1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Riemann tensor given the space/metric

  1. May 25, 2016 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant 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}##

    3. 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: May 25, 2016
  2. jcsd
  3. May 25, 2016 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Your metric tensor is not of the given form, the off diagonal components are linear and not quadratic in the coordinates.
     
  4. May 25, 2016 #3
    Yes, that is why I'm wondering. The book stated that the first space is flat and the second space has a nonzero curvature.
     
  5. May 26, 2016 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This just means you have to find you curvature tensor through other means, not that it is impossible to compute it.
     
  6. May 26, 2016 #5
    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.
     
  7. May 26, 2016 #6
    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)!
     
  8. May 27, 2016 #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.
     
  9. May 27, 2016 #8
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Riemann tensor given the space/metric
  1. Riemann Tensor (Replies: 30)

  2. Metric Tensor (Replies: 5)

Loading...