Riemann-Christoffel Covariant differentation

1. Jun 16, 2004

Jinroh

Hi

Given the Riemann-Christoffel Tensor :

$$R_{i{\text{ }},{\text{ }}jk}^l = \partial _j \Gamma _{ik}^l - \partial _k \Gamma _{ij}^l + \Gamma _{ik}^r \Gamma _{jr}^l - \Gamma _{ij}^r \Gamma _{kr}^l$$

I'm looking for the proof :

$$\nabla _t R_{i,rs}^l = \partial _{rt} \Gamma _{si}^l - \partial _{st} \Gamma _{ri}^l$$

Thanks for your help

2. Jun 16, 2004

Blackforest

Sorry I did not find the proof since yesterday evening when I saw your thread for the first time. But one thing is sure: the second equation is true because leading to the Bianchies identities used in general relativity. Blackforest

3. Jun 18, 2004

robphy

Last edited by a moderator: Apr 20, 2017
4. Jun 18, 2004

turin

You're using two different symbols, $$\partial$$ and $$\nabla$$, but I don't quite understand the distinction. Is $$\nabla$$ the covariant derivative?

And what's the difference between $$\partial _{r}$$ and $$\partial _{rt}$$? Is the second one just a shorthad for $$\partial _{r} \partial _{t}$$?

Last edited: Jun 18, 2004
5. Jun 22, 2004

DW

I don't think it is generally correct, so good luck.

6. Jun 22, 2004

robphy

The left-hand side is a tensor.
Is the right-hand side written in a special choice of coordinates?

7. Jun 22, 2004

DW

Right, the right hand side is what one gets for local free fall frames. I don't think the relation will hold in general. When I take the covariant derivative of the Riemann tensor for general coordinates I wind up with many terms on the right hand side a whole lot of which would have to subtract each other off for it to reduce to that and I don't see them doing that. The extra terms are products with Christophel symbols which vanish in local free fall, but otherwise are not zero.