Kerr Metric in Cartesian Coordinates

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?

• Cartesian coordinates don’t describe a space with curvature. There has to be something non-Minkowskian about the metric. – G. Smith Jul 28 '20 at 19:43
• But Cartesian coordinates describe the spatial Schwarzchild metric for example, is that because it is conformally flat? – Tom Jul 28 '20 at 20:12
• Please clarify. Are you talking about something in this table? – G. Smith Jul 28 '20 at 20:15

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

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

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 $$m \to 0$$.

• 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. – Tom Jul 28 '20 at 21:55