Practical measurements of rotation in the Kerr metric

[yuiop]

I'm not sure the derivatives you need to take will be well-defined if you restrict to a single ZAMO ring. The physical interpretation of the vorticity requires Lie transported vectors pointing to neighboring members of the congruence; but if you restrict to a single ZAMO ring, there are no "neighboring" members in any direction off the ring, so how can you take the appropriate derivatives in those directions?

I have tried to find a definition of Lie transported vectors on the internet without any luck. (Any help on that would be appreciated). I guess a lot depends on how the congruence is defined. If we have a congruence of members with equal orbital angular velocity, then we have a rigid congruence and members of such a congruence would be at rest with orthonormal basis vectors such that a vector at rest in this congruence that is initially pointing at the centre of the orbit will remain pointing at the centre of the orbit. Rindler is clearly using a rigid congruence to define vorticity and since his equations agree with other sources, it would would appear all definitions of vorticity are relative to a rigid congruence or orthonormal basis vectors. If the formal definition of vorticity is relative to Lie transported vectors and if the formal definition of Lie transported vectors is relative to a congruence of members with equal angular momentum, then things get very messy in the Kerr metric due to sheer. For what its worth I have calculated the averaged rotation of the non rigid equal momentum congruence in the Kerr metric using regular Newtonian fluid dynamics (i.e. the paddle wheel) to be:

$\frac{-4ma(r^2-ma^2/r)}{r^3(r^2+a^2+2ma^2/r)^2}$

This turns out to be a factor of approximately 4/3 larger than the ZAMO precession rate of:

$\frac{-ma(3r^2+a^2)}{r^3(r^2+a^2+2ma^2/r)}$

and approximately 4 times larger than the precession rate of gyroscope held by a static observer in the Kerr metric of:

$\frac{-ma}{r^3(1-2m/r)}$

for large r. I can show how I obtained this result if you are interested. Although this Newtonian fluid vorticity is not exactly equal to the GR precession rate, there does appear to be a close connection.

If your response to the above is that we just restrict all derivatives to the two-dimensional subspace of the chosen ZAMO ring, consider that, in order to Einstein synchronize two clocks on the same ZAMO ring, they have to exchange light signals, and those light signals will not be restricted to that ZAMO ring (unless it happens to be at the same radius as a photon orbit). So any physical interpretation of the theorem in question will implicitly have to rely on some extension of the congruence off the chosen ZAMO ring.

The paper referred to by WBN uses "a hollow toroidal tube of glass with perfectly reflecting internal walls" to transit the light signals and so restricts the light path to the ring. Could we not use such a reflecting tube to transmit the synchronisation signals and restrict the analysis to the ring only?

 

[WannabeNewton]

"Einstein synchronization" seems to be a weird term to describe this, since that term is usually taken to imply a constant round-trip light travel time. In other words, "Einstein synchronization" is usually taken to imply a rigid congruence, but the "comoving" congruence in FRW spacetime is of course not rigid, since it has nonzero expansion.

I certainly agree with you there.

I'm not sure the derivatives you need to take will be well-defined if you restrict to a single ZAMO ring. The physical interpretation of the vorticity requires Lie transported vectors pointing to neighboring members of the congruence; but if you restrict to a single ZAMO ring, there are no "neighboring" members in any direction off the ring, so how can you take the appropriate derivatives in those directions?

Right which is why I said we have to extend the vector field to a neighborhood of $\mathcal{R}$. But the problem for me is there is no unique way to perform such an extension. One extension gives us a vector field with vanishing vorticity (the ZAMO congruence) whereas the other gives us a vector field with non-vanishing vorticity (a Killing congruence which agrees with the ZAMO ring when restricted to $\mathcal{R}$). So one extension would give us a synchronizable vector field whereas the other wouldn't even though both yield the same exact ZAMO ring and vector field when restricted to $\mathcal{R}$.

The problem of course is the vanishing vorticity condition for synchronization requires as you said derivatives in all directions in space-time. This is no problem if we're talking about a vector field filling all of space-time but if we're talking about only the tangent field to $\mathcal{R}$ then it isn't enough to only have the derivatives along $\mathcal{R}$ we also need the derivatives perpendicular to $\mathcal{R}$ i.e. the radial and polar derivatives. This would require us to extend the tangent field off of $\mathcal{R}$ but there is no unique way to do so as stated so this is why I was unsure as to whether the vanishing vorticity condition for synchronization could actually apply to 2-dimensional submanifolds like $\mathcal{R}$ or could only apply to vector fields filling all of space-time.

Is my confusion apparent?

If your response to the above is that we just restrict all derivatives to the two-dimensional subspace of the chosen ZAMO ring, consider that, in order to Einstein synchronize two clocks on the same ZAMO ring, they have to exchange light signals, and those light signals will not be restricted to that ZAMO ring (unless it happens to be at the same radius as a photon orbit). So any physical interpretation of the theorem in question will implicitly have to rely on some extension of the congruence off the chosen ZAMO ring.

I was thinking, just like in the case of the rotating disk in flat space-time, we place concave mirrors along the circumference of the ring and exchange light signals along the ring using the mirrors. Granted the light signals will follow null curves that aren't geodesics but it still allows us to exchange signals between points of the ring alone.

 

[WannabeNewton]

I have tried to find a definition of Lie transported vectors on the internet without any luck. (Any help on that would be appreciated).

You'll find it in most standard GR texts and every differential topology text. Take a vector field $\xi^{\mu}$ and another vector field $\eta^{\mu}$. Then $\eta^{\mu}$ is Lie transported along $\xi^{\mu}$ if $\mathcal{L}_{\xi}\eta^{\mu} := \xi^{\nu}\partial_{\nu}\eta^{\mu} - \eta^{\nu}\partial_{\nu}\xi^{\mu} = 0$. This is a completely general definition with no reference to any frames (orthonormal basis vectors). If we are using a torsion-free connection then we can replace $\partial_{\mu}$ with $\nabla_{\mu}$.

Rindler is clearly using a rigid congruence to define vorticity and since his equations agree with other sources, it would would appear all definitions of vorticity are relative to a rigid congruence or orthonormal basis vectors.

Well vorticity is defined for any congruence without reference to any kind of frame and is simply $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\partial_{\alpha}\xi_{\beta}$.

But if you want to interpret vorticity in terms of rotation of Lie transported orthonormal basis vectors relative to local gyroscopes then you need a Killing congruence, or possibly just a rigid congruence although I have yet to see a proof that goes beyond Killing congruences.

If the formal definition of vorticity is relative to Lie transported vectors and if the formal definition of Lie transported vectors is relative to a congruence of members with equal angular momentum, then things get very messy in the Kerr metric due to sheer.

Thankfully neither of those are true :)

 

[PeterDonis]

there is no unique way to perform such an extension. One extension gives us a vector field with vanishing vorticity (the ZAMO congruence) whereas the other gives us a vector field with non-vanishing vorticity (a Killing congruence which agrees with the ZAMO ring when restricted to $\mathcal{R}$). So one extension would give us a synchronizable vector field whereas the other wouldn't even though both yield the same exact ZAMO ring and vector field when restricted to $\mathcal{R}$.

Yes, but that doesn't make them the same vector field or the same congruence. You are talking about two different congruences--the ZAMO congruence and the constant-angular-velocity Killing congruence--that just happen to have the same restriction to a particular submanifold ($\mathcal{R}$). So yes, there's no unique "vorticity" for the restriction to $\mathcal{R}$, because that restriction doesn't define a unique congruence--more precisely, it doesn't define "enough" of one to have a well-defined, unique vorticity.

this is why I was unsure as to whether the vanishing vorticity condition for synchronization could actually apply to 2-dimensional submanifolds like $\mathcal{R}$ or could only apply to vector fields filling all of space-time.

I think the actual range of application is somwhere in between. For example, consider the 3-dimensional submanifold of Kerr spacetime that we call the "equatorial plane". The restriction of the ZAMO congruence to this submanifold has a well-defined (vanishing) vorticity, and the Killing (i.e., constant angular velocity) congruence on this submanifold also has a well-defined (non-vanishing) vorticity. But this is still not all of spacetime; it's a 3-dimensional submanifold.

I think it's worth noting that the Newtonian case works the same way. You can't define vorticity for a Newtonian "fluid" that only moves in one spatial dimension (for example, flow in a really thin pipe modeled as a one-dimensional flow); you have to at least have two spatial dimensions involved (for example, flow restricted to a plane).

 

[yuiop]

In the Ashtekar paper linked by WBN, two examples for the Sagnac shift are given, which when restricted to the equatorial plane can be expressed as:

$\Delta \tau = 16\pi\frac{ ma}{r\sqrt{1-2m/r}}$ for a stationary ring in the Kerr case and

$\Delta \tau = 4\pi\frac{ \omega r^2}{\sqrt{1-\omega^2r^2}}$ for a rotating ring in the Minkowski case.

If the precession angular velocity measured at infinity relative to the orthonormal basis vectors is denoted by $\Omega$ then the two equations above can then be expressed as:

$\Delta \tau = 16\pi r^2 \Omega$ for a stationary ring in the Kerr case and

$\Delta \tau = 4\pi r^2 \Omega$ for a rotating ring in the Minkowski case.

I find this relationship interesting. I wonder if someone could find an expression for the equatorial Kerr case for arbitrary orbital angular velocity? I feel it will be enlightening.

 

[WannabeNewton]

So yes, there's no unique "vorticity" for the restriction to $\mathcal{R}$, because that restriction doesn't define a unique congruence--more precisely, it doesn't define "enough" of one to have a well-defined, unique vorticity.

Thanks Peter. The reason I asked is regarding the relationship between the Sagnac effect and clock synchronization relative to a time-like congruence, in the sense of local Einstein synchronization propagated throughout the congruence. In the Ashtekar paper, a stationary space-time is considered with time-like Killing field $t^{\mu}$ and a tube made of perfectly reflecting internal mirrors is considered such that each point of the tube follows an orbit of $t^{\mu}$-the paper terms this a 'Sagnac tube'. Clearly such a Sagnac tube is at rest in the gravitational field. Then the paper derives the following relationship between the Sagnac shift $\Delta \tau$ of prograde and retrograde light signals emitted from a half-silvered mirror placed somewhere inside the tube as measured by a clock comoving with the mirror and the twist $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}t_{\nu}\nabla_{\alpha}t_{\beta}$ of the time-like Killing field which also happens to be the tangent field to the Sagnac tube:

\Delta \tau = (\lambda_M)^{1/2}\int_{\Sigma} \lambda^{-3/2}\omega^{i}\epsilon_{ijk}dS^{jk}

where $\lambda = -t^{\mu}t_{\mu}$, $\lambda_M$ is this field restricted to the world-line $M$ of the mirror, and $\Sigma$ is the region bounded by a closed curve $C$ on the 2-manifold representing the tube.

Therefore for Sagnac tubes at rest in the gravitational field of the stationary space-time, the Sagnac shift vanishes if the twist (vorticity) vanishes. At the same time, clocks laid out along the circumference of the tube will be synchronizable in the above sense if the twist vanishes and in fact if the twist doesn't vanish, the time discontinuity accrued by attempting local Einstein synchronization around the tube will be exactly given by the Sagnac shift. Examples are clocks laid out along Sagnac tubes at rest in Kerr space-time and along peripheries of rotating disks in flat space-time.

Consider Kerr space-time again but a Sagnac tube whose points follow orbits of the Killing field $\tilde{\xi}^{\mu} = t^{\mu} + \omega_0 \psi^{\mu}$ with $\omega_0$ being a fixed/constant ZAMO angular velocity. Then in general $\tilde{\omega}^{\mu} = \epsilon^{\mu\nu\alpha\beta}\tilde{\xi}_{\nu}\nabla_{\alpha}\tilde{\xi}_{\beta}\neq 0$ but on the tube itself we have $L = \psi_{\mu}\tilde{\xi}^{\mu} = 0$ which, according to Proposition 3.2.2. in Malament's text, means there is no Sagnac effect for this tube. Therefore if the Sagnac tube is not at rest in the gravitational field but rather has some angular velocity then we can choose a tangent field for the tube, such as the Killing field above, for which the twist doesn't vanish but the Sagnac shift still does. Since the standard clocks following orbits of $\tilde{\xi}^{\mu}$ cannot be synchronized with one another in the usual sense if $\tilde{\omega}^{\mu} \neq 0$, does this mean there is no general relationship between clock synchrinization (or lack thereof) and the Sagnac shift if the Sagnac tube has an angular velocity relative to infinity and follows the Killing field $\tilde{\xi}^{\mu}$?

 

[PeterDonis]

Consider Kerr space-time again but a Sagnac tube whose points follow orbits of the Killing field $\tilde{\xi}^{\mu} = t^{\mu} + \omega_0 \psi^{\mu}$ with $\omega_0$ being a fixed/constant ZAMO angular velocity.

I think it's worth noting here that the *reason* the Sagnac tube has to follow orbits of the Killing congruence, not the (varying angular velocity) ZAMO congruence, is that the congruence describing the Sagnac tube must be rigid. Otherwise the tube won't work properly. So it's really a physical requirement that's being modeled by choosing a Killing congruence, which is what drives everything else.

Since the standard clocks following orbits of $\tilde{\xi}^{\mu}$ cannot be synchronized with one another in the usual sense if $\tilde{\omega}^{\mu} \neq 0$, does this mean there is no general relationship between clock synchrinization (or lack thereof) and the Sagnac shift if the Sagnac tube has an angular velocity relative to infinity and follows the Killing field $\tilde{\xi}^{\mu}$?

It looks that way to me, yes.

 

[WannabeNewton]

Peter it seems I spoke too soon. Take a Sagnac tube $\mathcal{R}$ in Kerr space-time with the tangent field $\xi^{\mu}_{\mathcal{R}} = t^{\mu} + \omega|_{\mathcal{R}}\psi^{\mu}$ where $\omega|_{\mathcal{R}} = -\frac{g_{t\phi}}{g_{\phi\phi}}$. Then two neighboring clocks on the tube will be (locally Einstein) synchronized if $\xi^{\mu}_{\mathcal{R}}dx_{\mu} = 0$ where $dx^{\mu}$ is the space-time displacement between the two (locally Einstein) simultaneous events on the world-lines of the clocks. Expanding this out we see that $dt \propto g_{t\phi} + \omega|_{\mathcal{R}}g_{\phi\phi} = 0$ and hence $\oint_{\gamma} dt = 0$ for any closed simultaneity curve $\gamma$ on $\mathcal{R}$. At the same time we know as well that the Sagnac shift vanishes for this tube.

This is entirely independent of how we extend $\xi_{\mathcal{R}}^{\mu}$ to all of space-time i.e. it doesn't matter whether we use the tangent field of the zero angular momentum congruence or that of the Killing congruence. So again we have clock synchronization and vanishing Sagnac shift occurring simultaneously. Also it would seem from this that, as an upshot, the condition $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ for clock synchronization really doesn't apply to single rings or Sagnac tubes due to the issue already discussed regarding the non-uniqueness of the extension $\xi^{\mu}$ of $\xi_{\mathcal{R}}^{\mu}$ to all of space-time. So I guess what we could say is the twist of $\xi^{\mu}$ is in general unrelated to the Sagnac shift for, and clock synchronization along, Sagnac tubes that have non-vanishing angular velocities relative to infinity (in this case the frame-dragging angular velocity) simply because there is no unique such twist we could even assign to the Sagnac tube when all we know is $\xi^{\mu}_{\mathcal{R}}$. But would it be fair to conclude that there is indeed a relationship between the Sagnac shift and clock synchronization or was this just a coincidence?

 

[PeterDonis]

Take a Sagnac tube $\mathcal{R}$ in Kerr space-time with the tangent field $\xi^{\mu}_{\mathcal{R}} = t^{\mu} + \omega|_{\mathcal{R}}\psi^{\mu}$ where $\omega|_{\mathcal{R}} = -\frac{g_{t\phi}}{g_{\phi\phi}}$.

This tube is restricted to $\mathcal{R}$, which means it's restricted to a 2-dimensional submanifold spanned by the $t, \phi$ coordinates, at constant $r, \theta$. Malament would call this the "striated orbit cylinder" at $r, \theta$. So "neighboring clocks on the tube" are just nearby striations on the cylinder, i.e., nearby worldlines in the 2-dimensional submanifold. But we've already seen that you can't define vorticity in a 2-dimensional submanifold anyway; all you need to synchronize clocks in this submanifold is for the worldlines describing the clocks to be orthogonal to some set of spacelike curves, which of course they are, as you show.

This is entirely independent of how we extend $\xi_{\mathcal{R}}^{\mu}$ to all of space-time

But that's only because we've restricted "neighboring clocks" to mean "neighboring in the $\phi$ direction" only. As soon as we try to consider "neighboring" clocks in any other spatial direction, we have to face the question of how to extend the congruence, and different answers to that question give different outcomes.

So again we have clock synchronization and vanishing Sagnac shift occurring simultaneously.

But only when we restrict to a 2-dimensional submanifold only. In other words, if we want this relationship between vanishing Sagnac shift and clock synchronization to hold, we have to restrict consideration to just the 2-dimensional submanifold that describes the Sagnac ring. We can't look at *any* clocks off the ring. That seems like a very strong restriction to me.

it would seem from this that, as an upshot, the condition $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$ for clock synchronization really doesn't apply to single rings or Sagnac tubes due to the issue already discussed regarding the non-uniqueness of the extension $\xi^{\mu}$ of $\xi_{\mathcal{R}}^{\mu}$ to all of space-time.

Yes.

would it be fair to conclude that there is indeed a relationship between the Sagnac shift and clock synchronization or was this just a coincidence?

It's not a coincidence, but it is a restricted definition of "clock synchronization". See above.

 

[yuiop]

...I wonder if someone could find an expression for the equatorial Kerr case for arbitrary orbital angular velocity? I feel it will be enlightening.

I think I have now figured out the Sagnac delay for arbitrary orbital angular velocity $\omega$ in the equatorial plane of the Kerr metric, using my own brute force method. The result is:

$\Delta \tau = \left(\frac{ 2 \pi r}{S_P-\omega r}-\frac{2\pi r}{S_R+\omega r}\right) \gamma^{-1}$

where $S_P$ and $S_R$ are the prograde(+) and retrograde(-) tangential coordinate speeds of light given by

$\frac{r \sqrt{a^2 +r^2-2mr} \pm 2ma}{r^2+a^2+2ma^2/r}$.

$\gamma= 1/\sqrt{1-\omega^2(a^2+r^2)-(2m/r)(1-a\omega)^2}$ is the time dilation factor for an observer at rest with the rotating Sagnac ring.

Unfortunately the equation does not simplify to anything pretty.
When a=0 and m=0, the equation simplifies to the Minkowski case:

$\Delta \tau = \frac{ 4\pi r^2\omega}{\sqrt{1-\omega^2r^2}}$

When a=0, the equation reduces to the Schwartzschild case:

$\Delta \tau = \frac{4\pi r^2 \omega}{\sqrt{1-2m/r-\omega^2 r^2}}$

For both the above metrics there is a simple relationship between the Sagnac delay and the gyroscope precession rate that is independent of $\omega$.

For the Kerr case, an angular velocity equal to the ZAMO velocity yields a zero Sagac delay as it should.

For arbitrary $\omega$ there is no simple relationship between Sagnac delay and gyroscope precession rate.

When $\omega=0$ the Sagnac delay is:

$\Delta \tau = \frac{8\pi ma}{r\sqrt{1-2m/r}}$

Unfortunately this last result is half that obtained by the Ashtekar paper. My equation passes all the other sanity checks, so I am not sure of the reason for the discrepancy?

 

[WannabeNewton]

yuiop based on the papers I've read, your result for the $\omega = 0$ case is correct, although most sources have the opposite sign which is almost certainly just due to you subtracting retrograde from prograde instead of the opposite so that isn't a problem. Maybe the discrepancy is in Ashtekar's definition of $A_0$ which he never explicitly writes down.

See for example equation (17) in this paper: http://arxiv.org/pdf/gr-qc/0510047.pdf

 

[yuiop]

See for example equation (17) in this paper: http://arxiv.org/pdf/gr-qc/0510047.pdf

I take it back. Your linked paper shows there is a reasonable simplification of the Sagnac delay equation in the Kerr metric, which in the notation I used earlier is

\Delta \tau = 4 \pi\left(\omega(r^2+a^2+2ma^2/r)-2ma/r\right)\gamma

 

[WannabeNewton]

Peter, while looking for more papers on inertial forces in rotating frames in curved space-time I happened upon a paper that more or less settles (in our favor) our previous discussion regarding vorticity being defined as rotation of connecting vectors relative to local gyroscopes (Fermi-transported frames). Take a look at section 3.1 of the following paper, and eq. (45) in particular: http://arxiv.org/pdf/1207.0465.pdf

In the paper's notation, $\omega_{\hat{i}\hat{j}}$ is the rotation of the frame of an observer following an orbit of a particular congruence expressed relative to a chose frame field of the congruence and $\Omega_{\hat{i}\hat{j}}$ is the vorticity of the congruence also expressed relative to the chosen frame field. The other quantities are self-explanatory. Consider now the ZAMO congruence in Kerr space-time; choose the rest frame of each ZAMO observer such that $\omega_{\hat{i}\hat{j}} = \Omega_{\hat{i}\hat{j}}$ i.e. lock the rotation of the axes of the frame to the local circulation due to the vorticity of the congruence. We know the vorticity of the congruence vanishes, that is $\Omega_{\hat{i}\hat{j}} = 0$, so this would be a non-rotating frame in the sense that it agrees with the local Fermi-transported frames.

However, as eq. (40) makes clear, the spatial projection $Y^{\hat{i}}$ of the connecting vectors between neighboring ZAMO observers does not have vanishing angular velocity relative to these rest frames precisely because we have non-vanishing shear, $\sigma_{\hat{i}\hat{j}} \neq 0$. The connecting vectors do in fact rotate in these frames even though the frames themselves have non-rotating spatial axes by definition as they are the frames adapted to the vanishing vorticity of the congruence. Note that in the vanishing vorticity adapted frame, $\frac{d}{dt}Y^{\hat{i}} = 0$ if and only if $\theta = \sigma_{\hat{i}\hat{j}} =0$ i.e. we must have a rigid congruence (although not necessarily a Killing congruence).

In the MTW exercise and previous calculations in this thread, the natural rest frames of the ZAMOs were taken to be the basis vectors adapted to the stationary and axial symmetries of Kerr space-time. As we've seen, these frames do rotate relative to local gyroscopes so they are not the frames adapted to the vanishing vorticity of the ZAMO congruence but rather they rotate relative to the vanishing vorticity adapted frame. This is not surprising of course because the spatial axes of the natural rest frame of a given ZAMO are rigidly locked onto the (non-ZAMO) neighboring observers following orbits of the time-like Killing field $\eta^{\mu} = \xi^{\mu} + \omega_0 \psi^{\mu}$ and since $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} \neq 0$ the natural frame would have to rotate relative to the vanishing vorticity adapted frame.

 

[PeterDonis]

The connecting vectors do in fact rotate in these frames even though the frames themselves have non-rotating spatial axes by definition as they are the frames adapted to the vanishing vorticity of the congruence.

Yes, I think this is true, but the interesting question to me is, *how* do the connecting vectors rotate relative to these frames? The key complication that I see here, due to the nonzero shear, is that tangential connecting vectors (i.e., those that, at some instant of a given ZAMO's proper time, point at neighboring ZAMOs that are at the same $r$ but slightly different $\phi$) will rotate *differently*, relative to the Fermi-Walker transported vectors, from the radial connecting vectors (those that, at some instant, point at neighboring ZAMOs at the same $\phi$ but slightly different $r$). Working out, by computation, exactly how they differ has been on my list of things to do in my copious free time for a while now.

 

[WannabeNewton]

You know I've been meaning to ask something. Sometime ago I started a similar thread on the topic of the ZAMO observers in Kerr space-time (perhaps you remember it) and the terminology "locally non-rotating observer" as applied to the ZAMOs confused the living hell out of me. Wald for example uses this terminology when characterizing the ZAMOs (c.f. exercise 7.3) and so does MTW (c.f. exercise 33.3).

But why does this terminology even exist for the ZAMOs? This terminology is usually applied with reference to the "natural" rest frames of the ZAMOs i.e. those frames we obtain by fixing the tetrads to the background time-like and axial symmetries. But as we know these frames do rotate relative to local gyroscopes-the vanishing vorticity of the ZAMOs is not enough to guarantee that the "natural" rest frames of the ZAMOs are non-rotating because the ZAMOs altogether do not follow orbits of a time-like Killing field; each ZAMO alone does follow an orbit of a time-like Killing field but the neighboring observers of this Killing field are not ZAMOs hence the Killing field has non-vanishing vorticity and as we've seen this leads to gyroscopic precession in the "natural" rest frame of a given ZAMO indicating that this frame does in fact rotate as determined by a local compass of inertia.

On the other hand if the terminology is used with reference to Fermi-Walker transported frames attached to ZAMOs (which are by definition non-rotating) then the terminology seems pointless when used specifically for ZAMOs because we can attach Fermi-Walker frames to any and all observers in arbitrary space-times.

 

[PeterDonis]

why does this terminology even exist for the ZAMOs?

I think it's because the term "non-rotating" can have different meanings. One meaning is simply "having zero angular momentum", which obviously applies to a ZAMO. Of course, just making that observation doesn't help with making sense of the other issues involved.

This terminology is usually applied with reference to the "natural" rest frames of the ZAMOs

Is it? I don't have references handy right now, but from what I remember of them, they don't explicitly say this. They simply say that the ZAMO congruence has zero vorticity; but the physical interpretation of zero vorticity for a non-rigid congruence is more complicated than it is for a rigid one. That's what makes this issue difficult to resolve.

the vanishing vorticity of the ZAMOs is not enough to guarantee that the "natural" rest frames of the ZAMOs are non-rotating because the ZAMOs altogether do not follow orbits of a time-like Killing field

I would say it's because the ZAMO congruence is not rigid, as above.

 

[yuiop]

... the interesting question to me is, *how* do the connecting vectors rotate relative to these frames? The key complication that I see here, due to the nonzero shear, is that tangential connecting vectors (i.e., those that, at some instant of a given ZAMO's proper time, point at neighboring ZAMOs that are at the same $r$ but slightly different $\phi$) will rotate *differently*, relative to the Fermi-Walker transported vectors, from the radial connecting vectors (those that, at some instant, point at neighboring ZAMOs at the same $\phi$ but slightly different $r$). Working out, by computation, exactly how they differ has been on my list of things to do in my copious free time for a while now.

This is my attempt to work out the averaged angular velocity of the ZAMO congruence (using the paddle wheel metaphor), to see how it relates to the gyroscope precession. Referring to the diagram below

the solid circle represents the paddle wheel with a radius of p and with its centre a distance r from the centre of the Kerr black hole. In the illustration, r is only about twice as big as p, to exaggerate the approximations I make, but normally r would be much greater than the radius of the paddle wheel which can be considered to be infinitesimal. Here I am considering the point of view of a rigid frame co-moving with a ZAMO at radius r, such that faster moving ZAMOs lower down are moving to the left and slower moving ZAMOs higher up are moving to the right. The first part of the calculation is to find the velocity of a given ZAMO at an arbitrary point on the perimeter of the paddle wheel. The distance of this ZAMO from the gravitational centre (C) is R and for the calculation I am using the approximation, R=(r+h). From simple trigonometry $h=-p \cos(\theta)$ so

$R=r -p \cos(\theta)$.

This value of R can be substituted into the ZAMO angular velocity:

$\frac{2ma}{(R^3+a^2R+2ma^2)}$

Multiplying the above by R gives the horizontal velocity (v) of the ZAMO at that radius. This is an approximation because the actual velocity tangential to r is (v*) as shown in the diagram, but the error is small for p<<R. The velocity of the ZAMO frame at altitude R is then subtracted to give the velocities as seen in the rigid frame co-moving with the representative ZAMO. The component of this velocity that is tangential to the paddle wheel perimeter (v') then has to be found and to a reasonable approximation this is $v' = -cos(\theta)*v$.

$v' =-\cos(\theta)\left(\frac{2maR}{(R^3+a^2*R+2ma^2)}-\frac{2maR}{(r^3+a^2r+2ma^2)}\right)$

This is the velocity component tangent to the paddles of the wheel at a given point on the perimeter of the wheel. This value is integrated all the way around the perimeter from $\theta=0$ to $\theta=2\pi$ to sum all the velocities and divided by $2\pi$ to obtain an averaged tangential velocity and further divided by p to obtain the angular velocity of the paddle wheel. Finally the limit is taken as the radius (p) of the paddle wheel goes to zero, because I am looking for the angular velocity of an infinitesimal paddle wheel:

$\Omega_p = \lim_{p \to 0} \left(\frac{1}{2 \pi p} \left( \int_0^{2 \pi} -cos(\theta)\left(\frac{(r-p \cos(\theta))2ma}{(r-p\cos(\theta))^3+a^2(r-p\cos(\theta))+2ma^2}-\frac{(r-p\cos(\theta))2ma}{r^3+a^2r+2ma^2}\right) d\theta \right)\right)$

and the final result is:

$\Omega_p = \frac {-mar (3r^2+a^2)}{(r^3+a^2r+2ma^2)^2}$

The known angular velocity of the gyroscopes in the ZAMO frame is:

$\Omega = \frac {-ma (3r^2+a^2)}{(r^3+a^2r+2ma^2)}$

Dividing $\Omega_p$ by $\Omega$ and plotting the result using $a=0.9$ and $m=c=1$ from r=0 to r=100 gives:

The red curve is the ratio of the paddle wheel angular velocity, relative to the ZAMO gyroscope precession angular velocity and is close to unity all the way down to near the Kerr ergosphere. This is a good match considering the simplifications I had to make. (The green curve is just the time dilation factor of the representative ZAMO for comparison purposes.)

The exact value of R is

$R= \sqrt{r^2+p^2-2pr \cos(\theta)}$

using the solution for a triangle with 2 known side lengths and a known included angle.

The exact value for the tangential velocity v' is

$v' = v*\frac{p-r \cos(\theta)}{\sqrt{r^2+p^2-2 r p \cos(\theta)}}$.

Unfortunately, the hardware and mathematical software I have available is not able to obtain a closed form solution for the integration step when I use the more complicated exact values. An exact solution also requires consideration of how coordinate velocities transform to local velocities and this is not straight forward in the Kerr metric. Nevertheless, this simplified result hints that the rest frame of the un-torqued gyroscopes is simply the averaged angular velocity of the local ZAMO congruence in an infinitesimal region.

P.S. If we equate the precession of the gyroscopes with the 'averaged' rest frame of the local ZAMO congruence, there is a potential complication due to the non rigidity of the congruence. If we had an actual paddle wheel in a fluid flow, the wheel automatically averages the flow, because parts of the wheel immersed in slow flowing fluid are sped up by parts of the wheel immersed in faster flowing fluid (and vice versa) because paddles on the perimeter of the wheel are rigidly connected to each other and the the wheel is forced to rotate a single common angular velocity. On the other hand, gyroscopes cannot be forced to assume a common angular velocity by a similar averaging mechanism, because there will then be a real torque on the slower gyroscopes, causing a torque reaction on the spin axis of the slower gyroscopes so that their spin axis does not remain orthogonal to the equatorial plane. I think this relates to the apparent *anisotropic* rotation of the gyroscopes that Peter is mulling over in the quote at the top of this post, which is at odds with the fact that Fermi transported vectors remain orthogonal to each other. Something does not seem quite right here.

 

[WannabeNewton]

Hey guys, sorry to drag this thread back into the light but I was rereading the "Sagnac Effect in General Relativity" paper by Ashtekar and Magnon (which I think you both have copies of at this point) and I got confused by a technical detail.

To start with, the authors take a space-time $(\mathcal{M},g_{ab})$ and consider a Sagnac tube $(\mu, h_{ab})$ in this space-time such that $\mu$ admits a time-like Killing field $t^{a}$ to which it is tangent. Note that the tangent field $t^{a}$ is only required to be a Killing field on $\mu$ i.e. $\tilde{\nabla}_{(a}t_{b)} = 0$ where $\tilde{\nabla}$ is the derivative operator on $\mu$ associated with its metric $h_{ab}$; hence $t^{a}$ need not necessarily be a Killing field on $\mathcal{M}$ itself. This means in particular that if we are in Kerr space-time we can let $\mu$ be the Sagnac tube with tangent field $t^{a} = \xi^a|_{\mu} + \omega|_{\mu}\psi^a|_{\mu}$, where $\omega|_{\mu}$ is the ZAMO angular velocity evaluated on this Sagnac tube and $\xi^a|_{\mu},\psi^a|_{\mu}$ are the stationary and axial Killing fields of the stationary source also evaluated on this Sagnac tube, since $t^a$ is a Killing field on $\mu$.

Now if I'm not mistaken, even though eq. (3) of the paper is derived in the specific context of a Sagnac tube at rest with respect to the stationary source i.e. $t^a = \xi^a$, the derivation of eq. (3) itself holds just as well for any $t^a$ which is a time-like Killing field on $\mathcal{M}$ (not just on $\mu$!) because from eq. (3) we have

$(\lambda_{M})^{1/2}\int _{\Sigma}\lambda^{-3/2}\omega^{a}\epsilon_{abc}dS^{bc}= (\lambda_{M})^{1/2}\int _{\Sigma}\lambda^{-2}\epsilon^{aefg}\epsilon_{abcd}t^dt_{e}\nabla_{f}t_{g}dS^{bc}= -6(\lambda_{M})^{1/2}\int _{\Sigma}\lambda^{-2}t^dt_{[b}\nabla_{c}t_{d]}dS^{bc}$

and furthermore $3t_{[b}\nabla_{c}t_{d]} = t_b t^d \nabla_c t_d - t_c t^d\nabla_b t_d -\lambda \nabla_b t_c = t_{[b}\nabla_{c]}\lambda - \lambda \nabla_{[b}t_{c]}$

hence $(\lambda_{M})^{1/2}\int _{\Sigma}\lambda^{-3/2}\omega^{a}\epsilon_{abc}dS^{bc}= 2(\lambda_{M})^{1/2}\int _{\Sigma}\nabla_{[a}(\lambda^{-1}t_{b]})dS^{ab} = \Delta \tau$.

Note that directly under eq. (3) in the paper, the authors state that the $\nabla_a$ appearing in the Sagnac shift is the derivative operator on $\mathcal{M}$ itself, and not just the derivative operator $\tilde{\nabla}_a$ on $\mu$. Hence the reason we need $t^a$ to be a Killing field on all of $\mathcal{M}$ and not just on $\mu$ is in the above calculation we needed to make use of $\nabla_{[a} t_{b]} = \nabla_a t_b$ under $\nabla_a$ and not under $\tilde{\nabla}_a$. This implies then that eq. (3) holds in Kerr space-time for the Sagnac tube given by the time-like Killing field $t^a = \xi^a + \omega\psi^a$ on $\mathcal{M}$, where as usual $\omega$ is a single ZAMO angular velocity held constant throughout space-time (note however that eq. (3) fails to hold for the ZAMO congruence for the aforementioned reason).

This brings me (finally!) to my question. We know that $\omega^a = \epsilon^{abcd}t_b \nabla_c t_d \neq 0$ for the above $t^a$. On the other hand we know from Malament's text that $\Delta \tau = 0$ should hold for this Sagnac tube since $L = t_a \psi^a = 0$ on $\mu$. This would require then that $\int _{\Sigma}\lambda^{-3/2}\omega^{a}\epsilon_{abc}dS^{bc} = 0$ for said Sagnac tube. I haven't yet attempted to show this explicitly by means of calculation but regardless, assuming I haven't made any mistakes above , is there a physically intuitive way to see why the above integral would necessarily have to vanish, even with $\omega^a \neq 0$? Do the symmetries of $\Sigma$ (see fig. 1) somehow make the flux of $\lambda^{-3/2}\omega^{a}$ through the Sagnac tube cancel out upon integration?

Thanks in advance.

 

[WannabeNewton]

Okay so if we take eq. (1) in the paper, for the above ZAMO Sagnac tube $\mu$ with tangent field $t^a = \xi^a + \omega|_{\mu}\psi^a$, and evaluate it along a curve $dt = d\theta = dr = 0$ on $\mu$, then $dS^a = \sqrt{g_{\phi\phi}}d\phi \psi^a$ and so $t_a dS^a \propto t_a \psi^a = L = 0$ hence $\oint_{S} \lambda^{-1}t_a dS^a = 0$ trivially which would then imply, as per Stokes' theorem as usual, that $\Delta \tau = \int _{\Sigma}\lambda^{-3/2}\omega^{a}\epsilon_{abc}dS^{bc} = 0$ but I still cannot see intuitively and physically why $\int _{\Sigma}\lambda^{-3/2}\omega^{a}\epsilon_{abc}dS^{bc} = 0$ directly without appealing to eq. (1) and Stokes' theorem; does anyone know of such a physically intuitive argument?

 

[WannabeNewton]

As an aside, I finished reading another paper that you guys might be interested in as it also deals with criteria of non-rotation, particularly a maximal acceleration criterion for non-rotation that we haven't discussed as of yet in this thread, and many of the calculations parallel what we have already calculated and discussed in this thread.

http://arxiv.org/pdf/gr-qc/9706029v2.pdf