Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 74 4 iii Solution Created 2026-10-03 Updated 2026-10-14
Let and use for the source radius, reserving for observation distance. Incompressible spherical potential flow satisfies . Its velocity potential, tending to zero at infinity, and velocity areThe unsteady Bernoulli equation givesThe surface pressure is independent of angle. Hence the integrated surface force is zero, , and the leading compact surface dipole contribution vanishes. This also makes in the notation of the preceding part.
The volume flux is , so the acoustic thickness noise contribution isThe leading part is linear in the radial oscillation. Keeping the quadratic terms explicitly gives .
For the acoustic quadrupole, use . To leading quadratic order,ThusKeeping the instantaneous lower limit instead gives , whose leading small-oscillation term is the expression just obtained. Although the stress moment is isotropic, its contracted second derivative is not zero: .
For a sinusoidal radius perturbation , the leading compact radiation moments of a pulsating spherical bubble areWriting and , the quadrupole amplitude relative to the linear monopole is . The monopole also has a quadratic correction, ; the volume-stress quadrupole is smaller than this quadratic monopole correction by .
In the small-amplitude compact regime, the linear monopole dominates, the leading dipole is zero, and the volume quadrupole is weaker and begins at twice the oscillation frequency. The compact formulas require as well as a far observer. The stated small surface Mach number alone does not imply compactness. The incompressible near field has a decaying tail, but its stress moment is dominated by distances of order ; an acoustic matching region is still needed outside it. These are leading compact contributions, not a claim to retain every finite-wavelength correction at the smaller quadrupole order.