Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 79 / 2

Write $A_{ij}={\partial }_j{u}_i$ and use incompressibility, $A_{ii}=0$. The displayed vector can be expressed asOn taking its divergence, ${\partial }_i{A}_{ij}=0$. Commuting the remaining partial derivatives giveswhich cancels the derivative of the second term of $D_i$. The derivatives falling on $u_k$ leaveIn homogeneous turbulence, averaging commutes with differentiation and $⟨D_i⟩$ is position independent, provided these moments exist. Consequently $⟨\mathrm{tr}(A^3)⟩=0$.

Using the strain-rate tensor and vorticity decomposition supplied in the question now gives the Betchov relation:This uses homogeneity and incompressibility; isotropy is unnecessary.

The principal strain rates $a,b,c$ are the eigenvalues of the symmetric strain-rate tensor. Incompressibility makes its trace zero, so $a+b+c=0$. The polynomial identitytherefore gives $a^3+b^3+c^3=3abc$. Since the trace of $S^3$ is the sum of the cubes of its eigenvalues, the Betchov relation becomes

In established three-dimensional turbulence, mean vortex stretching is positive: $⟨{\omega }_i{\omega }_j{S}_{ij}⟩>0$. The Betchov relation then gives $⟨abc⟩<0$. Bi-axial strain has two positive principal strain rates and one negative rate, hence $abc<0$; extension along one axis with compression along the other two instead gives $abc>0$.

Thus the cubic strain statistic favors bi-axial strain. Strictly, this is an amplitude-weighted preference: a negative average alone cannot prove that there are more events of one type, since a few strong events could dominate it. Positive mean vortex stretching is additional physical information, not a consequence of homogeneity alone.

Bi-axial strain spreads a nearly incompressible material volume along two directions while thinning it along the third, favoring sheet-like concentrations of vorticity. A sheet can then become unstable and roll up into vortex tubes. The Townsend-Betchov cascade cartoon uses this sequence to reconcile strain-favored sheets with the intense tubes observed at dissipative scales. Alignment of vorticity with the principal strain rates also affects the stretching; the determinant statistic does not specify the entire geometry.

For a Burgers vortex, integrating the axial vorticity gives the azimuthal velocity:The swirl contribution to the viscous dissipation per unit length and per unit density isFor the equality, expand the two squares using ${\omega }_z=u_{\theta }^{\prime }+u_{\theta }/r$; their integrated difference is proportional to $[u_{\theta }^2{]}_0^{\infty }$, which vanishes. Substitution of the Gaussian vorticity profile givesThis excess dissipation of a Burgers vortex is independent of kinematic viscosity at fixed strain $\alpha$ and circulation $\Gamma$. Multiply by the mass density if a dimensional power per length is wanted.

The qualification "excess" matters. The imposed uniform strain itself has viscous dissipation density $3\nu {\alpha }^2$ per unit mass, so its integral over an infinite cross-section diverges. One must subtract that background or restrict to a finite cross-section. In a core-sized area, its contribution is of order ${\nu }^2\alpha$, smaller than the swirl contribution by order $(\nu /\Gamma {)}^2$ as $\Gamma /\nu \to \infty$.

Meanwhile $\delta \propto {\nu }^{1/2}$ shrinks and the central vorticity $\Gamma /(\pi {\delta }^2)$ grows. The finite dissipation becomes concentrated in a narrow tube: a useful local model of internal intermittency and the turbulent dissipation anomaly, though not a proof that an entire turbulent flow consists of such vortices.