Locally non-rotating observers

[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.)
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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).

 

[WannabeNewton]

Ok ok. So if we imagine a given observer on the disk and two nearby observers, one separated by $e_{\phi}$ and one separated by $e_{r}$ then the rotation of these two relative to the axes of the given observer's F-W frame will be the vorticity, as usual. And the axes of the F-W frame themselves precess relative to the ticket booth/global inertial frame due to Thomas Precession so that is the rotation of the F-W frame relative to the global inertial frame. And the rotation of $e_r$ and $e_{\phi}$ relative to the global inertial frame is just in the sense that they constantly turn around about the center of the disk with the same angular velocity as the disk so as to remain fixed on the disk (like the usual unit vectors from classical mechanics)?

 

[PeterDonis]

So if we imagine a given observer on the disk and two nearby observers, one separated by $e_{\phi}$ and one separated by $e_{r}$ then the rotation of these two relative to the axes of the given observer's F-W frame will be the vorticity, as usual. And the axes of the F-W frame themselves precess relative to the ticket booth/global inertial frame due to Thomas Precession so that is the rotation of the F-W frame relative to the global inertial frame. And the rotation of $e_r$ and $e_{\phi}$ relative to the global inertial frame is just in the sense that they constantly turn around about the center of the disk so as to rotate along with the disk with the same angular velocity as the disk?

All correct.

 

[WannabeNewton]

Ok thanks just one more question: if we imagine the comoving frame as describing an observer who orients his spatial basis vectors so that they remain anchored to his $e_{\phi}$ neighbor and $e_{r}$ neighbor (as opposed to orienting them via F-W transport) then is there a reason why there is no Thomas Precession for this observer (hence the comoving frame)?

 

[PeterDonis]

if we imagine the comoving frame as describing an observer who orients his spatial basis vectors so that they remain anchored to his $e_{\phi}$ neighbor and $e_{r}$ neighbor (as opposed to orienting them via F-W transport) then is there a reason why there is no Thomas Precession for this observer (hence the comoving frame)?

Um, because he's not letting his basis vectors be Fermi-Walker transported?

Also, it's a little misleading to say "there is no Thomas Precession", because that could be taken to imply that the "comoving" basis vectors are somehow always pointing in the same direction with respect to infinity (e.g., that the $e_r$ vector is always pointing exactly at the ticket booth), which is obviously not the case. (At least, it's not if you are talking about a rotating "comoving" frame. If you're talking about a "hovering" frame, i.e., a frame that is moving inertially in Minkowski spacetime, then of course there's no Thomas precession because there's no rotation to begin with.)

It seems to me that the way you asked the question is backwards. You first decide how you are going to determine the orientation of a particular set of spatial basis vectors; then you see how they end up pointing with respect to infinity. The reason why they end up pointing a particular way with respect to infinity is that that's a consequence of how you decided to determine their orientation. If you Fermi-Walker transport them, you get one result (Thomas Precession); if you anchor them to the merry-go-round, you get a different result, because you chose a different method of orienting them.

It is true that Fermi-Walker transport is, in a sense, specially picked out as a way of determining the orientation of a set of spatial basis vectors, because it corresponds to "no spatial rotation" (strictly speaking, no rotation over and above the spacetime rotation needed to keep the vectors orthogonal to the 4-velocity). But that's no different than geodesic motion being specially picked out as a way of determining your 4-velocity, because it corresponds to "no proper acceleration". Asking why the "comoving" vectors don't behave like Fermi-Walker transported vectors seems to me to be like asking why non-geodesically moving objects behave differently than geodesically moving ones: because that's a consequence of the state of motion you chose. Fire your rockets and you feel acceleration; anchor your basis vectors to a rotating merry-go-round and they will have a nonzero vorticity.

 

[WannabeNewton]

I was asking physically, if the Langevin observer is still orbiting the center of the disk, why is there Thomas Precession with one kind of orientation setup and not with another. Either way we have to Lorentz boost from one consecutive comoving inertial frame to the next when talking about the Langevin observer so why does the Thomas precession effect only show up when we orient using F-W transport as opposed to orienting by keeping the spatial basis vectors anchored to the $e_r$ and $e_{\phi}$ vectors? How come with the second orientation setup, if a Lorentz boost is applied from one consecutive comoving inertial frame to the next there is no rotation effect induced by the noncommutativity of boosts along different directions? Because we're saying that the only rotation of the spatial basis vectors relative to infinity, with the latter orientation setup, is simply due to the fact that the spatial basis vectors must turn around about the center of the disk due to being anchored to $e_r$ and $e_{\phi}$.

 

[PeterDonis]

Either way we have to Lorentz boost from one consecutive comoving inertial frame to the next

But if you're Fermi-Walker transporting, the Lorentz boost is the *only* change that you make. If you're determining the orientation of the spatial vectors some other way that doesn't give the same result as F-W transport, then whatever process you use to orient the spatial vectors will cause them to undergo some *additional* change in addition to the change induced by the Lorentz boost.

why does the Thomas precession effect only show up when we orient using F-W transport as opposed to orienting by keeping the spatial basis vectors anchored to the $e_r$ and $e_{\phi}$ vectors?

Again, I think it's misleading to say that the Thomas precession "doesn't show up" when you are anchoring the spatial basis vectors to the merry-go-round. The spatial basis vectors are certainly rotating relative to infinity; and one way to view this, as I said above, is that they are undergoing Thomas precession, because of the Lorentz boost that gets induced, *plus* a further change induced by the fact that you are forcing their orientation to be fixed relative to the merry-go-round. On this view, the Thomas precession *is* there for the "comoving" vectors; it just has an additional change added on top of it. See further comments below.

How come with the second orientation setup, if a Lorentz boost is applied from one consecutive comoving inertial frame to the next there is no rotation effect induced by the noncommutativity of boosts along different directions?

How do you know there isn't? See below.

Because we're saying that the only rotation of the spatial basis vectors relative to infinity, with the latter orientation setup, is simply due to the fact that the spatial basis vectors must turn around about the center of the disk due to being anchored to $e_r$ and $e_{\phi}$.

I can see that it looks that way at first glance, but it's not that simple.

Consider: what does the term "spatial basis vector" actually mean? The standard meaning is that it is a vector that is orthogonal to the 4-velocity. But that means that just fixing the spatial direction of a vector as, for example, purely radial, does *not* fully determine a spatial basis vector; you also have to impose the orthogonality condition.

Now think about, say, the $e_r$ vector that is fixed to the merry-go-round. As the merry-go-round rotates, this vector, *if* we want it to remain part of the orthonormal basis of the Langevin observer, *cannot* just rotate spatially, from one radial direction to the next. It must *also* rotate in the "time" direction, in order to stay orthogonal to the 4-velocity. In other words, it must undergo a Lorentz boost *in addition* to the spatial rotation because of the change in the radial direction. But the direction of this Lorentz boost *also* changes, which means that there *is*, in fact, an underlying Thomas precession going on; it just gets masked by the fact that the additional spatial rotation is being applied.

Put another way, the spatial rotation that has to be applied to keep the $e_r$ vector pointing radially outward is *larger* than it would be if the Thomas precession didn't exist. That's why the vorticity of the Langevin congruence is larger than its angular velocity, by a factor of $\gamma$, corresponding to the fact that the underlying Thomas precession is there.

 

[WannabeNewton]

So basically there *is* a Thomas precession but in implicitly fixing the constraint that the spatial basis vectors always have to equal $e_r$ and $e_{\phi}$ (i.e. must always point radially and tangentially to the circle respectively) we have implicitly applied a spatial rotation from one instant to the next that is necessarily larger than normal (i.e. larger than what we would apply in the absence of Thomas precession) in order to "counter" the rotation due to the Thomas precession and keep the spatial basis always equaling $e_r$ and $e_{\phi}$ along the trajectory?

 

[PeterDonis]

So basically there *is* a Thomas precession but in implicitly fixing the constraint that the spatial basis vectors always have to equal $e_r$ and $e_{\phi}$ (i.e. must always point radially and tangentially to the circle respectively) we have implicitly applied a spatial rotation from one instant to the next that is necessarily larger than normal (i.e. larger than what we would apply in the absence of Thomas precession) in order to "counter" the rotation due to the Thomas precession and keep the spatial basis always equaling $e_r$ and $e_{\phi}$ along the trajectory?

That's the way it looks to me, yes.

 

[WannabeNewton]

Alrighty, thanks Peter!

 

[WannabeNewton]

Hey Peter, sorry to bring up an old thread but I came across something you might be interested in. This is regarding something we talked about earlier in this thread: the relationship between the non-vanishing twist of the time-like killing field $\xi$ in a stationary but non-static space-time and the precession of the frames of static observers in the space-time. I don't recall there being a definitive statement with proof about the relationship in this thread so I figured I should mention what follows.

Consider an observer following an orbit $\gamma$ of $\xi$ and an orthonormal triad $\{e_{i}\}$ which is Lie transported along $\gamma$ i.e. $\mathcal{L}_{\xi}e_i = 0$. Note that $g(e_i,u) = g(e_i,\xi) = 0$ all along $\gamma$ because $\mathcal{L}_{\xi}g(e_i,\xi) = 0 = \xi g(e_i,\xi) - g(\mathcal{L}_{\xi}e_i,\xi)- g(e_i,\mathcal L_{\xi}\xi) = \xi g(e_i,\xi)$. Denoting by $u$ the 4-velocity of the observer and by $a$ the 4-acceleration, the quantity $\omega_{ij} = g(\nabla_{u}e_i,e_j)$ can be shown to satisfy $\nabla_{u}e_i -g(e_i,a)u = \omega_{i}{}{}^{j}e_{j}$ i.e. $\omega_{ij}$ measures the failure of the spatial axes $\{e_i \}$ to get Fermi-Walker transported along $\gamma$. It can also be shown that $\omega_{ij} =(-g(\xi,\xi))^{-1/2}\nabla \xi^{\flat}(e_j,e_i)= \frac{1}{2}(-g(\xi,\xi))^{-1/2}d\xi^{\flat}(e_i,e_j)$.

The result of interest is that $\omega_{ij} = 0$ if and only if $\xi^{\flat}\wedge d\xi^{\flat} = 0$. So an observer at rest in a stationary space-time who Lie transports a set of spatial axes along his/her worldline will constitute a non-rotating frame if and only if the twist of the time-like killing field vanishes. The proof can be found in section 2.10.4 (p. 53) of General Relativity-Straumann which I believe you have an eBook copy of but if not then I can post the proof here.

 

[PeterDonis]

an observer at rest in a stationary space-time who Lie transports a set of spatial axes along his/her worldline will constitute a non-rotating frame if and only if the twist of the time-like killing field vanishes.

This makes sense, and it fits what we already know about Schwarzschild vs. Kerr spacetime; observers following orbits of the timelike KVF $\xi^a = \partial_t$ with Lie transported basis vectors (i.e., basis vectors that are constant in the global coordinate chart) are non-rotating in the former but not in the latter.

The proof can be found in section 2.10.4 (p. 53) of General Relativity-Straumann which I believe you have an eBook copy of

Yes, I do, I'll look it up, thanks!

 

[WannabeNewton]

Also Peter while we're at it I hope it's no trouble if I ask a question from a related section in Straumann's book; this is regarding what he says at the very top of page 53. Straumann takes an arbitrarily accelerating observer given by the worldline $\gamma(\tau)$ with a Lorentz frame $\{e_{\alpha}\}$ that is not necessarily Fermi-Walker transported i.e. $F_{u}e_{\alpha} = \omega^{\beta}{}{}_{\alpha}e_{\beta}$ where $F_{u}$ is the Fermi derivative along $\gamma$.

Then he derives the precession of a spinning top relative to this frame but to do this he first says that "for a spinning top we have $F_{u}S = 0$" where $S$ is the spin vector of the top. Why should the spin vector of the top necessarily get Fermi-Walker transported along $\gamma(\tau)$? Is this some universal property of spinning tops?

 

[PeterDonis]

Why should the spin vector of the top necessarily get Fermi-Walker transported along $\gamma(\tau)$?

I think that to be precise, he should have made explicit the assumption that whatever forces are applied to the top to make it follow the given worldline are applied in such a way that there is zero torque on the top (the simplest way to ensure that is to apply the forces at the top's center of mass). If there is zero torque on the top, then its spin vector will get Fermi-Walker transported for the same reason that a gyroscope's does.

 

[WannabeNewton]

Ok I think I drew myself into a corner of confusion again :p

If you take a look at section 2.10.3 on page 51, Straumann lists some properties of the Fermi derivative. In particular he says that if the vector field on $\gamma$, for our purposes $S$, is orthogonal to the 4-velocity $u$ of $\gamma$, then $F_u S = (\nabla_u S)_\perp$ where $\perp$ is the projection orthogonal to $\gamma$ of $(\nabla_u S)$. This holds true for the spin vector $S$ which already satisfies $g(S,u) = 0$. Since $S$ is purely spatial, the expression $(\nabla_u S)_\perp = 0$ would amount to saying that the spatial direction of $S$ relative to $\gamma$ is constant along $\gamma$ right? So wouldn't the observer described by $\gamma$ have to see $S$ always point in the same (spatial direction)? How then can the spin vector precess relative to this observer? I guess my confusion also extends to gyroscopes. Thanks!

 

[PeterDonis]

Since $S$ is purely spatial, the expression $(\nabla_u S)_\perp = 0$ would amount to saying that the spatial direction of $S$ relative to $\gamma$ is constant along $\gamma$ right?

Yes. But this just means that the Fermi derivative of $S$ is zero, i.e., that $S$ is being Fermi-Walker transported.

So wouldn't the observer described by $\gamma$ have to see $S$ always point in the same (spatial direction)?

Yes. But again, this is because $S$ is being Fermi-Walker transported; that's what the condition $(\nabla_u S)_\perp = 0$ ensures. (More precisely, the condition $F_u S = (\nabla_u S)_\perp = 0$ is what *defines* Fermi-Walker transport of $S$.)

How then can the spin vector precess relative to this observer?

By not being Fermi-Walker transported. That is, by having $F_u S = (\nabla_u S)_\perp \neq 0$. For example, applying a torque to a gyroscope does this.

 

[WannabeNewton]

But if you look at page 53 at the top, he says $F_u S = 0 = \frac{\mathrm{d} S^{i}}{\mathrm{d} \tau}e_i + S^{j}\omega^{i}{}{}_{j}e_{i}$ and then concludes that $\frac{\mathrm{d} S^{i}}{\mathrm{d} \tau} = S^{j}\omega_{j}{}{}^{i}$ and states this implies that $S$ has a precession relative to the spatial axes $\{e_i \}$ of the observer. The non-vanishing of the $\omega_{ij}$ implies that the observer's spatial axes are themselves not Fermi-Walker transported. The fact that $F_u S = 0$ at the same time as $S$ having a precession relative to the observer's spatial axes doesn't make sense to me since $F_u S =0$ should mean that $S$ has constant spatial direction relative to the observer.

Does $F_u S=0$ only imply that $S$ has constant spatial direction relative to this observer if the observer's own spatial axes are Fermi-Walker transported as well? In other words, does the statement "$F_u S = 0$ means $S$ has constant spatial direction relative to this observer" necessarily mean "$S$ has constant spatial direction relative to a set of Fermi-Walker transported spatial axes carried by the observer"? So if the observer had another set of spatial axes $\{e'_i \}$ such that $F_u e'_i = 0$, then $F_u S = 0$ necessarily means that $S$ has constant spatial direction relative to $\{e'_i \}$ but the spatial axes $\{e_i \}$ from above rotate relative to $\{e'_i \}$ and thus $S$ rotates relative to $\{e_i \}$? Thanks again.

 

[PeterDonis]

Does $F_u S=0$ only imply that $S$ has constant spatial direction relative to this observer if the observer's own spatial axes are Fermi-Walker transported as well?

Yes. I should have made that clear in my last post. It's easy to see that this must be the case, since the observer's own spatial axes themselves must be orthogonal to his 4-velocity, so the same equations that would apply to $S$ apply to them; in particular, $F_u \hat{e} = 0$, where $\hat{e}$ is a spatial basis vector, is equivalent to that basis vector being Fermi-Walker transported. But the change in the relative directions of $S$ and $\hat{e}$ along the worldline is just $( \nabla_u ( S \cdot \hat{e} ) )_{\perp} = ( \nabla_u S )_{\perp} \cdot \hat{e} + S \cdot ( \nabla_u \hat{e} )_{\perp} = \hat{e} \cdot F_u S + S \cdot F_u \hat{e}$, so it will be zero if both vectors are Fermi-Walker transported.

 

[WannabeNewton]

That makes perfect sense, thanks!

 

[WannabeNewton]

Hi Peter! I had some more questions crop up, except this time it's related to something I've been reading about global rotation in curved space-times. We start with a stationary axisymmetric space-time $(M,g_{ab})$ with the properties listed in section 7.1, and Theorem 7.1.1. in particular, in Wald (pp. 162-163).

The author then considers a one-dimensional ring (two-dimensional sub-manifold $\mathfrak{R}$ embedded in $M$) that is invariant under the isometries generated by the time-like killing field $t^a$ and the axial killing field $\phi^a$ i.e. the ring is axially symmetric about the axis of rotation of the space-time (the set of points on which $\phi^a = 0$) and the rotation looks the same at each "instant" as determined by $t^a$. The author then models $\mathfrak{R}$ as a congruence of time-like curves describing rigid rotation i.e. as a vector field $\xi^{a}$ such that $\nabla^{(b}\xi^{a)} = 0$. Since the ring is invariant under the isometries generated by $t^a$ and $\phi^a$, we must have that $\xi^a = t^a + k\phi^a$ where $k$ is a constant.

We know that $\xi^a$ is non-rotating at a single event in space-time if $\xi_{[c}\nabla_{b}\xi_{a]} = 0$ at that event. The author then defines the ring to be non-rotating if $\xi_{[c}\nabla_{b}\xi_{a]} = t_{[c}\nabla_{b}t_{a]} + kt_{[c}\nabla_{b}\phi_{a]} + k\phi_{[c}\nabla_{b}t_{a]} + k^2 \phi_{[c}\nabla_{b}\phi_{a]} = 0$ everywhere on $\mathfrak{R}$.

Now the author takes a gyroscope mounted at some point on the ring, with its axis initially tangent to the ring, and proves that the axis of the gyroscope remains tangent to the ring if and only if the above condition of non-rotation holds. Letting $\gamma$ be the worldline of the point the gyroscope is mounted on, the author says that the gyroscope has constant spatial direction relative to $\gamma$ by definition of being a gyroscope. Going back to what was said in a previous post, does this in reality mean that if an observer was comoving with the point described by $\gamma$ and had a set of Fermi-Walker transported spatial axes then the gyroscope would always maintain a constant spatial direction relative to these axes? This is the only way I can make sense of the statement "the gyroscope has constant spatial direction relative to $\gamma$" because if the observer instead chose one of his axes as the tangent field to the ring (i.e. $\phi^a$) and the ring was rotating then the gyroscope would rotate relative to this set of axes. So does the statement presuppose that the "constant spatial direction" of the gyroscope relative to $\gamma$ is really relative to a set of Fermi-Walker transported axes carried along $\gamma$?

Next the author takes $\hat{\xi^a}$ to be the normalization of $\xi^a$ and takes the usual spatial metric $h_{ab} = g_{ab} + \hat{\xi_a}\hat{\xi_b}$. He then says that $h^{a}{}{}_{b}\phi^b$ is the spatial direction of $\phi^a$ relative to $\gamma$. I don't see why this is. Usually when I think of a space-like vector solely indicating a direction in space, I think of a unit vector. However $h^{a}{}{}_{b}\phi^b$ is not of unit norm so I don't get why it would simply be the spatial direction of $\phi^a$ relative to $\gamma$.

Finally, the author says that the spatial direction of $\phi^a$ is not changing relative to $\gamma$ if $h^{a}{}{}_{b}\hat{\xi^c}\nabla_c(h^{b}{}{}_{d}\phi^d) = 0$. Now this is nothing more than Fermi-Walker transport of $h^{a}{}{}_{b}\phi^b$ along $\gamma$ but the intuitive picture isn't immediately clicking for me. $\hat{\xi^c}\nabla_c(h^{b}{}{}_{d}\phi^d)$ itself is the rate of change of the spatial direction of $\phi^a$ along $\gamma$, so if we want this to be constant why not simply have $\hat{\xi^c}\nabla_c(h^{b}{}{}_{d}\phi^d) = 0$? What does contracting with the spatial metric $h^{a}{}{}_{b}$ get for us physically?

Sorry for the long post and sudden onslaught of questions Peter, and thanks in advance!

 

[PeterDonis]

the author says that the gyroscope has constant spatial direction relative to $\gamma$ by definition of being a gyroscope. Going back to what was said in a previous post, does this in reality mean that if an observer was comoving with the point described by $\gamma$ and had a set of Fermi-Walker transported spatial axes then the gyroscope would always maintain a constant spatial direction relative to these axes?

I haven't had a chance to look in my copy of Wald, but I think this is what is meant, yes. At any rate, it *should* be what is meant, since it's correct.

$h^{a}{}{}_{b}\phi^b$ is not of unit norm so I don't get why it would simply be the spatial direction of $\phi^a$ relative to $\gamma$.

"Spatial direction" may be a confusing choice of words. $h^a{}_b$ is a projection tensor, so $h^a{}_b \phi^b$ is just the projection of $\phi^b$ into the hypersurface that is locally orthogonal to $\hat{\xi}^a$. So I would say that a better term would be the "spatial part" of $\phi^b$.

What does contracting with the spatial metric $h^{a}{}{}_{b}$ get for us physically?

It's possible that the second projection may be to pick out the "spatial part" of the rate of change. However, if the rate of change itself is zero, I'm not sure I see the point of projecting out the "spatial part". This one needs more thought.

 

[WannabeNewton]

Thanks for the reply Peter! I should note that I was only referring to Wald for the definition of a stationary, axisymmetric space-time and the desired properties of the associated killing fields of the space-time. The above discussion itself is from a different text ("Topics in the Foundations of General Relativity and Newtonian Gravitation Theory"-Malament).

 

[George Jones]

I haven't had much chance to think about this, and I'm on my way out, but ...

"Spatial direction" may be a confusing choice of words. $h^a{}_b$ is a projection tensor, so $h^a{}_b \phi^b$ is just the projection of $\phi^b$ into the hypersurface that is locally orthogonal to $\hat{\xi}^a$. So I would say that a better term would be the "spatial part" of $\phi^b$.

I agree.

It's possible that the second projection may be to pick out the "spatial part" of the rate of change.

I think so, too.

However, if the rate of change itself is zero, I'm not sure I see the point of projecting out the "spatial part". This one needs more thought.

Zero before the projection implies zero after the projection, but not the other way around.

 

[WannabeNewton]

Thanks for the reply George! Before I hit the hay, here are the relevant pages from the aforementioned text that this discussion comes from:

 

[George Jones]

Thanks for the reply George! Before I hit the hay, here are the relevant pages from the aforementioned text that this discussion comes from:

I, too, am about to hit the sack, though three hours earlier than.

This reminds of how a connection on an ambient space induces a connection on a hypersurface, e.g., in Lee's Riemannian Manifolds: An Introduction to Curvature. I don't have this home with me, so I am not sure how close the connection (pun intended) is.

 

[WannabeNewton]

Right so if you take a look at the third attachment, there is the statement "And $\phi^a$ is 'not changing (spatial) direction relative to $\gamma$' iff $h^{a}{}{}_{b}\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$" and then it states "This condition asserts that the spatial component of $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$ as determined relative to $\gamma$ vanishes." I don't see the relation between these two statements. Why is it not enough to just have $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$ in order to claim that "$\phi^a$ is not changing (spatial) direction relative to $\gamma$"? Why is it that we must contract $\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$ with the spatial metric $h_{ab} = g_{ab} - \hat{\eta_a}\hat{\eta_b}$ (using the author's metric signature convention) in order to have that constancy?

Also, as both of you noted, $h^{a}{}{}_{b}\phi^b$ is just the spatial part of $\phi^a$ relative to $\gamma$ and not necessarily just the spatial direction of $\phi^a$, again relative to $\gamma$, so isn't the requirement that $h^{a}{}{}_{b}\hat{\eta^m}\nabla_m(h^{b}{}{}_{n}\phi^n) = 0$ stronger than the requirement that the spatial direction of $\phi^a$ be constant along $\gamma$? It's bothering me because this ostensibly stronger requirement in the end does prove equivalent to the requirement that the congruence $\eta^a = \tilde{t^a} + k\phi^a$ satisfy $\eta_{[a}\nabla_b \eta_{c]} = 0$ everywhere on the "striated orbit cylinder" (1st and 2nd attachments) even though it seems to place more restrictions on $h^{a}{}{}_{b}\phi^b$ along $\gamma$ than just having constant spatial direction, this being what the author gives as the operational equivalence to $\eta_{[a}\nabla_b \eta_{c]} = 0$ via the gyroscope experiment.

Thanks!