{x}^\nu  \cr
    &= -(1-\Omega^2  {\rho}^2)\left( d {t}-\frac{\Omega {\rho}^2 }
            {1-\Omega^2  {\rho }^2 }d {\varphi}\right)^2 
            + d {\rho}^2 
        + \frac{ {\rho}^2}{1-\Omega^2  \rho^2}d {\varphi}^2 
            + d {z}^2. 
\label{tildeg}\end{aligned}$$ The norm of the Killing vector is given by $f=\xi_\mu \xi^\mu=-(1-\Omega^2  \rho^2)$, and the metric of orbit space defined by is $$\begin{aligned}
 ds_{(3)}^2 = h_{ij} dx^i dx^j 
        = d \rho^2 
        + \frac{\rho^2}{1-\Omega^2 \rho^2} d \varphi^2 
        +dz^2, \quad (i,j=1,2,3).
\label{met3}\end{aligned}$$ The singular point $\rho=1/|\Omega|$ of the three-dimensional metric is the light cylinder where the rotation velocity becomes the light velocity. For a stationary rotating string the Killing vector $\xi$ is timelike on $\Sigma$, then the string should stay in the region $ \rho < 1/|\Omega|$. The Nambu-Goto equation for the stationary rotating string whose world surface is tangent to $\xi$ in Minkowski spacetime is reduced to the three-dimensional geodesic equation in the metric $$\begin{aligned}
 d {\tilde s}_{(3)}^2 = \tilde h_{ij} d {x}^i d {x}^j 
                = -f h_{ij} d {x}^i d {x}^j 
        = (1-\Omega^2  {\rho }^2 ) d {\rho}^2 
        +  {\rho}^2 d {\varphi}^2 
        + (1-\Omega^2  {\rho }^2 ) d {z}^2 . 
\label{eff3metric} \end{aligned}$$ The world surface of the string is spanned by the geodesic curve, say $ {x}^k (\sigma)$, and the Killing vector $\xi$. Then, by 