Locally non-rotating observers

[WannabeNewton]

Right but then if we Fermi-Walker transport the spatial basis vectors of the static observer, how will that tell us he is self-rotating if Fermi-Walker transporting implies by definition that there is zero self-rotation? Sorry if I'm not being clear enough.

 

[PeterDonis]

Right but then if we Fermi-Walker transport the spatial basis vectors of the static observer, how will that tell us he is self-rotating if Fermi-Walker transporting implies by definition that there is zero self-rotation? Sorry if I'm not being clear enough.

Ah, I see your point. Yes, this is another area where it's difficult to find good terminology, because our intuitions don't really allow for this kind of behavior.

The Fermi-Walker transport definition of "no self-rotation" is local; locally, the observer is doing the best he can to keep his spatial basis vectors pointing "in the same direction".

But this local definition might not match a global definition of "no rotation", such as the one the observer at infinity would naturally use: he sees the hovering observer "spinning in place" because of frame dragging. Conversely, the hovering observer would see the distant stars spinning about his gyro-stabilized basis vectors. But that's not a local observation.

Here's another way of thinking about it: suppose the hovering observer were inside an "Einstein elevator" with no way to see out, so the only reference he has for "lack of rotation" is what he can measure inside the elevator. If the elevator's orientation is maintained by gyro-stabilization, there will be no inertial effects due to rotation inside it that the hovering observer can detect; for example, he will feel zero centrifugal force due to "rotation" of the elevator. (Note that we of course have to first factor out the effects of the nonzero proper acceleration of the elevator; the point is that being inside this elevator would be exactly like being inside an accelerated elevator in flat spacetime that was not rotating at all, with respect to flat spacetime.) Only if he opens a window and looks out at the distant stars will he see anything rotating.

Now consider a second hovering "elevator", whose orientation is stabilized by being attached to neighboring hovering observers, all belonging to the hovering congruence (think of its sides as being anchored by the "diamond" we talked about before). The hovering observer inside *this* elevator *will* be able to detect local effects of rotation, corresponding to the failure of the spatial vectors defining this elevator's orientation to be Fermi-Walker transported. For example, there will be a nonzero centrifugal force inside this elevator (again, after factoring out the effects of the elevator's proper acceleration--in other words, being inside this elevator would be like being inside an elevator in flat spacetime that was not only being linearly accelerated, but also spun about an axis at right angles to the direction of acceleration).

 

[WannabeNewton]

I'm not quite getting the physical significance of the two scenarios Peter. In the first scenario, are we basically saying that Fermi-Walker transport of the orientation will result in zero local self-rotation as defined (so if the observer were enclosed in an Einstein elevator and only had his gyros as a method of measuring rotation he wouldn't be able to detect anything) but globally he will see the stars rotate around him if he were able to peer out so there will be a non-zero global self-rotation relative to the stars? I can't seem to connect this to the second scenario though.

Thanks.

 

[PeterDonis]

In the first scenario, are we basically saying that Fermi-Walker transport of the orientation will result in zero local self-rotation as defined (so if the observer were enclosed in an Einstein elevator and only had his gyros as a method of measuring rotation he wouldn't be able to detect anything)

Yes. In fact, this is *always* how Fermi-Walker transport of the orientation works.

but globally he will see the stars rotate around him if he were able to peer out so there will be a non-zero global self-rotation relative to the stars?

Yes, although I'm not sure the term "self-rotation" is really a good one, since it seems to be confusing you. This "self-rotation" is not locally detectable, which is the key point.

I can't seem to connect this to the second scenario though.

In the second scenario, the elevator's spatial basis vectors are *not* being Fermi-Walker transported; instead they are locked to the "diamond" formed by neighboring members of the same congruence (in this case, the hovering congruence). The failure of Fermi-Walker transport is locally detectable as "self-rotation", even though the elevator is not "rotating" relative to the distant stars.

In other words, the point of the two scenarios together is to show that "no local self-rotation" and "no global rotation" are different and independent things; either one can be present without the other.

 

[WannabeNewton]

Oh ok. I guess my trouble is just in picturing what happens when Fermi-Walker transport fails. What I mean by that is, in the first scenario we can't detect local rotation in the sense that we Fermi-Walker transport the orientation (so my gyros won't detect anything) but if an observer at infinity saw my elevator he would see it twirling around in place. This is the kind of "self-rotation" that I normally picture. In the second scenario, the observer at infinity wouldn't see the elevator twirling around in place, but I can still detect local rotation due to failure to Fermi-Walker transport the orientation? Is there a way to picture what this kind of rotation would look like or is the best we can do say that I will be able to see a discrepancy using my gyros?

 

[PeterDonis]

in the first scenario we can't detect local rotation in the sense that we Fermi-Walker transport the orientation (so my gyros won't detect anything) but if an observer at infinity saw my elevator he would see it twirling around in place.

Yes.

This is the kind of "self-rotation" that I normally picture.

Yes, but bear in mind that you are picturing it based on a global observable, not on local measurements. It's intuitively appealing to do that, but it has limits, as you have seen.

In the second scenario, the observer at infinity wouldn't see the elevator twirling around in place

Right.

but I can still detect local rotation due to failure to Fermi-Walker transport the orientation?

Yes, but bear in mind that the way you detect it is not by looking for "rotation" relative to something outside the elevator, but by measuring different inertial behavior due to the rotation (centrifugal force). So this sense of the word "rotation" is different from what we are used to intuitively picturing--but it has the advantage of being purely local.

Is there a way to picture what this kind of rotation would look like or is the best we can do say that I will be able to see a discrepancy using my gyros?

I'm not sure how you would picture it as a "rotation", precisely, but you can certainly tie it to the different inertial behavior. In a previous post I suggested the analogy of an accelerating rocket in flat spacetime, that is also spinning about an axis perpendicular to the direction of acceleration. What would you expect to observe inside such a rocket, that would be different than inside a rocket that was just accelerating linearly, with no spinning? (In flat spacetime, the local and global notions of "spinning" match, since there's no frame dragging, so this scenario might be easier to picture intuitively.)

 

[WannabeNewton]

I'm not sure how you would picture it as a "rotation", precisely, but you can certainly tie it to the different inertial behavior. In a previous post I suggested the analogy of an accelerating rocket in flat spacetime, that is also spinning about an axis perpendicular to the direction of acceleration. What would you expect to observe inside such a rocket, that would be different than inside a rocket that was just accelerating linearly, with no spinning? (In flat spacetime, the local and global notions of "spinning" match, since there's no frame dragging, so this scenario might be easier to picture intuitively.)

Well if I have a sphere apparatus which has frictionless prongs sticking out with beads inserted through the prongs, and my feet are planted on the rotating elevator then I would be able to see the beads get flung outwards due to the centrifugal force, something I wouldn't be able to see if the elevator was only linearly accelerating. So I can detect this rotation without any reference to the stars. Then if I'm in curved space-time I can tell if I'm Fermi-Walker transporting my orientation (which is a local thing) by imagining myself in an elevator locally and doing the same experiment with the sphere apparatus? So if I do notice the beads get flung outwards then locally I can imagine myself as being in a rotating elevator in flat space-time in the above sense (with no reference to the stars or infinity or what have you)?

So are we saying that the static observer in Kerr space-time who is not locally non-rotating in the sense of non-vanishing twist will experience self-rotation in the sense that he sees stars rotate around him but if he were enclosed in an elevator locally and did the above experiment, he wouldn't notice anything i.e. he could claim his orientation is being Fermi-Walker transported (hence detect no centrifugal forces)? If so, is there a reason why the non-vanishing twist of the static congruence in Kerr space-time doesn't result in any centrifugal forces for the static observers in the above sense?

Thanks.

 

[PeterDonis]

Well if I have a sphere apparatus which has frictionless prongs sticking out with beads inserted through the prongs, and my feet are planted on the rotating elevator then I would be able to see the beads get flung outwards due to the centrifugal force, something I wouldn't be able to see if the elevator was only linearly accelerating. So I can detect this rotation without any reference to the stars.

Yes.

Then if I'm in curved space-time I can tell if I'm Fermi-Walker transporting my orientation (which is a local thing) by imagining myself in an elevator locally and doing the same experiment with the sphere apparatus? So if I do notice the beads get flung outwards then locally I can imagine myself as being in a rotating elevator in flat space-time in the above sense (with no reference to the stars or infinity or what have you)?

Yes.

So are we saying that the static observer in Kerr space-time who is not locally non-rotating in the sense of non-vanishing twist...

Meaning a ZAMO?

...will experience self-rotation in the sense that he sees stars rotate around him but if he were enclosed in an elevator locally and did the above experiment, he wouldn't notice anything i.e. he could claim his orientation is being Fermi-Walker transported (hence detect no centrifugal forces)?

Yes, *if* the elevator's orientation is in fact determined by Fermi-Walker transport . In other words, he can't just "claim" that the orientation is being Fermi-Walker transported; it has to actually be set up that way, physically (for example by using gyroscopes).

If so, is there a reason why the non-vanishing twist of the static congruence in Kerr space-time doesn't result in any centrifugal forces for the static observers in the above sense?

It *does* result in centrifugal forces for the static observers, *if* the orientation of their "elevator" is determined by being fixed relative to neighboring static observers, which is *not* the same as having it determined by Fermi-Walker transport. If a static observer's "elevator" is oriented using Fermi-Walker transport (for example, by using gyros), he won't detect any centrifugal force inside it, but if he looks out the window, he will see neighboring static observers rotating relative to it (as well as the distant stars).

 

[WannabeNewton]

I guess what's weird for me is how using different orientation schemes can result in no centrifugal force vs. a centrifugal force. When you say fix the orientation of a given static observer relative to neighboring static observers, do you mean something along the lines of taking an initial set of 3 spatial basis vectors associated with the given static observer and "anchoring" them to 3 neighboring static observers who are directly in front of the respective spatial basis vectors? If so, does this orientation fixing then allow us to detect a centrifugal force in an absolute sense using just some measuring apparatus because of the fact that the non-vanishing twist (which, most importantly, is a purely local phenomena) will cause the so "anchored" basis vectors to rotate along with the neighboring static observers as these observers rotate about the given static observer? I guess it's weird for me because it doesn't seem to be to be an "absolute" local detection of centrifugal force if we are orienting the spatial basis vectors based on other nearby observers (as opposed to using a device such as a gyroscope).

Also, with this kind of orientation fixing there will be no self-rotation relative to the stars in the sense that the spatial basis vectors will be "anchored" to the nearby static observers who themselves are fixed relative to the stars?

On the other hand if we fix the orientation of those initial spatial basis vectors via Fermi-Walker transport (i.e. via gyro-stabilizing) then we will notice self-rotation relative to the stars in the manner we discussed before but we won't detect any centrifugal forces by definition of Fermi-Walker transport (no rotation of the basis vectors)?

Thanks.

 

[PeterDonis]

When you say fix the orientation of a given static observer relative to neighboring static observers, do you mean something along the lines of taking an initial set of 3 spatial basis vectors associated with the given static observer and "anchoring" them to 3 neighboring static observers who are directly in front of the respective spatial basis vectors?

Yes.

If so, does this orientation fixing then allow us to detect a centrifugal force in an absolute sense using just some measuring apparatus because of the fact that the non-vanishing twist (which, most importantly, is a purely local phenomena) will cause the so "anchored" basis vectors to rotate along with the neighboring static observers as these observers rotate about the given static observer?

Yes.

I guess it's weird for me because it doesn't seem to be to be an "absolute" local detection of centrifugal force if we are orienting the spatial basis vectors based on other nearby observers (as opposed to using a device such as a gyroscope).

Well, the way we are "orienting" the spatial basis vectors is not just an abstraction: there has to be something physical that actually does the orienting. For example, the "diamond" we talked about in this thread could be an actual structure, or at least its diagonals could be: you could have actual physical rods sticking out from the fiducial observer and connected to neighboring observers in the same congruence, and you could use these rods to define the directions of the fiducial observer's spatial basis vectors.

Also, with this kind of orientation fixing there will be no self-rotation relative to the stars in the sense that the spatial basis vectors will be "anchored" to the nearby static observers who themselves are fixed relative to the stars?

Yes.

On the other hand if we fix the orientation of those initial spatial basis vectors via Fermi-Walker transport (i.e. via gyro-stabilizing) then we will notice self-rotation relative to the stars in the manner we discussed before but we won't detect any centrifugal forces by definition of Fermi-Walker transport (no rotation of the basis vectors)?

Yes.

In fact, you could even use both methods of "orientation" in the same scenario. For example, suppose members of the hovering congruence in Kerr spacetime set up connecting rods between them, defining one set of spatial basis vectors; and the observer in the center (the "fiducial" observer) also sets up gyroscopes to define a second set of spatial basis vectors. Then the two sets of basis vectors will rotate relative to each other, and the centrifugal force effects will be observed relative to the "connecting rod" set only.

 

[WannabeNewton]

So if we want to detect any kind of local rotation, we're going to have to use some physical orientation setup that manages to fail Fermi-Walker transport? And what this physical setup is would depend on what kind of local rotation we're looking for e.g. in this case the local rotation due to twist with the physical setup being 3 mutually perpendicular rods connected to 3 neighboring static observers?

 

[PeterDonis]

So if we want to detect any kind of local rotation, we're going to have to use some physical orientation setup that manages to fail Fermi-Walker transport?

To detect local rotation in the sense of locally measurable inertial effects, yes.

And what this physical setup is would depend on what kind of local rotation we're looking for e.g. in this case the local rotation due to twist with the physical setup being 3 mutually perpendicular rods connected to 3 neighboring static observers?

Yes, although I'm not sure I would put it quite this way. It's not that you're looking for local rotation; you're just taking some "natural" method of determining the orientation of spatial vectors (such as connecting to neighboring members of the same congruence), and finding that it leads to that system of spatial vectors having local rotation.

 

[WannabeNewton]

Alrighty. I think I'm out of questions for now lol. Thank you so much for the help Peter, it's much appreciated. That was fun :)

 

[atyy]

Does Fermi-Walker transport conceptually involve a congruence, since Fermi-Walker differentiation seems to be defined for a vector field, but not a single vector?

 

[PeterDonis]

To detect local rotation in the sense of locally measurable inertial effects, yes.

Actually, on thinking about this some more, there might be complications even here. Consider the Langevin congruence, as described here (I posted this link before but I'll post it again):

http://en.wikipedia.org/wiki/Born_coordinates

This congruence has nonzero twist, but note that the twist is *not* equal to the angular velocity of the congruence (this is all in flat spacetime, so there are no frame dragging or other effects to confuse things; as I said before, zero angular velocity and zero angular momentum are the same thing here). The twist is:

$$
\Omega = \frac{\omega}{1 - \omega^2 R^2}
$$

where $\omega$ is the angular velocity and $R$ is the radius from the axis of rotation. This means that if I set up the same kind of scenario we were talking about before, where the fiducial Langevin observer has a set of connecting rods to neighboring Langevin observers, defining one set of spatial vectors, and gyroscopes defining a second set of spatial vectors, *both* sets will be rotating as seen by an observer at infinity. The connecting rod set of vectors will be rotating, relative to infinity, with angular velocity $\omega$; but the gyrostabilized set of vectors will be rotating, relative to infinity, with angular velocity

$$
\omega - \Omega = \omega \left[ 1 - \frac{1}{1 - \omega^2 R^2} \right] = \frac{- \omega^2 R^2}{1 - \omega^2 R^2}
$$

This is a highly counterintuitive result; it says that Fermi-Walker transporting the spatial basis vectors along the worldline of the Langevin observer causes them to rotate in the *opposite* sense to the observer's rotation about the central axis of the congruence! But leave that aside for the moment (I'll come back to it in a follow-up post; I think it's due to the fact that the proper acceleration of this congruence is radially inward, instead of radially outward like the Kerr spacetime congruences we've been considering).

The key point is that the Fermi-Walker transported basis vectors are in fact rotating, relative to an observer at infinity, and this is flat spacetime, so this rotation should cause a nonzero centrifugal force to be measured even with reference to the Fermi-Walker transported basis vectors. So I might have been wrong before when I said Fermi-Walker transport always corresponds to no centrifugal force being locally detected. I need to think about this some more.

Sorry to throw a monkey wrench into the works when we had things settled pretty well.

 

[PeterDonis]

Does Fermi-Walker transport conceptually involve a congruence, since Fermi-Walker differentiation seems to be defined for a vector field, but not a single vector?

I'm not sure. I can come up with a way of defining Fermi-Walker transport that doesn't require a congruence, only a single worldline; but I'm not sure if it covers all the cases in which a Fermi-Walker derivative might come up.

Let me briefly describe the definition I just referred to, since it's also relevant to the monkey wrench I just threw in my last post. Take a very simple example, linear acceleration in flat Minkowski spacetime. Suppose I have a worldline with constant proper acceleration in the $x$ direction, so we can just look at the $t, x$ plane. At some event O on this worldline, the worldline's instantaneous 4-velocity in a chosen global inertial frame is $(u^t, u^x) = (1, 0)$. An infinitesimal interval later, at event P, the worldline has accelerated and now has 4-velocity $(\gamma, \gamma v)$.

The question is, how do we Fermi-Walker transport the spatial basis vector in the $x$ direction along this infinitesimal interval? Well, we could do this: first, parallel transport the "zweibein" of orthonormal basis vectors in the $t, x$ directions from event O to event P. Since this is a global inertial frame in flat Minkowski spacetime, parallel transport just leaves all vector components unchanged. So the parallel transported zweibein vectors will be $(1, 0), (0, 1)$ at P, just as they were at event O.

Now find a Lorentz transformation that takes the parallel transported 4-velocity (the $t$ vector in the zweibein) at P to the actual 4-velocity at P. Obviously this is just a Lorentz boost in the $x$ direction with velocity $v$, which takes the vector $(1, 0)$ to the vector $(\gamma, \gamma v)$.

Then apply the same Lorentz transformation to *all* the vectors that were parallel transported from O to P; so we boost the vector $(0, 1)$ in the $x$ direction with velocity $v$ to obtain the new spatial basis vector $(\gamma v, \gamma)$ at P. This will be the Fermi-Walker transported spatial basis vector at P; i.e., the Fermi-Walker transported zweibein at P will be $(\gamma, \gamma v), (\gamma v, \gamma)$. (In the fully general case, we would apply the Lorentz transformation to all four basis vectors that were parallel transported from O to P; here this is a no-op for the basis vectors in the $y$ and $z$ directions, so including them makes no real difference.)

I believe this definition can be generalized to all cases of Fermi-Walker transport along a single worldline--you just allow *any* arbitrary Lorentz transformation at event P, which covers all possible non-inertial worldlines--and obviously it does not require a congruence; it only requires knowledge of the 4-velocity at each event on the worldline itself. (The case of a geodesic worldline, where Fermi-Walker transport is the same as parallel transport, is just the case where the Lorentz transformation you apply to the transported vectors is the identity.)

 

[WannabeNewton]

If we attach rods from a given Langevin observer to neighboring Langevin observers then why would they rotate relative to infinity with $\omega$, which is the angular speed at which each Langevin observer is rotating around the center of the disk? Why does the rotation $\Omega$ of the rods about the given Langevin observer due to the twist not affect what is seen at infinity? The $\omega$ part comes from the fact that all the Langevin observers rotate around the center of the disk with that speed so an observer at infinity will see the rods being dragged around the disk at the same rate right but why will $\Omega$ not affect what is seen at infinity?

 

[PeterDonis]

If we attach rods from a given Langevin observer to neighboring Langevin observers then why would they rotate relative to infinity with $\omega$, which is the angular speed at which each Langevin observer is rotating around the center of the disk?

Because when the Langevin observer has made one complete rotation around the center of the disk, the rods have also made one complete rotation around the center of the disk. For example, pick a distant star in a particular spatial direction from the center of the disk, say the $x$ direction in the global inertial frame in which the center of the disk is at rest. Suppose that one connecting rod, the one directed radially outward, points directly at that star when the Langevin observer is just seeing it overhead (i.e., when he is just crossing the $x$ axis). Then each time the Langevin observer sees that star directly overhead (i.e., each time he crosses the $x$ axis), the connecting rod that was pointed at that star when he started with the star directly overhead will be pointed directly at that star again--and it won't be pointed at that star anywhere else in the course of his rotation.

Another way of putting this is that the connecting rods are, in effect, rigidly attached to the disk, so they must rotate, relative to infinity, with the same angular velocity as the disk.

Why does the rotation $\Omega$ of the rods about the given Langevin observer due to the twist not affect what is seen at infinity?

Because $\Omega$ is defined relative to Fermi-Walker transport, not relative to infinity. As above, you can derive how an observer at infinity will see the connecting rods rotate without even knowing $\Omega$.

 

[WannabeNewton]

Ok so if we attached a rod radially outward as you said from an observer $O(R_0)$ to another one $O(R_1)$ then the reason why an observer at infinity will see the rod rotate at exactly $\omega$ around the disk is because both $O(R_0)$ and $O(R_1)$ are rotating around the disk at the same angular speed $\omega$ relative to infinity so the observer at infinity will never see the rod rotate due to $O(R_1)$ overtaking $O(R_0)$ or vice-versa and will only see it rotate due to the equal rotation rate $\omega$ of said two observers around the disk relative to infinity? So is this in contrast with what happens in Kerr space-time for the ZAMOs because observers at different radii have different angular velocities around the black hole and so a rod radially attached initially from one observer to another will not only rotate around the black hole relative to infinity due to the overall orbital rotation but also rotate because the inner observer is overtaking the outer observer?

As a side note, when we say "the twist causes nearby observers to rotate about the given reference observer from the perspective of the reference observer" do we really always mean "the twist causes the nearby observers to rotate relative to the Fermi-Walker transported spatial basis of the reference observer"? I ask because we discussed this earlier in the thread and it makes physical sense to me but some of my texts (e.g. Malament's text and Wald's text) just say "rotating about" and don't make any mention of Fermi-Walker transport.

Thanks!

 

[PeterDonis]

the reason why an observer at infinity will see the rod rotate at exactly $\omega$ around the disk is because both $O(R_0)$ and $O(R_1)$ are rotating around the disk at the same angular speed $\omega$ relative to infinity

Yes.

so the observer at infinity will never see the rod rotate due to $O(R_1)$ overtaking $O(R_0)$ or vice-versa and will only see it rotate due to the equal rotation rate $\omega$ of said two observers around the disk relative to infinity?

Yes.

So is this in contrast with what happens in Kerr space-time for the ZAMOs because observers at different radii have different angular velocities around the black hole

Yes, this is a difference, but there are some caveats to how we interpret it. See below.

so a rod radially attached initially from one observer to another will not only rotate around the black hole relative to infinity due to the overall orbital rotation but also rotate because the inner observer is overtaking the outer observer?

Actually that is shear, not twist. To factor out the shear, just consider the leading and trailing observers in the ZAMO congruence, relative to the fiducial observer; i.e., the observers who are at the same $r$ but slightly larger and smaller $\phi$. The connecting rods to these observers will rotate, relative to infinity, at exactly the ZAMO angular velocity for that $r$.

when we say "the twist causes nearby observers to rotate about the given reference observer from the perspective of the reference observer" do we really always mean "the twist causes the nearby observers to rotate relative to the Fermi-Walker transported spatial basis of the reference observer"?

That's my understanding, yes.

 

[WannabeNewton]

I guess it was easier for me to picture the twist of the static observers in the Kerr space-time because I could just think about how a given static observer would be "self-rotating" relative to the stars and will also see nearby static observers fixed to the stars being spun around him and in this sense result in a non-vanishing twist. Here, the Langevin observers aren't "self-rotating" relative to infinity in any way, they are all just spinning around the cylindrical walls so if I were to put myself in the rest frame of a given Langevin observer I can't really imagine why I would see nearby Langevin observers rotate around me (i.e. rotate relative to my Fermi transported spatial basis vectors). Even if the observers at different radii were rotating at different angular velocities, this would only result in a shear (much like for the ZAMO congruence) and not a twist but regardless here we have no shear at all and just a twist but I can't really picture this twist. If I imagine myself in the frame of a given Langevin observer, I just picture all the other nearby Langevin observers as being at rest with me.

 

[PeterDonis]

I were to put myself in the rest frame of a given Langevin observer I can't really imagine why I would see nearby Langevin observers rotate around me (i.e. rotate relative to my Fermi transported spatial basis vectors).

It's hard for me to imagine too, which is why I am looking for alternate ways to think about Fermi-Walker transport, such as what I described a few posts back (parallel transport the basis vectors, then Lorentz transform them so the 4-velocity matches what it actually is at the new event).

 

[WannabeNewton]

Peter do you happen to have the text "Gravitation and Inertia" by Wheeler et al?

 

[PeterDonis]

Peter do you happen to have the text "Gravitation and Inertia" by Wheeler et al?

Yes. Why?

 

[atyy]

It's hard for me to imagine too, which is why I am looking for alternate ways to think about Fermi-Walker transport, such as what I described a few posts back (parallel transport the basis vectors, then Lorentz transform them so the 4-velocity matches what it actually is at the new event).

Thanks for that explanation a few posts back. It does seem that Fermi-Walker transport of a vector along a curve is possible without defining a congruence.

 

[WannabeNewton]

Ok because I was reading pages 237-238 where they talks about rotation of a congruence with respect to Fermi-Walker transport much like you did before and I just had two questions that cropped up:

On page 238 they say "This expression explicitly shows that...give the rate of change with time of the separation...between neighboring particles, that is, the motions and the rate of change of the dimensions of an infinitesimal volume element of the fluid relative to a local comoving Fermi frame". Are we supposed to take the local comoving Fermi frame to be a local fluid element that Fermi transports its spatial axes?

On the same page, slightly below, they say "In particular, for an infinitesimal spherical surface described by the fluid particles with $\delta x^{i}\delta x_{i} = \delta r^{2}$ in the Fermi frame...and $\omega_{ik}$ its angular velocity relative to the Fermi axes, that is, relative to local gyroscopes". Is the Fermi frame supposed to be attached to a local fluid element that is at the center of the spherical surface defined above by the other fluid elements? In particular, in the image at the bottom of the page, the picture depicting the rotation is showing two dots (fluid elements) on the surface of the sphere that are rotating relative to the spatial axes of a Fermi frame attached to a fluid element at the center of that sphere (with the spatial axes representing gyros) correct? I ask because I didn't know what the dot at the top of that rotating sphere was representing.

 

[TrickyDicky]

Hey guys. I have a question about two (possibly ostensibly) different definitions of a locally non-rotating observer that I have come across in my texts.

The first is specifically for stationary, axisymmetric space-times in which we have a canonical global time function $t$ associated with the time-like killing vector field. We define a locally non-rotating observer to be one who follows an orbit of $\nabla^{a}t$ i.e. his 4-velocity is given by $\xi^{a} = (-\nabla^{b}t\nabla_{b}t)^{-1/2}\nabla^{a}t = \alpha \nabla^{a}t$. Such an observer can be deemed as locally non-rotating because his angular momentum $L = \xi^{a}\psi_{a} = \alpha g_{\mu\nu}g^{\mu\gamma}\nabla_{\gamma}t\delta^{\nu}_{\varphi} = \alpha \delta^{\gamma}_{\nu}\delta^{t}_{\gamma}\delta^{\nu}_{\varphi} = \alpha \delta^{t}_{\varphi} = 0$ where $\psi^{a}$ is the axial killing vector field; these observers are also called ZAMOs for this reason. It might also be worth noting that the time-like congruence defined by the family of ZAMOs has vanishing twist $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = \alpha\epsilon^{ab[cd]}\nabla_{b}t\nabla_{(c}\nabla_{d)}t + \alpha^{3}\epsilon^{a[b|c|d]}\nabla_{(b}t \nabla_{d)}t\nabla^{e}t\nabla_{c}\nabla_{e}t = 0$.


The second definition I have seen is the much more general notion of Fermi-Walker transport. That is, if we choose an initial Lorentz frame $\{\xi^{a}, u^{a},v^{a},w^{a}\}$ and the spatial basis vectors evolve according to $\xi^{b}\nabla_{b}u^{a} = \xi^{a}u_{b}a^{b}$, $\xi^{b}\nabla_{b}v^{a} = \xi^{a}v_{b}a^{b}$, and $\xi^{b}\nabla_{b}w^{a} = \xi^{a}w_{b}a^{b}$, where $a^{b} = \xi^{a}\nabla_{a}\xi^{b}$ is the 4-acceleration, then the observer is said to be locally non-rotating.

My question is, to what extent are these two definitions equivalent (both mathematically and physically) whenever they can both be applied? In other words, in what sense is the qualifier 'rotation' being used in each case?

Let me elucidate my question a little bit. I know that for asymptotically flat axisymmetric space-times, there must exist a fixed rotation axis on which $\psi^{a}$ vanishes. Then the first definition tells us that the ZAMOs have no orbital rotation about this fixed rotation axis (for example no orbital rotation about a Kerr black hole); because these observers are at rest with respect to the $t = \text{const.}$ hypersurfaces, these observers are as close to stationary hovering observers as we can get in a space-time with a rotating source. We know however that such observers have an instrinsic angular velocity $\omega$; if we imagine a ZAMO holding a small sphere with frictionless prongs sticking out with beads through the prongs then a ZAMO should be able to notice his intrinsic angular velocity $\omega$ by seeing that the beads are thrown outwards along the prongs at any given instant. Is this correct?

Now, on the other hand, the second definition of local non-rotation (using Fermi-Walker transport) seems to be saying sort of the opposite. That is, it seems to be telling us which observers can carry such spheres and never see the beads get thrown outwards i.e. the orientation of the sphere will remain constant (the orientation will be represented by the spatial basis vectors of a Lorentz frame) so these are the observers who have no intrinsic angular velocity i.e. no self-rotation. This is why the two definitions confused me because they seem to be talking about two totally different kinds of non-rotation.

Thank you in advance.

WN, first sorry about my remark regarding the definition of observers I posted above, I was in a hurry and didn't pay much attention, second thanks for this discussion you and Peter are having, very instructive.
Now, if I may ask something related to the OP: why would you want to reconcile definitions about local (non)rotation that are built on different solutions of the EFE, therefore different geometries? The first definition refers to the Kerr geometry with axisymmetry, but no spherical symmetry, and the second that is valid for the more general spherical symmetry. In the first case total angular momentum is not even conserved since the spacetime is not spherically symmetric about any of its points, while it is in the second.

Actually it is a bit of a puzzle for me how exactly the expected physical consequences of the different and sometimes mutually incompatible GR solutions are chosen and considered to be mainstream physics or not in the absence of empirical confirmation.
For instance the Kerr stationary axisymmetric solution that gives rise to your first definition and that it seems logical that it should apply to rotating objects such as planets and stars, produces the prediciton of the frame-dragging effect, but this geometry seems to be not compatible with the more general spherical symmetry (that on the other hand is observationally quite confirmed and theoretically expected globally) of the Schwarzschild solution that predicts the geodetic effect, in as much as the effect demands stationarity of the spacetime(but non-staticity).

 

[WannabeNewton]

The first definition applies to any axially symmetric, stationary space-time so not just the Kerr space-time. You don't need spherical symmetry for angular momentum to be conserved, you just need axial symmetry. If axial symmetry is available then there exists an axial killing field $\psi^{a}$ and the quantity $u_{a}\psi^{a}$ is conserved when $u^{a}$ is a geodesic 4-velocity as usual i.e. $u^{a}\nabla_{a}(\psi_{b}u^{b}) = \psi_{b}u^{a}\nabla_{a}u^{b} + u^{(a}u^{b)}\nabla_{[a}\psi_{b]} = 0$. We define the angular momentum as $L = u^{a}\psi_{a}$.

The Fermi-Walker condition can be applied in any space-time whatsoever; there are no restrictions there. I wasn't exactly looking for the two definitions to be reconciled per say but rather wanted to see what their differences were i.e. what kind of non-rotation does Fermi-Walker refer to and what kind of non-rotation does following an orbit of $\nabla^{a} t$ refer to. Peter helped reconcile these differences for me in the thread (thanks again Peter ).

The latter refers to a congruence of observers, all following orbits of $\nabla^{a} t$, with vanishing twist which is equivalent to hypersurface orthogonality i.e. $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = 0\Leftrightarrow \xi_{[a}\nabla_{b}\xi_{c]} = 0$ for any tangent field $\xi^{a}$. A given observer in the congruence will not see locally separated observers rotate around him as his proper time passes so it is local non-rotation in that sense. Furthermore since $\nabla^{a}t$ happens to be everywhere orthogonal to $\psi^{a}$, the angular momentum of these observers also vanishes so that's an added bonus.

The former condition, that of Fermi-Walker transport, is more akin (locally) to non-rotating frames in SR in the sense that locally we can imagine the Fermi frame as an observer with 3 mutually perpendicular gyroscopes enclosed in an elevator in SR with the gyroscopes yielding null results for rotation of the elevator in the SR sense.

 

[TrickyDicky]

The first definition applies to any axially symmetric, stationary space-time so not just the Kerr space-time.

Sure, i was using it as an example in reference to the second part of my post.

You don't need spherical symmetry for angular momentum to be conserved, you just need axial symmetry.

The component of angular momentum $L = u^{a}\psi_{a}$ of a particle along the rotation axis is conserved with axial symmetry but not the total angular momentum of the particle.

The Fermi-Walker condition can be applied in any space-time whatsoever; there are no restrictions there. I wasn't exactly looking for the two definitions to be reconciled per say but rather wanted to see what their differences were i.e. what kind of non-rotation does Fermi-Walker refer to and what kind of non-rotation does following an orbit of $\nabla^{a} t$ refer to.

Ok, this answers my first question.

 

[WannabeNewton]

The component of angular momentum $L = u^{a}\psi_{a}$ of a particle along the rotation axis is conserved with axial symmetry but not the total angular momentum of the particle.

What does "total angular momentum" even mean when only $L$ is available? It's all there is.

 

[PeterDonis]

Ok because I was reading pages 237-238 where they talks about rotation of a congruence with respect to Fermi-Walker transport much like you did before and I just had two questions that cropped up

Oh, you meant you actually wanted me to dig out my copy and look at it? I don't have it handy right now but I think I remember it well enough to give some useful comments.

On page 238 they say "This expression explicitly shows that...give the rate of change with time of the separation...between neighboring particles, that is, the motions and the rate of change of the dimensions of an infinitesimal volume element of the fluid relative to a local comoving Fermi frame". Are we supposed to take the local comoving Fermi frame to be a local fluid element that Fermi transports its spatial axes?

I believe that's correct, yes.

On the same page, slightly below, they say "In particular, for an infinitesimal spherical surface described by the fluid particles with $\delta x^{i}\delta x_{i} = \delta r^{2}$ in the Fermi frame...and $\omega_{ik}$ its angular velocity relative to the Fermi axes, that is, relative to local gyroscopes". Is the Fermi frame supposed to be attached to a local fluid element that is at the center of the spherical surface defined above by the other fluid elements?

IIRC, yes, that's right.

 

[PeterDonis]

the Kerr stationary axisymmetric solution that gives rise to your first definition and that it seems logical that it should apply to rotating objects such as planets and stars, produces the prediciton of the frame-dragging effect

It seems logical, yes, but AFAIK nobody has actually proven that the vacuum region exterior to a rotating object (not a black hole) such as a planet or star actually is the Kerr geometry. The corresponding proof in the spherically symmetric case requires Birkhoff's theorem, and there is no analogue of Birkhoff's theorem for the non-spherical but axisymmetric case.

(Also, AFAIK in practice the full Kerr geometry is not used in a lot of cases; a sort of perturbation approach is used where the leading order Kerr terms are added on to an underlying Schwarzschild or PPN-type solution. See below.)

but this geometry seems to be not compatible with the more general spherical symmetry (that on the other hand is observationally quite confirmed and theoretically expected globally)

No, it isn't observationally confirmed; it's an approximation that works well in many cases but is well understood to be an approximation. Even disregarding all the other asymmetries of a rotating planet like the Earth, it's an oblate spheroid, not a sphere.

of the Schwarzschild solution that predicts the geodetic effect

The Schwarzschild solution does not predict the geodetic effect (by which I assume you mean frame dragging). You need to add something to it to represent the rotation of the central body. As I understand it, the way this is done in practice (for things like analyzing Gravity Probe B data) is to start with the Schwarzschild metric and add a perturbation to it to represent the expected frame dragging effects. The form of the perturbation is obtained by expanding the Kerr metric around $a = 0$ (i.e., by assuming that $a << M$).

 

[WannabeNewton]

Oh, you meant you actually wanted me to dig out my copy and look at it?

At least it's not as heavy as MTW xD :)

Thanks for the confirmations! Much appreciated. I just wanted to make sure I was interpreting the text correctly because "Gravitation and Inertia" is the only text I own wherein the term "rotate about" is actually defined rigorously in terms of Fermi frames, in the way you described earlier in the thread, so I wanted to make sure I wasn't fudging anything up in my head. All my other texts just say "rotate about" and leave it at that unfortunately.

Speaking of Wheeler, they go into detail on the theoretical and experimental aspects of the geodetic and Lense-Thirring effects in section 3.4.3.

 

[TrickyDicky]

The Schwarzschild solution does not predict the geodetic effect (by which I assume you mean frame dragging).

No, the geodetic effect is not frame dragging, look up de Sitter precession.

 

[TrickyDicky]

What does "total angular momentum" even mean when only $L$ is available? It's all there is.

Exactly, that is all there is in an axisymmetric space, to be compared with the angular momentum available that is conserved in classical static spherically symmetric scenario( of SR and Schwarzschild metrics for instance or of classical mechanics for that matter).