Locally non-rotating observers

[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.

 

[PeterDonis]

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.

And I think that's a matter of terminology, not physics. First of all, you have to be willing to use the term "gravitational field" for a phenomenon in flat spacetime; people's preferences differ about that.

Second, you have to be willing to abstract away significant differences between this "spacetime" and Kerr spacetime: to name just two, observers at rest at infinity are at rest in the chart on Kerr spacetime that has just a single cross term, the $dt d\phi$ term, whereas they are rotating in this chart; and the net proper acceleration of observers rotating about the hole in Kerr spacetime will be radially outward for any angular velocity less than "orbital velocity", whereas it is always radially inward for observers in this "spacetime".

Third, even if you abstract all that stuff away, the twist for corresponding observers will not always be the same--at least, not for any value of "corresponding" that I can see coming out of the comparison you are trying to make. The congruence of observers who are at rest in the chart you wrote down has a positive twist; the congruence of observers at rest in Kerr spacetime has a negative twist. So I don't really see what's supposed to be "the same" about the two scenarios when viewed this way.

But finally, as I said, this is all terminology to me. The physics of Fermi-Walker transport is clear: you take the transformation that carries the 4-velocity at one event to the 4-velocity at another event on the worldline, and apply that same transformation, and nothing else, to the spatial basis vectors. That recipe works for any congruence in any spacetime and expresses what all the different scenarios have in common. Basically it all boils down to the fact that the proper acceleration of a given worldline in a given spacetime is determined by the inertial structure of the spacetime--which states of motion are inertial and which are not. Change the spacetime and you change the inertial structure.

 

[WannabeNewton]

When I meant "akin to" I meant if we can explain it in a geometrical way using the above metric as opposed to a kinematical way via Thomas precession (sort of like talking about physical phenomena for Rindler observers using kinematics vs using a uniform static gravitational field). Sorry if I wasn't' clear enough, I didn't mean to come off as argumentative or anything.

But anyways, getting back to things in the global inertial frame, let's say we attach connecting vectors from a given Langevin observer $O$ to three nearby Langevin observers such that the connecting vectors initially line up with the respective spatial basis vectors of $O$ (which we take to be gyro-stabilized). We know that $O$ as well as the connected neighbors are all orbiting some point with the same angular velocity $\omega$ hence relative to an observer $O'$ at rest in the global inertial frame, the connecting vectors are being carried around rigidly in exactly the same manner so relative to $O'$, the connecting vectors also rotate around said point with an angular velocity $\omega$ in the same sense as the Langevin observers. Now we know that the twist of the Langevin congruence is given by $\omega^{a} = (1 - \omega^{2}r^{2})^{-1}\omega (\partial_{z})^{a}$ so the connecting vectors will start to rotate relative to the spatial basis vectors of $O$ with angular velocity $(1 - \omega^{2}r^{2})^{-1}\omega$ in the same sense as the orbital motion of the Langevin observers about said point and parallel to the axis of rotation of the Langevin observers (i.e. the z-axis). Hence relative to $O'$, the spatial basis vectors of $O$ will precess with an angular velocity $\Omega = \omega-(1 - \omega^{2}r^{2})^{-1}\omega$?

And are we saying that $\Omega$ will be seen by $O'$ as a self-rotation of $O$? And that one can view the mechanism responsible for the existence of such an $\Omega$ as coming from the fact that consecutive Lorentz boosts are non-commutative and result in rotations of consecutive comoving Lorentz frames following the circular trajectory of a given Langevin observer leaving us with a continuous self-rotation of a given Langevin observer relative to $O'$ due to the continuous application of consecutive Lorentz boosts (i.e. the Thomas precession)?

 

[PeterDonis]

When I meant "akin to" I meant if we can explain it in a geometrical way using the above metric as opposed to a kinematical way via Thomas precession (sort of like talking about physical phenomena for Rindler observers using kinematics vs using a uniform static gravitational field).

But changing frames doesn't change the metric; it just changes how the metric is expressed. I don't think you can convert a "kinematic" explanation to a "geometrical" explanation that way, since the geometry has not changed; you've just changed the way you look at it. (I would make similar remarks about looking at the Rindler congruence as realizing a "uniform static gravitational field"; surely you've seen enough threads here at PF by now where that causes more confusion than it solves. ) I still think this is a matter of terminology, and different people do have different preferences for that.

Sorry if I wasn't' clear enough, I didn't mean to come off as argumentative or anything.

No worries.

Hence relative to $O'$, the spatial basis vectors of $O$ will precess with an angular velocity $\Omega = \omega-(1 - \omega^{2}r^{2})^{-1}\omega$?

Not quite; you have to take time dilation into account. Let $\gamma^2 = 1 / (1 - \omega^2 r^2)$ for ease of writing. The angular velocity $\omega$ is relative to the static observer, i.e., to $O'$; but the twist $\omega^a = \gamma^2 \omega$ of the congruence is relative to $O$. Relative to $O'$, the congruence vectors rotate, compared to the gyro vectors, with angular velocity $\omega^a / \gamma = \gamma \omega$. So the gyro vectors, relative to $O'$, will rotate with angular velocity $\Omega = \omega - \gamma \omega = - ( \gamma - 1 ) \omega$.

And are we saying that $\Omega$ will be seen by $O'$ as a self-rotation of $O$?

You could view it that way, yes. $O'$ sees $O$ carrying around two sets of vectors: the congruence vectors, which $O'$ sees rotating (prograde) at $\omega$, just as $O$ himself does; and the gyro vectors, which $O'$ sees rotating (retrograde) at $\Omega$, i.e., they are spinning backwards as $O$ goes around forwards.

And that one can view the mechanism responsible for the existence of such an $\Omega$ as coming from the fact that consecutive Lorentz boosts are non-commutative and result in rotations of consecutive comoving Lorentz frames following the circular trajectory of a given Langevin observer leaving us with a continuous self-rotation of a given Langevin observer relative to $O'$ due to the continuous application of consecutive Lorentz boosts (i.e. the Thomas precession)?

I would say so, yes.

 

[WannabeNewton]

Not quite; you have to take time dilation into account.

Is this from transforming $\omega^{a} = \gamma^{2} \omega (\partial_{z})^{a}$ under a boost along the tangential direction from a frame instantaneously comoving with $O$ to the frame of $O'$?

I would say so, yes.

So it would be fair to say that a given Langevin observer himself sees his neighbors rotate around him i.e. have non-zero twist precisely because the Thomas precession induces a precession relative to the stars of the given Langevin observer's gyro-stabilized basis vectors?

Thanks Peter!

 

[PeterDonis]

Is this from transforming $\omega^{a} = \gamma^{2} \omega (\partial_{z})^{a}$ under a boost along the tangential direction from a frame instantaneously comoving with $O$ to the frame of $O'$?

Kinda sorta. Think of $O$ and $O'$ both measuring the elapsed time, by their clocks, between successive times that they pass each other. The time $O'$ measures will be a factor $\gamma$ larger than the time that $O$ measures. So the angular velocity that $O'$ measures will be $1 / \gamma$ times the angular velocity that $O$ measures. The same applies to any angular velocity measurement.

So it would be fair to say that a given Langevin observer himself sees his neighbors rotate around him i.e. have non-zero twist precisely because the Thomas precession induces a precession relative to the stars of the given Langevin observer's gyro-stabilized basis vectors?

Well, but the angular velocity that the Langevin observer sees his neighbors rotate around him, relative to his gyro vectors, is *different* from the angular velocity that he sees the distant stars rotate around him, relative to his gyro vectors. And both of these are different from the angular velocity that he sees his neighbors rotate around him, relative to the stars. The first angular velocity is the twist, $\gamma^2 \omega$; the second is $- \gamma \Omega = - ( \gamma^2 - \gamma ) \omega$; the third is $\gamma \omega$.

The second angular velocity is what I would expect the Langevin observer to attribute to Thomas precession: it causes his gyro vectors to precess relative to the distant stars. I don't know that I would expect him to attribute the first angular velocity to Thomas precession alone, although I could see him thinking of it as the difference between the second and third angular velocities: his gyro vectors spin backwards relative to the stars because of Thomas precession, and his congruence vectors spin forwards relative to the stars because he's on the rotating disk, and the twist is the difference between the two.

 

[WannabeNewton]

Kinda sorta. Think of $O$ and $O'$ both measuring the elapsed time, by their clocks, between successive times that they pass each other. The time $O'$ measures will be a factor $\gamma$ larger than the time that $O$ measures. So the angular velocity that $O'$ measures will be $1 / \gamma$ times the angular velocity that $O$ measures. The same applies to any angular velocity measurement.

Oh are we talking about the time dilation factor affecting the $\gamma^{2} \omega$ expression itself (not the twist vector)?

I don't know that I would expect him to attribute the first angular velocity to Thomas precession alone, although I could see him thinking of it as the difference between the second and third angular velocities: his gyro vectors spin backwards relative to the stars because of Thomas precession, and his congruence vectors spin forwards relative to the stars because he's on the rotating disk, and the twist is the difference between the two.

I guess I should have been clearer again. I meant like, the sole physical phenomena that eventually leads to the non-zero twist of the Langevin congruence is the Thomas precession?

 

[PeterDonis]

Oh are we talking about the time dilation factor affecting the $\gamma^{2} \omega$ expression itself (not the twist vector)?

It amounts to the same thing, doesn't it?

I guess I should have been clearer again. I meant like, the sole physical phenomena that eventually leads to the non-zero twist of the Langevin congruence is the Thomas precession?

No, I don't think so. Look at it this way: what's the point of having a set of gyro-stabilized vectors? To point at the same place relative to the distant stars. If the Thomas precession didn't exist, then the gyro vectors carried by the Langevin observer would do exactly that, which means that neighboring Langevin observers would rotate about him, with respect to his gyro vectors, at the same angular velocity as the distant stars. In other words, there would still be a non-zero twist. The Thomas precession just adds an extra rotation to the twist, because it makes the gyro vectors *not* keep pointing at exactly the same place relative to the distant stars.

I suppose one could take the view that the "twist" that would still be present if the Thomas precession didn't exist does not "count" as a twist, because it's purely due to the observer's rotation with the disk; so only the extra twist due to the Thomas precession is a "real" twist. But that's not the way the twist of a congruence is defined; it's defined as rotation relative to the gyro vectors, not rotation relative to the gyro vectors minus "rotation due to rotation", so to speak.

As I understand it, the standard definition of the twist is part of the standard kinematic decomposition of a congruence, which is what naturally arises when you project into the submanifold orthogonal to the congruence. I'm not aware of any alternative kinematic decomposition that tries to separate out a part due to "rotation relative to infinity" and an additional part due to effects like the Thomas precession.

 

[WannabeNewton]

It amounts to the same thing, doesn't it?

I would hope so

No, I don't think so. Look at it this way: what's the point of having a set of gyro-stabilized vectors? To point at the same place relative to the distant stars. If the Thomas precession didn't exist, then the gyro vectors carried by the Langevin observer would do exactly that, which means that neighboring Langevin observers would rotate about him, with respect to his gyro vectors, at the same angular velocity as the distant stars.

But I thought the Fermi-Walker transport was not relative to the fixed stars i.e. that it was to locally resemble a non-rotating frame in SR, in an absolute sense. The reason I'm having trouble is, if I put myself in the perspective of a given Langevin observer and there is no Thomas precession then I can't imagine why nearby Langevin observers would revolve around me. What would cause that if they were all moving around the disk in the exact same manner as me?

As I understand it, the standard definition of the twist is part of the standard kinematic decomposition of a congruence, which is what naturally arises when you project into the submanifold orthogonal to the congruence.

Yeah this is how I learned it (from Wald's text and Malament's text) but I don't really like that definition because I can never tell why the twist is happening physically.

 

[PeterDonis]

But I thought the Fermi-Walker transport was not relative to the fixed stars i.e. that it was to locally resemble a non-rotating frame in SR, in an absolute sense.

A non-rotating frame in SR *is* fixed relative to the fixed stars--at least, its spatial directions are. Consider linear acceleration in the $x$ direction in flat spacetime. Yes, technically the $\partial_x$ basis vector "rotates" in the spacetime sense, since it needs to stay orthogonal to the $\partial_t$ vector; but there is no additional rotation of any of the spatial vectors. (MTW discusses this in some detail in their section on Fermi-Walker transport.) So a gyroscope carried by the accelerating observer that starts out pointing in the $x$ direction, will always point in the $x$ direction: if there is a distant star in that direction, the gyro will stay pointed at it.

if I put myself in the perspective of a given Langevin observer and there is no Thomas precession then I can't imagine why nearby Langevin observers would revolve around me. What would cause that if they were all moving around the disk in the exact same manner as me?

Because you're implicitly adopting a different definition of what it means for the adjacent observers to "revolve around you" than the one that is used when defining the twist of a congruence. You're implicitly seeing everything relative to the observers in the congruence, who are fixed to the disk. You're imagining, say, going around a merry-go-round, with others near you also going around with it, and all of you at rest relative to each other. If you fix your gaze on one particular neighbor (say the one just a bit further outward, radially, than you), it will seem like he is indeed fixed in place relative to you, not rotating around you.

But suppose that you instead fix your gaze on an object that's stationary on the ground--say there's a ticket booth some distance off. (And suppose the merry-go-round is transparent, so you can keep looking at the booth during a complete revolution of the merry-go-round. Also assume that you're able to continuously pivot your feet to keep yourself facing directly at the booth.) Now it will seem like your neighbors are rotating around you, in the same sense as the merry-go-round is rotating. That is the twist of the congruence (without the Thomas precession--it would have to be a *very* fast merry-go-round for that to be noticeable anyway ).

 

[WannabeNewton]

A non-rotating frame in SR *is* fixed relative to the fixed stars--at least, its spatial directions are. Consider linear acceleration in the $x$ direction in flat spacetime. Yes, technically the $\partial_x$ basis vector "rotates" in the spacetime sense, since it needs to stay orthogonal to the $\partial_t$ vector; but there is no additional rotation of any of the spatial vectors. (MTW discusses this in some detail in their section on Fermi-Walker transport.) So a gyroscope carried by the accelerating observer that starts out pointing in the $x$ direction, will always point in the $x$ direction: if there is a distant star in that direction, the gyro will stay pointed at it.

Something like this then: http://postimg.org/image/6tpj77slj/ ? And thanks, I'll check out the MTW section that you referenced. So is it ok even in GR to locally always picture Fermi-Walker transport like this (not necessarily around a circle but as the spatial basis vectors always pointing in the same direction out into the distance throughout the trajectory)?

But suppose that you instead fix your gaze on an object that's stationary on the ground--say there's a ticket booth some distance off. (And suppose the merry-go-round is transparent, so you can keep looking at the booth during a complete revolution of the merry-go-round. Also assume that you're able to continuously pivot your feet to keep yourself facing directly at the booth.) Now it will seem like your neighbors are rotating around you, in the same sense as the merry-go-round is rotating. That is the twist of the congruence (without the Thomas precession--it would have to be a *very* fast merry-go-round for that to be noticeable anyway ).

Wow this makes things like $10^{100}$ times clearer! So my mistake was in picturing that my gaze will always be turning around so that I always face the center of the disk? This is what I was picturing before: http://postimg.org/image/al4ma06e9/ But if I picture it like that then my neighbors will certainly look fixed relative to me because the axes of the frame turn around exactly in accord with the turning of the connecting vector about the center of the disk so that there is no relative rotation right? But this isn't Fermi-Walker transport right? On the other hand if I continually face a constant direction off into the distance (by continually readjusting myself through Fermi-Walker transport) then I should see them rotate around me about the z-axis like this: http://postimg.org/image/83nt5n3bb/ where the right image is what it looks like from the origin of $O$'s Fermi-Walker frame (the red dot is the origin of the Fermi-Walker frame) yeah?

Unfortunately neither of the books I was using (Wald nor Malament) ever explained what it meant for the neighbors to rotate around me due to twist i.e.. they never said that there must be Fermi-Walker transport of my spatial basis vectors and that it should look like the above (if what I drew above is correct) and that the connecting vectors would rotate around the Fermi-Walker axes as above (again if I drew it correctly) so I guess that's why I was so confused.

Thanks a ton Peter!

 

[Mentz114]

Peter, a follow-up question. In the local frame of the Schwarzschild circular geodesic the vorticity is $\sqrt{m/r^3}$ around the $\theta$ axis, the Newtonian value. Does this mean the the observer is rotating ( on its own axis) exactly at the revolution frequency ?

 

[PeterDonis]

Something like this then: http://postimg.org/image/6tpj77slj/ ?

If there is no Thomas precession (or other effects in curved spacetime, see below), yes.

So is it ok even in GR to locally always picture Fermi-Walker transport like this (not necessarily around a circle but as the spatial basis vectors always pointing in the same direction out into the distance throughout the trajectory)?

If you ignore Thomas precession, and de Sitter precession if there is a mass (or black hole) present, and Lense-Thirring precession if the mass (or black hole) is rotating, yes. But those are all relativistic effects which are undetectable if the rotation is slow enough, and they can all be visualized (with one caveat, see below) as corrections to the "base" behavior, which is as you describe it in the above quote.

The caveat is that this way of picturing the "base" behavior, as far as I can see, depends on the spacetime being asymptotically flat, so there is a meaningful notion of "infinity" and "spatial directions at infinity". I'm not sure you could picture Fermi-Walker transport this way in, for example, a closed FRW spacetime, since there is no spatial infinity. (In a flat FRW spacetime, which is what we currently believe our universe is, you could, since there is a meaningful notion of spatial infinity in each flat spatial slice. I'm not sure about an open, i.e., negative spatial curvature, FRW spacetime.)

So my mistake was in picturing that my gaze will always be turning around so that I always face the center of the disk?

Yes.

But if I picture it like that then my neighbors will certainly look fixed relative to me because the axes of the frame turn around exactly in accord with the turning of the connecting vector about the center of the disk so that there is no relative rotation right?

Right.

But this isn't Fermi-Walker transport right?

Right.

On the other hand if I continually face a constant direction off into the distance (by continually readjusting myself through Fermi-Walker transport) then I should see them rotate around me about the z-axis like this: http://postimg.org/image/83nt5n3bb/ where the right image is what it looks like from the origin of $O$'s Fermi-Walker frame (the red dot is the origin of the Fermi-Walker frame) yeah?

Yes, exactly.

Thanks a ton Peter!

You're welcome!

 

[PeterDonis]

In the local frame of the Schwarzschild circular geodesic the vorticity is $\sqrt{m/r^3}$ around the $\theta$ axis, the Newtonian value.

Is it? I get

$$
\vec{\Omega} = \frac{\omega}{1 - \omega^2 r^2} \frac{1 - 3M / r}{\sqrt{1 - 2M / r}} \hat{e}_z
$$

which I think agrees with what is given in Bill_K's blog post (there's an extra factor of $\sqrt{1 - 2M / r}$ somewhere, but I think that's because I'm normalizing $\Omega$ instead of writing it in terms of components as he does). The key is that the vorticity includes Thomas precession and de Sitter precession, which are both corrections to the Newtonian value.

 

[Mentz114]

Is it? I get

$$
\vec{\Omega} = \frac{\omega}{1 - \omega^2 r^2} \frac{1 - 3M / r}{\sqrt{1 - 2M / r}} \hat{e}_z
$$

which I think agrees with what is given in Bill_K's blog post (there's an extra factor of $\sqrt{1 - 2M / r}$ somewhere, but I think that's because I'm normalizing $\Omega$ instead of writing it in terms of components as he does). The key is that the vorticity includes Thomas precession and de Sitter precession, which are both corrections to the Newtonian value.

What is $\hat{e}_z$ ? I'm not used to seeing a z-coordinate in the Schwarzschild spacetime.

The quantity I calculated is $\omega^a = \frac{1}{2}\epsilon^{abmi}u_b \omega_{mi},\ \omega_{mi}=\epsilon_{miab}\nabla^a u^b$. I should have hatted all the indexes because it's a frame field calculation.

In the coordinate basis I get

$\omega^a = -\frac{\sqrt{m}\,\left( r-6\,m\right) \,\left( r-2\,m\right) }{4\,\sqrt{r}\,{\left( r-3\,m\right) }^{2}}\ \partial_\theta$

I'll have a look at Bill K's blog.

I still don't understand if this number is related to the orbital period or the rotation of the axes of the transported frame, or both.

From BillK's blog

(Note that for a free particle following a geodesic, the acceleration is zero, a' = 0, and the orbital velocity is given by ω² = M/r³. We recognize this as Kepler's Law, "period squared goes as distance cubed." It's remarkable that in terms of the coordinate angular velocity, the circular orbits in the Schwarzschild field obey Kepler's Law exactly!)

This agrees with the $\omega$ I found in the local frame basis.

(Bill, your calculations are succinct and elegant. I don't know if you've seen the exact solution for the elliptical orbits, esp. Mercury, as given in arXiv:astro-ph/0305181v3 , in terms of the Weierstrass elliptical function $\wp$).

 

[WannabeNewton]

If you ignore Thomas precession, and de Sitter precession if there is a mass (or black hole) present, and Lense-Thirring precession if the mass (or black hole) is rotating, yes. But those are all relativistic effects which are undetectable if the rotation is slow enough, and they can all be visualized (with one caveat, see below) as corrections to the "base" behavior, which is as you describe it in the above quote.

Noted. Also, I was wondering again why it is things like Thomas Precession and Lense-Thirring precession are not eliminated by Fermi-Walker transport. Is it because Fermi-Walker transport is done by the observer i.e. it is under his operational control (in the merry go round example he constantly pivots his feet so that his spatial axes are Fermi-Walker transport as he orbits the center of the merry go round) whereas things like Thomas Precession and Lense-Thirring precession are out of his control as they are intrinsic properties of Lorentz transformations and stationary space-times respectively so he has no operational way of eliminating them by himself as an observer?

You're welcome!

Thanks again. I can't believe I was picturing it wrong this entire time. That really took a huge worry off of my chest :)

 

[PeterDonis]

What is $\hat{e}_z$ ? I'm not used to seeing a z-coordinate in the Schwarzschild spacetime.

Technically, it's a unit vector in cylindrical coordinates, but in the equatorial plane ($\theta = \pi / 2$), these are basically the same as the standard spherical coordinates; $\hat{e}_z$ is then just the "vertical" unit vector, the one pointing in the $\theta$ direction. I should have clarified that, sorry.

(Also, the vorticity is really an antisymmetric 2nd-rank tensor in the 3-space of constant coordinate time; but in 3-D space of course you can always convert an antisymmetric 2nd-rank tensor to a vector. A vector in the "vertical", $z$ or $\theta$ direction, is equivalent to a 2nd-rank antisymmetric tensor in the $r - \phi$ plane.)

From BillK's blog

This agrees with the $\omega$ I found in the local frame basis.

That $\omega$ is the angular velocity of rotation, not the vorticity. The vorticity is what Bill_K calls $\Omega$ in his blog post (actually his $\Omega$ is minus the vorticity because he is working in a rotating frame).

I still don't understand if this number is related to the orbital period or the rotation of the axes of the transported frame, or both.

I more or less summarized what the different angular velocities mean in my post #117, but I'll briefly recap some of that here, this time referring to Schwarzschild spacetime (where de Sitter precession as well as Thomas precession is present):

(1a) As seen from infinity, an observer orbiting the hole geodesically at radius $r$ orbits with angular velocity $\omega = \sqrt{M / r^3}$.

(1b) As seen by the observer orbiting the hole, if he keeps his line of vision pointed radially outward, the "fixed stars" at infinity are rotating about him with angular velocity $- \gamma \omega$ (i.e., in the opposite sense to the rotation seen at infinity in #1a), where $\gamma = 1 / \sqrt{1 - 2M / r - \omega^2 r^2}$. The factor of $\gamma$ is due to time dilation; the $2M / r$ term is usually called gravitational time dilation, and the $\omega^2 r^2$ term is the usual time dilation due to relative motion.

(2a) If such an observer is part of a congruence of observers, all of whom are circling the hole with the same angular velocity $\omega$ (note that this means members of the congruence at smaller and larger radial coordinates will *not* be moving on geodesics--think of them as all being on a rotating disk similar to the flat spacetime case, though here the disk has to have a hole in the center where the central mass is), then the observer can define a set of spatial vectors by sticking out connecting rods to neighboring members of the same congruence. These rods will be fixed in place on the rotating disk, and so this set of spatial vectors will also rotate, as seen from infinity, with angular velocity $\omega$. Call these spatial vectors the "congruence vectors".

(2b) As seen by the observer orbiting the hole at radius $r$, once again, the "fixed stars" at infinity will rotate, relative to the congruence vectors, with angular velocity $- \gamma \omega$.

(3b) If the orbiting observer also carries a set of gyroscopes, and uses them to define a second set of spatial vectors, then these vectors, if Newtonian physics were exactly correct, would always point in the same direction relative to infinity; i.e., they would not rotate at all relative to infinity. This would mean that, relative to the orbiting observer, the second set of vectors--call them the "gyro vectors"--would rotate relative to the congruence vectors with angular velocity $- \gamma \omega$. The vorticity of a congruence is standardly defined as the angular velocity of rotation of the congruence vectors relative to the gyro vectors, as seen by the orbiting observer, so it would be $\Omega_{Newton} = \gamma \omega$.

However, there are two relativistic effects that change this: Thomas precession and de Sitter precession. Thomas precession adds a retrograde component to the rotation of the gyro vectors, and de Sitter precession adds a prograde component; the net result is that the orbiting observer sees the gyro vectors rotate, relative to the congruence vectors, with angular velocity $- \Omega = - \gamma^2 \omega \left( 1 - 3M / r \right)$. The vorticity of the congruence is minus this, so it is $\Omega = \gamma^2 \omega \left( 1 - 3M / r \right)$.

This also means that, as seen by the orbiting observer, the "fixed stars" at infinity will rotate, relative to the gyro vectors, with angular velocity $\Omega - \gamma \omega = \omega \gamma \left[ \gamma \left( 1 - 3M / r \right) - 1 \right] = \omega \gamma \left( \gamma - 1 - 3 \gamma M \ r \right)$.

(3a) As seen from infinity, the congruence vectors rotate, relative to the gyro vectors, with angular velocity $\Omega / \gamma = \gamma \omega \left( 1 - 3M / r \right)$. That means the gyro vectors rotate, relative to infinity, with angular velocity $\omega - \Omega / \gamma = \omega \left[ 1 - \gamma \left( 1 - 3M / r \right) \right] = - \left( \gamma - 1 \right) \omega + 3 \gamma \omega M / r$. (If we flip the sign and add a factor of $\gamma$ for time dilation, we obtain the angular velocity given at the end of #3b, as we should.) The first term is the Thomas precession and the second is the de Sitter precession.

 

[PeterDonis]

Following on from my previous post, where I wrote everything except #1a in general terms, I'll fill in what things look like for $\omega = \sqrt{M / r^3}$, which was the specific value Mentz114 gave. (The equations I wrote, which are based on what's in Bill_K's blog post, are valid for any value of $\omega$, including non-geodesic as well as geodesic worldlines.)

For $\omega = \sqrt{M / r^3}$, we have $\gamma = 1 / \sqrt{1 - 3 M / r}$, and therefore the vorticity of the rotating congruence is $\Omega = \gamma^2 \omega \left(1 - 3M / r \right) = \omega = \sqrt{M / r^3}$. So Mentz114, the value you gave in your earlier post was correct for the vorticity; I hadn't checked the specific value for a geodesic orbit when I posted earlier, sorry about that.

However, as I noted in my previous post, the vorticity is an angular velocity relative to the rotating observer, not relative to infinity. An observer at infinity would see the congruence vectors rotating relative to the gyro vectors with angular velocity

$$
\frac{\Omega}{\gamma} = \sqrt{\frac{M}{r^3} \left( 1 - \frac{3M}{r} \right)}
$$

Since this is less than $\sqrt{M / r^3}$, this also means that the observer at infinity would see the gyro vectors rotating in the prograde direction, i.e., the de Sitter precession is larger than the Thomas precession so the net rotation of the gyro vectors for a geodesic orbit is positive.

 

[PeterDonis]

Also, I was wondering again why it is things like Thomas Precession and Lense-Thirring precession are not eliminated by Fermi-Walker transport.

Because relativity adds effects that our Newtonian intuition can't easily picture.

Is it because Fermi-Walker transport is done by the observer i.e. it is under his operational control (in the merry go round example he constantly pivots his feet so that his spatial axes are Fermi-Walker transport as he orbits the center of the merry go round)

Don't read too much into that description; I was describing the Newtonian version of Fermi-Walker transport, not the relativistically correct version, so I cheated by using an object fixed at "infinity" (the ticket booth) to define the way the observer was facing, instead of using a gyroscope. See further comments below.

whereas things like Thomas Precession and Lense-Thirring precession are out of his control as they are intrinsic properties of Lorentz transformations and stationary space-times respectively so he has no operational way of eliminating them by himself as an observer?

I think this is a possible point of view, yes. Take the merry-go-round example again, but this time with a gyroscope. You are going around the merry-go-round, and you are holding a gyroscope that starts out pointed radially outward, and pivoting so you are always facing the way the gyroscope is pointing. At the starting instant, when the gyro is pointed radially outward, it is also pointed directly at the distant ticket booth.

If Newtonian physics were exactly correct, the gyro would always stay pointed exactly at the ticket booth. However, Newtonian physics is *not* exactly correct. If we assume the merry-go-round spacetime is flat (so Thomas precession is the only relativistic effect present), then as we continue to face the way the gyro is pointing, we will see the ticket booth slowly rotating around us, in the same sense as the merry-go-round. (As I mentioned before, the merry-go-round would have to be spinning quite fast for us to actually notice this. )

We are not doing anything to make the gyro and the ticket booth get out of alignment; it just happens to be a fact about Minkowski spacetime that a gyro carried by a rotating observer behaves like this. Similar remarks would apply if there were a mass present (so de Sitter precession gets added in), and if the mass were rotating (so Lense-Thirring precession gets added in); we are just doing the best we can to keep a set of gyro-stabilized basis vectors, and the way they behave relative to objects fixed at infinity tells us about the structure of the spacetime we are in.

 

[WannabeNewton]

Thank you very much, explained awesomely as always :)! And the booth would appear to rotate in the same sense as the merry go round because the Thomas precession for that observer is retrograde with respect to the rotation of the merry go round in that case so the booth itself would appear in his own view to go the opposite way i.e. prograde?

I took a look at the MTW section you referenced by the way, it was quite helpful. I'll have to do the exercises in that section later so hopefully I don't have too many extra questions with regards to those problems xD. I wish there were more diagrams/pictures in that section though, which is something I'd never thought I'd say about MTW!

 

[PeterDonis]

the booth would appear to rotate in the same sense as the merry go round because the Thomas precession for that observer is retrograde with respect to the rotation of the merry go round in that case so the booth itself would appear in his own view to go the opposite way i.e. prograde?

Yes.