Covariant derivative of riemann tensor

In summary: The curvature 2-form is only closed in the torsion-free case. In summary, the conversation discusses the calculation of the curvature of a manifold using the Christoffel symbols and the Bianchi identities. The speaker provides a general formula for the covariant derivative and the Riemann tensor, as well as a specific formula for the Riemann tensor in terms of the Christoffel symbols. The conversation also touches on the proof for the equation Rab;a=1/2gabR,a and the use of the second Bianchi identity. The conversation concludes with a request for help with the tedious calculation and a discussion of the torsion-free case.
  • #1
solveforX
19
0
what would Rabcd;e look like in terms of it's christoffels?

or Rab;c
 
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  • #2
Well [tex]\bigtriangledown \gamma R^{\alpha }_{\beta \mu \nu } = \frac{\partial R^{\alpha }_{\beta \mu \nu }}{\partial x^{\gamma }} + \Gamma ^{\alpha }_{\sigma \gamma }R^{\sigma }_{\beta \mu \nu } - \Gamma^{\sigma }_{\gamma \beta }R^{\alpha }_{\sigma \mu \nu } - \Gamma ^{\sigma }_{\gamma \mu }R^{\alpha }_{\beta \sigma \nu } - \Gamma ^{\sigma }_{\gamma \nu }R^{\alpha }_{\beta \mu \sigma }[/tex] by applying the general formula for the covariant derivative. If you want to see what it looks like in terms of the christoffel symbols use the respective formula for the Riemann tensor [tex]R^{\alpha }_{\beta \mu \nu } = \frac{\partial \Gamma ^{\alpha }_{\beta \nu } }{\partial x^{\mu }} - \frac{\partial \Gamma ^{\alpha }_{\beta \mu } }{\partial x^{\nu }} + \Gamma ^{\alpha }_{\sigma \mu }\Gamma ^{\sigma }_{\beta \nu} - \Gamma ^{\alpha }_{\sigma \nu }\Gamma ^{\sigma }_{\beta \mu }[/tex] but it will look pretty messy.
 
  • #3
OK thank you. I saw in a youtube lecture though that Rab;a=1/2gabR,a. I cannot find the proof for this. can anyone help me with this problem?
 
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  • #4
That's standard textbook material. But you don't need the textbook itself, just know what the tensors look like in terms of the torsion-less connection. So pick up your pencil and a sheet of paper & do it.
 
  • #5
I've tried over and over again. If someone could post a solution I'd really appreciate it.
 
  • #6
Could you give a link to the video =D?
 
  • #7
about 1:05:44 in
 
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  • #8
It essentially boils down to a really long but not hard calculation but its easy to make mistakes with all the indeces. If you're intent on doing it then try [tex]\bigtriangledown _{\mu }R^{\mu \nu } = \bigtriangledown _{\mu }(g^{\mu \alpha }g^{\nu \beta }R_{\alpha \beta }) = g^{\mu \alpha }g^{\nu \beta }\bigtriangledown _{\mu }R^{\sigma }_{\alpha \sigma \beta }[/tex] and work down to the christoffel symbols and try to get the expression that equals [itex]\frac{1}{2}\bigtriangledown _{\mu }(g^{\mu \nu }R)[/itex] with R expanded, formula - wise,up to the christoffel symbols. You can try googling or looking through your texts but most texts (at least the ones I own) simply skip the tedious calculations and also just state the bianchi identities which are related to what Susskind is talking about in the video.
 
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  • #9
yeah I've seen the method for the contraction of bianchi identities to arrive at the einstein tensor but I was determined to use this method too. It's really tedious and i just can't find the expression. I guess I'm just forced to believe the equation but thanks so much for the help.
 
  • #10
The second Bianchi identity for the Riemann tensor (torsion-less manifold, so that the curvature 2-form is closed) by double contraction with the covariantly constant metric tensor immediately yields

[tex] \nabla^{\mu}\left(R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R\right) = 0 [/tex]
 
  • #11
dextercioby said:
The second Bianchi identity for the Riemann tensor (torsion-less manifold, so that the curvature 2-form is closed) by double contraction with the covariantly constant metric tensor immediately yields

[tex] \nabla^{\mu}\left(R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R\right) = 0 [/tex]

The curvature 2-form is not closed, generally speaking. Everything else you say is correct, though. The second Bianchi identity

[tex]\nabla_{[\lambda} R_{\mu\nu]}{}^\rho{}_\sigma = 0[/tex]
is not the exterior derivative of the curvature 2-form. (I used symmetries [itex]R^\rho{}_{\sigma\mu\nu}[/itex] to make the formula more legible). Here is a case where index notation can be deceptive.
 
  • #12
Yes, you are right. I stand corrected.
 

1. What is the covariant derivative of the Riemann tensor?

The covariant derivative of the Riemann tensor is a mathematical concept used in differential geometry to describe how a tensor field changes as it is transported along a curve in a curved space. It is denoted by ∇aRbcde, where ∇a represents the covariant derivative operator.

2. How is the covariant derivative of the Riemann tensor calculated?

The covariant derivative of the Riemann tensor is calculated using the Christoffel symbols, which represent the components of the connection on a curved space. The formula for the covariant derivative of the Riemann tensor is: ∇aRbcde = ∂aRbcde + ΓbaeRdc – ΓcaeRdb + ΓbadRec – ΓcadReb, where Γabc are the Christoffel symbols.

3. What is the physical significance of the covariant derivative of the Riemann tensor?

The covariant derivative of the Riemann tensor is used in Einstein's theory of general relativity to describe the curvature of spacetime. It is a key component in the mathematical framework of the theory and is used to calculate the geodesic equation, which describes the path of a free-falling object in a curved space. It is also used in the Einstein field equations, which relate the curvature of spacetime to the distribution of matter and energy.

4. How does the covariant derivative of the Riemann tensor relate to the concept of parallel transport?

The covariant derivative of the Riemann tensor is closely related to the concept of parallel transport. Parallel transport is the idea of transporting a vector or tensor along a curve on a manifold while keeping it parallel to itself. The covariant derivative of the Riemann tensor measures how a tensor field changes as it is transported along a curve, taking into account the curvature of the space.

5. Can the covariant derivative of the Riemann tensor be extended to higher dimensions?

Yes, the covariant derivative of the Riemann tensor can be extended to higher dimensions, such as 4-dimensional spacetime. In fact, it is a fundamental concept in the study of higher-dimensional manifolds and is used in various areas of mathematics and physics, including string theory and quantum gravity.

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