Practical measurements of rotation in the Kerr metric

[WannabeNewton]

Equation 47 in the paper.

So there are definitely some interpretational variations here.

Thanks! The flurry of differences regarding the decomposition of gyroscopic precession is making my head spin

This is true, but the difference is just a correction factor of $\gamma = \sqrt{1 / \left( 1 - 2M / r - \omega^2 r^2 \right)}$ multiplying equation 4 in the paper you linked to, which doesn't resolve the discrepancy.

Ah ok I see now what you were originally referring to. I'm burnt out right now but let me see what I can come up with (of course knowing you, you'll probably figure it out before I do )

 

[yuiop]

...
I'm not sure how to resolve this discrepancy.

Since we are looking for an error, Mentz mentioned their might be one in the original paper back in post #77 which I have quoted below:

I have a query about the paper referred to earlier ( http://arxiv.org/pdf/1210.6127v4.pdf). It has a section on the Kerr metric at the top of the second column on page 4. I think the frame basis and cobasis vectors given are wrong, in that they do not give the $d\phi^2$ term in the metric above. This is easy to see because the contributions to the $d\phi^2$ term come from $e^\hat{t}$ and $e^\hat{\phi}$, i.e. $\left( \frac{2\,a\,m\,r\,{\sin\left( \theta\right) }^{2}}{\sqrt{\Sigma}\,\sqrt{\Sigma-2\,m\,r}} \right)^2 + \left(\frac{\sin\left( \theta\right) \,\sqrt{\Delta}\,\sqrt{\Sigma}}{\sqrt{\Sigma-2\,m\,r}} \right)^2 = \frac{{\sin\left( \theta\right) }^{2}\,\left( \Delta\,{\Sigma}^{2}+4\,{a}^{2}\,{m}^{2}\,{r}^{2}\,{\sin\left( \theta\right) }^{2}\right) }{\Sigma\,\left( \Sigma-2\,m\,r\right) }$.

Have I made a mistake, or missed something ?

 

[yuiop]

which can be rewritten, using $\omega = \sqrt{M / r^3}$, as

$$
\Omega = \frac{3}{2} \frac{M}{r} \omega
$$

However, if we look at the details of the derivation (leading up to equation 48 in the paper), we see that the formula as I have just written it is actually generally applicable to any angular velocity, not just a geodesic orbit, at least in the weak field approximation (we substitute $\omega r$ for $v$ in the general formulas in the paper and use $\nabla \phi = M / r^2$ in the Newtonian limit). So this latter formula should be compared with the general formula from the other paper,

$$
\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}
$$

I'm not sure how to resolve this discrepancy...

As mentioned before the equation $\Omega = \omega \frac{1 - 3M / r}{1 - 2M / r - \omega^2 r^2}$ implies a geodesically orbiting gyroscope remains locked on the distant stars which is obviously not correct.

It turns out that this is the expression for the rotation of a gyroscope axis relative to the LTSB vectors in terms of the proper time of the orbiting gyroscope. This is effectively the rate at which a gyroscope must rotate relative to the LTSB vectors (in the opposite sense to the orbital angular velocity) in order to remain fixed fixed on the distant stars, but because this is in proper time, it fails to do so when corrected for coordinate time.

When we apply the correction gamma factor $\gamma = \sqrt{1 / \left( 1 - 2M / r - \omega^2 r^2 \right)}$ (as mentioned by WBN) the correct expression becomes:

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

so that now both $\Omega$ and $\omega$ are coordinate angular velocities measured by an observer at infinity.

For a geodesic orbit, $\omega^2 = M/r^3$ and when substituted into the above equation we get:

$\Omega = \omega \sqrt{1 - 3M / r}$.

What we want is the precession rate relative to the distant stars rather than the LTSB vectors , so we must subtract the above expression from the orbital velocity to obtain:

$\Omega_{stars} = \omega \left(1-\sqrt{1 - 3M / r} \right)$

When we carry out a series expansion of $\Omega_{stars}/\omega$ we get:

$\Omega_{stars}/\omega = (3 M)/(2 r)+(9 M^2)/(8 r^2)+(27 M^3)/(16 r^3)+(405 M^4)/(128 r^4) ...+O(M^7)$

which makes it clear that the equation in the Gravity Probe B paper:

$$
\Omega = \frac{3}{2} \frac{M}{r} \omega
$$

is just an approximation of $\Omega_{stars}$ as defined above.

 

[PeterDonis]

It turns out that this is the expression for the rotation of a gyroscope axis relative to the LTSB vectors in terms of the proper time of the orbiting gyroscope. This is effectively the rate at which a gyroscope must rotate relative to the LTSB vectors (in the opposite sense to the orbital angular velocity) in order to remain fixed fixed on the distant stars, but because this is in proper time, it fails to do so when corrected for coordinate time.

Ah, of course. So really the formula (for an arbitrary orbit, not necessarily a geodesic one) should read, using our earlier notation,

$$
\Omega = \gamma^2 \omega_{\infty} \left( 1 - \frac{3M}{r} \right)
$$

where $\Omega$, without the subscript $\infty$, is relative to the proper time of the orbiting object. To convert this to a frequency relative to infinity, we apply the correction factor of $\gamma^{-1}$ to obtain

$$
\Omega_{\infty} = \gamma \omega_{\infty} \left( 1 - \frac{3M}{r} \right)
$$

For an orbiting object, $\gamma = 1 / \sqrt{1 - 3M / r}$, and the rest follows. Cool!

Btw, this means the notation in the paper WBN linked to (which is apparently fairly common since a paper by Rindler is referenced that gives the same result) is rather misleading; by stating the result for an orbiting object as $\Omega = \omega$ it invites the incorrect interpretation that I made. Using the $\infty$ subscript makes it clear that $\Omega = \omega_{\infty}$ is comparing apples and oranges; the "apples to apples" result is $\Omega_{\infty} = \gamma^{-1} \omega_{\infty}$.

 

[Mentz114]

Since we are looking for an error, Mentz mentioned their might be one in the original paper back in post #77 which I have quoted below:

I corresponded with the second author (梁為傑) and we concluded that the frame and coframe bases are correctly written in the paper. I disagree with the components of the vorticity vector, but get the same value for $\Omega=\left( \omega^\mu \omega_\mu \right)^{(1/2)}$ as they do.

 

[WannabeNewton]

which makes it clear that the equation in the Gravity Probe B paper:
...
is just an approximation of $\Omega_{stars}$ as defined above.

Awesome thanks! That clears up everything.

 

[WannabeNewton]

I thought I should bring something up that I completely missed at the inception of this thread. Consider the congruence of ZAMOs in Kerr space-time with 4-velocity field $u$; the most natural frame field for this congruence is: u = e_{t} = e^{-\nu}(\partial_t + \omega \partial_{\phi})\\ e_{r} = e^{-\mu}\partial_{r}\\ e_{\theta} = e^{-\lambda}\partial_{\theta}\\ e_{\phi} = e^{-\psi}\partial_{\phi} where $e^{2\nu} = \frac{\rho^2 \Delta}{\Sigma^2}, e^{2\mu} = \frac{\rho^2}{\Delta}, e^{2\psi} = \frac{\Sigma^2}{\rho^2}\sin ^2\theta, e^{2\lambda} = \rho^2, \omega = \frac{2Ma r}{\Sigma^2}$ in Boyer–Lindquist coordinates.

For future convenience, let $\eta = \partial_t + \omega \partial_{\phi}$ and let $\psi^{\mu} = (\partial_{\phi})^{\mu}$. We know that $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ i.e. that the vorticity satisfies $\omega^{\mu} = \epsilon^{\mu\nu\gamma\delta}\eta_{\nu}\nabla_{\gamma}\eta_{\delta} = 0$. However keep in mind that $\nabla_{(\mu}\eta_{\nu)} \neq 0$, that is, the congruence of ZAMOs is not rigid. This is because the angular velocity $\omega$ is not constant across the entire congruence but rather only along the worldlines of individual ZAMOs.

Note that $\mathcal{L}_{u}\psi^{\mu} = 0$ and $\psi_{\mu}a^{\mu} = \psi_{\mu}u^{\mu} = 0$ where $a^{\mu}$ is the 4-acceleration of the ZAMOs.

Therefore $h^{\mu}{}{}_{\nu}u^{\gamma}\nabla_{\gamma}(h^{\nu}{}{}_{\delta}\psi^{\delta}) = u^{\gamma}\nabla_{\gamma}\psi^{\mu} = F_{u}\psi^{\mu}$ where $h_{\mu\nu}$ is the spatial metric relative to the ZAMOs and $F_{u}\psi^{\mu}$ is the Fermi-Walker derivative along the worldline of an individual ZAMO. Note also that $F_{u}e^{\mu}_{\phi} = e^{-\psi}F_{u}\psi^{\mu} + \psi^{\mu}(u^{\nu}\partial_{\nu}e^{-\psi}) = e^{-\psi}F_{u}\psi^{\mu}$.

Recall from Malament's text that $h^{\mu}{}{}_{\nu}u^{\gamma}\nabla_{\gamma}(h^{\nu}{}{}_{\delta}\psi^{\delta}) = 0$ if and only if $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ (see p.223: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf). If we apply this proposition to the congruence of ZAMOs and use the results from the previous paragraph then $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} = 0$ would imply that $F_{u}e_{\phi} = 0$.

In other words, a gyroscope carried by a ZAMO observer would not precess relative to the $e_{\phi}$ axis of the natural frame we've attached to this ZAMO. But from exercise 33.4 in MTW, such a gyroscope precesses relative to the $e_{r}$ axis of this frame with an angular velocity $\Omega^{r} \propto \partial_{\theta}\omega / \rho$. If the gyroscope precesses relative to $e_r$ then it also has to precess relative to $e_{\phi}$.

There's no contradiction of course because the proposition in Malament's text required $\eta$ to be a Killing field i.e. it required the congruence described by $\eta$ to be rigid everywhere and not just when restricted to a single ring (see p.221 and p.223). The fact that the congruence of ZAMOs is not rigid (as mentioned before) lends to a gyroscopic precession in the natural ZAMO frame defined above. From exercise 33.4 in MTW we in fact see that the gyroscopic precession in this frame exists only because $\omega$ is not constant everywhere across the congruence which is exactly what prevents the congruence of ZAMOs from being rigid.

I just thought I should make note of that because I totally missed it the first time I read Malament's text.

 

[PeterDonis]

from exercise 33.4 in MTW, such a gyroscope precesses relative to the $e_{r}$ axis of this frame with an angular velocity $\Omega^{r} \propto \partial_{\theta}\omega / \rho$.

But note that in the equatorial plane, $\theta = \pi / 2$, we have $\partial_{\theta} \omega = 0$, so this precession goes to zero in the equatorial plane, which is where we've been doing most of our analysis in this thread. It is definitely worth remembering, though, that Kerr spacetime is only axisymmetric, so an analysis restricted to the equatorial plane leaves out significant physics.

 

[WannabeNewton]

I do have a related comment though. On p.221, Malament states the following: "We want to think of the ring as being in a state of rigid rotation, i.e., rotation with the distance between points on the ring remaining constant. So we are further led to restrict attention to just those congruences of timelike curves on $\mathcal{R}$ that are invariant under all isometries generated by $\tilde{t}^a$. Equivalently (moving from the curves themselves to their tangent fields), we are led to consider future-directed timelike vector fields on $\mathcal{R}$ of the form $(\tilde{t}^a+ k \phi^a)$, where $k$ is a number."

Here $\mathcal{R}$ represents the world tube of the ring. However I have a slight problem with this definition of rigid rotation of the ring. The definition starts out by saying that the congruence of time-like curves forming $\mathcal{R}$ must be invariant under all isometries generated by the time-like killing field $\tilde{t}^a$ which is fine but note that even the ring we obtain by restricting the congruence of ZAMOs in Kerr space-time to a single radius $R$ satisfies this condition because the tangent field to such a ring would be $\eta^a = \tilde{t}^a + \omega \phi^a$, where $\frac{2MaR}{\Sigma^2}$, hence $\mathcal{L}_{\tilde{t}}\eta^{a} = \mathcal{L}_{\tilde{t}}\tilde{t}^a + \omega \mathcal{L}_{\tilde{t}}\phi^a + \phi^a \tilde{t}^a \partial_{a}\omega =0$.

But as we know $\omega$ is not constant everywhere in space-time i.e. $\partial_{\theta} \omega, \partial_{r}\omega \neq 0$ identically. However the definition quoted above then subsequently says that the ring satisfies rigid rotation if its tangent field is of the form $\eta^a = \tilde{t}^a + k \phi^a$ where $k$ is constant everywhere in space-time*. This would disqualify the ring in the previous paragraph from being rigid because $\omega$ is not constant everywhere in space-time, only on the ring itself. However we know for a fact that the ring from the previous paragraph satisfies Born rigidity. Isn't the definition quoted above a much stronger condition than simply requiring that the ring exhibit rigid rotation? By having $k$ be constant everywhere in space-time, the definition is requiring that all rings formed by $\eta^a = \tilde{t}^a + k \phi^a$ be rigidly rotating with respect to one another right?

*It isn't clear from the wording whether $k$ is meant to be constant just on $\mathcal{R}$ or everywhere in space-time beccause all that's stated is "$k$ is a number". However it is made evident later on, albeit indirectly, that $k$ is constant everywhere in space-time because Malament writes the zero vorticity condition $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ as $\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + k\tilde{t}_{[a}\nabla_{b}\phi_{c]} + k\phi_{[a}\nabla_{b}\tilde{t}_{c]} + k^2 \phi_{[a}\nabla_{b}\phi_{c]} = 0$ (p.222) which clearly assumes that $k$ is constant everywhere in space-time otherwise there would be a term $\tilde{t}_{[a}\phi_b \nabla_{b]} k$.

 

[PeterDonis]

But as we know $\omega$ is not constant everywhere in space-time i.e. $\partial_{\theta} \omega, \partial_{r}\omega \neq 0$ identically.

It isn't for the ZAMO congruence, correct. But you could pick another congruence where $\omega$ *was* constant everywhere (subject to some limitations--see below).

By having $k$ be constant everywhere in space-time, the definition is requiring that all rings formed by $\eta^a = \tilde{t}^a + k \phi^a$ be rigidly rotating with respect to one another right?

Yes. More precisely, if you have a congruence that includes worldlines with different $r$ and/or $\theta$, the congruence will only be rigid if the "rings" at different $r$ and $\theta$ are all rigidly rotating with respect to each other, which means that the angular velocity $\omega$ (which is set by the constant $k$ in the 4-velocity field of the congruence) must be the same for all "rings". This in turn implies that at most one of these "rings" can be a ring of ZAMOs, i.e., only at at most one $(r, \theta)$ in the congruence will $\omega$, the angular velocity of the entire congruence, be equal to the ZAMO angular velocity $\Omega ( r, \theta )$, which of course, as the notation shows, varies with $r$ and $\theta$. (Actually, technically, there could be two such values of $(r, \theta)$ if the congruence was symmetric about the equatorial plane, since the ZAMO angular velocity is a function of $\sin^2 \theta$ and $\sin \theta = \sin \left( \pi - \theta \right)$.)

Of course such a rigid congruence will also be limited in spatial extent, since for fixed $\omega$ there will be a "boundary" in the $( r, \theta )$ plane at which the tangential velocity reaches the speed of light, so the congruence can't extend further outward than this boundary.

 

[yuiop]

But as we know $\omega$ is not constant everywhere in space-time i.e. $\partial_{\theta} \omega, \partial_{r}\omega \neq 0$ identically.

$\omega$ can be constant everywhere (within boundaries) if we do not insist that all the "other" rings also be ZAMO's, but I am not sure that has been specified. I think that is basically what Peter is saying.

However the definition quoted above then subsequently says that the ring satisfies rigid rotation if its tangent field is of the form $\eta^a = \tilde{t}^a + k \phi^a$ where $k$ is constant everywhere in space-time*. This would disqualify the ring in the previous paragraph from being rigid because $\omega$ is not constant everywhere in space-time, only on the ring itself. However we know for a fact that the ring from the previous paragraph satisfies Born rigidity. Isn't the definition quoted above a much stronger condition than simply requiring that the ring exhibit rigid rotation? By having $k$ be constant everywhere in space-time, the definition is requiring that all rings formed by $\eta^a = \tilde{t}^a + k \phi^a$ be rigidly rotating with respect to one another right?

Condition (3.2.1) on page 222 is a condition that must hold if the ring is to qualify as non rotating by the CIR criterion, so it does not explicitly state that this is a condition that must hold on $\mathcal{R}$ for the ring to qualify as having rigid rotation. I think the qualifier "must hold on $\mathcal{R}$" may be significant. See below.

*It isn't clear from the wording whether $k$ is meant to be constant just on $\mathcal{R}$ or everywhere in space-time because all that's stated is "$k$ is a number". However it is made evident later on, albeit indirectly, that $k$ is constant everywhere in space-time because Malament writes the zero vorticity condition $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ as $\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + k\tilde{t}_{[a}\nabla_{b}\phi_{c]} + k\phi_{[a}\nabla_{b}\tilde{t}_{c]} + k^2 \phi_{[a}\nabla_{b}\phi_{c]} = 0$ (p.222) which clearly assumes that $k$ is constant everywhere in space-time otherwise there would be a term $\tilde{t}_{[a}\phi_b \nabla_{b]} k$.

Page 231 makes it clear that k is a function of r (it gives an example in the Godel spacetime) so for k to be a constant, r must equal $\mathcal{R}$. This is also implied in the qualifier to condition (3.2.1) that it "must hold on $\mathcal{R}$".

Page 237 states that if we have two concentric rings that have Born rigid rotation with respect to each other, then if one ring qualifies as non-rotating by a given criteria, then the other ring can fail the non-rotating test using the same criteria. This is certainly true in the equatorial plane in the Kerr metric.

P.S. I hope I am not speaking out of turn

 

[WannabeNewton]

This in turn implies that at most one of these "rings" can be a ring of ZAMOs, i.e., only at at most one $(r, \theta)$ in the congruence will $\omega$, the angular velocity of the entire congruence, be equal to the ZAMO angular velocity $\Omega ( r, \theta )$, which of course, as the notation shows, varies with $r$ and $\theta$.

Ah right so let's say we have a ring of ZAMOs in Kerr space-time with some angular velocity $\omega_0 = \omega|_{r_0,\theta_0}$. We then consider the timelike congruence with tangent field $\eta^a = \tilde{t}^a + \omega_0 \phi^a$. This is of course not the congruence of ZAMOs but the ring generated by $\eta^a$ at $(r_0,\theta_0)$ will match the respective ring of ZAMOs. Then all the properties we derive for $\eta^a$ will also apply to the respective ring of ZAMOs simply by restricting $\eta^a$ to $(r_0,\theta_0)$.

In particular if we find that $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ then we know Fermi-transport of the spatial direction $\hat{\varphi}^{a} = \frac{h^{a}{}{}_{m}\phi^n}{\left \| h^{a}{}{}_{m}\phi^n\right \|}$ of $\phi^a$ holds: $F_{\hat{\eta}}\hat{\varphi}^a = h^{a}{}{}_{b}\hat{\eta}^m \nabla_m \hat{\varphi}^b = 0$. This then implies that there is no gyroscopic precession along the worldline of any ZAMO (relative to the natural ZAMO frame) if the ZAMO belongs to the ring of ZAMOs at $(r_0,\theta_0)$.

If we let $\xi^a = \tilde{t}^a + \omega \phi^a$ represent the tangent field to the ZAMO congruence, where $\omega = \frac{2M a r}{\Sigma^2}$ is now the non-constant angular velocity of the ZAMO congruence, then we know that $\xi_{[a}\nabla_b \xi_{c]} = 0$. However this does not (necessarily) imply that $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ right? Otherwise all rings of ZAMOs in Kerr space-time would satisfy the property that gyroscopes mounted on the rings don't precess relative to the natural ZAMO frame which isn't true according to MTW.

In fact $\xi_{[a}\nabla_b \xi_{c]} = 0$ implies that \tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + \omega\tilde{t}_{[a}\nabla_{b}\phi_{c]} + \omega\phi_{[a}\nabla_{b}\tilde{t}_{c]} + \omega^2 \phi_{[a}\nabla_{b}\phi_{c]} + \tilde{t}_{[a}\phi_b \nabla_{c]} \omega = 0 whereas \eta_{[a}\nabla_{b}\eta_{c]} = \tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + \omega_0\tilde{t}_{[a}\nabla_{b}\phi_{c]} + \omega_0\phi_{[a}\nabla_{b}\tilde{t}_{c]} + \omega_0^2 \phi_{[a}\nabla_{b}\phi_{c]}
So $\eta_{[a}\nabla_{b}\eta_{c]} = 0$ if and only if $(\tilde{t}_{[a}\phi_b \nabla_{c]} \omega )|_{r_0,\theta_0} = 0$. I don't know what ring of ZAMOs in Kerr space-time actually satisfies this.

 

[WannabeNewton]

$\omega$ can be constant everywhere (within boundaries) if we do not insist iff we insist that all the "other" rings also be ZAMO's...

Sure but I was talking about the ZAMO congruence in particular.

Condition (3.2.1) on page 222 is a condition that must hold if the ring is to qualify as non rotating by the CIR criterion, so it does not explicitly state that this is a condition that must hold on R for the ring to qualify as having rigid rotation.

That condition can't hold unless the congruence is rigid. See below.

Page 231 makes it clear that k if a function of r (it gives an example in the Godel spacetime) so for k to be a constant, r must equal $\mathcal{R}$. This is also implied in the qualifier to condition (3.2.1) that it "must hold on $\mathcal{R}$".

Given a congruence with tangent field $\eta^a = \tilde{t}^a + k \phi^a$, $k$ cannot vary across this congruence; this is what it means for the entire congruence to be rigid. The condition $\tilde{t}_{[a}\nabla_{b}\tilde{t}_{c]} + k\tilde{t}_{[a}\nabla_{b}\phi_{c]} + k\phi_{[a}\nabla_{b}\tilde{t}_{c]} + k^2 \phi_{[a}\nabla_{b}\phi_{c]} = 0$ can clearly only hold if the congruence is rigid otherwise there would be terms involving derivatives of $k$. So for each choice of $\eta^a$ there is a choice of $k$ that is constant across the entire congruence associated with $\eta^a$. What the author shows in Godel space-time is that there are different such choices of $k$ for different values of $r$ which result in the associated $\eta^a$ being non-rotating according to the CIR criterion.

There may be some overlap here with post #132.

 

[yuiop]

Sure but I was talking about the ZAMO congruence in particular.

You seem to be talking about a congruence of ZAM observers with different r, whereas the paper is (possibly) restricting itself to a congruence of ZAM observers with equal r that constitute the congruence of smooth timelike curves on the two dimensional submanifold $\mathcal{R}$ or what they call the striated orbit cylinder.

That condition can't hold unless the congruence is rigid. See below.

A congruence of ZAMO's with unequal r in the Kerr metric, cannot have rigid rotation, so there would be no way for a ring to qualify as non rotating by the CIR criteria if we insist on this stricter definition.

Sometimes it is useful to restrict ourselves to a ring of constant r. It seems to me that we can have a definition of Born rigid motion for a ring by reference only to events on the ring. While we can have a definition of Born rigid rotation for a disc if the disc has constant angular rotation, it is not possible to have Born rigid motion for a disc that has angular acceleration. A ring on the other hand can have Born rigid motion, even when the ring is undergoing angular acceleration, so consideration of a ring is less restrictive.

Just my 2 cents.

 

[WannabeNewton]

...whereas the paper is (possibly) restricting itself to a congruence of ZAM observers with equal r that constitute the congruence of smooth timelike curves on the two dimensional submanifold $\mathcal{R}$ or what they call the striated orbit cylinder.
...
Sometimes it is useful to restrict ourselves to a ring of constant r. It seems to me that we can have a definition of Born rigid motion for a ring by reference only to events on the ring. While we can have a definition of Born rigid rotation for a disc if the disc has constant angular rotation, it is not possible to have Born rigid motion for a disc that has angular acceleration. A ring on the other hand can have Born rigid motion, even when the ring is undergoing angular acceleration, so consideration of a ring is less restrictive.

I think you're forgetting one small but important detail: it isn't enough to just have $\eta^a$ on $\mathcal{R}$ itself. If we only had $\eta^a$ on the cylinder then $\eta_{[a}\nabla_b \eta_{c]} = 0$ would have no meaning because we need to know the behavior of $\eta^a$ in a neighborhood of the cylinder in order to take derivatives of the form $\eta_{[a}\nabla_b \eta_{c]} = 0$. It isn't enough to just have $\eta^a$ on the cylinder itself.

For this reason, if a ring has angular velocity $\omega_0$ we simply consider the vector field $\eta^a = \tilde{t}^a + \omega_0 \phi^a$ on the entirety of space-time so that we can take derivatives off of the cylinder, ergo the impetus for my qualms in post #129 and the posts following it.

 

[yuiop]

So far in this thread we have been a bit vague about the direction of precession relative to the direction of orbital velocity or relative to the direction of rotation of the gravitational body in the Kerr metric.

Could someone confirm that if what we want is the rotation $\Omega_{gyr}$ in proper time, of a gyroscope axis relative to the LTSB vectors transported along with gyroscope, then we should use the opposite sign to that indicated in this paper.

If that is the case then:

In the Kerr metric $\Omega_{gyr} = \frac{-ma}{r^3(1-2m/r)}$.

In the Schwarzschild metric $\Omega_{gyr} = \frac{-\omega(1-3m/r)}{(1-2m/r-\omega^2r^2)}$

In the Minkowski metric $\Omega_{gyr} = \frac{-\omega}{(1-\omega^2r^2)}$.

In general for large r and small $\omega$, the $\Omega_{gyr}$ is in the opposite sense to the orbital angular velocity $\omega$ or to the angular velocity $a$ of the gravitational body. This should guarantee that in asymptotically flat spacetime in the Newtonian regime, an untorqed gyroscope continues to point more or less at the same distant stars. Does that seem right?

 

[yuiop]

I think you're forgetting one small but important detail: it isn't enough to just have $\eta^a$ on $\mathcal{R}$ itself. If we only had $\eta^a$ on the cylinder then $\eta_{[a}\nabla_b \eta_{c]} = 0$ would have no meaning because we need to know the behavior of $\eta^a$ in a neighborhood of the cylinder in order to take derivatives of the form $\eta_{[a}\nabla_b \eta_{c]} = 0$. It isn't enough to just have $\eta^a$ on the cylinder itself.

I will bow to your vastly greater knowledge on this topic

 

[WannabeNewton]

Does that seem right?

Keep in mind that $\Omega$ as defined in the paper is only the magnitude of the gyroscopic precession relative to a connecting vector. The direction of rotation is contained in the vorticity vector $\omega^{\mu}$ but yes if we want the gyroscopic precession relative to a connecting vector then we need to look at $-\omega^{\mu}$ because $\omega^{\mu}$ itself is the rotation of the connecting vector relative to local torque-free gyroscopes.

I will bow to your vastly greater knowledge on this topic

Sorry I wasn't trying to disagree with you or anything. I agree with you on all your points. I was just trying to clarify things for myself. Thanks for the help :)

EDIT: In other words, your statement in the following quote made it clear to me that the author's definition of rigid rotation for a single ring by use of a vector field $\eta^a = \tilde{t}^a + k\phi^a$ spanning all of space-time has no loss of generality in it so as long as we stick to talking about a single ring because we can simply construct a different $\eta^a$ for each ring we wish to analyze:

$\omega$ can be constant everywhere (within boundaries) if we do not insist that all the "other" rings also be ZAMO's, but I am not sure that has been specified. I think that is basically what Peter is saying.

However such a mode of analysis would have us be cautious for even though the ZAMO congruence is itself vorticity free, it doesn't mean all the rings of ZAMOs satisfy the compass of inertia criterion because each such ring corresponds to a different $\eta^a$. That's what I was trying to say in post #132.

 

[yuiop]

Keep in mind that $\Omega$ as defined in the paper is only the magnitude of the gyroscopic precession relative to a connecting vector. The direction of rotation is contained in the vorticity vector $\omega^{\mu}$ but yes if we want the gyroscopic precession relative to a connecting vector then we need to look at $-\omega^{\mu}$ because $\omega^{\mu}$ itself is the rotation of the connecting vector relative to local torque-free gyroscopes.

Thanks for the confirmation!

Sorry I wasn't trying to disagree with you or anything. I agree with you on all your points. I was just trying to clarify things for myself. Thanks for the help :)

No probs. I assume in your final conclusion is similar to Peter's position, that for a set of concentric rings that have rigid angular motion, "that at most one of these "rings" can be a ring of ZAMOs", (in the Kerr metric).

 

[yuiop]

In the Kerr metric $\Omega_{gyr} = \frac{-ma}{r^3(1-2m/r)}$.

In the Schwarzschild metric $\Omega_{gyr} = \frac{-\omega(1-3m/r)}{(1-2m/r-\omega^2r^2)}$

In the Minkowski metric $\Omega_{gyr} = \frac{-\omega}{(1-\omega^2r^2)}$.

Assuming the above is correct, then:

In the Minkowski metric $\Omega_{gyr} = -\gamma\omega$,

where $\gamma$ is the time dilation factor. This is the purely kinematic Thomas precession. (Strictly speaking it is the Wigner precession, as Thomas precession is the accumulation of Wigner precession over a complete orbit.)

In the Schwarzschild metric $\Omega_{gyr} = -\gamma\omega + 3\gamma m/r$.

Expressed like this, the total precession is a combination of Thomas precession and another effect (geodetic?) that operates in the opposite direction. I think this is something that Peter was referring to in a earlier post.

Basically I am trying to break down the components of the total precession. The precession in the Kerr metric is a bit more complex. (That is for later.)

 

[WannabeNewton]

In the Minkowski metric $\Omega_{gyr} = -\gamma\omega$,

where $\gamma$ is the time dilation factor. This is the purely kinematic Thomas precession. (Strictly speaking it is the Wigner precession, as Thomas precession is the accumulation of Wigner precession over a complete orbit.)

Politically correct now are we ? Haha just kidding. But yes I agree with that result. Note that it makes sense physically; we can relate it back to our discussion of static observers in Kerr space-time in the following manner:

Imagine a flat disk in Minkowski space-time rotating with constant angular velocity $\Omega$ relative to an inertial observer hovering at the center of the disk. By transforming to the frame corotating with the disk, the metric on the disk is given by $ds^2 = -\gamma^{-2}dt^2 + 2r^2 \Omega dt d\phi + r^2 d\phi^2 + dr^2$ where $\gamma^{-2} = 1 - \Omega^2 r^2$. The vector field $\xi = \gamma \partial_t$ represents the congruence of observers sitting on the disk; notice that $\xi$ is Born rigid because it represents a Killing congruence. It has a vorticity $\omega = \gamma^2 \Omega \partial_z$. If a given static observer $O$ following an orbit of $\xi$ defines a set of spatial basis vectors by Lie transport along his worldline then these spatial basis vectors will be fixed relative to the origin in the sense that they will have no precession relative to the origin; this is a consequence of $\xi$ being Born rigid. Now if $O$ defines a different set of spatial basis vectors by Fermi transport along his worldline then these spatial basis vectors will rotate relative to the Lie transported ones with an angular velocity $-\gamma^2 \Omega \partial_{z}$ which is also the gyroscopic precession relative to the origin (due to the Thomas-Wigner precession).

 

[yuiop]

First, I would like to correct a couple of errors in my post #140. I started with these equations (in various coordinate systems) for the gyroscopic precession relative to the instantaneous orbital radial vector, in the proper time of a gyroscope 'orbiting' in the equatorial plane:

In the Kerr metric $\Omega_{gyr} = \frac{-ma}{r^3(1-2m/r)}$.

In the Schwarzschild metric $\Omega_{gyr} = \frac{-\omega(1-3m/r)}{(1-2m/r-\omega^2r^2)}$

In the Minkowski metric $\Omega_{gyr} = \frac{-\omega}{(1-\omega^2r^2)}$.

I think we are all agreed on the above equations. If I replace the time dilation factor for the respective metrics with the symbol $\gamma$, the equations for the precession rate in terms of coordinate time, relative to a distant fixed point are:

In the Kerr metric $\Omega_{gyr} = \gamma ma/r^3$.

In the Schwarzschild metric $\Omega_{gyr} = \omega -\gamma \omega +3 \gamma \omega * m/r$

In the Minkowski metric $\Omega_{gyr} = \omega -\gamma \omega$.

(In post #140 I omitted to state that I had switched to coordinate time.)

The equation for the Kerr metric does not really belong in this group, because it only applies to precession of a static gyroscope. Does anyone happen to know (or have a reference to) a more general equation for the precession of a gyroscope that has has orbital velocity $\omega$ in the Kerr metric?

 

[WannabeNewton]

Does anyone happen to know (or have a reference to) a more general equation for the precession of a gyroscope that has has orbital velocity $\omega$ in the Kerr metric?

See the MTW reference in post #127.

 

[yuiop]

Earlier we assumed that the rotational direction of the gyroscopic precession was simply the negative of $\Omega$ as defined in this paper http://arxiv.org/pdf/1210.6127v4.pdf.

However, various sources relating to Gravity Probe B, show diagrams showing the frame dragging effect to cause Lense-Thirring precession to co-rotate with the rotating gravitational body, while we have have been assuming counter-rotation. This includes the diagram on page 1 of the above linked paper which has a similar diagram to this:

Now the paper gives the precession as $\Omega = (\omega^\alpha \omega_\alpha)^{1/2}$. Could it be that the ambiguity as to the direction of precession is due to the fact that a square root has two solutions, one positive and one negative? Given this possible ambiguity, how do we independently determine the precession direction?

 

[WannabeNewton]

Consider, for simplicity, a static observer in Kerr space-time located in the plane $\theta = 0$. According to the paper you just linked, the radial component of the vorticity is $\omega^r = -\frac{2mra}{(r^2 + a^2)^2}$. Therefore $\omega^r_{\text{gyro}} = -\omega^r = \frac{2mra}{(r^2 + a^2)^2}$ represents the precession of a gyroscope carried by the static observer (force applied to center of mass of gyroscope so that it remains torque-free) relative to the radial axis of the observer's natural rest frame (the one whose spatial axes are fixed with respect to the distant stars). As you can see, the gyroscopic precession is in the prograde direction. Maybe you were assuming that $\omega^{\mu}$ would be a priori positive but as you can see that need not be true.

 

[yuiop]

Consider, for simplicity, a static observer in Kerr space-time located in the plane $\theta = 0$. According to the paper you just linked, the radial component of the vorticity is $\omega^r = -\frac{2mra}{(r^2 + a^2)^2}$. ...

Isn't the equatorial plane normally represented by $\theta = \pi/2$ in the normal Kerr metric (which equation 32 appears to be). If that is the case, then by equation 35, $\omega^r = 0$. Do you mean some other plane other than the equatorial plane?

Also, I was assuming that $\Omega$ is defined in a consistent way in equations 39, 52 and 58 in that paper. Maybe that is not the case.

 

[WannabeNewton]

Yes, the plane $\theta = 0$, which slices a sphere into left and right hemispheres; I chose it as an example in order to simplify the general expression for the vorticity of the static congruence in Kerr space-time (without making it vanish of course). I don't recall having mentioned the equatorial plane. There is no precession relative to the distant stars for a static gyroscope in the equatorial plane.

 

[yuiop]

Yes, the plane $\theta = 0$, which slices a sphere into left and right hemispheres; I chose it as an example in order to simplify the general expression for the vorticity of the static congruence in Kerr space-time (without making it vanish of course). I don't recall having mentioned the equatorial plane. There is no precession relative to the distant stars for a static gyroscope in the equatorial plane.

Are you sure? I thought Lense-Thirring precession was due the rotation of the gravitational body and still occurred even if the gyroscope is static in the equatorial plane in the Kerr metric. Figure (1b) on page one of the paper illustrates that situation. Perhaps you mean there is no precession around the radial axis ($\omega^r=0$) when the gyroscope is static in the equatorial plane, but in that case $\omega^\theta \ne 0$).

 

[WannabeNewton]

Perhaps you mean there is no precession around the radial axis ($\omega^r=0$) when the gyroscope is static in the equatorial plane, but in that case $\omega^\theta \ne 0$).

Sorry, yes that's what I meant.

Also, I was assuming that $\Omega$ is defined in a consistent way in equations 39, 52 and 58 in that paper. Maybe that is not the case.

Yes the definition is the same throughout the paper: $\Omega = (\omega^{\mu}\omega_{\mu})^{1/2}$ where $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}u_{\nu}\nabla_{\alpha}u_{\beta}$; if it helps, in the rest frame of $u^{\mu}$ this reduces to $\vec{\omega} = \vec{\nabla}\times \vec{u}$ which is just the usual curl from vector calculus. However the interpretation of $\omega^{\mu}_{\text{gyro}} = -\omega^{\mu}$ as the gyroscopic precession relative to the distant stars only holds if $u^{\mu}$ describes the 4-velocity field of a family of observers that is at rest with respect to the distant stars.

 

[WannabeNewton]

I'm still uneasy with the stipulations of the local compass of inertia measurement of ring non-rotation described in the notes. Say we have a ring of ZAMOs in Kerr space-time with angular velocity $\omega_0 := \frac{2Ma r_0}{\Sigma_0^2}$; the world tube $\mathcal{R}$ of the ring is a topological cylinder and has a tangent field $\eta^a |_{\mathcal{R}} = \tilde{t}^a|_{\mathcal{R}} + \omega_0 \phi^a |_{\mathcal{R}}$ where $\tilde{t}^a$ is the time-like Killing field and $\phi^a$ is the axial Killing field. Now the ring is non-rotating according to a local compass of inertia if $\eta_{[a}\nabla_b\eta_{c]}|_{\mathcal{R}} = 0$.

However it's clearly not enough to just know $\eta^a|_{\mathcal{R}}$ when evaluating $\eta_{[a}\nabla_b\eta_{c]}|_{\mathcal{R}}$, as already noted, because this expression involves space-time derivatives in directions off of $\mathcal{R}$. Therefore we need to extend $\eta^a|_{\mathcal{R}}$ to the entirety of space-time (or at the least a neighborhood of $\mathcal{R}$) in order to even make sense of $\eta_{[a}\nabla_b\eta_{c]}|_{\mathcal{R}} = 0$.

But there's no unique way to extend $\eta^a|_{\mathcal{R}}$ to all of space-time. We can for example extend it to the vector field $\eta^a = \tilde{t}^a + \omega_0 \phi^a$ or we can extend it to the vector field $\tilde{\eta}^a = \tilde{t}^a + \omega \phi^a$ where $\omega = \frac{2Ma r}{\Sigma^2}$. Both of these result in the same tangent field when restricted to the world tube of our chosen ring of ZAMOs but the former describes a congruence of rings all rigid with respect to one another (on top of being rigid themselves) whereas the latter is the ZAMO congruence (which describes a congruence of rings that are themselves rigid but not rigid with respect to one another).

As we know $\tilde{\eta}_{[a}\nabla_b \tilde{\eta}_{c]} = 0$ everywhere but $\tilde{\eta}^a$ is not rigid (hence not a Killing field) because $\omega$ varies across space-time so the proposition about the equivalence of non-rotation according to the local compass of inertia to non-rotation according to a gyroscope mounted on the ring ($h^{a}{}{}_{b}\hat{\eta}^c\nabla_{c}(h^{b}{}{}_{d}\psi^{d}) = 0$ i.e. Fermi-transport) cannot be applied. So the compass of inertia criterion can't be applied here based on the stipulations of the notes.

On the other hand $\eta^a$ is a Killing field (hence rigid) because $\omega_0$ is constant everywhere in space-time by construction. But is there actually a value of $\omega_0$ for which $\eta_{[a}\nabla_b\eta_{c]} = 0$? In other words is there a ring of ZAMOs in Kerr space-time that actually qualifies as non-rotating according to the compass of inertia criterion? Unless I'm missing something obvious, $\tilde{\eta}_{[a}\nabla_b \tilde{\eta}_{c]} = 0$ does not trivially imply that $\eta_{[a}\nabla_b\eta_{c]} = 0$ (see post #132).

So as you can see my uneasiness with $\eta_{[a}\nabla_b\eta_{c]} = 0$ as a definition of non-rotation for a single ring arises from $\eta_{[a}\nabla_b\eta_{c]}$ depending not only on the behavior of a given ring but also on the behavior of neighboring rings formed by the integral curves of the extended vector field $\eta^a$; in particular, various calculations and propositions in sections 3.2 and 3.3 of the notes rely explicitly on $\eta^a$ being a Killing field i.e. they rely on the neighboring rings being in rigid rotation with respect to one another. So even though $\eta^a$ and $\tilde{\eta}^a$ both describe the same ring of ZAMOs, that is, $\eta^a|_{\mathcal{R}}$, only the former can actually be used in sections 3.2 and 3.3 of the notes.

In other words, this definition of ring non-rotation seems to require more physics than just the physics of the given ring that we wish to analyze (the properties of space-time itself included in the physics of the given ring of course). Or am I missing something obvious again?