Schwarzschild and Kerr solution in $(x, y, z, t)$ coordinates?

Question

The Schwarzschild solution ('simple' black holes) and the Kerr solution (rotating black holes) are very well known in General Relativity.

The coordinates which are used to describe them are mostly spherical symmetric $(r, \phi, \theta, t)$ for the Schwarzschild or axisymmetric $(r, \phi, z, t)$ for the Kerr metric. There are also some coordinates which move along with a free-falling observer.

How do the solutions look like in $(x, y, z, t)$ - coordinates?

We need to put the point $(x=0, y=0, z=0, t=0)$ somewhere outside the event horizon. The space(time) within the black hole (inside the event horizon) isn't of interest. We can simply set it to 'not defined'.

EDIT: I start to understand that what I want is most probably not possible in Kerr metric (due to inevitable $g_{0i} \neq 0$ therein).

Answer 1

The transformation rule from spherical to cartesian is

$ \rm dr=\frac{dx \ x+dy \ y+dz \ z}{r}$

$ \rm d\theta=\frac{dx \ x \ z+dy \ y \ z-dz \left(x^2+y^2\right)}{r^2 \sqrt{x^2+y^2}}$

$ \rm d\phi=\frac{dy \ x-dx \ y}{x^2+y^2}$

$ \rm r=\sqrt{x^2+y^2+z^2}$

$ \rm \theta =\arctan \left(z, \ \sqrt{x^2+y^2}\right)$

Schwarzschild in cartesian Droste $\rm \{t,x,y,z\}$ coordinates:

$$g_{\mu \nu}^\rm D=\left( \begin{array}{cccc} \rm 1-\frac{r_s}{r} & 0 & 0 & 0 \\ 0 & \rm \frac{r_s \ x^2}{r^2 \ (r_s-r)}-1 & \rm \frac{r_s \ x \ y}{r^2 \ (r_s-r)} & \rm \frac{r_s \ x \ z}{r^2 \ (r_s-r)} \\ 0 & \rm \frac{r_s \ x \ y}{r^2 \ (r_s-r)} & \rm \frac{r_s \ y^2}{r^2 \ (r_s-r)}-1 & \rm \frac{r_s \ y \ z}{r^2 \ (r_s-r)} \\ 0 & \rm \frac{r_s \ x \ z}{r^2 \ (r_s-r)} & \rm \frac{r_s \ y \ z}{r^2 \ (r_s-r)} & \rm \frac{r_s \ z^2}{r^2 \ (r_s-r)}-1 \\ \end{array} \right)$$

Schwarzschild in cartesian Raindrop $\rm \{\bar{\tau},x,y,z\}$ coordinates:

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g_{\mu \nu}^\rm R=\left( \begin{array}{cccc} 1-\frac{\rm r_s}{\rm r} & -\rm \frac{\sqrt{r_s} \ x}{\rm r^{3/2}} & -\rm \frac{\sqrt{r_s} \ y}{\rm r^{3/2}} & -\rm \frac{\sqrt{r_s} \ z}{\rm r^{3/2}} \\ -\rm \frac{\sqrt{r_s} \ x}{\rm r^{3/2}} & -1 & 0 & 0 \\ -\rm \frac{\sqrt{r_s} \ y}{\rm r^{3/2}} & 0 & -1 & 0 \\ -\rm \frac{\sqrt{r_s} \ z}{\rm r^{3/2}} & 0 & 0 & -1 \\ \end{array} \right)$$

Schwarzschild in cartesian Finkelstein $\rm \{{T},x,y,z\}$ coordinates:

$$\ \ \ \ \ \ \ g_{\mu \nu}^\rm F=\left( \begin{array}{cccc} \rm 1-\frac{r_s}{r}& -\rm \frac{r_s \ x}{\rm r^2} & -\rm \frac{r_s \ y}{\rm r^2} & -\rm \frac{r_s \ z}{\rm r^2} \\ -\rm \frac{r_s \ x}{\rm r^2} & -\rm \frac{r_s \ x^2}{\rm r^3}-1 & -\frac{\rm r_s \ x \ y}{\rm r^3} & -\frac{\rm r_s \ x \ z}{\rm r^3} \\ -\frac{\rm r_s \ y}{\rm r^2} & -\rm \frac{r_s \ x \ y}{\rm r^3} & -\rm \frac{r_s \ y^2}{\rm r^3}-1 & -\rm \frac{r_s \ y \ z}{\rm r^3} \\ -\rm \frac{r_s \ z}{\rm r^2} & -\frac{\rm r_s \ x \ z}{\rm r^3} & -\rm \frac{r_s \ y \ z }{\rm r^3} & -\rm \frac{r_s \ z^2}{\rm r^3}-1 \\ \end{array} \right)$$

The Kerr metric gets pretty bloated in cartesian form, so I wouldn't recommend that, but the transformation rule from pseudospherical to cartesian and back can be found here at the bottom of the page.