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Numerical Relativity: Components of the Lapse Function?

  1. Oct 16, 2009 #1


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    I have a technical question about numerical relativity, hopefully someone can help.

    In the usual 3 + 1 decomposition in NR, the four gauge freedoms are expressed via the lapse, [tex]\alpha[/tex] and three shift components [tex] \beta^i [/tex]. In a finite element numerical scheme, each grid point will have a value for these 4 components.

    Now, looking at evolution equations on the other hand, I see terms that want to take the covariant derivative of the lapse function, which I can't understand since it is a scalar, and hence the covariant derivative reduces to the regular derivative.

    To be explicit, take for instance the ADM evolution equation for the extrinsic curvature. The textbook I have ("Elements of Numerical Relativity" by Carles Bona and Carlos Palenzuela-Luque), write this down as

    [tex] (\partial_t - L_{\beta} ) K_{ij} = -\alpha_{j;i} + \alpha [ ....] [/tex]

    I've left the rest of the equation out for simplicity (note that [tex]L_{\beta}[/tex] is the Lie Derivative, I couldn't work out how to make the nice curly L with the tex tags). See that both [tex]\alpha[/tex] and [tex]\alpha^i[/tex] appear which I don't understand. It would make sense if this book was using [tex]\alpha^i[/tex] to denote the shift vector, but as you can see from the LHS (and it made clear in the book) it uses [tex]\beta^i[/tex] for this.

    Any ideas? I'm just replacing the Covariant derivative with the regular one for [tex]\alpha[/tex] but maybe the equation is actually telling me something very different?
  2. jcsd
  3. Oct 16, 2009 #2


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    I think I found the answer, comparing to other literature it looks like they were meaning

    [tex]\alpha_i \equiv \partial_i \alpha[/tex]

    that makes sense, because the covariant derivative of that would not be 'trivial', i.e. reduce simply to the regular derivative.
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