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

[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.

 

[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.