the $t$ transformation.

Answer 1

I figured this out following Sec. 10.3 of these notes. I'll briefly summarize here.

First, we observe few basic facts about FRW metric. Using the same notations in the OP, consider the observer $O$ which sits at the origin of coordinate space: $x^\mu _O(t)=(t,\boldsymbol 0)$. It is easy to see that:

1. $O$ is free-falling, i.e. $x^\mu _O$ is a geodesic.
2. The cosmological time $t$ coincides with $O$'s proper time along the worldline $x^\mu$.

Therefore, it should be possible to construct a locally inertial system in which $O$ is stationary at the origin of space. The time $T$ and space $\boldsymbol X$ coordinates of such a system may be constructed as follows: we imagine that $O$ emits a light signal at cosmological time $t_1$, which is reflected at time $t$ by a mirror with comoving coordinates $\boldsymbol x$, and reabsorbed by $O$ at time $t_2$. In the locally inertial frame, the coordinates of the reflection event are: $$T=\frac{t_1+t_2}{2}\\ X=\frac{t_2-t_1}{2},$$where $X=\vert \boldsymbol X \vert$. These equations are correct because $t=\tau$, the proper time of our observer $O$.

In order to express $T$ and $X$ in terms of the cosmological coordinate $t$ and $x$, we observe that the two light-signals emitted by $O$ travel along null geodesics, $\text d s^2 =0$, implying: $$x=\intop _{t_1} ^t \frac{\text d t'}{a(t')}=\intop _{t} ^{t_2} \frac{\text d t'}{a(t')}.$$ Expanding these equations to quadratic order, and using the definitions of $T$ and $X$ above, one finds: $$t\approx T-\frac{H_0X^2}{2}\\ x\approx X-H_0 X T$$ up to cubic terms. Finally, if the observer $O$ chooses to align his space axes with those of the comoving frame, we have $\frac{\boldsymbol X}{X}= \frac{\boldsymbol x}{x}$, and the coordinate change of the OP is found.

Answer 2

I calculate the line element:

Case I:

$\left[ \begin {array}{c} t\\ x\end {array} \right]\mapsto \left[ \begin {array}{c} T-1/2\,H_{{0}}{X}^{2}\\ X- H_{{0}}XT\end {array} \right] $

Line element with $V=H_0\,X$

$ds^2\mapsto \left( 1-{V}^{2}a \left( t \right) \right) {{\it dT}}^{2}+ \left( -2 \,V+2\,Va \left( t \right) -2\,Va \left( t \right) H_{{0}}T \right) { \it dX}\,{\it dT}+{{\it dX}}^{2} \left( {V}^{2}-a \left( t \right) +2 \,a \left( t \right) H_{{0}}T-a \left( t \right) {H_{{0}}}^{2}{T}^{2} \right) $

For $H_0=0$

$ ds^2=\left( 1-{V}^{2}a \left( t \right) \right) {{\it dT}}^{2}+ \left( -2 \,V+2\,Va \left( t \right) \right) {\it dX}\,{\it dT}+{{\it dX}}^{2} \left( {V}^{2}-a \left( t \right) \right) \qquad (1) $

Case II

$\left[ \begin {array}{c} t\\ x\end {array} \right]\mapsto \left[ \begin {array}{c} T-H_{{0}}{X}^{2}\\ X-H_{{0 }}XT\end {array} \right] $

Line element with $V=H_0\,X$

$ds^2\mapsto \left( 1-{V}^{2}a \left( t \right) \right) {{\it dT}}^{2}+ \left( -4 \,V+2\,Va \left( t \right) -2\,Va \left( t \right) H_{{0}}T \right) { \it dX}\,{\it dT}+{{\it dX}}^{2} \left( 4\,{V}^{2}-a \left( t \right) +2\,a \left( t \right) H_{{0}}T-a \left( t \right) {H_{{0}}}^{2}{T}^{2 } \right) $

For $H_0=0$

$ds^2= \left( 1-{V}^{2}a \left( t \right) \right) {{\it dT}}^{2}+ \left( -4 \,V+2\,Va \left( t \right) \right) {\it dX}\,{\it dT}+{{\it dX}}^{2} \left( 4\,{V}^{2}-a \left( t \right) \right) \qquad (2)$

Case III:

$\left[ \begin {array}{c} t\\ x\end {array} \right]\mapsto \left[ \begin {array}{c} T-VX+1/2\,{\frac {{V}^{2}}{H_{{0}}}} \\X-VT\end {array} \right] $

but $V=H_0\,X$ so we get:

$\left[ \begin {array}{c} t\\ x\end {array} \right]\mapsto \left[ \begin {array}{c} T-1/2\,H_{{0}}{X}^{2}\\ X- H_{{0}}XT\end {array} \right] $

This is the transformation case I, so you don't have new line element!

Comparing equation (1) with (2) , we have "the same" line element for $H_0=\dot{a}(0)=0$

Answer 3

In relativity, velocity is direction/angle and elapsed time is distance, so if you have different objects moving at different velocities from a common point for the same amount of time (as measured by their intrinsic evolution), they'll end up on a sphere of constant distance from the starting point, not in a plane as they would in a Newtonian world with absolute time. Spheres are extrinsically curved in the background space, and that's why the FRW constant-time spacial slices are curved. (This picture is complicated by the fact that at the critical density, the spatial slices are intrinsically flat – but they're still extrinsically curved.)

Standard FRW coordinates are essentially polar coordinates. The surface of the earth, idealized as a sphere of radius 1, has a Euclideanized FRW metric, $\text d s^2 = \text d t^2 + a(t)^2\text d x^2$, where $x$ is longitude, $t$ is latitude ($0$ at the south pole, say), and $a(t) = \sin t$. If you place yourself at a location in the southern hemisphere and compare the local lines of latitude and longitude to a local almost-Cartesian grid made of parallel lines (i.e. great circles), you'll see that the longitude lines are straight (also great circles) but differ in slope from your grid by an amount proportional to the difference of latitude, while the latitude lines are parallel to your grid at the origin but separate quadratically from them with longitude. Both effects are zero at the equator and have the opposite sign in the northern hemisphere, suggesting that they're proportional to $H$. I don't know how to immediately see the correct values of the constant factors, but I'd guess 1 for the linear term and ½ for the quadratic term, and I'd be right.

The approach below the dividing line in the question in on the wrong track. The coordinates aren't related by a Lorentz boost or any approximation of one.