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I have checked online and the Kerr metric never seems to be given in Cartesian coordinates (although there is a conversion factor from Cartesian to Boyer-Lindquist coordinates). Is there some reason for this, or would the metric become prohibitively complicated if one tried to switch to Cartesian coordinates?

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  • $\begingroup$ Cartesian coordinates don’t describe a space with curvature. There has to be something non-Minkowskian about the metric. $\endgroup$ – G. Smith Jul 28 '20 at 19:43
  • $\begingroup$ But Cartesian coordinates describe the spatial Schwarzchild metric for example, is that because it is conformally flat? $\endgroup$ – Tom Jul 28 '20 at 20:12
  • $\begingroup$ Please clarify. Are you talking about something in this table? $\endgroup$ – G. Smith Jul 28 '20 at 20:15
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You can find the Kerr metric in pseudo-Cartesian coordinates for instance in "The Kerr Spacetime", by Wiltshire et al, as

\begin{eqnarray} ds^2 = &&-dt^2 + dx^2 + dy^2 + dz^2 \\ &+& \frac{2mr^3}{r^4 + a^2 z^2} \left[ dt + \frac{r(x dx + y dy)}{a^2 + r^2} + \frac{a(y dx - x dy)}{a^2 + r^2} + \frac{z}{r} dz \right]^2 \end{eqnarray}

with

\begin{equation} x^2 + y^2 + z^2 = r^2 + a^2(1 - \frac{z^2}{r^2}) \end{equation}

and for the angular coordinates,

\begin{eqnarray} x &=& (r \cos \phi + a \sin \phi) \sin \theta\\ y &=& (r \sin \phi - a \cos \phi) \sin \theta\\ z &=& r \cos \theta \end{eqnarray}

This gives the appropriate limits, of giving the Schwarzschild metric in Cartesian coordinates for $a \to 0$, and the Minkowski metric for $m \to 0$.

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  • $\begingroup$ If one writes the spacelike Kerr metric (ie. the part for $t-0$, in that case can one write this metric in Cartesian coordinates: this is for a spacelike hypersurface in this case. $\endgroup$ – Tom Jul 28 '20 at 21:55

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