- #1
Andrew Kim
- 12
- 6
What 20 index combinations yield Riemann tensor components (that are not identically zero) from which the rest of the tensor components can be determined?
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##
Andrew Kim said: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.
The Riemann tensor is a mathematical object used in differential geometry to describe the curvature of a manifold. It is named after the mathematician Bernhard Riemann and is a fundamental concept in Einstein's theory of general relativity.
The Riemann tensor has 20 independent components in 4 dimensions. These components describe the curvature of space-time in terms of the gravitational field. They are used to calculate the gravitational force between masses and to understand the behavior of matter in the presence of curved space-time.
The Riemann tensor is calculated using the Christoffel symbols, which are derived from the metric tensor. The metric tensor describes the distance between points in a manifold, and the Christoffel symbols describe how this distance changes as you move along different paths in the manifold. These symbols are then used to calculate the Riemann tensor.
The Riemann tensor is of great significance in physics, particularly in Einstein's theory of general relativity. It is used to describe the curvature of space-time and how this curvature affects the behavior of matter and energy. It is also used in other areas of physics, such as quantum field theory and string theory.
Yes, there are several real-world applications of the Riemann tensor. It is used in the study of black holes, gravitational waves, and the behavior of matter in the early universe. It is also used in engineering, particularly in the design of space vehicles and satellites. Additionally, the Riemann tensor has applications in computer science, such as in image processing and pattern recognition.