Physics Asked on September 30, 2021
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?
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 $m to 0$.
Correct answer by Slereah on September 30, 2021
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