TITLE: What's the "actual" shape of a black hole accretion disk?
QUESTION [16 upvotes]: [Warning: I have no expertise in general relativity, so this question might not be very rigorous]
More and more often we come across science popularization articles like this one which show beautiful images of (simulations of) a black hole surrounded by its accretion disk:

Of course, what we see is the effect of how the trajectories of light rays escaping the accretion disk are bent by the effect of space-time curvature, which is very high in the proximity of the black hole and more or less zero away from it.
This results in a sort of "artifact" in which there seems to be two or more rings of falling bright matter conjoined together, or one disk presented from two or more sides maybe by some gravitational lensing.
This is a phenomenologically correct thing: we -physical beings- can only perceive that matter through the light it emits, and the light is subject to the laws of general relativity. 
But what's the "actual" shape of the accretion disk?
Let me try to make sense of the above question.
First of all, the topology. The accretion material can be seen as a subset of 3-space (minus the black hole singularity), or we can probably consider the level sets of a "brightness function".
Second, the metric. I can imagine that in a given instant the whole picture is a leaf of a constant time foliation of spacetime (does such foliation exist in the case of a black hole?), which is a Riemannian manifold diffeomorphic to 3-space (minus the singularity). So the accretion disc is in fact a metric space with the induced distance. Can this metric space be isometrically (or with some limited amount of distortion) embedded in flat 3-space? How would it look like?
What I would like to know -I guess- is how this object would be perceived by a non-physical being that is able to "see" the metric of the space slice all at once without being bounded by the need of probing things by relativistic light rays. 
In other words, I want to do away with the general relativity aspects, and just keep the visualization of a subset of a Riemannian 3-manifold aspect, which may not be so crazy after all:  
 

I think I also have a more precise question, whose answer may indicate that the above question does or does not make much sense:

Are light rays geodesics also for the induced Riemannian metric on the constant-time slices? 

If they are, my previous questions are probably not very interesting (e.g. already answered by the first picture).

REPLY [5 votes]: This is not really a complete answer, but would not have fit in a comment. Your description of the topology and the time slices contains some incorrect assumptions.
Let's take the simplest example of a black hole spacetime, which is the Schwarzschild spacetime (a black hole with zero charge and zero angular momentum). This spacetime is split into interior and exterior regions, with an event horizon in between. The event horizon is a null surface, not a timelike surface, so it doesn't act like a stationary boundary. It would be more accurate to think of it as if it were moving at the speed of light. (In the maximal analytic extension of the spacetime, there are actually two more regions, for a total of four, but these are not present in a black hole that forms by gravitational collapse.)
The exterior is stationary, meaning that it has a timelike Killing vector. An observer whose world-line is parallel to this vector observes a gravitational field that doesn't change over time, and can be considered to be at rest relative to the black hole. Because the Schwarzschild spacetime is also not rotational, such observers can synchronize their clocks, and we therefore have a preferred time foliation. However, this preferred time foliation does not exist for rotating black holes, which are what really exist in our universe. (The simulation in the OP is of a rotating black hole.)
But the interior region is not stationary. Therefore the preferred time foliation can't be extended into the interior. What this tells us is that there is never any well-defined way to talk about what is happening "now" inside a black hole. You can extend your spacelike surface of simultaneity into the interior, but there is no preferred way to do so. We can say that the infalling matter never reaches the horizon, but we can also say that it does. (According to an infalling observer, the matter passes through the horizon and reaches the singularity in milliseconds in the case of a solar-mass black hole.) I would interpret this as meaning that your question doesn't actually have a meaningful answer if you try to talk about anything but the exterior region.

I can imagine that in a given instant the whole picture is a leaf of a constant time foliation of spacetime (does such foliation exist in the case of a black hole?), which is a Riemannian manifold diffeomorphic to 3-space (minus the singularity).

The singularity is analogous to a spacelike surface, not a timelike one. A spacelike surface of simultaneity can be chosen either to intersect the singularity or not to intersect it. Therefore there are at least two different topologies possible for a surface of simultaneity.

Are light rays geodesics also for the induced Riemannian metric on the constant-time slices?

This question only makes sense on the exterior (and only for a non-rotating black hole), where there is a preferred foliation. There you can project out the time dimension. In that region, I think the answer is no. Geodesics of the spacetime are in general not geodesics after this type of projection. For example, the orbit of the earth is an ellipse under this type of projection, and that's not a geodesic of the spatial metric. I think it's pretty easy to see why we shouldn't expect the geodesic property to be preserved under projection. Different spacetime geodesics can be tangent to one another after projection, and you can't have different geodesics with the same tangent vector where they intersect.