Does the Sagnac Effect Indicate Rotation in Non-Rotating Gyroscope Frames?

[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?

 

[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!

 

[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.

 

[bahamagreen]

Lie theoretic obstruction imposed by the Frobenius integrability theorem?

 

[WannabeNewton]

Lie theoretic obstruction imposed by the Frobenius integrability theorem?

What?

 

[bahamagreen]

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

 

[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.

 

[PhilDSP]

Lie theoretic obstruction imposed by the Frobenius integrability theorem?

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

There's a link there to
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.

 

[WannabeNewton]

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

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.

 

[TrickyDicky]

... the sagnac effect is not an entirely local effect.

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

 

[TrickyDicky]

the gyroscopic notion of rotation in general disagrees with the sagnac shift notion of rotation.

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

 

[WannabeNewton]

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.

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.

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.

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.

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

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.

 

[Matterwave]

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.

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.

 

[WannabeNewton]

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

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.

 

[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.

 

[WannabeNewton]

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.

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

 

[TrickyDicky]

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

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 which they shouldn't coincide.

 

[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/Inertial_navigation_system
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

 

[WannabeNewton]

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.

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?

That being said, your statement...

So they are measuring different things(torque vs. angular velocity) and there are situations in which they shouldn't coincide

...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.

 

[TrickyDicky]

What does the ability to measure gyroscopic precession have to do with being able to Einstein synchronize globally in a rotating frame?

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).

 

[PeterDonis]

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'm not sure "coincidental" is the right word, because I think you can predict to what extent the two notions will agree based on which sorts of "precession" effects exist, relative to our intuitive "baseline" case of Newtonian physics, for which the two notions agree exactly.

First, consider the case of Minkowski spacetime. Do the two notions agree in this case? Not completely, because even in Minkowski spacetime there is Thomas precession, so for $\omega \neq 0$ there will be a mismatch between rotation relative to local gyroscopes and rotation relative to a Sagnac ring, where by "mismatch" we mean that although both will be present (whereas neither is present for $\omega = 0$), they will not be the same (whereas in the Newtonian case, with no Thomas precession, they *are* the same).

In Schwarzschild spacetime, in addition to Thomas precession, we have geodetic precession. What difference does that make? Well, now the difference between the two rotations is more complicated, because the geodetic precession changes the $r$ dependence of the behavior of local gyroscopes. (In fact, as of course you've observed, there is a value of $r$, the photon sphere, at which the gyroscope behavior is independent of $\omega$, which does not happen in Minkowski spacetime.)

In Kerr spacetime, in addition to the above, we have frame dragging, which means that the two rotations are no longer the same even for $\omega = 0$. It also means that there are now two photon spheres instead of one (for co-rotating and counter-rotating photons), and therefore the behavior of the difference in the two rotations is more complicated still.

When I look at the above patterns, I don't see "coincidence"; I see a fairly orderly progression from the Newtonian case. Of course, all of these spacetimes have a high degree of symmetry that no real spacetime has; and relaxing those symmetries makes the correspondence between the two rotations even more complicated still, to the extent that it even has meaning. (For example, what about spacetimes that aren't even asymptotically flat?) But I don't think it's a coincidence that, when particular symmetries exist, particular correspondences between the two notions of rotation also exist.

 

[WannabeNewton]

Thanks Peter. But then, if it isn't coincidental, how does one understand on a more fundamental level the relationship between the two notions of ring rotation? I mean it is pretty clear that the intuitive picture of the gyroscopic notion of rotation as the turning of $\hat {\phi}$ relative to a vector $\hat {n}$ at each point on the ring pointing always at the same fixed star at infinity correctly yielding the rings that actually are rotating according to the local gyroscopes everywhere except $r=3M$ is purely coincidental because the geodetic and Thomas precessions also affect these gyroscopes on rings not on the photon sphere but not enough to cause the gyroscopes to remain parallel to $\hat {\phi}$. On the other hand I will admit that the consistency in Schwarzschild spacetime between the intuitive picture of the Sagnac effect in terms of the angular velocity of the ring relative to spatial infinity and its formal definition in terms of ing angular momentum is not coincidental as it has all to do with the fact that Schwarzschild is static.

But more importantly, is there a physically intuitive way to understand the consistency between the two notions of ring rotation in any static spacetime on any ring that isn't on a photon sphere, if not for coincidence? In other words, is there a more fundamental relationship between these two notions that naturally decouples in non-static but stationary spacetimes and even on photon spheres in static ones? In my post above I couldn't think of anything beyond coincidence seeing as how these are two rather different notions of ring rotation and seeing as how the link between the intuitive picture and motivation behind the gyroscopic definition and its formal definition is fragile even in static spacetimes whereas the same is not true of the Sagnac effect definition.

 

[PeterDonis]

if it isn't coincidental, how does one understand on a more fundamental level the relationship between the two notions of ring rotation?

I'm not sure there is a general way of doing it that applies to any spacetime; as I noted before, there have to be certain symmetries, or maybe "properties" is a better term since asymptotic flatness is one of the key things needed (as is evident from the fact that we keep bringing in fixed stars at spatial infinity as points of reference). And given those symmetries/properties, the relationship is given by how the two notions of rotation relate to the symmetries/properties.

the intuitive picture of the gyroscopic notion of rotation as the turning of $\hat {\phi}$ relative to a vector $\hat {n}$ at each point on the ring pointing always at the same fixed star at infinity correctly yielding the rings that actually are rotating according to the local gyroscopes everywhere except $r=3M$ is purely coincidental because the geodetic and Thomas precessions also affect these gyroscopes on rings not on the photon sphere but not enough to cause the gyroscopes to remain parallel to $\hat {\phi}$.

But this "intuitive picture", at least if I'm understanding you correctly, does *not* give vectors that are non-rotating according to local gyroscopes, even for rings not on the photon sphere, precisely because of geodetic and Thomas precessions. Again, for simplicity, consider Minkowski spacetime, where there is no geodetic precession, only Thomas precession (and therefore no photon sphere). A vector $\hat{n}$ that always points at the same fixed star at infinity is *not* non-rotating according to local gyroscopes for a ring with $\omega \neq 0$, because of Thomas precession; whereas, according to our Newtonian intuitions, a vector $\hat{n}$ that always points at the same fixed star at infinity *should* also be non-rotating according to local gyroscopes, even if $\omega \neq 0$. In other words, Thomas precession adds an *extra* feature that is not present in our Newtonian intuitive picture. (Geodetic precession then just adds a further extra feature.)

So I think it's a bit misleading to just look at the single question of "rotating" vs. "non-rotating" rather than looking at the behavior in more detail. For example, consider a ring that is just outside the photon sphere, i.e., $r = 3M + \epsilon$. For this ring, $\hat{\phi}$ precesses slightly relative to local gyroscopes, but only slightly. So even though, strictly speaking, the ring is rotating by the gyroscope criterion for $\omega \neq 0$, it is only very slightly rotating, so to speak. In other words, instead of just looking at whether $\hat{\phi}$ is or is not precessing relative to local gyroscopes, we need to look at *how fast* it is precessing, i.e., the actual twist rather than just zero vs. nonzero twist.

The question then becomes, what is the analogue of the "twist" (or angular velocity of precession) for the Sagnac effect? I.e., for a ring which is rotating according to the Sagnac criterion, how fast is it rotating by that criterion? I think looking at this may help to shed some light on the relationship between the two criteria and why it isn't coincidental. Unfortunately I don't have time to work through the math right now, so I'm leaving that as an exercise for the reader.

 

[WannabeNewton]

The question then becomes, what is the analogue of the "twist" (or angular velocity of precession) for the Sagnac effect? I.e., for a ring which is rotating according to the Sagnac criterion, how fast is it rotating by that criterion? I think looking at this may help to shed some light on the relationship between the two criteria and why it isn't coincidental.

Well for the rotation of the ring relative to local gyroscopes the precession angular velocity is given by $\omega^{\alpha} = -\frac{1}{2}(\eta^{\lambda}\eta_{\lambda})^{-1}\epsilon^{\alpha\beta\gamma\delta}\eta_{\beta}\nabla_{\gamma}\eta_{ \delta}$. For the Sagnac effect, which measures the angular momentum of the ring, the analogous quantity is the Sagnac shift $\Delta \tau = -2(-\eta^{\lambda}\eta_{\lambda})^{1/2} \oint_S (\eta^{\lambda}\eta_{\lambda})^{-1}\eta_{\alpha}dS^{\alpha}$ where $S$ is a closed curve in the world-tube of the ring. If we take $S$ as an integral curve of the axial Killing field $\psi^{\mu}$, whose spatial projection onto the global rest frame of spatial infinity is just the circle which $\hat{\phi}$ is tangent to, then $\Delta \tau = [2(-\eta^{\lambda}\eta_{\lambda})^{-1/2}(\psi^{\mu}\psi_{\mu})^{-1/2}\Delta S](\eta^{\alpha}\psi_{\alpha})$ where $\Delta S$ is the length of $S$. Note that $L = \eta^{\alpha}\psi_{\alpha}$ is the angular momentum up to normalization: the larger the angular momentum, the larger the Sagnac shift. In Schwarzschild space-time this reduces to $\Delta \tau = 4\pi\omega r^2\sin^2\theta (-\eta^{\lambda}\eta_{\lambda})^{-1/2} = 4\pi\omega r^2(1 - \frac{2M}{r} - \omega^2 r^2)^{-1/2}$ on the equatorial plane. We also have $\Omega \equiv (\omega^{\alpha}\omega_{\alpha})^{1/2} = \omega(1 - \frac{3M}{r})(1 - \frac{2M}{r} - \omega^2 r^2)^{-1}$.

These two expressions certainly reflect the fact that in an arbitrary static axisymmetric space-time, $L = 0 \Leftrightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0 \Leftrightarrow \omega = 0$ if the ring is not on the photon sphere i.e. when $\nabla_{\mu}(\xi^{\alpha}\xi_{\alpha}/\psi^{\beta}\psi_{\beta}) \neq 0$. In fact we can prove that if $\nabla_{\mu}(\xi^{\alpha}\xi_{\alpha}/\psi^{\beta}\psi_{\beta}) \neq 0$ then $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0 \Leftrightarrow L = 0$ and if $\nabla_{\mu}(\xi^{\alpha}\xi_{\alpha}/\psi^{\beta}\psi_{\beta}) = 0$ then $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0, \forall \omega$ but $L \neq 0$ unless $\omega = 0$. In particular, if $\omega \neq 0$ then the Sagnac shift never vanishes but the gyroscopic precession will vanish on the photon sphere even if $\omega \neq 0$ and will not vanish off of it but will retain extremely small values close to it. Apart from that, I cannot discern any other relationship between $\Delta \tau$ and $\Omega$. Let me ask then, what is it about being static that leads to this "almost" equivalence of $L$ and $\omega^{\alpha}$?

There is certainly a clear link between the intuitive picture of the Sagnac effect, in terms of angular velocity relative to infinity, and the rigorous definition in terms of angular momentum. Here it is easy to see that in static space-times, the two will always agree precisely because of the lack of frame-dragging. In stationary but non-static space-times they will not always agree and the reason is also clear: frame-dragging decouples angular velocity relative to infinity and angular momentum. In other words in static space-times if a ring is rotating relative to infinity it will also rotate relative to the local space-time geometry on purely physical grounds. But I do not have a clear link between the angular velocity relative to infinity of the ring and the rotation of the ring relative to local gyroscopes based solely upon being static i.e. what is it about being static that mandates $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0 \Leftrightarrow \omega = 0$ off of a photon sphere but $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0, \forall \omega$ on the photon sphere? If an explanation can be given for that then I think it would be straightforward to find a physically intuitive link between the Sagnac effect and gyroscopic precession based upon the space-time being static.

I ask because on a more fundamental level, in terms of the local rest frame of the ring, I cannot intuitively come up with any physically meaningful relationship between the Sagnac effect and gyroscopic precession. In this respect let me quote something I said earlier:

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.

Thanks.

 

[PeterDonis]

In Schwarzschild space-time this reduces to $\Delta \tau = 4\pi\omega r^2\sin^2\theta (-\eta^{\lambda}\eta_{\lambda})^{-1/2} = 4\pi\omega r^2(1 - \frac{2M}{r} - \omega^2 r^2)^{-1/2}$ on the equatorial plane. We also have $\Omega \equiv (\omega^{\alpha}\omega_{\alpha})^{1/2} = \omega(1 - \frac{3M}{r})(1 - \frac{2M}{r} - \omega^2 r^2)^{-1}$.

What about the corresponding values for Kerr spacetime (in the equatorial plane)?

 

[WannabeNewton]

What about the corresponding values for Kerr spacetime (in the equatorial plane)?

I'll compute them using Mathematica or find papers that have them and get back to you on that.

In the meanwhile I was looking at the following. Again let $\eta^{\alpha} = \xi^{\alpha} + \omega \psi^{\alpha}$ be the tangent field to the ring. We can then write the ring angular momentum as $L = \eta_{\phi} = g_{t\phi} + \omega g_{\phi\phi}$.

We then have that
\omega^{\mu} \propto \epsilon^{\mu\beta\gamma\delta}\eta_{\beta}\partial_{\gamma}\eta_{ \delta} \\= \epsilon^{tr\theta\phi}[\eta_t (\partial_r L \delta^{\mu}_{\theta} - \partial_{\theta}L\delta^{\mu}_r) + L(\partial_{\theta}\eta_t \delta^{\mu}_r - \partial_r \eta_t \delta^{\mu}_{\theta})] \\= \epsilon^{tr\theta\phi}[\omega g_{tt}(\partial_r g_{\phi\phi} \delta^{\mu}_{\theta} - \partial_{\theta}g_{\phi\phi}\delta^{\mu}_r)+\omega g_{\phi\phi}(\partial_{\theta}g_{tt} \delta^{\mu}_r - \partial_r g_{tt} \delta^{\mu}_{\theta})]\\+\epsilon^{tr\theta\phi}(g_{tt} - \omega^2 g_{\phi\phi})(\partial_r g_{t\phi}\delta^{\mu}_{\theta} - \partial_{\theta}g_{t\phi}\delta^{\mu}_r) + \epsilon^{tr\theta\phi}g_{t\phi}[\partial_{\theta}(g_{tt} - \omega^2 g_{\phi\phi})\delta^{\mu}_r - \partial_{r}(g_{tt} - \omega^2 g_{\phi\phi})\delta^{\mu}_{\theta} ]
We can rewrite the first term in brackets in the very last equality as $g_{\phi\phi}\zeta L_{\text{static}} (\hat{\zeta}_{\theta}\delta^{\mu}_r - \hat{\zeta}_r \delta^{\mu}_{\theta})$ where $\zeta_{\mu} = \partial_{\mu}(\xi^{\alpha}\xi_{\alpha}/\psi^{\beta}\psi_{\beta})$, $\zeta = (g_{\mu\nu}\zeta^{\mu}\zeta^{\nu})^{1/2}$, $\hat{\zeta}_{\mu} = \zeta_{\mu}/\zeta$, and $L_{\text{static}} = \omega g_{\phi\phi}$ is the angular momentum the ring would have in a static space-time. Note that $\zeta_{\mu}$ is space-like since $\xi^{\mu}\zeta_{\mu} = 0$ so $\zeta_{\mu} = 0 \Leftrightarrow \zeta = 0$. Then, in a static space-time, $\zeta = 0$ on a photon sphere and only on a photon sphere.

So we finally have
\omega^{\mu} \propto \epsilon^{tr\theta\phi} g_{\phi\phi}\zeta L_{\text{static}} (\hat{\zeta}_{\theta}\delta^{\mu}_r - \hat{\zeta}_r \delta^{\mu}_{\theta})\\+\epsilon^{tr\theta\phi}(g_{tt} - \omega^2 g_{\phi\phi})(\partial_r g_{t\phi}\delta^{\mu}_{\theta} - \partial_{\theta}g_{t\phi}\delta^{\mu}_r) + \epsilon^{tr\theta\phi}g_{t\phi}[\partial_{\theta}(g_{tt} - \omega^2 g_{\phi\phi})\delta^{\mu}_r - \partial_{r}(g_{tt} - \omega^2 g_{\phi\phi})\delta^{\mu}_{\theta} ]
The proportionality is just $-\frac{1}{2}(\eta^{\mu}\eta_{\mu})^{-1}$. There are two cases of interest:

If we are in a static space-time then $g_{t\phi} = 0$ identically and we just have $\omega^{\mu}\propto \epsilon^{tr\theta\phi} g_{\phi\phi}\zeta L_{\text{static}} (\hat{\zeta}_{\theta}\delta^{\mu}_r - \hat{\zeta}_r \delta^{\mu}_{\theta})$; in particular $\Omega = (\omega^{\mu}\omega_{\mu})^{1/2} \propto \zeta L_{\text{static}}$. We see then that in a static space-time the precession relative to $\hat{\phi}$ of a comoving gyroscope on the ring is necessarily proportional to the angular momentum of the ring as well as to $\zeta$, which again vanishes if and only if we are on a photon sphere. This explains the mathematical relationship between the ring gyroscopic precession, the ring angular momentum, and the photon sphere(s) in a static space-time. However, as of yet I do not have a physically intuitive explanation for why $\Omega \propto L_{\text{static}}$ in a static space-time.

The other case of interest is when $L = 0$ on the ring in a stationary space-time i.e. when we are on a ZAM ring. Then $\omega^{\mu} \propto \epsilon^{tr\theta\phi}[\eta_t (\partial_r L \delta^{\mu}_{\theta} - \partial_{\theta}L\delta^{\mu}_r)$ on the ZAM ring; note $\partial_{\mu} L \neq 0$ because we are considering a time-like Killing field $\eta^{\mu}$ which is tangent only to a single ZAM ring, it is not the ZAM congruence itself. This expression for $\omega^{\mu}$ in effect says that the precession relative to $\hat{\phi}$ of a comoving gyroscope on the ZAM ring is given by the curl of the angular momentum. I again don't immediately have a physically intuitive explanation of this.

 

[PeterDonis]

What about the corresponding values for Kerr spacetime (in the equatorial plane)?

I just remembered that Bill_K has a blog post with handy formulas for $\Omega$ in Minkowski, Schwarzschild, and Kerr spacetime:

https://www.physicsforums.com/blog.php?b=4448

His formula for Kerr spacetime is (rearranged somewhat to match how you gave the Schwarzschild formula):

$$
\Omega = \left[ \omega \left( 1 - \frac{3M}{r} ( 1 - a \omega ) \right) + \frac{M a}{r^3} \left( 1 - a \omega \right)^2 \right] \left[ 1 - \frac{2M}{r} \left( 1 - a \omega \right)^2 - \omega^2 \left( r^2 + a^2 \right) \right]^{-1}
$$

Unfortunately I don't have a handy source for the Sagnac time delay in Kerr spacetime.

 

[milankecman]

My hypothesis:Sagnac effect is the measuring the angle of aberration by interference of light.

Like this:http://www.sciencechatforum.com/viewtopic.php?f=39&t=26841

 

[WannabeNewton]

Peter would you agree with the following progression of statements:

-In Newtonian mechanics, the rate at which the circle tangent $\hat{\phi}$ precesses relative to spatial infinity, that is relative to the telescope vector $\hat{n}$ pointing at a fixed guide star in the sky, is just $\frac{d\phi}{dt} = \omega$ which is of course the rate at which the ring rotates relative to spatial infinity. In Newtonian mechanics we know $\hat{n}$ constitutes a local gyroscope axis at each point on the ring.

-In a static space-time, $\hat{\phi}$ still precesses at the rate $\omega$ with respect to $\hat{n}$ but $\hat{n}$ no longer constitutes a local gyroscope axis comoving with $\hat{\phi}$. Instead the local gyroscope axis $\hat{S}$ itself precesses relative to $\hat{n}$ according to Fermi-transport $\frac{d \vec{S}}{d\tau} = \vec{\tilde{\Omega}}\times \vec{S}$. Now because the space-time is static, the source has no angular momentum $J$ and hence no angular velocity $\omega_J = \frac{J}{M}$. The only angular velocity involved in a static space-time, with regards to the rigidly rotating ring, is that of the ring itself, $\omega$; in other words it naturally follows from the space-time being static that the precession of the local gyroscope axis relative to $\hat{n}$ is due purely to the kinematics of the ring, in this case its rotation. Hence $\tilde{\Omega} \propto \omega$ and so the precession $\Omega$ of $\hat{\phi}$ relative to local gyroscopes, which is $\omega - \tilde{\Omega}$, must satisfy $\Omega \propto \omega$.

At the same time, because the space-time is static the angular momentum $L$ of the ring satisfies $L \propto \omega$ because again the source angular momentum is zero so as far as rotational parameters go $L$ can only depend on $\omega$, that is on the kinematics of the ring, by dimensional analysis. Therefore $\Omega \propto L$ naturally in a static space-time. So $\Omega$ vanishes whenever $L$ vanishes and if $L$ vanishes then $\Omega$ does too unless we are on a photon orbit wherein $\tilde{\Omega} = \omega$ for reasons we discussed in my "Null geodesics and rotation" thread. So to sum up, in a static space-time the lack of source rotation forces the precession of local gyroscopes on a ring with respect to spatial infinity as well as the angular momentum of the ring to be proportional to the angular velocity of the ring with respect to spatial infinity.

Therefore the two notions of rotation always agree except at the photon orbits where the geodetic and Thomas precession are strong enough to make the gyroscopic precession exactly equal to the ring angular velocity, thus making the ring locally non-rotating according to comoving gyroscopes even if it is rotating in the sense of having an angular momentum which in static space-times is equivalent to having an angular velocity. So in static space-times the reason the non-rotation definitions $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ and $L = 0$ disagree on photon orbits is due entirely to static space-time curvature (geodetic precession) and special relativistic kinematics (Thomas precession) causing disagreement between the local gyroscope axis and the telescope axis pointing at a fixed distant star in the sky.

-In stationary space-times the source now has a non-vanishing angular momentum $J$. We know that $\vec{A} = (0,0,A_{\phi})$ defines a gravitomagnetic vector potential induced by $J$, where $A_{\phi} \equiv g_{t\phi}$. Now neither non-rotation in the sense of the Sagnac effect, with $L = 0$, nor non-rotation relative to local gyroscopes, with $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$, agree with non-rotation relative to spatial infinity with $\omega = 0$. This is easiest to see in a weak gravitational field wherein the precession of a ring comoving gyroscope relative to spatial infinity is given by $\vec{\tilde{\Omega}} = \omega r \hat{\phi}\times (-\frac{1}{2}\vec{a} + \frac{3}{2}\vec{\nabla}\varphi) + \vec{\nabla}\times \vec{A}$ where $\varphi$ is the Newtonian potential. The non-normalized ring angular momentum is $L = \omega g_{\phi \phi} + A_{\phi}$.

We see that the angular momentum differs from its value in a static space-time by $\vec{A}$ whereas the gyroscopic precession differs from its value in a static space-time by $\vec{\nabla}\times \vec{A}$ which is is to say the Sagnac shift $\Delta \tau$ is influenced directly by the gravitomagnetic vector potential, whereas the gyroscopic precession relative to spatial infinity is influenced only by its curl which makes sense since the latter measures a torque whereas the former measures rotation relative to the local space-time geometry in the sense of non-equivalent local $\hat{\phi}$ and $-\hat{\phi}$ directions. As a result, $L = 0$ does not imply $\Omega = 0$ where $\Omega$ is the precession of $\hat{\phi}$ relative to local gyroscopes; in a stationary space-time $\tilde{\Omega}$, and hence $\Omega$, are no longer proportional to $L$.

Does that constitute a more or less complete description/explanation of the situation? Thanks.

 

[PeterDonis]

Peter would you agree with the following progression of statements

Yes, with a couple of minor comments:

In a static space-time, $\hat{\phi}$ still precesses at the rate $\omega$ with respect to $\hat{n}$

With respect to infinity, yes; but with respect to the local observer, the rate is $\gamma \omega$, where $\gamma$ is the "time dilation factor", e.g., $\gamma = \left( 1 - \omega^2 r^2 \right)^{-1/2}$ in Minkowski spacetime. This doesn't change the proportionalities, but it should be noted since the formulas written previously in this thread included the extra factor of $\gamma$. (Note that an additional factor of $\gamma$ appears when, for example, you evaluate the Thomas precession##.)

So $\Omega$ vanishes whenever $L$ vanishes and if $L$ vanishes then $\Omega$ does too unless we are on a photon orbit

I think you mean if $\Omega$ vanishes then $L$ vanishes too unless we are on a photon orbit. I.e., $L$ vanishing implies $\Omega$ vanishing always, and $\Omega$ vanishing implies $L$ vanishing except on a photon orbit.

 

[WannabeNewton]

With respect to infinity, yes; but with respect to the local observer, the rate is $\gamma \omega$, where $\gamma$ is the "time dilation factor", e.g., $\gamma = \left( 1 - \omega^2 r^2 \right)^{-1/2}$ in Minkowski spacetime. This doesn't change the proportionalities, but it should be noted since the formulas written previously in this thread included the extra factor of $\gamma$.

Yes, good point.

I think you mean if $\Omega$ vanishes then $L$ vanishes too unless we are on a photon orbit. I.e., $L$ vanishing implies $\Omega$ vanishing always, and $\Omega$ vanishing implies $L$ vanishing except on a photon orbit.

Sorry, that is indeed what I meant.

There is one more thing I was hoping to get clarified. Now it is easy to see that in a stationary axisymmetric space-time, a congruence of observers with 4-velocity field $u^{\alpha} = \gamma(\xi^{\alpha} + \tilde{\omega} \psi^{\alpha})$ has vanishing vorticity if and only if it has vanishing angular momentum, where now $\tilde{\omega} = \tilde{\omega}(r,\theta)$ so that $\mathcal{L}_{u}\tilde{\omega} = 0$ but $\nabla_{\mu}\tilde{\omega} \neq 0$.

Indeed $L = 0 \Rightarrow \omega = -\frac{g_{t\phi}}{g_{\phi\phi}}\Rightarrow u^{\alpha} = \gamma \nabla^{\alpha} t \Rightarrow u_{[\alpha}\nabla_{\beta}u_{\gamma]} = 0$. Conversely $u_{[\alpha}\nabla_{\beta}u_{\gamma]} = 0 \Rightarrow u^{\alpha} = \gamma \nabla^{\alpha}t$ for some global time function $t$ so $L = \gamma g_{\alpha\beta}g^{\alpha \gamma} \nabla_{\gamma}t \psi^{\beta} = \delta^{t}_{\phi} = 0$ using a coordinate system adapted to both $\psi^{\alpha}$ and $\nabla^{\alpha} t$, which can always be constructed since they commute.

So there should be a physically intuitive way to understand $L = 0$ in terms of the vanishing vorticity $u_{[\alpha}\nabla_{\beta}u_{\gamma]} = 0$ of the ZAM congruence. We already know that $L =0$ is equivalent to $\Delta \tau = 0$, where $\Delta \tau$ is as usual the Sagnac shift or clock desynchronization, so we have a physically intuitive way to understand it already; indeed $\Delta \tau = 0$ just means that the observer is non-rotating relative to the local space-time geometry. But I was just curious as to whether one could physically interpret non-rotation relative to the local space-time geometry in terms of $u_{[\alpha}\nabla_{\beta}u_{\gamma]} = 0$ seeing as how they are mathematically equivalent. Do you happen to have any ideas?

By the way, recall we had a similar discussion in an older thread. There we noted that even though $u_{[\alpha}\nabla_{\beta}u_{\gamma]} = 0$, this does not imply that the $\{e_i \} \equiv \{e_r, e_{\theta},e_{\phi}\}$ local rest frame of the ZAM ring does not precess relative to comoving gyroscopes; here $e_{\phi} = \hat{\phi}$. As we've seen, what determines the local non-rotation of the local rest frame $\{e_i \}$ is whether or not $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ where $\eta^{\alpha} = \xi^{\alpha} + \omega \psi^{\alpha}$ with $\omega$ a constant angular velocity ($\nabla_{\mu}\omega = 0$) which equals the angular velocity of a single ZAM ring, the one to which the above local rest frame is attached. I've attached a discussion from Frolov and Novikov if it helps at all.

Thank you very much in advance.

 

[PeterDonis]

I've attached a discussion from Frolov and Novikov if it helps at all.

It does help, at least in raising what might be a better question. The question is: what do the principal axes of deformation look like for a ZAMO in Kerr spacetime in the equatorial plane? Frolov and Novikov say that a local gyroscope does not rotate relative to those axes (at least if I'm understanding them right), so visualizing those axes should help in visualizing the local gyroscope behavior. By "visualize" here I really mean visualize relative to spatial infinity: do the principal axes of deformation keep on pointing at the same distant star, for example? I think that's been a key missing piece in previous discussions: we don't have a convenient visualization of how a local gyroscope for a ZAMO in Kerr spacetime points relative to a fixed object at infinity, the way we do for a local gyroscope in Minkowski or Schwarzschild spacetime.

 

[WannabeNewton]

Peter I agree that a more in-depth analysis of the discussion in Frolov and Novikov will greatly aid us in understanding the ZAMO precession effects. But before that I just wanted to sum up my understanding so far to see which elements you find flawed. We know the following. If $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ everywhere on space-time then all observers following orbits of $\eta^{\alpha}$ can Einstein synchronize their clocks with one another by setting them to the time $t$ given by $\eta^{\alpha} = \gamma \nabla^{\alpha} t$, this holding precisely because $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ on the entirety of space-time. This is true whether or not $\eta^{\alpha}$ is a Killing field and is simply because in the coordinate system adapted to $\eta^{\alpha}$, which has global time coordinate $t$, we have between any two infinitesimally separated Einstein simultaneous events $0 = dx^{\alpha}\eta_{\alpha} = \gamma dx^{\alpha} \nabla_{\alpha} t = \gamma dt$ so $\oint dt = 0$ trivially along any curve of Einstein simultaneous events relative to $\eta^{\alpha}$ in this coordinate system.

Denote by $\mu$ the worldtube of a possibly rotating ring. Then, if $\eta^{\alpha}$ is a Killing field, $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu} = 0$ implies that a gyroscope comoving with any point on the ring does not precess relative to the ring's local rest frame there; however it does not imply Einstein synchronization of clocks laid out around the ring. Indeed as we've seen $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\mu}$ cannot by itself characterize Einstein synchronization on the ring because the clock desynchronization is given by $\Delta \tau = \lambda^{1/2}\int_{\Sigma} \lambda^{3/2}\epsilon_{\alpha\beta\gamma}\omega^{\alpha}d\Sigma^{\beta\gamma}$, where $\omega^{\alpha}$ is the twist of $\eta^{\alpha}$ and $\Sigma$ is the interior of a closed curve in $\mu$, so that only $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{\Sigma} = 0$ suffices in Einstein synchronizing the clocks around the ring. Again this exact relationship between $\Delta \tau$ and $\omega^{\alpha}$ holds only if $\eta^{\alpha}$ is a Killing field; if $\eta^{\alpha}$ isn't a Killing field then only the first paragraph above holds.

Intuitively the reason for gyroscopic precession on the ring being given by $\omega^{\alpha}|_{\mu} = \epsilon^{\mu\nu\gamma\delta}\eta_{\nu}\nabla_{\gamma}\eta_{\delta}|_{\mu}$ but the Sagnac shift being given by $\Delta \tau = \lambda^{1/2}\int_{\Sigma} \lambda^{3/2}\epsilon_{\alpha\beta\gamma}\omega^{\alpha}d\Sigma^{\beta\gamma}$ is the same as that in the Aharonov-Bohm effect. Therein we send two charged particles through adjacent slits such that the local magnetic field is zero everywhere along their paths but with a solenoid in between their trajectories in which a non-zero magnetic field exists; the phase shifts of the associated matterwaves, and resulting interference upon impinging on a screen, is directly given by the flux of the magnetic field in the solenoid through the interior of the closed loop defined by the particles' trajectories whereas the precession of their spins, modulo the precession due to kinematic effects, depends only on the local magnetic field which in this case vanishes. In the case of the rotating ring, if we imagine going to the coordinate system corotating with the ring then $\omega^{\alpha}$ defines a gravitomagnetic field in these coordinates so even if $\omega^{\alpha}|_{\mu} = 0$ it's entirely possible for it to have a non-vanishing flux $\Delta \tau = \lambda^{1/2}\int_{\Sigma} \lambda^{3/2}\epsilon_{\alpha\beta\gamma}\omega^{\alpha}d\Sigma^{\beta\gamma}$ through $\Sigma$ if $\omega^{\alpha}|_{\Sigma} \neq 0$ identically.

In other words the gyroscopic precession only couples to the local gravitomagnetic field since in the corotating coordinate system its at rest at a fixed point and as such is only subject to the gravitomagnetic torque there, whereas the Sagnac shift involves light beams that traverse entire closed circuits around the ring and so are affected globally by the flux of the gravitomagnetic field through the enclosed region due to Stokes' theorem. If $\eta^{\alpha}$ is a Killing field and $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ everywhere then both the Sagnac shift and gyroscopic precession vanish relative to rings described by $\eta^{\alpha}$ but now we're in a static space-time by definition in which case the rings also have no angular velocity relative to spatial infinity so there's nothing surprising there.

In the other extreme if we are looking at the tangent field $t^{\alpha}$ of the ZAMO congruence, which isn't a Killing field, then $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ everywhere which means there is no Sagnac shift and comoving clocks are Einstein synchronizable globally but the ZAMO rings still rotate relative to local gyroscopes as we've seen already. According to Frolov and Novikov, these local gyroscopes do not precess relative to the principal axes of the shear tensor $\sigma_{\alpha\beta}$ of $t^{\alpha}$. If true, then I think the only physically intuitive way to understand $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ is in fact through the vanishing of the Sagnac shift and the ability to globally Einstein synchronize clocks of this tangent field; the fact that gyroscopes don't precess relative to the principal axes of $\sigma_{\alpha\beta}$ doesn't by itself tell us anything meaningful about the rotation of the ring through the space-time symmetries and in fact I don't even think it gives us a meaningful notion of ring rotation at all; in other words I don't think it directly relates back to the Sagnac effect in any physically intuitive way, for what possible relation could there be between the Sagnac effect and the principal axes of $\sigma_{\alpha\beta}$?

Would you agree with all this or are there points that you feel are wrong? Thanks in advance.

 

[PeterDonis]

Would you agree with all this

This all looks OK to me; I have a couple of comments/questions.

(1) For a spacetime which is stationary but not static, must it be the case that any congruence satisfying the ZAMO condition (i.e., zero Sagnac shift) must have nonzero shear? (This is certainly the case for Kerr spacetime.) If so, that would be a connection between the Sagnac shift and the shear--basically, if zero Sagnac shift requires nonzero angular velocity, it must also require nonzero shear.

(2) I like the analogy with the Aharonov-Bohm effect, because it makes clear the distinction between local effects and nonlocal effects.

 

[WannabeNewton]

(1) For a spacetime which is stationary but not static, must it be the case that any congruence satisfying the ZAMO condition (i.e., zero Sagnac shift) must have nonzero shear?

I do believe so. The congruences we are interested in are representations of circular trajectories with angular velocities $\omega(r,\theta)$ and tangent fields of the form $\eta^{\alpha} = \xi^{\alpha} + \omega(r,\theta)\psi^{\alpha}$ in a stationary, axisymmetric space-time. If $\nabla_{\mu}\omega = 0$ i.e. if $\omega(r,\theta) = \text{const.}$ throughout space-time then $\eta^{\alpha}$ would be a Killing field. If it is also a ZAMO tangent field then we know $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ but if the space-time possesses a Killing field with vanishing twist it must necessarily be static so to be non-static the ZAMO tangent field has to have $\nabla_{\mu}\omega \neq 0$ which in turn makes $\sigma_{\alpha\beta}\neq 0$.

(This is certainly the case for Kerr spacetime.) If so, that would be a connection between the Sagnac shift and the shear--basically, if zero Sagnac shift requires nonzero angular velocity, it must also require nonzero shear.

Thank you, this is a very good point. So there is definitely some physical connection between the non-zero shear of the ZAMO congruence and the rotation of the space-time. I'll write up a more detailed post after working though this potential connection in more detail.

 

[WannabeNewton]

Hi Peter. So here's what I've found so far. If you take a look here, https://www.physicsforums.com/showpost.php?p=4823868&postcount=7, you'll see basically a restatement of the claim in Frolov and Novikov that since $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ for the ZAMO congruence, the local frame $\{e_{\alpha}\}$ chosen for any integral curve of $t^{\alpha}$ with spatial axes aligned with the principal axes of $\sigma_{\alpha\beta}$ is non-rotating relative to gyroscopes i.e. it is Fermi-transported; see also http://postimg.org/image/gaokzoapr/. This certainly makes sense intuitively since the principal axes are left invariant under $\sigma_{\alpha\beta}$ hence they rotate rigidly through $\omega^{\alpha} = \epsilon^{\alpha\beta\gamma\delta}t_{\beta}\nabla_{\gamma}t_{\delta}$ which for the case at hand vanishes so the principal axes should be non-rotating, leading to $\{e_{\alpha}\}$ being Fermi-transported. My problem is I can't show rigorously that $F_u e_{\alpha} = 0$ is indeed the case, where $u^{\alpha} = (-t_{\beta}t^{\beta})^{-1/2}t^{\alpha}$. Do you happen to know of a proof?

But let's take it for granted for now that the principal axes $\{e_{\alpha}\}$ do indeed define a Fermi-transported frame when $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ as claimed in Frolov and Novikov and those other books in the thread I linked. Refer now to eq. (51) in http://users.ugent.be/~nvdbergh/workshop/info/Costa_Natario Gravito-magnetic analogies.pdf which gives the rate of change, relative to some frame, of the spatial part $Y^{\hat{i}}$ of a connecting vector $X^{\alpha}$ between neighboring observers in a congruence i.e. the relative velocity of a neighboring observer in the chosen frame of a fiducial observer.

In the case of a Fermi-transported frame, such as the principal axes above, and a vorticity-free and expansion congruence such as the ZAMO congruence at hand, we have for the relative velocity $\nabla_u Y^{\hat{i}} = \sigma^{\hat{i}}{}{}_{\hat{j}}Y^{\hat{j}}$. But $\sigma^{\hat{i}}{}{}_{\hat{j}}$ is diagonal in this frame so $\nabla_u Y^{\hat{i}}$ is clearly parallel to $Y^{\hat{i}}$ i.e. neighboring observers of the ZAMO congruence do not rotate relative to the fiducial ZAMO observer in this frame and since the frame itself is non-rotating relative to comoving gyroscopes, we can say the neighboring observers do not rotate relative to gyroscopes carried by the fiducial observer.

Does that sound about right? If so then we will have a nice intuitive picture that brings everything together, at least I think.

I will post my thoughts on this subsequently so that this one does not get too long.

 

[WannabeNewton]

If the above is right, then when it comes to the ZAMOs, we can say that they are locally non-rotating in two senses. First, we know $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0 \Leftrightarrow L = 0$ the latter being itself equivalent to a vanishing Sagnac shift, so they are locally non-rotating in the already mentioned sense of finding the local $\hat{\phi}$ and $-\hat{\phi}$ directions equivalent. But if the statement in Frolov and Novikov etc. are true then the vanishing vorticity also means the ZAMOs are locally non-rotating in the sense that there exists for each ZAMO a Fermi-transported frame, particularly the one with the principal axes of $\sigma_{\alpha\beta}$, in which neighboring ZAMOs do not rotate.

And these two notions are related of course. The $L = 0$ notion basically says that the ZAMOs are stationary with respect to the local rotation of space-time i.e. they allow themselves to be "dragged" by the local rotation so as to be at rest relative to it. In order to do this, the father out a ZAMO is from the source the smaller the angular velocity relative to spatial infinity since the local space-time rotation drops off as $1/r$. Because of this, a connecting vector between neighboring ZAMOs would appear to rotate relative to spatial infinity because of the difference in angular velocities that lends itself to a non-vanishing $\sigma_{\alpha\beta}$.

But the difference in angular velocities is a global relative rotation whereas it is $L = 0$ that is local non-rotation, in the above sense, so one could argue that in a local inertial frame momentarily comoving with a ZAMO it is the fact that all neighboring ZAMOs also have $L = 0$ that should determine the (non)rotation of neighboring ZAMOs relative to this momentarily comoving local inertial frame. To put it differently, if all the ZAMOs are at rest with respect to the local space-time rotation then it would seem natural that neighboring ZAMOs do not rotate in a momentarily comoving local inertial frame attached to a fiducial ZAMO. And if it is indeed true that $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0 \Leftrightarrow L = 0$ implies the principal axes of $\sigma_{\alpha\beta}$ define a Fermi-transported frame then we exactly have that there exists such a momentarily comoving local inertial frame wherein the neighboring ZAMOs do not rotate and it is the one in which the irrelevant shearing due to difference in angular velocities is factored out as it physically only represents rotation relative to spatial infinity, at least I think that's the correct physical picture.

But of course the ZAMOs are locally rotating in the sense discussed previously in the thread i.e. $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} \neq 0$, where $\eta^{\alpha}$ is the time-like Killing field which coincides with anyone ZAM ring so that the natural local rest frame of the ZAM ring, determined entirely by the stationary and axial symmetries of the space-time, rotates relative to comoving gyroscopes.

 

[PeterDonis]

Does that sound about right?

I'm not sure because I'm not sure how the nonzero shear of the ZAMO congruence comes in. The principal axes of the shear tensor are non-rotating relative to local gyroscopes, but the connecting vectors between neighboring ZAMOs are not aligned along the principal axes. I'm not sure how that's being captured.

 

[WannabeNewton]

I'm not sure because I'm not sure how the nonzero shear of the ZAMO congruence comes in. The principal axes of the shear tensor are non-rotating relative to local gyroscopes, but the connecting vectors between neighboring ZAMOs are not aligned along the principal axes. I'm not sure how that's being captured.

Right but if you see the equation I linked in that paper by Costa, you will see that the "time" evolution of any connecting vector between neighboring ZAMOs in the chosen frame is given solely by the shear tensor expressed in that frame in an equation of the form $\dot Y ^i = \sigma ^i _{}{} _j Y^j$. If the frame has its spatial axes aligned with the principal axes then the shear tensor will be diagonal so the connecting vector will not rotate in this frame because $\dot Y^i \propto Y^i$ i. e. the relative velocities between neighboring ZAMOs in this frame have no components perpendicular to the connecting vectors so there are no relative rotations between ZAMOs in this frame as a result.

 

[PeterDonis]

If the frame has its spatial axes aligned with the principal axes then the shear tensor will be diagonal so the connecting vector will not rotate in this frame because $\dot Y^i \propto Y^i$ i. e. the relative velocities between neighboring ZAMOs in this frame have no components perpendicular to the connecting vectors so there are no relative rotations between ZAMOs in this frame as a result.

Ah, I see; with the axes oriented this way, the shear shows up entirely as changes in *length* of the connecting vectors, with no changes in *direction* of those vectors.

 

[WannabeNewton]

Ah, I see; with the axes oriented this way, the shear shows up entirely as changes in *length* of the connecting vectors, with no changes in *direction* of those vectors.

Yes indeed. And I meant to ask, what are your thoughts on my comments above, at the top of this page? Thanks!

 

[PeterDonis]

what are your thoughts on my comments above, at the top of this page?

Everything looks correct mathematically, as far as I can tell. I'm still trying to visualize how it all works.

 

[WannabeNewton]

Actually Peter I think I spoke too soon

So we have $\nabla_u Y^{\hat{i}} = \sigma^{\hat{i}}{}{}_{\hat{j}}Y^{\hat{j}}$. When the matrix is diagonal as is the case in the basis of principal axes, we have $\nabla_u Y^{\hat{i}} = \sigma^{i}{}{}_{i}Y^{\hat{i}}$ (no implied summation). But $\nabla_u Y$ (as a vector) is in general still not parallel to $Y$ (as a vector) because $\sigma^1{}{}_1 \neq \sigma^2{}{}_2 \neq \sigma^3{}{}_3$ in general.

It only holds when $Y$ is an eigenvector of $\sigma_{\alpha\beta}$ which is just the statement that if the connecting vector is aligned with a principal axis then it doesn't rotate relative to comoving gyroscopes when there is vanishing vorticity (this is assuming that an eigenvector of $\sigma_{\alpha\beta}$ is actually Lie transported by the congruence so as to be a connecting vector, which I haven't yet verified). In other words $\nabla_u Y$ can still have a component corresponding to an instantaneous rotational velocity if it isn't aligned with a principal axis. As such, I don't get how Sachs and Wu, in http://postimg.org/image/gaokzoapr/, come to the conclusion that if we choose a basis for a fiducial congruence observer in which the shear tensor is diagonalized (principal axes) then neighboring observers do not rotate relative to this observer. How do they come to this conclusion?

Intuitively it just doesn't make sense to me because the ZAMOs are supposed to be as close as possible to a state of rest with respect to space-time geometry in a rotating space-time since they allow themselves to exactly follow the local rotation of space-time. This leads to them all having vanishing angular momentum. As such I don't get why physically a given ZAMO would see neighboring ZAMOs rotate relative to gyroscopes that he carries if the neighbors happen not to be aligned with a principal axis of the shear tensor. How can the neighbors rotate relative to the fiducial ZAMOs when they are meant to all define being at rest relative to the local space-time geometry? But unless I'm making another mistake, this unfortunately ruins the nice intuitive picture of local non-rotation of ZAMOs in terms of vanishing vorticity as opposed to the picture we already had in terms of the Sagnac effect and Einstein synchronization.

 

[PeterDonis]

It only holds when $Y$ is an eigenvector of $\sigma_{\alpha\beta}$ which is just the statement that if the connecting vector is aligned with a principal axis then it doesn't rotate relative to comoving gyroscopes when there is vanishing vorticity

Which would apply to the ZAMO congruence in Kerr spacetime, but...

(this is assuming that an eigenvector of $\sigma_{\alpha\beta}$ is actually Lie transported by the congruence so as to be a connecting vector, which I haven't yet verified).

I'm not sure if this is true of the ZAMO congruence in Kerr spacetime.

I don't get how Sachs and Wu, in http://postimg.org/image/gaokzoapr/, come to the conclusion that if we choose a basis for a fiducial congruence observer in which the shear tensor is diagonalized (principal axes) then neighboring observers do not rotate relative to this observer. How do they come to this conclusion?

I'm not sure. I need to re-read several papers on this that have been linked to in the course of these discussions; unfortunately, I'm not sure how soon I'll have time to do that.

Intuitively it just doesn't make sense to me because the ZAMOs are supposed to be as close as possible to a state of rest with respect to space-time geometry in a rotating space-time since they allow themselves to exactly follow the local rotation of space-time. This leads to them all having vanishing angular momentum. As such I don't get why physically a given ZAMO would see neighboring ZAMOs rotate relative to gyroscopes that he carries if the neighbors happen not to be aligned with a principal axis of the shear tensor. How can the neighbors rotate relative to the fiducial ZAMOs when they are meant to all define being at rest relative to the local space-time geometry?

Because the local spacetime geometry has shear in it; its "rate of rotation" is not constant. I'm not sure there's a way to completely "factor out" the shear, because it's part of the geometry.

unless I'm making another mistake, this unfortunately ruins the nice intuitive picture of local non-rotation of ZAMOs in terms of vanishing vorticity as opposed to the picture we already had in terms of the Sagnac effect and Einstein synchronization.

Perhaps it will help to go back to some basic questions:

(1) Consider two ZAMOs in Kerr spacetime at different radial coordinates (in the equatorial plane, for simplicity). Both observe zero Sagnac effect in light signals confined to their particular ZAMO rings. They have different angular velocities. Can they Einstein synchronize their clocks? In other words, if each ZAMO emits a light signal at the same coordinate time $t_1$ (and assume that both are at the same angular coordinate $\varphi_1$ at this instant, for simplicity), will each signal reach the other ZAMO's worldline at the same coordinate time $t_2$? Looking at the math that describes the light signals' worldlines might be helpful.

(2) Consider two ZAMOs in Kerr spacetime at the *same* radial coordinate (again in the equatorial plane), but different angular coordinates $\varphi_1$ and $\varphi_2$. Again, both observe zero Sagnac effect, and both have the same angular velocity, but the instantaneous directions of their velocities are different. Can they Einstein synchronize their clocks? Again, looking at the math that describes the light signals' worldlines might be helpful.

 

[WannabeNewton]

I'm not sure. I need to re-read several papers on this that have been linked to in the course of these discussions; unfortunately, I'm not sure how soon I'll have time to do that.

That's fine Peter, please take your time. I'm in no rush to get this answered .

Because the local spacetime geometry has shear in it; its "rate of rotation" is not constant. I'm not sure there's a way to completely "factor out" the shear, because it's part of the geometry.

I was reading more of "Vorticity and Vortex Dynamics"-Wu et al and based on the statements made there (see top of http://books.google.com/books?id=P5...orticity principal axes strain tensor&f=false) and based on the statements made by Chestermiller in the other thread I linked, it would seem there is only one way to think about vorticity when we have non-vanishing shear if we want to consider all neighboring ZAMOs and not just the rotation of the principal axes themselves: if we take a fiducial ZAMO and consider an infinitesimal sphere of neighboring ZAMOs then the vorticity measures the average rotation of this sphere. I think yuiop mentioned this in a previous thread as well.

So to sum up $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ implies that the neighboring ZAMOs have no average angular velocity relative to the (gyroscopes carried by) fiducial ZAMO. Perhaps this is what Sachs and Wu meant when they said there is no "overall rotation" of the neighbors about the principal axes.

Now, I definitely agree with you that the shear $\sigma_{\alpha\beta}$, of the ZAMO congruence, associated with the non-uniform local rotation of the space-time is an intrinsic property of the local space-time geometry itself. As such, do you think there's a notion of local non-rotation based upon $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0 \Rightarrow$ zero average rotation of connecting vectors between neighboring ZAMOs that has a physically intuitive connection/relation to the notion that ZAMOs are non-rotating relative to the local space-time geometry (zero angular momentum), akin to what I described in post #37? They're mathematically equivalent so I feel like there has to be some physical connection/relation between them.

Thanks! In the meanwhile I'll start thinking about the questions you've posed regarding Einstein synchronization.

 

[PeterDonis]

it would seem there is only one way to think about vorticity when we have non-vanishing shear if we want to consider all neighboring ZAMOs and not just the rotation of the principal axes themselves: if we take a fiducial ZAMO and consider an infinitesimal sphere of neighboring ZAMOs then the vorticity measures the average rotation of this sphere.

This was my understanding following the previous threads, but consider this: if there is zero average rotation of the connecting vectors (zero vorticity), then there is a function $f(V)$, where $V$ is a continuous parameter that labels the connecting vectors, and $f(V)$ gives the rotation of the vector labeled by $V$, and the average value of $f(V)$ over all $V$ is zero.

Then, if there is nonzero shear, there will be values of $V$ for which $f(V)$ is positive, and values of $V$ for which $f(V)$ is negative; and by the mean value theorem, there must therefore be at least one value of $V$ for which $f(V) = 0$, i.e., there must be at least one connecting vector that, individually, has zero rotation. (Actually, I think there must be at least two, because $f(V)$ is a periodic function of $V$.)

So if the average rotation is zero, there must be at least one (or two if I'm right above) connecting vector(s) whose individual rotation is zero. The obvious hypothesis is that these are the vectors oriented along one of the principal axes of the shear tensor.

 

[WannabeNewton]

Shouldn't there be three such connecting vectors, since there are three principal axes of the shear tensor? For example when a sphere is deformed into an ellipsoid there are three connecting vectors from the center of the sphere which remain non-rotating and they are the ones which are aligned with the semi-major and two semi-minor axes of the ellipsoid of deformation.

Regardless, consider again the Killing field $\eta^{\alpha} = \xi^{\alpha} + \omega \psi^{\alpha}$ corresponding to a family of observers all of whom undergo circular orbit with the same angular velocity $\omega$ as a single ZAMO. At the location of the ZAMO, the gravitomagnetic potential $\eta_{\alpha}$ vanishes because the ZAMO is non-rotating relative to the local space-time geometry. But $\eta_{\alpha}$ is non-vanishing on the locations of neighboring observers because they are rotating relative to the local space-time geometry; they resist the local rotation of space-time to an extent. As a result, if we consider connecting vectors between neighboring observers of $\eta^{\alpha}$ they will be subject to a torque coming from the curl $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}$ of the gravitomagnetic potential.

The situation is not unlike a rigid paddle wheel centered in a circulating fluid which flows with increasing angular velocity along the length of each paddle. Because the fluid flows faster at higher points along each paddle, there will be a torque on the paddle from the resulting curl of the fluid velocity field and the paddle wheel will start to rotate about its pivot. In the case of $\eta^{\alpha}$, the circulating fluid corresponds essentially to the rotation of the space-time which decreases in rate as we move away from the source. Connecting vectors between neighbors of $\eta^{\alpha}$ form a rigid set of spatial axes which then undergo rotation under $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}$ for the same reason the paddle wheel rotates in the fluid, which is not exactly the curl $\xi_{[\alpha}\nabla_{\beta}\xi_{\gamma]}$ of the space-time rotation itself since there are also kinematic precession effects involved in the case of $\eta^{\alpha}$, but the idea remains the same. In the coordinate system corotating with $\eta^{\alpha}$ we of course know that $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}$ manifests itself as the precession of a gyroscope at rest in the coordinate system.

Switching gears a bit, consider now the ZAMO congruence with tangent field $t^{\alpha}$. We know each individual ZAMO is at rest with respect to the local space-time geometry, manifested by their having identically vanishing angular momentum, and as you noted there is an intrinsic "shearing" in the space-time due to the non-uniformity of its local rotation and the ZAMOs are subject to this "shearing", but nothing more. This means $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ i.e. if we consider an infinitesimal sphere of ZAMOs surrounding a fiducial ZAMO, the sphere will get sheared into an ellipsoid but it will not undergo any rigid rotation relative to a momentarily comoving local inertial frame since the average angular velocity of each ZAMO in the sphere vanishes with respect to comoving gyroscopes (the average velocity being the same as the actual velocity of the principal axes).

So the way to visualize $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ with the presence of the intrinsic "shearing" in the space-time, and the way to consider it a notion of local non-rotation, is through the fact that the swam of neighboring ZAMOs do not rotate as a whole relative to the fiducial ZAMO's gyroscopes. Resorting again to the example of the sphere being deformed into an ellipsoid, while connecting vectors from the center of the sphere to its surface certainly rotate during the deformation, the sphere itself as a whole does not rotate about the $x,y,z$ axes.

If we wanted to contrast intuitively the $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ and $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} \neq 0$ situations then we could say that in the latter case the infinitesimal sphere will rotate rigidly as a whole whereas in the former case the infinitesimally sphere will not rotate rigidly but it will unavoidably get deformed due to the intrinsic "shearing" in the space-time; this is one of two senses in which the ZAMO congruence is locally non-rotating (the other being in terms of the Sagnac effect). I don't think there is any notion of locally non-rotating we can come up with for the ZAMO congruence that involves relative motion between only a pair of neighboring ZAMOs, as opposed to average relative motion of an entire swarm of neighboring ZAMOs.

Would you agree with all of the above? Thanks in advance!

 

[WannabeNewton]

Can they Einstein synchronize their clocks?

Just to clarify, this is a different notion of Einstein synchronization than what I had mentioned before and what is usually meant in the literature. The method in mind is better termed local Einstein synchronization wherein infinitesimally neighboring observers Einstein synchronize their clocks (using null geodesics in the local rest space) and this local synchronization is propagated along simultaneity curves associated with one-parameter families of pairwise infinitesimally neighboring observers. If the observers define a congruence for which the tangent field satisfies $t_{[\alpha}\nabla_{\beta}t_{\gamma]} = 0$ then this local Einstein synchronization will be well-defined i.e. transitive. Of course this requires the observers to set their clocks to read the time $t$ satisfying $u^{\mu} = \gamma \nabla^{\mu}t$ so it isn't Einstein synchronization in the sense one uses the term in SR which is something to bear in mind.

That being said, the answer to both your questions is "yes".

To see this, write the ZAMO tangent field as $t^{\alpha} = \xi^{\alpha} + \omega(r,\theta) \psi^{\alpha}$ where $\omega = -\frac{g_{t\phi}}{g_{\phi\phi}}$ is the ZAMO angular velocity. Consider the coordinate transformation $\phi' = \phi - \omega(r,\theta)t$ or $\phi = \phi' + \omega(r',\theta')t'$, with all other coordinates unchanged. In these coordinates, we clearly have $t^{r'} = t^{\theta'} = 0$, $t^{0'} = \frac{\partial t'}{\partial x^{\mu}}t^{\mu} = t^{0}$, and $t^{\phi'} = \frac{\partial \phi'}{\partial x^{\mu}}t^{\mu} = \frac{\partial \phi'}{\partial t} + \frac{\partial \phi'}{\partial \phi}\omega = 0$. Therefore in this coordinate system all the ZAMOs have fixed spatial coordinates i.e. they are at rest in this comoving coordinate system.

Furthermore, $g_{0'i'} = \frac{\partial x^{\mu}}{\partial t'}\frac{\partial x^{\nu}}{\partial x^{i'}}g_{\mu\nu} = \frac{\partial t}{\partial t'}\frac{\partial x^{\nu}}{\partial x^{i'}}g_{t\nu} + \frac{\partial \phi}{\partial t'}\frac{\partial x^{\nu}}{\partial x^{i'}}g_{\phi\nu} = \frac{\partial \phi}{\partial x^{i'}}(g_{t \phi} +\omega g_{\phi\phi}) = 0$.

But $\partial_{0'}g_{\mu'\nu'} \neq 0$ in this coordinate system. In particular, $g_{r'r'} = g_{rr} + (\partial_{r'}\omega)^2 t'^2 g_{\phi\phi}$ and similarly for $g_{\theta'\theta'}$, with $g_{0'0'} = g_{tt} + 2\omega g_{\phi t} + \omega^2 g_{\phi\phi}$ and $g_{\phi'\phi'} = g_{\phi\phi}$ as usual.

So in the coordinate system comoving with the ZAMOs, $g_{\mu'\nu'}$ is completely diagonal but time-dependent. What this means is, if we choose two ZAMO observers $O,O'$ say at fixed respective locations $(r_0,\phi_0)$ and $(r_1,\phi_1)$ on the equatorial plane in this coordinate system, then the coordinate time interval $\Delta t$ it takes for a light signal from either $O$ or $O'$ to reach the other depends on what instant $t$ the light signal was emitted but it doesn't depend on which of the two emitted it i.e. $\Delta t$ is the same for $O$ and $O'$ given that they emit the signals at the same time $t$. If instead we had say $O$ send a light signal to $O'$ who upon reception of the light signal reflected it back to $O$ then the two time intervals will clearly not be equal when $r_0 \neq r_1$ and this is precisely because the distance between radially separated ZAMOs is not constant in time.

Thanks in advance!

 

[WannabeNewton]

Peter I think I finally have it all sorted out. To start with, note that for any time-like congruence of the form $\eta^{\alpha} = \xi^{\alpha} + \omega(x^{\mu})\psi^{\alpha}$ we can always write $\omega^{\alpha} \propto \epsilon^{\alpha\beta\gamma\delta}\eta_{\beta}\nabla_{\gamma}\eta_{ \delta} = \epsilon^{\alpha t \beta \phi}(\eta_t \partial_{\beta}L - L \partial_{\beta}\eta_t)$ where $L$ is the reduced orbital angular momentum as usual.

Consider this first in flat space-time where $\eta_t = -\eta^t = -1$ identically so that $\omega^{i} \propto \epsilon^{ i j \phi} \partial_{j}L$ with $L = r^2 \omega$. So, in other words, if we imagine a fiducial observer from the (rigidly orbiting) Born congruence and a swarm of neighboring observers of the congruence then the swarm as a whole rotates rigidly about the fiducial observer precisely because the orbital angular momentum varies across the swarm and this certainly makes sense intuitively for the Born congruence wherein the spatial variation of the angular momentum across the congruence is exactly what leads to a collective relative orbital velocity of the members of the swarm about the fiducial observer.

We then come back to Kerr space-time, considering a congruence which contains (at least) one ZAMO. At the location $(r,\theta)$ of the ZAMO we know $L = 0$ so $\omega^i|_{(r,\theta)}\propto - \epsilon^{ i j \phi}\eta_t \partial_{j}L|_{(r,\theta)}$. If the congruence is one which rotates rigidly with the ZAMO, so that $\nabla_{\mu}\omega = 0$, then we know $\nabla_{\mu}L \neq 0$ and hence $\omega^i |_{(r,\theta)}\neq 0$ which basically means that the local swarm of congruence observers itself rotates rigidly around the fiducial ZAMO because the observers have non-zero angular momentum i.e. they have orbital rotation relative to the local space-time geometry whereas the ZAMO doesn't.

On the other hand for the ZAMO congruence $\nabla_{\mu}L = 0$ and as a result $\omega^i|_{(r,\theta)} = 0$. Put differently, because all of the ZAMOs have vanishing orbital rotation relative to the local space-time geometry, the swarm of ZAMOs surrounding the fiducial ZAMO has no overall rigid rotation, just as in the flat space-time case. So for the ZAMO congruence the statement $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ amounts to a notion of non-rotation in the sense that all ZAMOs have no orbital rotation relative to the local space-time geometry which is of course the same as saying $L = 0$ for all the ZAMOs. It's just harder to conceptualize in this case because $\nabla_{\mu}\omega \neq 0$.

As an aside, something worth pointing out, if we consider instead the case of rigidly orbiting observers following integral curves of the associated time-like Killing field $\eta^{\alpha}$ in any stationary axisymmetric space-time then the condition $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_{(r,\theta)} = 0$, for a specific $(r,\theta)$, is equivalent to the non-rotation of the associated observer's acceleration vector relative to comoving gyroscopes which is actually equivalent to the observer having extremal acceleration $\frac{\partial a}{\partial \omega} = 0$
(c.f. Page 1997) which I thought was a very nice way of understanding intuitively what $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]}|_R = 0$ means kinematically.

What do you think?

 

[PeterDonis]

for any time-like congruence of the form $\eta^{\alpha} = \xi^{\alpha} + \omega(x^{\mu})\psi^{\alpha}$ we can always write $\omega^{\alpha} \propto \epsilon^{\alpha\beta\gamma\delta}\eta_{\beta}\nabla_{\gamma}\eta_{ \delta} = \epsilon^{\alpha t \beta \phi}(\eta_t \partial_{\beta}L - L \partial_{\beta}\eta_t)$ where $L$ is the reduced orbital angular momentum as usual.

I think this is a good way of breaking it down, since it makes clear what has to happen to have nonzero vorticity: angular momentum has to vary over the congruence, or the time component of the 4-velocity has to vary over the congruence (and the two variations can't cancel each other). As you show, all of the cases we've been discussing pop easily out of this equation.

I wonder if there is a similarly intuitive way of writing the expansion and shear; this would be particularly helpful for the ZAMO congruence in Kerr spacetime, because of the nonzero shear.