# Sagnac effect and gyroscopes

1. Jun 28, 2014

### WannabeNewton

As you may already know, given a time-like congruence describing some extended body with world-tube $\mu$ embedded in space-time, there are various different characterizations of what it means for this extended body to be non-rotating. Of particular interest for me is the setting of non-rotation criteria as discussed in the following paper: http://scitation.aip.org/content/aip/journal/jmp/16/2/10.1063/1.522521 Now I don't know how many of you will be able to access the article as it is pay-walled and I have university access but I will try to summarize the parts of the paper relevant to my question, as well as spell out the details of the calculations that the paper completely left out.

Consider a stationary axisymmetric asymptotically flat space-time with time-like and axial Killing fields $\xi^{\mu},\psi^{\mu}$ and let $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$ be another time-like Killing field. Let $\mu$ be a 2-dimensional time-like integral manifold of $\eta^{\mu}$ i.e. each point of the 2-manifold $\mu$ follows an integral curve of $\eta^{\mu}$; then $\mu$ represents the world-tube of what the authors call a "Sagnac tube" and we can think of $\omega$ as the angular velocity of the Sagnac tube relative to spatial infinity (the Sagnac tube can be imagined as a 1-dimensional axisymmetric ring with perfectly reflecting internal walls, surrounding some isolated body). Finally, let $\lambda = -\eta_{\mu}\eta^{\mu}$.

Consider now a half-silvered mirror placed at some point on the Sagnac tube, with world-line $\gamma$, and two null curves $C_+, C_-$ in $\mu$ representing corotating and counterrotating light beams emerging from the mirror; let $k^{\mu},k'^{\mu}$ represent null vector fields on $\mu$ chosen so as to be tangent to $C_+, C_-$. We can always choose $k^{\mu},k'^{\mu}$ such that $k^{\mu}\eta_{\mu} = k'^{\mu}\eta_{\mu} = -1$ by an appropriate normalization. Because $\mu$ is 2-dimensional, all 2-forms on $\mu$ are proportional; letting $\tilde{\nabla}_{\mu}$ be the derivative operator on $\mu$ we then have that $\tilde{\nabla}_{[\mu}k_{\nu]} = \varphi k_{[\mu}\eta_{\nu]}$ hence $-\frac{1}{2}\varphi = k^{\mu}\eta^{\nu}\tilde{\nabla}_{[\mu}k_{\nu]} = -\frac{1}{2}k^{\mu}k^{\nu}\tilde{\nabla}_{\mu}\eta_{\nu} =0$, where the final equality comes from $\eta^{\mu}$ being a Killing field. So $\tilde{\nabla}_{[\mu}k_{\nu]} = 0$ meaning $\oint _{C} k_a dS^a$ is independent of the closed curve $C$ in $\mu$, and similarly for $k'^{\mu}$. Ashtekar and Magnon then show that $\Delta \tau = 2(\lambda|_{\gamma})^{1/2}\oint_C \lambda^{-1}\eta_{\mu}dS^{\mu}$ where $\Delta \tau$ is the Sagnac shift between the corotating and counterrotating light beams upon arrival back at the mirror. From Stokes' theorem we thus have that $\Delta \tau = 2(\lambda|_{\gamma})^{1/2}\int_{\Sigma} \nabla_{[\mu}(\lambda^{-1}\eta)_{\nu]}dS^{\mu\nu}$ where $\Sigma$ is the interior of $C$. Finally, let $\epsilon_{\mu\nu\alpha} = \lambda^{-1/2}\epsilon_{\mu\nu\alpha\beta}\eta^{\beta}$ and $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\eta_{\nu}\nabla_{\alpha}\eta_{\beta}$.

We then have

$(\lambda|_{\gamma})^{1/2}\int _{\Sigma}\lambda^{-3/2}\omega^{\mu}\epsilon_{\mu\nu\alpha}dS^{\nu\alpha}= (\lambda|_{\gamma})^{1/2}\int _{\Sigma}\lambda^{-2}\epsilon^{\mu \gamma \delta \sigma}\epsilon_{\mu \nu\alpha\beta}\eta^{\beta} \eta_{\gamma}\nabla_{\delta}\eta_{\sigma}dS^{\nu\alpha}= -6(\lambda|_{\gamma})^{1/2}\int _{\Sigma}\lambda^{-2}\eta^{\beta} \eta_{[\nu}\nabla_{\alpha}\eta_{\beta]}dS^{\nu\alpha}$

and furthermore $3\eta^{\beta} \eta_{[\nu}\nabla_{\alpha}\eta_{\beta]} = \eta_{\nu} \eta^{\beta} \nabla_{\alpha} \eta_{\beta} - \eta_{\alpha} \eta^{\beta}\nabla_{\nu} \eta_{\beta} -\lambda \nabla_{\nu} \eta_{\alpha} = \eta_{[\nu}\nabla_{\alpha]}\lambda - \lambda \nabla_{[\nu}\eta_{\alpha]}$ since $\eta^{\mu}$ is a Killing field

hence $(\lambda|_{\gamma})^{1/2}\int _{\Sigma}\lambda^{-3/2}\omega^{\mu}\epsilon_{\mu\nu\alpha}dS^{\nu\alpha}= 2(\lambda|_{\gamma})^{1/2}\int _{\Sigma}\nabla_{[\mu}(\lambda^{-1}\eta_{\nu]})dS^{\mu\nu} = \Delta \tau$

Now $\omega^{\mu} = 0$ means that gyroscopes do not precess in the rest frame of the Sagnac tube (since $\eta^{\mu}$ is a Killing field, it makes sense to talk about the rest frame of the entire Sagnac tube). From the above, $\omega^{\mu} = 0 \Rightarrow \Delta \tau = 0$ so it would seem that when dealing with time-like Killing fields, non-rotation relative to local gyroscopes implies non-rotation in the sense of vanishing Sagnac shift. But according to pp.231-232 of the notes http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf, in Godel space-time there exists a Sagnac tube which is non-rotating relative to local gyroscopes ("CIR" or "compass of inertia on the ring" criterion for non-rotation) but rotating according to the Sagnac effect ("ZAM" or "zero angular momentum" criterion for non-rotation); the $k$ in the notes is the $\omega$ in $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$. But this example is clearly at odds with the result above that $\omega^{\mu} = 0 \Rightarrow \Delta \tau = 0$ so what's going on? Why is there this apparent contradiction?

Last edited by a moderator: Jun 29, 2014
2. Jun 30, 2014

### WannabeNewton

It was pointed out to me that the implication $\omega^{\mu} = 0 \Rightarrow \Delta \tau = 0$ in Ashtekar's paper requires $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\Sigma} = 0$ whereas the criterion for non-rotation relative to local gyroscopes attached to the Sagnac tube developed in section 3.2 of the GR notes linked above requires $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ so, the two conditions being nonequivalent, there is no issue at all-it was just a silly mistake on my part.

However that still begs the question: to what extent are the vanishing gyroscopic precession on the Sagnac tube and transitivity of Einstein clock synchronization on the Sagnac tube equivalent? For starters, how does one define clock synchronization on a Sagnac tube? In "General Relativity for Mathematicians"-Sachs and Wu, transitivity of Einstein synchronization for a congruence $\eta^{\mu}$ defined on all of space-time is given by $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ which would be a propagation of local Einstein time through a simultaneity connection with vanishing holonomy along a closed simultaneity curve consisting of events in the vicinity of any and all observers filling space-time and following orbits of $\eta^{\mu}$.

But if we are only interested in clock synchronization on a single Sagnac tube then how should one define clock synchronization? Would $\eta_{[\alpha}\tilde{\nabla}_{\beta}\eta_{\gamma]}|_{\mu} = 0$ suffice, where $\tilde{\nabla}_{\mu}$ is the projected derivative operator on $\mu$ associated with its 2-metric? And if so, is this definition equivalent to the usual one in terms of the Sagnac shift? Try as I may, I keep running into simple conceptual issues when trying to answer any of these questions myself in settings that aren't as simple as the rotating disk in flat space-time wherein all of the above questions can be answered through explicit computation in the rotating coordinates. Thanks in advance!

Last edited: Jun 30, 2014
3. Jul 16, 2014

### WannabeNewton

I've made just a little bit more headway in my question so I've quickly sorted through what I know and what I would like to know just to make things more coherent:

Clearly $\eta_{[\alpha}\tilde{\nabla}_{\beta}\eta_{\gamma]} = 0$ won't tell us anything about clock synchronization or the Sagnac effect because it's a 3-form on the 2-manifold $\mu$ so it will always vanish irrespective of the physics. One might guess that the next best thing then is $\tilde{\nabla}_{[\alpha}\eta_{\beta]} = 0$ as a possible definition of clock synchronization and vanishing Sagnac shift solely on the ring (Sagnac tube). However it can easily be shown that $\tilde{\nabla}_{[\alpha}\eta_{\beta]} = 0$ also always holds for $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$ above.

Indeed $\tilde{\nabla}_{[\alpha}\eta_{\beta]} = \tilde{\nabla}_{\alpha}\eta_{\beta} = \varphi \eta_{[\alpha}\psi_{\beta]}$ since it's a differential form of top degree on $\mu$. We thus have $\eta^{\beta}\psi^{\alpha}\tilde{\nabla}_{\alpha}\eta_{\beta} = \frac{1}{2}\varphi[(\psi^{\alpha}\eta_{\alpha})^2 - (\eta_{\alpha}\eta^{\alpha})(\psi_{\beta}\psi^{\beta})]$ but $\eta^{\beta}\psi^{\alpha}\tilde{\nabla}_{\alpha}\eta_{\beta} = \eta^{\beta}\eta^{\alpha}\tilde{\nabla}_{\alpha}\psi_{\beta} = 0$ hence $\varphi = 0$ since $(\psi^{\alpha}\eta_{\alpha})^2 - (\eta_{\alpha}\eta^{\alpha})(\psi_{\beta}\psi^{\beta}) > 0$. Therefore $\tilde{\nabla}_{[\alpha}\eta_{\beta]} = 0$.

Coming then to my question, in general it does not hold that $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ implies Einstein synchronization of clocks laid out along the ring. Note that $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ is evaluated with the full space-time derivative operator $\nabla_{\mu}$ throughout space-time and then evaluated on $\mu$ so this is completely different from $\eta_{[\alpha}\tilde{\nabla}_{\beta}\eta_{\gamma]} = 0$ which is a trivial statement and uses the induced derivative operator $\tilde{\nabla}_{\mu}$ on $\mu$. As mentioned, $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ is equivalent to the statement that a local rest frame attached to any point on the ring with spatial axes fixed to the circular orbit of this point on the ring will not rotate relative to a momentarily comoving local inertial frame. So even if we have such a local rest frame attached to each point on the ring, the non-rotation of each frame relative to comoving gyroscopes does not guarantee Einstein synchronization, particularly its transitivity, because it's still possible for the Sagnac shift to be non-vanishing on the ring: $\Delta \tau \neq 0$ on $\mu$.

Is there any intuitive way to understand this? The math is straightforward but I can't seem to understand physically why this disconnect exists in general.

4. Jul 16, 2014

### bahamagreen

Lie theoretic obstruction imposed by the Frobenius integrability theorem?

5. Jul 16, 2014

### WannabeNewton

What?

6. Jul 16, 2014

### bahamagreen

Look at the "Transforming to the Born chart" section of Wiki's "Born Coordinates" page...

7. Jul 16, 2014

### Matterwave

If I were to venture a guess at the physical implications of the difference between the two standard definitions of local rotation, it would be that the sagnac effect is still not completely local, because its world tube encompasses a non zero volume of space time. Whereas gyroscopic motion depends on Born rigidity idealized so that every point on the gyroscope can be described as in some kind of rigid motion with respect to the other points, the sagnac effect is not an entirely local effect.

8. Jul 17, 2014

### PhilDSP

I'm not familiar with that particular theorem but it sounds possibly related to the concept of the local to global differences in gauge calculations where phase factors are affected by topological obstructions. In other words, the topological regions are no longer simply connected. See Berry Phase for a more concrete application of these concepts.

Here's an accessible related article:
http://en.wikipedia.org/wiki/Geometric_phase

http://en.wikipedia.org/wiki/Riemann_curvature_tensor

Which gives a link and invocation of, you guessed it: integrability obstruction

Local to global topological relations may be one manifestation of non-locality (to relate this to Matterwave's post). Another way of putting this is that symmetries may exist in a global setting that are broken locally.

Last edited: Jul 17, 2014
9. Jul 17, 2014

### WannabeNewton

No it has absolutely nothing to do with that. That is a statement about how non-vanishing twist of a time-like vector field prevents a global space-like foliation by it which is totally unrelated to the question of why the gyroscopic notion of rotation in general disagrees with the sagnac shift notion of rotation.

10. Jul 17, 2014

### TrickyDicky

According to wikipedia the Sagnac effect "is evidently a global effect".

11. Jul 17, 2014

### TrickyDicky

If it were so, wouldn't it affect the technology based on Sagnac like ring laser gyroscopes, etc?

12. Jul 17, 2014

### WannabeNewton

The problem with this is that we are talking about rotation of an entire ring. So yes while the Sagnac effect is a quasi-local one, so is the notion of non-rotation of the entire ring relative to a gyroscope in the local rest frame. More precisely, imagine having an observer sitting on the ring carrying a gyroscope. The observer initially orients the gyroscope along the local axis tangent to the ring. He then determines if the ring is rotating or not by seeing if the gyroscope precesses relative to the local tangential axis. In order for the ring to be non-rotating, he must determine that the gyroscope does not precess relative to the local tangential axis at every single point on his worldline. This translates over to $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ i.e. we must know that the twist of $\eta^{\mu}$ vanishes everywhere on $\mu$. This is certainly quasi-local. The sagnac shift vanishing is equivalent to the statement that $\psi_{\alpha}\eta^{\alpha}|_{\mu} = 0$ which is definitely quasi-local but I don't see how it is any more non-local than $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$.

Or have I misunderstood what you were saying? Are you using the notions of local and quasi-local in different senses than I am? Are you talking about the fact that rotation of the local tangential axis relative to a comoving gyroscope happens instant by instant, point by point on an integral curve of $\mu$ whereas the Sagnac shift is only measured after complete circuits around some closed curve in $\mu$? I might have misconceptions of my own actually regarding the quasi-locality of gyroscopic non-rotation of the entire ring vs. the quasi-locality of sagnac effect non-rotation so do take what I say above with caution.

The Sagnac effect actually manifests itself in a way entirely analogous, in fact equivalent, to the Aharanov-Bohm effect in QM. This is quite easy to show in fact and you can find a lot of papers on the relation between the Aharanov-Bohm effect and the Sagnac effect. See e.g. Anandan (1981). So it is definitely true that the Sagnac effect itself has direct relation to topological obstructions due to non-trivial fundamental groups of time-like submanifolds of space-time. In the calculations mentioned in the OP, the Sagnac effect in fact arises due to the fact that the cylindrical worldtube of the ring is not simply connected.

But I am unsure as to whether this has relation to why in general the gyroscopic notion of rotation differs from the Sagnac effect notion of rotation in stationary but non-static space-times (I probably should have mentioned that in static space-times the two always agree whenever the ring rotates outside of the photon sphere). It's definitely a good suggestion though so thank you. Like I said I can't immediately see the relation between topological obstructions and the difference in these two notions of rotation but I'll try to think more about it. If you have any more suggestions, possibly related to the Aharanov-Bohm analogy, do mention them, thanks!

Your suggestion is in fact quite similar to the discussion in Statchel (1981) "Globally stationary but locally static space-times: A gravitational analog of the Aharonov-Bohm effect"; I'll read the paper in more detail and see if I can find anything that offers any insight as to why these two notions of rotation differ in e.g. Kerr space-time.

A ring-laser gyroscope is a purely Sagnac effect based apparatus. It has nothing to do with the kind of gyrocopes associated with the notion of non-rotation given by $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$; the latter involves mechanical gyroscopes which are spherical so as to eliminate tidal torques from coupling to the Riemann tensor and sufficiently small so as to be on length scales over which the Riemann tensor is uniform. The use of the term "gyroscope" is ring-laser gyroscopes is a misnomer as the apparatus only measures the Sagnac effect and has no relation to the mechanical gyroscopes one uses to measure rotation through precession.

13. Jul 17, 2014

### Matterwave

I was making the second point, that the Sagnac shift is measured after complete circuits around closed curves.

If I am moving around a central mass in a ring, I can carry with me a sagnac tube and a set of gyroscopes. My guess is that if I just let the gyroscope ride along the ring, and the sagnac tube ride along the ring, then I get, as you calculated, two different notions of rotation. But if the gyroscopes are not simply moving around the ring, but are also carried (and this is very difficult now because physical gyroscopes cannot move along null curves like the light in a sagnac tube) in some specific way around the sagnac tube AS it they move around the ring (think like epicycles) then I should recover a similar notion of rotation.

This is, of course, just my guess. How to implement this calculation is not clear to me, specifically because gyroscopes cannot travel along light like curves.

It may not be possible to implement this calculation for this reason.

14. Jul 18, 2014

### WannabeNewton

Yes the Sagnac effect certainly can only be measured after the light signals complete their retrograde and prograde circuits.
I do think this has some relevance at the least. More specifically, the derivation of the Sagnac shift, not only for light signals but for matter waves as well, assumes that in the local inertial frame momentarily comoving with the light source on the possibly rotating ring at the event at which the prograde and retrograde light signals are emitted, said light signals have the exact same speed. This is tantamount to assuming Einstein clock synchronization in this local inertial frame. By axial symmetry this must be true at every point on the ring. In other words in a momentarily comoving local inertial frame, prograde and retrograde light signals are on completely equal footing. On the other hand, the angular velocity of rotation of the local rest frame of the ring at any given event on it is by definition measured relative to a momentarily comoving local inertial frame so gyroscopic precession is always measured at the level of a local inertial frame whereas the Sagnac effect cannot be measured at this level.

I'm trying to see how to extrapolate from this some conclusion about the relationship, or lack thereof, between the condition $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ on the ring and the condition of vanishing Sagnac shift for Einstein clock synchronization on the ring.
But the fact that $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\Sigma} = 0$ in the interior of the ring (more precisely, in the space-like interior $\Sigma$ of a closed curve $S = \partial \Sigma$ in $\mu$) implies the vanishing of the Sagnac effect has me still quite confused on an intuitive level and prevents me from quickly reaching a conclusion. What's worse is this is only a sufficient condition and not a necessary one for clock synchronization on the ring e.g. for a ring in Kerr space-time with the angular velocity of a zero angular momentum observer we have $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\Sigma} \neq 0$ everywhere in the interior of the ring but the Sagnac effect still vanishes on this ring by construction.

Finally let me again note that in a static axisymmetric space-time, if $\nabla_{\mu}(\psi^{\alpha}\psi_{\alpha}/\xi_{\beta}\xi^{\beta})|_{\mu}\neq 0$, that is if there are no circular null geodesics on the ring, then $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ if and only if the Sagnac shift vanishes. Therefore in stationary axisymmetric space-times like Kerr, there should be an intuitive explanation for the disagreement between the gyroscopic notion of rotation and that of the Sagnac shift based upon the fact that the source is now rotating i.e. based upon the presence of frame dragging.

As you can see my thoughts are quite muddled at the moment; I will try to write something more coherent once I clear up the confusions of mine that I spelled out in this post.

15. Jul 22, 2014

### TrickyDicky

As you know the apparent disconnect you refer to between the gyroscopic and Sagnac concepts of non-rotation is related to the conventionality of Einstein clock sync. You say you understand it from the mathematical point of view. I'm not sure I understand what the purely physical doubt would be.

16. Jul 22, 2014

### WannabeNewton

Hi, thanks. I don't think I know that actually. In what sense is the apparent disconnect due to conventionality of Einstein synchronization?

17. Jul 22, 2014

### TrickyDicky

Maybe I read too much into what you wrote.
Rotating observers in closed paths can't in general trust Einstein synchronization for geometrical reasons. Linearly accelerated observers can and they can measure gyroscopic effects like Thomas rotation, while according to theory Sagnac interferometers should not measure any Sagnac effect. So they are measuring different things(torque vs. angular velocity) and there are situations in wich they shouldn't coincide.

18. Jul 23, 2014

### PhilDSP

This is really a fascinating study, delving into it further. It might be worth comparing 3 different compass systems, all of which are used very heavily by the transportation industry: inertial compasses, gyroscopic compasses and ring laser compasses.

Inertial compasses employ an angular accelerometer to measure angular velocity. The current orientation of the compass’s mass is determined by integrating all prior orientation changes. From that, the newly applied angular force translates into the new orientation of the ship. Nothing rotates except the ship about the compass.

Gyroscopic compasses are similar except that the mass of the compass rotates quickly and therefore the Coriolis force enhances the directional stability of the internal mass. So there are 2 rotations involved. Apparently, the particular orientation of the spin axis with respect to the ship has no bearing on the compass results (if you remove any related mechanical advantages in the operation of the gimbal and float system).

Ring lasers making use of the Sagnac effect have no moving parts. The single rotation involved is again the ship about the compass. However there are 2 paths of light traveling in opposite directions so the rotation is made with respect to 2 different frames of reference.

Gyrocompasses are based on the Foucault pendulum and function in the same way. The addition of the Coriolis force to the force of gravity results in a continuous exchange of momentum between the Earth and pendulum bob. Even though topologically a holonomy exists due to the rotation of the Earth, classical modeling handles the situation perfectly well. My guess is that the angular changes are instantaneously so minute that there is no need to invoke optical or quantum mechanical arguments using SU(2) modeling as you would for the Sagnac effect. Classically, Maxwell’s displacement current deviates imperceptibly from how it would if the Earth were not rotating.

With the ring laser, forces or a change of momentum are not invoked. Phase shift of the light signal is determined by the enclosed angle of the circuit area times photon spin (1). Similarly the phase shift of the Gyrocompass is the enclosed angle of the travel area. Conversely in classical terms for the ring laser, for each light path I believe Maxwell’s displacement current would be required to deviate substantially from how it would if the ship were not rotating.

http://en.wikipedia.org/wiki/Gyrocompass
http://en.wikipedia.org/wiki/Foucault_pendulum
http://en.wikipedia.org/wiki/Ring_laser_gyroscope
http://en.wikipedia.org/wiki/Geometric_phase

Last edited: Jul 23, 2014
19. Jul 23, 2014

### WannabeNewton

What does the ability to measure gyroscopic precession have to do with being able to Einstein synchronize globally in a rotating frame? An observer riding on the ring doesn't need to have his clocks Einstein synchronized with all the other clocks on the ring in order to measure the precession of his spatial axes relative to comoving gyroscopes. He just needs his own clock. A linearly accelerated observer's worldline has zero torsion and if the worldline is given by the orbit of a Killing field the observer will therefore measure no precession of gyroscopes relative to his spatial axes. If the observer's worldline is not the orbit of a Killing field then none of the above applies so it's irrelevant. Furthermore the Sagnac effect derived above is on integral submanifolds of Killing fields that are diffeomorphic to a cylinder so the calculation does not apply to congruences of linearly accelerating observers even if the congruence is a stationary one.

My original question was: Why is the transitivity of Einstein synchronization on the ring, which is equivalent to a vanishing Sagnac shift on the ring, entirely unaffected by the rotation of the ring measured through gyroscopic precession, given that the ring can be rotating relative to local gyroscopes even if it is non-rotating according to the Sagnac effect?

...Actually helped me, I think, understand intuitively why things are as they are. The vanishing Sagnac shift is tantamount to the statement that the local $\hat{\phi}$ and $-\hat{\phi}$ directions are equivalent for an observer on the ring; intuitively, global Einstein synchronization on the ring will be achievable exactly when this is obtained for in this case we can consider the ring as being non-rotating relative to the local space-time geometry. On the other hand, $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ means, as mentioned, that the ring is non-rotating relative to local gyroscopes i.e. it means that the local $\hat{\phi}$ direction does not precess relative to a comoving gyroscope because the precession $\omega^{\alpha}$ of $\hat{\phi}$ satisfies $\omega^{\alpha} \propto \epsilon^{\alpha\beta\gamma\delta}\eta_{\beta}\nabla_{\gamma}\eta_{ \delta}$.

Now intuitively we think of these as being equivalent because when we imagine a ring that is rigidly rotating about an axis we do so in the extended rigid rest frame of spatial infinity and in it we see $\hat{\phi}$ turning continuously as the ring rotates and at the same time we see prograde light signals taking longer to complete a circuit than retrograde signals because the former has to catch up whereas the latter gets caught up with. Let me emphasize that both of these intuitive pictures are in terms of the rotation of the ring in the rest frame of spatial infinity so when we picture the continuous turning of $\hat{\phi}$ along the ring we are doing so relative to a comoving space-like vector $\hat{n}$ that always points towards a fixed star at spatial infinity and when we picture the Sagnac effect we are doing so through the angular velocity of rotation of the ring relative to spatial infinity.

In Schwarzschild space-time, the only value of the angular velocity $\omega$ relative to spatial infinity in $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$ for which the associated ring is non-rotating relative to the local space-time geometry, i.e. who find $\hat{\phi}$ and $-\hat{\phi}$ to be locally equivalent, is $\omega = 0$ because there is no frame-dragging from a non-rotating source; thus in this case the intuitive picture of the Sagnac effect remains consistent. But even in Schwarzschild space-time the intuitive picture of ring rotation relative to local gyroscopes that we have at spatial infinity is not an accurate one because geodetic and Thomas precession will prevent the local gyroscope direction from being $\hat{n}$. This is exemplified most dramatically by the fact that at the photon sphere $r = 3M$ the geodetic and Thomas precessions cause a gyroscope tangent to $\hat{\phi}$ initially to remain tangent to it i.e. $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{r = 3M} = 0$; therefore even though we imagine $\hat{\phi}$ as turning continuously relative to $\hat{n}$ due to the ring's rotational angular velocity relative to spatial infinity, $\hat{\phi}$ is not precessing relative to local gyroscopes so no matter what $\omega$ is, the ring is non-rotating relative to local gyroscopes. However any ring at $r = 3M$ with $\omega \neq 0$ is of course rotating in the sense of yielding a non-vanishing Sagnac shift since $L = \eta^{\mu}\psi_{\mu} = \omega \psi^{\mu}\psi_{\mu}$.

In the local rest frame of the ring at $r = 3M$, the gyroscope is fixed with respect to $\hat{\phi}$ and in Kerr space-time, when considering the local rest frame of a zero angular momentum ring, $\hat{\phi}$ rotates rigidly relative to a comoving gyroscope but this as Matterwave mentioned is purely local in the sense that this measures the instantaneous angular velocity, or torque, of $\hat{\phi}$ relative to the gyroscope about a local axis perpendicular to the instantaneous plane of rotation of $\hat{\phi}$, which is itself a subspace of the local simultaneity surface of the single integral curve $\gamma$ of $\eta^{\mu}$ to which this local rest frame belongs.

In other words, again as Matterwave stated, the gyroscopic precession constitutes a measurement of torque that can be entirely described and calculated in the local simultaneity surface of and along $\gamma$. Propagation of Einstein synchronization across the ring on the other hand requires transport from the local simultaneity surface of $\gamma$ at a given instant to the local simultaneity surface of the neighboring local rest frame of the ring at an infinitesimally simultaneous instant i.e. we move across a curve of simultaneity points belonging to a continuous family of different simultaneity surfaces; the time gap accrued in propagating Einstein synchronization around the ring in this way is exactly what the Sagnac shift measures,which is based upon the angular momentum of the ring itself, and it is clear from this that the measurement is quasi-local as opposed to local because the null curves of the prograde and retrograde light signals will move through every integral curve of $\eta^{\mu}$ on the ring meaning it cannot be described entirely in terms of $\gamma$ and its local simultaneity surface.

So as you said there is certainly no a priori reason to expect the gyroscopic notion of ring rotation to have any relation to the Sagnac effect notion of ring rotation and thus to Einstein synchronization on the ring. That these two notions of rotation agree, at least for the most part, in Schwarzschild space-time and at the same time conform to our intuitive pictures of them is merely coincidental. I hate leaving physics discussions on the note of "coincidental" but in this case I think it is safe to conclude as such.

20. Jul 23, 2014

### TrickyDicky

It was just an example of a situation with gyroscopic rotation but no Sagnac rotation. I just meant that for that to happen Einstein sync must be a convention, otherwise the disconnect wouldn't arise(to begin with there wouldn't even be a Sagnac effect as you say in your question).