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I am trying to find the inverse of the metric

$$\mathrm ds^2 = \rho^2~\mathrm d\theta^2 -2a\sin^2\theta ~\mathrm dr~d\varphi + 2~\mathrm dr~\mathrm du + \rho^{-2}\left[\left(r^2+a^2\right)^2 -\Delta a^2\sin^2\theta\right]\sin^2\theta~\mathrm d\varphi^2-2a\rho^{-2}\left(2mr-e^2\right)\sin^2\theta~\mathrm d\varphi~\mathrm du-\left[1-\rho^{-2}\left(2mr-e^2\right)\right]~\mathrm du^2 $$

The inverse was given in Global Structure of the Kerr Family of Gravitational Fields as

$$(\partial/\partial s)^2 = \rho^{-2}(\partial/\partial\theta)^2+ 2\rho^{-2}\left(r^2+ a^2\right)(\partial/\partial r)(\partial/\partial u)+ 2\rho^{-2}a(\partial/\partial r)(\partial/\partial\varphi)+ 2\rho^{-2}a(\partial/\partial u)(\partial/\partial \varphi)+ \rho^{-2}\sin^2\theta(\partial/\partial u)^2 + \rho^{-2}\sin^2\theta(\partial/\partial \varphi)^2+ \rho^{-2}\Delta(\partial/\partial r)^2$$

I know how to find the inverse of the metric $g_{\mu\nu}$ but they seem to give different components to the components presented in the inverse metric. Does anyone know what method was used to find the inverse?

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  • $\begingroup$ This is the inverse of the metric as to find the geodesics. If you read the paper carefully, you will see that the Hamiltonian of geodesics is being derived, which has the form: $H = \frac{1}{2}g^{ab} p_{a} p_{b}$, where $g^{ab}$ is the inverse metric. $\endgroup$
    – Dr. Ikjyot Singh Kohli
    Commented Sep 16, 2016 at 16:50
  • $\begingroup$ Do you mean that you calculated the inverse metric and you got a different result? In this case, you should probably show us how you did it. $\endgroup$
    – DelCrosB
    Commented Sep 16, 2016 at 16:53
  • $\begingroup$ @IkjyotSinghKohli, What is the equation used to find the inverse metric? $\endgroup$
    – gbd
    Commented Sep 16, 2016 at 17:31
  • $\begingroup$ @gbd Hi. Since the Kerr metric has a non-diagonal cross-term, it is slightly more complicated than the usual diagonal metric cases. However, it is a common problem, see: roma1.infn.it/teongrav/VALERIA/TEACHING/… specifically after Eq. 19.10 $\endgroup$
    – Dr. Ikjyot Singh Kohli
    Commented Sep 16, 2016 at 17:47
  • $\begingroup$ Watch out. that Valeria writeup uses the Lindquist metric, not the same the OP is using, which is harder (less block symmetric) $\endgroup$
    – Bob Bee
    Commented Sep 16, 2016 at 20:51

1 Answer 1

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It's

$g_{ab}g^{bc} = \delta_a^c$

Where repeated indexes are summed, and $\delta$ is the Kronecker delta function

In abstract notation it is $g * g^{-1} = I$

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    $\begingroup$ What I did is I took all the components from the first metric then inserted them in a matrix and found the inverse of that matrix. Is this the correct method? $\endgroup$
    – gbd
    Commented Sep 16, 2016 at 17:44
  • $\begingroup$ @gbd Yes. That is the common thing to do. $\endgroup$
    – Dr. Ikjyot Singh Kohli
    Commented Sep 16, 2016 at 17:47
  • $\begingroup$ @IkjyotSinghKohli, I tried the method you suggested but I still get the wrong result. $\endgroup$
    – gbd
    Commented Sep 16, 2016 at 18:50
  • $\begingroup$ @gbd okay. I will look at it a bit later when I have some more time... $\endgroup$
    – Dr. Ikjyot Singh Kohli
    Commented Sep 16, 2016 at 18:51
  • $\begingroup$ I think you should explain how you inverted the matrix. There must be a mistake in what you do since there is no other way to solve your problem than actually calculate the inverse matrix. $\endgroup$
    – DelCrosB
    Commented Sep 16, 2016 at 19:06

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