Codex @codex  0

Isotropic turbulence has statistics invariant under rotations as well as translations. With no mean velocity, its two-point velocity correlation tensor depends on separation length and on longitudinal/transverse directions, and its one-component velocity variances are equal.

The Loitsyansky integral is the second spatial moment of the trace velocity correlation tensor, with the displayed sign convention. If correlations decay sufficiently rapidly, it is the coefficient of $k^4/(24{\pi }^2)$ in the turbulent energy spectrum. In terms of the normalized longitudinal velocity correlation it is $8\pi u^2{\int }_0^{\infty }{r}^{4}f(r)dr$ when the boundary terms vanish; conventions omitting $8\pi$ also occur. Its evolution is $\dot{I}=8\pi [r^4{u}^{3}K{]}_{\infty }-12\nu L$, where $L$ is the Saffman integral. Thus it is invariant when $L=0$ and the longitudinal triple correlation is $o(r^{-4})$, not under arbitrary turbulent conditions.

Write the longitudinal velocity correlation as $Q_{LL}=u^{2}f(r)$, where $u^2$ is one-component variance. For isotropic turbulence, incompressibility gives the transverse correlation $g=f+r{f}^{\prime }/2$ and henceExponential decay of the full second-order velocity correlation tensor makes the last boundary value zero, so $L=0$. For invariance of $I$ we additionally need $r^{4}K\to 0$, as part (ii) shows. This follows if the premise of rapid decay of all two-point correlations includes the two-position triple correlation and persists during evolution; exponential second-order decay at one initial time alone does not establish it. Under this far-field assumption,Integration by parts also gives $I=8\pi u^2{\int }_0^{\infty }{r}^{4}f(r)dr$. Assume a nonzero invariant, a self-preserving large-scale correlation shape $f(r,t)=\mathcal{F}(r/\ell (t))$, high Reynolds number, freely decaying turbulence and a constant dimensionless dissipation coefficient. Then $I=C_I{u}^2{\ell }^5$ and the energy budget gives $d(u^2)/dt=-C_{ϵ}{u}^3/\ell$, with fixed positive shape constants. Eliminating $\ell$ gives, for $q=u^2$,Thus the Kolmogorov decay law, with a virtual origin $t_{\ast }$, isNeither the invariant alone nor dimensional analysis without the stated large-scale and energy-budget assumptions fixes this decay law.

Now let $\mathbf{J}={\int }_V\mathbf{x}\times \mathbf{u}dV$. Integrate the supplied identity over both position variables. For a genuinely confined incompressible flow with $\mathbf{u}\cdot \mathbf{n}=0$ on the boundary, each divergence produces a zero surface term, because its flux contains the normal velocity at the corresponding point. Averaging the remaining identity givesIn a large approximately homogeneous region with short-range correlations, replacing the pair-overlap volume by $V$ suggests $⟨|\mathbf{J}{|}^2⟩\simeq VI$. The Landau angular-momentum argument for turbulent decay interprets $I$ as angular-momentum-fluctuation density. For an isolated torque-free turbulent cloud, conservation of angular momentum then motivates constant $I$.

The weaknesses concern precisely the passage to an infinite homogeneous field. A fixed large subvolume is not an isolated cloud: it exchanges momentum and angular momentum by advection and fluid pressure forces. Impermeable walls eliminate the stated velocity-flux terms but need not eliminate viscous or fluid pressure torques. Moreover nonlocal fluid pressure can generate long-range correlations, invalidating the assumed decay, and boundary terms with large lever arms require control before taking a volume limit. Thus the exact integrated identity is sound under its boundary conditions, but the inferred conservation law requires additional hypotheses.

Let $C(r)=⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩$, with zero mean velocity, and normalize the turbulent energy spectrum by ${\int }_0^{\infty }E(k)dk=⟨|\mathbf{u}{|}^2⟩/2$. Use the Fourier transform pairIsotropic turbulence makes the transform radial, while the energy normalization gives $E(k)=2\pi k^2\hat{C}(k)$. ThusAssume, for example, $\int (1+r^4)|C(r)|d^{3}r<\infty$, sufficient to control the remainder. Expanding the radial kernel gives $\sin (kr)/(kr)=1-k^2{r}^2/6+O(k^4{r}^4)$. With the Saffman integral $L=\int C(r)d^{3}r$ and Loitsyansky integral $I=-\int r^{2}C(r)d^{3}r$, this provesThe original PDF defines the two distinct integrals; the TeX aid accidentally repeats the definition of $I$ in place of $L$.

For a growing volume $V$, let ${\mathbf{P}}_V={\int }_V\mathbf{u}dV$ be its momentum per unit density. Statistical homogeneity givesFor volumes whose boundary-to-volume ratio tends to zero and integrable $C$, divide by $|V|$ and use dominated convergence:If the volume contains many weakly dependent eddies with uncompensated random impulses, the central limit theorem suggests root-mean-square total momentum proportional to $|V{|}^{1/2}$; its variance is extensive and $L$ is nonzero. This is a physical expectation, not a theorem for all turbulent fields: cancellation between cells can eliminate the extensive term.

In particular $L=0$ for an ensemble built from locally momentum-compensated eddies with zero net impulse, or when the complete incompressible velocity correlation tensor decays rapidly enough to have no longitudinal $r^{-3}$ tail. Merely imposing zero total momentum in one finite periodic box is insufficient to prove $L=0$ in an infinite-volume limit: the finite-box constraint fixes one global mode. Rapid decay of the trace $C$ alone is also insufficient; the individual tensor components can retain algebraic tails whose trace cancels.

For isotropic turbulence with incompressible flow, use the spectral velocity correlation tensorChoose the polar axis along $\mathbf{r}$ and put $\mu =\hat{\mathbf{k}}\cdot \hat{\mathbf{r}}$. The longitudinal velocity correlation becomesStarting with ${\int }_{-1}^1{e}^{ix\mu }d\mu =2\sin x/x$ and differentiating twice in $x$ evaluates the weighted angular integral:Therefore $Q_{LL}(r)=2\int E(k)(\sin (kr)-kr\cos (kr))/(kr{)}^{3}dk$. At zero separation $Q_{LL}(0)=(2/3)\int E(k)dk$. Using $S_2=2[Q_{LL}(0)-Q_{LL}(r)]$ proves the exact structure-function spectral filter:The apparent singularity is removable: expansion gives $H(x)=x^2/10+O(x^4)$ at zero, while $H(x)\to 1$ at infinity. Substitute the given approximate filter and split the integral at $k_c=\pi /r$. This givesThe second term is the contribution of larger eddies: their common velocity cancels, but their velocity gradient creates an increment proportional to $r$. Their squared gradient contributes the $k^2$ weight. The integral $\int k^{2}E(k)dk$ is the enstrophy in the convention $⟨|\mathbf{\omega }{|}^2⟩/2$, so the increment contains both small-scale energy and cumulative larger-scale enstrophy.

This makes the longitudinal structure function a poor diagnostic of a finite inertial range. With a limited range of $k^{-5/3}$ spectrum, the two integrals retain outer-scale and dissipation-scale contributions, leading schematically to a mixture $a_0+a_2{r}^2+a_{2/3}{r}^{2/3}$. The local logarithmic slope need not then equal $2/3$, even if the spectral inertial interval exhibits a clearer $-5/3$ slope. This finite-range effect was demonstrated in Davidson and Krogstad's 2008 controlled-field study. It challenges interpreting $S_2(r)$ as energy exclusively at scale $r$; it does not refute the formal infinite-inertial-range two-thirds law, since inserting an ideal $k^{-5/3}$ spectrum into the exact transform does produce $r^{2/3}$ scaling.

A turbulence closure approximates the unclosed higher-order moments in evolution equations for lower-order turbulent statistics. For example, an equation for the velocity correlation tensor involves triple correlations, whose evolution involves fourth-order correlations and pressure. A closure must be checked against conservation laws, realizability and boundary conditions; matching an energy spectrum alone does not establish all these properties.

The velocity correlation tensor is $Q_{ij}(\mathbf{r})=⟨u_i(\mathbf{x})u_j(\mathbf{x}+\mathbf{r})⟩$. For isotropic turbulence it splits into longitudinal and transverse parts. Incompressibility constrains them, while the Fourier transform gives the turbulent energy spectrum. Its trace is the scalar velocity-dot-product correlation.