 Solution

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.