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The Kármán-Howarth equation evolves two-point correlations in homogeneous isotropic turbulence. In trace form it is ${\partial }_{t}C=r^{-2}{\partial }_r[r^{-1}{\partial }_r(r^4{u}^{3}K)]+2\nu r^{-2}{\partial }_r(r^{2}C\prime )$. Integrating with radial volume weight reduces conservation of the Saffman integral to vanishing nonlinear and viscous boundary fluxes.

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.

Put $F(r)=r^4{u}^{3}K(r)$ and retain the scalar trace correlation $C(r)$. For radial functions the Laplacian is ${\nabla }^{2}C=r^{-2}(r^2{C}^{\prime }{)}^{\prime }$. Multiplying the trace Kármán-Howarth equation by the radial volume element and integrating givesRegularity at the origin removes the lower endpoints. The stated triple-correlation tail gives $F=u^3(a+b/r+\cdots )$ and hence $F^{\prime }/r\to 0$. Rapid enough decay of the trace correlation removes the viscous endpoint. ThereforeFor the Loitsyansky integral, multiplying by $-4\pi r^4$ gives the nonlinear contributionThe given tail removes $r{F}^{\prime }$ but permits a nonzero limit of $F$. The viscous contribution isHere the stronger weighted endpoints vanish under the assumed sufficiently rapid correlation decay. Combining the terms yieldsThe conserved Saffman integral measures the density of large-volume momentum fluctuations established in part (i). Internal interactions redistribute momentum without changing that leading extensive variance when forcing and far-field momentum fluxes are absent. This is a statistical expression of linear momentum conservation, not the assertion that kinetic energy is conserved during viscous decay.

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.

A large spherical region contains $N\sim V_R/{\ell }^3$ energy-containing eddies. If their hydrodynamic impulses have zero mean, finite variance and sufficiently short-range correlations, the mean-square sum grows as $N$, rather than $N^2$. In the ideal independent case,The spherical impulse-momentum relation makes the integrated velocity proportional to this sum for isolated enclosed structures. In a full field, external dipolar velocities and boundary effects can change the coefficient; they do not alter the volume scaling under these random-impulse statistics. ThusIt is a random walk law for momentum fluctuations, not an extensive mean flow. Although the mean impulse is zero by isotropy, its variance need not vanish. The Saffman integral measures this large-volume momentum-fluctuation density. If every eddy had exactly zero impulse or correlations enforced systematic cancellation, this leading density could vanish and a different small-wavenumber spectrum could result.

Let $C(r)=⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩$ in zero-mean isotropic turbulence. Define the scalar Saffman integral $L={\int }_{{\mathbb{R}}^3}C(r)d^{3}r=4\pi {\int }_0^{\infty }{r}^{2}C(r)dr$. The spectral transform isFor an integrable $r^{2}C(r)$, dominated convergence applied to $\sin (kr)/(kr)$ givesA further $O(k^4)$ expansion requires an additional correlation moment and is not automatic from finite $L$. The normalization agrees with $\int Edk=⟨|\mathbf{u}{|}^2⟩/2$.

For a control volume $V$ put ${\mathbf{P}}_V={\int }_V\mathbf{u}dV$. Statistical homogeneity gives the exact finite-volume identityFor geometrically similar volumes increasing without bound, the overlap fraction $|V\cap (V-\mathbf{r})|/|V|$ tends to one at every fixed separation. If $C$ is absolutely integrable, dominated convergence givesThe printed finite-large-volume expression is the corresponding asymptotic approximation, not an exact identity for an arbitrary finite volume. For instance, positive nonzero $C$ with a finite correlation length gives an overlap fraction smaller than one and hence a strictly smaller finite-volume value. The limit also shows $L\ge 0$; a nonzero Saffman integral is positive.

Write $u^2=⟨u_x^2⟩$ for the one-component variance and $Q_{LL}=u^{2}f(r)$. Isotropy gives $Q_{ij}=u^2[(f-g)n_i{n}_j+g{\delta }_{ij}]$, and incompressibility gives $g=f+r{f}^{\prime }/2$. Taking the trace yieldsTherefore, with a regular correlation at the origin,For $L>0$, the longitudinal velocity correlation has the tail $u^{2}f(r)\sim L/(4\pi r^3)$. Its physical source is the velocity induced far from a localized eddy with nonzero hydrodynamic impulse. The hydrodynamic Biot-Savart kernel gives the dipolar far fieldThe absence of a vorticity monopole and the surviving impulse dipole give the $r^{-3}$ power. Random eddy impulses thus produce long-range velocity correlations despite the vorticity itself being concentrated in localized structures. Incompressibility also explains how a distant pressure response can redistribute velocity without transporting local vorticity there.

There is no inconsistency with integrability of the trace correlation. At leading order, $g\sim -f/2$, soThe longitudinal and transverse dipolar tails cancel in the trace. The scalar Saffman integral can remain finite even though individual tensor components have long-range, direction-dependent tails.

The Saffman integral is $L=\int ⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩d^{3}r$. With integrable trace correlation it equals the large-volume limit of $⟨|{\int }_V\mathbf{u}dV{|}^2⟩/|V|$. It is nonnegative and measures momentum-fluctuation density. Under appropriate unforced far-field flux conditions it is invariant.

A finite nonzero Saffman integral gives $E(k)=L{k}^2/(4{\pi }^2)+o(k^2)$ in the standard isotropic spectrum normalization. This follows by expanding the radial spectral transform at small wavenumber. A further analytic term requires additional correlation moments.