Intuitive understanding of Thomas precession

[WannabeNewton]

Actually, it is worse than that. Lie transported vectors do not correspond with orthonormal basis vectors in flat space time for the same reason.

Sure they do. Following the notation of post #35, consider the time-like vector field $K = \partial_t + \omega \partial_{\phi}$ in flat space-time with polar coordinates $(t,r,\phi)$; here $\nabla \omega =0$ identically and $\omega^2 r^2 < 1$ in order to ensure that $K$ is time-like. Physically of course $K$ describes a rigidly rotating disk since $K$ is a Killing field. Now consider an observer sitting on the disk and attach to the observer the orthonormal spatial basis vectors $\{e_r,e_{\phi}\} := \{e_i\}$ where $e_r = \partial_r$ and $e_{\phi} = \frac{1}{r}\partial_{\phi}$ as usual. It's easy to see that $\{e_i\}$ is Lie transported by $K$ since $K$ and $e_{r}$ are not functions of the coordinates at all and $e_{\phi}$ is only a function of $r$ whereas $K$ flows in the $\partial_t$ and $\partial_{\phi}$ directions. This is not surprising of course because if the observer happens to initially draw arrows of unit length pointing in the radial and azimuthal directions respectively at this initial location, and we imagine two other observers initially sitting at the tips of these arrows, then they will remain sitting at the tips of these arrows-this is in fact what it means for the $\{e_i\}$ to be Lie transported by $K$. This happens of course because the disk is rotating rigidly.

Let's now go back to Kerr space-time to put things in perspective. Let's use the same form for $K$ as above i.e. $K = \partial_t + \omega \partial_{\phi}$ where $\omega$ now corresponds to the angular velocity of some ZAM ring with $\nabla \omega = 0$ so that we have an entire set of concentric rings all rotating with the same angular velocity $\omega$, hence they all rigidly rotate with respect to one another; only one of these rings however will correspond to the chosen ZAM ring. With that in mind, focus entirely on the chosen ZAM ring and forget all about the other rings; imagine an observer sitting on this ring and attach to him the orthonormal triad $\{e_r,e_{\theta},e_{\phi}\} := \{e_i\}$ depicted in your orbiting lab diagram. Are the $\{e_i\}$ Lie transported by $K$? Yes because $K$ is not a function of any of the coordinates and the $\{e_i\}$ are only functions of $r$ and $\theta$ whereas $K$ flows in the $\partial_t$ and $\partial_{\phi}$ directions just like before.

If we now have this observer carry a (torque-free) gyroscope along with him, will it precess relative to any of these axes? Letting $\Omega = \Omega^i e_i$ be the precession angular velocity of the gyroscope with respect to each of the axes, we have from section 2.10.4 of Straumann's text that $\Omega = -\frac{1}{2}\langle K, K\rangle^{-1}\star (K^{\flat}\wedge dK^{\flat})$ where as mentioned in the previous post $\star (K^{\flat}\wedge dK^{\flat})$ is the vorticity of $K$. In other words the gyroscopic precession relative to $\{e_i\}$ is the negation of the vorticity of $K$. Hence there will be no gyroscopic precession relative to the axes in your diagram only if we can find a ZAM ring in Kerr space-time with an angular velocity $\omega$ such that the vorticity of $K = \partial_t + \omega \partial_{\phi}$ vanishes identically.

To that aim let $\tilde{K} = \partial_{t} + \tilde{\omega}\partial_{\phi}$, where $\tilde{\omega} = \frac{2Ma r}{\Sigma^2}$ in standard BL coordinates; this represents the ZAMO congruence. Because the ZAMO congruence has vanishing vorticity, we have

$0 = \epsilon^{\mu\nu\alpha\beta}\tilde{K}_{\nu}\partial_{\alpha}\tilde{K}_{\beta} \\= \epsilon^{\mu\nu\alpha\beta}t_{\nu}\partial_{\alpha}t_{\beta}+\epsilon^{\mu\nu\alpha\beta}t_{\nu}\partial_{\alpha}(\tilde{\omega}\phi_{\beta})+\tilde{\omega}\epsilon^{\mu\nu\alpha\beta}\phi_{\nu}\partial_{\alpha}t_{\beta}+\tilde{\omega}\epsilon^{\mu\nu\alpha\beta}\phi_{\nu}\partial_{\alpha}(\tilde{\omega}\phi_{\beta}) \\= \epsilon^{\mu\nu\alpha\beta}K_{\nu}\partial_{\alpha}K_{\beta} + \epsilon^{\mu\nu\alpha\beta}t_{\nu}\phi_{\beta}\partial_{\alpha}\tilde{\omega} + \tilde{\omega}\epsilon^{\mu\nu\alpha\beta}\phi_{\nu}\phi_{\beta} \partial_{ \alpha } \tilde{\omega} \\= \epsilon^{\mu\nu\alpha\beta}K_{\nu}\partial_{\alpha}K_{\beta} + \epsilon^{\mu\nu\alpha\beta}g_{\nu t}g_{\alpha \phi}\partial_{\beta}\tilde{\omega} = \epsilon^{\mu\nu\alpha\beta}K_{\nu}\partial_{\alpha}K_{\beta} +(g_{tt} - \omega^2g_{\phi\phi})^{1/2}(g_{\phi\phi})^{1/2}(\frac{1}{\rho}\partial_{\theta}\tilde{\omega}e_{r}\delta^{\mu}_{r} -\frac{\Delta^{1/2}}{\rho}\partial_{r}\tilde{\omega} e_{\theta}\delta^{\mu}_{\theta})$

I skipped a lot of steps in that very last equality so let me know if you want me to explain that last step in more detail. Using $\Omega = -\frac{1}{2}\langle K, K\rangle^{-1}\star (K^{\flat}\wedge dK^{\flat})$ we then get $\Omega = \frac{1}{2}\frac{(g_{\phi\phi})^{1/2}}{(g_{tt} - \omega^2g_{\phi\phi})^{1/2}}(\frac{1}{\rho}\partial_{\theta}\tilde{\omega}e_{r}-\frac{\Delta^{1/2}}{\rho}\partial_{r}\tilde{\omega} e_{\theta})$.

So a gyroscope carried by an observer riding on a ZAM ring will not precess relative to the Lie transported orthonormal triad $\{e_r,e_{\theta},e_{\phi}\}$ attached to this observer if and only if the ZAM ring is situated at an $(r,\theta)$ coordinate such that $\Omega(r,\theta) = 0$.

By the way, compare the above expression for $\Omega$ with the one given by MTW in exercise 33.4

 

[WannabeNewton]

Alright I got around to the rest of your questions.

Are you saying the explanation and mathematical calculations given by mathages are incorrect?

It is needlessly cumbersome and adds no pedagogical value.

We have defined the angular 'orbital' velocity and centrifugal acceleration to be constant, so the only possible time varying oscillating quantity is the precession rate.

No. What's varying harmonically are the components of the spin vector. If you take a unit vector in the Cartesian plane fixed to the origin and have it sweep out a circle at a constant angular velocity then it's components clearly vary harmonically relative to the Cartesian axes even though the precession rate of the vector is constant. The same thing is happening to the spin vector of the gyroscope axis: it's components relative to the orthonormal frame $\{e_i\}:= \{e_r,e_{\phi}\}$ naturally adapted to the circular orbit are varying harmonically at a constant precession rate relative to said frame.

Shouldn't that be more quickly?

Yes. The gyroscopic precession relative to the distant stars is $\frac{d\phi'}{dt} = -\gamma \omega$ whereas the precession of the instantaneous tangential axis $e_{\phi}$ relative to the distant stars is given by $\frac{d\phi}{dt} = \omega$ and $\gamma > 1$ so the gyroscope precesses faster than does $e_{\phi}$ relative to the distant stars but in the retrograde sense since $\frac{d\phi'}{dt} < 0$ whereas $\frac{d\phi}{dt}> 0$. If the gyroscope axis was initially aligned with $e_{\phi}$ then after one complete orbit it will have advanced past $e_{\phi}$ in the retrograde sense i.e. $e_{\phi}$ is $\frac{2\pi}{\omega}$ periodic whereas the gyroscope axis is $\gamma^{-1}\frac{2\pi}{ \omega} < \frac{2\pi}{\omega}$ periodic so if initially aligned with $e_{\phi}$, the gyroscope axis will, after one complete orbit, have swept out the same angle as $e_{\phi}$ after one complete orbit and then some in the retrograde sense.

In the section where you address precession in the Schwarzschild metric there appears to be a mistake in your derivation.

No, Bill's derivation is perfectly correct and one that you can find in many standard GR texts.

Alternatively you can consider a single circular orbit in Schwarzschild space-time with angular velocity $\omega$ as well as the Killing field $K = \partial_t + \omega \partial_{\phi}$ of which the circular orbit's world line is an integral curve, and compute $\Omega = -\frac{1}{2}\langle K, K \rangle ^{-1} \star (K^{\flat} \wedge dK^{\flat})$ as in the previous post. This will give the same answer for the gyroscopic precession.

 

[PeterDonis]

So far your equation for the Kerr precession passes all the basic checks with flying colours, but the problem comes when solving:

Ω = γ²ω[r - 3M(1 - aω)]/r + γ²Ma(1 - aω)²/r³ =0 for ω,

because this should yield the ZAMO angular velocity, but that does not seem to be the case.

You're right, it isn't. That's because all of Bill_K's results, including the ones for Kerr spacetime, are for a rigid congruence, i.e., for a congruence of circular orbits (in the equatorial plane) all with the same angular velocity. So his results don't apply to the ZAMO congruence, which has angular velocity varying with r (and theta, if you're not in the equatorial plane).

I've been trying to work out the Lie derivatives for the ZAMO congruence from scratch on the side, but it's a hefty computation and has been going slowly. I hope to post results here at some point, to help shed some light on the question of how the nonzero shear affects the physical interpretation of vorticity.

 

[WannabeNewton]

So his results don't apply to the ZAMO congruence, which has angular velocity varying with r (and theta, if you're not in the equatorial plane).

In what sense? If we use a time-like Killing field for which one of the integral curves describes a ZAMO orbit then wouldn't the derived gyroscopic precession apply to this ZAMO orbit? This is what I did in post #36 and got the same answer as MTW for the gyroscopic precession relative to the ZAMO frame naturally adapted to this orbit.

But that doesn't change what you said of course because setting $\Omega = 0$ as yuiop did would still be incorrect since the time-like Killing field would not (necessarily) have vanishing vorticity.

 

[PeterDonis]

But that doesn't change what you said of course because setting $\Omega = 0$ as yuiop did would still be incorrect since the time-like Killing field would not (necessarily) have vanishing vorticity.

Exactly. The congruence that has vanishing vorticity is the full ZAMO congruence, including circular orbits at different values of $r$ which have different angular velocities. Bill_K's equations don't apply to that congruence, so you can't use them to compute its vorticity and show that it's zero.

You *can* use his equations to compute the vorticity of a rigid congruence that happens to have one circular orbit that "matches" the ZAMO angular velocity at its radius; that's basically what yuiop did, and his results should match the ones you derived by other means (and which are given in the MTW exercise). As you have seen, such a congruence does *not* have zero vorticity. However, it does have the advantage that its Lie transported basis vectors remain orthonormal, so the vorticity has an easy physical interpretation in terms of the relative precession of Lie transported basis vectors and Fermi-Walker transported basis vectors. This, as we have seen, is not true for the ZAMO congruence, for which the Lie transported basis vectors are not, in general, orthonormal.

 

[WannabeNewton]

As you have seen, such a congruence does *not* have zero vorticity. However, it does have the advantage that its Lie transported basis vectors remain orthonormal, so the vorticity has an easy physical interpretation in terms of the relative precession of Lie transported basis vectors and Fermi-Walker transported basis vectors.

Right and this, I think, explains the issue with the statement that even for an arbitrary time-like vector field the vorticity measures the rotation of Lie transported spatial basis vectors relative to local gyroscope axes. Let's say we take a spatial basis $\{e_i \}$ such that $\mathcal{L}_K e_i = 0$ with $K$ being the time-like vector field. If $K$ isn't a Killing field or a geodesic field then the angles between $e_1$, $e_2$, and $e_3$ need not necessarily remain constant when being carried by the flow of $K$; subject to $\mathcal{L}_K e_i = 0$ the angles between them can certainly vary continuously along the flow so as long as $K$ is neither a Killing field nor a geodesic field. Hence if we consider another spatial basis $\{e'_i \}$ such that $F_K e'_i = 0$ i.e. a set of gyroscope axes then it doesn't matter if $K^{\flat} \wedge dK^{\flat} = 0$ because if the elements of $\{e_i \}$ are rotating relative to one another then they will also precess relative to $\{e'_i \}$ irrespective of $K^{\flat} \wedge dK^{\flat} = 0$.

 

[yuiop]

... No, Bill's derivation is perfectly correct and one that you can find in many standard GR texts.

This is the section of Bill's derivation that concerns me:

Supposing that Sμ is Fermi-Walker transported, we get evolution equations for each component:
dS_r/dτ = γω(r - 3M)S_φ (before it was da/dτ = γω b)
dS_φ/dτ = -γ³ω(r - 3M)/r² (before it was db/dτ = -γ³ω a)

Combining these:

d²S_r/dτ² + γ⁴ω²(r - 3M)/r² S_r = 0,

Let me show you my calculation and perhaps you can show me the error in my method and how Bill got his result.

Given dS_r/dτ = γω(r - 3M)S_φ and dS_φ/dτ = -γ³ω(r - 3M)/r² :

dS_r/dS_φ = (dS_r/dτ)/(dS_φ/dτ) = (γω(r - 3M)S_φ)/(-γ³ω(r - 3M)/r²) = -r²/γ² S_φ

d²S_r/dS²_φ = -r²/γ²

d²S_r/dτ² = (d²S_r/dS_φ²)*(dS²_φ/dτ²) = (-r² / γ²)*(-γ⁶ω²(r - 3M)²/r⁴) = -γ⁴ω²(r - 3M)²/r²

d²S_r/dτ² + γ⁴ω²(r - 3M)²/r² = 0

 

[WannabeNewton]

d²S_r/dτ² = (d²S_r/dS_φ²)*(dS²_φ/dτ²) = ...

This is not how the second derivative chain rule works in general; you got lucky here but see below.

I do see now that there is a typo in Bill's calculation: the equation for $\dot{S}_{\phi}$ is missing the factor of $S_r$.

 

[yuiop]

This is not how the second derivative chain rule works in general; you got lucky here but see below.

Ah! Thanks for the reminder ;)

I do see now that there is a typo in Bill's calculation: the equation for $\dot{S}_{\phi}$ is missing the factor of $S_r$.

The (r-3M) factor in his result should be squared, right?

Also, please confirm that Bill's result(s) are the negative of the precession relative to the Lie transported vectors, rather than relative to the orthonormal $e_r$ and $e_{\phi}$ vectors (in proper time).

 

[PeterDonis]

The (r-3M) factor in his result should be squared, right?

I don't think so; that factor is due to the de Sitter precession. I get the same result as Bill_K doing the computation the laborious way using components and connection coefficients.

Also, please confirm that Bill's result(s) are the negative of the precession relative to the Lie transported vectors, rather than relative to the orthonormal $e_r$ and $e_{\phi}$ vectors (in proper time).

They are precession relative to Lie transported vectors [edit: which are, as WBN notes, the same as $\hat{e}_r$ and $\hat{e}_{\phi}$ for this case] (not sure about the sign), but as I noted in a previous post, those are Lie transported vectors for a rigid congruence with constant angular velocity [edit: which is why they are the same as $\hat{e}_r$ and $\hat{e}_{\phi}$ for this case]. For the Kerr spacetime case, this is a *different* congruence from the ZAMO congruence, which has angular velocity varying with $r$. The Lie transported vectors for a rigid congruence remain orthonormal; the Lie transported vectors for the ZAMO congruence, which is non-rigid, do not, which raises the issues that WBN described a few posts ago.

 

[WannabeNewton]

Here yuiop take a look at this: http://books.google.com/books?id=Vc...ic precession schwarzschild spacetime&f=false

Also, please confirm that Bill's result(s) are the negative of the precession relative to the Lie transported vectors, rather than relative to the orthonormal $e_r$ and $e_{\phi}$ vectors (in proper time).

In this case they are one and the same however make note of what Peter said in post #40.

 

[WannabeNewton]

Also, please confirm that Bill's result(s) are the negative of the precession relative to the Lie transported vectors, rather than relative to the orthonormal $e_r$ and $e_{\phi}$ vectors (in proper time).

yuiop, it might help if you think of this as follows. Imagine you're an observer in circular orbit in Minkowski, Schwarzschild, or Kerr space-time. Now picture yourself inside of a negligibly small cubical elevator say at one of its corners. At your corner you've placed an ideal clock that marks off the time of events in your vicinity and you've also bolted down three unit length telescopes one pointing at the corner left-adjacent to you, one pointing at the corner right-adjacent to you, and one pointing to the corner directly above you-notice that by bolting them down like this you're forcing them to keep pointing at the same respective corners of the elevator at each event read off by the clock as well as forcing them to remain perpendicular to one another. Your 4-velocity $u$ can be associated with the clock and the spatial basis vectors $\{e_r,e_{\theta},e_{\phi}\}$ naturally adapted to the circular orbit can be associated with the telescopes.

What then does it mean to have the telescopes bolted down to the elevator so as to keep them permanently pointing at the respective corners of the elevator as well as keep them perpendicular to one another? For simplicity assume the elevator size is such that the telescopes span the distance between you and the respective corners of the elevator. Your world line $\gamma_u$ and your clock are described by $u$ but the fronts of the telescopes have their own world lines $\gamma_r, \gamma_{\theta}, \gamma_{\phi}$ so all together we have four world lines that are jointly described by some time-like vector field $K$. Then the act of bolting down the telescopes amounts to Lie transport of the above spatial basis vectors representing the telescopes such that the flow of $K$ preserves the angles between said vectors i.e. $\mathcal{L}_{K} e_i = 0$ as well as $\mathcal{L}_{K}\langle e_i, e_j \rangle = 0$. Does $\mathcal{L}_{K}e_i = 0 \Rightarrow \mathcal{L}_{K}\langle e_i, e_j \rangle = 0$ for any and all time-like vector fields $K$? No. However this implication will necessarily hold if $K$ is a Killing field i.e. if $\mathcal{L}_{K}g = 0$ where $g$ is the metric tensor which is of course what actually determines angles between vectors that lie in the same tangent space. For a circular orbit we have $u = \partial_t |_{{\gamma_u}} + \omega_u\partial_{\phi}|_{{\gamma_u}}$, where $\omega_u$ is the angular velocity of the orbit, so $K$ will necessarily be of the form $K = \partial_t + \omega_u \partial_{\phi}$.

Now say you have a gyroscope at your corner that's pointing in some arbitrary direction and is free to rotate-no external torques act on the gyroscope. Then you can tell if the elevator is rotating by seeing if the gyroscope precesses relative to the walls of the elevator and by our above construction this is the same thing as seeing if the gyroscope precesses relative to the telescopes i.e. $\{e_r,e_{\theta},e_{\phi}\}$ hence if the above mathematical conditions are met then, as shown in Straumann's text, the precession angular velocity of the gyroscope will be given by $\Omega = -\frac{1}{2}\langle K,K \rangle^{-1}\star (K^{\flat}\wedge dK^{\flat})$ where $\star (K^{\flat}\wedge dK^{\flat})$ is the vorticity of $K$. So that's basically what's going on between vorticity and gyroscopic precession.

 

[yuiop]

I don't think so; that factor is due to the de Sitter precession. I get the same result as Bill_K doing the computation the laborious way using components and connection coefficients.

I was not disputing the result Bill got for the Schwarzschild case as it is the same as the result from other sources. I was questioning how he got the right result with an incorrect intermediate step. Let me see if I can do the intermediate step correctly this time ;)

Given $\frac{dS_{r}}{d\tau} = \gamma\omega(r-3m)S_{\varphi}$ and $\frac{dS_{\varphi}}{d\tau} = -\gamma^3 \omega(r-3m)/r^2$,

$\frac{d^2S_{r}}{d\tau^2} = \frac{d}{d\tau}\left(\frac{dS_{r}}{d\tau}\right) =\frac{d}{d\tau}\left(\gamma\omega(r-3m)S_{\varphi}\right)=\frac{dS_{\varphi}}{d\tau}\frac{d}{dS_{\varphi}} \left(\gamma\omega(r-3m)S_{\varphi}\right) =\frac{dS_{\varphi}}{d\tau}\left(\gamma\omega(r-3m)\right) = -\gamma^2 \omega^2(r-3m)^2/r^2$

This is not the result Bill obtained. His harmonic oscillation method depends on S_r being present in the intermediate result and it is not. I want to be sure that Bill's method for the Schwarzschild case is correct, so that I can have confidence in his result for the Kerr case using the same method. I cannot find an independent source that verifies Bill's result for the Kerr case.

They are precession relative to Lie transported vectors [edit: which are, as WBN notes, the same as $\hat{e}_r$ and $\hat{e}_{\phi}$ for this case] (not sure about the sign), but as I noted in a previous post, those are Lie transported vectors for a rigid congruence with constant angular velocity [edit: which is why they are the same as $\hat{e}_r$ and $\hat{e}_{\phi}$ for this case]. For the Kerr spacetime case, this is a *different* congruence from the ZAMO congruence, which has angular velocity varying with $r$. The Lie transported vectors for a rigid congruence remain orthonormal; the Lie transported vectors for the ZAMO congruence, which is non-rigid, do not, which raises the issues that WBN described a few posts ago.

I was more interested in what Bill's Kerr result is relative to. Is it relative to the $\hat{e}_r$ and $\hat{e}_{\phi}$ orthonormal vectors or is it relative to the Lie transported vectors? As has been pointed out several times in this thread, the Lie transported vectors do not remain orthogonal and the angular velocity of these vectors cannot be represented by a single number. You mentioned earlier that we might be able to represent the Lie transported vectors by a sort of average angular velocity so that we might be able to quantify this quantity. After looking up some stuff on the angular momentum of fluid elements in a vortex I came upon a method that does what you suggest. Imagine a paddle wheel on its side in the flow, so that all the paddles are immersed. The paddle wheel acquires an angular velocity that is representative of the average angular momentum of the fluid elements of the local flow. We could paint orthogonal axes on this paddle wheel and these axes would of course remain orthogonal and (may?) represent the averaged Lie transported basis vectors. For an irrotational flow, such that an axis on the paddle wheel always points at a fixed point at infinity, the velocity of the flow has to be inversely proportional to the orbital radius. as far as I can tell, the congruence of ZAMOs in the Kerr metric has a velocity that is inversely proportional to $r^3$ to a good approximation outside the photon orbit.

If we represent the equation for the orbital velocity $\omega_{o}$ of a congruence as a function of radius so that $\omega_o = k*r^p$ where k is a constant then the rotational velocity $\omega_L$ of our paddle wheel can be approximated by $\omega_L = \omega_o(1+p)$. For a rigid flow, such that $\omega_o$ is independent of radius, so that p=0, the result is $\omega_L = \omega_o$ and an axis on the paddle wheel that points at the centre of the vortex continues to point at the centre. If $\omega_o$ is inversely proportional to r such that p=-1, the result is that $\omega_L = 0$ and the paddle wheel does not rotate relative to the distant stars. If $\omega_o$ is inversely proportional to $r^2$ such that p=-2, the result is that $\omega_L = -\omega_o$ and the paddle wheel rotates in the opposite sense to the vortex rotation. For the Kerr case, assuming an approximation of p=-3, the averaged angular velocity of the Lie transported basis vectors relative to a fixed point at infinity, is approximately twice the ZAMO orbital angular velocity and retrograde by this estimation.

This is of course an approximation and does not always hold near the centre of the vortex. For more information on the ideas outlined here, see

http://maxwell.ucdavis.edu/~cole/phy9b/notes/fluids_ch3.pdf
http://en.wikipedia.org/wiki/Vortex
http://www2.math.umd.edu/~jcooper/math241/curl.pdf

Hopefully the above references will help you come up with a more rigorous mathematical definition.

 

[WannabeNewton]

I was more interested in what Bill's Kerr result is relative to. Is it relative to the $\hat{e}_r$ and $\hat{e}_{\phi}$ orthonormal vectors or is it relative to the Lie transported vectors?

Again, they are the same. We don't need to use the ZAMO congruence in order to determine the gyroscopic precession along a single ZAMO orbit. We can use a Killing field to which this single orbit belongs, Lie transport $\{e_r,e_{\theta},e_{\phi}\}$, and calculate the gyroscopic precession along this orbit using the usual methods. I've already done this in post #36.

 

[PeterDonis]

I was questioning how he got the right result with an incorrect intermediate step.

As WBN pointed out earlier, there's a typo in there:

Given $\frac{dS_{r}}{d\tau} = \gamma\omega(r-3m)S_{\varphi}$ and $\frac{dS_{\varphi}}{d\tau} = -\gamma^3 \omega(r-3m)/r^2$

There should be a factor of $S_r$ on the RHS of the second equation.

 

[PeterDonis]

As has been pointed out several times in this thread, the Lie transported vectors do not remain orthogonal

You left out a key qualifier: the Lie transported vectors of the ZAMO congruence do not remain orthogonal. Bill_K's computation did not use the ZAMO congruence; it used a congruence with constant angular velocity (the Killing congruence that WBN mentioned). The angular velocity of the ZAMO congruence varies with $r$ (and $\theta$ if you're outside the equatorial plane).

 

[PeterDonis]

We don't need to use the ZAMO congruence in order to determine the gyroscopic precession along a single ZAMO orbit.

Strictly speaking, you don't need to use *any* congruence to compute the precession itself; you can just use Fermi-Walker transport along the single worldline. The only role the congruence plays is in defining the Lie transported vectors relative to which the precession is expressed (and strictly speaking, you don't need a congruence for that either, any set of orthonormal basis vectors defined all along the worldline will do).

 

[PeterDonis]

You mentioned earlier that we might be able to represent the Lie transported vectors by a sort of average angular velocity so that we might be able to quantify this quantity.

Yes, I'm still working on that. But one key observation, following on from the discussion of Bill_K's computation, is that changing the congruence we use to define the Lie transported vectors doesn't change the behavior of the gyroscopes (the Fermi-Walker transported vectors) at all; it only changes the Lie transported vectors, which means it changes the *relative* behavior of the F-W vectors with respect to the Lie transported vectors.

In the ZAMO case in Kerr spacetime, I think that the Lie transported $\hat{e}_{\phi}$ will be the same as the $\hat{e}_{\phi}$ of the rigid congruence Bill_K used (since that vector already points towards the neighboring ZAMO in the same circular orbit). So the difference between the two congruences (ZAMO vs. rigid) is in the behavior of the $\hat{e}_r$ Lie transported vectors; in the rigid congruence, they stay orthogonal to $\hat{e}_{\phi}$, whereas in the ZAMO congruence, they don't.

If this is correct, then the results Bill_K and WBN derived will correctly give the precession of gyroscopes relative to $\hat{e}_{\phi}$ in the ZAMO case. What still needs to be computed (and what I'm working on) is the behavior of the $\hat{e}_r$ vectors in the ZAMO congruence. Your paddle wheel analogy is interesting and I'll look at the links you gave to see if it helps in the computation I'm doing.

 

[WannabeNewton]

Strictly speaking, you don't need to use *any* congruence to compute the precession itself; you can just use Fermi-Walker transport along the single worldline.

True, I should have clarified that I was referring to a calculation of the gyroscopic precession that made use of the vorticity. This has the disadvantage of requiring a Killing field and Lie transport of a frame by the Killing field but it has the advantage of being much simpler computationally.

The only role the congruence plays is in defining the Lie transported vectors relative to which the precession is expressed (and strictly speaking, you don't need a congruence for that either, any set of orthonormal basis vectors defined all along the worldline will do).

Well mathematically the Lie derivative is only defined for two vector fields so what we're really Lie transporting is a frame field along a Killing field. Only the Fermi-derivative and the covariant derivative require a single curve (which is another reason for why we should be cautious of the relationship between vorticity and gyroscopic precession in the case of non-Killing fields).

 

[PeterDonis]

mathematically the Lie derivative is only defined for two vector fields

Ah, yes, you're right, you need some specification of at least one other vector field (other than the 4-velocity).

 

[yuiop]

I could use some expert assistance here! With reference to the attached paper n gyroscope precession by Rindler, would someone be kind enough to calculate $e^{\varphi}$, $\omega_{3,1}$, $h^{11}$ and $h^{33}$ from the canonical form of the Kerr metric? Rindler does the calculations on page 12 but he specialises the results to the specific case of a grid rotating at the geodesic angular velocity. I need the more general case for any angular velocity (but still remaining restricted to the equatorial plane).

This would assist greatly in comparing Rindler's results with results I have obtained from other sources.

P.S. If nothing else, just the solution for $\omega_{3,1}$ would be a help. I think i can figure the others out for myself. (maybe).

 

[yuiop]

This paper http://www.ias.ac.in/jarch/jaa/20/103-120.pdf might be of interest. it gives a general equation (35) for the precession of a gyroscope in the Kerr metric. there appears to be a typo in the denominator of the equation as it does not reduce to the Schwarzschild case (36) when a=0. They give the denominator as:

$\left[1-(r^2+a^2)\omega^2 - \frac{2M(a\omega)^2}{r}\right]$

I think it is intended to be the Kerr time dilation factor for an object in a circular orbit and should actually be:

$\left[1-2M/r -(r^2+a^2)\omega^2 - \frac{2M(a\omega)^2}{r}+\frac{4Ma\omega}{r}\right]$

 

[PeterDonis]

would someone be kind enough to calculate $e^{\varphi}$, $\omega_{3,1}$, $h^{11}$ and $h^{33}$ from the canonical form of the Kerr metric?

You can read them off from equation 14, which is the Kerr metric in canonical form for arbitrary $\omega$. I get

$$
e^{2 \psi} = 1 - \omega^2 \left( a^2 + r^2 \right) - \frac{2M}{r} \left( 1 - a \omega \right)^2
$$

(the above is actually given as equation 15)

$$
h^{11} = h^{rr} = 1 - \frac{2M}{r} + \frac{a^2}{r^2}
$$

$$
h^{33} = h^{\phi \phi} = \frac{r^2 - 2 m r + a^2}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2}
$$

$$
w_3 = w_{\phi} = \frac{\omega \left( a^2 + r^2 \right) - \left( 2 a M / r \right) \left( 1 - a \omega \right)}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2}
$$

Then we need to differentiate the last equation with respect to $r$ to get

$$
w_{3, 1} = w_{\phi , r} = \frac{2 \omega r + \left( 2 a M / r^2 \right) \left( 1 - a \omega \right)}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2} - \frac{\omega \left( a^2 + r^2 \right) - \left( 2 a M / r \right) \left( 1 - a \omega \right)}{\left[ 1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2 \right]^2} \left[ - 2 \omega^2 r + \frac{2M}{r^2} \left( 1 - a \omega \right)^2 \right]
$$

If I've done the algebra right, this simplifies to

$$
w_{3, 1} = w_{\phi , r} = \frac{2 \omega r - 2 \omega M \left[ 2 + \left( a^2 + r^2 \right) / r^2 \right] \left( 1 - a \omega \right) \left( 1 - 2 a \omega \right) + \left( 2 a M / r^2 \right) \left( 1 - a \omega \right)}{\left[ 1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2 \right]^2}
$$

which doesn't look very enlightening, so it's quite possible that I've made an algebra error.

 

[yuiop]

... which doesn't look very enlightening, so it's quite possible that I've made an algebra error.

Thanks Peter! That is really awesome! :thumbs:

The first easy check is for the $\omega=0$ case and I get $\frac{ma}{r^3(1-2m/r)}\frac{2(r^2-2mr-a^2)}{(1-2m/r)}$ so there might be a problem there. I will check it all out thoroughly. You have given me a solid start. Thanks again!

 

[yuiop]

$$
h^{33} = h^{\phi \phi} = \frac{r^2 - 2 m r + a^2}{1 - \omega^2 \left( a^2 + r^2 \right) - \left( 2 M / r \right) \left( 1 - a \omega \right)^2}
$$

From what I can glean from the Rindler paper, $h^{33}$ should be inverted, which simplifies things a lot due to cancellations.

With that change, I now get the correct result for the w=0 case. I will check the rest later. Thanks again for your help. Thanks to you, I think I understand the paper a whole lot better now!

 

[PeterDonis]

From what I can glean from the Rindler paper, $h^{33}$ should be inverted

Hm, yes, it looks like what I wrote down is $h_{33}$ (and $h_{11}$ as well), so if you want the inverse it would be $h^{33} = 1 / h_{33}$ (since the metric is diagonal).

 

[yuiop]

Hm, yes, it looks like what I wrote down is $h_{33}$ (and $h_{11}$ as well), so if you want the inverse it would be $h^{33} = 1 / h_{33}$ (since the metric is diagonal).

I think your $h^{11}$ is OK. It is already the inverse of the factor in the canonical form of the Kerr metric.

I put it all into maple software (worksheet attached) and this is the end result with a little manual simplification:

$$
\Omega = \frac{2}{r^3}\frac{ma\omega^2(a^2+3r^2)+ma(1-2a\omega)+\omega (1-3m/r)r^3 }{1 - \omega^2 \left( a^2 + r^2 \right) - \frac{2M}{r} \left( 1 - a \omega \right)^2}
$$

Other than the factor of 2, the result agrees with other sources for $\omega=0$ case and the ZAMO angular velocity. The result also collapses correctly to the Schwarzschild and Minkowski forms for $a=0$ and $m=o$ respectively.

Rindler consistently leaves out the factor of 2 that appears in the derivative of $w_3$ in all his derivations. I notice that a factor of 2 is also left out when reading off $g_{t\phi}$ from the Kerr metric in most references so it seems standard practice, but I am not sure of the reason why. Can anyone enlighten me?

 

[PeterDonis]

Rindler consistently leaves out the factor of 2 that appears in the derivative of $w_3$ in all his derivations. I notice that a factor of 2 is also left out when reading off $g_{t\phi}$ from the Kerr metric in most references so it seems standard practice, but I am not sure of the reason why. Can anyone enlighten me?

If you're reading $g_{t \phi}$ off the line element, you have to divide by 2 because the line element counts both $g_{t \phi}$ and $g_{\phi t}$, which are equal since the metric is symmetric, but both have to be counted in the line element because the summation $ds^2 = g_{ab} dx^a dx^b$ includes "cross terms" twice.

 

[yuiop]

If you're reading $g_{t \phi}$ off the line element, you have to divide by 2 because the line element counts both $g_{t \phi}$ and $g_{\phi t}$, which are equal since the metric is symmetric, but both have to be counted in the line element because the summation $ds^2 = g_{ab} dx^a dx^b$ includes "cross terms" twice.

Is it a similar argument for why Rindler leaves out the factor of 2 for the $w_{3,1}$ term?

 

[PeterDonis]

Is it a similar argument for why Rindler leaves out the factor of 2 for the $w_{3,1}$ term?

I don't think so, because if I'm understanding the formulas right, $w_{i, j}$ is not a symmetric tensor whose off-diagonal terms appear twice in a "line element" type of expression; $w_i$ appears in the canonical line element, but only once, not "double counted", since it only has one index and it's contracted with $dx^i$.

 

[yuiop]

I don't think so, because if I'm understanding the formulas right, $w_{i, j}$ is not a symmetric tensor whose off-diagonal terms appear twice in a "line element" type of expression; $w_i$ appears in the canonical line element, but only once, not "double counted", since it only has one index and it's contracted with $dx^i$.

I think I have an explanation. I suspect Rindler's equation $\Omega = \epsilon^{\varphi}(h^{11}h^{33}\omega^2{[3,1]})^{1/2}$ is actually the curl (or vorticity?) and the rotational velocity is 1/2 that value. See line (11) of this paper. (It mentions my favourite object, the paddle wheel ).

 

[yuiop]

Section 3 of the paper linked in the last post mentions that a paddle wheel with a single blade would rotate unevenly which is logical in the fluid context. However, it takes me back to an issue that was possible unresolved in these threads. If we have two independent gyroscopes in at the same location in the Kerr metric, one with its spin axis parallel to $e_{\phi}$ (tangential to a circular orbit) and the other with its spin axis parallel to $e_r$, (so pointing outward from the gravitational body centre), would the $e_r$ gyroscope precess faster relative to $e_r$ than the other gyroscope precesses relative $e_{\phi}$? I ask this, because it seems the gyroscope parallel to $e_{\phi}$ is subject to much less sheer or frame dragging than the other gyroscope. Ihis is pretty much the picture painted by Ohanian. If any of the above is true, it would of course imply the gyroscopes would not remain orthogonal to each other. Is there any law that says untorqued gyroscopes that are initially orthogonal must remain orthogonal in any kind of gravitational field, if not subjected to torque forces?

 

[WannabeNewton]

If we have two independent gyroscopes in at the same location in the Kerr metric, one with its spin axis parallel to $e_{\phi}$ (tangential to a circular orbit) and the other with its spin axis parallel to $e_r$, (so pointing outward from the gravitational body centre), would the $e_r$ gyroscope precess faster relative to $e_r$ than the other gyroscope precesses relative $e_{\phi}$?

They must precess at the same rate because they always remain orthogonal. See below.

Is there any law that says untorqued gyroscopes that are initially orthogonal must remain orthogonal in any kind of gravitational field, if not subjected to torque forces?

Yes. Fermi-transport preserves orthogonality. This is why we often just speak of a compass of inertia instead of referring to individual gyroscopes.

I ask this, because it seems the gyroscope parallel to $e_{\phi}$ is subject to much less sheer or frame dragging than the other gyroscope.

Well frame dragging is not exactly the same thing as shear but they're related in this case so no harm there. However the gyroscopes aren't the things subjected to the shear but rather the connecting vectors are the objects subjected to the shear. The connecting vectors are by definition the vectors Lie transported by the congruence whereas the gyroscopes are Fermi-transported by the congruence. You can think of the connecting vectors as representing the paddles of the wheel as I already explained in the other thread.

 

[yuiop]

They must precess at the same rate because they always remain orthogonal. See below.

I have started a new thread just to satisfy myself there are no counter examples in a relatively easily analysed situation (hopefully).

The connecting vectors are by definition the vectors Lie transported by the congruence whereas the gyroscopes are Fermi-transported by the congruence. You can think of the connecting vectors as representing the paddles of the wheel as I already explained in the other thread.

Earlier in these threads it was stated that $\Omega$ was the precession of the Lie Transported Basis Vectors relative to a set of gyroscopes and that the negative of this was the precession on the gyroscopes relative to the LTBVs. Has this now been retracted? Some sources state that $-\Omega$ is the precession of the gyroscopes relative to the orthonormal basis vectors $e_r$ and $e_{phi}$ which is a different kettle of fish. This latter viewpoint means all the talk of LTBVs was irrelevant to the issue of gyroscope precession. The former viewpoint means we need to quantify the rotation of the LTBVs relative to a fixed point at infinity or to the the orthonormal basis vectors, if we want to quantify the precession of the gyroscopes relative to a fixed point at infinity.

Finally just for clarity, consider this thought experiment in Kerr spacetime. We have an orbiting lab. To measure rotation, the lab operators has a device in the centre with lightly sprung weights that essentially measure centrifugal force. The weights are thrown outwards if the device is rotating. They also have some large weights that they can position at the out limits of the lab. If the lab is rotating, when they bring the weights to the centre of the lab, the intrinsic spin of the lab will increase. The lab has small orientation thrusters that can correct its rotation. They adjust the rate of rotation of the lab until it has no rotation according to the two lab devices. They can place the large weights wherever they like in the lab and there will be no detectable change. This to me, would be a sensible way of defining locally non rotating. Now if we introduce some gyroscopes into this lab, will they precess relative to the the lab? Intuition would say no, but I just want to be sure.

 

[WannabeNewton]

Earlier in these threads it was stated that $\Omega$ was the precession of the Lie Transported Basis Vectors relative to a set of gyroscopes and that the negative of this was the precession on the gyroscopes relative to the LTBVs. Has this now been retracted?

No that's perfectly fine but you cannot have a Lie transported set of spatial basis vectors if there is shear. Shear by definition is the deformation of connecting vectors, which are Lie transported vectors that are not spatial basis vectors (if they were then the only way they could be Lie transported is if there is no shear as already mentioned).

Some sources state that $-Omega$ is the precession of the gyroscopes relative to the orthonormal basis vectors $e_r$ and $e_{phi}$ which is a different kettle of fish. This latter viewpoint means all the talk of LTBVs was irrelevant to the issue of gyroscope precession.

We've already addressed this multiple times in the thread. See posts #38-41 as well as posts #47 and #49. Is the math posing a hurdle?

Now if we introduce some gyroscopes into this lab, will they precess relative to the the lab? Intuition would say no, but I just want to be sure.

What you've basically done is define a Fermi-transported laboratory (so no they won't). Making the centrifugal forces vanish in a given coordinate system is equivalent to having the coordinate system be Fermi-transported. Another way to do this is to have the observer in the laboratory place a mirror up against any of the walls such that the plane of the mirror is orthogonal to the line joining the mirror to the center of the laboratory-if the observer emits a beam of light from the center towards the mirror then it will be reflected back to the center if and only if the laboratory is non-rotating. However this definition of "locally non-rotating" is applicable to any observer whatsoever in any space-time. It's not the same sense in which "locally non-rotating" is being used when referring to zero angular momentum observers. In the latter case "locally non-rotating" is referring to the vanishing angular momentum which consequently makes the Sagnac effect vanish for these observers.