= Solution
Let $C(r)=\langle\boldsymbol u(\boldsymbol x)\cdot\boldsymbol u(\boldsymbol x+\boldsymbol r)\rangle$, with zero mean <velocity>, and normalize the <turbulent energy spectrum> by $\int_0^\infty E(k)dk=\langle|\boldsymbol u|^2\rangle/2$. Use the <Fourier transform> pair
$$
C(\boldsymbol r)=\int e^{i\boldsymbol k\cdot\boldsymbol r}\widehat C(\boldsymbol k)d^3k,
\qquad\widehat C(\boldsymbol k)=\frac1{(2\pi)^3}\int e^{-i\boldsymbol k\cdot\boldsymbol r}C(\boldsymbol r)d^3r.
$$
<Isotropic turbulence> makes the transform radial, while the energy normalization gives $E(k)=2\pi k^2\widehat C(k)$. Thus
$$
E(k)=\frac{k^2}{4\pi^2}\int C(r)\frac{\sin(kr)}{kr}d^3r.
$$
Assume, for example, $\int(1+r^4)|C(r)|d^3r<\infty$, sufficient to control the remainder. Expanding the radial kernel gives $\sin(kr)/(kr)=1-k^2r^2/6+O(k^4r^4)$. With the <Saffman integral> $L=\int C(r)d^3r$ and <Loitsyansky integral> $I=-\int r^2C(r)d^3r$, this proves
$$
\boxed{E(k)=\frac{Lk^2}{4\pi^2}+\frac{Ik^4}{24\pi^2}+O(k^6).}
$$
The 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 $\boldsymbol P_V=\int_V\boldsymbol u\,dV$ be its <momentum> per unit density. <Statistical homogeneity> gives
$$
\langle|\boldsymbol P_V|^2\rangle
=\int C(\boldsymbol r)|V\cap(V-\boldsymbol r)|d^3r.
$$
For volumes whose boundary-to-volume ratio tends to zero and integrable $C$, divide by $|V|$ and use dominated convergence:
$$
\boxed{L=\lim_{|V|\to\infty}\frac{\langle|\boldsymbol P_V|^2\rangle}{|V|}\ge0.}
$$
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.
Back to article page
