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the transformation from Boyer-Lindquist $\{t,r,\theta,\phi\}$ coordinates to Kerr-Schild $\{t',r,\theta,\phi' \}$coordinates can be written as $$ dt' = dt + \frac{2 M r}{r^2 - 2 M r + a^2} dr, \;\;\; d\phi'=d\phi + \frac{a}{r^2 - 2 M r + a^2} dr. \tag{1} $$ or more explicitly by $$ t' = t + 2 M \int \frac{r dr}{r^2 - 2 M r + a^2}, \;\;\; \phi' = \phi + a \int \frac{dr}{r^2 - 2 M r + a^2}. \tag{2} $$ Now I understand that these are equivalent. Since the simply differentiating (2) leads to (1).

The expressions (2) leave room for constants. I was wondering how we set these these constants. Do we perhaps want that $t'$ and $t$ are equal at some special point $r = C$, where $C = \text{const.}$? The same for $\phi'$ and $\phi$. But what choice of $C$ is a "reasonable one".

Edit: We probably want $t'$ and $\phi'$ to be real, that fixed the imaginary parts of $C$. But not the real ones.

Nitaa a
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    Those constants are just the initial t and φ, you can set them to 0 since the metric is independend of those 2 coordinates. Since the transformation is done via the differentials the constants won't even show up though. – Yukterez Oct 29 '23 at 19:01
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    . . . leads to (1). I couldn't edit it as it is two single characters. – m4r35n357 Oct 29 '23 at 19:58
  • @Yukterez I understand that. But I was unsure about the domain of the new coordinates, which seems important if you want to calculate quantities like the area of the Black-Hole in the Kerr-Schild coordinates. – Nitaa a Oct 30 '23 at 21:37

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