How do I evaluate the variation of the riemann tensor with respect the metric tensor?
Answers
Answered by
0
I am not doing standard general relativity. I am not interested in varying the Einstein-Hilbert action to get the Einstein field equations, I am interested in finding an explicit formula for specifically the following quantity:
δRμνρσδgαβ=?δRμνρσδgαβ=?
I am interested in a completely general result for this quantity
Given that I am not looking to vary the Einstein Hilbert action, manipulations with the Palatini identity to get it in the form of a total derivative that won't contribute to the action or some argument like that isn't what I am looking for. I need a general result for this quantity
δRμνρσδgαβ=?δRμνρσδgαβ=?
I am interested in a completely general result for this quantity
Given that I am not looking to vary the Einstein Hilbert action, manipulations with the Palatini identity to get it in the form of a total derivative that won't contribute to the action or some argument like that isn't what I am looking for. I need a general result for this quantity
Answered by
0
❤️❤️__
__❤️❤️
The covariant derivative of the Riemann tensor and the metric tensor is well defined, you can compute both and then divide them to get the answer. However, if you multiply and divide with: ∂xγ∂xγ, in Gaussian normal coordinates, the derivative of the metric tensor would yield a null result, thus making the partial derivative of the Riemann tensor tend to infinity.
The covariant derivative of the Riemann tensor and the metric tensor is well defined, you can compute both and then divide them to get the answer. However, if you multiply and divide with: ∂xγ∂xγ, in Gaussian normal coordinates, the derivative of the metric tensor would yield a null result, thus making the partial derivative of the Riemann tensor tend to infinity.
Similar questions