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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
    Commented Aug 6, 2016 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
    Commented Aug 7, 2016 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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  • $\begingroup$ If you can find an orthogonal system at each point, why can't you combine these systems into a global system which is orthogonal at each point? I think of Eucliden 2-plane. After defining the radial coordinate r, you define (at each point) an angular coordinate which is orthogonal to r, and then combine them all. This angular coordinate will point at a different direction at each point, but that's okay. $\endgroup$
    – Arik
    Commented Aug 4, 2021 at 20:16
  • $\begingroup$ @Arik You can only do that when your spacetime has no curvature. Think about trying to do the same thing on the Earth: it's impossible because Earth's surface is curved. $\endgroup$
    – knzhou
    Commented Aug 4, 2021 at 20:22
  • $\begingroup$ I can do it on the Earth just the same. I first define the 'radial' coordinate (actually the latitude, but it is actually a radial coordinate), and then, at each point, I define the 'phi' angular coordinate (the longitude). I works perfectly. (How do you write that 'at' sign in the beginnig of the comment? it doesn't work for me) $\endgroup$
    – Arik
    Commented Aug 4, 2021 at 21:08
  • $\begingroup$ Apparently, any metric is diagonalizable; just not necessarily by a coordinate basis. $\endgroup$
    – Vincent Thacker
    Commented Aug 6, 2021 at 16:46

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