# Question on the transformation from Boyer-Lindquist to Kerr-Schild coordinates, for a modified Kerr metric

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

• Not every metric admits KS form. Have you checked that yours does? Commented Jan 16 at 13:03
• @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? Commented Jan 16 at 15:33
• I don't know how to perform this check Have a look at references from this answer or maybe at this paper. Commented Jan 16 at 16:44
• 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? Commented Jan 16 at 16:48
• Commented Jan 17 at 23:22