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Determinant g = g_{00} det |g_{ij}|

  1. Jun 27, 2006 #1
    If we let latin alphabets {i,j,k,...} denote the spatial indices, and the greek ones run from 0 to 3, then I've seen the following

    [tex]det| g_{\mu \nu} | = g_{00} det | g_{ij} |[/tex]

    both in Landau's "Classical theory of fields", as well as ADM's paper on the initial value formulation of GR ("Dynamics of General Relativity", Arnowitt et al.) I don't quite understand it though, or am I reading it right?

    Specifically, what about the terms that involve [itex]g_{01}, g_{02}, g_{03}[/itex] when we do the co-factor expansion? Do they somehow vanish by symmetry?

    In (2+1) dimensions I got the determinant to be

    [tex]det| g_{\mu \nu} | = g_{00} det | g_{ij} | - g_{10}(g_{01} g_{22} - g_{02} g_{21}) + g_{20} (g_{01} g_{12} - g_{11} g_{02})[/tex]

    How could the last four terms cancel out?
     
  2. jcsd
  3. Jun 27, 2006 #2

    pervect

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    [itex]g_{\mu\nu}[/itex] can be any symmetric matrix.

    If we try for an example

    [tex]
    \left[ \begin {array}{cccc} m&1&2&3\\\noalign{\medskip}1&4&5&6\\\noalign{\medskip}2&5&7&8\\\noalign{\medskip}3&6&8&9\end {array}
    \right]
    [/tex]

    we find the determinant is -m-2, which is not a multiple of m.

    So I don't think your statement can be correct as written. Start reading the fine print :-).

    (Is the ADM paper online anywhere?)
     
  4. Jun 28, 2006 #3
    It is equation 3.12 of "The Dynamics of General Relativity", Arnowitt, Deser and Misner:

    http://arxiv.org/abs/gr-qc/0405109

    Perhaps someone could explain to me if I am mis-reading it.
     
  5. Jun 28, 2006 #4

    HallsofIvy

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    That is correct assuming that the metric tensor is of the form
    [tex]\left[ \begin {array}{cccc} g_{00}&0&0&0\\\noalign{\medskip}0&g_{11}&g_{12}&g_{13}\\\noalign{\medskip}0&g_{21}&g_{22}&g{23}\\\noalign{\medskip}0&g_{31}&g_{32}&g_{33}\end {array} \right] [/tex], the "three plus one dimensional decomposition of the Einstein field" referred to in the text.
     
  6. Jun 28, 2006 #5

    George Jones

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    When a metric is static, coordinates can be chosen such that the components of the metric take the form given by HallsofIvy.

    I don't think equations (3.10) and (3.12) of the ADM paper give this.
     
  7. Jul 1, 2006 #6
    I have indeed mis-read the ADM paper. What it says is

    [tex]g^{00} = \frac{\textrm{det} |g_{ij}|}{\textrm{det} |g_{\mu \nu}|}[/tex]

    where [itex]\textrm{det} |g_{ij}|[/itex] is the determinant of the space-space part of the metric and [itex]\textrm{det} |g_{\mu \nu}|[/itex] is the determinant of the whole metric tensor.

    This follows from Cramer's rule.
     
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