Locally non-rotating observers

[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.

 

[Mentz114]

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.)
...
...

Thanks for your posts. I'll study them, and BillK's stuff and that should clear it up for me.

 

[PeterDonis]

part (c) of the following diagram: http://postimg.org/image/z69htcfdz/

Something has been nagging at me since I looked at this diagram, and after looking at Bill_K's Kerr spacetime results and working through some computations of my own, I realized what it is: I got one thing backwards in previous posts in this thread, and made one misinterpretation which hasn't affected much of what I've said in this thread, but which is still, I think, worth mentioning.

First, what I got backwards: the twist of the "hovering" congruence in the equatorial plane in Kerr spacetime is positive, not negative!

Mathematically, this is evident from Bill_K's Kerr formula; if you plug in $\omega = 0$, you still have a nonzero term, $\gamma^2 M a / r^3$. The sign of this term is the same as the sign of the Thomas precession term $\gamma^2 \omega$, so it represents a retrograde precession of the gyro vectors relative to the "rotating" frame (since Bill_K's results are all given relative to the rotating frame). I put "rotating" in quotes because in this case, of course, the frame is not actually rotating relative to infinity; but the point is that his result indicates that gyroscopes carried by hovering observers in Kerr spacetime will precess in the retrograde direction. That means the congruence vectors, the ones that are fixed to always point at neighboring members of the same congruence, are rotating in the prograde direction relative to the gyro vectors, which is a positive twist.

The picture WN linked to indicates the same thing, but although, as I said, something nagged at me when I looked at it, I didn't fully catch on at the time. (What finally twigged it for me was that in trying to compute the acceleration and twist of the hovering congruence by a different route than Bill_K's route, I kept getting opposite signs from what I was expecting for the covariant derivatives of the radial and tangential frame field vectors.) The little balls to the right and left of the big ball are like observers hovering in the equatorial plane of the hole, and those balls are rotating in the retrograde direction (as opposed to the balls above and below the big ball, corresponding to observers hovering on the rotation axis of the hole, which are rotating in the prograde direction). The rotation of each little ball is the same as the rotation of the gyro vectors of the corresponding observer.

Now for the thing I was misinterpreting. I referred to the ZAMO congruence a number of times in this thread, and said that it has a twist of zero. I'm no longer sure that's true, since I haven't been able to get my computations to give that result; but I'll save that for a separate post some time. The key thing here is that I realized that, in all my previous posts in this thread about the ZAMO congruence, I had been implicitly thinking of it as a "rotating" congruence like the Langevin congruence in flat spacetime. But it isn't, because any "rotating" congruence like the Langevin congruence has the same angular velocity for all worldlines in the congruence, and the ZAMO congruence does not. (Put another way, any "rotating" congruence like the Langevin congruence must have zero shear, and the ZAMO congruence has nonzero shear.)

That means that you can't, for example, compute the twist of the ZAMO congruence by plugging in the ZAMO angular velocity (which is just $- g_{t \phi} / g_{\phi \phi}$ into Bill_K's formula, because that formula assumes a Langevin-type congruence with zero shear. (More precisely, interpreting Bill_K's $\Omega$ as the vorticity of a congruence assumes a Langevin-type congruence with zero shear. I don't think any of Bill_K's actual computations require assuming a congruence at all; that's the point of his opening comments about particle properties and absolute derivatives.)

 

[WannabeNewton]

Now for the thing I was misinterpreting. I referred to the ZAMO congruence a number of times in this thread, and said that it has a twist of zero. I'm no longer sure that's true, since I haven't been able to get my computations to give that result; but I'll save that for a separate post some time.

Hi Peter. I need to read the rest of your post in detail but this caught my eye so I'll respond to it now. If by ZAMO congruence you mean as usual the congruence with tangent field given by $\xi^{a} = (-\nabla^{b}t\nabla_{b}t)^{-1/2}\nabla^{a}t$ then I showed this in post #1: $\omega^{a} = \epsilon^{abcd}\xi_{b}\nabla_{c}\xi_{d} = \alpha\epsilon^{ab[cd]}\nabla_{b}t\nabla_{(c}\nabla_{d)}t + \alpha^{3}\epsilon^{a[b|c|d]}\nabla_{(b}t \nabla_{d)}t\nabla^{e}t\nabla_{c}\nabla_{e}t = 0$; the first term vanishes because $\nabla_{a}$ commutes on scalar fields hence we have antisymmetric indices being contracted with symmetric indices and similarly the second term vanishes because $\epsilon^{abcd}$ is antisymmetric over $b,d$ and $\nabla_{b} t\nabla_{d} t$ is symmetric over $b,d$.

 

[PeterDonis]

If by ZAMO congruence you mean as usual the congruence with tangent field given by $\xi^{a} = (-\nabla^{b}t\nabla_{b}t)^{-1/2}\nabla^{a}t$

Yes, the congruence whose worldlines are orthogonal to the surfaces of constant coordinate time (and therefore have zero angular momentum--ZAMO stands for zero angular momentum observers).

I showed this in post #1

Ah, that's right. I'm glad not *all* of my intuitions were wrong.

But there are still some interesting issues of interpretation here. I'll save that for a follow-up when I've done some more computations.

 

[WannabeNewton]

Mathematically, this is evident from Bill_K's Kerr formula; if you plug in $\omega = 0$, you still have a nonzero term, $\gamma^2 M a / r^3$. The sign of this term is the same as the sign of the Thomas precession term $\gamma^2 \omega$, so it represents a retrograde precession of the gyro vectors relative to the "rotating" frame (since Bill_K's results are all given relative to the rotating frame). I put "rotating" in quotes because in this case, of course, the frame is not actually rotating relative to infinity; but the point is that his result indicates that gyroscopes carried by hovering observers in Kerr spacetime will precess in the retrograde direction.

What rotating frame is being talked about here? Also, is it still ok to say that the gyroscopes carried by the hovering observers also precess relative to the distant stars? I ask because I think the picture I linked from Wheeler's text (that you quoted above) is depicting how the small hovering balls are spinning as seen from the distant stars, based on the way the diagram is drawn.

 

[PeterDonis]

What rotating frame is being talked about here?

In the first section (that derives the results for flat spacetime), he sets up a set of basis vectors that move with the rotating observer. That's the rotating frame I'm talking about. The two basis vectors that are involved in the vorticity are what he calls $e_r$ and $e_\phi$, and which point radially outward and tangentially in the direction of motion of the observer.

(Of course, in the case of a hovering observer, these basis vectors aren't actually moving, since the observer doesn't move--where "move" here means relative to an observer at infinity.)

Also, is it still ok to say that the gyroscopes carried by the hovering observers also precess relative to the distant stars?

Yes (for the Kerr spacetime case); they just precess in the opposite direction to the one I was originally imagining.

 

[WannabeNewton]

In the first section (that derives the results for flat spacetime), he sets up a set of basis vectors that move with the rotating observer. That's the rotating frame I'm talking about.

So when you say the spatial axes of the comoving frame (as defined in Bill's blog) are rotating do you mean rotating relative to the spatial axes of the observer's Fermi-Walker frame? That is, if we imagine that the spatial axes of the observer's Fermi-Walker frame are always pointing in the same direction relative to the fixed stars, throughout the circular orbit, then the spatial axes of the comoving frame defined in Bill's blog (which always point radially outward and tangential to the circle and hence constantly readjust their directions relative to the same fixed stars) will be rotating relative to the spatial axes of the Fermi-Walker frame.

 

[PeterDonis]

So when you say the spatial axes of the comoving frame (as defined in Bill's blog) are rotating do you mean rotating relative to the spatial axes of the observer's Fermi-Walker frame?

No, relative to infinity--more precisely, relative to the "fixed" asymptotically flat background at infinity. (I.e., relative to the "fixed stars".) Actually, mathematically this shows up as the axes rotating relative to the global coordinate chart; the angular velocity $\omega$ is the angular velocity of Bill_K's basis vectors relative to this chart. See below.

That is, if we imagine that the spatial axes of the observer's Fermi-Walker frame are always pointing in the same direction relative to the fixed stars

But they aren't, so rotating relative to the fixed stars is not the same as rotating relative to the Fermi-Walker transported spatial axes. The rotation of Bill_K's basis vectors is relative to the global coordinate chart, which means relative to the fixed stars. The F-W transported vectors are also, in general, rotating relative to the global coordinate chart; Bill_K's $\Omega$ gives the difference between the F-W transported vectors' rotation and the rotation of the basis vectors he defines. The fact that $\Omega$ is, in general, different from $\omega$ is how his formalism illustrates the fact that the F-W transported vectors do not, in general, stay pointing in the same direction relative to the fixed stars.

 

[WannabeNewton]

Oh are you talking about after taking the effect of Thomas precession into account? I totally forgot about that, my bad!

Also, I guess I just don't get what you mean by rotating relative to the global inertial frame in the case of the comoving frame as compared to the rotation induced by Thomas precession on the F-W frame. For the F-W frame, the Thomas precession induces a rotation of the F-W frame in the sense that the spatial axes of the F-W frame start precessing relative to the global inertial frame i.e. gyroscopic precession. On the other hand, the comoving frame is rotating relative to the global inertial frame in the sense that if we fix a star in the global inertial frame that say the radial axis of the comoving frame is pointing towards then at the next instant the radial axis will of course be pointing in a different direction because the comoving frame has to readjust its axes so that the radial axis always stays radial to the circle. Is that what you mean?

 

[PeterDonis]

Also, I guess I just don't get what you mean by rotating relative to the global inertial frame in the case of the comoving frame as compared to the rotation induced by Thomas precession on the F-W frame.

Go back to the merry-go-round example. The rotating frame that Bill_K defines is fixed to the merry-go-round and rotating with it; his vector $e_r$ always points directly outward along a radial line painted on the merry-go-round, and his vector $e_{\phi}$ always points tangentially in the direction of the merry-go-round's rotation, at right angles to $e_r$. You can think of them as two arrows painted on the floor of the merry-go-round.

The distant ticket booth is at rest in the global inertial frame (strictly speaking the ticket booth would be "at infinity"). The fact that, if you carry a gyro with you while going around on the merry-go-round, it will not stay pointed exactly at the ticket booth, is Thomas precession. The angular velocity of rotation of the arrows painted on the floor of the merry-go-round, relative to the direction the gyro is pointing, is the vorticity $\Omega$ (note that it is an angular velocity as measured by you, going around on the merry-go-round). The angular velocity of the merry-go-round relative to the ticket booth, i.e., relative to the global inertial frame, is $\omega$ (think of someone standing in the ticket booth and timing the intervals when a mark painted on the side of the merry-go-round passes his line of vision).