What is the Riemann Curvature Tensor for Flat and Minkowski Space?

AI Thread Summary
All components of the Riemann curvature tensor are zero for flat and Minkowski space due to the constancy of the metric components. The Christoffel symbols, which are derived from the metric, vanish identically in these spaces. This leads to the conclusion that the Riemann curvature tensor, calculated using the standard formula, is also zero. The discussion emphasizes the straightforward nature of this problem in differential geometry. Overall, the key takeaway is that flat and Minkowski spaces exhibit no curvature.
Elliptic
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Homework Statement



Show that all components of Riemann curvarue tensor are equal to zero for flat and Minkowski space.

Homework Equations


The Attempt at a Solution


## (ds)^2=(dx^1)^2+(dx^2)^2+...+(dx^n)^2 \\
R_{MNB}^A=\partial _{N}\Gamma ^{A}_{MB}-\partial _{M}\Gamma ^{A}_{NB}+\Gamma ^{A}_{NC}\Gamma_{MB}-\Gamma ^{C}_{NB}\Gamma_{MC} ##
 
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It's a straightforward problem. What are the Christoffel symbols for Minkowski space-time in the standard coordinates?
 
##
\Gamma^{K}_{MK}=\frac{1}{2}\left(\partial_Mg_{ML}+\partial_Ng_{ML}-\partial_Lg_{MN} \right ) ##
 
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Yes but can you tell me why?
 
Elliptic said:
##
\Gamma^{K}_{MK}=\frac{1}{2}\left(\partial_Mg_{ML}+\partial_Ng_{ML}-\partial_Lg_{MN} \right ) ##
Ok but you don't need to edit your posts, you can just reply to my subsequent posts (it makes it easier to keep track of who's saying what). Ok so you know the Christoffel symbols vanish identically because the metric components are constant. So what does that say about the Riemann curvature tensor based on the usual formula?
 
That Riemann curvarue tensor is equal to zero?
 
Yeahp that's it! :)
 
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Thanks. I have another problem, but i must respect the rules of forum and post a new thread.
 
Alrighty :)
 
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