A rigidly rotating stretched solution is
$$
X^0=A\tau,
\qquad
X^1=A\cos\tau\cos\sigma,
\qquad
X^2=A\sin\tau\cos\sigma,
$$
with all other coordinates constant. Each coordinate obeys the wave equation and $X'=0$ at $\sigma=0,\pi$. Directly, $\dot X\cdot X'=0$ and
$$
\dot X^2+X'^2=-A^2+A^2\cos^2\sigma+A^2\sin^2\sigma=0,
$$
so both Virasoro constraints hold. Its energy and planar angular momentum are
$$
M=P^0=\pi TA,
\qquad
J=J^{12}=TA^2\int_0^\pi\cos^2\sigma,d\sigma
=\frac{\pi TA^2}{2}.
$$
Consequently $J=M^2/(2\pi T)=\alpha'M^2$, the classical leading open-string <Regge trajectory>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 306 / 1 / d / Solution

{c}
{title2=$J=\alpha'M^2+a$}
{wiki}

A Regge trajectory is an approximately linear relation between angular momentum and squared mass. A classical rotating open string obeys $J=\alpha'M^2$, with quantum effects adding an intercept.


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 Incoming links: Regge trajectory