Undergrad Metric Tensor: Symmetry & Other Constraints

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The discussion centers on the mathematical constraints of the metric tensor, emphasizing that while symmetry is essential, other conditions like the signature (1,3) or (3,1) are also crucial. It is clarified that the determinant of the full metric must be negative, but specific component conditions, such as g_{00}, can vary under different coordinate choices. Examples are provided, including null coordinates in Minkowski spacetime and static coordinates in de Sitter spacetime, illustrating the flexibility in metric definitions. The conversation also touches on the importance of continuity and differentiability for curvature tensors. Overall, the metric tensor's properties are complex and can be influenced by the chosen coordinate system.
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What are the mathematical constraints on the metric?
Aside from being symmetric, are there any other mathematical constraints on the metric?
 
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Physical interpretation requires some other features like
g_{00}>0, g=det(g_{ik})<0
in (+---) 0123 convention.
 
Last edited:
anuttarasammyak said:
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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PeterDonis said:
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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Nugatory said:
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.
 
PeterDonis said:
He's welcome to correct me if I am wrong.
I intended i,k=0,1,2,3. Thanks.
 
anuttarasammyak said:
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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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.
 
anuttarasammyak said:
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
Isn't there also a requirement that the metric be continuous and twice differentiable in order that the curvature tensors are well behaved?
 
  • #11
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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