Why is the Gödel Universe Rotating

[GRDoer]

Homework Statement

Consider the Godel Metric in spherical coordinates as on page 6 here;
$$ds^2=4a^2\left[-dt^2+dr^2+dz^2-(\sinh^{4}(r)-\sinh^{2}(r))d\phi^2+2\sqrt{2}\sinh^{2}(r)dt d\phi)\right]$$

This is a solution to Einstein's Equations if we have $a=\frac{1}{2\sqrt{2\pi\rho}}$ and $\Lambda =-\frac{1}{2a^2}$, where $\Lambda$ is the cosmological constant and $\rho$ is the density of the uniform dust with which we fill the universe.

It is often stated that this represents a rotating universe, but to me that is not immediately obvious, and so I want to prove it. To do so I'm considering an observer stationary with respect to the Godel coordinates at some point $(r,\phi,z)$ and equipped with a gyroscope. If the universe is rotating, the gyroscope should obviously precess. I want to find the rate of precession.

Homework Equations

Einstein's Equations and those mentioned throughout the post.

The Attempt at a Solution

The observer is stationary, so we have $-1=\vec{u}\cdot\vec{u}=g_{\alpha\beta}u^{\alpha}u^{\beta}=g_{tt}(u^{t})^{2}=-4a^{2}(u^{t})^{2}$, so $u^{t}=\frac{1}{\sqrt{2}a}$ and so $\vec{u}=(\frac{1}{\sqrt{2}a},0,0,0)$.
$a$ is constant so clearly $\vec{a}=0$, so the gyroscope is in free fall. This means its spin, $\tilde{S}$ undergoes parallel transport; $\nabla_{\vec{u}}\tilde{S}=0$. Spin is a 1-form so we have:
$$\frac{S_{\alpha}}{dt}=\Gamma^{\gamma}_{\alpha\beta}S_{\gamma}u^{\beta}u$$
This means that $s_{z}$ and $s_{t}$ are constant, and as we must have $\tilde{S}\cdot\vec{u}=0$, $s_{t}=0$.
The resulting ODES are;
$$\frac{dS_{r}}{dt}=\frac{\sqrt{2}}{a\sinh(2r)}S_{\phi}$$
$$\frac{dS_{\phi}}{dt}=-\frac{1}{2\sqrt{2}a}\sinh(2r)S_{r}$$

They have general solution:
$$S_{r}(t)=-2i(A\exp({\frac{it}{a\sqrt{2}}})-B\exp({-\frac{it}{a\sqrt{2}}}))$$
$$S_{\phi}(t)=\sinh(2r)(A\exp({\frac{it}{a\sqrt{2}}})+B\exp({-\frac{it}{a\sqrt{2}}}))$$

I'm not sure how to go from here to the angular velocity/momentum associated with the spin.
The spin 1-form is given component wise by $s_{\alpha}=\frac{1}{2}\epsilon_{\alpha\beta\gamma\delta}J^{\beta\gamma}u^{\delta}$, which let's us use the above result to find the components of $J^{\alpha\beta}$, and I believe I should find something along the lines of $\frac{dS}{dt}=\Omega\times S$ for some angular velocity $\Omega$, but classical mechanics never was my strong suit so I'm not sure of the specifics. Honestly the whole concept of spin and precession in GR is doing my head in.
I think the result should be $2\sqrt{\pi\rho}$, which I have seen while researching online, but yeah, I have no idea how to prove it.
I guess the crux of my question is how do we relate $S_{\alpha}$ to the precession velocity of the gyroscope of which it is the spin?
Any pointers, general or specific would really help, thanks.​

 

[GRDoer]

I just noticed a few typos in the above. Throughout the t's should be $\tau$'s, theyre the proper time of the observer. Also $u^{t}=\frac{1}{2a}$ not $\frac{1}{\sqrt{2}a}$. I already made one edit, and apparently I can't make more so I'll post this here.

Also, if I try to evaluate $\Omega\times S$ for an arbitrary angular velocity $\Omega$ and equate this to $\frac{ds_{i}}{d\tau}$, where $i\in\{1,2,3\}$, I get

$$(\frac{2}{a\sqrt{2}\sinh(2r)}S_{\phi}(\tau),\frac{-\sinh(2r)}{2\sqrt{2}a}S_{r}(\tau),0)=(-\Omega_{z}S_{\phi},\Omega_{z}S_{r},\Omega_{r}S_{\phi}-\Omega_{\phi}S_{r})$$

Which due to the $\sinh(2r)$ factor cannot be true for a general position.
I don't know where I'm going wrong. Maybe I've solved the equations of motion wrong somehow, maybe I'm misunderstanding precession, or maybe I need to transform the result to the locally flat frame of the observer (which I have no idea how to do), or something else entirely.