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For certain metrics in general relativity, the metric tensor $g_{{\alpha}{\beta}}$ is not a diagonal matrix. For example, the Alcubierre metric is given by

$$ds^2 = -dt^2 + [dx - V_s(t) f(r_s) dt]^2 + dy^2 + dz^2.$$

The matrix corresponding to this metric is

$$ g_{{\alpha}{\beta}}=\begin{pmatrix} V_s(t)^2f(r_s)^2 - 1 & -V_s(t) f(r_s) &0 &0 \\ -V_s(t) f(r_s) & 1 & 0& 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

which is not diagonal. On the other hand, the Schwarzchild metric does not have cross terms: $${ds}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

I expect that one could find a coordinate system where the Alcubierre metric is diagonalized. However, in coordinate systems where the off-diagonal elements exist, is there a deeper meaning to what the cross-terms represent?

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  • $\begingroup$ If the metric tensor is not diagonalized then the determinant can possibly vanish at a given point, allowing for the existence of a degenerate coordinate system. $\endgroup$ – hebetudinous Aug 6 '16 at 22:16
  • $\begingroup$ Why would you expect the metric tensor to always be diagonal? The metric tensor is such that the scalar product $\sigma(u,v) = g_{\alpha\beta}u^{\alpha}v^{\beta}$; unless you're always on an orthogonal basis, there is no requirement for the diagonal terms to vanish. $\endgroup$ – gented Aug 7 '16 at 0:24
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At a single point, off-diagonal elements of the metric tensor simply say that your coordinate system is not orthogonal. For example, if you use $\hat{x}$ and $\hat{x} + \hat{y}$ as your basis in the plane, the resulting metric has a cross term.

Since the metric tensor is symmetric, you can always diagonalize it at a single point, so that's not very interesting. The interesting question is to see if you can find a system of coordinates which diagonalize the metric at every point; this has a physical meaning.

Suppose you could find a coordinate system where there were no $dt dx^i$ cross terms and no time-dependence. Then the metric is invariant under time reversal, since time only appears through $dt^2$. This works for the Schwarzschild metric, since the black hole is just sitting there doing nothing.

Conversely, if your situation is not static, you can't diagonalize the metric everywhere. For example, in the Kerr metric for a rotating black hole, there's a $dt d\phi$ cross term from the rotation. In your metric, there is a $dt dr$ cross term due to the motion of the Alcubierre drive/"warp bubble".

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