Is observing (superluminal) velocity independent of choice of coordinates for asymptotically flat spacetimes?

Question

Alcubierre's warped metric in ($1+1$)D is typically given in the form of:

$$ds^2 = -dt^2 + [dx-v(t)f(r) dt]^2 \ \ .$$

Then it is nicely discussed how manipulating:

$$-dt^2 + [dx-v(t) f(r) \ dt]^2=0$$

leads to:

$$\frac{dx}{dt}=1+v(t)f(r)$$

and hence the claimed property of travel that appears superluminal to a remote observer (sitting away far enough to be in what is approximately Minkowski spacetime).

At first glance, this derivation seems to rely on the coordinates chosen to present the metric as in the above. Is this also the case in other coordinates? I am asking this because no matter what, the remote observer far outside the "bubble" sits in (approximately) Minkowski spacetime, meaning there are coordinate transformations that will not change this, such as going to synchronous coordinates. So it appears to me, that the remote observer, to a degree, does not care what coordinates we choose as it only changes the metric far away from it, while the observer remains in Minkowski spacetime.

Does this mean that the observation of superluminal travel is independent of the choice of coordinates for the Alcubierre metric, or more generally is the observed velocity independent of coordinate choice for asymptotically flat spacetimes, where the observer is in the approximately flat region?

Answer 1

First of all remember that, contrary to special relativity, in general relativity measurements which are not local lose meaning, due the many possible ways to parallel transport a vector on a curved manifold.

Given two stars A and B with proper distance D and an observer located in A, we are really interested in measuring how much time it takes for the drive to go from A to B and back (the space around A and B is approximately flat). In this way the warp drive will have travelled a proper distance D (observer independent quantity, for every choice of coordinates) in a proper time $\tau$ (again, observer independent) as measured by the local observer.

It's worth emphasising that in special relativity we know that the time measured by the person making the round trip can be made arbitrarily small if his speed approaches $c$. However, the fact that in general relativity one can actually make such a round trip in an arbitrarily short time as measured by an observer that remained at rest is the non trivial feature of the warp drive.

What you write above is a lightlike trajectory ($ds^2 = 0$), but the actual reason why the warp is superluminal is simply that $v(t)=v_s=dx_s/dt$, interpreted as the warp velocity, is arbitrary large and same for the coordinate acceleration $a$. Moreover for the spaceship's trajectory $x=x_s$ we have

$$d\tau = dt$$

that is proper time equals coordinate time. Since coordinate time of the ship is also equal to the proper time of observers in A (A is in flat space), we conclude that the spaceship suffers no time dilation as it moves. Then you can show that

$$t \simeq \tau \simeq 2 \sqrt{\frac{D}{a}} $$

which is a coordinate independent statement showing that the measured time can be arbitrarily short, since $a$ is arbitrary.