is given by $$\begin{aligned}
\int\sqrt{\rho_0^2 d\varphi^2 
        +(1-\Omega^2\rho_0^2)d z^2}     
    =\int_{ z}^{ z+\Delta z} d z 
    = \Delta z. 
\end{aligned}$$ It shows that all helical strings have the same proper length as the straight string with the same $ z$-interval.

In the inertial reference frame , the helical strings are described by $$\begin{aligned}
    &\bar \rho=\rho_0, \\
    &\bar \varphi = \Omega (\bar z+\bar t). 
\label{Helsol2} 
\end{aligned}$$ The solutions consist of a down-moving wave of the angular frequency $\Omega$ and the amplitude $\rho_0$ given by with the circular polarization. The pitch of helix, equivalently the wave length, is given by $2\pi \Omega^{-1}$. The two-dimensional world surface of helical string has another Killing vector $\partial_{\bar z}+\Omega\partial_{\bar\varphi}$ which is tangent to it in addition to $\partial_{\bar t}+\Omega\partial_{\bar\varphi}$. It means the world surface of helical string is a homogeneous space.

### Planar strings 

In the case of $\ell = 0$, it is convenient to use the rigidly rotating Cartesian coordinate $$\begin{aligned}
     x= \rho\cos\varphi, \quad 
    y= \rho \sin\varphi.
\label{cartesian}
\end{aligned}$$ From we can set $\varphi=0$ ($ y=0$) without loss of generality. Then, from and we have $$\begin{aligned}
    q^2\left(\frac{d x}{d z}\right)^2 = 1-q^2-\Omega^2 x^2. 
\label{eq_planar}
\end{aligned}$$ The strings are confined in $ x- z$ plane, so we call these 