Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 78 1 Solution Created 2026-10-03 Updated 2026-10-14
Let the fluid approach a uniform, quiescent state at infinity. Use the outward unit normal vector , pointing from the object into the fluid. Write for the Heaviside step function, , and . With the viscous stress tensor , defineThus is the Lighthill stress tensor, and is the surface loading exerted on the fluid. The total force on the object has the opposite sign. Extend by zero inside the object. Since the boundary is fixed and impermeable, , and the distributional mass conservation and momentum conservation equations areFor example, the surface contribution to the momentum equation is . Differentiating the first equation in time and subtracting the divergence of the second gives the fixed-body Ffowcs Williams-Hawkings equation:There is no acoustic thickness noise: a fixed impermeable boundary injects no mass. Apply the retarded acoustic Green function , where . The supplied surface-delta identity converts the loading term to a surface integral. For an exterior observer, the required integral equation isThis acoustic analogy is an exact rearrangement of the fluid equations; source approximations enter later. In the distant linear acoustic region, .
The acoustic far field requires observer distance much larger than the source extent and , so derivatives of the retarded phase dominate derivatives of . The acoustic compact-source approximation requires , so the source's differential propagation delays can be neglected. These conditions concern different lengths and neither implies the other. Put , , and define the loading and stress momentsThe leading compact acoustic far field isThe surface term is an acoustic dipole; the volume term is an acoustic quadrupole. The plus sign of the loading term follows from the minus sign of its spatial divergence and the derivative of retarded time.
For the power comparison, assume low Mach number , high Reynolds number, eddy size , characteristic time , and stress magnitude . Keep the prescribed turbulent region and its leading volume stress comparable when inserting the object; neglect changes of source statistics and exceptional cancellations of the leading moments. Compactness follows from . Since and the outgoing acoustic intensity is , the volume power isIf the object's size is comparable to and its fluctuating resultant loading is , thenAngular factors and dimensionless loading coefficients have been omitted. The acoustic dipole and acoustic quadrupole interference vanishes in the leading power integrated over a full sphere, by odd angular parity; individual directions can show interference.
A size threshold also requires a loading model. For a small nonseparating object of size in an approximately inviscid, smoothly varying incident eddy, a spatially uniform pressure has zero resultant on its closed surface. The first surviving loading is the pressure-gradient or added mass force, . Under these explicitly stated assumptions,The two mechanisms become comparable at ; for smaller objects the cases with and without the object have the same leading order of power. This is the small-body loading-noise threshold. It is not a geometry-independent size law: a separated drag force fluctuating on the original eddy time instead gives and . If both force and time are set by body-scale eddies, , their loading power is and comparison with the original volume source gives . Each estimate states which source time and force it uses.
The physical distinction is momentum conservation. Unforced turbulence in a fluid supplies internal stresses, whose net-force contribution cancels and leaves acoustic quadrupole radiation. The supported object exchanges momentum with the fluid and the external support, permitting a time-dependent resultant force and acoustic dipole radiation. One fewer retarded derivative makes the loading radiation more efficient at low Mach number.