- #1
space-time
- 218
- 4
We know how the curvature of a vector V or a manifold is depicted by the following formula:
dx[itex]\mu[/itex]dx[itex]\nu[/itex][∇[itex]\nu[/itex] , ∇[itex]\mu[/itex]]V
Now we know that the commutator is simply the Riemann tensor. My question here is:
How do you actually apply that vector V to the Riemann tensor? Here is an example of what I mean:
I know that the covariant derivative ∇j(Viei) = [(∂Vi)/(∂xj) + [itex]\Gamma[/itex]ikjVk]ei
Now having established this, would I distribute the vector Vi to the terms of my Riemann tensor or would I multiply the terms of the tensor by Vk?I also have one more question:
What extra information does the Riemann tensor tell you that the Ricci tensor does not (from a physical standpoint)? I know that the Ricci is a contraction of the Riemann and that the Riemann has a lot more elements than the Ricci, but what physical meaning do all of those other terms possesses that is not present in the Ricci tensor?
dx[itex]\mu[/itex]dx[itex]\nu[/itex][∇[itex]\nu[/itex] , ∇[itex]\mu[/itex]]V
Now we know that the commutator is simply the Riemann tensor. My question here is:
How do you actually apply that vector V to the Riemann tensor? Here is an example of what I mean:
I know that the covariant derivative ∇j(Viei) = [(∂Vi)/(∂xj) + [itex]\Gamma[/itex]ikjVk]ei
Now having established this, would I distribute the vector Vi to the terms of my Riemann tensor or would I multiply the terms of the tensor by Vk?I also have one more question:
What extra information does the Riemann tensor tell you that the Ricci tensor does not (from a physical standpoint)? I know that the Ricci is a contraction of the Riemann and that the Riemann has a lot more elements than the Ricci, but what physical meaning do all of those other terms possesses that is not present in the Ricci tensor?