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Proving the weyl tensor is zero problem

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that all Robertson - Walker models are conformally flat.


    2. Relevant equations
    Robertson Walker Metric: [itex]ds^{2}=a^{2}(t)\left(\frac{dr^{2}}{1-Kr^{2}}+r^{2}(d\theta^{2}+(sin\theta)^{2}d\phi^{2} )\right)-dt^{2}[/itex]

    Ricci Tensor: [itex]R_{\alpha\beta}=2Kg_{\alpha\beta}[/itex]

    Ricci Scalar: [itex]R=8K[/itex]

    Weyl Tensor: [itex]C_{\alpha\beta\gamma\delta}=R_{\alpha\beta\gammaδ}-\frac{1}{2}(g_{\alpha\gamma}R_{\beta\delta}-g_{\alpha\delta}R_{\beta\gamma}-g_{\beta\gamma}R_{\alpha\delta}+g_{\beta\delta}R_{\alpha\gamma}) + \frac{R}{6}(g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma})[/itex]

    3. The attempt at a solution
    In order for the models to be conformally flat the weyl tensor must vanish, therefore that is what I have tried to show. By subbing in the values for the Ricci tensor and the Ricci scalar (both of which were given in a lecture by my professor) I arrived at the following expression:

    [itex]C_{\alpha\beta\gammaδ}=R_{\alpha\beta\gamma\delta}-\frac{2}{3}K(g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma})[/itex]

    However, as you can see I am left with the Riemann tensor undefined and I cannot show the weyl tensor to be zero. Any help is greatly appreciated, thanks!
     
  2. jcsd
  3. Oct 9, 2012 #2
    I'm pretty sure your textbook contains the definition of the Riemann tensor, and of the other tensors. Look it up. (You will only need to know the metric to calculate it.) Then, the Ricci tensor is related to the Riemann one through
    [itex]R_{ij} = R^k_{\, ikj} [/itex],
    and, finally, the Ricci scalar is simply the trace of the Ricci tensor.

    So well, I'd recommend you start working from the definitions. It is probably more instructive to derive those forms of the Ricci tensor and scalar yourself, and see if you get the same forms for this specific metric.
     
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