# Practical measurements of rotation in the Kerr metric

1. Dec 20, 2013

### yuiop

and gave this reference:
Until WBN mentioned it, I had never given any thought to the difference between these methods of measuring rotation, so I would like to explore those ideas further here, particularly in relation to the Kerr metric.

Consider a *large* stationary thin ring in the equatorial plane of a Kerr black hole, centred on the rotation axis of the KBH. 'Stationary' here is defined as $dr=d\phi=d\theta=0$ as measured in standard B-L Kerr coordinate system as defined in Eq1 here. The ring is hollow with a reflective inner surface such that a light signal can be sent all the way around the inside of ring in either direction and so can be used as a large Sagnac device. A quick calculation reveals that if the gravitational body has non zero angular momentum, then the time for light to travel in opposite directions inside the ring is unequal and the Sagnac device indicates the 'stationary' ring is rotating. The coordinate angular velocity of the light inside the ring can be described by an equation of the form $d\phi/dt = a \pm \sqrt{b}$ where a and b are functions of radius r. If there is a radius where the 'stationary' ring indicates non rotation, then the value of b is zero for that radius. The outermost solution turns out to be $r=GM + \sqrt{GM-\alpha^2}$ where $\alpha$ is the angular momentum per unit mass of the gravitational body. (This equation is equally valid for any Sagnac ring centred on the rotation axis of the gravitational body and lying in a plane parallel to the equatorial plane.) This radius also happens to be the outer event horizon or 'static limit' where no object can maintain constant radius. Therefore there is no location outside the event horizon of a Kerr black hole, where a 'stationary' Sagac ring indicates zero rotation according to the Sagnac effect. (Additionally no ring can be rotationally stationary (i.e $d\phi = 0$) in the Kerr metric, within the ergosphere of the KBH.).

If 3 axis gyroscopes are attached to a large 'stationary' Sagnac ring, would they indicate any rotation at all? How is this calculated?

Now consider a large equatorial Sagnac ring that is rotating with angular velocity:

$$\Omega = - \frac{g_{t \phi}}{g_{\phi \phi}}$$

This is the angular velocity (at that radius) at which an inertial reference frame is said to 'dragged' by the rotating black hole. Would the large Sagnac device still record non zero rotation?

For reference, define 'distant stars' as being stationary with respect to the Kerr metric and so far away that they have negligible effect on measurements local to the KBH. What would the small gyroscopes attached to the ring indicate? As I understand it, when a small box containing a 3 axis gyroscope indicates zero rotation, the box will be rotating relative to a local part of the ring and rotating on the opposite sense to the rotation of the black hole and rotating relative to distant stars. This in turn implies that no global notion of "zero rotation" can be defined using local gyroscopes. Is this correct?

Also, as I understand it, a particle in free fall and will have its angular velocity $(d\phi/dt)$ increased or decreased until it matches $\Omega = - g_{t \phi}/g_{\phi\phi}$. Now if the natural orbital angular velocity of a particle in a circular orbit with radius r is not equal to $\Omega = - g_{t \phi}/g_{\phi \phi}$ then it seems to follow that there is no such thing as a natural stable circular orbit outside a KBH except maybe at a certain critical radius. Does that make any sense?

Sorry for all the questions. Basically I would like to know under what conditions (preferably with an equation) would gyroscopes attached to a large ring indicate zero rotation and when would a large Sagnac ring indicate zero rotation?

Last edited: Dec 21, 2013
2. Dec 20, 2013

### Staff: Mentor

For Kerr spacetime, the two criteria are the same, and the "zero rotation" condition is that, with respect to an observer at infinity, a ring at radial coordinate $r$ is rotating around the hole, in the same sense as the hole's rotation, with angular velocity

$$\Omega (r) = - g_{t \phi} / g_{\phi \phi} = \frac{2 M \alpha}{r^3 + \alpha^2 \left( r + 2M \right)}$$

(Note that the plane of rotation is the hole's "equatorial plane", i.e., $\theta = \pi / 2$. Also note, btw, that the standard coordinate labeling for Boyer-Lindquist coordinates has $\phi$ as the angular coordinate corresponding to the rotational Killing vector field associated with the hole, whereas you appear to be using $\theta$ to label that coordinate.) So the Sagnac ring that is being "dragged along with the hole" will show zero rotation by both criteria. (This state of rotation is often called the "ZAMO" condition, for "Zero Angular Momentum Observer", because the angular momentum of an object rotating with this angular velocity about the hole is zero.)

3. Dec 21, 2013

### Staff: Mentor

No, it won't; the relationship between angular velocity and angular momentum is affected by the value of $\Omega$ at the particle's orbital radius (see my previous post), but the "dragging" due to the hole's rotation doesn't exert any force that will change the particle's angular velocity or force it to be a certain value.

What is true, however, is that the closer the particle's orbit is to the hole's horizon, the smaller the range of possible angular velocities is, and that range is always "centered" on $\Omega (r)$. Inside the ergosphere (i.e., at radial coordinates smaller than the static limit), the possible range of angular velocities no longer includes zero; i.e., it is impossible to remain stationary with respect to infinity. But there is still *some* allowed range; the angular velocity is only forced to be $\Omega (r)$ in the limiting case of an "orbit" exactly at the hole's horizon (which isn't actually physically possible, of course).

4. Dec 21, 2013

### yuiop

Sorry, I meant to use the same notation as in the Wikipedia link. I have now corrected the notation in the OP.
I have deleted that part. I have checked it again and there was an error in that calculation.

I will respond to your other responses when I have digested them ;)

5. Dec 21, 2013

### Staff: Mentor

Don't forget to correct the subscripts in the metric coefficient expressions as well.

6. Dec 21, 2013

### Staff: Mentor

Last edited: Dec 21, 2013
7. Dec 21, 2013

### yuiop

Section 3.2 of the paper linked by WBN compares the Zero Angular Momentum (ZAM) and Compass of Inertia on the Ring (CIR) criteria and seems to conclude in proposition 3.3.4. (No-Go Theorem) that they are not the same in the Kerr metric, but I am not sure if I have read it right as the paper is a bit too technical for me.

If a gyroscope is attached to the ring as in Figure 3.2.3 of that paper, it seems inevitable that it will precess relative to the tangent of the ring when the ring is rotating. To a first approximation the gyroscope will be pointing to the 'distant stars' and on top of that there will be Thomas precession plus whatever the rotating KBH adds to the equation. Additionally the Wikipedia article mentions that:
This implies that when the 'ice scater' is not rotating locally (when compared to a local gyroscope or when using WBN's beaded porcupine device*), she is rotating relative to the gravitational body, the distant stars and the ring.

*

Last edited: Dec 21, 2013
8. Dec 21, 2013

### WannabeNewton

Hey yuiop! Peter already gave a very detailed reply so let met just add one overarching point regarding one of your questions, that of differentiating the Sagnac effect test from the gyroscope test in a general stationary axisymmetric space-time:

As you already noted in your post, non-rotation of the ring as per the Sagnac effect is equivalent to the ring having zero angular momentum. On the other hand, if we mounted a gyroscope on the ring then non-rotation of the ring as per the gyroscope means that the tangent field to the ring is Fermi-Walker transported along the worldline of the gyroscope; this is equivalent to the time-like congruence formed by the worldlines of the points on the ring having zero vorticity i.e. said time-like congruence must be irrotational.

The paper I linked you gives Godel space-time as an example wherein the Sagnac effect test and gyroscope test fail to be equivalent for rings in different regions of Godel space-time.

For Kerr space-time, as Peter noted, they will be equivalent in well-behaved regions. This is because the ring would be described by the time-like congruence $u^{\mu} = \alpha \nabla^{\mu}t$ where $t$ is the time-coordinate in the coordinate system used above and $\alpha$ is the normalization factor. The angular momentum is defined as $L = \psi_{\mu}u^{\mu}$ where $\psi^{\mu}$ is the axial killing field. We have $L = \alpha g_{\mu\nu}\psi^{\mu}g^{\nu\gamma}\nabla_{\gamma}t = \alpha \delta^{\gamma}_{\mu}\psi^{\mu}\nabla_{\gamma}t = \alpha\delta^{t}_{\phi} = 0$. Furthermore since this time-like congruence is the gradient of a scalar field, it is trivially irrotational i.e. $u_{[\gamma}\nabla_{\mu}u_{\nu]} = 0$. Hence the Sagnac effect test and gyroscope test would be equivalent in the well-behaved regions.

A ring at constant coordinate radius $R$ will then have the angular velocity $\Omega = \frac{u^{\phi}}{u^{t}} = \frac{g^{\phi \phi}}{g^{tt}} = -\frac{g_{t\phi}}{g_{\phi\phi}} = \frac{2MRa}{(R^2 + a^2)^2 - a^2\Delta \sin^2\theta}$ relative to a stationary observer at spatial infinity (see Schutz pp. 313-314). As Peter noted, this frame dragging effect can equally characterize the state of non-rotation.

EDIT: And to clarify, $\Omega$ is in the same sense as the rotation of the black hole, which is given by $J = Ma$.

Last edited: Dec 21, 2013
9. Dec 21, 2013

### WannabeNewton

The No-Go Theorem basically states that any general notion of non-rotation in stationary axisymemtric space-times obeying the three specific conditions given in the paper cannot exist in Kerr space-time. It's a different discussion from that of the previous section wherein the equivalence of the ZAMO and compass of inertia tests are discussed; there the counter-example given is Godel space-time not Kerr space-time.

10. Dec 21, 2013

### yuiop

As I understand it, if a particle with initially zero angular momentum is dropped from infinity, it will not free fall along a radial path, but will acquire a horizontal component with $d\phi/dt \ne 0$ (well before reaching the ergosphere). This implies that a rocket hovering at r with $dr=d\phi=d\theta=0$ will not only require vertical thrust in the radial direction to maintain $dr/dt=0$, but will also require a small horizontal thrust in order to maintain $d\phi/dt =0$.

11. Dec 21, 2013

### WannabeNewton

Oh before I forget, I really recommend purchasing the following book if you want detailed discussions of rotation, gyroscopic precession by means of frame dragging, de-Sitter effect etc. in the context of GR: https://www.amazon.com/Gravitation-Inertia-Ignazio-Ciufolini/dp/0691033234

As you can see there are used copies for like 7 bucks so it's totally worth it. It's one of my most favorite GR texts.

12. Dec 21, 2013

### WannabeNewton

This is true. The static observer so mentioned has a 4-velocity $u^{\mu} = \alpha \xi^{\mu}$ where $\xi^{\mu}$ is the time-like killing field and $\alpha$ is the normalization factor. If one computes $a^{\mu} = u^{\gamma}\nabla_{\gamma}u^{\mu}$ then one finds that not only is there a radial component to the acceleration (which is the vertical rocket thrust you mentioned) but also a polar component to the acceleration (which is the horizontal rocket thrust you mentioned). See p.14 of the following: http://arxiv.org/pdf/0704.0986v1.pdf

13. Dec 21, 2013

### yuiop

If Santa doesn't deliver, I'll treat myself

Last edited by a moderator: May 6, 2017
14. Dec 21, 2013

### Staff: Mentor

No, it isn't. Read the paper you linked to carefully: the components of the proper acceleration required to remain static are in the $r$ and $\theta$ directions; there is no component in the $\phi$ direction.

The $\theta$ component is only nonzero outside the "equatorial plane", so if we adopted cylindrical coordinates the only component would be radial--meaning radial in the cylindrical sense, away from the axis of the cylinder. That is, the $\theta$ component is really a coordinate artifact; it's there because we are trying to use spherical coordinates in a spacetime that has only axial symmetry.

[STRIKE]So a freely falling observer will fall inward in the "radial" direction in the axial sense--i.e., radially inward towards the axis.[/STRIKE] [Edit: not sure the previous statement is true, see follow-up in post #18.] He will *not* acquire any component in the tangential (i.e., $\phi$) direction.

Last edited: Dec 21, 2013
15. Dec 21, 2013

### Staff: Mentor

This is not correct; as I said in my reply to WN, there is a nonzero $\theta$ component (if the particle is outside the equatorial plane), but no $\phi$ component. As I noted there, what this is really saying is that the particle free-falls radially inward in the axial rather than spherical sense, i.e., towards the axis of rotation of the hole.

The effect that the "dragging" of spacetime by the hole has is that as the particle free-falls, it will acquire a spin, in the opposite sense to the rotation of the hole. The paper WN linked to notes this on pp. 14-15, though they don't put it in quite those terms; but that's what the comments about the "gravitational Larmor's theorem" mean.

16. Dec 21, 2013

### WannabeNewton

Yes I said there is an acceleration in the polar direction, not the azimuthal direction and if I recall yuiop was using the convention that $\phi$ was polar.

17. Dec 21, 2013

### Staff: Mentor

Not after I pointed out that that isn't the usual convention. See posts #2, #4, and #5.

18. Dec 21, 2013

### Staff: Mentor

On looking again at the paper WN linked to, I'm not sure this is true, except possibly in the initial stages of free-fall for a object that starts from rest far enough away from the axis. An obvious case where it can't be true is an object that is free-falling inward *on* the axis: it certainly won't stay put, so free-fall in general can't be purely in the "radial" direction in the axial (as opposed to spherical) sense.

19. Dec 21, 2013

### WannabeNewton

I was hoping you could clarify this point a bit more. We have (using the regular coordinate conventions for the polar and azimuthal angles) $p^{\phi} = g^{\phi t}p_{t} + g^{\phi\phi}p_{\phi}$ and $p^{t} = g^{t t}p_{t} + g^{\phi t}p_{\phi}$. If we drop in a particle with $p_{\phi} = \psi^{\mu}p_{\mu} = L = 0$ from infinity then $\omega = \frac{d\phi}{dt}= \frac{p^{\phi}}{p^{t}} = \frac{g^{t\phi}}{g^{\phi\phi}}$ so the observer has an angular velocity $\omega$ relative to an observer at infinity when dropped with zero angular momentum $L = 0$ from infinity. Could you explain how to interpret this properly then? Thanks!

Ah ok, missed that sorry!

20. Dec 21, 2013

### Staff: Mentor

At infinity, $\Omega = 0$, so the particle dropped from infinity starts out with zero angular velocity relative to infinity. I.e., at infinity, zero angular velocity is the same state of motion as zero angular momentum. Since there is no $\phi$ component induced in his motion by free-fall, his angular velocity relative to infinity remains zero all the way in. But since zero angular velocity (note that this is *orbital* angular velocity, $d \phi / dt$, to be distinguished from *intrinsic* angular velocity, i.e., spin) is *not* the same as zero angular momentum at a finite radius, the free-falling particle will, as I noted before, acquire a nonzero angular momentum, i.e., a spin, in the opposite sense to the hole's rotation, induced by "frame dragging".

(More precisely, this is what will happen for free fall in the "equatorial plane"; I'm not sure the induced angular momentum will always have the same sign. For example, a particle free-falling inward along the axis of rotation will, if I am reading sources like MTW correctly, acquire angular momentum in the *same* sense as the hole's rotation.)

21. Dec 21, 2013

### WannabeNewton

I agree that initially, that is at infinity, $\Omega= 0$ but as the particle nears finite radii won't $\Omega$ start increasing in the same sense as the hole's rotation?

I'm a bit confused by this. If we started off with $L = 0$ at an initial event at infinity for the freely falling particle then won't $L = 0$ at every event on the particle's worldline since $L$ is a conserved quantity?

22. Dec 21, 2013

### Staff: Mentor

Yes, $\Omega$ does, but there is nothing forcing the angular velocity of the free-falling observer to always be equal to $\Omega$. The only constraint is that, if the free-falling observer starts out with zero angular momentum, then its initial angular velocity must be equal to $\Omega$ at the radius where it starts. So if it starts at infinity, its initial angular velocity will be equal to $\Omega$ at infinity, i.e., zero.

I agree there's something weird here; but the paper you linked to makes clear that there is no $\phi$ component induced in free-fall motion. I'll have to think about this some more.

23. Dec 21, 2013

### WannabeNewton

Ok but if we agree with the usual result that $L = \psi^{\mu}p_{\mu}$ is a conserved quantity in Kerr space-time, so that if $L = 0$ initially on the particle's trajectory then $L = 0$ for all proper time along the particle's worldline, then won't $\Omega = -\frac{g_{t\phi}}{g_{\phi\phi}}$ always equal the angular velocity $\frac{d\phi}{dt}$ of said freely falling particle as per post #19? In other words, $\Omega$ will only fail to be the angular velocity of the freely falling particle if the particle goes from initially having vanishing angular momentum to having a non-zero angular momentum along its trajectory right? But this is what doesn't make sense to me since as noted $L$ is a conserved quantity along this trajectory.

24. Dec 21, 2013

### Staff: Mentor

That seems plausible, but it's not consistent with the paper you linked to, which, as I noted, makes clear that the only components of free-fall motion that change are the $r$ and $\theta$ components; the $\phi$ component does not change. So we must be missing something; I'm just not sure what.

25. Dec 21, 2013

### WannabeNewton

Thanks. I'll have to reread the paper more carefully in a bit but at a cursory glance of the paragraph at the very bottom of p.14, I think the difference in form between the angular velocity of the freely falling observer at the very bottom of p.14 of the paper and the angular velocity of the freely falling observer who starts at infinity with zero angular momentum is in the initial conditions of the free fall motion.

We were talking about an observer freely falling from infinity with an initial angular momentum $L = 0$. On the other hand, if I read the bottom of p.14 correctly, the paragraph is instead talking about an observer who is initially static at some fixed spatial coordinate using a rocket to hover in place and then at some instant switches off the rocket thrusters so as to go into free fall; the initial conditions of the free fall motion in this case would be the state of the static observer at the very instant the rocket thrusters are switched off. In particular, the initial angular momentum would not be zero since the static observer has non-zero angular momentum and I think this is what changes the form of the angular velocity between the two aforementioned cases.