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Use target-space signature $(-,+,+,+)$ and conformal gauge $g_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$. Dots and primes denote differentiation with respect to $\tau$ and $\sigma$. Varying the Polyakov action with respect to the string embedding map gives the wave equation; varying the worldsheet metric sets the worldsheet stress tensor to zero. Explicitly,With ${\sigma }^{\pm }=\tau \pm \sigma$ and ${\partial }_{\pm }=\frac{1}{2}({\partial }_{\tau }\pm {\partial }_{\sigma })$, the two Virasoro constraints are $({\partial }_{+}X{)}^2=({\partial }_{-}X{)}^2=0$.

A fixed spatial circle would sweep out a cylindrical string worldsheet. In static coordinates its string embedding map is $X=(t,R\cos \theta ,R\sin \theta ,0)$ and its induced worldsheet metric is $\mathrm{diag}(-1,R^2)$. The radial part of the Nambu–Goto equations of motion is proportional to $R^{-2}{\partial }_{\theta }^2(R\cos \theta ,R\sin \theta )=-(\cos \theta ,\sin \theta )/R$, which is nonzero. Thus a nonzero-radius cylinder cannot solve the equations. Motion of labels around this same cylinder is a worldsheet diffeomorphism and cannot alter this conclusion.

The familiar pulsating circular string makes the obstruction particularly explicit. Put $X^0=\kappa \tau$ and $X^1+i{X}^2=r(\tau )e^{i\sigma }$. The wave equation and Virasoro constraints reduce toA constant $r=R\ne 0$ violates the first equation. Even rotating the labels, $X^1+i{X}^2=R{e}^{i(\sigma +\Omega \tau )}$, cannot help: the wave equation requires ${\Omega }^2=1$, whereas $\dot{X}\cdot X^{\prime }=R^2\Omega$ requires $\Omega =0$.

Nor does a rigid rotation of the circular plane rescue the usual circular ansatz in three spatial dimensions. Write its spatial part as $A(\tau )\cos \sigma +B(\tau )\sin \sigma$, with $A^2=B^2=R^2$ and $A\cdot B=0$. The wave equation gives $\ddot{A}=-A$ and $\ddot{B}=-B$. The constant norms give $A\cdot \dot{A}=B\cdot \dot{B}=0$ and ${\dot{A}}^2={\dot{B}}^2=R^2$. The mixed Virasoro constraint, together with the derivative of $A\cdot B=0$, gives $\dot{A}\cdot B=A\cdot \dot{B}=0$. Differentiating $A\cdot B$ twice gives $\dot{A}\cdot \dot{B}=0$. This would require four mutually orthogonal nonzero spatial vectors in ${\mathbb{R}}^3$. This obstruction is specific to the stated spatial dimension; a rigid circular string in four spatial dimensions has more room.

To express the Virasoro constraints in modes, extend the string oscillators byDifferentiating the two chiral expansions gives ${\partial }_{-}{X}^{\mu }=\sqrt{{\alpha }^{\prime }/2}{\sum }_n{\alpha }_n^{\mu }{e}^{-in{\sigma }^{-}}$ and the corresponding expression with tildes for ${\partial }_{+}X$. Equating every Fourier coefficient of their squares to zero yieldsReality of the string embedding map means ${\alpha }_{-n}={\alpha }_n^{\ast }$ and ${\tilde{\alpha }}_{-n}={\tilde{\alpha }}_n^{\ast }$. In particular, define the classical string level operators by $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$ and $\tilde{N}={\sum }_{n>0}{\tilde{\alpha }}_{-n}\cdot {\tilde{\alpha }}_n$. The zero-mode Virasoro constraints becomeConsequently the classical bosonic string mass spectrum isThe equality of the two levels is closed-string level matching. There is no string intercept here: that shift comes from quantum normal ordering, not from these classical equations.

For the compact boson, closed-string periodicity permits a winding number $w\in \mathbb{Z}$, so $X^3(\sigma +2\pi )-X^3(\sigma )=2\pi wR$. Let $q$ be the momentum in this direction. Its zero-mode expansion and the corresponding chiral momenta areThus replace the equal left and right zero modes by ${\tilde{\alpha }}_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_L^3$ and ${\alpha }_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_R^3$; the nonzero string oscillators remain integer-moded. Classically $q$ is continuous. Quantizing the compact boson makes momentum wavefunctions single-valued around its circle, giving $q=n/R$ with $n\in \mathbb{Z}$.

If $k^a$, $a=0,1,2$, is the noncompact momentum and $M^2=-k^a{k}_a$, the two zero-mode Virasoro constraints giveAveraging and subtracting gives the momentum and winding modes and compact-circle closed-string level matching formulasThe latter becomes $\tilde{N}-N+nw=0$ on imposing quantum momentum quantization. The sign follows from assigning tildes to the left mover ${\sigma }^{+}$.

An explicit winding-supported circular string, valid classically for every $R>0$, isThe noncompact projection is a circle of constant radius. Every coordinate obeys the wave equation. Its compact winding is $w=1$, and a direct calculation givesThus both Virasoro constraints hold and the induced worldsheet metric is $2R^2\mathrm{diag}(-1,1)$, not a degenerate cylinder. The compact motion cancels the mixed stress that spoiled rotation of labels in flat space. Here $q=-R/{\alpha }^{\prime }$, $N=0$ and $\tilde{N}=R^2/{\alpha }^{\prime }$ satisfy the classical compact-circle closed-string level matching condition. Requiring this particular classical momentum to equal an individual quantum eigenvalue would impose an additional radius-dependent condition; that is not required to construct the classical solution.

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.