string case, $\ell=0$, the rotating velocity is physical, that is, the velocity is orthogonal to the string because the planar string is confined in $ x - z$ plane. Then the planar string does not move along the rotation axis. It is consistent with the fact that the planar string consists of a standing wave as mentioned before.

In order to see the effective equation of state[@VandS] for the stationary rotating strings, we transform the reference system such that the averaged value of momentum $\braket{P}$ vanishes. Using the fact that $$\begin{aligned}
    \braket{T^{\bar t\bar t}-T^{\bar z\bar z}} ~\mbox{and}~ 
    \braket{T^{\bar t\bar t}T^{\bar z\bar z}-(T^{\bar t\bar z})^2} 
\end{aligned}$$ are invariant under the Lorentz transformation along the $\bar z$-axis, we obtain effective line density, $\tilde\mu$, and effective tension, $\tilde {\cal T} $, as $$\begin{aligned}
    &\tilde\mu         =\frac{\mu}{2|q|}
                \left[{1-\ell^2+q^2}
                +\sqrt{(1-q-\ell)(1-q+\ell)(1+q-\ell)(1+q+\ell)}\right],\\
  &\tilde {\cal T}         =\frac{\mu}{2|q|}
                \left[{1-\ell^2+q^2}
                -\sqrt{(1-q-\ell)(1-q+\ell)(1+q-\ell)(1+q+\ell)}\right].
\end{aligned}$$ In general, it holds that $\tilde \mu\tilde {\cal T}= \mu^2$ and $\tilde \mu \geq \tilde {\cal T}$. In the case of helical strings, there exists no inertial reference system such that $\braket{P}$ vanishes because a single wave moves with the velocity of light along the rotating axis.

Conclusion
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We study stationary rotating Nambu-Goto strings in Minkowski spacetime. It 