Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 49 3 a iii Solution Created 2026-10-03 Updated 2026-10-07
Variation of the inverse metric relation gives . Hence the metric variation of scalar curvature isPut . The connection variation gives the contractionsUsing metric compatibility and the Palatini identity,Renaming dummy indices givesNo interchange of covariant derivatives on a tensor is needed in this derivation, so no hidden curvature-commutator term is discarded.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 309 2 b Solution Created 2026-10-03 Updated 2026-10-06
Use spatial indices , set , and assume is smooth. The inverse metric has and . The Christoffel symbols of the Levi-Civita connection follow from :All other components vanish. Although diverges, its affine connection has a smooth limit on compact subsets:This is the Newtonian connection from an exponential lapse. It is torsion-free, and its geodesic equation, using as an affine parameter when , is . Thus has the role of a Newtonian gravitational potential. There is no finite limiting nondegenerate Lorentzian metric in these fixed coordinates; tends to , a rank-one tensor. The limiting connection preserves and the contravariant spatial tensor with components , , which describe the degenerate temporal/spatial structures of this limit.
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 4 14E Solution Created 2026-09-24 Updated 2026-10-06
The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy is constant along its solutions. Constant curves are also geodesics.
For the Poincare half-plane model, translation in gives the conserved quantity , and the energy is . If , the nonconstant geodesics are vertical lines, with . If ,Writing and integrating yields . Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.
For the two specified points, the circle centre is and the radius is five:For example, an affine unit-speed parametrization is , . The endpoints on the real axis are and , outside the hyperbolic surface.
The upper-half-plane geodesic through the two specified points
. Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 2 38B a Solution Created 2026-09-24 Updated 2026-09-25
For coordinates , the Christoffel symbols of the Levi-Civita connection arewhere is the inverse metric.
For the two-dimensional hyperbolic metric in polar coordinates,Only has a nonzero derivative, namelyIt follows that the nonzero connection coefficients are
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 2 38B d Solution Created 2026-09-24 Updated 2026-09-25
Contracting the Riemann curvature tensor gives the nonzero components of the Ricci tensor:The Ricci scalar is the contraction with the inverse metric. ThereforeIt is independent of both coordinates, so the hyperbolic plane has constant negative scalar curvature.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 309 1 a ii Solution Created 2026-09-24 Updated 2026-09-25
Write and . Varying with fixed endpoints and then choosing proper time as parameter, for which , gives the Euler-Lagrange equationExpanding the derivative, raising the free index with the inverse metric, and using the symmetry of gives the geodesic equationThese are the Christoffel symbols of the Levi-Civita connection.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 357 1 a Solution Created 2026-09-24 Updated 2026-09-25
The inverse metric condition isKeeping only terms linear in the metric perturbation givesThus the linearized inverse metric iswhere indices on are raised with the Minkowski metric.
