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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.

Actions and equations. Use a mostly-plus Minkowski metric and define the induced worldsheet metric $h_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$. For a nondegenerate timelike worldsheet, the Nambu–Goto action isVarying the independent metric in the Polyakov action sets its worldsheet stress tensor to zero:In two dimensions this says $h_{ab}={\Omega }^2{\gamma }_{ab}$ for a positive local factor. The factor drops out of $\sqrt{-\gamma }{\gamma }^{ab}$, and substitution gives the Nambu–Goto action. Conversely, any nondegenerate induced metric solves the auxiliary-metric equation up to a Weyl transformation. This classical equivalence of Polyakov and Nambu–Goto actions is a statement about classical embeddings; quantum equivalence additionally requires treatment of the metric measure and anomaly.

Variation of $X^m$ in the metric action, followed by elimination of $\gamma$, gives the Nambu–Goto equations of motionA closed string has periodic $X$ and no spatial endpoint variation.

For a background metric $G$, Kalb–Ramond field $b$, and dilaton $\Phi$, one consistent Lorentzian convention isThe orientation fixes the two-form sign; here the Lorentzian curvature convention is chosen so Wick rotation gives the positive Euclidean dilaton term $S_{E,\Phi }=\int \sqrt{\gamma }\Phi R^{(2)}/(4\pi )$. This avoids hiding the convention in the topology argument. For constant vacuum value ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{E,\Phi }={\Phi }_0\chi$. A connected closed oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$, soThe connected vacuum amplitude has a string genus expansion ${\sum }_{g\ge 0}{g}_s^{2g-2}{\mathcal{A}}_g$; disconnected vacuum diagrams exponentiate the connected sum. Each additional handle supplies a factor $g_s^2$. This is dilaton Euler-characteristic weighting.

The rotating circle. For the specified embedding, the squared spatial speed and tangent length are both $1/2$, and their spatial inner product is zero. Including $X^0=t$ therefore givesThe metric is constant. Its equation reduces to $\ddot{X}-X^{\prime \prime }=0$, satisfied by the left- and right-moving trigonometric components and the linear time component. On a constant-time slice with $0\le \sigma <2\pi$, the proper length of a string isIt is time-independent.

Let $R(\theta )$ be an ordinary two-dimensional rotation matrix. Rotate the first coordinate pair by $R(t)$ and the second by $R(-t)$. In these time-dependent Cartesian coordinates both pairs become $(\cos \sigma ,\sin \sigma )/2$. A further fixed orthogonal change of basis givesThus every spatial slice is a planar circle of radius $1/\sqrt{2}$; its plane rotates in the ambient four-space. This is a rigid circular string in four spatial dimensions, not a pulsating circle. The rotating axes establish its spatial shape, not a transformation to an inertial spacetime frame.

The momentum density obtained from the Nambu–Goto action is $P_m=-T\sqrt{-h}{h}^{ta}{\partial }_a{X}_m$. Here $P_0=-T$, hence the conserved target-space energy isThe material velocity is transverse to the tangent and has magnitude $1/\sqrt{2}$. Its Lorentz factor is $\sqrt{2}$, so the excess over $TL$ is kinetic energy. The spatial circle being stationary in rotating axes does not eliminate this energy.

Constraints and endpoints. Varying the multipliers in the Nambu-Goto phase-space action imposesThese are first-class constraints generating normal and tangential worldsheet diffeomorphisms. They remove the two longitudinal embedding degrees of freedom; they are not extra physical force laws. Hamilton's equations areFor an open string, integration by parts leaves the spatial boundary variationAllowed variational open-string boundary conditions must make this vanish for every permitted endpoint displacement, and be preserved by the evolution. The bracketed expression is the open-string endpoint momentum flux. For free displacements, set that flux to zero. In a boundary-preserving gauge with $u=0$ and finite nonzero $e$, this is the Neumann boundary condition $X^{\prime m}=0$ at each end. This shows that free-end string boundary conditions are consistent.

At such an end, $C=0$ gives $P^2=0$, and $\dot{X}=eP$ gives ${\dot{X}}^2=0$. For a nontrivial endpoint trajectory with nonzero time velocity, its spatial speed is therefore one. This null motion of a free string endpoint follows from the boundary condition together with the constraints, not merely from the bulk wave equation. The induced metric may degenerate right at a free end; the result is understood as the endpoint limit or in the Polyakov description.

For a fixed spatial $p$-plane, split coordinates into $X^A$ along its $(p+1)$-dimensional worldvolume, including time, and $X^i$ transverse to it. Set $X^{\prime A}=0$ and $X^i=y^i$ at the ends. Tangential variations are free and their momentum flux vanishes; normal variations vanish, so their flux need not vanish. Evolution preserves the fixed normal values with $P^i=0$ at the boundary in the same gauge. These mixed Neumann boundary conditions and Dirichlet boundary conditions consistently restrict the endpoint worldlines to the plane.

The massless open-string excitations then separate into a gauge vector along the worldvolume and transverse scalar fields. A scalar displacement changes the corresponding $y^i$; its interpretation is a fluctuation of the plane's embedding. These worldvolume fields from open-string massless states support the interpretation as a dynamical planar D-brane. In the bosonic theory the still lower ground state is a tachyon, indicating an unstable brane; the interpretation does not require pretending that this tachyon is a stable massless field. Appropriate superstring sectors can remove that instability.

Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.