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

[WannabeNewton]

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.

Thanks for the reply Peter! There is actually such a relation but before I go into that I thought the following would help even further clarify what was discussed above: http://math.stackexchange.com/questions/428839/irrotational-vortices

Note the first animation corresponds to a congruence of observers for which the angular velocity goes like $\omega \sim 1/r^2$ which is essentially what happens with the ZAMO congruence at least in the equatorial plane. The second animation on the other hand corresponds to a (rigid) congruence of observers who all orbit with the same angular velocity such as, in our case, the Killing field which agrees with a single ZAMO orbit. Each paddle wheel represents the local swarm of observers surrounding the fiducial observer centered on the paddle wheel, in the chosen congruence.

Lifting these animations to Kerr space-time or even Minkowski space-time is of course not entirely accurate because the animations assume comoving gyroscopes do not precess relative to spatial infinity which is not true in our case but the basic intuition is still correct, we just have to lift it to a completely local description in terms of comoving gyroscopes in Kerr space-time instead of relative to spatial infinity.

Anyways, I realized I made a HUGE mistake before; if I spotted it before I could have saved us a lot of posts. It is not true that $L = 0 \Leftrightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$; rather it is only true that $L = 0 \Rightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$, the converse does not hold. If it did then in flat space-time the only irrotational congruence would be a rigid inertial congruence but this is clearly not true as the $\omega \sim 1/r^2$ congruence above is also irrotational.

This actually clarifies things for us greatly. Indeed from the expression for $\omega^{\alpha}$ from the previous post we see that there are two trivial cases of congruences for which $\omega^{\alpha} = 0$: when $L = 0$ everywhere in space-time and when $L = \eta_t$. The $L = 0$ case in flat space-time is just that of inertial observers whereas the $L = \eta_t$ case corresponds to $\omega = \frac{L}{g_{\phi\phi}} \sim 1/r^2$ which we know from before.

So $L = 0 \Rightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ is simply the statement that if a ZAMO carried a paddle wheel along with the flow of the ZAMO congruence then the paddle wheel will not rotate relative to comoving gyroscopes because the ZAMOs are all locally non-rotating, in the sense that they all have no local orbital rotation relative to the space-time geometry. But the converse as we have seen is not true.

Indeed $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ simply means that the paddle wheel carried along the flow locally doesn't rotate but that doesn't mean each flow line can't have an orbital rotation relative to the local space-time geometry, it just means the orbital rotation, and hence shear, varies along the congruence in such a way that the paddle wheel doesn't rotate. It's just that for the ZAMOs there is no local orbital rotation at all so the paddle wheel won't rotate locally as a result i.e. $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$.

I'm sorry for that critical mistake I made. But does this clarify things at all? I'll talk about the shear in a separate post.

 

[PeterDonis]

Lifting these animations to Kerr space-time or even Minkowski space-time is of course not entirely accurate because the animations assume comoving gyroscopes do not precess relative to spatial infinity which is not true in our case but the basic intuition is still correct, we just have to lift it to a completely local description in terms of comoving gyroscopes in Kerr space-time instead of relative to spatial infinity.

The animations also don't illustrate nonzero shear; to show that we would need each little circle to have at least two diameters displayed, and the diameters would have to change either length or direction as the circles revolved. I thought I remembered seeing such an animation somewhere, but I haven't been able to find it.

from the previous post we see that there are two trivial cases of congruences for which $\omega^{\alpha} = 0$: when $L = 0$ everywhere in space-time and when $L = \eta_t$. The $L = 0$ case in flat space-time is just that of inertial observers whereas the $L = \eta_t$ case corresponds to $\omega = \frac{L}{g_{\phi\phi}} \sim 1/r^2$ which we know from before.

Yes. The obvious next question is how this works in Schwarzschild and Kerr spacetime--i.e., how do we characterize the $L = \eta_t$ case in those spacetimes? (Of course the $L = 0$ case is easy.)

does this clarify things at all? I'll talk about the shear in a separate post.

I think we've got the general relationship between angular momentum and vorticity clear, yes. Looking forward to the post on shear.

 

[WannabeNewton]

The animations also don't illustrate nonzero shear; to show that we would need each little circle to have at least two diameters displayed, and the diameters would have to change either length or direction as the circles revolved. I thought I remembered seeing such an animation somewhere, but I haven't been able to find it.

I can't seem to find anything other than: http://en.wikipedia.org/wiki/Vorticity#Examples and page 7-8 of http://books.google.com/books?id=JB...wSG7oHoCg&ved=0CD0Q6AEwBQ#v=onepage&q&f=false and page 4 of http://www.pma.caltech.edu/Courses/ph136/yr2011/1114.2.K.pdf

The last one by Kip Thorne in particular is very useful.

Note that in the $L = \eta_t$ case for both Schwarzschild and Kerr space-times, $\partial_{\mu}L \neq 0$ so even if the angular momentum varies across the congruence we can have vanishing vorticity. Combining this with the discussion in Kip Thorne's notes above, I want to correct a statement I repeatedly made prior. I don't think it is accurate to say "$L = 0 \Rightarrow \eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ intuitively because the ZAMOs are all locally non-rotating relative the space-time geometry."

This intuition seems to be wrong because locally non-rotating relative to the space-time by definition means the Sagnac shift vanishes for any given ZAMO observer, which is by itself a quasi-local effect that only depends on the properties of a single ZAMO at a single $(r,\theta)$. On the other hand $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ means that an entire local swarm of neighboring ZAMOs doesn't rotate as a whole relative to comoving gyroscopes because, as explained by Kip Thorne, the neighboring ZAMO along $\hat{\phi}$ rotates outwards whereas the neighboring ZAMO along $\hat{r}$ rotates inwards by the same amount leading to an average velocity of zero around the fiducial ZAMO (in the equatorial plane this means the principal axes of the shear tensor are $\pm \frac{1}{\sqrt{2}}(\hat{r} + \hat{\phi})$.

Therefore, intuitively, the fact that each individual ZAMO has no orbital rotation relative to the local space-time geometry isn't by itself what makes the vorticity vanish because this is a statement about the motion of a given ZAMO relative to the local geometry and not the (average) motion of neighboring ZAMOs relative to one another which is what vorticity depends on. Indeed even though the ZAMOs are non-rotating relative to the local geometry, the individual ZAMOs in the local swarm still rotate relative to the fiducial ZAMO. But they do so in a way that makes the rotation averaged over the entire swarm vanish so that the swarm itself is non-rotating.

To put it another way, there are essentially two separate effects of having $L = 0$ for the entire congruence. There is the fact that since $L = 0$ for any given ZAMO, the Sagnac shift vanishes for that specific ZAMO and thus the ZAMO is non-rotating relative to the local space-time geometry; this involves the ZAMO and the ZAMO alone, in relation to the space-time. There is also the fact that since $L = 0$ everywhere on the congruence, the angular velocity of the congruence varies across space-time as $\omega = -\frac{g_{t\phi}}{g_{\phi\phi}}$ which is exactly sufficient to write the tangent field as $\eta^{\alpha} = \gamma \nabla^{\alpha}t$ meaning $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$.

In other words this is a local differential relationship because what it's roughly saying is $\partial_{\mu}\omega$ is such that, in the equatorial plane, the rotation of a ZAMO radially separated locally from a fiducial ZAMO exactly cancels out the rotation of a ZAMO azimuthally separated so as to yield a vanishing average rotation relative to the fiducial ZAMO. Indeed $\sigma_{\alpha\beta} = \omega \psi_{(\alpha}\partial_{\beta)}\omega$ is the shear of the congruence and the relative rotation of a spatial connecting vector $Y^{\alpha}$ between neighboring ZAMOs is given simply by $F_{u}Y^{\alpha} = \sigma^{\alpha}{}{}_{\beta}Y^{\beta}$ where $F_u$ is the Fermi-derivative along the congruence. So these are the two separate effects of $L = 0$ for the entirety of $\eta^{\alpha}$.

Would you agree or do you think the physical description in the previous post is more accurate? Sorry for yet another long post and thanks in advance!

 

[PeterDonis]

the neighboring ZAMO along $\hat{\phi}$ rotates outwards whereas the neighboring ZAMO along $\hat{r}$ rotates inwards by the same amount

I'm not sure "outwards" and "inwards" are the right words here. As Thorne puts it, the neighboring member of the congruence along $\hat{\phi}$ has *positive* angular velocity, relative to the fiducial member (i.e., its relative rotation is in the same sense as the overall rotation of the congruence, relative to spatial infinity), while the neighboring member of the congruence along $\hat{r}$ has *negative* angular velocity (i.e., its relative rotation is in the *opposite* sense to the overall rotation of the congruence), with the same magnitude, so the two average to zero. I'm not sure that "positive" equates to "outwards" here, or "negative" to "inwards"; "forwards" and "backwards" would seem better, although even those might not convey it properly.

(in the equatorial plane this means the principal axes of the shear tensor are $\pm \frac{1}{\sqrt{2}}(\hat{r} + \hat{\phi})$.

Yes.

Therefore, intuitively, the fact that each individual ZAMO has no orbital rotation relative to the local space-time geometry isn't by itself what makes the vorticity vanish because this is a statement about the motion of a given ZAMO relative to the local geometry and not the (average) motion of neighboring ZAMOs relative to one another which is what vorticity depends on.

Yes. I think this is an important point. It corresponds to the important distinction between (1) a set of orthonormal basis vectors carried by a single ZAMO, and the relationship between these and a set of vectors oriented by gyroscopes, and (2) a set of connecting vectors between a fiducial ZAMO and neighboring ZAMOs, and the relationship between *these* and a set of vectors oriented by gyroscopes.

Would you agree or do you think the physical description in the previous post is more accurate? Sorry for yet another long post and thanks in advance!

I don't know that your previous description was wrong, exactly; but the clarifications in this latest post are very helpful.

 

[WannabeNewton]

I'm not sure that "positive" equates to "outwards" here, or "negative" to "inwards"; "forwards" and "backwards" would seem better, although even those might not convey it properly.

Right, fair point.

It corresponds to the important distinction between (1) a set of orthonormal basis vectors carried by a single ZAMO, and the relationship between these and a set of vectors oriented by gyroscopes, and (2) a set of connecting vectors between a fiducial ZAMO and neighboring ZAMOs, and the relationship between *these* and a set of vectors oriented by gyroscopes.

I agree with you but I just wanted to clarify one thing. When I said $L = 0$ for a single ZAMO implies the ZAMO has no orbital rotation relative to the local space-time geometry I meant it in terms of the Sagnac effect. So this would be distinct from the sense of orbital rotation in (1) in rotating space-times as we have seen earlier. But certainly both of these share the same distinction from (2) that you made as both (1) and $L = 0$ for a single ZAMO are characterized entirely in terms of the worldline of said ZAMO whereas (2) fundamentally requires a swarm of ZAMOs.

 

[WannabeNewton]

Hey Peter. I think we've fairly well grounded our intuition regarding the rotational effects in Kerr space-time at this point but I thought it would still be worthwhile to include some excerpts from "Black Holes: The Membrane Paradigm"-Thorne et al because the authors basically give lucid descriptions of what we've already established in this thread. I've attached the images. Kip Thorne uses the term FIDO in place of ZAMO but they're the same thing.

Also the calculation verifying that for the Killing field $\eta^{\alpha} = \xi^{\alpha} + \omega \psi^{\alpha}$ with proper acceleration $a^2$ we have that $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0 \Leftrightarrow \frac{\partial a^2}{\partial \omega} = 0$ can be found in sections 4 and 5 of http://arxiv.org/pdf/gr-qc/9706029v2.pdf. Actually the author shows that $F_u a^{\alpha} = 0 \Leftrightarrow \frac{\partial a^2}{\partial \omega} = 0$ where $F_u$ is the Fermi derivative along $u^{\alpha} = \gamma \eta^{\alpha}$. However we know $a^{\alpha}$ only has components along $e_{\theta}$ and $e_{r}$ and furthermore $\mathcal{L}_{\eta}a^{\alpha} = 0$ so we can use $u^{\alpha}, e_{\phi}$ along with the unit vector along $a^{\alpha}$ and a unit vector orthogonal to these three as a Lie transported local Lorentz frame along the worldline. Since the tetrad has to rotate rigidly relative to comoving gyroscopes, if $F_u a^{\alpha} = 0$ then $F_u e_{\alpha} = 0$ for the entire tetrad which is equivalent to $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0$ since this is a Killing field.

Lastly, I read in http://link.springer.com/article/10.1007/BF00763554 that another property of the ZAMOs is that an observer freely falling radially from infinity passing by a ZAMO does so with only a radial velocity in the ZAMO's natural rest frame, which is a measurement of non-rotation. I thought it was pretty cool that the purely radial velocity of the freely falling observer in the ZAMO frame depends on the fact that the ZAMO has no orbital rotation relative to the local space-time geometry even though the natural rest frame of a ZAMO does rotate relative to comoving gyroscopes. In Newtonian mechanics and even in flat space-time, where both of these notions of rotation agree, it instead obtains that an inertial particle moving purely radially in the inertial frame will get deflected from its radial motion due to Coriolis forces in the rotating frame of an observer in circular motion.

I should note that I haven't verified the statement in the paper yet but the verification should be an easy calculation that I'll work out in the class I have in about an hour and then post it here

 

[WannabeNewton]

Hi Peter! I apologize for suddenly bringing this thread back to life but I had a quick question crop up when reading the following paper: http://journals.aps.org/prd/abstract/10.1103/PhysRevD.36.1045

I've attached the relevant paragraph. As you can see, the paper states that in the ZAMO frame (4.8), there are no Coriolis-type effects. I've been trying to understand what exactly the paper means by this.

On the one hand if we consider a test particle dropped from rest at infinity with no initial angular momentum then we know that as the particle falls in space-time it picks up an angular velocity $\frac{d\varphi}{dt} = \omega_{\text{ZAMO}}$ so that in the coordinate system comoving with the ZAMOs the particle falls radially i.e. it is not deflected in these coordinates which in a sense means there are no Coriolis effects; however these coordinates are not rigid so to what extent this sense of Coriolis force aligns with that of Newtonian mechanics is unclear to me. Furthermore this is clearly a property of the comoving ZAMO coordinates and not of the ZAMO frame.

On the other hand, the frame (4.8) rotates with an angular velocity $\Omega$ relative to local gyroscopes as we know already. Thus if we consider the local "proper coordinate system" or "proper reference frame" of a single ZAMO observer constructed from the frame (4.8), c.f. MTW sec. 13.6, then there is a Coriolis force on the test particle from above just as it passes through the origin of these coordinates and this Coriolis force is clearly exactly the same (conceptually) as that in Newtonian mechanics.

So do you think the paper is talking about the absence of angular deflection in the comoving ZAMO coordinates in the sense described above or is it talking about something else? If it is the former do you think it actually makes sense to call it an absence of Coriolis-like effects in the ZAMO frame in light of the paragraph directly above?

Thanks in advance.

 

[PeterDonis]

do you think the paper is talking about the absence of angular deflection in the comoving ZAMO coordinates in the sense described above or is it talking about something else?

"Angular deflection in the comoving ZAMO coordinates" basically equates to absence of angular deflection with respect to a rigid frame whose spatial axes are the principal axes of the shear tensor, i.e., the "local gyroscope" frame, correct? If so, I think that's the intent (but see below), but I agree that, since the actual ZAMO congruence is not rigid, and the intuitive Newtonian effects of rotation, such as Coriolis, really only make sense relative to a rigid frame, it would be nice if all these references would be more precise about exactly what they are defining things relative to.

do you think it actually makes sense to call it an absence of Coriolis-like effects in the ZAMO frame in light of the paragraph directly above?

I think you are right that frame 4.8 is rotating relative to the local gyroscope frame, and if so, I don't understand why the paper designates frame 4.8 as the "non-rotating" frame instead of the local gyroscope frame (i.e., the frame defined by the principal axes of the shear tensor). They seem to me to be two different frames, and it seems to me that only the latter can be described as having zero Coriolis effect. Perhaps I'm missing something, but that's how it looks to me on a quick reading.

 

[WannabeNewton]

Thanks for the reply!

"Angular deflection in the comoving ZAMO coordinates" basically equates to absence of angular deflection with respect to a rigid frame whose spatial axes are the principal axes of the shear tensor, i.e., the "local gyroscope" frame, correct?

Actually when I said that, I meant the trajectory of the freely falling particle in the coordinates in which all ZAMOs are at rest (congruence adapted coordinates) will have no angular component i.e. it will be purely radial. But it is certainly true that a local Lorentz frame with gyroscope axes attached to a ZAMO will have no Coriolis forces in it precisely in the Newtonian sense as it is a non-rotating frame.

However I think these are two distinct concepts because the former depends only on the kinematics of the ZAMO congruence whereas the latter depends on what local frame we attach to each ZAMO; furthermore the former is a global notion and, I think, is misleading to be associated with Coriolis effects at least in the Newtonian sense whereas the latter is a local notion (occurs at the origin of the local Fermi coordinates of a single ZAMO) and does align with Newtonian coriolis forces. Of course it is true that for a congruence of the form $\eta^{\alpha} = \xi^{\alpha} + \omega \psi^{\alpha}$ in a stationary axisymmetric spacetime we have $\eta_{[\alpha}\nabla_{\beta}\eta_{\gamma]} = 0 \Leftrightarrow \omega = \omega_{\text{ZAMO}} \Leftrightarrow$ freely falling particle dropped initially radially from infinity keeps falling radially in comoving congruence adapted coordinates, but conceptually I believe the effects are distinct for the reasons above.

Perhaps I'm missing something, but that's how it looks to me on a quick reading.

I don't think you are missing anything. I looked at some other papers and it seems to me that there are just misconceptions present in them simply due to poor terminology and careless wording.