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Ricci curvature

  1. May 22, 2014 #1
    Ansatz metric of the 4 dimensional spacetime:

    [itex]ds^2=a^2 g_{ij}dx^i dx^j + du^2[/itex] (1)

    where:

    Signature: [itex]- + + +[/itex]

    Metric [itex]g_{ij} \equiv g_{ij} (x^i) [/itex] describes 3 dimensional AdS spacetime

    [itex]i,j = 0,1,2 = 3[/itex] dimensional curved spacetime indices

    [itex]a(u)=[/itex] warped factor

    [itex]u = x^D = x^3[/itex]

    [itex]D = 3 =[/itex] number of spacial dimensional


    Now I have to proof that

    [itex]R_{ij} = [(\frac{a'}{a})' + 3 (\frac{a'}{a})^2 - \frac{\Lambda_3}{a^2}] a^2 g_{ij}[/itex]

    [itex]R_{33} = -3 [(\frac{a'}{a})' + (\frac{a'}{a})^2][/itex]

    [itex]R = 6 [(\frac{a'}{a})' + 2 (\frac{a'}{a})^2] - \frac{3\Lambda_3}{a^2}[/itex]

    where

    [itex]R_{ij} =[/itex] the Ricci curvature of metric (1)

    [itex]R =[/itex] the Ricci scalar of metric (1)

    [itex]a' = \frac{∂a}{∂u}[/itex]


    My steps to calculate [itex]R_{ij}[/itex]:

    • calculating [itex]R_{\mu\nu}[/itex], where [itex]\mu,\nu = 0,1,2,3 = 4[/itex] dimensional curved spacetime indices

    • finding that [itex]R_{\mu\nu} = ... R_{ij} [/itex] (failed)

    • Subtituting [itex]R_{\mu\nu} = ... R_{ij}[/itex] to [itex]R_{\mu\nu} = \Lambda_D g_{\mu\nu} [/itex] (failed)

    Furthermore I can't find [itex]R_{33}[/itex] and [itex]R[/itex]


    What are the right steps to find [itex]R_{ij}, R_{33}, R[/itex] ?
     
  2. jcsd
  3. May 23, 2014 #2

    Matterwave

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    The trouble you are probably running into (I can't say for sure since I can't see all of your calculations) is that in order to relate the 4-dimensional curvatures to the 3-dimensional curvatures, terms with the extrinsic curvature of the submanifold will begin to appear.

    For example, see the relations of Gauss and Codacci (one of which basically tell you that the 4-dim Riemann tensor restricted to the hypersurface is equal to the 3-dim Riemann tensor plus some extrinsic curvature terms). You don't want to calculate the 4-dimensional Ricci tensor and try to just naively restrict it to the 3-surface and assume that you have arrived at the 3-dimensional Ricci tensor. You should calculate the 3-dimensional Ricci tensor separately, using the covariant derivative defined on your hypersurface.
     
  4. May 23, 2014 #3
    Oh okay, but I've calculated both the 4-dimensional Ricci tensor and the 3-dimensional Ricci tensor separately. Here is my calculation:
     

    Attached Files:

  5. May 23, 2014 #4

    Matterwave

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    I'm going to guess, from the forms of the equations that you have to prove...that this problem would probably be a lot easier to approach from using Einstein's field equations.

    I also just realized that I sort of just assumed you wanted the 3-D Ricci tensor on your hypersurface, but maybe you really do just want the 3x3 sub components of the 4-D Ricci tensor? You were not explicit in defining ##R_{ij}##
     
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