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[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:

enter image description here

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:

enter image description here


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).

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  • $\begingroup$ If the black hole is not currently accumulating any mass, then the local geometry of space time should be constant in the time dimension. Even if it is sucking in matter that mass is probably trivial compared to the mass of the black hole, so the geometry should change very slowly. But this ignores the time dilation, so I'm not sure whether that would change things. $\endgroup$
    – quarague
    Commented Sep 26, 2019 at 14:25
  • $\begingroup$ I think the accretion discs can have a variety of shapes, depending on spin and the materials in the disc. arxiv.org/abs/1511.04445 arxiv.org/abs/1304.8106 But perhaps you want a clean answer in a simple situation? $\endgroup$ Commented Sep 26, 2019 at 18:46
  • $\begingroup$ But if I'm reading the Kerr solution correctly, I think if you're just talking about light then there appears to be an accretion solid torus, provided the light is allowed to have all possible direction vectors. i.e. much of the trapped light does not have a closed orbit. $\endgroup$ Commented Sep 26, 2019 at 19:30
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    $\begingroup$ As to the model you linked to, the accretion disc is a disc with a hole inside. The actual NASA webpage has more discussion and other simulations with different views. $\endgroup$ Commented Sep 26, 2019 at 21:33
  • $\begingroup$ Is this really a math question? It sounds more like a theoretical physics question. (For example, several arXiv preprints have been cited in the comments and the answer; none of them are in a math category.) $\endgroup$ Commented Sep 27, 2019 at 10:23

2 Answers 2

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The "shape" of an accretion disc is the spatial profile of the gas density $\rho$ at a given time $t$. Here is a recent calculation, arXiv:1810.0083. This image shows what a distant observer, outside of the event horizon, would measure. Matter is falling into the black hole without ever crossing the event horizon, falling ever more slowly as it approaches the horizon.

The density profile is shown on a logarithmic color scale (high density = red = $10^2$ g/cm$^3$, low density = blue = $10^{-7}$ g/cm$^3$), at a time $t=59\cdot 10^3 r_g/c$ after core collapse, with $r_g=GM/c^2$ the gravitational radius of the black hole (one half the Schwarzschild radius, which is the radius of the event horizon, of order $10^{13}$ m). The profile is a slice through the $x$$z$ plane, when the $z$-axis is aligned with the rotational axis of the spinning black hole. (The density is rotationally symmetric in the $x$$y$ plane.) The black lines are magnetic field lines.

The blue channel of relatively low density at the center appears because the lack of centrifugal forces along the rotational $z$-axis allows the gas to accrete much more readily than along the equatorial $x$$y$ plane.

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  • $\begingroup$ Thank you for your answer. I guess this answers my -admittedly, not completely rigorous- main question, at least in a qualitative (and numerical) way. $\endgroup$
    – Qfwfq
    Commented Sep 28, 2019 at 22:08
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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.

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  • $\begingroup$ Thank you for setting me straight on the general relativity (Not that I am able to understand everything of your very well written answer now, of course, but at least you made me aware of some wrong assumptions :) ). Incidentally, I realized that I was assuming the accretion disk is a structure sitting outside the event horizon, so I didn't consider it affected by the geometry that happens inside (though, I admit, my phrasing -especially the "3-space minus a point" bit- was not very consistent....). $\endgroup$
    – Qfwfq
    Commented Sep 28, 2019 at 22:13

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