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?

  • $\begingroup$ Cartesian coordinates don’t describe a space with curvature. There has to be something non-Minkowskian about the metric. $\endgroup$
    – G. Smith
    Commented Jul 28, 2020 at 19:43
  • $\begingroup$ But Cartesian coordinates describe the spatial Schwarzchild metric for example, is that because it is conformally flat? $\endgroup$
    – Tom
    Commented Jul 28, 2020 at 20:12
  • $\begingroup$ Please clarify. Are you talking about something in this table? $\endgroup$
    – G. Smith
    Commented Jul 28, 2020 at 20:15
  • $\begingroup$ The metric tensor would have cross terms all over the place and the other tensors and the equations of motion would be so long that you could fill many screens with them and therefore it would become practically impossible to keep track of all the terms, while in pseudospherical coordinates they are relatively elegant. So best thing is doing all the calculations in r,θ,φ and only converting the output to x,y,z $\endgroup$
    – Yukterez
    Commented Aug 19, 2023 at 21:26

1 Answer 1


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}


\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 $m \to 0$.

  • $\begingroup$ 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
    Commented Jul 28, 2020 at 21:55

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