Riemann-Christoffel Covariant differentation

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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
 
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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
 
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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}?
 
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Jinroh said:
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

I don't think it is generally correct, so good luck.
 
The left-hand side is a tensor.
Is the right-hand side written in a special choice of coordinates?
 
robphy said:
The left-hand side is a tensor.
Is the right-hand side written in a special choice of coordinates?

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.
 

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