Constraints on Metric Tensor

  • #1

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What are the mathematical constraints on the metric?

Main Question or Discussion Point

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

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  • #2
anuttarasammyak
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Physical interpretation requires some other features like
[tex]g_{00}>0, g=det(g_{ik})<0[/tex]
in (+---) 0123 convention.
 
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  • #3
PeterDonis
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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.
 
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  • #4
Nugatory
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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?
 
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  • #5
PeterDonis
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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.
 
  • #7
PeterDonis
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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.
 
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  • #8
anuttarasammyak
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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.
 
  • #9
PeterDonis
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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.
 
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  • #10
Ibix
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Isn't there also a requirement that the metric be continuous and twice differentiable in order that the curvature tensors are well behaved?
 
  • #11
vanhees71
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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)##).
 

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