Calculating proper volume in the Alcubierre spacetime I'm trying to calculate the proper volume of a portion of the alcubierre spacetime to see how it compares to the euclidean volume element. As I understand it, the proper volume element in cartesian coordinates is usually found with:
$dV=\sqrt{det\gamma}dxdydz$
Where I'm confused is that $det\gamma$ in this case is just $1$, yet I feel like a hypersurface $\Sigma$ of constant $t$ would have larger volume elements behind the spaceship  than in front.
The metric, written in terms of $c=1$:
$ds^2=(v^2f^2-1)dt^2-2vfdxdt+dx^2+dy^2+dz^2$
Comoving local tetrad:
$e_{(t)}^t=1$,   $e_{(t)}^x=vf$,   $e_{(y)}^y=e_{(z)}^z=1$
Static local tetrad:
$e_{(t)}^t=\frac{1}{\sqrt{1-v^2f^2}}$, $e_{(x)}^t=\frac{vf}{\sqrt{1-v^2f^2}}$, $e_{(x)}^x=\sqrt{1-v^2f^2}$, $e_{(y)}^y=e_{(z)}^z=1$
Is it possible to find the proper volume with the above objects?
 A: Typically, when one derives 3-volumes in a metric, you need to define some sort of foliation for the metric in the timelike variable.  Then, each leaf of your foliation is labeled by a function $\tau(t,x,y,z)$ such that $\tau$ is constant on each leaf.  In your case, you want spacelike surfaces, so $\nabla_{a}\tau$ should be timelike.  Define $\alpha^{2} = -\nabla_{a}\tau\nabla^{a}\tau$, and the unit vector $n_{a} = \alpha \nabla_{a}\tau$.
Then, the 3-metric is $\gamma_{ab} = g_{ab} + n_{a}n_{b}$.
If you do a coordinate change from t to $\tau$, then $\gamma$ is simply the 3x3 spacelike block of the metric.  So yes, if you just choose the $t$ coordinate from your original metric as the time coordinate, the leaves are indeed all spatially flat.
But, if you choose, $t = \tau + G(x)$, then we have $dt = d\tau + G'(x)dx$, and the metric becomes:
$$ds^{2} = -(1 - (vf)^{2})d\tau^{2} - 2d\tau dx\left(G'(1-(vf)^{2}) + vf\right) + dx^{2}\left(1 - G'^{2}(1-(vf)^{2}) - 2vfG' \right) +dy^{2} + dz^{2} $$
Which certainly looks like a mess.  But the operative point is that we are free to choose $G(x)$ to be anything we want.  And in particular, we kill the diagonal term if we choose:
$$G(x) = \int dx\frac{-vf}{1-(vf)^{2}}$$
And our metric reduces to:
$$ds^{2} = -(1-(vf)^{2})d\tau^{2} + \left(1 + \frac{(vf)^{2}}{1-(vf)^{2}}\right)dx^{2} + dy^{2} + dz^{2}$$
And now, if we foliate on $\tau = $ Constant, it's clear that the 3-determinant of our leaves will be $1 + \frac{(vf)^{2}}{1-(vf)^{2}} = \frac{1}{1-(vf)^{2}}$
But note that this is an even function in $x -> -x$
