= Solution
Put $W=\langle\omega^2\rangle$, $C=\langle|\nabla\times\boldsymbol\omega|^2\rangle$ and $M=\langle G^3\rangle$, where $G=\partial_xu_x$. The small-$r$ <Taylor series> give
$$
F=u^2-\frac{Wr^2}{30}+\frac{Cr^4}{840}+\cdots,\qquad S_3=Mr^3+\cdots.
$$
In 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 $\nu C/15$. Hence
$$
-\frac{\dot W}{30}=\frac76M+\frac\nu{15}C,
\qquad
\boxed{\frac12\dot W=-\frac{35}{2}\langle G^3\rangle-\nu C}.
$$
Comparing with the <mean enstrophy balance> identifies the mean <vortex stretching>:
$$
\langle\omega_i\omega_jS_{ij}\rangle=-\frac{35}{2}\langle G^3\rangle.
$$
Also $S_2=2(u^2-F)=Wr^2/15+\cdots$, whereas differentiability gives $S_2=\langle G^2\rangle r^2+\cdots$. Thus $\langle G^2\rangle=W/15$. Homogeneity makes $\langle G\rangle=0$, so the <longitudinal velocity-gradient skewness> is $S_0=M/(W/15)^{3/2}$. Substitution gives
$$
\boxed{\langle\omega_i\omega_jS_{ij}\rangle=-\frac7{6\sqrt{15}}S_0\langle\omega^2\rangle^{3/2}}.
$$
Positive 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-79-gradient-skewness.png]
{title=Illustrative negatively skewed longitudinal velocity-gradient density}
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