# B Frame dragging and co-rotating shells

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1. May 28, 2017

If I understand correctly, a massive enough hollow sphere rotating in space will rotate the minkowski spacetime inside the shell such that a pair of balls held together by a string and rotating around their mutual center and at the same angular velocity as the shell will experience no (or at least less) tension on the string. If this is correct, then would the following be true?

Imagine two such spheres and balls/string with a common axis of rotation. The spheres are co-rotating and the balls/string are also co rotating (and rotating with the spheres) and aligned so a ball of one is directly above the ball of the other. There is a laser attached to the ball of one and aimed at the ball of the other such that if the two spheres were not rotating, the laser would strike the second ball through a small hole in each sphere.

The spheres are separated by some distance.

The question is, if the spheres are co-rotating will the laser beam strike the second ball.

The subject line says counter rotating but should have said co rotating

2. May 28, 2017

### Staff: Mentor

Not quite. If the shell is rotating, spacetime inside it is not Minkowski spacetime. There is no such thing as a "rotating Minkowski spacetime".

It's true that there will be some nonzero angular velocity for the balls and string at which the tension in the string will be zero (whereas if the shell were not rotating the angular velocity at which the string tension was zero would be zero). But I'm not sure if that angular velocity is identical to the angular velocity of the shell.

With one sphere inside the other? I.e., both concentric but with different radius?

Meaning, the two sets of balls/string have different angular velocities? But both sets of balls/string are inside the inner sphere?

If my assumptions above are correct, then the spacetime geometry inside the inner sphere will be affected by the rotation of both spheres. If the spheres are co-rotating, the effects of their rotation will add, though they might not add linearly. I have not tried to do the math to see. But in any case, there will be only one spacetime geometry inside the inner sphere.

Also, if the balls and string are rotating with different angular velocities, then of course the laser won't hit the second ball. That has nothing to do with the spacetime geometry; it's a simple consequence of the different angular velocities.

I've fixed that.

3. May 28, 2017

Sorry, I actually meant the two sphere's are stacked like tomatoes on a shush-kabob so the laser would exit one sphere go through outside space, then enter the second sphere and then the ball. Also both spheres would have the same angular velocity and would also have the same angular velocities as both sets of balls/string

I thought it was. IIRC DaleSpam said in another thread that even though there is gravity inside the shell there is no curvature inside the shell. Isn't this Minkowski Spacetime? So spacetime cannot be said to be rotating relative to outside the shell? Is there another way to phrase what the spacetime is doing inside relative to outside? Perhaps its OK to say they have a different spacetime geometry?

Also I'm assuming the mass of the sphere's is so great that the balls/string could have very nearly the same angular velocity as the shell and show no rotational forces.

4. May 28, 2017

### pervect

Staff Emeritus
As other posters have mentioned, the space inside the rotating hollow sphere isn't Minkowskii. I think the first thing to clarify is the concept of a rotating frame of reference. Talking about "space rotating" is frankly confusing, it's hard to figure out how you might be interpreting those words. Talking about a rotating frame of reference makes more sense to me, and, with a bit of background discussion, I hope will make sense to you, too. Then we will be in the happy position of a post that makes sense to both of us, rather than the unfortunate position where a post that makes sense to oly one of us and does not make any sense to the other.

How can we tell if a frame of reference is rotating or not, in a quantitative way that allows us to tell "how fast" it's rotating? A simple test is to imagine a penduluum or a gyroscope that's moving without any external forces, and to ask if the gyroscope or pendulum rotates with respect to our frame of reference. A Newtonian mathematical description of what causes this rotation is the fictitious coriolis force, or the non-fictitious coriolis acceleration.

With this insight that it's the coriolis acceleration that tells us how fast a frame of reference is rotating, we can start to analyze the rotating hollow sphere.

We start out with defining a specific frame of reference. We'll use one anchored to the fixed stars. An example of such a frame of reference in astronomical use is the "International Celestial Referece Frame", the ICRF. Anchoring frames of refrerence to distant objects is a time-honored idea with a lot of history, fairly easy to understand (I hope), and has empirical realizations that we actually use in practice. So it's a good thing to use to describe our problem.

Now that we've defined our frame of reference, we can ask - does this frame of reference, anchored to the fixed stars, rotate inside a massive hollow rotating sphere? The Newtonian answer would be no. The GR answer is yes, it does.

Wiki gives the amount of this rate as the constant $d1$ times $\omega$ in <this link> under the description of the Lense-Thiring effect. The relevant formula is $\frac{4GM}{3Rc^2}\omega$, where M is the mass of the rotating hollow sphere, R is it's radius, $\omega$ is the angular frequency at which the hollow sphere rotates with repsect to the fixed stars, and G and c are physical constants.

$\frac{4GM}{3Rc^2}$ will be a very small number for realistic values of M and R. To see this, calculate it's value for a large M, say a million kg, and as small an R as you think is reasonable for a million kg object - a small R maximizes the number. You'll get a tiny number, the details depend on what you think is "reasonable".

So we can see that our frame of reference rotates, even though it's anchored to the fixed stars, rotates because a gyroscope or penduluum placed inside the rotating massive sphere changes it's orientation (precesses) with respect to our frame of reference. Not only that, we have a number that tells us how fast the frame of reference rotates.

When we look at the rate of rotation, we see that it is very, very small. The frame of reference anchored to the fixed stars is nearly non-rotating. The rotating massive hollow object is not even close to being non-rotating - your problem specification assumes that it is, but this idea is somewhere between misleading and wrong.

Last edited: May 30, 2017
5. May 28, 2017

### Staff: Mentor

If both spheres are rotating, the holes can't possibly stay lined up for more than an instant.

I suspect he was talking about the case of a non-rotating shell. If the shell is rotating, spacetime inside the shell is not Minkowski spacetime.

I'm not sure I understand. Making the masses of the spheres (shells) larger increases the curvature of spacetime inside them, which means it increases the effects of the spheres on the balls/strings.

6. May 29, 2017

The spheres are rotating together, same angular velocity, same direction.
I don't think there was any mention of rotation so that could be the case. I find it interesting though that there is a difference between rotating and non-rotating for an ideal situation where there is 100% frame dragging inside the shell.
OK maybe I have everything wrong. I thought the gravity increases but there remains no curvature even if the sphere is rotating. In addition there would be uniform frame dragging throughout the inside of the shell and the balls could rotate with the shell and not feel rotational forces.

7. May 29, 2017

Thanks for the clarification. OK so even for a vastly massive shell there will be very little frame rotation. I will modify my original question to accommodate this. the balls/string in both shells are allowed to rotate with the same angular velocity as the frame inside the shell. In other words the string has no tension, but the balls are still rotating relative to the ICRF. The holes in the shells are now rings instead of holes so the light is not blocked under any circumstance. The question is if a laser is pointed directly from one ball in one sphere to the ball in the other sphere, and considering that there is a gap between spheres, will the laser strike the 2nd ball. Assume the laser was aligned when the spheres were not rotating so the light would indeed hit the other ball during alignment.

The idea behind my question is that the laser, before reaching the hole (ring) in the sphere will travel in a straight line if viewed from an observer sitting on top of the laser since to this person the balls are not rotating. In other words the laser will follow a path parallel to the axis of rotation. However when the laser exits the sphere, and because it is now in the frame outside the sphere (0 rotation relative to ICRF) to the observer the laser would spiral about the axis until it entered the 2nd sphere. where again it would be parallel to the axis from the observers point of view, so it would miss the ball by virtue of the gap between the spheres.

It seems to me that this is what would happen but I would like to know for sure.

8. May 29, 2017

### Staff: Mentor

As pervect pointed out in his post, you have to be very careful about what "rotating" and "non-rotating" mean. See below.

That is not correct. If the sphere is rotating, spacetime inside it is curved, not flat.

The presence of frame dragging, which is not present inside a non-rotating shell, is because spacetime is curved, not flat. There is no frame dragging in flat spacetime.

I don't know if the frame dragging is uniform inside the shell, though. As I said before, I have not tried to work through the math. I have also not been able to find a reference online that gives a metric for this case.

As I said before, I also don't know that the angular velocity that makes the tension in the string between the balls zero (which is what "not feel rotational forces" means, see below) is the same as the angular velocity of the shell. I would need to work through the math, which is not a simple exercise.

However, we can at least see some qualitative features without doing the detailed math. First, consider the case of a non-rotating spherical shell. By "non-rotating" here, we mean, heuristically, that the shell has zero angular velocity relative to an observer at rest at infinity. (A more precise mathematical treatment would define an angular momentum tensor for the shell in terms of the asymptotically flat metric of the spacetime and show that it was zero.) Spacetime inside the shell is flat. That means if we have two balls connected by a string, and we let them all float freely inside the shell, starting them out at mutual rest (and at rest relative to the shell, for simplicity) with the string under zero tension, the string will stay under zero tension, and the balls and string will all stay at rest relative to each other. In this state, the angular velocity of the balls, relative to the shell, will be zero. And since the shell has zero angular velocity relative to infinity, the balls will also have zero angular velocity relative to infinity.

But there is another way we can measure "rotation", which is purely local. Suppose we put a set of three gyroscopes on each ball, with the gyroscopes set to point in three mutually perpendicular directions. One direction is "straight up", perpendicular to the plane in which the balls and the string lie. Another direction is along the string. The third direction is perpendicular to the first two. Now, as the balls float freely, the gyroscope directions will all remain the same, not just relative to the shell, but relative to the balls and string.

Now suppose we start the balls and string rotating around their common center of mass (which will be at the midpoint of the string, if we assume constant density for the balls and constant mass per unit length for the string). In other words, we leave the center of mass of the balls/string motionless relative to the shell, but give the balls a nonzero angular velocity about that center of mass, and therefore a nonzero angular velocity relative to the shell and to infinity. This will have the following effects:

(1) There will be nonzero tension in the string.

(2) The gyroscope directions will change, relative to the balls and string. More precisely, the two gyroscopes in the plane of the balls and string will change direction--to an observer on one ball, looking along the string towards the other ball, the gyroscope that started out pointing along the string will point in a direction that gradually changes relative to the string, and the other gyroscope in that plane will stay perpendicular to it, so it will also point in a direction that gradually changes. The gyroscopes give a local definition of "non-rotating", and so really what their changing direction relative to the string is telling us is that the balls and string are rotating, in a local sense (relative to local gyroscopes).

(Note that there are other complications here which I won't go into right now, to do with the relationship between the two senses of "rotating", local relative to the gyroscopes and global relative to an observer at rest at infinity. In Newtonian physics the two would be the same, but in relativity they are not. The various effects that cause them to be different go by names like "Thomas precession", "de Sitter precession", and "Lense-Thirring precession". The last of the three is also called "frame dragging", and we'll discuss that a bit below, but a full discussion would require an "I" or possibly even "A" level thread.)

Now let's consider the case of a rotating shell. First we need to specify what we mean when we say the shell is rotating. As above, we mean, heuristically, that the shell has nonzero angular velocity relative to an observer at rest at infinity. A more precise mathematical treatment would say that the angular momentum tensor of the shell, defined in terms of the asymptotically flat metric of the spacetime, is nonzero.

Next we put the balls and string inside the shell, along with the gyroscopes on each ball, and ask what the tension in the string will be and in what directions the gyroscopes will point, for a given angular velocity of the balls and string relative to an observer at rest at infinity. (Note that this will no longer be the same as the angular velocity of the balls and string relative to the shell.) We will find, generally speaking, the following:

(1) There will be some nonzero angular velocity at which the tension in the string will be zero and the gyroscope directions will not change relative to the direction of the string, the same as for zero angular velocity in flat spacetime. (As I said above, I do not know if this angular velocity will the same as the angular velocity of the shell relative to infinity; I think it won't, but I haven't done the math.) At this angular velocity, the balls and string will be "non-rotating" locally (relative to the gyroscopes), but globally, relative to infinity, they will be rotating. This is due to "frame dragging" by the shell, and it indicates, as noted before, that spacetime inside the shell is not flat.

(2) At an angular velocity larger than the one found in #1 above, the tension in the string will be nonzero and the gyroscope directions will change, relative to the string, the same way they did in the flat spacetime case above for any nonzero angular velocity.

(3) At an angular velocity smaller than the one found in #1 above (which includes zero angular velocity, and also negative angular velocities, i.e., rotation in the opposite sense relative to infinity), the tension in the string will be nonzero and the gyroscope directions will change, relative to the string, in the opposite way from #2 above. This would be similar to the behavior in flat spacetime for a nonzero angular velocity in the opposite sense.

9. May 29, 2017

### Staff: Mentor

If the laser is pointed along the axis of rotation of the shell, then yes, it will travel parallel to the axis, because there is no frame dragging in that direction. But if the laser is pointed in any other direction, not parallel to the axis of rotation of the shell, then its path will look curved in the spatial coordinates you are using, because of the curvature of spacetime.

This is not correct. The rotation of the spheres causes frame dragging outside the spheres as well as inside them. Just as the Earth, for example, causes frame dragging on satellites in orbit above it (as was confirmed by Gravity Probe B).

10. May 29, 2017

There is a distance between the spheres. If this distance is great enough then wouldn't this be the case?

11. May 29, 2017

### Staff: Mentor

Wouldn't what be the case? If the distance between the spheres is very large, spacetime near the midpoint between them will be flat to a good approximation (i.e., frame dragging would be negligible); but closer to each sphere spacetime will not be flat (i.e., frame dragging will not be negligible).

12. May 30, 2017

Thanks PeterDonis and pervect for the very detailed and thoughtful answers. Introducing gyroscopes into the picture is great. I was awake most of last night thinking about this and finally realized that just by virtue of spacetime being curved you can get rotational motion. In the same way that just releasing an object on Earth causes the object to accelerate toward the Earth not because spacetime is moving but because it is curved, a curved spacetime inside a rotating shell as we've been discussing can cause an object to rotate relative to a frame outside the sphere. So in addition to the concept of a rotating spacetime being invalid, it is not even necessary as the curvature itself is all that's needed to rotate a frame. And an object rotating at the same angular velocity as that frame will feel no rotational forces.

With regard to the frame-dragging formula in pervect's post, it looks like it might be possible to cause the balls/string to rotate faster than the shell if the mass of the shell were great enough and the radius was small enough. which further enlightens me to the fact that it is only the curvature of spacetime and not any kind of meaningless spacetime rotation that is causing the spinning. Very interesting stuff.

13. May 30, 2017

### Staff: Mentor

As I've said repeatedly, I don't think that this is the case, and I don't think you should keep on asserting it unless you have the math to back it up. It is true that there will be some nonzero angular velocity at which "no rotational forces" are felt (a better technical way to put it is that the motion of a system with this angular velocity will have zero vorticity). But this angular velocity does not have to be the same as the angular velocity of the "frame" (the rotating shell in your example).

As a quick example, the "angular velocity of frame dragging" on the Gravity Probe B satellites, in low Earth orbit, was much, much smaller than the angular velocity of Earth's rotation.

As another example, in Kerr spacetime, the spacetime around a rotating black hole, the "angular velocity of frame dragging" is only equal to the angular velocity of the hole itself at the hole's horizon. Anywhere outside the horizon, the "angular velocity of frame dragging" is smaller, and it decreases as you get further away from the horizon.

14. May 30, 2017

### pervect

Staff Emeritus
No, we must have R be greater than the Scwharzschild radius associated with the mass M, i.e.$R > 2GM/c^2$, else our hollow sphere of radius R would be black hole, and a hollow black hole doesn't make any sense. This makes the maximum value of the constant ${4GM}/{3Rc^2}$ equal to 2/3.

15. Jul 6, 2017

### Yukterez

The event horizon is not at R=2, it is at R=√(r²+a²sin²θ) with r=1+√(1-a²) and "a" being the spin parameter of the central mass. With maximal a=1, r=1 this gives R=√2 at the equator and R=1 at the poles.

Only if the laserbeam is travelling exactly in the center of the rotational axis, if it travels only parallel to, but not exactly along the axis the beam will also be frame dragged.