Metric Tensor: Symmetry & Other Constraints

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
10 replies · 2K views
quickAndLucky
Messages
32
Reaction score
3
TL;DR
What are the mathematical constraints on the metric?
Aside from being symmetric, are there any other mathematical constraints on the metric?
 
Physics news on Phys.org
Physical interpretation requires some other features like
[tex]g_{00}>0, g=det(g_{ik})<0[/tex]
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.
 
  • Like
Likes   Reactions: anuttarasammyak
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?
 
  • Like
Likes   Reactions: anuttarasammyak
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.
 
  • Like
Likes   Reactions: 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.
 
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.
 
  • Like
Likes   Reactions: anuttarasammyak
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)##).