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Relation between det(spacetime metric) and det(spatial metric)

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  1. Sep 19, 2013 #1
    I have a metric g on spacetime and a spatial metric ##\gamma## such that the components of g can be written in matrix form as

    $$ g_ {\alpha, \beta} = \begin{pmatrix} g_{00} & g_{0 j} \\
    g_{i 0} & \gamma_{ij} \end{pmatrix} $$

    where ##i,j = 1,2,3## and ##\alpha = 0,1,2,3##. Now I want to find a relation between the determinant of ##g## and the determinant of ##\gamma## expressed in terms of the components of ##g##. Using Cramer's rule I get

    $$ det(g_{\alpha, \beta}) \equiv g = \frac{C_{00}}{g^{00}}$$

    where ##C_{00}## is the cofactor of the (0,0)-element of the matrix above; i.e. it is ##det(\gamma_{ij}) = \gamma##. But in order to find the full relation I will still need to find the (0,0) element of the inverse. What is the best way to go about doing this? Do I really have to solve the full set of equations ##g^{\alpha \lambda}g_{\lambda \beta} = \delta^{\alpha}_{\ \ \ \beta}##? Or is there a better way?
     
  2. jcsd
  3. Sep 19, 2013 #2
    I'm assuming you mean ##\displaystyle \mathrm{g}_{\alpha,\beta}=\begin{bmatrix} g_{0,0} & g_{0,1} & g_{0,2} & g_{0,3} \\ g_{1,0} & \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ g_{2,0} & \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ g_{3,0} & \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}##, right?

    I'm probably misinterpreting something here, but I'm not seeing how you're applying Cramer's Rule.

    If we were talking about finding the determinant of the given matrix, you'd get $$\det(\mathrm{g}_{\alpha,\beta})=g_{0,0}~\det\begin{bmatrix}\gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}-g_{1,0}~\det\begin{bmatrix} g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}+g_{2,0}~\det\begin{bmatrix} g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}-g_{3,0}~\det\begin{bmatrix}g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3}\end{bmatrix}$$ by Laplace Expansion.

    Given, I'm not really knowledgeable in general relativity (or differential geometry for that matter :tongue:) so I could be mistaking this determinant for something else, or misunderstanding your notation.
     
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