- #1
Nabigh R
- 11
- 0
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} ##
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} ##