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The Schwarzschild metric ansatz is given by $$ds^2=-A(r)dt^2+B(r)dr^2+r^2d \Omega^2$$ where upon applying the Einstein Field Equations $$G_{\mu\nu}=\kappa T_{\mu\nu}$$ we obtain the normal Schwarzschild metric $$ds^2=\left(1-\frac{r_s}{r}\right)dt^2+\left(1-\frac{r_s}{r}\right)^{-1}dr^2+r^2d \Omega^2.$$

Following this same notion, what would be the metric ansatz of the kerr metric in line element form such that it can solved using the EFEs to obtain the Kerr solution? Note: Im not trying to obtain the Kerr solution from the Schwarzchild solution. The goal is to find the metric ansatz used to obtain the kerr solution.

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  • $\begingroup$ Maybe use oblate spheroidal coordinates? I don't remember having ever seen a derivation of the Kerr metric à la Schwarzschild, only the original paper by Kerr. $\endgroup$ Commented Apr 30, 2022 at 17:51
  • $\begingroup$ This answer may be useful physics.stackexchange.com/questions/150446/… $\endgroup$
    – Eletie
    Commented Apr 30, 2022 at 18:02

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Pedagogical derivations (the ones I found by googling for a few minutes) seem to prefer the use of "oblate spheroidal coordinates" (OSC) as a good starting point to describe static, axially-symmetric space-time. In the following I will repeat the discussion of Sec. 7 of Saul A Teukolsky 2015 Class. Quantum Grav. 32 12400. The oblate spheroidal coordinates $(r,\theta,\phi)$ are related to Cartesian ones by $$ \begin{align} x&\equiv\sqrt{r^2+a^2}\sin\theta\cos\phi,\\ y&\equiv\sqrt{r^2+a^2}\sin\theta\sin\phi\\ z&\equiv r\cos\theta. \end{align} $$ The flat Minkowski metric in OSC reads $$ \mathrm{d}s^2= -\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2=-\mathrm{d}t^2+\frac{\Sigma}{r^2+a^2}\mathrm{d}r^2+\Sigma\,\mathrm{d}\theta^2+(r^2+a^2)\,\mathrm{d}\phi^2,\tag{1} $$ with $\Sigma\equiv r^2+a^2\cos^2\theta$. This can be rewritten as $$ \begin{align} \mathrm{d}s^2&= -\frac{r^2+a^2}{\Sigma}(\mathrm{d}t-a\sin^2\theta\,\mathrm{d}\phi)^2+\frac{\Sigma}{r^2+a^2}\mathrm{d}r^2+\Sigma\,\mathrm{d}\theta^2+\\&\qquad+\frac{\sin^2\theta}{\Sigma}((r^2+a^2)\,\mathrm{d}\phi-a\,\mathrm{d}t)^2,\tag{2} \end{align} $$ which includes cross terms $\mathrm{d}t\,\mathrm{d}\phi$ which cancel each other for the flat Minkowski metric. The Kerr Metric has such cross terms describing the Lense–Thirring "dragging of inertial frames". The idea is now to make some modifications to Eq. (2) to try to derive the Kerr metric. One good starting point is introducing two metric potentials $Z(r)$ and $F(r)$ in the first $\frac{r^2+a^2}{\Sigma}\rightarrow \frac{r^2+a^2-Z(r)}{\Sigma}$ and last term $r^2+a^2\rightarrow F(r)$ in Eq. (2) which only depend on $r$. This leads to the ansatz $$ \begin{align} \mathrm{d}s^2&= -\mathrm{d}t^2 + \frac{\Sigma}{F(r)}\mathrm{d}r^2 + \Sigma\,\mathrm{d}\theta^2 + (r^2+a^2)\sin^2\theta\, \mathrm{d}\phi^2+\frac{Z(r)}{\Sigma}(\mathrm{d}t- a\sin^2\theta\,\mathrm{d}\phi)^2. \end{align} $$ Plugging this into the vacuum Einstein equations $G_{\mu\nu}=0$ leads to two non-trivial equations for $F(r)$ and $Z'(r)$ which can be solved leading to the Kerr metric in Boyer–Lindquist form/coordinates (see Saul A Teukolsky 2015 Class. Quantum Grav. 32 12400 and references therein for further details).

I think this approach is a reasonable compromise for a pedagogical derivation: using OSC for static, axially-symmetric space-time is intuitively reasonable. How to modify it/make the ansatz not so much. Without knowing about Lense–Thirring cross terms it is not natural how to get from Eq. (1) to Eq. (3) and the reshuffling from Eq. (1) to Eq. (2) can also look a bit artificial. But writing down a static, axis-symmetric metric with general coefficients depending on $r$ and $\theta$ leads to complicated systems of partial and algebraic equations which can not be solved with simple methods/by hand.

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