Help Covariant Derivative of Ricci Tensor the hard way.

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Discussion Overview

The discussion revolves around the calculation of the covariant derivative of the Ricci tensor, specifically attempting to replicate the method used by Einstein. Participants explore various approaches, including direct calculations and references to the Bianchi identities.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents two expressions for the covariant derivative of the Ricci tensor, noting discrepancies in their formulation.
  • Another participant points out an indexing error in the second expression, suggesting that free indices should not be used as summation indices.
  • A different participant references the second Bianchi identity as a method to derive the result, implying a more straightforward approach.
  • One participant expresses a desire to adhere to Einstein's original method, arguing that using the Bianchi identities is not in line with that approach.
  • Another participant suggests that Einstein's lack of access to symbolic algebra programs would have influenced his methodology, contrasting it with modern techniques.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to calculate the covariant derivative of the Ricci tensor. There are competing views on whether to use the Bianchi identities or to follow Einstein's original approach.

Contextual Notes

There are unresolved issues regarding the proper treatment of indices in the expressions presented, as well as the implications of using modern mathematical tools versus historical methods.

nobraner
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I am trying to calculate the covariant derivative of the Ricci Tensor the way Einstein did it, but I keep coming up with

[itex]\nabla_{μ}[/itex]R[itex]_{αβ}[/itex]=[itex]\frac{∂}{∂x^{μ}}[/itex]R[itex]_{αβ}[/itex]-2[itex]\Gamma^{α}_{μ\gamma}[/itex]R[itex]_{αβ}[/itex]

or


[itex]\nabla_{μ}[/itex]R[itex]_{αβ}[/itex]=[itex]\frac{∂}{∂x^{μ}}[/itex]R[itex]_{αβ}[/itex]-[itex]\Gamma^{α}_{μ\gamma}[/itex]R[itex]_{αβ}[/itex]-[itex]\Gamma^{β}_{μ\gamma}[/itex]R[itex]_{αβ}[/itex]

Any help will be much appreciated.
 
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Your second expression is almost right, except you have some trouble with indices. If alpha and beta are free indices on the left, you should not use them as summation indices on the right. Use gamma as your summation index, and then alpha and beta should remain free indices.
 
Then, how did Einstein get

[itex]\nabla_{μ}[/itex]R[itex]_{αβ}[/itex]=[itex]\frac{1}{2}[/itex]g[itex]_{αβ}[/itex][itex]\nabla_{μ}[/itex]R
 
Try the second Bianchi identity, contract it twice and you get the result.
 
Thanks for the advice, but anyone can do it that way. I'm trying to do it the way Einstein did it; the hard way. Einstein didn't know about the Bianchi Identities.
 
You can put the definition of the Ricci and scalar curvature tensor in term of metric...
 
Einstein didn't have symbolic algebra programs, I would guess. But that's what I'd use.
 

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