Metric Tensor: Symmetry & Other Constraints

[quickAndLucky]

TL;DR Summary

What are the mathematical constraints on the metric?

Aside from being symmetric, are there any other mathematical constraints on the metric?

 

[anuttarasammyak]

Physical interpretation requires some other features like
$$g_{00}>0, g=det(g_{ik})<0$$
in (+---) 0123 convention.

 

[PeterDonis]

Physical interpretation requires some other features

Not the ones you state. It is perfectly possible to have a metric tensor that violates your conditions, if the coordinates are chosen appropriately.

Physically, the metric tensor must have a (1, 3) signature (or (3, 1) if we choose the opposite signature convention), but that in no way requires the condition you impose on the particular components.

 

[Nugatory]

but that in no way requires the condition you impose on the particular components.

That’s clearly true for the sign of $g_{00}$, but for the statement about the determinant?

 

[PeterDonis]

for the statement about the determinant?

The determinant of a 3-submatrix is not constrained. The determinant of the full metric is, but I don't think that's the determinant that the poster in post #2 meant, since he used the indexes $ik$, which usually means just the "spatial" indexes. He's welcome to correct me if I am wrong.

 

[anuttarasammyak]

He's welcome to correct me if I am wrong.

I intended i,k=0,1,2,3. Thanks.

 

[PeterDonis]

I intended i,k=0,1,2,3.

Ah, ok. Then your constraint on the determinant is correct, but your constraint on $g_{00}$, as noted, is not.

 

[anuttarasammyak]

Now I know for an example in rotating system $g_{00}<0$ for region r > c / $\omega$ where no real body cannot stay still to represent coordinate (r,$\phi$). Thanks.

 

[PeterDonis]

Now I know for an example in rotating system $g_{00}<0$ for region r > c / $\omega$ where no real body cannot stay still to represent coordinate (r,$\phi$). Thanks.

There are plenty of examples. Just a few off the top of my head:

Null coordinates in Minkowski spacetime, and the various kinds of null charts in curved spacetime (for example, Eddington-Finkelstein, Kerr-Schild).

Painleve coordinates in Schwarzschild spacetime, at and inside the event horizon.

Static coordinates in de Sitter spacetime, at and outside the cosmological horizon.

 

[Ibix]

Isn't there also a requirement that the metric be continuous and twice differentiable in order that the curvature tensors are well behaved?

 

[vanhees71]

I think the correct statement is already made, i.e., the pseudometric (I insist on calling it NOT metric, because it's not positive definite, it's the fundamental form of PSEUDO-Riemannian manifold and not a Riemannian one, and that's very important physics wise since it allows for defining a causality structure of spacetime) must be non-degenerate and have the signature (1,3) (west-coast convention) or (3,1) (east-coast convention). That means the the components $g_{\mu \nu}$ form a real symmetric $4 \times 4$ matrix with 1 positive and 3 negative (or 3 positive and 1 negative) eigenvalue. Consequently the determinant $\mathrm{det}(g_{\mu \nu})<0$. Since GR is covariant unders general diffeomorphisms of the coordinates at any point in spacetime you can choose "Galilean coordinates", such that in this one point $(g_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$ (or $(g_{\mu \nu})=\mathrm{diag}(-1,1,1,1)$).