Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    where

    [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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Last edited by a moderator: Apr 23, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kerr solution as complex transformation
Loading...