Andrew Kim said:
I'm not doing these as homework problems. I was just inspired by a question (that I answered correctly, thank you very much) asking how many independent components there were. Here's how I solved the problem while users on this thread were answering me with unhelpful questions or giving me hints instead of the whole answer. I was working with the coordinates ##{t,r,\theta,\phi}## in an arbitrary metric in GR, the only restriction being that the space time was static and spherically symmetric.
I knew I could list all of the components by starting with ##R_{tttt}## and then treating each index like a digit in base 4, and I realized that the first one that wasn't identically zero was ##R_{trtr}##. ##R_{trt\theta}## and ##R_{trt\phi}## followed.
I then calculated ##R_{trr\alpha}##, starting with
##R_{trr\theta}## since I already knew ##R_{trrt}## by antisymmetry in the first pair of indices. I continued to calculate ##R_{tr\alpha\beta}## with the following rule:
1. Given an ##\alpha##, the first independent component is given by ##\beta = \alpha + 1##.
Then I began the set of components ##R_{t\theta\alpha\beta}##. Because of the symmetry of the pairs of indices, I had already calculated ##R_{t\theta tr}##, so I started those calculations with ##R_{t\theta t\theta}##.
I then established the next 2 rules of calculation:
2. For ##R_{\mu\nu\alpha\beta}##, given a ##\mu,\nu##, where ##\nu## but not ##\mu## has just been incremented, the first independent component has indices given by ##\alpha=\mu,\beta=\nu##, since all the previous components are either identically zero or have been calculated before by symmetry in the pairs of indices.
3. For ##R_{\mu\nu\alpha\beta}##, given a ##\mu## that has just been incremented, the first independent component has indices given by ##\alpha=\mu,\beta=\nu=\mu+1##, since all the previous components are either identically zero or have been calculated before by symmetry in the pairs of indices.
These rules reduced my calculations to 21 independent components (I never used the bianchi identity, since it would only eliminate 1 calculation, which is nothing compared to the previous 20). Rule 1 tells us how to increment the third index. Rule 2 tells us how to increment the second. Rule 3 tells us how to increment the third. The independent components, and the rules that yielded each one, were:
##trtr## (Rule 3)
##trt\theta##
##trt\phi##
##trr\theta## (Rule 1)
##trr\phi##
##tr\theta\phi## (Rule 1)
##t\theta t\theta## (Rule 2)
##t\theta t\phi##
##t\theta r\theta## (Rule 1)
##t\theta r\phi##
##t\theta \theta\phi## (Rule 1)
##t\phi t\phi## (Rule 2)
##t\phi r\theta## (Rule 1)
##t\phi r\phi##
##t\phi\theta\phi## (Rule 1)
##r\theta r\theta## (Rule 3)
##r\theta r\phi##
##r\theta \theta\phi## (Rule 1)
##r\phi r\phi## (Rule 2)
##r\phi\theta\phi## (Rule 1)
##\theta\phi\theta\phi## (Rule 3)
For any readers interested in my application of these, I used the following metric for a spherically symmetric, static space time in GR:
##ds^2 = -e^{2\Phi}dt^2+e^{2\lambda}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2## for arbitrary functions ##\Phi (r), \lambda (r)##
And got the following components of the Riemann tensor:
##R^t_{rtr}=-\Phi'' + \Phi'(2\lambda'-1)-(\Phi')^2##
##R^t_{\theta t\theta}=-r\Phi'e^{-2\lambda}##
##R^t_{\phi t\phi}=R^t_{\theta t\theta}\sin^2 \theta##
##R^r_{\theta r\theta} = re^{-2\lambda}##
##R^r_{\phi r\phi}=R^r_{\theta r\theta}\sin^2 \theta##
##R^{\theta}_{\phi\theta\phi}=\Big(\dfrac{\cot\theta}{r}-\csc^2\theta-e^{-2\lambda}+2\Big)\sin^2\theta##
You can think of this as a symmetric 6x6 matrix, if you regard the pairwise entries (t,r) (t,##\theta##), (t,##\phi##), (r,##\theta##), (r,##\phi##), (##\theta##,##\phi##) as labelling the rows and columns of the matrix. With a bit of adjustment in notation and the wedge product (denoted by ^), you can regard these pairs as bivectors: notationally we replace (t,r) with t^r. So you can regard the Riemann as a 6x6 symmetric matrix of bivectors.
This is 36 components, which have 6 unique diagonal elements, and 6*5 off-diagional elements which are duplicated twice, because the matrix is symmetric. This gives the total of 21.
It's also helpful to note that there is a duality relationship between ##t \wedge r## and ##\theta \wedge \phi##, the hodge dual relationship, which breaks the six bivectors into two sets of three. This eventually gives you a very useful breakdown of the Riemann sometimes known as the Bel decomposition, but you'll find a description of it in, say, MTW's "Gravitation" without using this name. The end result is that you can regard the Riemann as a set of three 3x3 matrices, as in the diagram below:
$$ \begin{bmatrix} E & H \\ H & T \end{bmatrix} $$
Given that you've singled out a time direction by ##\hat{t}##, these matrices are essentially spatial, represented by the ##\hat{x} \wedge \hat{t}, \hat{y} \wedge \hat{t}, \hat{z} \wedge \hat{t}##, and their hodge duals. This is especially useful in an orthonormal basis, which I'm not sure if you're familiar with.
E and T are both symmetric - six components each. H has no special symmetry, so it has 9 components. Physically, E represents the "Electrogravitic" part of tensor, which represents static tidal forces. H represents the "Magnetogravitic" part of the tensor, it represents the magnetogravitic forces, which can be regarded as being linearly proportional to velocity. T is the "Topogravitic" part of the tensor, it can be regarded as representing the spatial part of the space-time curvature.
