Set $\psi =e^{ikx+\mu {\chi }_1+O({\mu }^2)}$. Substitution into the Helmholtz equation gives the first Rytov approximation equationFor the forward parabolic approximation, omit ${\partial }_x^2{\chi }_1$. With zero perturbation at $x=0$, the Fresnel propagator then givesA jointly Gaussian random field makes ${\chi }_1$ a centered complex Gaussian variable. The complex Gaussian exponential moment therefore gives the first-Rytov coherent meanThe second moment here has no complex conjugation. It cannot be replaced by the positive intensity variance $\mathbb{E}[|{\chi }_1{|}^2]$.

For example, let $\hat{C}(u,v;\mathbf{\kappa })$ be the transverse spectrum of the covariance between depths $u$ and $v$. The Fourier multiplier of the Fresnel propagator is $e^{-i(x-u)|\mathbf{\kappa }{|}^2/(2k)}$, soThis supplies the requested mean in terms of medium statistics. For the longitudinal white-noise model, $\hat{C}=\delta (u-v)\Phi$, which reduces the double longitudinal integral to one.

The exponential is the mean of the first-order logarithmic model. It is not a complete second-order perturbation expansion of the true mean: the omitted logarithmic term ${\mu }^2{\chi }_2$ can itself have a nonzero mean. Strictly at first order, $\mathbb{E}[\psi ]=e^{ikx}+O({\mu }^2)$. As an independent check, the closed white-noise parabolic model gives exactlyfor plane-wave input. A truncated first-Rytov covariance expression with transverse diffraction need not equal this exact mean; omitted logarithmic terms account for the difference.

For a general colored medium and unit plane-wave input $E(0,\mathbf{s})=1$, let $U_W(x,0)$ be the propagator of ${\partial }_x=i{\Delta }_{\perp }/(2k)+ik\mu W(x,\mathbf{s})$. Then the solution isA time-ordered operator exponential defines $U_W$. Its expectation depends on the full longitudinal covariance; the initial Gaussian and transverse-stationarity assumptions do not specify a universal closed expression.

With the stated longitudinal Gaussian white noise and transverse statistical homogeneity, the moment depends only on $\mathbf{\rho }={\mathbf{s}}_1-{\mathbf{s}}_2$. Hence $({\Delta }_1-{\Delta }_2)\Gamma =0$. The closed equation in the previous part and initial value $\Gamma (0)=1$ giveIn particular $\Gamma (x,0)=1$: the ensemble mean intensity remains constant for plane-wave input, even though coherent attenuation in a white-noise random medium reduces the mean complex field.

Use a consistent transverse power spectrum convention,For an isotropic spectrum, angular integration gives the isotropic transverse covariance spectrum, $C(\rho )=(2\pi {)}^{-1}{\int }_0^{\infty }\Phi (\kappa )J_0(\kappa \rho )\kappa d\kappa$. ThereforeThe supplied two-dimensional Fourier transform definition has angular factor $2\pi$, whereas the displayed radial transform pair omits it. The latter is a normalized Hankel transform pair. If its radial spectrum is denoted ${\Phi }_H=\Phi /(2\pi )$, use ${\Phi }_H$ in the integral and omit the prefactor $1/(2\pi )$. These are equivalent conventions; mixing them changes the attenuation rate.

Write $\psi =e^{ikx}E(x,\mathbf{s})$, $\mathbf{s}=(y,z)$. In the weak-contrast approximation $n^2\simeq 1+2\mu W$, neglecting $E_{xx}$ relative to $k{E}_x$ gives the parabolic wave equationFor the mutual coherence $\Gamma (x;{\mathbf{s}}_1,{\mathbf{s}}_2)=\mathbb{E}[E_1{E}_2^{\ast }]$, differentiating gives the exact moment equation within this wave approximation,This is an evolution equation but not a closed one. Even a jointly Gaussian random field does not make $W$ independent of $E$, which depends on its history. Transverse statistical homogeneity alone therefore does not close the second moment. A covariance or longitudinal Markov approximation is additional information.

Under the white-in-$x$ assumption introduced in the next part, interpret $Wdx=d{\mathcal{B}}_x$, a longitudinal Brownian field with transverse covariance $C$, and use the Stratonovich integral as the smooth-medium limit. Conversion to the Itô integral yieldsThe product rule with quadratic variation gives the closed mutual coherence in a white-noise parabolic medium equation,The two attenuation drifts contribute $-k^2{\mu }^{2}C(0)$ and the cross variation contributes $+k^2{\mu }^{2}C({\mathbf{s}}_1-{\mathbf{s}}_2)$. The coefficient $C$ is the white-noise covariance intensity; it replaces, rather than literally retains, the original finite pointwise variance of $W$.

With constant permeability, the electric equation reduces to $({\partial }_x^2+{\partial }_z^2+k^2(z))E_y=0$. Insert the oblique-wave dependence $E_y=e_y(z)e^{-i{k}_0\sin {\theta }_{0}x}$ and set $s=z-z_0$. The inverse-square refractive-index profile givesSubstituting $e_y=\sqrt{|s|}w(\beta |s|)$ gives the Bessel equation of order $\nu$. On either side of the profile singularity, when ${\beta }^2>0$,Equivalently use Hankel functions for an incoming/outgoing wave basis. If ${\beta }^2<0$, write $\gamma =\sqrt{-{\beta }^2}$ and replace $J_{\nu },Y_{\nu }$ by modified Bessel functions $I_{\nu },K_{\nu }$ with argument $\gamma |s|$. For $\beta =0$, the solutions are $|s{|}^{1/2+\nu }$ and $|s{|}^{1/2-\nu }$, with a logarithmic second solution in the repeated-root case $\nu =0$. The constants on each side require incident-wave and boundary or matching data; the supplied singular profile alone does not choose them.

Use time dependence $e^{i\omega t}$ and consider fields independent of $y$, with $\mathbf{E}=\hat{\mathbf{y}}{E}_y(x,z)$. The source-free Maxwell's equations are $\nabla \times \mathbf{E}=-i\omega \mu \mathbf{H}$ and $\nabla \times \mathbf{H}=i\omega \epsilon \mathbf{E}$. The first givesSubstitute these into the $y$ component of the second equation. The TE wave equation in a stratified magnetodielectric isBecause its coefficients depend only on $z$, ${\partial }_x{E}_y$ obeys the same equation. Since it is proportional to $\mu H_z$, substitution and division by $\mu$ giveFor a conserved transverse wavenumber $k_x$, replace ${\partial }_x^2$ by $-k_x^2$ to obtain the corresponding ordinary Helmholtz equations.

If “propagating along the $z$ axis” is read as strict normal incidence with no $x$ dependence, then $H_z=0$ identically. The nonzero magnetic component is $H_x$, and its equation isThe $H_z$ equation above applies to the oblique transverse electric polarization used in the next part and also includes the zero normal-incidence solution. This distinguishes the named component from the usual transverse magnetic field.

Write $h=\log \psi$ and $q=2ϵh_{\theta }$, with $ϵ>0$. Direct differentiation givesThus the Cole-Hopf transformation with this negative-flux convention maps the heat equation to the stated viscous Burgers equation. Conversely its residual is independent of $\theta$ and can be removed by multiplying $\psi$ by a function of $Z$.

Convolving each initial exponential with the Gaussian heat kernel givesThe resulting finite-exponential Cole-Hopf solution isIt is a softmax function average of the constant states, so $q_1\le q\le q_N$ and $q_{\theta }={\mathrm{Var}}_w(q_j)/(2ϵ)\ge 0$.

For $N=2$, put $\Delta =q_2-q_1>0$ and $\overline{q}=(q_1+q_2)/2$. Equality of the two exponents occurs atThereforeThis is an increasing viscous shock from $q_1$ to $q_2$, moving at speed $-\overline{q}$ and with transition width of order $ϵ/\Delta$. That speed agrees with the Rankine-Hugoniot condition for flux $-q^2/2$; its concavity makes the increasing jump the entropy shock.

For general $N$, use the shock frame $\theta =-(q_1+q_N)Z/2+\xi$. The leading $Z$ coefficient in $A_j$ is $q_j(q_j-q_1-q_N)/(4ϵ)$. The endpoint coefficients coincide, and every interior coefficient is smaller by $(q_j-q_1)(q_N-q_j)/(4ϵ)>0$. The extremal-state asymptotics of exponential Burgers solutions therefore reduce the large-$Z$ profile to the same tanh shock with states $q_1,q_N$ and centreAt a fixed $\theta$, rather than in the moving frame, the endpoint of largest $|q_j|$ dominates. If $q_N=-q_1$, both endpoint rates coincide and the limiting shock is stationary. Intermediate plateaus disappear in the moving-frame limit.

Use polar coordinates with $-\pi <\theta <\pi$; the two faces of the screen are $\theta =\pm \pi$. Take $0<{\theta }_0<\pi$, so the incident wave approaches from the upper half-plane, and retain the specified time dependence $e^{i\omega t}$. Outgoing cylindrical waves then have radial factor $e^{-i{k}_{0}r}$. Define the normalized Fresnel integralThe exact Neumann Sommerfeld half-plane diffraction solution for the total spatial field isOne way to verify the Helmholtz equation is to use parabolic coordinates $\xi =\sqrt{2r}\cos (\alpha /2)$, $\eta =\sqrt{2r}\sin (\alpha /2)$. Each term has the form $e^{i{k}_0({\xi }^2-{\eta }^2)/2}\mathcal{F}(\pm \sqrt{k_0}\xi )$. The operator is $({\xi }^2+{\eta }^2{)}^{-1}({\partial }_{\xi }^2+{\partial }_{\eta }^2)$, and ${\mathcal{F}}^{\prime \prime }(s)+2is{\mathcal{F}}^{\prime }(s)=0$ makes its Helmholtz residual zero. At $\theta =\pm \pi$, the two Fresnel arguments coincide, while the derivatives of both the arguments and plane-wave phases cancel in pairs. Hence ${\partial }_{\theta }\Phi =0$, which is the required normal boundary condition.

Away from the Fresnel transition regions, integration by parts gives the Fresnel tail expansionwhere $H$ is the Heaviside step function. The same expansion follows from the complementary-error-function asymptotics in NIST DLMF. Because $s_{\pm }^2=k_{0}r[1+\cos (\theta \pm {\theta }_0)]$, the tails in both terms have outgoing phase $e^{-i{k}_{0}r}$. The scattered far field is consequentlyThe first term cancels the incident geometrical optics field in its shadow, the second is its reflected plane wave and the last is the diffracted edge wave. The expansion is nonuniform at $\theta ={\theta }_0-\pi$ and $\theta =\pi -{\theta }_0$; the exact Fresnel expression resolves those regions. Incidence from below follows by reflecting $\theta$ and ${\theta }_0$.

Let the observer have cylindrical coordinates $(r,\theta ,z)$ and total distance $R=\sqrt{r^2+z^2}$. At retarded emission time $\tau$, the source-observer distance isThe retarded time relation $t=\tau +|\mathbf{x}-\mathbf{y}(\tau )|/c_0$ consequently becomesSet $\Omega \tau =\theta -\pi /2+\Theta$. ThenThe observer-direction Mach number at emission is $M_r=\dot{\mathbf{y}}(\tau )\cdot \mathbf{n}/c_0=m\cos \Theta$. For a constant strength $q$ and $m<1$, the single retarded root has $1-M_r>0$. Substituting into the moving-source field gives the rotating acoustic point source resultIn particular, on the rotation axis $m=0$ and the result is $q/(4\pi c_0^{2}R)$.

For a genuinely varying source strength, replace $q$ by $q(\tau )$ in the last formula. If $m>1$, the retarded relation need not be one-to-one: each retarded root contributes with $|1-M_r|$ in the denominator. The general result is a sum of $q({\tau }_j)|{\dot{\Theta }}_j|/(4\pi c_0^2\Omega R)$. Dropping the absolute value and the root sum requires the subsonic assumption. At $1-M_r=0$ the ideal point-source expression has a caustic; finite source size and bandwidth regularize that idealization.

Let $R=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/R$ and $t_R=t-R/c_0$. Convolving the source terms in the Lighthill acoustic analogy with the retarded acoustic Green function, and moving their spatial derivatives outside the integral, givesThe integrals have compact spatial support. For an acoustically compact source, its size $l$ satisfies $\omega l/c_0\ll 1$, so the source-dependent retardation is negligible at leading order. Also take $R\gg l$. Define $S_{ij}=\int T_{ij}{d}^{3}y$ and $F_i=\int f_i{d}^{3}y$. In the radiation terms the spatial derivatives act on $t_R$, with ${\partial }_{x_i}{t}_R=-n_i/c_0$. Derivatives of $1/R$ produce smaller near-field terms. Thus the far-field acoustic force and stress moments giveUsing $n_i=x_i/R$ gives the equivalent powers of $R$ requested. The force term is an acoustic dipole; the Lighthill stress tensor produces an acoustic quadrupole.

For a small body in low-speed flow, put $M=u/c_0\ll 1$. A typical unsteady force has size $F=O({\rho }_0{u}^2{l}^2)$ and time scale $l/u$, whereas $S=O({\rho }_0{u}^2{l}^3)$. ThereforeThese are the acoustic dipole Mach-number scaling and compact acoustic quadrupole Mach-number scaling. Since $p^{\prime }=c_0^2{\rho }^{\prime }$ and outgoing acoustic intensity is of order $p^{\prime 2}/({\rho }_0{c}_0)$, their sound powers scale asThus an unsteady force on the body can produce the dominant low-Mach number sound. If the integrated force or its time derivative vanishes, that dipole contribution is absent. Force on the body and force on the fluid have opposite signs.

Separate a smooth large-scale velocity $U_L$ from a remaining smaller-scale component $w_L$. If the large-scale variation length is much greater than $r$, their longitudinal increment has the formFor a positive integer $p$, the leading expansion of its structure function isThus even a velocity component at scales much larger than $r$ enters through its gradients and through mixed moments with the finer field. It is not generally legitimate to discard those mixed moments: large-scale strain can organize and modulate the finer motions. For example, at fourth order the expansion containsin addition to the small-scale fourth moment and the generally present odd mixed terms. Even if the components were independent and symmetric, the displayed mixed second-moment term would survive. This explicitly exhibits large-scale contributions to structure functions beyond second order.

The implication for K62 refined similarity is that an observed structure function need not measure a purely scale-local velocity fluctuation. Dividing it by $({ϵ}_{r}r{)}^{1/3}$ may still leave dependence on outer-scale gradients, correlations or finite-range cutoffs. The universal conditional coefficients ${\beta }_p$ and the identification of a measured exponent with the lognormal intermittency model therefore require additional locality and scale-separation tests. Finite-range mixed contributions can imitate or obscure anomalous scaling. This is an objection to treating the 1962 dimensional argument as an unconditional law for measured increments; it is not, by itself, a proof that an asymptotic refined similarity regime cannot exist.

The motivation for Kolmogorov refined similarity hypothesis is internal intermittency: instantaneous energy dissipation is spatially uneven, so scaling every velocity increment by the global mean dissipation can mix regions of very different activity. Define the coarse-grained energy dissipation over a region of size $r$ byIts mean is $ϵ$ in homogeneous turbulence. The refined hypothesis asserts that, at high local Reynolds number and within the inertial range, the normalized incrementhas universal conditional statistics independent of ${ϵ}_r$ and of $r$. In particular, $⟨V_r^p\mid {ϵ}_r⟩={\beta }_p$ for finite moments. Taking a conditional expectation and then averaging givesHere ${ϵ}_r$ is the paper's ${ϵ}_{AV}(r)$. Signed odd moments need not have positive coefficients; the third-order forward-cascade coefficient is negative. The signed first moment is zero in homogeneous flow, so one does not assign it a nonzero scaling law. Absolute moments are often used when discussing general positive-order scaling exponents.

The refined hypothesis alone does not specify the distribution of ${ϵ}_r$. Kolmogorov's 1962 estimate supplements it with a lognormal intermittency model. Let $Y_r=\ln ({ϵ}_r/ϵ)$ be Gaussian with variance ${\sigma }_r^2=\mu \ln (\ell /r)+\mathrm{constant}$. The condition $⟨{ϵ}_r⟩=ϵ$ fixes its mean at $-{\sigma }_r^2/2$. The Gaussian exponential moment then gives, with $q=p/3$,Combining with the refined moment formula yields, for the nonzero moments described by that model,The intermittency parameter $\mu$ must be supplied by dissipation statistics; it is not determined by dimensional analysis. The formula preserves ${\zeta }_3=1$. Its extrapolation to arbitrarily high orders is not justified: the quadratic correction eventually predicts pathological exponents and is one limitation of the lognormal model, distinct from the refined hypothesis itself.

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.

The first similarity hypothesis of Kolmogorov 1941 theory states that at sufficiently large Reynolds number, the small-scale statistics in the equilibrium range are locally universal and determined by the mean turbulent kinetic energy dissipation rate $ϵ$ and kinematic viscosity $\nu$, together with the separation or wavenumber being measured. Large-scale forcing details are excluded only at scales well below the integral scale of turbulence.

These parameters define the Kolmogorov microscalesDimensional analysis gives universal functions $F$ and $G$ withThe dimensional factors follow because $S_2$ has units of velocity squared, whereas $E$ has units of velocity squared times length. The universal equilibrium forms include viscous effects; they do not impose a power law through the dissipation range.

The second similarity hypothesis asserts independence of kinematic viscosity at scales large compared with $\eta$ but still small compared with the integral scale. In this inertial range, $S_2$ can depend only on $ϵ$ and $r$. Balancing time and length dimensions forces powers $2/3$ and $2/3$. The same calculation for $E$, using $k$ instead of $r$, forces powers $2/3$ and $-5/3$. HenceThe constants $C_2$ and $C_K$ are universal within the hypotheses and the stated normalization. They are related by the exact structure-function spectral filter, rather than by assigning all energy to one sharp scale.

In this question $\Delta v$ is the velocity increment in the direction of the separation, so $S_2(r)=⟨(\Delta v{)}^2⟩$ is a longitudinal velocity structure function. This interpretation is necessary for the kernel in part (iii); the full vector increment has a different kernel.

A velocity component that varies very little over distance $r$ is almost the same at the two sample points, so its common part cancels from their difference. A Fourier component acquires the difference factor $e^{i\mathbf{k}\cdot \mathbf{r}}-1$. Its squared magnitude is $2[1-\cos (\mathbf{k}\cdot \mathbf{r})]$, of order $(kr{)}^2$ when $kr\ll 1$ and of order one after averaging rapid phases when $kr\gg 1$. Thus a longitudinal structure function attenuates large eddies and retains smaller eddies, but is a smooth filter rather than an exact scale cutoff.

The heuristic cutoff $k_c=\pi /r$ marks wavelengths of order $2r$. If motions below that cutoff are treated as perfectly coherent and those above it as uncorrelated at the two points, isotropy allocates one third of their velocity variance to the longitudinal component. Since the turbulent energy spectrum integrates to one half of total velocity variance, the uncorrelated high-wavenumber component contributesorThe factor $3/4$ follows from the component and energy normalizations. The approximation discards the finite gradient of eddies larger than $r$; part (iii) quantifies that missing contribution.

Use $S_C(r)=⟨(C^{\prime }-C{)}^2⟩$ for the scalar structure function, and assume statistical homogeneity as well as isotropy. Expanding the square givesIf the zero-mean scalar decorrelates at large separation, thenThe decorrelation assumption is necessary: a spatially constant, zero-ensemble-mean random scalar is isotropic but has $S_C=0$ at every separation.

For the small-separation limit assume mean-square differentiability. Taylor expansion gives $C^{\prime }-C=r_i{\partial }_{i}C+o(r)$ in mean square. Isotropy implies $⟨{\partial }_{i}C{\partial }_{j}C⟩={\delta }_{ij}⟨|\nabla C{|}^2⟩/3$. Define the scalar dissipation rate in the half-variance convention, ${ϵ}_c=\alpha ⟨|\nabla C{|}^2⟩$, so that diffusion removes $⟨C^2/2⟩$ at rate ${ϵ}_c$. It follows thatIf dissipation is instead defined as the loss rate of the full variance, the coefficient must be adjusted by a factor of two.

In the inertial-convective scalar range, $\max (\eta ,{\eta }_c)\ll r\ll \ell$, suppose scalar variance cascades locally at an approximately constant flux ${ϵ}_c$, while velocity increments have the Kolmogorov 1941 theory turnover time $t_r\sim {ϵ}^{-1/3}{r}^{2/3}$. The variance associated with scale $r$ divided by that turnover time is of order its transfer rate, giving the Obukhov-Corrsin theory resultThe assumptions include a passive scalar, high Reynolds number, local transfer, adequate scale separation, and statistical equilibrium or sufficiently slow variation of the supply. Isotropy alone does not provide a constant scalar flux.

For weak scalar diffusion, $\alpha \ll \nu$, the velocity is smooth below $\eta$, with characteristic strain rate $s\sim (ϵ/\nu {)}^{1/2}$. Diffusion rate $\alpha /r^2$ balances this at the Batchelor scalar microscaleIn the viscous-convective scalar range, the derivative of $S_C$ is assumed to depend only on ${ϵ}_c$, $s$ and $r$. Their dimensions forceIntegration from the diffusion cutoff givesThus, well within a broad logarithmic range,The additive matching contribution is not exactly zero; the formula is a leading range estimate, not a boundary condition at $r={\eta }_c$. Using $d{S}_C/dr$ isolates the contribution acquired on changing separation. The structure function itself contains scalar variance accumulated from smaller scales and must retain the inner cutoff through its integration constant. If one assumed that $S_C$ depended only on ${ϵ}_c,s,r$ without that cutoff, dimensional analysis would incorrectly give an $r$-independent value proportional to ${ϵ}_c/s$. This explains the use of the derivative with respect to $r$; the TeX aid changes that derivative to $t$ in its final sentence, but the original PDF consistently uses $r$.

Finally let $\alpha \gg \nu$. Balancing diffusion with the inertial eddy rate gives the Corrsin scalar microscale ${\eta }_c={\eta }_C\sim ({\alpha }^3/ϵ{)}^{1/4}\gg \eta$. For $\eta \ll r\ll {\eta }_c$, diffusion is faster than the eddy turnover rate. The scalar is smooth to leading order on those separations, even though the velocity still has inertial-range fluctuations. The inertial-diffusive scalar range therefore hasThis continues the scalar's leading gradient-controlled quadratic behaviour; it is not a continuation of the inertial-convective two-thirds law below its diffusion cutoff.

Measure the cloud radius about its moving centre, so uniform sweeping does not count as spreading. In the inertial range, an eddy of size $R$ has velocity difference $u_R\sim (ϵR{)}^{1/3}$ and turnover time $t_R\sim R/u_R$. A scale-local relative diffusivity is thereforegiving the estimateIt requires separation from the molecular and viscous cutoffs and from the integral scale of turbulence, high Reynolds number, approximately steady energy supply, and scale-local relative motion. Mean shear, walls, particle inertia or coherent large-scale strain can invalidate this simple diffusivity model. At very short times a pair retains its initial relative velocity, so a diffusion closure is not yet appropriate.

Let $s(t)=⟨|\delta \mathbf{x}(t){|}^2⟩$. A self-similar pair-separation distribution, together with the relative-diffusivity estimate, givesSelf-similarity is needed to replace a separation-dependent moment such as $⟨r^{4/3}⟩$ by a fixed multiple of $s^{2/3}$; stationarity alone does not do that. Integrating this model givesAfter loss of initial-separation memory, it reduces to the Richardson pair dispersion lawThe required time window is $t\gg t_0\sim (r_0^2/ϵ{)}^{1/3}$ while the separation remains far below the integral scale. The stated stationarity and inertial-range separation conditions alone cannot imply the displayed $t^3$ law at every time: a nonzero release separation already gives $s(0)=r_0^2$, whereas $gϵt^3$ vanishes at zero. At sufficiently short times, $\delta \mathbf{x}(t)-\delta {\mathbf{x}}_0\simeq \delta {\mathbf{u}}_{0}t$, so its mean-square change is ballistic and remembers $r_0$.