Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 73 / 1

Work with an incompressible Newtonian fluid of constant kinematic viscosity. The vorticity equation is $D\mathbf{\omega }/Dt=(\mathbf{\omega }\cdot \nabla )\mathbf{u}+\nu {\nabla }^2\mathbf{\omega }$. In the axisymmetric Burgers vortex the swirl does not advect the axial vorticity, and the stretching rate is ${\partial }_z{u}_z=\alpha$. Put $s(t)={\delta }^2(t)$ and $W={\omega }_z$. The Gaussian profile hasSubstitution into ${\partial }_{t}W+u_r{\partial }_{r}W=\alpha W+\nu {\nabla }^{2}W$ matches both the constant and $r^2$ terms precisely whenFor every finite positive initial $s_0$,The strain compresses the cross-section and stretches its vorticity, while kinematic viscosity spreads the core. The attracting radius is the balance of these effects; its relaxation time is ${\alpha }^{-1}$. Total circulation stays fixed because stretching increases the vorticity as the cross-sectional area decreases.

At the steady core radius, the swirl speed is of order ${\Gamma }_0/\delta$, and its velocity gradient tensor is of order ${\Gamma }_0/{\delta }^2$. Its viscous dissipation integrated over a cross-section therefore scales asThis is per unit length and per unit density. More precisely, for the swirl contribution, integration by parts relates the integrated strain viscous dissipation to $\nu \int W^{2}dA$; the boundary term vanishes for the $1/r$ swirl velocity. ThusMultiply by density for power per physical length. Here the azimuthal velocity's circulation $\Gamma$ is ${\Gamma }_0$, as follows by integrating $W$ over the cross-section. The result is finite as $\nu \to 0$ with strain and circulation held fixed. Local core viscous dissipation grows like ${\nu }^{-1}$ while its area shrinks like $\nu$. For ${\Gamma }_0/\nu \gg 1$, the swirl gradients dominate the background strain within the core.

This finite quantity is the excess viscous dissipation due to the vortex. The uniform straining flow alone dissipates $3\nu {\alpha }^2$ per unit volume, and integrating that over an infinite cross-section diverges. The total viscous dissipation of the complete idealized flow is therefore not a finite tube power unless the background is subtracted or spatially bounded.

A Burgers vortex sheet instead smooths a fixed velocity jump $U$ across a layer of thickness ${\delta }_s\sim \sqrt{\nu /\alpha }$. Its viscous dissipation per sheet area is of orderA tube with fixed circulation supports finite viscous dissipation in a vanishingly small core; a sheet with fixed velocity jump does not. This makes strained vortex tubes plausible intense dissipative structures at high Reynolds number. It does not assert that sheets are absent from turbulence: sheets can carry inertial range organization and become unstable, producing finer vortical structures.

For unforced incompressible flow, multiply the vorticity equation by ${\omega }_i$. The antisymmetric part of the velocity gradient tensor drops out of ${\omega }_i{\omega }_j{\partial }_j{u}_i$, leavingStatistical homogeneity removes mean divergences. Integration by parts, together with $\nabla \cdot \mathbf{\omega }=0$, givesThe last equality holds for the averages, not pointwise. It follows from the usual curl/divergence identity after the divergence terms have averaged to zero. Vortex stretching produces enstrophy, while viscous diffusion destroys it.

Let $T=\ell /u$, $ϵ\sim u^3/\ell$, and use the Kolmogorov microscales $\eta =({\nu }^3/ϵ{)}^{1/4}$ and ${\tau }_{\eta }=(\nu /ϵ{)}^{1/2}$. Since $⟨{\omega }^2⟩=ϵ/\nu$, a slowly changing mean enstrophy has a time derivative of order $ϵ/(\nu T)$. The usual small-scale estimate givesThe relative imbalance is consequently ${\tau }_{\eta }/T\sim (\nu /u\ell {)}^{1/2}={\mathrm{Re}}^{-1/2}$, yieldingThe estimate assumes an established high-Reynolds-number cascade and integral-time evolution of the mean, not a rapid initial transient. In a forced flow the exact balance includes $⟨\mathbf{\omega }\cdot \nabla \times \mathbf{f}⟩$; forcing smooth on the integral scale of turbulence is smaller than the leading small-scale terms.

Use $A=S+\Omega$, where $S$ is the rate-of-strain tensor and $\Omega$ is antisymmetric. For incompressible flow $\mathrm{tr}S=0$. Since ${\Omega }^2=(\mathbf{\omega }{\mathbf{\omega }}^T-{\omega }^{2}I)/4$, taking a trace givesThis uses the positive convention $R=\mathrm{tr}{A}^3/3$ specified here; a convention with the opposite sign for $R$ must not be mixed into the calculation. For a compressible flow there is an additional term $-3{\omega }^2\mathrm{tr}S/4$, so the displayed incompressible formula is not literally valid for every flow.

One can exhibit the required divergence directly. DefineDifferentiating, terms containing ${\partial }_i{u}_i$ vanish, and the two terms with a second derivative cancel after relabeling indices. Thereforein homogeneous turbulence with finite differentiable moments, or under periodic averaging. This is the Betchov relation. The preceding trace identity then impliesThe principal strain rates are the real eigenvalues of the symmetric matrix $S$. Incompressibility gives $a+b+c=0$, and the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ givesNo isotropy is required for this averaged tensor identity: statistical homogeneity and incompressibility are the key assumptions.

The high-Reynolds-number balance gives positive mean vortex stretching. The Betchov relation therefore gives $⟨abc⟩<0$. To connect this to a fixed longitudinal velocity derivative, additionally assume isotropic turbulence, or average over the direction of that derivative. For a unit vector $\mathbf{n}$, the derivative is $d={\mathbf{n}}^{T}S\mathbf{n}$. Uniform directional averaging, with $\mathrm{tr}S=0$, givesFor example these identities follow by averaging products of direction cosines on a sphere. Hence in isotropic turbulenceand division by the positive derivative variance to the power $3/2$ gives negative longitudinal velocity-gradient skewness. Homogeneity alone does not fix the orientation statistics and does not establish this sign for every chosen coordinate direction.

Order the principal strain rates as $a\ge b\ge c$. Except at degeneracies, $a>0$ and $c<0$, so $abc<0$ corresponds to $b>0$: two extensional directions and one compressive direction, or bi-axial strain. The negative average of $abc$ says that such configurations dominate this cubic amplitude-weighted statistic over axial-extension configurations with $b<0$. In the usual physical interpretation, comparably intense strain events are therefore biased toward bi-axial extension.

An unweighted frequency claim needs more information. As a counterexample to deducing it solely from the cubic average, assign strain eigenvalues $(1,1,-2)$ to 10% of samples and $(-0.1,-0.1,0.2)$ to 90%, with randomly oriented eigenvectors. Then $⟨abc⟩=-0.1982<0$, despite bi-axial extension occurring in only 10% of the samples. This example addresses what the moment inequality can infer, not a construction of a complete turbulent velocity field. A measured frequency preference is a further statistical observation.

The geometric significance is that a fluid element stretches in two directions and contracts in the third, becoming sheet-like while preserving volume. Distributed vorticity is swept into thin layers and its component along an extensional direction is amplified. The associated shear layers can subsequently roll up and produce vortex tubes. Thus the weighted strain bias helps explain sheet formation in the energy cascade even though the most intense dissipative structures can be tubes.

In the Townsend-Betchov cascade cartoon, bi-axial strain sweeps distributed vorticity into extended sheets at the upper part of the inertial range. These are predominantly inviscid structures: their shape and instability arise from advection and strain rather than local viscous smoothing. Strong velocity variation across a thin sheet permits Kelvin-Helmholtz instability, roll-up and three-dimensional deformation. The resulting small vortex tubes are stretched into still finer cores.

At the dissipative end, stretching and kinematic viscosity balance in Burgers-like cores. A core of strain rate $s$ has radius of order $(\nu /s{)}^{1/2}$; taking $s\sim (ϵ/\nu {)}^{1/2}$ gives the Kolmogorov length $({\nu }^3/ϵ{)}^{1/4}$. Circulation and shrinking cross-section allow finite integrated tube viscous dissipation. Thus sheet formation and tube formation play different roles, rather than the entire cascade being a succession of identical geometrical eddies.

The cartoon challenges a literal, uniformly self-similar, step-by-step cascade with independent interactions only between neighboring scales. Large-scale sheet formation can generate much smaller scales through an instability, and the resulting structures retain coherent spatial organization. A branching model that assumes every scale-transfer step has the same geometry and loses information about the parent flow can miss this mechanism. Closures that erase the relevant phase correlations also have difficulty representing its concentrated gradients and intermittency. The objection is to these dynamical assumptions, not to conservation of energy or the existence of a mean energy flux through scales. The cartoon alone is not a proof that all spectral transfer is nonlocal.