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 $(r, \phi, z, t)$ f...
What do you mean by $(x,y,z,t)$-coordinates? Cartesian coordinates do not exist on a curved spacetime, so you'll need to explain precisely what $x,y,z,$ and $t$ are meant to be.
I'm not perfectly sure what you mean but the Droste coordinates from the answer below look quite like that what I'm looking for (not 100% sure yet though). Probably one could say I'm looking for a projection on cartesian grid. A diffeomorphism: Cartesian Grid (x, y, z, t) on the one side and the $g_{\mu\nu} $on the other side. Since the $g_{\mu\nu} $ I'm interested in shall be "friendly" (no singularities, even no coordinate singularities, no inevitable $g_{0i} \neq 0$ as in Kerr metric. At the end I want at each point in space a volume (det $g_{ij} $) and a "velocity of time" $g_{00}$.
By Cartesian coordinates, one usually refers to a system of coordinates such that the unit vectors $\partial/\partial x^i$ are orthonormal. Such a system cannot exist in curved spacetime. This is precisely analogous to the fact that you can't put Cartesian coordinates on the surface of a sphere. You can use coordinates called $x$ and $y$ if you want, but those are simply names; there is simply no way to give them the properties that familiar, flat-space Cartesian coordinates have.
For example, you can apply the same transformation which you'd use to translate spherical coordinates in flat space to cartesian coordinates, but the resulting $(x,y,z,t)$ are not going to be Cartesian - in particular, they will no longer by orthogonal, which is presumably one of the properties you are trying to preserve.
The sphere has only extrinsic curvature (it is a curved 2D object in 3D). The spacetime in 4D GR, however, has only intrinsic curvature. That's an inportant difference...
Furthermore, I'm looking for the projection into cartesian grid.
I don't know why you think that, the 2-sphere with the standard metric $\mathrm ds^2 = \mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2$ has intrinsic curvature as a 2D surface.
Can you explain to me how you might do this "projection" on the 2-sphere, given polar coordinates $(\theta,\phi)$? How do I "project" them onto a cartesian coordinate grid?
For example: I put the plane (uncurved, plane cartesian grid) I want to project on right below the sphere. With that, I chose one point (it's a choice, but it's an arbitrary point) as the "zero" of my cartesian projection grid. Then, I project all the other points on the sphere on my grid by preserving the angles. It's a conformal projection: The shape of objects stays the same as the angles are preserved.
... lengths are not preserved using this projection.
I think you are confused about what extrinsic curvature is. Extrinsic curvature arises when you embed a space inside of a larger ambient space; intrinsic curvature is a property of the space itself, with no reference to an embedding. The 2-sphere possesses intrinsic curvature in its own right, and extrinsic curvature when you embed it in 3D Euclidean space. Similarly, the Schwarzschild spacetime has intrinsic curvature, and would have extrinsic curvature if you embedded it in a higher-dimensional Euclidean space too.
In any case - it was far from clear that you were talking about a conformal projection of the Schwarzschild spacetime. There are many such projections you could perform in general, but at least that specification narrows down the list.
J. Murray wrote: "one usually refers to a system of coordinates such that the unit vectors ∂/∂xi are orthonormal. Such a system cannot exist in curved spacetime. " The first sentence is true, the second is only true for extrinsic curvature I'm convinced. Wrong for only intrinsic curvature. Probably, the best I can do is to start a new question. :-)
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.
Discussion on question by MartyMcFly:…
Discussion on question by MartyMcFly:…