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Second fundamental form and Mean Curvature

  1. Jun 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Metric ansatz:
    [itex]
    ds^{2} = e^{\tilde{A}(\tilde{\tau})} d\tilde{t} - d\tilde{r} - e^{\tilde{C}(\tilde{\tau})} dΩ
    [/itex]

    where: [itex]d\tilde{r} = e^{\frac{B}{2}} dr[/itex]

    2. Relevant equations

    How to calculate second fundamental form and mean curvature from this metric?

    3. The attempt at a solution

    Metric tensor:

    [itex]g_{00}= e^{\tilde{A}(\tilde{\tau})} [/itex]
    [itex]g_{11}= -1 [/itex]
    [itex]g_{22}= - e^{\tilde{C}(\tilde{\tau})} [/itex]
    [itex]g_{33}- e^{\tilde{C}(\tilde{\tau})} sin^2 θ [/itex]

    Second fundamental form:

    [itex]h_{ij}= g_{kl} \Gamma^{k}_{ij} n^{l}[/itex]

    where:

    [itex]i, j, k, l = 0, 1, 2, 3[/itex]

    [itex]n^{l} = [/itex]normal vector [itex]= (0, 1, 0, 0)[/itex]

    so:

    [itex]n^{0} = 0[/itex]
    [itex]n^{1} = 1[/itex]
    [itex]n^{2} = 0[/itex]
    [itex]n^{3} = 0[/itex]

    Second fundamental form:

    [itex]h_{ij}= diag (\frac{1}{2}e^{\tilde{A}}\tilde{A'}, 0, \frac{1}{2}e^{\tilde{C}}\tilde{C'}, sin {θ} cos {θ}) [/itex]

    Mean curvature:

    [itex]h = g^{ij}h_{ij}= \frac{1}{2}\tilde{A'}-\frac{1}{2}\tilde{C'}-e^{-\tilde{C}}cot {θ}[/itex]
     
    Last edited: Jun 14, 2013
  2. jcsd
  3. Jun 17, 2013 #2
    Your idea seems correct, but I don't understand what you're doing. Is there any r-dependence in the metric? If not, then all the relevant Christoffel symbols vanish and the extrinsic curvature should be zero (all r=const hyperslices are flat)
     
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