Kerr solution as complex transformation

1. Jan 6, 2008

dpidt

I've been reading Ray D'Inverno's book about general relativity, and when he derives the Kerr metric he uses a trick with a complex transformation. I can follow the derivation, but I have no clue why it works. Anyone care to explain?

The derivation is:

They start with schwarzchild in Eddington-finklestein coordinates and use a null basis:

$$g^{a b} = l^a n^b + n^a l^b - m^a \overline{m}^b - \overline{m}^a m^b$$

where

$$l^a = \delta^a_1$$
$$n^a = -\delta^a_0 - \frac 1 2 [1-m(r^{-1}+\overline{r}^{-1})]\delta^a_1$$
$$m^a = \frac 1 {\sqrt{2} \overline{r}} (\delta^a_2 + \frac i {sin\theta} \delta^a_3)$$
$$\delta^a_0 = \partial_v$$
$$\delta^a_1 = \partial_r$$
$$\delta^a_2 = \partial_{theta}$$
$$\delta^a_3 = \partial_{phi}$$

Then they transform
$$v' = v + i a cos\theta$$
$$r' = r + i a cos\theta$$

and end up with:

$$l^a = \delta^a_1$$
$$n^a = -\delta^a_0 - \frac 1 2 (1 - \frac {2 m r'} {r'^2 + a^2 cos^2 \theta})\delta^a_1$$
$$m^a = \frac 1 {\sqrt{2} (r'+i a cos \theta)} (-i a sin \theta (\delta^a_0 + \delta^a_1) + \delta^a_2 + \frac i {sin\theta} \delta^a_3)$$

Then they use

$$g^{a b} = l^a n^b + n^a l^b - m^a \overline{m}^b - \overline{m}^a m^b$$

to get the Kerr metric.

2. Jan 7, 2008

George Jones

Staff Emeritus
Last edited by a moderator: Apr 23, 2017