- #1

binbagsss

- 1,254

- 11

I'm having trouble understanding this.

**What does the curvature describe otherwise? I've only heard of it being the metric itself.**

Also, I think, 'in GR we are mot concerned with the Christoffel connection' comes from having the fundamental theorem of Riemannian geometry holding , so we get that simple formula relating the connection to the metric. The conditions of the metric in this theorem are that it be 'symmetric, differentiable and non-degenerate'.

So, the unextended Schwarzschild metric has a singulaity at r=2GM, it's determinant is in indeterminate form, and is degenerate. So doesn't this mean that at a curavture singularity, because the conditions for the fundamental theorem of Riemannian geometry to hold are not satisfied, the connection can not be given by the metric, and if I have interpreted the above statement correctly, the curvature computed from the Riemannian tensor will not be that associated with the metric? What does it represent then?

**Thanks.**