• Support PF! Buy your school textbooks, materials and every day products Here!

Riemann tensor given the space/metric

  • #1
132
4

Homework Statement


Given two spaces described by
##ds^2 = (1+u^2)du^2 + (1+4v^2)dv^2 + 2(2v-u)dudv##
##ds^2 = (1+u^2)du^2 + (1+2v^2)dv^2 + 2(2v-u)dudv##

Calculate the Riemann tensor

Homework Equations


Given the metric and expanding it ##~~~g_{τμ} = η_{τμ} + B_{τμ,λσ}x^λx^σ + ...##
We have the Riemann tensor, ##R_{τρμν} = B_{τν,ρμ} + B_{ρμ,τν} - B_{ρν,τμ} - B_{τμ,ρν}##

The Attempt at a Solution


I'm having problems calculating the B's for the above spaces.
There is also one space given by ##ds^2 = (1+(ax+cy)^2)dx^2 + (1+(by+cx)^2)dy^2 + 2(ax+cy)(by+cx)dxdy## where I know how to compute the B's. Since in 2D, the Riemann tensor has only one component, that is ##R_{1212}##. For ##g_{11}## we have ##B_{1122} = c^2##, for ##g_{22}## we have ##B_{2211} = c^2##, for ##g_{12} = g_{21}## we have ##B_{1212} = ½(ab + c^2)##

For the above spaces, it seems every B is 0. I can't seem to understand why.
 
Last edited:

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,446
6,251
Your metric tensor is not of the given form, the off diagonal components are linear and not quadratic in the coordinates.
 
  • #3
132
4
Your metric tensor is not of the given form, the off diagonal components are linear and not quadratic in the coordinates.
Yes, that is why I'm wondering. The book stated that the first space is flat and the second space has a nonzero curvature.
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,446
6,251
Yes, that is why I'm wondering. The book stated that the first space is flat and the second space has a nonzero curvature.
This just means you have to find you curvature tensor through other means, not that it is impossible to compute it.
 
  • #5
132
4
This just means you have to find you curvature tensor through other means, not that it is impossible to compute it.
I tried finding the geodesic equation and read off the Christoffel symbol then derive the Riemann tensor through the Christoffel symbol. It worked but it's insanely long, that is why I'm thinking of another way to do it.
 
  • #6
17
7
In a 2 dimensional (semi)-Riemannian manifold there is only ##\mathbf{one}## linearly independent component of the Riemann tensor (I think in Wald's General Relativity there's a proof of this). That should make it easier (##\simeq## shorter)!
 
  • #7
132
4
In a 2 dimensional (semi)-Riemannian manifold there is only ##\mathbf{one}## linearly independent component of the Riemann tensor (I think in Wald's General Relativity there's a proof of this). That should make it easier (##\simeq## shorter)!
Yes I already stated that in my first post. My question is the method that I've done in post#1 is not working but I've done it in another way (post#5). Maybe there is another way.
 
  • #8
17
7
Yes I already stated that in my first post. My question is the method that I've done in post#1 is not working but I've done it in another way (post#5). Maybe there is another way.
Oh, sorry, it seems i missed that.
You can also compute the components of ##R^{a}{}_{bcd}## in a orthonormal frame, via the connection one-forms and Cartan's structure equations.
 

Related Threads for: Riemann tensor given the space/metric

Replies
0
Views
3K
Replies
4
Views
4K
Replies
4
Views
760
Replies
2
Views
623
Replies
2
Views
582
Replies
0
Views
706
Replies
1
Views
8K
  • Last Post
2
Replies
42
Views
6K
Top