EDIT: I get a different expression than those given in the referred questions that are supposed to answer it. Therefore I want to know if my result is correct.
I have tried to find the line element of a rotating disk. My question is of my reasoning and the resulting expression (6) are correct.
The laboratory frame is denoted by $L$. The frame that comoves with a rotating disk is called $R$. The line element is expressed in cylindrical coordinates in the lab frame $L$ by \begin{equation}\label{tauL} d\tau^2 = (c\,dt_L)^2 - (r\, d\theta_L)^2 - dr^2 -dz^2.\hspace{2cm} (1) \end{equation} The $r$ and $z$ coordinates coincide for both frames and therefore the subscript $L$ (and $R$, in later equations) is omitted for these variables. The disk rotates with angular velocity $\omega$.
Clocks are synchronized by sending a cylindrical light pulse from the center outwards, which means that all clocks in $L$ on a circle show the same time, and all times in $R$ on a circle are equal. Clocks in $L$ and $R$, however, can run at different rates.
To find the expression for the line element in the comoving frame $R$, we pick two other frames at some distance $r$ from the center/axis of the disk, at arbitrary angles and times. Their $x$-axes are tangential to the circle, the $y$-axes point in the radial direction, and the $z$-axes are perpendicular to the disk plane. One of these frames, denoted by a subscript $1$, is in rest w.r.t. the lab frame, while the other, frame $2$, is comoving with the disk. At certain times the spatial origins of these two frames colocate. We can then choose the time and space origins of $1$ and $2$ to be zero and look at the relation between infinitesimal changes in their coordinates. The Lorentz Transformation (LT) predicts: \begin{equation} dx_1 = \gamma\left\{dx_2+v\,dt_2\right\}, \; \gamma=(1-v^2/c^2)^{-0.5}, \; v=\omega r.\hspace{2cm}(2) \end{equation} This translates to an expression relating infinitisemal changes in the coordinates of $L$ and $R$: \begin{equation}\label{thetaL} rd\theta_L = \gamma\left\{r\,d\theta_R+v\,dt_R\right\}.\hspace{2cm}(3) \end{equation} However, the LT expression \begin{equation} dt_1 = \gamma\left\{dt_2+(v/c^2)dx_2\right\}\hspace{2cm}(4) \end{equation} does not translate directly to the relation between infinitesimal changes in frames $L$ and $R$, because clocks are synchronized differently in the rotating frames than they are in inertial frames for which the LT applies. The result is that the second term in this equation is synchronized away in frames $L$ and $R$, resulting in \begin{equation}\label{tL} dt_L = \gamma\, dt_R.\hspace{2cm}(5) \end{equation} Substituting (3) and (5) into (1) leads to the following expression for the line element in $R$: \begin{equation}\label{tauR} d\tau^2 = (c\,dt_R)^2 - \gamma^2 (r\, d\theta_R)^2 - 2\gamma^2 (v/c) (r\,d\theta_R)(c\,dt_R) - dr^2 -dz^2.\hspace{2cm}(6) \end{equation} By comparing (1) and (6), we can make a few observations:
When an observer in the rotating frame $R$ measures the circumference at a constant time ($dt_R=0$), he will find a larger value (by a factor $\gamma$) than an observer in the lab frame $L$ at a constant time ($dt_L=0$) in his frame.
The time interval measured between two events at constant angle $\theta_R$ is shorter according to an observer in $R$ than according to an observer in $L$, because in $L$ the term $(r\,d\theta_L)^2$ is not zero. Setting $d\theta_R = 0$ in (3) and using (5) gives $r\,d\theta_L = v\,dt_L$.
The time difference between light beams going round in the clockwise and counterclockwise directions (Sagnac effect) can be calculated by setting (6) to zero and solving for $dt_R$. This leads to $\Delta t = 4\pi r^2 \gamma^2 \omega/c^2$.
