Computation of the extrinsic curvature tensor for a warp drive metric.

Question

In Miguel Alcubierre's renowned paper discussing a "warp drive" metric, he discusses the extrinsic curvature. Here is an extract.

My questions are quite trivial to someone who understands the material, but I need to ensure I am not mislearning!

1. I understand that the lapse function is equal to 1. I also understand how to expand a covariant derivative, however the only way I see the connections cancelling is if they are symmetric, is this the case?

2. Does the the partial derivative of the three metric with respect to $t$ disappear when reduced due to the fact the three metric contains no such $t$ terms?

Thanks in advance, I hope this is specific enough.

Answer 1

In the Alcubierre warp drive, the spatial metric $\gamma_{ij}$ is a flat Cartesian metric--e.g. $\gamma_{ij} = \delta_{ij}$.

The connections cancel because $D_i$ denotes an intrinsic covariant derivative inside the 3d slice--which again, is Cartesian flat, so there are no connection terms.

Similarly, $g_{ij} = \gamma_{ij} = \delta_{ij}$, so $\partial g_{ij}/\partial t = 0$.