For context, I'm currently reading chapter 4 of Franklin's 'Advance Mechanics and General Relativity'. I'm having trouble understanding the proof of:
- vanishing of riemann curvature tensor at every point implies constant metric tensor. The book proves this uses the PDE argument from Dirac's 'General Theory of Relativity'
- existence of normal coordinates
I know literally nothing about PDE and manifolds, hence this proof like I'm 5 request.

Cheers

On Friday, August 11, 2023 at 4:56:01 PM UTC-7, Kevin S wrote:
> For context, I'm currently reading chapter 4 of Franklin's 'Advance Mechanics and General Relativity'. I'm having trouble understanding the proof of:
> - vanishing of riemann curvature tensor at every point implies constant metric tensor. The book proves this uses the PDE argument from Dirac's 'General Theory of Relativity'
> - existence of normal coordinates
> I know literally nothing about PDE and manifolds, hence this proof like I'm 5 request.
>
> Cheers

I think of a Riemannian curvature tensor as sort of like a circle, centered on the
center of mass of the local gravity well, i.e. it's a gravity well. You're familiar with
gravity as proportional to distance inverse square, as the curvature or 1/R goes to
zero as R or the distance goes to infinity, the resulting force vector goes to zero.

Then ds^2 is ds^2 and the metric is the metric.

So, in a usual gravity well it never vanishes, but two objects that are about the
same in deep space, don't share a gravity well and just fall together in gravity.