B Curvature of space vs. curvature of spacetime

  • #51
pervect said:
Do you have a reference? I'm not familiar with your argument about six arbitrary functions. I can, however, follow that there are 10 independent numbers in a metric. This excludes any considerations of diffeomorphisms, so it's not necessarily in conflict with what you said (though I'm not qutie following what you said). And the tensor transformation rules give us 16 possible linear transformations (a 4x4 matrix) , which gives us more than enough degrees of freedom to transform away all the components of the metric tensor at a single point. Which means that specifying the metric tensor alone at a single point can't tell us anything physical, as we can always find a coordinate system in which the metric tensor is diag(-1,1,1,1).

The argument "with a little imagination" made by Melgrin isn't quite rigorous enough for me to want to defend, though I thought it was interesting.

The conclusion that we can transform the metric to diag(-1,1,1,1) and also make all it's first-order derivatives vanish at a single point also seems reasonably obvious on physical grounds from the existence of Riemann normal coordinates.
Sorry, I'm not following this at all. Do you have any references or can you explain further?
One reference, to Spivak, is given in post #31, for the specific application to sectional curvature. The most succinct online reference I can find is to math stack exchange:

https://math.stackexchange.com/ques...a-metric-have-on-a-psuedo-riemannian-manifold
 
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  • #52
PAllen said:
One reference, to Spivak, is given in post #31, for the specific application to sectional curvature. The most succinct online reference I can find is to math stack exchange:

https://math.stackexchange.com/ques...a-metric-have-on-a-psuedo-riemannian-manifold

After some thought, I have to agree that the original article I quoted is probabl flawed. If we really had 10 degrees of freedom in the metric tensor, the metric tensor could be made all zero - there would be no constraints. But that isn't correct. I do think the idea of counting the degrees of freedom of the metric, it's first derivatives, and the second derivatives offers an explanation of how the curvature tensor winds up with more degrees of freedom than the metric does. But I have to agree that the detailed analysis appears to need a bit more work than the paper(s) I was looking at, and I'm not familiar enough with the topic to fix the argument.
 
  • #53
pervect said:
After some thought, I have to agree that the original article I quoted is probabl flawed. If we really had 10 degrees of freedom in the metric tensor, the metric tensor could be made all zero - there would be no constraints. But that isn't correct. I do think the idea of counting the degrees of freedom of the metric, it's first derivatives, and the second derivatives offers an explanation of how the curvature tensor winds up with more degrees of freedom than the metric does. But I have to agree that the detailed analysis appears to need a bit more work than the paper(s) I was looking at, and I'm not familiar enough with the topic to fix the argument.
You may find the following thread of interest. Focus only on the OP and page 4 when Ben Niehoff joins the discussion, and some final references Atty found ( a little of page 3 might be interesting, but much before the last page just goes in circles).

https://www.physicsforums.com/threads/independent-components-of-the-curvature-tenso.530857/
 
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