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Curvature Scalar in 2-d

  1. Apr 3, 2013 #1
    Where should I start from to show that curvature scalar (RiemannScalar) is

    2[itex]\frac{R_1212}{det (g_μ√)}[/itex]

    ?
     
  2. jcsd
  3. Apr 3, 2013 #2
    I am asked to show it by the symmetries of the Riemann tensor by the way.
     
  4. Apr 3, 2013 #3
    Ok here is my thoughts, I try to stay away from the connection coefficients. So, I don't write the R tensors in form of [itex]\Gamma[/itex]'s. So I am trying:

    R= g[itex]^{αβ}[/itex] R[itex]_{αβ}[/itex]
    = g[itex]^{αβ}[/itex] R[itex]^{c}[/itex][itex]_{αcβ}[/itex]
    = g[itex]^{αβ}[/itex] g[itex]^{αb}[/itex] R[itex]_{bαcβ}[/itex]

    but now I can't have the R[itex]_{αβαβ}[/itex] form. Since it is 2-d, I put α=1 and β=2, but the c and b contractions doesn't give me what I want.

    How can get that 2R[itex]_{1212}[/itex] ?
     
    Last edited: Apr 3, 2013
  5. Apr 3, 2013 #4

    Bill_K

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    Science Advisor

    remember that gαβ is the inverse matrix of gαβ. And it's easy to take the inverse of a 2 x 2 matrix.

    Write out completely the individual components of gαβ in terms of the gαβ components and see what you get.
     
  6. Apr 3, 2013 #5

    pervect

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    Staff Emeritus
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    My thoughts are to start by looking at the only completely anti-symmetric 2-form , which must have components

    [tex]
    \left[ \begin{array}{cc}
    0 & R \\
    -R & 0 \\
    \end{array} \right]
    [/tex]

    Now we know that R_abcd = 0 if a=b or c=d by the anti-symmetry properties, and also that R_abcd = R_bacd and that R_abcd = -R_abdc

    This, and a little thought, gives us the value of all the components of R, which can be described as an anti-symmetric 2d array of two-forms, i.e. it looks like the array above, but the members of the array are the anti-symmetric two-forms.

    Next we just have to compute the contractions to get the Riemann tensor and scalar, which I'm too lazy to do by hand.
     
  7. Apr 3, 2013 #6
    Thank you very much.

    Yes I rewrited everything with g lower index.

    I am not sure about that symmetry:

    R[itex]^{αβ}[/itex][itex]_{αβ}[/itex]= - R[itex]^{βα}[/itex][itex]_{βα}[/itex]

    But I feel I am close to it. Thank you again.
     
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