= 1$, we find from that $\rho$ should take a constant value $\rho_0$, where $$\begin{aligned}
    \rho_0 = \rho_{max}=\rho_{min} 
        =\frac{\sqrt{\ell}}{\Omega}.
\label{amp}
\end{aligned}$$ When $q\neq 0$, using and we see that solutions are described by $$\begin{aligned}
    &\rho =\rho_0, \quad
  &\varphi =\Omega z.
\label{Helsol} 
\end{aligned}$$ We call these solutions [*a helical strings*]{}. The shapes of helical strings are shown in Fig.\[FigHel\]. In the limit $\ell =0$ i.e., $q=1$, the solutions reduce to the straight string solution.

![Helical strings: $ \ell + q =1  (q \neq 0)$. The three-dimensional snapshots are given in the left panel, and the projection of strings to $x-y$ plane, i.e., projected curves, are given in the middle, where $x, y$ are defined by . The dashed circle in the middle figure represents the light cylinder $\rho=1/\Omega$. In each figure, the strings with the parameters $ (\ell, q)=(0,1)$, $(9/10, 1/10)$, $(1/2, 1/2)$, $(1/10, 9/10)$ are shown. The parameters on $\ell-q$ plane are plotted in the right panel. []{data-label="FigHel"}](helical.eps){width="12cm"}

We can calculate the proper length of snapshot of the string with respect to the metric as $$\begin{aligned}
        \int ds = \int\sqrt{(1-\Omega^2\rho^2)d\rho^2 + \rho^2 d\varphi^2 
        +(1-\Omega^2\rho^2)d z^2}. 
\end{aligned}$$ For the helical strings , the proper length for $z$-interval $z \sim  z+\Delta z$ 