7
$\begingroup$

I saw three different depictions:

Never ending 'hole':

hole

Point:

point

Paraboloid:

paraboloid

Also, do black holes rip space-time, if not, is there any other way to?

P.S. I'm not sure if the last one is a point as well.

$\endgroup$
7
$\begingroup$

These diagrams try to depict a slice of spacetime, but likely confuse it with a gravity well. A gravity well diagram depicts with an altitude for each location how much potential energy a particle in that location would have. There is also likely some confusion with the classic rubber sheet analogy used to describe gravity (plus, in many media, graphical designers not using the right shape to illustrate an article). A spacetime diagram instead tries to depict how spacetime is curved, not how much energy you would have in a particular spot.

The closest we can get to that for a non-rotating black hole is Flamm's paraboloid: $$z(r)=2\sqrt{r_s(r-r_s)}$$ where $r_s$ is the Schwarzschild radius. This is a surface in 3D space that corresponds to a 2D slice of the black hole spacetime in the equatorial plane for a given time. Distances between points measured along the surface correspond to distances measured between points in the black hole spacetime.

Flamm's paraboloid by AllenMcC, CC BY-SA3.0 via Wikimedia (By AllenMcC., CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=3871398)

However, the paraboloid only represents the spatial curvature and not the temporal curvature. It represents a single "moment in time" and a particle moving in spacetime will not stay on the paraboloid. One could imagine other paraboloids stacked on top of the first one with a distance corresponding to the time distance $dt^2$, but since this varies with radius the picture becomes pretty unclear immediately.

Also, note that this parabolid ends at the event horizon $r_s$. It is however entirely valid to consider the full parabolid $$z(r)^2=4r_s(r-r_s),$$ but the meaning is subtle: the lower half does not correspond to the interior of the black hole, but a so-called Einstein-Rosen bridge which is a form of wormhole.

To really understand the structure of spacetime Penrose diagrams are useful because they actually try to show the topology of (a part of) spacetime. For a black hole they make it clear that the singularity is not pointlike, neither a circle or some other space-like shape, but a space-like thing similar to a particular moment in time. This is why it is so unavoidable and hard to illustrate.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Overall the answer is fine and well written. However, there are two mistakes: 1. The lower half of the paraboloid doesn't correspond to the interior of the black hole! Note that it is just an "upside-down copy" of the upper half, and it, too, is valid only for $r>r_s$. So what does it correspond to? The alternative universe (region III in the maximally extended Penrose diagram). In fact, this full paraboloid is tightly connected to the Einstein-Rosen bridge connecting the two universes. 2. The Schwarzschild singularity is not timelike, it is spacelike. $\endgroup$ – Whyka Sep 3 '19 at 12:20
  • $\begingroup$ @Whyka - Thank you! I have corrected my errors. Especially the latter is downright embarrassing. $\endgroup$ – Anders Sandberg Sep 4 '19 at 19:52
7
$\begingroup$

The trouble with all these images is that they rely on intuition formed by studying curved two dimensional spaces, while general relativity deals with curved space-time, and it is the inclusion of time component that is responsible for such defining feature of black hole as horizon.

So, a more useful picture of a black hole spacetime would be an image of a waterfall, in which the space itself is falling toward a single point. Material objects are being carried by this flow and can move within it according to rules of special relativity. The closer to the singularity point the flow gets the greater its speed. On the surface of black hole horizon the velocity of the flow exceeds the speed of light, making it impossible for any object carried inside to escape.

Image from Andrew Hamilton's page

This flow analogy (that could be made mathematically precise) is presented in a paper:

  • Andrew J. S. Hamilton & Jason P. Lisle (2008) “The river model of black holes” Am. J. Phys. 76 519-532, arXiv:gr-qc/0411060.

Andrew Hamilton also has a webpage with various visualizations and renderings of black holes (the above image was taken from the waterfall section).

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

In GR (general relativity) the physical (proper) time and distances are not given directly by the coordinates, but must be computed from the metric. The Schwarzschild solution describes a static and spherically symmetric mass distribution. The existence of an event horizon defines a black hole.

In Schwarzschild coordinates the radial distance is a circumferential distance (divided by $2 \pi$), not the proper radial distance as measured by a local observer. We have
$$dl = \frac{dr}{\sqrt{1 - r_s/r}}$$ where:
$c = G = 1$ natural units
$r$ is the coordinate radial distance
$l$ is the proper radial distance
$r_s = 2M$ is the Schwarzschild radius, and
$M$ is the black hole mass.

Note: The Schwarzschild radius defines the event horizon which disconnects the interior of a black hole from the exterior spacetime.

The pictures in the question are embedded diagrams which should help to visualize the structure of a curved space, even if it is not what a black hole looks like, as sometimes misunderstood in popular level discussions.

In the case of a black hole, the diagram reduces progressively the radius having as limit the Schwarzschild radius, that is the event horizon. The meaning is that the Schwarzschild coordinates can not describe the spacetime beyond the horizon. To cross the horizon you have to change coordinates, e.g. to the Eddington–Finkelstein coordinates.
Note: The last picture does not end up in a point, but in the horizon radius.

Black holes do not rip spacetime, as the horizon is absolutely crossable, even if it is not possible to come back.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Well. General relativity says that the event horizon can be crossed and time would keep running precisely the same way. General relativity also says that the density of a black hole is infinite because its volume is formally zero, as it is a singularity. Of course, for a static black hole, the singularity is a point, whereas for a rotating black hole, it is a ring (regardless, the volume is still zero, $R_{sch}=\frac {2GM}{c^2}$). If an individual is positioned outside of the black hole and is observing a clock moving toward the event horizon, then the clock would appear to move slower and in fact, wouldn't ever actually reach the black hole because time gets infinitely slow at the event horizon. If an individual is attached to the clock, then everything would look precisely the same, and time would run just as the individual is used to.

Edit: Black holes don't rip spacetime necessarily. But there is a phenomenon called geodesic incompleteness, in reference to the singularity. This is where all geodesics end (geodesic paths are straight lines that become elliptical orbits or hyperbolic trajectories under the effect of gravity. Paths that carry on indefinitely are called complete geodesics, and those that stop abruptly are called incomplete geodesics). This is the point at which the maths seems to just break down entirely.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "Its volume is formally zero": when speaking about black holes, we usually mean everything that's beyond the event horizon, not just the singularity. In fact, for an observer inside the black hole (within the event horizon), $t$ is the spacial coordinate and thus the spacial volume is infinite! $\endgroup$ – Whyka Sep 3 '19 at 12:28
  • $\begingroup$ "Geodesic paths are straight lines that become elliptical orbits or hyperbolic trajectories under the effect of gravity": that is only true in specific simple cases, and of course one should solve the geodesic equation to get the actual geodesic curve. $\endgroup$ – Whyka Sep 3 '19 at 12:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.