Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 73 / 3

Consider an unforced incompressible flow with localized vorticity and sufficient smoothness for integration by parts. Define the hydrodynamic impulse per unit density by $\mathbf{I}=\frac{1}{2}\int \mathbf{x}\times \mathbf{\omega }dV$, denoting the vector impulse by $\mathbf{I}$ to distinguish it from the later scalar $L$. The vorticity equation can be writtenThe supplied identity, integrated over all space, gives $\frac{1}{2}\int \mathbf{x}\times \nabla \times \mathbf{F}dV=\int \mathbf{F}dV$, because $\mathbf{F}$ is localized and its surface terms vanish. ThereforeThe second integral is a vanishing surface integral. The vector identity $\mathbf{u}\times \mathbf{\omega }=\nabla (|\mathbf{u}{|}^2/2)-(\mathbf{u}\cdot \nabla )\mathbf{u}$ and incompressibility reduce the first to surface terms proportional to the quadratic velocity. Those vanish for $\mathbf{u}=O(R^{-3})$. Thus

For a large spherical control volume define ${\mathbf{P}}_R={\int }_{V_R}\mathbf{u}dV$. The impulse relation is $\mathbf{I}=(3/2){\lim }_{R\to \infty }{\mathbf{P}}_R$ for this localized field. The factor is important: impulse is not simply the integral of velocity, since an $R^{-3}$ velocity tail has a nonvanishing contribution in the impulse integration by parts involving a factor of $R$.

Using pressure per unit density $\Pi$ and the Navier-Stokes equation givesThe advective flux is $O(R^{-4})$, the pressure force is $O(R^{-1})$, and the viscous term is $O(\nu R^{-2})$ under the regular far-field derivative decay. All vanish as $R\to \infty$. The localized eddy cannot change its total impulse without a net external momentum input. Pressure redistributes momentum internally but its resultant force on an increasingly distant control surface tends to zero. At finite $R$, boundary forces need not vanish exactly; the all-space limit is the invariant statement.

Let $C(r)=⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩$ in zero-mean isotropic turbulence. Define the scalar Saffman integral $L={\int }_{{\mathbb{R}}^3}C(r)d^{3}r=4\pi {\int }_0^{\infty }{r}^{2}C(r)dr$. The spectral transform isFor an integrable $r^{2}C(r)$, dominated convergence applied to $\sin (kr)/(kr)$ givesA further $O(k^4)$ expansion requires an additional correlation moment and is not automatic from finite $L$. The normalization agrees with $\int Edk=⟨|\mathbf{u}{|}^2⟩/2$.

For a control volume $V$ put ${\mathbf{P}}_V={\int }_V\mathbf{u}dV$. Statistical homogeneity gives the exact finite-volume identityFor geometrically similar volumes increasing without bound, the overlap fraction $|V\cap (V-\mathbf{r})|/|V|$ tends to one at every fixed separation. If $C$ is absolutely integrable, dominated convergence givesThe printed finite-large-volume expression is the corresponding asymptotic approximation, not an exact identity for an arbitrary finite volume. For instance, positive nonzero $C$ with a finite correlation length gives an overlap fraction smaller than one and hence a strictly smaller finite-volume value. The limit also shows $L\ge 0$; a nonzero Saffman integral is positive.

A large spherical region contains $N\sim V_R/{\ell }^3$ energy-containing eddies. If their hydrodynamic impulses have zero mean, finite variance and sufficiently short-range correlations, the mean-square sum grows as $N$, rather than $N^2$. In the ideal independent case,The spherical impulse-momentum relation makes the integrated velocity proportional to this sum for isolated enclosed structures. In a full field, external dipolar velocities and boundary effects can change the coefficient; they do not alter the volume scaling under these random-impulse statistics. ThusIt is a random walk law for momentum fluctuations, not an extensive mean flow. Although the mean impulse is zero by isotropy, its variance need not vanish. The Saffman integral measures this large-volume momentum-fluctuation density. If every eddy had exactly zero impulse or correlations enforced systematic cancellation, this leading density could vanish and a different small-wavenumber spectrum could result.

Write $u^2=⟨u_x^2⟩$ for the one-component variance and $Q_{LL}=u^{2}f(r)$. Isotropy gives $Q_{ij}=u^2[(f-g)n_i{n}_j+g{\delta }_{ij}]$, and incompressibility gives $g=f+r{f}^{\prime }/2$. Taking the trace yieldsTherefore, with a regular correlation at the origin,For $L>0$, the longitudinal velocity correlation has the tail $u^{2}f(r)\sim L/(4\pi r^3)$. Its physical source is the velocity induced far from a localized eddy with nonzero hydrodynamic impulse. The hydrodynamic Biot-Savart kernel gives the dipolar far fieldThe absence of a vorticity monopole and the surviving impulse dipole give the $r^{-3}$ power. Random eddy impulses thus produce long-range velocity correlations despite the vorticity itself being concentrated in localized structures. Incompressibility also explains how a distant pressure response can redistribute velocity without transporting local vorticity there.

There is no inconsistency with integrability of the trace correlation. At leading order, $g\sim -f/2$, soThe longitudinal and transverse dipolar tails cancel in the trace. The scalar Saffman integral can remain finite even though individual tensor components have long-range, direction-dependent tails.

Multiply the Kármán-Howarth equation for $C(r,t)$ by $4\pi r^2$ and integrate in $r$. Integration by parts gives the exact boundary-flux expressionAt the origin smooth isotropic correlations give $K=O(r)$ or smaller and $C^{\prime }=O(r)$, so both boundary terms vanish. At infinity the usual regular power-law decay $K=O(r^{-4})$, with the corresponding derivative decay, makes the first term tend to zero. A sufficiently smooth integrable trace correlation makes $r^2{C}^{\prime }\to 0$ as well. Under these physical far-field conditions,The exact required conditions are the limits of these two boundary fluxes. A bare big-$O$ condition on $K$ without control of its derivatives is not by itself a rigorous sufficient condition: the smooth tail $K(r)=r^{-4}\sin (r^2)$ is $O(r^{-4})$, but $r^{-1}(r^{4}K{)}^{\prime }=2\cos (r^2)$ has no zero limit. The intended power-law assumption excludes such increasingly rapid far-field oscillations; this qualification is needed in a literal proof.

Physically, $L$ is the large-volume momentum variance divided by volume. Internal advective and pressure forces exchange momentum among eddies but do not provide a net impulse to the whole unforced field. For a large spherical region, the variance balance is driven by its boundary momentum flux. The stated correlation decay makes that contribution negligible per unit volume as the region grows. Thus the conservation law preserves momentum-fluctuation density, not kinetic energy. kinematic viscosity can remove kinetic energy while leaving this integral invariant.

Let $\mathcal{E}=⟨|\mathbf{u}{|}^2⟩/2$ be the turbulent kinetic energy per unit mass. In addition to $L>0$ and high Reynolds number, use the usual self-preserving large-eddy decay assumption: the normalized large-scale correlation has a fixed shape, sowhere $C_L>0$ is constant. For freely decaying high-Reynolds-number turbulence, the cascade viscous dissipation law is $ϵ=C_{ϵ}{\mathcal{E}}^{3/2}/\ell$. The kinetic-energy balance therefore givesPut $A=C_{ϵ}{C}_L^{1/3}{L}^{-1/3}$. Integrating,The constant $t_{\ast }$ is a virtual time origin. This is the Saffman decay law. The energy decreases while the integral scale of turbulence grows so that the conserved large-volume momentum variance remains constant.

The exponent does not follow from the invariant alone: it also uses unforced decay, one evolving large-eddy length and an asymptotically constant cascade viscous dissipation coefficient. A changing normalized spectral shape or a confining boundary can invalidate that closure. Moreover $\mathrm{Re}\sim {\mathcal{E}}^{1/2}\ell /\nu \propto (t-t_{\ast }{)}^{-1/5}$, so at fixed kinematic viscosity the high-Reynolds-number decay regime eventually ends; the $-6/5$ law is not a claim about the ultimate viscous limit.