Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 79 (Hydrodynamic Turbulence)

The zeroth law of turbulence states that at fixed outer velocity $u$ and integral scale of turbulence $\ell$, the mean viscous dissipation per unit mass approaches a nonzero finite value as the Reynolds number $\mathrm{Re}=u\ell /\nu$ becomes large. In an established forward energy cascade, this gives $ϵ\sim C_{ϵ}{u}^3/\ell$, with $C_{ϵ}$ of order one. The small-scale velocity gradients become large enough to offset decreasing kinematic viscosity; this is the turbulent dissipation anomaly.

At the Kolmogorov microscales, the local Reynolds number is of order one, $v\eta /\nu \sim 1$, and the viscous dissipation is $ϵ\sim \nu (v/\eta {)}^2$. Solving these balances givesSubstituting the outer-scale estimate, and omitting order-one constants, givesThe associated eddy turnover time is $\eta /v=(\nu /ϵ{)}^{1/2}$.

Local isotropy of turbulence means that joint small-scale velocity increment statistics are approximately unchanged by rotations, even when the large-scale flow is anisotropic. The universal equilibrium range contains separations much smaller than the integral scale of turbulence: their adjustment times are short compared with the outer eddy turnover time, and their normalized statistics are assumed independent of the detailed large-scale forcing. It includes the inertial range and the dissipation range.

The Kolmogorov first similarity hypothesis says that these locally isotropic statistics depend only on $r$, mean viscous dissipation $ϵ$ and kinematic viscosity $\nu$. Dimensional analysis therefore expresses the second-order longitudinal structure function in terms of the Kolmogorov microscales:The universality of $F$ is an assumption of the theory, not a consequence of dimensional analysis alone.

The Kolmogorov second similarity hypothesis removes dependence on kinematic viscosity when $\eta \ll r\ll \ell$. Only $ϵ$ and $r$ remain, so the Kolmogorov two-thirds law isThe same reasoning gives the predicted order-$P$ longitudinal structure function:These are signed moments for integer $P$; for noninteger orders, use $⟨|\Delta v{|}^P⟩$. In particular, the signed third-order coefficient is negative, ${\beta }_3=-4/5$, in the Kolmogorov four-fifths law.

External intermittency is the alternation of turbulent and non-turbulent fluid at a point as a fluctuating interface passes it. Internal intermittency in the equilibrium range instead describes uneven, bursty concentrations of small-scale velocity gradients and viscous dissipation within turbulent fluid. Integral-scale intermittency is modulation of the energy-containing motions and their energy supply. It is non-universal because it retains information about forcing, boundary conditions and the history of the large-scale flow.

The Landau intermittency counterexample challenges the unconditional universality asserted by Kolmogorov 1941 theory. Suppose turbulent populations have different conditional mean viscous dissipation rates ${ϵ}_a$. Even if each obeys the same conditional scaling, averaging over populations giveswhich generally differs from ${\beta }_P{r}^{P/3}⟨{ϵ}_a{⟩}^{P/3}$. Thus a coefficient normalized by the global mean depends on the large-scale intensity distribution. The third-order case escapes this particular objection because the power of ${ϵ}_a$ is one.

The Kolmogorov refined similarity hypothesis replaces global mean viscous dissipation by coarse-grained energy dissipation ${ϵ}_{\mathrm{AV}}(r)$. It assumes thathas universal conditional statistics at sufficiently large local Reynolds number. If the conditional $P$th moment is ${\beta }_P$, the law of total expectation givesThe refined similarity hypothesis does not itself prescribe the distribution of coarse-grained energy dissipation. Scale dependence of those moments can change the predicted exponents. As before, noninteger orders require absolute velocity increments.

A passive scalar $\theta$ is transported by the velocity field without feeding back on it:Take its mean to be zero, and define the scalar dissipation rate by the half-scalar variance convention $\chi =\kappa ⟨|\nabla \theta {|}^2⟩$. In a statistically equilibrated energy cascade of scalar fluctuations, this is also the flux of half-scalar variance toward small scales.

The Kolmogorov two-thirds law gives a typical velocity increment $\delta u_r\sim (ϵr{)}^{1/3}$ and hence eddy turnover time ${\tau }_r\sim r/\delta u_r\sim {ϵ}^{-1/3}{r}^{2/3}$. Assuming local scalar transfer on this same time scale, $\chi \sim ⟨(\Delta \theta {)}^2⟩/{\tau }_r$. The Obukhov-Corrsin theory therefore gives the scalar analogue:This applies to separations at which both direct forcing and molecular diffusion are negligible and the transporting velocity increments lie in the inertial range. For very large or very small ratios $\nu /\kappa$, the scalar and velocity cutoff scales differ, so this common range must be checked. Defining $\chi$ as dissipation of the full scalar variance instead simply changes the convention for $C_{\theta }$.

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.

Let $u^2$ denote one-component velocity variance, $F=u^{2}f$ the longitudinal velocity correlation, and $S_p=⟨(\Delta v{)}^p⟩$ the signed longitudinal structure functions. Homogeneity gives $F=u^2-S_2/2$, while the third-order convention here is $u^{3}K=S_3/6$. Thus the Kármán-Howarth equation becomesThe mean kinetic energy per unit mass is $3u^2/2$, so its decay gives ${\partial }_t{u}^2=-2ϵ/3$. In the universal equilibrium range, the small-scale variation ${\partial }_t{S}_2$ is negligible at leading order, hence ${\partial }_{t}F\simeq -2ϵ/3$. Integrating from zero and using regularity givesThe Kolmogorov equation for structure functions is thereforeFor finite local unsteadiness the right side has the additional term $-3r^{-4}{\int }_0^r{s}^4{\partial }_t{S}_2(s,t)ds$; dropping it is the local-equilibrium approximation used here.

In the inertial range the viscous term is also negligible, leaving the Kolmogorov four-fifths law, $S_3(r)=-4ϵr/5$. Its coefficient and sign follow from the exact energy balance and Kármán-Howarth equation, rather than from dimensional similarity. It supplies a firm third-order constraint on Kolmogorov 1941 theory and measures the forward energy cascade, while leaving the second- and higher-order exponents open to internal intermittency corrections.

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.

Combining the Kolmogorov two-thirds law with the Kolmogorov four-fifths law gives the velocity increment skewness in the inertial range:The constant-skewness turbulence closure assumes this same value throughout the universal equilibrium range, including the dissipative crossover. This is an extra modeling assumption; neither similarity law establishes it near $r=0$.

Write $A=\beta (15\beta {)}^{1/2}$, $B=(15\beta {)}^{3/4}$, $S_2=A{v}^{2}h$ and $r=B\eta x$. Since $\nu =v\eta$ and $ϵ\eta =v^3$, substituting $S_3=-4S_2^{3/2}/(5{\beta }^{3/2})$ into the Kolmogorov equation for structure functions givesDividing by $2B/5$ yields the normalized constant-skewness structure-function equation:At small $x$, $h\sim x^2$, recovering $S_2\sim ϵr^2/(15\nu )$ for differentiable velocity increments. At large $x$, the dominant balance gives $h\sim x^{2/3}$, recovering the assumed Kolmogorov two-thirds law.

This turbulence closure is inexpensive: a single ordinary differential equation interpolates between the two regimes while respecting the local third-order balance. Its prescribed skewness cannot independently predict scale-dependent derivative statistics, transient spectral evolution or internal intermittency.

EDQNM closure instead evolves the turbulent energy spectrum through modeled spectral triad interactions. Its quasi-normal fourth-order approximation is supplemented by eddy damping and a Markovian treatment of correlation memory. It retains more information about scale-dependent transfer and evolving spectra, at the cost of spectral integrals and modeled damping times. It is still approximate and does not automatically describe coherent structures or internal intermittency. The simple constant-skewness model is therefore useful as an equilibrium interpolation; EDQNM closure addresses a broader spectral evolution problem.