The Kerr metric in Boyer-Lindquist coordinates is given by

\begin{align} ds^2 &= -\left[ \frac{r^2 + a^2 \cos^2(\theta) - 2mr}{r^2+ a^2 \cos^2(\theta)} \right] dt^2 -\frac{4mra \sin^2(\theta)}{r^2 + a^2 \cos^2(\theta)} dt d\phi \\ & + \left[ \frac{r^2 + a^2 \cos^2(\theta)}{r^2 - 2mr + a^2} \right] dr^2 + \left(r^2 + a^2 \cos^2(\theta) \right) d\theta^2 \\ & + \left[ r^2 + a^2 +\frac{2mra^2 \sin^2(\theta)}{r^2 + a^2 \cos^2(\theta)} \right] \sin^2(\theta) d\phi^2 \end{align}

I want to find proper transformation to Cartesian

\begin{align} ds^2 &= -dt^2 + dx^2 + dy^2 + dz^2 \\ & + \frac{2mr^3}{r^4+ a^2 z^2} \left[ dt + \frac{r(xdx+ydy)}{a^2+ r^2} + \frac{a(ydx - xdy)}{a^2+ r^2} + \frac{z}{r}dz \right]^2 \end{align}

First of all I know how to deal with in the case of $m=0$. i.e., \begin{align} ds^2 &= -dt^2 + \frac{r^2+ a^2 \cos^2(\theta)}{r^2 + a^2} dr^2 + (r^2+ a^2 \cos^2(\theta)) d\theta^2 + (r^2+ a^2) \sin^2(\theta) d\phi^2 \end{align} In this case taking spheroidal coordinates $x=\sqrt{r^2+ a^2} \sin(\theta) \cos(\phi), y=\sqrt{r^2+ a^2} \sin(\theta)\sin(\phi), z=r\cos(\theta)$ this reduces to \begin{align} ds^2 = - dt^2 + dx^2 + dy^2 + dz^2 \end{align} What about $m\neq 0$? In this case, after plugging spherodical coordinates into the metric, I have different forms.

How to convert Boyer-Lindquist coordinates to Cartesian coordinates in the case of $m\neq 0$?

I know after some consecutive transformation: Cartesin -> Original Kerr(Edington) -> time shift -> Boyer-Lindquist...

Here the reason I post this question is I want to write transformation rule $\frac{dx'^{\mu}}{dx^{\nu}}$ on these two coordinates.


1 Answer 1


With the relation between two coordinates is given by \begin{align} & d\tilde{t} = dt + \frac{2mr}{r^2+a^2 -2mr} dr \\ &dx= \sin(\theta)\left( \cos(\bar{\phi}) - \frac{a \left( r \sin(\bar{\phi}) + a \cos(\bar{\phi}) \right)}{r^2+a^2 -2mr} \right)dr \\ &\qquad \qquad + \cos(\theta) \left( r \cos(\bar{\phi}) - a \sin(\bar{\phi}) \right) d\theta - \left( r\sin(\bar{\phi}) + a \cos(\bar{\phi}) \right) \sin(\theta)d \phi \\ & dy =\sin(\theta) \left( \sin(\bar{\phi}) + \frac{a \left( r \cos(\bar{\phi}) - a \sin(\bar{\phi}) \right)}{r^2+ a^2 -2mr} \right)dr \\ & \qquad \qquad + \cos(\theta) \left( r \sin(\bar{\phi}) + a \cos(\bar{\phi}) \right) d\theta + \sin(\theta) \left(r\cos(\bar{\phi}) - a \sin(\bar{\phi}) \right) d {\phi} \\ & dz = \cos(\theta) dr -r\sin(\theta) d\theta \end{align} where $\bar{\phi} = \phi + \int\frac{a}{r^2+a^2 -2mr} dr$, we have desired results.


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