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Put $W=⟨{\omega }^2⟩$, $C=⟨|\nabla \times \mathbf{\omega }{|}^2⟩$ and $M=⟨G^3⟩$, where $G={\partial }_x{u}_x$. The small-$r$ Taylor series giveIn the Kármán-Howarth equation, the coefficient of $r^2$ on the left is $-\dot{W}/30$. On the right, the triple-correlation term contributes $7M/6$ and the viscous term contributes $\nu C/15$. HenceComparing with the mean enstrophy balance identifies the mean vortex stretching:Also $S_2=2(u^2-F)=W{r}^2/15+\cdots$, whereas differentiability gives $S_2=⟨G^2⟩r^2+\cdots$. Thus $⟨G^2⟩=W/15$. Homogeneity makes $⟨G⟩=0$, so the longitudinal velocity-gradient skewness is $S_0=M/(W/15{)}^{3/2}$. Substitution givesPositive mean vortex stretching therefore requires negative skewness. A centered Gaussian distribution has zero third moment, so the derivative statistics cannot be Gaussian. This conclusion uses the physically positive mean stretching, not homogeneity alone.

The probability density function has mean zero but a longer or stronger negative tail, representing intermittent compression. The illustration uses a standardized Gaussian mixture distribution solely to sketch this sign of skewness; it is not turbulence simulation data or a fitted universal distribution.

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.