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).

  • $\begingroup$ I thought (x, y, z, t) would make clear that I'm looking for a conformal projection with 90° angles of the coordinate axes. Sorry that that wasn't the case. $\endgroup$
    – MartyMcFly
    Nov 20, 2023 at 8:06
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – rob
    Nov 20, 2023 at 18:00

1 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.

  • $\begingroup$ @safesphere wrote: "the original Schwarzschild paper defined spacelike r²=x²+y²+z²" - He just used a large letter R instead of a small r and the letter α for the Schwarzschild radius rₛ. His small r was just an arbitrary coordinate, see here at equation (14), but I already told you that in countless other comments. For the cartesian projection see here and here, for the Schwarzschild metric simply set spin a=0. $\endgroup$
    – Yukterez
    Nov 17, 2023 at 9:04
  • $\begingroup$ @safesphere wrote: "Your math measures not the distance to the origin" - Strawman argument, I never claimed that. The distance from the horizon to the center is velocity depended (in the frame of free falling observers with c√[rs/r] it is exactly 2GM/c² by the way, see here). {r,θ,φ} and {x,y,z} are coordinates, the correlation between coordinates and the physical distance with a given set of rulers (stationary, free falling, accelerating, or whatever coordinates you use) is in the corresponding line element, no matter if {r,θ,φ} or {x,y,z} $\endgroup$
    – Yukterez
    Nov 17, 2023 at 17:37
  • 1
    $\begingroup$ @safesphere wrote: »I still hope that you are one of the good guys who can see through this propaganda« - If we're talking about politics I am, but here we're talking about physics, why would there be some evil propaganda surrounding black holes? - safesphere wrote: »This site is full of the mainstream nonsense that the falling observer perspective is somehow “more important”« - Who said it was more important, both have equal rights. The stationary has the right to observe nothing cross the horizon due to time dilation as the freefalling has the right to cross the horizon, it is what it is $\endgroup$
    – Yukterez
    Nov 17, 2023 at 19:04
  • $\begingroup$ @safesphere wrote: »The Cartesian coordinates in the Schwarzschild solution correspond to his r« - You could also project it that way, but then you'd lose the circumference fidelity since his small r gives noneuclidean values for gθθ and gφφ. Nowadays we use coordinates with proper circumference (since that is the same for stationary and radially moving rulers) and project that into cartesian, but if you want to do it differently then you can also write an answer and see how it sells. $\endgroup$
    – Yukterez
    Nov 17, 2023 at 19:34
  • 1
    $\begingroup$ In the Droste coordinates one could write each time $r^2(r_s-r)$ in the denominator and omit the minus. $\endgroup$
    – MartyMcFly
    Nov 19, 2023 at 6:11

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