- #1
dpidt
- 1
- 0
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.
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.
