Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 306 / 1

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

The canonical momentum equation is $\dot{X}=eP+u{X}^{\prime }$. Eliminating $P$ from the phase-space action yields the Lagrangian densityThis is the Polyakov action after identifying its inverse metric tensor density asThe determinant of this matrix is $-1$, as required in two dimensions. The Weyl transformation leaves the matrix unchanged, so $e,u$ encode the metric modulo its conformal factor.

Define the induced worldsheet metric by $g_{\mu \nu }={\eta }_{mn}{\partial }_{\mu }{X}^m{\partial }_{\nu }{X}^n$. Varying the independent inverse metric tensor in the Polyakov action givesConsequently $g_{\mu \nu }=f{\gamma }_{\mu \nu }$, where $f=\frac{1}{2}{\gamma }^{\rho \sigma }{g}_{\rho \sigma }$. For a nondegenerate Lorentzian string worldsheet with the usual compatible time orientation, take $f>0$, or ${\gamma }_{\mu \nu }=e^{2\omega }{g}_{\mu \nu }$. The undetermined function is precisely Weyl invariance. Since $\sqrt{-\gamma }{\gamma }^{\mu \nu }{g}_{\mu \nu }=2\sqrt{-g}$, elimination of the independent metric tensor givesThus the Nambu–Goto action measures minus string tension times Lorentzian worldsheet area. The metric elimination holds in the nondegenerate interior; null endpoints are understood as boundary limits.

In conformal gauge, ${\gamma }_{\mu \nu }=e^{2\omega }\mathrm{diag}(-1,1)$. The Weyl invariance of the Polyakov action removes $\omega$, leavingIts complete variation isFix the string embedding map on the initial and final time slices. The interior Euler-Lagrange field equation is the wave equationAt each spatial endpoint the remaining variation vanishes under either Dirichlet boundary conditions, $\delta X^m=0$, or Neumann boundary conditions, $X^{m\prime }=0$, independently for each target-space component. Thus a fixed endpoint or a free endpoint makes the boundary contribution vanish. Fixing a spatial point throughout time imposes Dirichlet boundary conditions on its spatial coordinates, while the time coordinate may retain Neumann boundary conditions.

The independent metric tensor equation must also be retained after gauge fixing. Its Virasoro constraints areTogether with the wave equation, these ensure equivalence to the Nambu–Goto action; varying only the already fixed metric would lose these Virasoro constraints.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.