How do you find the inverse of metric tensor when there are off-diagonals?(adsbygoogle = window.adsbygoogle || []).push({});

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} ##

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# Finding inverse metric tensor when there are off-diagonal terms

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