= Solution
Use <Minkowski spacetime> with signature $(-,+,\ldots,+)$, and write the <string tension> as $T=1/(2\pi\alpha')$. If $X^\mu(\tau,\sigma)$ is the <string embedding map>, its <induced worldsheet metric> is $\gamma_{ab}=\partial_aX\cdot\partial_bX$. The <Nambu–Goto action> is
$$
S=-T\int d\tau\,d\sigma\,\sqrt{-\det\gamma}.
$$
The <first variation> uses $\delta\gamma_{ab}=2\partial_{(a}X_\mu\partial_{b)}\delta X^\mu$ and therefore gives
$$
\delta S=-T\int\sqrt{-\gamma}\,\gamma^{ab}\partial_aX_\mu\partial_b\delta X^\mu\,d\tau\,d\sigma.
$$
An <integration by parts> yields the bulk <Nambu–Goto equations of motion> and the endpoint term:
$$
\boxed{\partial_a\!\left(\sqrt{-\gamma}\,\gamma^{ab}\partial_bX^\mu\right)=0},
\qquad
\Pi^a_\mu=-T\sqrt{-\gamma}\,\gamma^{ab}\partial_bX_\mu,
\qquad
\left.\Pi^\sigma_\mu\delta X^\mu\right|_{\partial\Sigma}=0.
$$
For a <closed string>, periodicity cancels the endpoint term. For a freely moving <open string>, arbitrary endpoint variations require zero momentum flux, which is the <free-end string boundary condition>.
Choose orthogonal conformal coordinates on the nondegenerate part of the <worldsheet>, so that $\gamma_{\tau\sigma}=0$ and $\gamma_{\tau\tau}+\gamma_{\sigma\sigma}=0$. The <Nambu–Goto equations of motion> then become a <wave equation>, with the accompanying <Virasoro constraints>:
$$
\ddot X-X''=0,\qquad \dot X\cdot X'=0,\qquad \dot X^2+X'^2=0.
$$
Here a dot and a prime mean $\partial_\tau$ and $\partial_\sigma$. The last two equations are essential: a solution of the <wave equation> alone need not be a relativistic <fundamental string> motion.
For a free-ended <open string>, take $0\leq\sigma\leq\pi$ and $a>0$, and set
$$
X^0=a\tau,\qquad X^1=a\cos\sigma\cos\tau,\qquad X^2=a\cos\sigma\sin\tau,
$$
with the other coordinates constant. Every component satisfies the <wave equation>, and
$$
\dot X^2=-a^2\sin^2\sigma,\qquad X'^2=a^2\sin^2\sigma,\qquad \dot X\cdot X'=0.
$$
Moreover $X'=0$ at both endpoints, giving the <Neumann boundary condition>. At fixed target time $t=a\tau$, the spatial image is the straight segment
$$
\boldsymbol X=r\bigl(\cos(t/a),\sin(t/a)\bigr),\qquad -a\leq r\leq a.
$$
Thus its <angular velocity> is $1/a$. The speed at coordinate $r$ is $|r|/a$, so the endpoints have the <null motion of a free string endpoint>. The <induced worldsheet metric> degenerates there; the conserved densities below have finite limits, so the endpoint degeneracy causes no divergent charge.
In these coordinates $\Pi^{\tau\mu}=T\dot X^\mu$. The <target-space Noether charges of a string> give the <energy>, <momentum> and <angular momentum>:
$$
E=\int_0^\pi T\dot X^0\,d\sigma=\pi Ta,\qquad
\boldsymbol P=T\int_0^\pi\dot{\boldsymbol X}\,d\sigma=0,
$$
$$
J_{12}=T\int_0^\pi\left(X^1\dot X^2-X^2\dot X^1\right)d\sigma
=Ta^2\int_0^\pi\cos^2\sigma\,d\sigma=\frac{\pi Ta^2}{2}.
$$
Since this is the <centre-of-momentum frame>, the rest <mass> is $M=E$. Eliminating $a$ proves the <free-ended straight-string Regge relation>:
$$
\boxed{J=\frac{M^2}{2\pi T}=\alpha'M^2\quad\text{(free-ended open string)}}.
$$
A spacetime <Lorentz transformation> of the solution changes its frame, not this rest-frame relation.
There is also a folded <closed string> version: use the same formula over $0\leq\sigma\leq2\pi$. The spatial segment is then covered twice, with folds at $\sigma=0,\pi$, and the <closed-string mode expansion> is periodic. Each charge is twice its <open string> value, so
$$
M=2\pi Ta,\qquad J=\pi Ta^2,\qquad
\boxed{J=\frac{M^2}{4\pi T}=\frac{\alpha'M^2}{2}\quad\text{(folded closed string)}}.
$$
The folds also have degenerate induced metric and are understood through the finite canonical densities or a limiting smooth motion. The two classical <Regge trajectories> have different slopes because the <closed string> has two coincident branches. No quantum <string intercept> enters this classical calculation.
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