Locally non-rotating observers

[PeterDonis]

all the derivations of the formula for the frame-dragging I've seen (certainly those initially done by Einstein himself around 1917 -see "The meaning of relativity" pages107-109- , and I guess by Thirring and Lense), are based simply on the linearization of the EFE in the weak field applied to a self-rotating source.

The same comment would apply here; the weak field linearized approximation to a self-rotating source should have leading order terms which are just the weak field linearized approximation to a non-rotating source. At least, that's the way it seems to me, but I'll need to look at MTW's discussion (since WannabeNewton has already checked Wald and found little to chew on) to see if there are subtleties I'm missing.

 

[PeterDonis]

this gyroscope is Fermi-Walker transported along this observer's worldline (since Fermi-Walker transport is equivalent to parallel transport for geodesics) and hence we can think of $\omega^{\mu}$ as representing the rotation of nearby static observers inside the shell relative to this gyroscope?

Yes, this looks right. But note that $\omega^{\mu}$ will be the rate of rotation of the nearby static observers relative to the gyroscope, with respect to the central observer's proper time, *not* with respect to coordinate time. I left that part out of our previous discussion, and I think it makes a difference. See below.

So the central observer in the congruence will see nearby static observers rotate around him whilst being fixed to the distant stars, much like we imagined before with the static observers in Kerr space-time?

Yes.

On the other hand the quantity $\frac{\mathrm{d} \vec{S}}{\mathrm{d} t} = \frac{4}{3}\frac{M}{R}\vec{\omega}\times \vec{S}$ represents the rate at which an observer at infinity sees the aforementioned gyroscope spin in place (here $\vec{\omega}$ is the angular velocity of the shell)?

Yes, but as your notation makes clear, this rate is relative to coordinate time, *not* the central observer's proper time, which is why it's a different number than $\omega^{\mu}$, even though both are referring to the same thing (well, opposite signs of the same thing).

The reason I bring this up is that, as I mentioned above, I ignored the difference between coordinate time and proper time in our previous discussion, and putting it back in makes a difference. For example, consider the ZAMO and hovering congruences in Kerr spacetime. We talked about comparing two sets of basis vectors for each type of observer: a set of Fermi-Walker transported vectors (gyroscopes), and a set of vectors fixed by connecting rods to neighboring members of the same congruence. Consider what we said about them:

For the ZAMO congruence, the two sets of vectors remain aligned (assuming that they start out aligned); that means both of them rotate at the same rate relative to an observer at infinity. We said this rate is "the same" as the angular velocity of the ZAMO about the black hole, but actually it is different depending on who is observing it: the ZAMO will get a different actual number than the observer at infinity, because of time dilation (here a combination of the "redshift factor" from the $g_{tt}$ metric coefficient, and the SR time dilation due to nonzero tangential velocity).

For the hovering congruence, the two sets of vectors will *not* remain aligned (even if they start out aligned); they rotate relative to each other. But again, because of time dilation (here just the $g_{tt}$ "redshift factor"), the rate at which the hovering observer sees the neighboring static observers rotate around him, compared to his gyroscopes (which is a rate with respect to his proper time), will *not* be the same as the rate at which an observer at infinity sees the hovering observer's gyroscopes spinning in place (which is a rate with respect to coordinate time).

 

[WannabeNewton]

Ah, right twist is based on the proper time read by the observer in the congruence we're using as a reference whereas the other is the coordinate time as read by the observer at infinity so gravitational and kinematical time dilation will cause discrepancies. Thanks for pointing that out Peter, and thanks for clarifying the scenario with the static congruence inside the thin rotating shell. I was just using that scenario because it resembled the static observer scenario in Kerr space-time except it was somewhat easier to picture.

Thanks again!

 

[WannabeNewton]

Peter sorry to bother you again but I had a question that I meant to ask before but forgot to. If we take locally non-rotating to mean that the twist $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0$ for a time-like congruence $\xi^{a}$ then we know this implies that $\xi^{a} = \alpha \nabla^{a}\beta$ for smooth functions $\alpha,\beta$ but $L = \xi^{a}\psi_{a} = \alpha \psi_{a}\nabla^{a}\beta \neq 0$ in general so there is no equivalence between locally non-rotating in the sense of vanishing twist and zero angular momentum for an arbitrary twist free congruence. So is it just a coincidence that for the $\nabla^{a} t$ congruence we have local non-rotation in the sense that $\omega^{a} = \alpha \epsilon^{abcd}\nabla_{b} t \nabla_{c}(\alpha \nabla_{d}t) = 0$ as well as vanishing angular momentum $L = \alpha\psi_{a}\nabla^{a}t = 0$ or is there a physical connection between the two?

I ask because in Malament's GR text, local non-rotation is simply defined as vanishing twist of a congruence (so that the worldlines of the observers in the congruence are unfurled) and no mention of angular momentum is made whereas in Wald's text local non-rotation is specifically defined as being an observer in the $\nabla^{a} t$ congruence in which case $L = 0$ follows suit; I don't really get why Malament defines it in a much more general manner whereas Wald sticks to that specific case. I can tell intuitively that defining local non-rotation as $\omega^{a} = 0$ gives us a congruence which is hypersurface orthogonal which is quite important physically whereas I can't immediately see the relevance/use of the fact that for the specific case of the hypersurface orthogonal congruence given by $\nabla^{a} t$, we have $L = 0$ as well.

 

[TrickyDicky]

The same comment would apply here; the weak field linearized approximation to a self-rotating source should have leading order terms which are just the weak field linearized approximation to a non-rotating source. At least, that's the way it seems to me, but I'll need to look at MTW's discussion (since WannabeNewton has already checked Wald and found little to chew on) to see if there are subtleties I'm missing.

Well yes, when I say linearization I'm rather referring to the post-Newtonian expansion or the lowest order approximation from the Newtonian limit so it goes beyond linearization proper, sorry about the confusion. For instance one of the parameters of the parametrized post-Newtonian formalism is frame draging per angular momentum unit =Δ1.
The Lense-Thirring frame-dragging effect formula that was used in textbooks before the mathematical derivation from the Kerr geometry was available in the sixties is the one based in the gravitomagnetic field of the earth, that is basically the post-Newtonian approximation above mentioned, the weak field, low speed approximation of the GR equations of motion in the presence of angular momentum.

 

[PeterDonis]

Peter sorry to bother you again

No problem, I'm not bothered. I can talk about this stuff indefinitely.

there is no equivalence between locally non-rotating in the sense of vanishing twist and zero angular momentum for an arbitrary twist free congruence.

See below.

So is it just a coincidence that for the $\nabla^{a} t$ congruence we have local non-rotation in the sense that $\omega^{a} = \alpha \epsilon^{abcd}\nabla_{b} t \nabla_{c}(\alpha \nabla_{d}t) = 0$ as well as vanishing angular momentum $L = \alpha\psi_{a}\nabla^{a}t = 0$ or is there a physical connection between the two?

I think there might be a connection; see below.

I can tell intuitively that defining local non-rotation as $\omega^{a} = 0$ gives us a congruence which is hypersurface orthogonal which is quite important physically whereas I can't immediately see the relevance/use of the fact that for the specific case of the hypersurface orthogonal congruence given by $\nabla^{a} t$, we have $L = 0$ as well.

What other hypersurface orthogonal timelike congruences are there? Can you find one that has a nonzero $L$? I suspect there aren't any, which means that any twist free vector field $\alpha \nabla^a \beta$ must have $\beta = k t$ with $k$ constant. Which really means we set the coordinate chart up the right way for $t$ to have that property.

 

[TrickyDicky]

I looked at MTW and it is actually quite well explained in the PPN formalism in pages 1117-9. So if we expect both effects(de sitter precession and frame-dragging) to show up we must at the same time ignore the cross terms d\phi/dt (de Sitter effect) as it is done in Fermi-Walker transport, and not ignore them (frame dragging). Hmmm...well, one can say it is just an approximation after all, but if we refer to the actual GR solutions we must at the same time rely for the last effect on a geometry that explicitly rules out staticity (Kerr), and on a static geometry that that demands the cross terms to vanish for computation of the de Sitter effect. I don't get the logic behind this.

 

[Bill_K]

I don't get the logic behind this.

I have a blog which discusses precession in all its aspects: Thomas precession in flat space, de Sitter precession in Schwarschild, and Lense-Thirring in Kerr. The combined precession rate contains terms which can easily be identified with each of these effects.

 

[WannabeNewton]

What other hypersurface orthogonal timelike congruences are there? Can you find one that has a nonzero $L$? I suspect there aren't any, which means that any twist free vector field $\alpha \nabla^a \beta$ must have $\beta = k t$ with $k$ constant.

Well like in the Schwarzschild space-time, $\nabla^{a}t$ is everywhere time-like for $r > 2M$ since $g_{\mu\nu}\nabla^{\mu}t\nabla^{\nu}t = -(1 - \frac{2M}{r})^{-1}$ which is less than zero for $r > 2M$. Similarly, the function $f(t,r) = t + \frac{1}{2}[2M\ln (r - 2M) + r]$ gives us the vector field $\xi^{\mu} = \nabla^{\mu}f = g^{\mu t} + \frac{1}{2}g_{rr}g^{\mu r}$ and we have that $g_{\mu\nu}\xi^{\mu}\xi^{\nu} = g^{tt} + \frac{1}{4}g_{rr} = -\frac{3}{4}(1 - \frac{2M}{r})^{-1}$ which, much like the norm of $\nabla^{a} t$, is less than zero for $r > 2M$ hence everywhere time-like for $r > 2M$ just like $\nabla^{a} t$. $\xi^{\mu}$ is of course twist free but $\xi^{a} \neq \alpha \nabla^{a}t$ for some smooth function $\alpha$ however $L = \psi_{a}\xi^{a} = 0$ still holds so it would be possible to have twist free time-like vector fields in the space-time that are not colinear with $\nabla^{a} t$ but still have vanishing angular momentum if we included e.g. a radial component.

However if we want a twist free vector field that is time-like and has $L \neq 0$ then it would need to be non-orthogonal to $\psi^{a}$. I can't immediately find such a vector field in the Schwarzschild space-time (the computation got quite messy) nor can I prove one doesn't exist (what I tried to do was assume there existed a vector field $\xi^{a}$ such that $\xi^{a}\psi_{a}\neq 0$, which in the Schwarzschild space-time comes down to $\xi^{a}$ having a component along $\psi^{a}$, and then showing that $\xi^{a}$ can't satisfy both $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0$ and $\xi_{a}\xi^{a} < 0$ but it got insanely messy really fast and I can't immediately think of an easier/more elegant approach that would hold not just for Schwarzschild space-time but also in full generality for any stationary, axisymmetric space-time).

Anyways, the main reason I asked was in the specific case that we do have both vanishing twist (so local non-rotation) and vanishing angular momentum, is there a way to picture the vanishing angular momentum in terms of the vanishing twist? So in the case of the $\nabla^{a} t$ congruence in Kerr space-time, is there a way to visualize vanishing angular momentum for the observers in the congruence in terms of the fact that the congruence is twist free? I just can't really visualize vanishing angular momentum on its own so that's why I asked the question in the first place, to see if there was a way to physically connect them (since they are both true for the $\nabla^{a} t$ congruence) because unlike vanishing twist and angular velocity (as observed at infinity) which I can visualize, I can't seem to visualize vanishing angular momentum.

Thanks Peter!

 

[TrickyDicky]

Kerr space-time can exhibit both geodetic precession and Lense-Thirring precession. Are you asking how an experiment can distinguish between the two?

I can't think of any conditions under which we would expect to observe frame dragging but *not* the geodetic effect. In Kerr spacetime both effects are predicted.

Ok, but let's see an example of what I mean. Since it is easy to agree that the most general notion of locally non-rotating is the one given by Fermi-Walker transport that is valid for any GR solution by virtue of the orthonormal vectors of the frame field, let's stick to this definition and apply it to the actual physical gyroscopes of the satellite Gravity probe B, my questions:

-are the gyroscopes locally non-rotating in the sense of the Fermi-Walker transport (so that they can only precess due to effects of the central mass/energy but not due to rotation of the source) or not?

-Can they be both locally rotating and non-rotating at the same time?

-If they are locally non-rotating in the Fermi-Walker sense (gyroscopically stabilized), how can they measure the rotating effects of frame-dragging?

-If they are not Fermi-Walker stabilized how can they measure the full de Sitter precession?

 

[Bill_K]

-are the gyroscopes locally non-rotating in the sense of the Fermi-Walker transport (so that they can only precess due to effects of the central mass/energy but not due to rotation of the source) or not?
-Can they be both locally rotating and non-rotating at the same time?
-If they are locally non-rotating in the Fermi-Walker sense (gyroscopically stabilized), how can they measure the rotating effects of frame-dragging?
-If they are not Fermi-Walker stabilized how can they measure the full de Sitter precession?

They're locally nonrotating, which by definition means Fermi-Walker transported. If you read the blog I referred to, which discusses all three effects together, you won't need to ask silly questions like this.

 

[TrickyDicky]

They're locally nonrotating, which by definition means Fermi-Walker transported. If you read the blog I referred to, which discusses all three effects together, you won't need to ask silly questions like this.

Bill, by my experience silly questions are usually the best in science.
I wasn't able to elucidate this by reading your blog, that's why I asked. And you didn't answer my third question, can you?

 

[Bill_K]

-If they are locally non-rotating in the Fermi-Walker sense (gyroscopically stabilized), how can they measure the rotating effects of frame-dragging?

Ok, here's a description of Gravity Probe B. Measurement comes from comparison of the locally nonrotating frame as determined by the onboard gyroscopes versus the "fixed stars". It's the relative motion that counts - whether you regard the gyroscopes as fixed (F-W frame) and the stars as precessing, or that GR (frame dragging) causes the gyroscopes themselves to precess doesn't matter.

 

[TrickyDicky]

Measurement comes from comparison of the locally nonrotating frame as determined by the onboard gyroscopes versus the "fixed stars".

Right, this gives us the de Sitter precession but again, locally non-rotating according to Fermi-Walker transport by definition (according to Wikipedia) "defines a reference frame such that all curvature in the frame is due to the presence of mass/energy density and not to arbitrary spin or rotation of the frame." How can it measure frame-dragging due to rotation of the frame?

It's the relative motion that counts - whether you regard the gyroscopes as fixed (F-W frame) and the stars as precessing, or that GR (frame dragging) causes the gyroscopes to precess doesn't matter.

Well, this is fine and doesn't affect the issue at hand, but I have heard you say a few times that rotation is absolute and independent of the distant stars. Now you are Machian suddenly?

 

[Bill_K]

Right, this gives us the de Sitter precession but again, locally non-rotating according to Fermi-Walker transport by definition (according to Wikipedia) "defines a reference frame such that all curvature in the frame is due to the presence of mass/energy density and not to arbitrary spin or rotation of the frame." How can it measure frame-dragging due to rotation of the frame?

The point my blog is supposed to illustrate, even skipping the mathematics, is that all three precessions are aspects of one phenomenon. You can illustrate it using a path in flat space, or in Schwarzschild, or in Kerr. You can point to terms in the result and call them de Sitter or whatever, but the distinction is merely circumstantial.

Well, this is fine and doesn't affect the issue at hand, but I have heard you say a few times that rotation is absolute and independent of the distant stars. Now you are Machian suddenly?

The local nonrotating frame depends only on the particle's world line. I only meant to say that Gravity Probe B measures the difference between the two frames.

 

[WannabeNewton]

-If they are locally non-rotating in the Fermi-Walker sense (gyroscopically stabilized), how can they measure the rotating effects of frame-dragging?

You Fermi-Walker transport your basis vectors along your worldline in order to gyrostabilize them (so that there are no spurious rotations). Then you can calculate how the basis vectors precess relative to infinity/the fixed stars. In one of the previous posts I brought up the scenario of an observer at rest at the center of a thin rotating shell who parallel transports along his worldline a purely spatial vector; you end up calculating how the vector precesses relative to infinity/the fixed stars while it is being parallel transported along the central observer's worldline. This is basically a very simple case of the general calculation in Bill's blog.

 

[TrickyDicky]

You Fermi-Walker transport your basis vectors along your worldline in order to gyrostabilize them (so that there are no spurious rotations). Then you can calculate how the basis vectors precess relative to infinity/the fixed stars. In one of the previous posts I brought up the scenario of an observer at rest at the center of a thin rotating shell who parallel transports along his worldline a purely spatial vector; you end up calculating how the vector precesses relative to infinity/the fixed stars while it is being parallel transported along the central observer's worldline. This is basically a very simple case of the general calculation in Bill's blog.

Right, I have no problem with this but I can only see how de Sitter precessiob is obtained with this setting because I interpret the wiki definition above mentioned as not allowing to detect frame-dragging because we make sure the frame only can be affected by non-rotational curvature effects. I don't see this bit answered.
And just to be clear I'm GR's number one fan, this is not any kind of challenge to anything (in case anyone has that impression), I truly don't know how to make this out.

 

[WannabeNewton]

Right, I have no problem with this but I can only see how de Sitter precessiob is obtained with this setting because I interpret the wiki definition above mentioned as not allowing to detect frame-dragging because we make sure the frame only can be affected by non-rotational curvature effects.

Which wiki definition (I can't seem to find anything above) and where in it do we make sure the frame can only be affected by non-rotational curvature effects?

 

[Bill_K]

Right, this gives us the de Sitter precession but again, locally non-rotating according to Fermi-Walker transport by definition (according to Wikipedia) "defines a reference frame such that all curvature in the frame is due to the presence of mass/energy density and not to arbitrary spin or rotation of the frame." How can it measure frame-dragging due to rotation of the frame?

I agree this sentence from Wikipedia is poorly worded. By "curvature in the frame". what they mean is "change of the basis vectors from one point to another". By "the presence of mass/energy density" they mean any and all nearby masses, such as a Kerr source. (No, this is not Machian!) By "arbitrary spin or rotation of the frame", what they're referring to is that the F-W frame has, in a sense, the minimum rotation required. Any other frame you might think to define has "more".

 

[WannabeNewton]

... because I interpret the wiki definition above mentioned as not allowing to detect frame-dragging because we make sure the frame only can be affected by non-rotational curvature effects. I don't see this bit answered.

Having seen the wiki linked by Bill, I think I see what you mean. Recall that we can think of Fermi-Walker transport as locally describing an observer standing inside an Einstein elevator in flat space-time, with the spatial basis vectors representing gyroscopes carried by the observer that detect zero rotation of the elevator (say in contrast with an elevator that is kicked at the side and set spinning about its vertical axis). So locally the Fermi-Walker frame is the SR equivalent of a non-rotating frame and globally (i.e. along the entire worldline) it is the closest possible thing we can get to the SR equivalent of a non-rotating frame. It doesn't eliminate frame dragging from the source itself.

 

[TrickyDicky]

Thanks Bill and WN, now I get it.

 

[WannabeNewton]

I have a related question though. Say we are in a global inertial frame in Minkowski space-time and we transform to a frame rotating with angular velocity $\omega$. We can write the metric in this frame as $ds^{2} = -(1 - \omega^2 r^2)dt^{2} + 2\omega r^{2}d\phi dr + dr^2 + r^2 d\phi ^2 + dz^2$. We can interpret this as a space-time through the equivalence principle. The congruence of static observers in this space-time has a tangent field $\xi^{a} = (1 - \omega^2 r^2)^{-1/2}(\partial_{t})^{a}$. The twist of this congruence is given by $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = (1 - \omega^2 r^2)^{-1}\omega (\partial_{z})^{a}$.

Now the static observers are the observers who must be at rest with respect to the rotating frame so in the global inertial frame these are just the observers circling around the origin of the rotating frame with angular velocity $\omega$ (within the allowed domain i.e. within the null barrier of the cylinder). Because all these observers are circling the origin with the same angular speed, if I go to the frame of a given static observer in this congruence I can't understand why I would see nearby observers in the congruence rotate around me about an axis parallel to the z-axis. I imagine just seeing them sitting in place because they are all circling around at the same rate as me. The only way I can imagine seeing observers rotate around me about an axis parallel to the z-axis is if there is frame dragging in this space-time that causes me to have self-rotation about an axis parallel to the z-axis (i.e. precess about an axis parallel to the z-axis whilst staying in place in this space-time). So my question comes down to: does there exist frame dragging in the space-time obtained from the rotating frame via the equivalence principle that causes the static observers in this space-time to self-rotate?

 

[Bill_K]

does there exist frame dragging in the space-time obtained from the rotating frame via the equivalence principle that causes the static observers in this space-time to self-rotate?

To a first approximation (small rω), the nonrotating basis for these orbiting observers would continue to be the x, y, z axes, the same as the global IRF. In this frame, your nearest neighbors make a small circle about you with angular velocity ω. As you get farther out (larger rω) the [STRIKE]Lense-Thirring [/STRIKE] Thomas precession starts to take effect. Farther out still, it becomes larger and larger, and eventually it dominates. You can't get any farther out than rω = 1, where the world lines are no longer timelike.

 

[WannabeNewton]

To a first approximation (small rω), the nonrotating basis for these orbiting observers would continue to be the x, y, z axes, the same as the global IRF. In this frame, your nearest neighbors make a small circle about you with angular velocity ω. As you get farther out (larger rω) the Lense-Thirring precession starts to take effect. Farther out still, it becomes larger and larger, and eventually it dominates. You can't get any farther out than rω = 1, where the world lines are no longer timelike.

Thanks for the response Bill! I don't quite get what you mean in the beginning there. If the Lense-Thirring precession on me only starts to really take effect for larger $r\omega$ then what causes my nearest neighbors to make small circles around me from my perspective for small $r\omega$ (me being one of the members of the congruence of static observers in the aforementioned space-time)? I thought it was the Lense-Thirring precession that made me see my nearest neighbors rotate around me (which from infinity would be seen as me just precessing in place)?

EDIT: did you mean that for small $r\omega$, the Lense-Thirring precession is not nearly as dramatic as for larger $r\omega$ and $\omega^{a} = (1 - r^2\omega^{2})^{-1}\omega (\partial_{z})^{a}\approx \omega(\partial_{z})^{a}$ so the Lense-Thirring precession for small $r\omega$ will just have me see my nearest neighbors rotate around me about an axis parallel to the z-axis at approximately $\omega$ but the effect itself will not be that dramatic?

 

[PeterDonis]

As you get farther out (larger rω) the Lense-Thirring precession starts to take effect.

Is there Lense-Thirring precession in Minkowski spacetime? I understood WN's scenario to be set in Minkowski spacetime.

As I understand it, the nonzero twist of the congruence WN describes (which we called the "Langevin congruence" earlier in this thread) is due to the fact that the frame field of the congruence, with spatial vectors fixed to point at neighboring members of the congruence, does not Fermi-Walker transport those spatial vectors (or at least, it doesn't F-W transport the ones that are in the plane of rotation).

Bill_K derives the F-W transported vectors in his blog, and they show Thomas precession, which is slightly retrograde. So the twist of this congruence is the difference between the F-W transported vectors (which, relative to infinity, rotate slighly in the retrograde direction) and the vectors which are locked to neighboring members of the congruence (which, relative to infinity, rotate in the prograde direction). Earlier in this thread, I believe we derived a formula for the twist which showed that it is *larger* than the angular velocity of rotation, which makes sense in the light of what I just said: the twist is the sum of the angular velocity of rotation and the Thomas precession (since that is in the opposite direction to the rotation).

 

[WannabeNewton]

But then what mechanism causes the spatial basis vectors of the static observers in the space-time associated with the rotating frame to precess relative to the distant stars whilst the static observers remain in place in the space-time (or equivalently, what mechanism causes the connecting vectors from a given static observer to neighboring static observers to rotate relative to the Fermi-Walker transported basis vectors of the given static observer)?

I ask because the orbital motion of these observers around the origin of the rotating frame relative to the global inertial frame cannot by itself lead to a given such observer in the congruence seeing neighboring observers in the congruence rotate around him because they all orbit with the same angular velocity so if there was no extra mechanism involved, the given observer should just see the neighboring observers sitting in place and not rotating around him. I can only imagine the twist being non-zero if each observer has some kind of added precession relative to the distant stars that say is induced as a side effect of the orbital motion relative to the global inertial frame.

 

[PeterDonis]

But then what mechanism causes the spatial basis vectors of the static observers in the space-time associated with the rotating frame to precess relative to the distant stars whilst the static observers remain in place in the space-time (or equivalently, what mechanism causes the connecting vectors from a given static observer to neighboring static observers to rotate relative to the Fermi-Walker transported basis vectors of the given static observer)?

I think you're getting mixed up again about which rotation is which. Take things one by one:

(1) As seen by an observer at infinity, the spatial vectors defined by the Langevin congruence (i.e., the ones locked to point to neighboring members of the congruence--call these the "congruence vectors") obviously rotate, in the same sense and with the same angular velocity as the rotation of the observers themselves.

(2) Therefore, the Langevin observers will see the fixed stars rotating, relative to their congruence vectors, in the opposite sense to #1. (The angular velocity they measure will differ from that measured by the observer at infinity because of time dilation.)

(3) If a Langevin observer happens to carry with him a set of gyroscopes, whose orientations are Fermi-Walker transported (call these the "gyro vectors"), an observer at infinity will see these vectors rotating in the *opposite* sense to the Langevin observer's own rotation. This is the Thomas precession.

(4) Therefore, the Langevin observer will see the fixed stars rotating, relative to his gyro vectors, in the opposite sense to #3 (i.e., in the same sense as his own rotation, as seen by an observer at infinity). Again, the angular velocity he measures will be different from the Thomas precession angular velocity measured at infinity, because of time dilation.

(5) Therefore, a Langevin observer who carries both congruence vectors and gyro vectors, will see them rotating, relative to each other, at an angular velocity which is the sum of those from #2 and #4. (And an observer at infinity will see the two sets of vectors rotating, relative to each other, at an angular velocity which is the sum of those from #1 and #3.)

I ask because the orbital motion of these observers around the origin of the rotating frame relative to the global inertial frame cannot by itself lead to a given such observer in the congruence seeing neighboring observers in the congruence rotate around him because they all orbit with the same angular velocity so if there was no extra mechanism involved, the given observer should just see the neighboring observers sitting in place and not rotating around him.

Relative to the congruence vectors, he does. Don't confuse the congruence vectors with the gyro vectors. The twist is the rotation of the congruence vectors relative to the gyro vectors; by itself it doesn't tell you anything about rotation relative to infinity. For that you need to know how at least one of the two sets of vectors (congruence vectors or gyro vectors) rotates relative to infinity.

 

[WannabeNewton]

(1) As seen by an observer at infinity, the spatial vectors defined by the Langevin congruence (i.e., the ones locked to point to neighboring members of the congruence--call these the "congruence vectors") obviously rotate, in the same sense and with the same angular velocity as the rotation of the observers themselves.

Why would these congruence vectors rotate relative to infinity if the observers are fixed in place relative to infinity (I was talking about the static observers in the space-time associated with the rotating frame through the equivalence principle by the way).

Don't confuse the congruence vectors with the gyro vectors. The twist is the rotation of the congruence vectors relative to the gyro vectors; by itself it doesn't tell you anything about rotation relative to infinity. For that you need to know how at least one of the two sets of vectors (congruence vectors or gyro vectors) rotates relative to infinity.

Right the twist is relative to the spatial axes of the Fermi-Walker frame. I'm asking what physical mechanism is causing this twist i.e. why are the congruence vectors rotating relative to the spatial axes of the Fermi-Walker frame? In the case of the static observers in the Kerr space-time, it was frame dragging. Are you saying here it's the Thomas Precession due to the orbital motion of the observers in the global inertial frame? But if we look at this in terms of the space-time associated with the rotating frame, wherein the Langevin observers are static observers, why can't we view this as frame dragging?

 

[PeterDonis]

Why would these congruence vectors rotate relative to infinity if the observers are fixed in place relative to infinity

I was confused, I thought you were talking about the congruence vectors for the Langevin congruence. Obviously the static congruence vectors, those for observers fixed in place relative to infinity, in flat spacetime don't rotate at all. But Langevin observers aren't fixed in place relative to infinity; that's true regardless of which frame you work in (see further comments below).

(I was talking about the static observers in the space-time associated with the rotating frame through the equivalence principle by the way).

Now I'm confused again. What is "the space-time associated with the rotating frame through the equivalence principle"? There's only one spacetime, which I thought in your latest question to Bill_K was Minkowski spacetime. Do you mean a "rotating frame" in which the Langevin observers are at rest? Changing frames doesn't change the physics, so if we do calculations in such a rotating frame (actually, Bill_K's blog post does use such a frame), we should get the same answer as in a global inertial frame. In such a rotating frame, observers fixed at infinity are rotating, so the Langevin observers (who are fixed in this frame) are still rotating relative to infinity.

Right the twist is relative to the spatial axes of the Fermi-Walker frame. I'm asking what physical mechanism is causing this twist i.e. why are the congruence vectors rotating relative to the spatial axes of the Fermi-Walker frame?

It seems to me that this is a two-part question. The first part is easy: the congruence vectors are fixed to point at neighboring members of the congruence, so if that's not equivalent to Fermi-Walker transport, the congruence vectors will rotate relative to the gyro vectors.

The second part is: what physical mechanism causes the Fermi-Walker transported vectors to precess the way they do? (I.e., to undergo Thomas precession, in the retrograde direction.) The standard explanation (at least, it seems to me to be reasonably standard) is that the boost applied to the rotating observer is continually changing direction, and successive Lorentz boosts in different directions produce a spatial rotation. I have a hard time visualizing how this works so I can't really comment further.

In the case of the static observers in the Kerr space-time, it was frame dragging. Are you saying here it's the Thomas Precession due to the orbital motion of the observers in the global inertial frame?

Yes, see above.

But if we look at this in terms of the space-time associated with the rotating frame, wherein the Langevin observers are static observers, why can't we view this as frame dragging?

That's a question of terminology, not physics. I personally wouldn't use the term "frame dragging" because to me it suggests that a rotating mass is present; calling the effect of changing frames "frame dragging" doesn't seem to fit. (Another way of putting this would be to say that I view "frame dragging" as the name for something invariant; but the Thomas precession, as seen in a global inertial frame, is not usually called "frame dragging", so your interpretation would make "frame dragging" coordinate-dependent.) But different people have different intuitions about such things.

 

[WannabeNewton]

Peter I was talking about the observers at rest in the gravitational field associated with $ds^2=-(1-\omega^2 r^2)dt^2 +2\omega r^2 dt d\phi +dr^2 +r^2 d\phi ^2 +dz^2$ and if we can ascribe the twist of these observers to a property of said gravitational field akin to the scenario in Kerr spacetime.