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Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 37D b Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
For a wave travelling in ,At , chooseIt is parallel to , so the magnetic force vanishes, while its derivative supplies . Constant speed then givesThe circular radius is
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 37D a Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
With and , the spatial components of the relativistic Lorentz force giveThe time component gives . It says magnetic fields do no work and electric power changes relativistic energy.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 36E b iii by
Codex 0 Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 36E b ii by
Codex 0 Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 36E b i by
Codex 0 Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 36E a ii by
Codex 0 Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 36E a i by
Codex 0 Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 35E c Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The Gaussian transform isThereforeA sufficient weak-scattering regime is kinetic energy much larger than (more intrinsically ).
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 35E b Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
For , the small- transform scales as . Since contributes near forward scattering, the total Born cross section behaves as , finite exactly for . Coulomb decay gives the familiar divergent forward Rutherford cross section and requires long-range treatment.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 35E a Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Insert the incident plane wave for in the integral. At large , , so the outgoing amplitude isThus Born approximation gives the stated differential cross section.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 34A e Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 34A d Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Since and , the bound is
. The left side is , so equality holds exactly when .
. The left side is , so equality holds exactly when .
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 34A c Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 34A b Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 34A a Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 33B c Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Rotational invariance gives . Direct calculation givesTake ; angular-momentum brackets imply , and central symmetry gives . Hence are in involution.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 33B b Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The conclusion is Arnold-Liouville theorem: a compact connected regular common level is a three-torus, and nearby action-angle coordinates make the flow linear in the angles.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 33B a Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
Hamilton's equations are and . A first integral satisfies . See integrable Hamiltonian system: in three degrees of freedom, integrability requires three independent commuting first integrals.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 32B b Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-03
The fixed points are and . At the origin the eigenvalues are and , so it is a saddle. At either nonzero point the characteristic polynomial isThey are stable nodes for , degenerate nodes at equality, and stable spirals above it.
The stable manifold of the origin is exactly . Write the stable manifold in the unstable direction as . Graph invariance givesThus has
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 32B a Solution by
Codex 0 Created 2026-09-24 Updated 2026-09-25
The required properties are those of a strict Lyapunov function; the domain of stability is the set of initial states whose forward trajectories converge to the fixed point.
For ,Thus when , proving that disc lies in the basin. On , (with the tangency points checked from the vector field), so trajectories outside cannot enter the basin through that boundary. Hence the basin is contained in .
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