2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Sep 16 '16 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 Sep 16 '16 at 16:53
  • $\begingroup$ @IkjyotSinghKohli, What is the equation used to find the inverse metric? $\endgroup$ – gbd Sep 16 '16 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 Sep 16 '16 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 Sep 16 '16 at 20:51
4
$\begingroup$

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 $g * g^{-1} = I$

$\endgroup$
  • 1
    $\begingroup$ 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 Sep 16 '16 at 17:44
  • $\begingroup$ @gbd Yes. That is the common thing to do. $\endgroup$ – Dr. Ikjyot Singh Kohli Sep 16 '16 at 17:47
  • $\begingroup$ @IkjyotSinghKohli, I tried the method you suggested but I still get the wrong result. $\endgroup$ – gbd Sep 16 '16 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 Sep 16 '16 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 Sep 16 '16 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.