Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 76 2 c Solution Created 2026-10-03 Updated 2026-10-06
Use the same physical normalization for both fields. Part (b) has , so the common energy flux prefactor cancels. The differential cross-section for scalar wave scattering isThe far-field pattern has dimensions of length in this normalization, so the differential cross-section for scalar wave scattering has dimensions of area per unit solid angle. Only the scattered field is used in this ratio; interference between the incident and scattered fields is a different contribution to the total field's energy flux.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 321 2 a Solution Created 2026-10-03 Updated 2026-10-06
Let be the kinetic energy per unit volume. Dot the momentum equation with . The Coriolis acceleration does no work because . Incompressibility turns the advection and pressure terms into divergences, yieldingThus is the energy flux and is the source in the requested kinetic-energy form.
To decide about conservation of energy, absorb the ordinary radial shearing-sheet tidal potential into the density. Its material derivative cancels the work term:For this is the usual conservative Jacobi energy in a shearing sheet, subject to vanishing boundary energy flux. For , the extra body force has nonzero curl, , so no scalar potential can absorb it. A potential would necessarily introduce a vertical force , absent from the model. Therefore this vertical-shear model does not generically conserve the standard total mechanical energy: its maintained shear/irradiation background can supply or remove energy. This is the energy balance of the incompressible vertical-shear model.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 314 2 a Solution Created 2026-10-03 Updated 2026-10-05
Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio , and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is , and the specific enthalpy is . Integrating the local conservation laws across a thin stationary control volume makes the mass flux , momentum flux , and energy flux continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state must therefore not be imposed on both sides. For flow from to , both signed velocities are negative; the flux equations and ratios remain valid.
In the Strong-shock Rankine-Hugoniot conditions, tends to zero. Put . The first two flux balances give and to leading order. The energy balance becomesDiscarding the unchanged-state root gives the compressive shock compression ratio and the required limits:
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 345 1 a Solution Created 2026-10-03 Updated 2026-10-05
Let denote the fluid pressure perturbation divided by the reference mass density, and let be the buoyancy perturbation. The buoyancy frequency satisfies . Linearization requires small boundary slope , small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes ; boundary slope alone is insufficient when . The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.
The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as . Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber , this gives the dispersion relationWhen , define . The radiation condition selects , because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is , its phase velocity is , and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination above the horizontal satisfies .
When , set . Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over . There is no upward time-averaged energy flux, because and are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity .
For the response is the evanescent wave with ; the ratios should not be used. At the cutoff , the bounded harmonic solution has , : it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.
Propagating and evanescent responses
. The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.
Past exam of the mathematics course of the University of Cambridge 2019 ib Paper 4 7A Solution Created 2026-09-24 Updated 2026-09-29
In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to ; taking the positive-frequency branch gives .
For , , and , the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 345 1 d Solution 2026-09-28
Setwhere is the magnitude of the internal-wave ray slope and is the bottom slope. A rightward ray remains in the triangular basin precisely when its bottom reflection is subcritical internal-wave reflection, namely . Therefore
Conservation of frequency and of the component of the wavevector tangent to the slope gives the focusing power of internal-wave reflectionwhere . In this range , so the reflected wavelength is shorter by the factor . The reflected normal group velocity is smaller by ; conservation of normal energy flux therefore increases the wavelength-averaged energy density by
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 39C a Solution Created 2026-09-24 Updated 2026-09-25
The one-dimensional mass flux and momentum flux are respectivelyIn a steady flow, the conservation laws make both the mass flux and the given energy fluxconstant. Wherever , their ratio is therefore constant:where is the internal energy per unit volume of the ideal gas. This is the required steady-flow energy integral.
Surface albedo 2026-10-07
Surface albedo is the fraction of incident solar energy flux reflected by a surface, averaged over the specified spectrum and illumination geometry. The locally absorbed fraction is . It is distinct from geometric albedo, which measures brightness in a particular viewing configuration.
