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Write the total velocity as $\mathbf{U}=\Omega y\hat{\mathbf{x}}+\mathbf{u}$ and the magnetic field as $B_0\hat{\mathbf{x}}+\mathbf{b}$. For solenoidal fields the resistive induction equation is $({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{U}+\eta {\nabla }^2\mathbf{B}$. The first-order smoothing approximation drops the fluctuating terms $\mathbf{b}\cdot \nabla \mathbf{u}-\mathbf{u}\cdot \nabla \mathbf{b}$, retaining shear exactly. The uniform field is not stretched by the background shear because ${\partial }_x(\Omega y)=0$. The other retained stretching terms are $\mathbf{b}\cdot \nabla (\Omega y\hat{\mathbf{x}})=\Omega b_y\hat{\mathbf{x}}$ and $B_0{\partial }_x\mathbf{u}$. HenceThe same phase cancellation as in part (i), together with ${\nabla }^2{e}^{i\mathbf{k}\cdot \mathbf{x}}=-K^2{e}^{i\mathbf{k}\cdot \mathbf{x}}$, proves that a response can be taken as $\mathrm{Re}(\mathbf{c}(t)e^{i\mathbf{k}(t)\cdot \mathbf{x}})$, whereIf $\mathbf{k}\cdot \mathbf{c}=0$ initially, its derivative is $-\eta K^2\mathbf{k}\cdot \mathbf{c}$, so magnetic solenoidality is preserved.

For $j=y,z$ the response equation is ${\dot{c}}_j+\eta K^2{c}_j=i{k}_x{B}_0{v}_j$. Its integrating-factor solution isThe velocity amplitudes and wavevector vary on the shear timescale. In the fast-diffusion limit $\eta (k_x^2+k_y^2)\gg |\Omega |$, and after the free transient, the integral is concentrated within time $(\eta K^2{)}^{-1}$ of $t$. Replacing the slowly varying factors there by their current values gives the quasistatic magnetic response of a helical shearing wave:Complex conjugation introduces a minus sign in $c_j^{\ast }$. With precisely the EMF normalization specified in the question,A literal spatial average of the product of the two real Fourier waves would be half this quantity; no extra factor of $1/2$ belongs in the question's defined $\mathcal{E}$.

Use $A=k_x^2+k_z^2$ and part (i) to obtain $\mathcal{E}(t)\simeq -B_0{H}_{0}A/[\eta (A+{\Omega }^2{k}_x^2{t}^2{)}^2]$. For the time integral, take the driven response without a free magnetic transient and suppose the fast-diffusion approximation holds through the contributing interval, in particular at $t=0$; $\eta k_x^2\gg |\Omega |$ is a sufficient uniform condition. For $\Omega k_x\ne 0$, substitute $z=|\Omega k_x|t/\sqrt{A}$. Since ${\int }_{-\infty }^{\infty }(1+z^2{)}^{-2}dz=\pi /2$, this givesThe integral identity follows immediately from $z=\tan \chi$, reducing it to ${\int }_{-\pi /2}^{\pi /2}{\cos }^2\chi d\chi$. If $k_x=0$, the forcing and $H$ vanish and the driven EMF is zero. If $\Omega =0$ but $H_0\ne 0$, the leading EMF is constant rather than time-integrable; the finite integral relies on nonzero shear.

Let $K^2=|\mathbf{k}{|}^2=k_x^2+k_z^2+{\Omega }^2{t}^2{k}_x^2$. For the complex wave before taking its real part,since ${\dot{k}}_y=-\Omega k_x$. The asserted velocity equation therefore follows directly from $\dot{\mathbf{v}}+\Omega v_y\hat{\mathbf{x}}=-i\mathbf{k}q$. To verify the pressure amplitude, solenoidality is $\mathbf{k}\cdot \mathbf{v}=0$. Its derivative requiresand hence $q=2i\Omega k_x{v}_y/K^2$, as claimed.

The transverse amplitude equations areBoth coefficients are real. Differentiating $v_y{v}_z^{\ast }-v_z{v}_y^{\ast }$ makes the terms proportional to $b|v_y{|}^2$ cancel, while the remaining terms give $a(v_y{v}_z^{\ast }-v_z{v}_y^{\ast })$. Therefore the helicity invariant of an inviscid shearing wave obeysHere $H$ is real because the difference in its definition is purely imaginary. Its sign is preserved, and shear reduces its magnitude away from $t=0$.

Work in the laboratory frame, with fluid at rest at infinity and the instantaneous sphere centre at the origin. In the Unscaled Papkovich–Neuber representation, take the harmonic potentialsDifferentiating the two terms gives the translating sphere in Stokes flow:Here $\mathbf{n}=\mathbf{x}/r$. The potentials are harmonic outside the sphere, so they satisfy the Stokes equations. At $r=a$, the two coefficients of $\mathbf{nn}$ cancel and the coefficient of $I$ is one, proving the no-slip boundary condition $\mathbf{u}=\mathbf{V}$; the field decays at infinity. Its leading part is a Stokeslet of force $6\pi \mu a\mathbf{V}$ exerted on the fluid.

Let $\hat{\mathbf{z}}$ point upwards, $\delta =V_1-V_2$, and $D=a_2{V}_2-a_1{V}_1$. Each isolated sphere moves with velocity $-V_i\hat{\mathbf{z}}$. To leading order its correction is the other's incident Stokeslet velocity:The forces generating these fields are the fixed sedimenting loads, so their leading strengths are set by the isolated speeds. Subtraction yields the leading interaction of two sedimenting spheres:Since $\hat{\mathbf{z}}=\cos \theta \hat{\mathbf{r}}-\sin \theta \hat{\mathbf{\theta }}$, this reduces toThe stipulated higher-order errors can be appended to both equations; they do not enter this leading calculation.

For the same-sign case, interchange labels if necessary to arrange $\delta >0,D>0$, and put $c=3D/4$. The vertical separation $z=r\cos \theta$ satisfiesIf sphere one is initially above sphere two, its vertical order reverses; in every case $z\to -\infty$, implying $r\to \infty$. For a noncollinear trajectory there is no collision in the point-force equations. Indeed, divide the two polar equations and integrate:At side-by-side passage, $\theta =\pi /2$, the closest separation is the positive root of $\delta r^2+2cr=C$. Before passage $\dot{r}<0$ and after passage $\dot{r}>0$, proving overtaking and unbounded subsequent separation. This passing invariant for unequal point-force spheres describes a physical far-field passage only when the closest separation also greatly exceeds the sphere radii. Exactly collinear approach instead reaches the near-contact regime, where this leading approximation cannot describe two finite spheres passing through one another.

For opposite signs, choose $\delta >0,D<0$ and consider the invariant configuration with sphere one vertically above sphere two, $\theta =0$. The radial equation isIf $r>r_{\ast }$, separation decreases; if $r<r_{\ast }$, it increases. Thus it approaches this equilibrium along the vertical line. Linearization gives $\dot{\delta r}=-(\delta /r_{\ast })\delta r$. But a small tilt obeys $\dot{\theta }=(\delta /(2r_{\ast }))\theta$, so the vertical bound pair in the point-force sedimentation model is stable only against vertical perturbations and is a saddle in the full separation dynamics. The opposite vertical orientation has the reversed stability. Reversing gravity reverses both $\delta$ and $D$, leaves $r_{\ast }$ unchanged, and reverses every trajectory; radial attraction becomes repulsion. This is fully consistent with kinematic reversibility of Stokes flow.

There is an important restriction in the printed data. Its hierarchy $|D|\ll (a_1+a_2)|\delta |$ implies $r_{\ast }\ll a_1+a_2$, so the proposed equilibrium would overlap the physical spheres and lie outside the well-separated approximation. With that hierarchy there is no justified physical far-field bound pair. The intended far-field equilibrium instead requires the first inequality reversed, $|D|\gg (a_1+a_2)|\delta |$. The formal equilibrium and its vertical stability above show the intended mechanism, while also identifying the missing validity and transverse-stability qualifications.

For a force $\mathbf{F}\delta (\mathbf{x})$ acting on an unbounded incompressible fluid, the Stokeslet and its pressure areThey solve $-\nabla p+\mu {\nabla }^2\mathbf{u}+\mathbf{F}\delta =0$ and $\nabla \cdot \mathbf{u}=0$. Approximate a slender body by a line distribution of these point forces on its centreline. At a point labelled by arclength $s$,To extract the large logarithm, write $s^{\prime }=s+\ell$ and approximate $\mathbf{X}(s+\ell )-\mathbf{X}(s)\simeq \ell \mathbf{t}$, $\mathbf{f}(s+\ell )\simeq \mathbf{f}(s)$, where $\mathbf{t}={\mathbf{X}}^{\prime }$ is the unit tangent vector. The near-line contribution has kernel $(I+\mathbf{tt})/(8\pi \mu |\ell |)$. Cut it off at the wire radius $R$ and a macroscopic length $L$. The two sides of the line giveThe far contribution and changes in the cutoff add terms without the leading logarithm. Since $(I+\mathbf{tt}{)}^{-1}=I-\mathbf{tt}/2$, inversion proves the leading slender-body force densityThis is force on the fluid; the fluid's force on the wire has the opposite sign. It gives twice the resistance to transverse motion as to tangential motion at this order.

The stated arclength describes the circular helix $\mathbf{X}=(a\cos \theta ,a\sin \theta ,b\theta )$. The PDF omits the second $a$ in the coordinates, so the literal printed curve and its arclength are inconsistent unless $a=1$. First use the circular interpretation consistent with the given arclength. Put $q=\sqrt{a^2+b^2}$ and $c=2\pi \mu /\ln (L/R)$. ThenRotation about $z$ has velocity $\mathbf{v}=\Omega (-a\sin \theta ,a\cos \theta ,0)$, so $\mathbf{t}\cdot \mathbf{v}=\Omega a^2/q$. The axial force density is $f_z=-c\Omega a^{2}b/q^2$. Integrating over one turn givesFor translation, $\mathbf{v}=W\hat{\mathbf{z}}$ and $\mathbf{f}=cW(2\hat{\mathbf{z}}-(b/q)\mathbf{t})$. Only the tangential term contributes to axial couple. Since $(\mathbf{X}\times \mathbf{t}{)}_z=a^2/q$, its density is $-cW{a}^{2}b/q^2$, givingThese are the off-diagonal coefficients of the axial resistance matrix of a slender helix. Their equality does not depend on this particular geometry: with symmetric local resistance $D=c(2I-\mathbf{tt})$, the rotation-induced axial force is $\int \hat{\mathbf{z}}\cdot D(\hat{\mathbf{z}}\times \mathbf{X})ds$, whereas the translation-induced axial torque is $\int (\hat{\mathbf{z}}\times \mathbf{X})\cdot D\hat{\mathbf{z}}ds$. They agree pointwise by symmetry of $D$. The full Stokes-flow statement is the Lorentz reciprocal theorem.

For completeness, retaining the literal elliptic helix instead gives $s^{\prime }=\sqrt{a^2{\sin }^2\theta +{\cos }^2\theta +b^2}$ and the axial coupling of an elliptic helixThus reciprocity holds for the literal curve as well; only the geometric coefficient and arclength change. In all these formulas the equal and opposite force or couple on the wire reverses the displayed sign.

The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.

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.

The concern is that the rapid-decay assumptions need not be preserved by the dynamics. In a constant-density incompressible flow, the divergence of the Navier-Stokes equation gives the pressure Poisson equation for incompressible flow, with $\pi =p/\rho$:For a localized eddy, inversion with the free-space Green function and two integrations by parts yieldAt large separation the differentiated kernel has leading form $(3n_i{n}_j-{\delta }_{ij})/r^3$. Thus an anisotropic stress moment of an individual eddy produces a quadrupolar fluid pressure of order $r^{-3}$, even if the velocity source is spatially localized. Correlating that fluid pressure with local $u_x^2$ allows $⟨u_x^2{p}^{\prime }⟩=O(r^{-3})$. Isotropy may cancel a mean multipole but does not automatically cancel such a mixed fourth-order statistic.

The triple-correlation evolution contains a pressure-gradient term of the form $-⟨u_x^2{\partial }_{x^{\prime }}{\pi }^{\prime }⟩$, so these long-range pressure correlations in turbulence can generate a longitudinal triple-correlation tail $K\sim a{r}^{-4}$. With $L=0$, part (ii) then gives $\dot{I}=8\pi u^{3}a$, which need not vanish. This is the substance of the objection to deducing exact conservation from initially short velocity correlations. Batchelor and Proudman's original long-range calculation concerned anisotropic turbulence; extending it to isotropic decay requires the appropriate higher-order correlations and allows cancellations. A permitted tail is not a proof that its coefficient stays appreciable in fully developed turbulence.

This possibility appealed to workers constructing heuristic turbulence closures. A closure that generated a changing low-wavenumber $k^4$ coefficient and a decay exponent different from $10/7$ could then be viewed as representing a pressure-mediated effect, rather than as automatically violating a universal invariant. The closure's far-field coefficient still needs physical validation; closure-generated nonconservation does not itself prove the true dynamics nonconserving.

For evidence available before this exam, the 2006 simulations of Ishida, Davidson and Kaneda found that $I$ approached an approximately constant value after a transient when the box was much larger than the integral scale of turbulence and the Reynolds number was large. The energy decay was correspondingly close to $u^2\sim t^{-10/7}$. The observed conclusion was approximate late-time conservation in that regime, indicating weak net long-range triple-correlation transfer, rather than exact conservation for every initial condition, box size or Reynolds number.

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 $r_0$.

The Richardson constant might be universal if the inertial-range relative dynamics, including the normalized separation distribution, become independent of release details and forcing at sufficiently high Reynolds number. This is an additional universality hypothesis. As evidence available before 2008, Boffetta and Sokolov's numerical study estimated a coefficient near $0.55$ and reported agreement with experimental estimates, alongside small deviations from ideal Richardson scaling. Bourgoin and collaborators' 2006 experiment showed substantial dependence on initial separation in the accessible laboratory regime. Values of order $0.5$ are supported in suitable regimes, but finite-time fits do not establish one universally observed $t^3$ coefficient for all releases.

Since the particle starts at the origin, its Lagrangian trajectory satisfies $\mathbf{X}(t)={\int }_0^t\mathbf{v}(s)ds$. Differentiating the mean-square displacement givesHere we set $\tau =t-s$ and assume finite second moments so that averaging and integration commute. Under Lagrangian stationarity write the autocorrelation as $R_L(\tau )$. Uniform fluid-tracer sampling in homogeneous incompressible turbulence gives $R_L(0)=⟨|\mathbf{u}{|}^2⟩$. A second integration yields the exact Taylor turbulent dispersion formulaFor times short compared with velocity decorrelation, continuity at zero gives $R_L(\tau )\simeq R_L(0)$ throughout the integral. ThusThe particle retains nearly its initial velocity: this is ballistic transport.

If $R_L$ is integrable, then at long times the displacement derivative approaches $2{\int }_0^{\infty }{R}_L(\tau )d\tau =2R_L(0)t_L$, using the Lagrangian integral time. ThereforeThis is diffusive transport: successive long-time displacements lose memory of one another. In an isotropic three-dimensional diffusion equation, $⟨X^2⟩=6D_{T}t$, so $D_T=⟨|\mathbf{u}{|}^2⟩t_L/3$. The shorthand $t\ll t_L$ and $t\gg t_L$ presumes that $t_L$ is representative of the decorrelation time. With strongly oscillatory correlations, the area-based integral time need not characterize every crossover; the exact integral formula remains the appropriate starting point.

The three questions concern different statistics. Taylor turbulent dispersion follows the absolute displacement of one marked fluid element from its release position, or the spreading of the centre of a cloud. Its input is the velocity autocorrelation along a Lagrangian trajectory. It relates short-time ballistic motion to long-time effective diffusion. This is not the shear-flow mechanism ordinarily called Taylor dispersion.

Richardson pair dispersion instead follows the separation of two fluid elements, or the growth of a cloud relative to its moving centre. Common sweeping by an eddy much larger than the cloud contributes to absolute displacement but largely cancels from relative displacement. The relative diffusivity depends on separation; the classical inertial-range prediction is a $t^3$ mean-square separation regime after memory of the release separation is lost.

Obukhov-Corrsin theory, also called Kolmogorov-Obukhov-Corrsin theory, concerns spatial fluctuations of a passive scalar, their transfer across scales and their destruction by molecular diffusion. Its objects are scalar structure functions and scalar spectra, with an inertial-convective prediction controlled by scalar-variance flux and the turbulent energy supply. Thus Taylor describes one-particle transport, Richardson describes relative spreading, and Obukhov-Corrsin describes scale-by-scale scalar mixing. Following a cloud's location is not the same as predicting its internal concentration fluctuations.

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.