Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 306 1 i Solution Created 2026-10-03 Updated 2026-10-05
Use the Minkowski metric and units with speed of light one. The contractions are and ; and 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 physical degrees of freedom per point.
In Monge gauge, and . Write the transverse canonical variables as and . Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, , its proper length element is , and . Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has . Monge gauge is a local choice on a string embedding map for which is a valid coordinate; it need not cover folded strings or all endpoint configurations.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 306 1 iv Solution Created 2026-10-03 Updated 2026-10-05
Let , with , and let . Direct differentiation gives , while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is . The configuration solves the Nambu–Goto action equations in its interior.
Place the fixed end at . The first free endpoint has at . For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval , since that interval was a coordinate convention. The proper length is measured on a constant- slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.
All points have the same phase , hence the segment rotates rigidly with angular velocity . At radius , its speed is . Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: and the Virasoro constraints imply at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.
In this conformal gauge, , so the energy density per is . Equivalently, the energy density per proper length is . 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 , . Their proper length, energy and angular momentum are times the values above, and . The displayed answer uses the implicit unfolded-segment convention.