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How do I find the proper spacelike $x$-coordinate on a constant $t$ hypersurface in the two dimensional Alcubierre manifold? The metric is given by

$$\mathrm{d}s^2=(v^2f^2-1)\mathrm{d}t^2-2vf\mathrm{d}x\mathrm{d}t+\mathrm{d}x^2.$$

The determinant of the spatial metric is just $1$, suggesting spatial flatness, but space is expanding, so I would think proper distance would vary at different points on a hypersurface.

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  • $\begingroup$ It might be worth pointing out that all one-dimensional manifolds are flat. Notice that, in one dimension, $\mathrm{d}s^2 = f(x)^2 \mathrm{d}x^2$, and one can always just redefine coordinates to get rid of the $f(x)$ factor and get $\mathrm{d}s^2 = \mathrm{d}y^2$ ($y$ solves $\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)$). $\endgroup$
    – Níckolas Alves
    Commented Jul 31, 2022 at 23:56
  • $\begingroup$ I'd also ask "what happens to two freely falling observers at $t=0$, one at $x=0$ and the other at $x=L$, and both with, at $t=0$, ${\dot x} = {\dot y} = {\dot z} = 0$? Do they stay a proper distance $L$ away from each other forever? What is your spacelike geodesic measuring? $\endgroup$
    – Zo the Relativist
    Commented Aug 1, 2022 at 0:04
  • $\begingroup$ I guess I assumed that the proper distance between the two observers would change, while the coordinate distance would not. $\endgroup$
    – user345249
    Commented Aug 1, 2022 at 1:02
  • $\begingroup$ Jerry: I am basically thinking about a proper 3-volume measurement in 3 spatial dimensions (although my post asked about just one). $\endgroup$
    – user345249
    Commented Aug 1, 2022 at 2:03
  • $\begingroup$ @user345249: something like that is going to depend, intensely, on what time coordinate you choose. I don't think the $t$ above is the time coordinate you want. At minimum, I suspect you would want to change coordinates to some variable $T(t,x)$, replace $t$ with $T$, and pick $T$ in such a way that it diagonalizes the metric. $\endgroup$
    – Zo the Relativist
    Commented Aug 1, 2022 at 18:52

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