Calculating proper volume in the Alcubierre spacetime

Question

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?

Answer 1

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 $x -> -x$

Answer 2

You are mistaking the volume element for the expansion volume, which is defined as proportional to the trace of the extrinsic curvature tensor. Verify equation (11) on page 5 of Alcubierre's 1994 paper The warp drive: hyper-fast travel within general relativity. For the Alcubierre warp drive, the expansion volume $\theta$ is

$$\theta = \frac{x-x_s}{r_s} \sqrt{\left(\frac{\partial\beta}{\partial y}\right)^2 + \left(\frac{\partial\beta}{\partial z}\right)^2}$$

where $\beta = - v_s(t) f(r_s)$ is the shift vector, $v_s(t)$ is the warp bubble center velocity, $f(r_s)$ is the regulating form function of the bubble, and $r_s$ is the Euclidean distance connecting the center of the bubble $(x_s,0,0)$ to a generic point in spacetime located at each spatial hypersurface

$$r_s(t) = \sqrt{(x-x_s)^2 + y^2 + z^2} $$

PS: Zo the Relativist demonstrated that the warp bubble has an event horizon behavior due to a coordinate singularity. Alcubierre and Lobo demonstrated similarly in Warp drive basics.