yuiop said:
Will the axis of a gyroscope held by a static observer at the same radius also rotate once per year as observed by the distant observer, or is not as simple as that?
I hope I'm understanding your use of the term "rotation" correctly here in that "rotation of the ZAMO gyroscope" refers to the orbital rotation of the gyroscope around the central Kerr BH relative to infinity, since it has no spin rotation (as you noted its angle is constant wrt the instantaneous spatial axes of a ZAMO frame), whereas "rotation of the static gyroscope" refers to the spin rotation of the gyroscope relative to infinity.
If so, it isn't as simple as what you stated but it can still be described mathematically at the surface level. Consider the entire family of static observers in Kerr space-time; the congruence of worldlines of these observers has, as noted earlier, the 4-velocity field ##\xi^{\mu} = \frac{1}{\sqrt{-g_{tt}}}\delta^{\mu}_{t} = \gamma \delta^{\mu}_t##.
Consider an observer ##O## in this family. We attach to ##O## a
frame, which, by definition, is a choice of orthonormal basis wherein the time-like basis vector is simply the 4-velocity of ##O##. The particular frame we attach to ##O## is given by ##\{\xi^{\mu}, \eta^{\mu}_1, \eta^{\mu}_2, \eta^{\mu}_3 \}## where ##\eta^{\mu}_{i}## are spatial basis vectors setup in a special way: ##O## has three infinitesimally neighboring static observers separated from him along the directions ##\eta^{\mu}_1,\eta^{\mu}_2,\eta^{\mu}_3## of this frame and each of ##\eta^{\mu}_1,\eta^{\mu}_2,\eta^{\mu}_3## points from him to the respective neighbors. In order to force the ##\eta^{\mu}_{i}## to remain locked to the respective neighbors everywhere along his worldline, ##O## uses what's called Lie transport: ##\mathcal{L}_{\xi}\eta^{\mu}_{i} = \xi^{\gamma}\nabla_{\gamma}\eta^{\mu}_{i} - \eta^{\gamma}_{i}\nabla_{\gamma}\xi^{\mu} = 0##. We have thus defined a frame for ##O##; let's call this a
static frame.
*Of course there's a physical reasoning behind this choice of frame and it relates to the way in which we aim to measure the
rotation of the frame. Imagine we are in flat space-time and we have a rotating disk. Two superimposed observers ##O## and ##O'## sit at the center of the disk; ##O## belongs to a family of observers, the rest of whom are situated at all other points on the disk. Imagine that relative to ##O'## all these other observers have instantaneous tangential velocities ##\vec{v} = \vec{\omega}\times \vec{r}## where ##\vec{\omega}## is the angular velocity of the disk and ##\vec{r}## the position vector to a given observer, whereas relative to ##O## all these other observers are stationary on the disk. Then clearly ##O## is spinning in place relative to ##O'##.
Now if we attach to ##O## the frame ##\{\xi^{\mu}, \eta^{\mu}_1, \eta^{\mu}_2, \eta^{\mu}_3 \}## from above then the Lie transport condition ##\mathcal{L}_{\xi}\eta^{\mu}_{i} = 0## that locked the spatial basis vectors to neighboring observers is the exact same thing as having the observers sitting at all other points on the disk remain stationary relative to ##O##. On the other hand ##O'## has an inertial frame attached to him given by some ##\{e^{\mu}_0, e^{\mu}_1,e^{\mu}_2,e^{\mu}_3 \}## and because this is an inertial frame, the ##e^{\mu}_{i}## are physically realized by torque-free gyroscopes. In other words this frame constitutes what we call a non-rotating frame. Above we said that the observers sitting at all other points on the disk have tangential velocities relative to ##O'##; this is exactly the same thing as saying that the spatial basis vectors ##\eta^{\mu}_{i}## of ##O##'s frame rotate relative to the gyroscopes ##e^{\mu}_{i}##. This measures the rotation of ##O##'s frame relative to the non-rotating frame of ##O'## by means of torque-free gyroscopes.*
Coming back to the congruence of static observers in Kerr space-time, we measure the rotation of a static frame almost identically to what was described above. The main difference now is that we must work with
locally non-rotating frames so we superimpose a locally non-rotating observer on our chosen static observer ##O## (meaning they share the same worldline) and the rotation of the static frame attached to ##O## (more precisely, the rotation of the ##\eta^{\mu}_{i}##) is, at each event, measured relative to the torque-free gyroscopes of the locally non-rotating frame attached to the locally non-rotating observer superimposed on ##O##. We define the
vorticity 4-vector ##\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta}## as the rotation of the ##\eta^{\mu}_{i}## (and hence of the static frame). This is just a relativistic generalization of the vorticity 3-vector from fluid mechanics. It's magnitude is simply ##\omega = \sqrt{\omega^{\mu}\omega_{\mu}}## and the units of ##\omega## are ##[\omega] = \frac{1}{s}##.
The periodicity of ##\omega## is with respect to a clock carried by a locally non-rotating observer sharing the same worldline as our static observer ##O## so in order to get ##\omega_{\infty}## we simply tack on the "gamma factor" ##\gamma = \frac{dt}{d\tau}## where ##d\tau## is the proper time along the shared worldline of ##O## and a locally non-rotating observer. Doing this we get ##\omega_{\infty} = \gamma^{-1}\omega##.
Letting ##\Omega## be the angular velocity of a ZAMO observer about the central Kerr BH, we can then compute ##\frac{\omega_{\infty}}{\Omega}## in order to compare the two rotation rates. I haven't explicitly done the calculation yet but just looking at the form of ##\omega^{\mu}## from the paper I linked earlier, a priori I can't see any reason why we would get ##\frac{\omega_{\infty}}{\Omega} = 1##.
Sorry for the incredibly long post but I just wanted to describe, once and for all, what we are really measuring when it comes to the rotation of static frames.
. The comoving frame basis I'm using mixes ##t## and ##\phi## so I'm not sure what 'comoving' means here.
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