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Kerr solution as complex transformation

  1. Jan 6, 2008 #1
    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:

    [tex]g^{a b} = l^a n^b + n^a l^b - m^a \overline{m}^b - \overline{m}^a m^b[/tex]


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

    Then they transform
    [tex] v' = v + i a cos\theta [/tex]
    [tex] r' = r + i a cos\theta [/tex]

    and end up with:

    [tex]l^a = \delta^a_1 [/tex]
    [tex]n^a = -\delta^a_0 - \frac 1 2 (1 - \frac {2 m r'} {r'^2 + a^2 cos^2 \theta})\delta^a_1 [/tex]
    [tex]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) [/tex]

    Then they use

    [tex]g^{a b} = l^a n^b + n^a l^b - m^a \overline{m}^b - \overline{m}^a m^b[/tex]

    to get the Kerr metric.
  2. jcsd
  3. Jan 7, 2008 #2

    George Jones

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    Chris Hillman referred to this derivation as "rather mysterious," and he gives a few more details. See posts #4 and #5 in this thread.
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