Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 3 36D iii Solution Created 2026-09-24 Updated 2026-10-05
Let be the smooth nonzero tangent along a regular null curve, and . Metric compatibility gives . In a two-dimensional Lorentzian manifold, the orthogonal complement of a nonzero null vector is its own span; this follows directly in the preceding null basis. Therefore .
Consequently every regular null curve is a geodesic up to reparametrization. To obtain an affine parameter , write and solve . Then . A solution exists locally and along any nonsingular parameter interval. The original parameter need not be affine; the literal assertion of zero covariant acceleration for every parameter would be false.