# Does the twist of spacetime by a black hole increase over time?

As a black hole rotates, it twists spacetime in the direction of the rotation...

Does this "twisting up spacetime" add up over time in any measurable way? Or do I take a metaphor too literal?

What I mean by "adding up" is if we'd assume that spacetime is filled with a grid, then this grid would be deformed. Such a grid is also visible in a lot of illustrations. If we now twist this grid more and more, the grid lines would get closer and closer. But then again, I might take a visualization trick too literal and this grid does not exist.

Please excuse my possible layman terms, physics is not my area of expertise.

• I think (as another layperson) it's not that the act of rotating is continuously twisting spacetime more and more. Rather the presence of a rotating black hole makes spacetime twisted (compared to the hypothetical spacetime were the black hole not present), and the amount of "twist" depends on the rotation rate (not on "how many times it has rotated in the past").
– Ben
Jul 31, 2021 at 1:30
• If you want to stay with this very basic image - spacetime is not "fixed" to the black hole with a thumb tack, imagine it like a rubber band strung around the hole that gets warped through friction and at some point starts sliding. Jul 31, 2021 at 14:46
• Whose time? Time is already twisted. Jul 31, 2021 at 17:29

The spacetime around a rotating black hole (Kerr metric) is stationary, this means that you can choose a coordinate system where the metric doesn't depend on the time coordinate. In layman's terms this is equivalent to say that if I look at the black hole now, I measure its gravitational field at different points in space and make any other measurement I want, I can then come back later, do the same measurements and obtain the same results.

By the way you are imagining the spacetime grid, it seems that by measuring the amount of twisting, it should be possible to determine how long the black hole has been spinning. It is not the case.

An analogy may be a water vortex. Imagine a water vortex that is perfectly smooth and unchanging. Like this one, but even smoother. You don't see any ripple. If you didn't know it, you might think that water was still. You can measure the pressure and density of the water at different points and find that it doesn't change with time. But if you put a ball in the water, then you immediately see that it is dragged away.

In the same way, the spacetime is stationary just like the water, but anything that enters the ergosphere will be dragged away.

You could say that water is made of molecules, and they are swirling, so you could be able in principle to detect their motion. This is where the analogy breaks down. Spacetime is not made of any substance, and it is truly stationary.

• Best regards from Sicily 😀 Aug 8, 2021 at 12:38

Yes, you are taking the metaphor too literally. When we visualize spacetime like a flexible rubber sheet there is an accidental and erroneous impression that a twist will lead to spacetime getting changed over time. In the mathematical fomulation the twisting is all about how distances and angles are affected as you move through spacetime, but the spacetime is stationary - at each point in time (as seen by remote observers) it is the same.

• I think this answer is wrong. Or at least blows through some subtlety that can be important. The sentence referenced introduces frame dragging, which is real and does have measurable impacts on surrounding particles. Jul 30, 2021 at 13:01
• @Brick But the frame dragging effect does not vary with time - using the OP's language, the spacetime does not become more twisted in the future. Jul 30, 2021 at 13:26
• @J.Murray It depends on what coordinate you use. The "grid" could well be interpreted as a choice of coordinate, and then, depending on what you choose, it may well get twisted and these effects do depend on (coordinate) time. Jul 30, 2021 at 13:28
• @Brick Okay, fine - then there exists an asymptotically timelike Killing vector field $\xi$. If we choose a chart in which $\xi=(1,0,0,0)$ then the metric and the frame-dragging effects do not change with time. Of course if we choose a different time coordinate then that will change, but the same could be said of any spacetime; the Minkowski metric varies with time if we choose $t^* = t + t_0 \sin(\omega t_0)$, after all. Jul 30, 2021 at 13:44
• I converted to a full answer. If you can dissuade me of the fuller version, I'll happily take the correction and delete the answer. @J.Murray Jul 30, 2021 at 16:33

The twisting is real in a sense, but it's more subtle than your picture suggests. The other answers, on the other hand, are correct in a technical sense but, I think, missing the big picture.

Imagine that you set up very far from the black hole and start sending out probes toward the black hole at regular intervals (as measured on your local clock). You hope to use these to mark out a coordinate system between you and the black hole. Each probe is left to coast after the initial impulse that sends it on its way, and each gets the same initial velocity starting at the same distance from your station (both, again, measured in your frame). While they remain far from the black hole, the spacetime is essentially flat, and they behave as you expect - staying on line with the black hole and equally spaced (in both coordinate and proper distance). Let's say you send out 1 per hour at 1 mile per hour. Then if you check each hour, your system extends one mile closer to the black hole, but you have one probe at the 1 mile mark, one at the 2 mile mark, etc. Still as measured by you in your reference frame.

Once they get close enough to the black hole, however, the black hole's effects will be observable on the probes. In the case of a Kerr black hole, they will start to orbit and separate in distance. If you keep sending out a stream of these hoping that they will be reference points on your grid, this twisting and stretching will be observable to you at your distanced position. This is a consequence of the fact that there is no globally inertial frame of reference in this spacetime. (Of course, in their own respective comoving frames, they are each still coasting inertially.)

You could try to overcome this by putting thrusters on your probes that fire in just the right way to "cancel" the effects of the spacetime curvature so that you get the regular grid that you wanted, as viewed by you at a distance. You can even hold them at a fixed coordinate distance if you have enough fuel and strong enough thrusters so that they appear stationary - That is until you try to put one in the ergosphere where no amount of thrust will be enough to avoid them appearing to you to orbit or inside the event horizon, where no amount of thrust can prevent their inward fall. Except in these special regions, you now have a situation where you view the probes to be at fixed coordinate positions in your frame. (But, correspondingly, each individual probe has to expend energy and is accelerating relative to a momentarily comoving reference frame.)

This coordinate system marked by the probes with thrusters is adapted to the black hole spacetime in way analogous to taking a corotating frame to study a planetary orbit. The thrust that you need to have each probe provide is equal and opposite to the psuedo forces associated with the non-inertial nature of the frame.

The other answers are pointing to special coordinates - along the lines of the second case - where any "twisting" is not evident in the form of the metric. Not all spacetimes have that, so it is something remarkable about Kerr and a few others. For theoretical coordinates, unlike our hypothetical physical probes that had size and mass, you can extend the system further toward the black hole. We can mark the mathematical coordinates however we want even if no physical item would be capable of station-keeping a constant-coordinate position. The answers are also, correctly, pointing to the fact that for the free-falling probes of the first scenario, each subsequent probe experiences the same gravitational effects as the ones before it when it passes by a corresponding point in space. Equally, the probes with thrusters in the second scenario, once they achieve their station, do not need a time-varying thrust to keep station. (Though each probe in the sequence needs a different thrust than the ones adjacent to it.) In that sense that spacetime is not dynamical such that some sort of "twisting" has a cumulative effect on the curvature.

• This is a perfectly fine description of the Kerr spacetime. My point was that my understanding of the question is answered by your final sentence. Nobody was arguing that the frame-dragging effects aren't real, and in the coordinates I was talking about the "twisting" is evident by the existence of a non-zero $(t,\phi)$ cross term. However, the question seems to be asking whether the frame dragging effect will be more dramatic next week than it is today, to which the straightforward answer is no. Jul 31, 2021 at 11:29

The worldline of a point traveling in a circle is a helix. If you have a mass traveling in a circle, it will drag spacetime with it. Some of the curvature of its worldline will transferred to spacetime. This is called "frame dragging". The more mass is traveling in a circle, the more frame dragging. The frame dragging doesn't build up over time. Whatever mass-worldline-curvature there is produces a particular amount of frame dragging, not a certain amount of frame dragging per second.