EDIT:
Based on the mistake pointed below, I have computed the (now hopefully) correct metric and the differential equations for the geodesics.
I have the following co-ordinate transformation,
\begin{align} t & = t' \\ x' & = r \cos(\theta - \omega t) \\ y' & = r \cos(\theta - \omega t) \\ z' & = z \end{align}
Physically, this respresents rotations of co-ordinates, which can be most easily descrived in the cylindirical coordinate systems. If we drop the time dependent terms in the argument for the trignometric functions, the metric is that of the cylindirical polar coordinates,
\begin{align} ds^2 = -dt^2 + dr^2 + r^2 d \theta^2 + dz^2, \end{align}
where we have used the Minkowski metric to convert to metric tensor in the cylindrical polar coordinates. Now adding the time dependent terms, we move to the $(t', r', \theta', z')$ coordinates such that
\begin{align} t' & = t \\ r' & = r \\ \theta' & = \theta - \omega t \\ z' & = z, \end{align}
and
\begin{align} ds'^2 = (-1 + r^2 \omega^2)dt'^2 + dr'^2 + r'^2 d \theta'^2 + dz'^2 + r^2 \omega ( dt' d \theta' + d \theta' dt') .\end{align}
I now wish to calculate the most general solution of the geodesic. Using the only non vanishing Christoffel symbols
\begin{align} \Gamma^{r'}_{t' \theta'} & = \Gamma^{r'}_{\theta' t'} = -r \omega \\ \Gamma^{r'}_{\theta' \theta'} & = - r \\ \Gamma^{\theta'}_{t' r'} = \Gamma^{\theta'}_{r' t'} & = \frac{\omega}{r} \\ \Gamma^{\theta'}_{r' \theta'} = \Gamma^{\theta'}_{\theta' r'} & = \frac{1}{r}, \\ \end{align}
I get the following differential equations (where I have dropped the primes, except for the one on $\theta$):
\begin{align} \frac{d^2 t}{d \beta} & = 0 \\ \frac{d^2 r}{d \beta} - 2 r \omega \Bigg ( \frac{d t}{d \beta} \Bigg ) \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) - r \Bigg ( \frac{d \theta'}{d \beta} \Bigg )^2 & = 0 \\ \frac{d^2 \theta'}{d \beta} + \frac{2 \omega}{r} \Bigg ( \frac{d t}{d \beta} \Bigg ) \Bigg ( \frac{d r}{d \beta} \Bigg ) + \frac{1}{r} \Bigg ( \frac{d r}{d \beta} \Bigg ) \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) & = 0 \\ \frac{d^2 z}{d \beta^2} & = 0 \end{align}
Since $t' = t$ is an independent coordinate, we can choose $\beta = t$. From the last DE, we have $z = At + B$, and the second and third differential equations become
\begin{align} \frac{d^2 r}{d \beta} - 2 r \omega \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) - r \Bigg ( \frac{d \theta'}{d \beta} \Bigg )^2 & = 0 \\ \frac{d^2 \theta'}{d \beta} + \frac{2 \omega}{r} \Bigg ( \frac{d r}{d \beta} \Bigg ) + \frac{1}{r} \Bigg ( \frac{d r}{d \beta} \Bigg ) \Bigg ( \frac{d \theta'}{d \beta} \Bigg ) & = 0 \end{align}
I am unsure how to solve these horrible looking coupled differential equations. I am not very good at DE's as of yet; it'd be great if someone could comment on my approach to the question (up till the point where I have calculated the DE's) and comment/hint on the difficulty of solving the equations. Or perhaps even answer the question after some comments have been exchanged.
