Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 79 / 3 / ii

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.