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From Kerr metric, we do know that there exist a function with the form of:

$$\Delta = r^2 - 2 M r + a^2 \tag{1}.$$

Following $[1]$, I did understand the coordinate transformation from Boyer-Lindquist (BL) to Kerr-Schild (KS) coordinates, (BL$\to$ KS), for the Kerr spacetime:

$$ dt' = dt + \frac{2 M r}{r^2 - 2 M r + a^2} dr = dt + \frac{2 M r}{\Delta} dr , d\phi'=d\phi + \frac{a}{r^2 - 2 M r + a^2} dr = d\phi + \frac{a}{\Delta} dr. \tag{2}. $$

Now, suposse we have a general version of $\Delta$ function, such as:

$$\Delta' = Kr^2 - 2 M r + a^2 + f(r) \tag{3}.$$

Where, $K$ is a constant. My question is:

The tranformation (BL$\to$KS) would have its form precisely as: $$ dt' = dt + \frac{2 M r}{\Delta'} dr , d\phi'= d\phi + \frac{a}{\Delta'} dr \tag{4}? $$


$[1]$ Coordinate transform of Kerr metric to Kerr-Schild coordinates

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  • $\begingroup$ Not every metric admits KS form. Have you checked that yours does? $\endgroup$
    – A.V.S.
    Commented Jan 16 at 13:03
  • $\begingroup$ @A.V.S. No because I don't know how to perform this check. Nevertheless, what do you think? Do you think $(4)$ is a suitable form? $\endgroup$
    – M.N.Raia
    Commented Jan 16 at 15:33
  • $\begingroup$ I don't know how to perform this check Have a look at references from this answer or maybe at this paper. $\endgroup$
    – A.V.S.
    Commented Jan 16 at 16:44
  • $\begingroup$ what do you think? This depends on what are you trying to accomplish. Does your metric represent some specific solution? Are you looking for KS form with a curved seed metric? $\endgroup$
    – A.V.S.
    Commented Jan 16 at 16:48
  • $\begingroup$ related: physics.stackexchange.com/a/772647/24093 $\endgroup$
    – Yukterez
    Commented Jan 17 at 23:22

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