 Solution

Use Minkowski spacetime with signature $(-,+,\dots ,+)$, and write the string tension as $T=1/(2\pi {\alpha }^{\prime })$. If $X^{\mu }(\tau ,\sigma )$ is the string embedding map, its induced worldsheet metric is ${\gamma }_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$. The Nambu–Goto action isThe first variation uses $\delta {\gamma }_{ab}=2{\partial }_{(a}{X}_{\mu }{\partial }_{b)}\delta X^{\mu }$ and therefore givesAn integration by parts yields the bulk Nambu–Goto equations of motion and the endpoint term: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: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 $0\le \sigma \le \pi$ and $a>0$, and setwith the other coordinates constant. Every component satisfies the wave equation, andMoreover $X^{\prime }=0$ at both endpoints, giving the Neumann boundary condition. At fixed target time $t=a\tau$, the spatial image is the straight segmentThus 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:Since this is the centre-of-momentum frame, the rest mass is $M=E$. Eliminating $a$ proves the free-ended straight-string Regge relation: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 $0\le \sigma \le 2\pi$. 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, soThe 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.