Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding inverse metric tensor when there are off-diagonal terms

  1. Jan 25, 2014 #1
    How do you find the inverse of metric tensor when there are off-diagonals?
    More specifivally, given the (Kerr) metric,
    $$ d \tau^2 = g_{tt} dt^2 + 2g_{t \phi} dt d\phi +g_{rr} dr^2 + g_{\theta \theta} d \theta^2 + g_{\phi \phi} d \phi^2 + $$
    we have the metric tensor;
    $$ g_{\mu \nu} = \begin{pmatrix}
    g_{tt} & 0 & 0 & g_{t \phi} \\
    0 & g_{rr} & 0 & 0 \\
    0 & 0 & g_{\theta \theta} & 0 \\
    g_{\phi t} & 0 & 0 & g_{\phi \phi} \\
    \end{pmatrix} $$
    In "A First Course in General Relativity", Schutz say that since the only off-diagonal element involves ##t## and ##\phi##, the contravariant components ##g^{rr}## and ##g^{\theta \theta}## are given by ##g^{rr} = \frac{1}{g_{rr}}## and ##g^{\theta \theta} = \frac{1}{g_{\theta \theta}}##. And then invert the matrix
    \begin{pmatrix}
    g_{tt} & g_{t \phi} \\
    g_{\phi t} & g_{\phi \phi} \\
    \end{pmatrix}
    to find ##g^{tt}##, ##g^{\phi t}## and ## g^{\phi \phi} ##. I don't get why we can do that. Is it some kind on generalised version for the inverse of a diagonal matrix.
    If ## A = \begin{pmatrix}
    a_{11} & 0 & 0 & 0 \\
    0 & a_{22} & 0 & 0 \\
    0 & 0 & a_{33} & 0 \\
    0 & 0 & 0 & a_{44} \\
    \end{pmatrix} ##
    then ## A^{-1} = \begin{pmatrix}
    \frac{1}{a_{11}} & 0 & 0 & 0 \\
    0 & \frac{1}{a_{22}} & 0 & 0 \\
    0 & 0 & \frac{1}{a_{33}} & 0 \\
    0 & 0 & 0 & \frac{1}{a_{44}} \\
    \end{pmatrix} ##
     
  2. jcsd
  3. Jan 25, 2014 #2

    Mentz114

    User Avatar
    Gold Member

  4. Jan 25, 2014 #3
    Thanks Mentz :-D I know I can get ##g^{\mu \nu}## by inverting the matrix representation of ##g_{\mu \nu}##. But what I want to know is reasoning Schutz used to simplify the problem of finding the inverse of a ##4 \times 4## matrix to that of finding the inverse of a ##2 \times 2## matrix :-)
    If there is a general theorem or something that allows it, then it sure can save me a lot of work.
     
  5. Jan 25, 2014 #4

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    It's perhaps easier to see by writing the coordinates in a different order:
    $$ g_{\mu \nu} = \begin{pmatrix}
    g_{tt} & g_{t \phi} & 0 & 0 \\
    g_{\phi t} & g_{\phi \phi} & 0 & 0 \\
    0 & 0 & g_{\theta \theta} & 0 \\
    0 & 0 & 0 & g_{rr} \\
    \end{pmatrix} $$
    and the observation that if you can decompose a matrix into smaller sub-matrices
    $$ \textbf{A} = \left( \begin{array}{c|c}
    \textbf{P} & \textbf{0} \\
    \hline
    \textbf{0} & \textbf{Q}
    \end{array} \right) $$
    then it inverts as
    $$ \textbf{A}^{-1} = \left( \begin{array}{c|c}
    \textbf{P}^{-1} & \textbf{0} \\
    \hline
    \textbf{0} & \textbf{Q}^{-1}
    \end{array} \right) $$
    (then finally put the coordinates back into the original order).
     
  6. Jan 25, 2014 #5
    Thanks a lot Greg. That's just what I was looking for. Just saw blockwise inversion theorem of matrices on Wikipedia. Since it didn't occur me to change the order of coordinates, I didn't make the connection. Thanks again :approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook