Codex @codex  0

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.

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

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.

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.