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Use fixed spatial coordinates in a stationary heterogeneous Earth and define the operatorHere the elastic stiffness tensor has its minor and major symmetries, and the physical body force per unit volume is $q_i=\rho f_i$. The retarded elastodynamic Green tensor $G_i^j(\mathbf{x},\mathbf{\xi },t)$ is the displacement in direction $i$ due to a unit physical force impulse in direction $j$ at $\mathbf{\xi }$:It satisfies the actual Earth's homogeneous boundary conditions, including zero traction on the free surface and continuity of displacement and traction at bonded internal interfaces. Integration through $t=0$ gives $\rho (\mathbf{x}){\partial }_t{G}_i^j(0^{+})={\delta }_i^j\delta (\mathbf{x}-\mathbf{\xi })$, with $G(0^{+})=0$. No spherical symmetry is required.

For zero initial conditions, form the convolutionApplying $\mathcal{L}$ to this expression uses the defining Dirac delta function to give $\rho (\mathbf{x})f_i(\mathbf{x},t)$. Causality gives zero initial conditions, and each Green column supplies the required boundary conditions. The positive-energy uniqueness argument then identifies this expression with the physical response. The source factor is $\rho (\mathbf{\xi })$ because $f$ is force per unit mass, whereas the Green tensor was normalized to unit force.

A force dipole is a limiting pair of opposite point forces separated by a small vector. Put $+F_j$ at $\mathbf{\xi }+\epsilon \mathbf{d}/2$ and $-F_j$ at $\mathbf{\xi }-\epsilon \mathbf{d}/2$, and hold $D_{jk}=\epsilon F_j{d}_k$ fixed as $\epsilon \to 0$. Expansion of the two Dirac delta functions givesIts resultant force is zero, and integration by parts gives its radiated displacement as ${\int }_0^t{D}_{jk}(\tau ){\partial }_{{\xi }_k}{G}_i^j(\mathbf{x},\mathbf{\xi },t-\tau )d\tau$. Thus the source derivative of the elastodynamic Green tensor represents a dipole with force direction $j$ and separation direction $k$.

For the surface source choose a normal $\mathbf{n}$ from the minus side to the plus side and defineThe same normal is used in both tractions. If the elastic coefficients are continuous through the surface, the distributional gradient of a piecewise smooth displacement is ${\partial }_j{u}_i=({\partial }_j{u}_i{)}_{\mathrm{reg}}+b_i{n}_j{\delta }_{\Sigma }$. Its stress therefore has a singular part $c_{ijkl}{b}_k{n}_l{\delta }_{\Sigma }$. Taking the divergence includes the regular stress jump, so a field satisfying the source-free equations on both sides has equivalent sourceIn particular, the signs follow from the chosen jump convention, not from an unspecified orientation. The elastodynamic surface-jump representation is consequentlywith the Green tensor evaluated at $(\mathbf{x},\mathbf{\xi },t-\tau )$. The first term is a surface distribution of forces, and the second is a surface distribution of dipoles.

One can derive the same formula without multiplying discontinuous material coefficients by a surface delta. Apply integration by parts on the two sides of $\Sigma$ to the field $u$ and the reciprocal Green field $v$. Major symmetry cancels the volume strain terms; the spatial identity iswhen the temporal terms are handled by the retarded time convolution. More explicitly, take $v_i(\mathbf{\xi },\tau )=G_i^a(\mathbf{\xi },\mathbf{x},t-\tau )$, extend the zero-initial-data solution causally to $\tau <0$, and integrate over all source times. Integration by parts cancels the two temporal terms, with zero endpoint terms; the causal supports subsequently reduce the representation to $0\le \tau \le t$. The Green impulse contributes $-u_a(\mathbf{x},t)$ on the left. The two outward face normals are $\mathbf{n}$ and $-\mathbf{n}$, so their sum on the right is $-[u_i]t_i(v)+v_i[t_i]$. Moving it to the other side gives $[u_i]t_i(v)-v_i[t_i]$. The same identity applied to two force Green fields with interchanged source and receiver proves the reciprocal relation $G_i^j(\mathbf{x},\mathbf{\xi },t)=G_j^i(\mathbf{\xi },\mathbf{x},t)$; hence the displayed surface formula follows. On a bonded material interface the Green displacement and common Green traction are continuous: $[u_i]t_i(v)$ is then the unambiguous version of the dipole term, evaluated with the matching one-sided coefficients. External free-surface terms vanish because both tractions vanish.

For a fault in isotropic material, the Lamé parameters giveTangential slip obeys $\mathbf{b}\cdot \mathbf{n}=0$, and ordinary fault traction is continuous across the two faces. Therefore the source is a double-couple fault source withFor a uniform patch of area $A$ with $\mathbf{b}=b{\mathbf{e}}_1$ and $\mathbf{n}={\mathbf{e}}_3$, its integrated seismic moment tensor isThe $13$ dipole is a force in the slip direction separated in the normal direction; the $31$ dipole is a normal force separated in the slip direction. Their equal and opposite torques cancel. The symmetric trace-free seismic moment tensor has eigenvalues $M_0,-M_0,0$, with zero resultant force and torque. A temporal stress drop on a fault is not a jump of traction between its two simultaneously matching faces; it does not add a single-force layer to this tangential-slip representation.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.

For a homogeneous lossless elastic solid, insert the plane wave $u_i=\mathrm{Re}\{U{a}_i{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}\}$ into the equations of linear elasticity. Put $\mathbf{k}=\kappa \mathbf{n}$, $|\mathbf{n}|=1$, and $v=\omega /\kappa$. The resulting eigenvalue problem isThe minor and major symmetries make this acoustic tensor a real symmetric matrix. For a stable material it is positive-definite, so its three positive eigenvalues give three real positive phase velocities and therefore three slownesses $s_a=1/v_a$. Coincident eigenvalues are possible, so there need not be three distinct speeds. For distinct eigenvalues the displacement polarization are orthogonal; in an anisotropic solid they are generally quasi-longitudinal and quasi-transverse rather than exactly longitudinal and transverse. Stability, or at least strong ellipticity in elasticity, is required for the assertion about positive real speeds.

The elastic slowness surface is the locus of the three vectors $\mathbf{p}=\mathbf{n}/v_a(\mathbf{n})=\mathbf{k}/\omega$. Its equation isFor each wavefront normal there are three radial sheets, which may meet at degeneracies. To obtain the energy flux directly, multiply the elastic equation by ${\dot{u}}_i$. Forthe symmetry of stress gives ${\partial }_{t}W={\partial }_j({\sigma }_{ij}{\dot{u}}_i)$. Thus the instantaneous elastic-wave energy flux is $F_j=-{\sigma }_{ij}{\dot{u}}_i$. With a real branch polarization $\mathbf{a}$, harmonic averaging givesNow differentiate $c_{ijkl}{p}_j{p}_l{a}_k=\rho a_i$ along a regular slowness surface sheet and contract with $a_i$. The terms containing $d\mathbf{a}$ cancel by symmetry and the original eigenvalue equation, leavingEvery tangent $d\mathbf{p}$ is consequently perpendicular to $⟨\mathbf{F}⟩$: the energy flux is normal to the slowness surface. At a degenerate point this statement is made on the limiting branches, rather than assuming a unique normal.

For completeness, equality of the mean kinetic and strain energies gives $⟨W⟩=\rho {\omega }^2|U{|}^2|\mathbf{a}{|}^2/2$. Differentiating the acoustic tensor equation in the wavevector variables similarly givesThus the group velocity is the elastic energy velocity. The homogeneous elastic dispersion relation is degree one in $\mathbf{k}$, so differentiating $\omega (\lambda \mathbf{k})=\lambda \omega (\mathbf{k})$ at $\lambda =1$ gives $\mathbf{p}\cdot \mathbf{g}=1$.

The elastic wave surface is the unit-time locus of these group velocities. From an impulsive point source, the propagating wavefronts at time $t$ are its sheets scaled to $\mathbf{x}=t\mathbf{g}$. A particular source may fail to excite a mode in some directions, and its amplitudes require the source polarization and geometrical spreading; the wave surface determines the arrival geometry. It does not specify all near-field motion between the radiative fronts. Geometrically it is the envelope of the planes $\mathbf{p}\cdot \mathbf{x}=1$: tangent variation gives $d\mathbf{p}\cdot \mathbf{x}=0$, so $\mathbf{x}$ is the appropriately normalized normal to the slowness surface.

In a symmetry-plane cross-section write $\mathbf{p}=r(\theta )\mathbf{n}$, $\mathbf{n}=(\cos \theta ,\sin \theta )$ and $\mathbf{t}=(-\sin \theta ,\cos \theta )$. The two envelope conditions giveThe numerator is the signed curvature numerator of the slowness surface cross-section. At a simple inflection point, $h=v+v^{\prime \prime }=0$ and $h^{\prime }\ne 0$. Then ${\mathbf{g}}^{\prime \prime }=h^{\prime }\mathbf{t}$ and ${\mathbf{g}}^{\prime \prime \prime }=h^{\prime \prime }\mathbf{t}+2h^{\prime }{\mathbf{t}}^{\prime }$ are linearly independent. These are precisely the local conditions for an ordinary cusp. The slowness curvature and wave-surface cusps relation explains why a smooth nonconvex slowness surface can generate a cusped wavefront and multiple ray arrivals.

The example below uses a positive orthotropic stiffness in normalized units: $\rho =1$, $C_{11}=C_{22}=C_{33}=4$, $C_{44}=C_{55}=1.2$, $C_{66}=1$, and all normal off-diagonal stiffnesses zero. In the symmetry plane $n_3=0$, its two in-plane acoustic tensor eigenvalues are $v_{\pm }^2=(5\pm \sqrt{5+4\cos 4\theta })/2$, and its out-of-plane shear speed is $\sqrt{1.2}$. The in-plane shear sheet is nonconvex and produces the marked cusps; the construction uses an actual stable elastic tensor.

An elastic deformation is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous stress is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the displacement field $u_i$ gives the infinitesimal strain tensor $e_{ij}=(u_{i,j}+u_{j,i})/2$. The Cauchy stress tensor is defined by traction $t_i={\sigma }_{ij}{n}_j$ on a plane of unit normal $\mathbf{n}$; absence of body couples makes ${\sigma }_{ij}={\sigma }_{ji}$.

Let $\mathcal{E}(e)$ be the recoverable elastic energy per unit mass. The local mechanical power per unit volume isbecause the symmetric Cauchy stress tensor annihilates the antisymmetric part of the velocity gradient. Conservation of energy during the deformation gives $\rho \dot{\mathcal{E}}={\sigma }_{ij}{\dot{e}}_{ij}$. Since $\dot{\mathcal{E}}=(\partial \mathcal{E}/\partial e_{ij}){\dot{e}}_{ij}$ for arbitrary symmetric strain rates,Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.

Take a stress-free reference state of mass density ${\rho }_0$. The Taylor series of the elastic energy starts quadratically:Using $\rho ={\rho }_0+O(|e|)$ gives ${\sigma }_{ij}=c_{ijkl}{e}_{kl}+O(|e{|}^2)$, the relation of linear elasticity. If the reference state has initial stress, the first-order expansion also retains that constant stress. Symmetry of stress and strain gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of elastic energy gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the elastic stiffness tensor is a symmetric map on a six-dimensional strain space: there are $6\cdot 7/2=21$ independent coefficients.

For twofold elastic symmetry, choose the symmetry axis as $x_3$. Rotation through $\pi$ is $Q=\mathrm{diag}(-1,-1,1)$, and transforms the infinitesimal strain tensor into $Qe{Q}^T$. Components $e_{11},e_{22},e_{33},e_{12}$ are even and $e_{23},e_{13}$ are odd. Invariance of elastic energy excludes every coupling between these parity classes. In the engineering strain ordering $(e_{11},e_{22},e_{33},2e_{23},2e_{13},2e_{12})$, with corresponding stress ordering $({\sigma }_{11},{\sigma }_{22},{\sigma }_{33},{\sigma }_{23},{\sigma }_{13},{\sigma }_{12})$, the symmetric stiffness matrix has the formIts symmetric even block has ten independent entries and its symmetric odd block has three. Hence a twofold axis leaves $13$ independent coefficients. No invariance under other rotation angles has been assumed.

Because $s$ is arc length, $|d\mathbf{x}/ds|=1$. Thus $\mathbf{y}=\mathbf{n}/\alpha$, where $\mathbf{n}$ is the unit tangent to the elastic ray. In an isotropic medium this is also the wavefront normal. Consequently $\mathbf{y}$ is the vector slowness, of magnitude $1/\alpha$; locally $\mathbf{y}=\nabla T=\mathbf{k}/\omega$, with $T$ the travel time and $\mathbf{k}$ the wavevector.

For depth-dependent wave speed $\alpha (z)$, the ray equation gives $d{y}_x/ds=d{y}_y/ds=0$. The horizontal component of $d\mathbf{x}/ds$ therefore has a fixed direction, and the elastic ray lies in the vertical plane determined by that direction and its initial point. A vertical elastic ray is the limiting case. If $i$ is the angle to the vertical, its conserved horizontal slowness givesAt a horizontal interface, matching the tangential wavevector at fixed frequency gives the same invariant on both sides: this is Snell's law.

For radial wave speed $\alpha (r)$, take the cross product of position with slowness. The ray equations givesince the second vector is radial. A nonzero constant $\mathbf{L}=\mathbf{x}\times \mathbf{y}$ is perpendicular to every position on the elastic ray, so the ray lies in a fixed plane through the centre. Radial rays obey the same conclusion with a nonunique plane. The spherical elastic ray invariant is thereforewhere $i$ is measured from the radial direction. Across a spherical interface the radius is common and tangential slowness is continuous, so this form of Snell's law survives an interface discontinuity.

In a homogeneous sphere a P wave follows the straight chord between the two surface points. If their principal angular separation is $0\le \Delta \le \pi$, its length is $2R\sin (\Delta /2)$. For constant P wave speed $c_P$,Now put $a=R^{\prime }$ and denote the mantle and core P wave speeds by $c_m>c_c$. Equal incidence and reflection angles put the core-boundary reflection point on the angular bisector. Each mantle leg has length $L$ given by $L^2=R^2+a^2-2Ra\cos (\Delta /2)$. Hence the core-reflected travel time isThe stated domain is essential. At the reflection point the incoming line approaches from the mantle only if $R\cos (\Delta /2)\ge a$. Otherwise it has already crossed the core boundary. The mantle-only direct chord also exists only for $\Delta \le {\Delta }_g$, with $T_D=2R\sin (\Delta /2)/c_m$. In this interval,so the reflected arrival is later. Its intercept is $2(R-a)/c_m$, whereas the direct intercept is zero. At grazing incidence the branches meet at $T=2\sqrt{R^2-a^2}/c_m$, and both have slope $dT/d\Delta =a/c_m$. A continuation of the homogeneous-sphere direct formula beyond this endpoint is not a mantle-only arrival in the modified model.

The core-transmitted elastic rays consist of three straight legs. Their impact distances in mantle and core are $b_m=p{c}_m$ and $b_c=p{c}_c$, respectively. At core entry,Since $c_c<c_m$, the elastic ray bends towards the normal on entry and away from it on exit. The mantle leg subtends $i_m-i_R$ at the centre, and the core chord subtends $\pi -2i_c$. Thus the total angular sweep along the transmitted path is $2(i_m-i_R)+\pi -2i_c$. This sweep need not equal the principal separation if it exceeds $\pi$; the diagram uses a path with sweep below $\pi$.

A simple fluid reduces to a linear viscoelastic fluid when its deformation and rotation over the relevant memory interval are small enough to linearize the constitutive history about an isotropic relaxed state. Small amplitude is essential; small frequency is not. For example, arbitrarily high-frequency oscillation can still be linear if its strain amplitude is sufficiently small. The relaxation modulus $G(t)$ is the shear-stress response to a unit small step of shear strain.

Choose the harmonic convention $e^{i\omega t}$. If the imposed shear rate is $\mathrm{Re}(\hat{\dot{\gamma }}{e}^{i\omega t})$, the corresponding shear stress is $\mathrm{Re}({\eta }^{\ast }(\omega )\hat{\dot{\gamma }}{e}^{i\omega t})$. Substitution into the constitutive convolution gives the complex viscosity:For a nondecaying elastic part this formula is understood through the limit from a positive Laplace damping parameter. The complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, so $G^{\prime }$ is the storage modulus, $G^{\prime \prime }$ the loss modulus, and ${\eta }^{\ast }=G^{\prime \prime }/\omega -i{G}^{\prime }/\omega$ in this convention.

Creep is the progressive deformation produced by a held stress. For a small step $S_{0}H(t)$ of shear stress, define the creep compliance $J$ by $\gamma (t)=S_{0}J(t)$ from a relaxed state. Writing hats for Laplace transforms, the same constitutive law gives $\hat{S}=p\hat{G}\hat{\gamma }$. Since $\hat{S}=S_0/p$, this yieldsHere $H$ is the Heaviside step function. Inversion determines the full creep response, including any instantaneous elastic strain jump. Equivalently, the causal relaxation/creep convolution includes that jump through ${\int }_{0^{-}}^{t}G(t-s)dJ(s)=1$ for $t>0$.

For the Maxwell fluid, $\hat{G}(p)=G_0/(p+1/\tau )$. Thus the two material responses areThe creep response is an elastic jump followed by constant-rate viscous deformation. The zero-frequency shear viscosity is $\eta =G_0\tau$.

For the wall problem, take $v=u(y,t)e_x$ and denote the shear component of stress by $S$. The incompressibility constraint is satisfied, and the advective part of the material derivative vanishes because no field depends on $x$. With no imposed longitudinal pressure gradient, momentum conservation and the linear Maxwell law giveThe initial conditions are $u=S=0$ in $y>0$; the no-slip boundary condition is $u(0,t)=UH(t)$, with decay as $y\to \infty$. Transforming these equations givesChoose the analytic square-root branch in $\mathrm{Re}p>0$ with $\mathrm{Re}q>0$, so that the spatially increasing solution is excluded. The Maxwell start-up shear layer consequently hasThe Bromwich inversion formula now givesThe vertical Bromwich contour is directed upward and lies to the right of all singularities, including the origin and the branch points at $0$ and $-1/\tau$. A cut joining these branch points is a possible continuation into the left half-plane; inversion only requires the stated right-half-plane branch.

The Newtonian limit holds $\eta =G_0\tau$ fixed while $\tau \to 0$, so $G_0\to \infty$. Then $q(p)\to \sqrt{\rho p/\eta }$. With $\nu =\eta /\rho$, the inverse Laplace transform gives the complementary error function profileThis is diffusive momentum penetration in a Newtonian fluid, with thickness of order $\sqrt{\nu t}$.

The elastic limit instead holds $G_0$ fixed while $\tau \to \infty$. Write $c=\sqrt{G_0/\rho }$. On the chosen branch, $q(p)\to p/c$, so the Laplace transform time-shift rule yieldsAn elastic shear front travels at speed $c$: the material ahead is at rest, while that behind moves at the wall speed. Indeed the limiting displacement is $d(y,t)=U(t-y/c)H(t-y/c)$, whose shear strain is $d_y=-(U/c)H(t-y/c)$. A nonzero pointwise velocity gradient at the front is interpreted distributionally, or by smoothing the ideal wall start.

The use of linear viscoelasticity is conditional, rather than automatically justified for arbitrary $U$. The material derivative simplifying to a time derivative does not remove the nonlinear constitutive effects of finite shear strain. In this Maxwell problem the rapid elastic response supplies the natural strain scale $U/c=U\sqrt{\rho /G_0}$, which must be small for linearization through start-up. Away from the front, the strain accumulated over an effective memory interval must likewise be small. In the late diffusive regime its characteristic size is $\tau U/\sqrt{\nu t}$, which decreases with time. Large absolute wall displacement $Ut$ alone is not a disqualification for a fluid with finite memory; it is relative deformation during that memory that matters. In the elastic limit, where memory does not decay, the explicit strain behind the front confirms the condition $U/c\ll 1$. Without such a small-strain restriction, the computed field is a formal linear response and need not describe an arbitrary simple fluid.

An isotropic simple fluid has local stress determined by the deformation history of the material element. A viscometric flow is one whose relative deformation history is that of steady simple shear flow, up to rigid changes of frame. At a given element the shear rate is constant along its history, although it can differ between elements. This includes the locally simple-shear kinematics used in standard viscometric measurements.

Use the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$. The first two Rivlin-Ericksen tensors areIn the axes of a steady shear $v_x=sy$, the relative deformation gradient over a lag $q$ is $F=I+qK$, since $K^2=0$. The material history is determined by $s$, and its tensor symmetries exclude $xz$ and $yz$ stress components. The remaining symmetric stress consists of an isotropic part, one shear component, and two independent normal-stress differences. The relevant matrices areThus $I,A_1,A_1^2,A_2$ span the allowed stress components. Material isotropy and reversal of the shear show that their scalar coefficients depend on $|s|=(A_1:A_1/2{)}^{1/2}$. Define the viscometric functions bySolving for the matrix coefficients givesThis is an exact steady-viscometric representation, rather than an assumption that the fluid is everywhere a second-order fluid. The shear viscosity $\mu$ measures the shear resistance, while ${\psi }_1$ and ${\psi }_2$ measure the first and second normal-stress differences generated by shear. Their zero-rate values are understood as limits when these exist. An isotropic pressure changes neither normal difference.

For a flow in the lubrication approximation with gap $H$, longitudinal scale $\ell \gg H$ and speed $U$, the shear-rate scale is $U/H$. The viscous shear force per unit volume scales as $\mu U/H^2$. Quadratic viscoelastic normal stresses scale as ${\psi }_i{U}^2/H^2$; their longitudinal gradients, and the comparable mixed-component corrections, contribute forces of scale ${\psi }_i{U}^2/(H^2\ell )$. Therefore the normal-stress smallness in lubrication criterion isEquivalently, the thin-gap aspect ratio must multiply the shear-based elastic parameter to give a small quantity: $(H/\ell )|{\psi }_i|\dot{\gamma }/\mu \ll 1$. Normal stresses need not themselves be smaller than the shear stress; it is their effect on the leading force balance that matters. If the flow changes on an additional short time scale $T$, the $D_t{A}_1$ term also requires $|{\psi }_1|/(\mu T)\ll 1$ for its neglect. Material coefficients should be evaluated at the characteristic shear rate.

Now take positive cylinder rotation, measure $\theta$ counterclockwise from the rightward horizontal, and let $y=r-a$ increase outward through the film. The tangent has upward component $\cos \theta$, so the tangential gravitational acceleration is $-g\cos \theta$. To leading order in $h/a$, neglect inertia and higher-order hydrostatic or capillary pressure gradients, and write $U=\Omega a$. Tangential force balance and the shear-free free surface givehenceThe wall shear stress is $-\rho gh\cos \theta$. It opposes rotation on the uphill side and assists it on the downhill side.

For a power-law fluid, ${\sigma }_{\theta y}=k|u_y{|}^{n-1}{u}_y$. Its inversion must retain the sign:Integrating with the no-slip boundary condition $u(0)=U$ yieldsThe volume flux per unit axial length is $Q={\int }_0^{h}u(y)dy$. The integrated bracket is $[(n+1)/(2n+1)]h^{(2n+1)/n}$, so the power-law coating on a rotating cylinder hasAt $\cos \theta =0$ the gravitational term vanishes by continuity. Steady mass conservation makes $Q$ independent of $\theta$.

On the uphill half, put $D=(\rho g\cos \theta /k{)}^{1/n}$ and $m=(2n+1)/n$. The local flux is $Q(h)=Uh-(D/m)h^m$, withIt has a unique maximum atA positive coating must carry the same steady flux at every angle. The tightest local bound occurs at $\cos \theta =1$, not on the downhill side, where the flux increases monotonically with thickness. ConsequentlyFor $0<Q<Q_{\max }$ the thinner positive root on the uphill side connects through $\cos \theta =0$ to the unique downhill root, giving a periodic leading-order coating profile. At the limiting flux the two uphill roots meet at $\theta =0$; it is the maximum flux of this lubrication model, with a possible limiting corner resolved by neglected higher-order effects. For $n=1$ the check is $Q_{\max }=(2/3)U\sqrt{kU/(\rho g)}$. All these conclusions require the resulting film to remain thin enough for the lubrication approximation and for the neglected normal stresses and inertia to satisfy their scaling restrictions.

A large molecular Schmidt number $\mathrm{Sc}=\nu /\kappa$ separates the velocity's smallest smooth length scale from the scalar's smaller diffusive length scale. Follow a Lagrangian trajectory $X(t)$ and measure displacement $x$ from it. Within a sufficiently small scalar packet, Taylor expansion of the smooth velocity field givesThe moving frame removes uniform sweeping, leaving a linear random velocity gradient tensor. Thus the local scalar equation is the advection-diffusion equation with velocity $Cx$, and averaging over realizations of $C$ gives the corresponding ensemble average. The approximation requires the packet to lie within the velocity's locally smooth region. Its thin, diffusive direction can satisfy this condition even when its long direction eventually becomes too large for one global linearization; applications to a spatially varying flow then use local packet descriptions. The statistical conclusions below refer to the linear-flow model and the usual mixing assumptions, not to every conceivable trace-free random matrix process.

Write $Q=\int \chi dA$ and assume sufficiently rapid decay to discard boundary terms. Integration by parts givesFor $J_{ij}=\int x_i{x}_j\chi dA$, integrate the advective term once and the diffusive term twice. Since ${\partial }_k(x_i{x}_j)={\delta }_{ki}{x}_j+{\delta }_{kj}{x}_i$ and $\Delta (x_i{x}_j)=2{\delta }_{ij}$, one obtainsThese are mass conservation and the evolution of the unnormalized second moment tensor. The first moment obeys $\dot{M}=CM$, so a centered packet stays centered.

For a positive Gaussian scalar packet in a linear incompressible flow, $B$ must be a positive-definite matrix. Diagonalizing its symmetric matrix and applying the Gaussian integral givesThe factor $Q^2$ in the last identity matters because $J$ has not been divided by the scalar mass. Substitution into the advection-diffusion equation also givesThese identities agree with the second moment equation and prove that the Gaussian family is preserved. For an initially negative packet the same analysis applies after replacing $\chi$ by $-\chi$ and taking its positive mass.

To see the roles of stretching and diffusion explicitly, let the deformation gradient $F$ solve $\dot{F}=CF$, $F(0)=I$. Incompressibility gives $\det F=1$. Differentiating $F^{-1}J{F}^{-T}$ integrates the moment equation:Without diffusion, $J=FJ(0)F^T$ and $\det J$ is constant. In a statistically stationary, mixing flow with positive Lyapunov exponent $\lambda$, the two stretching exponents are $\lambda$ and $-\lambda$. Writing the ordered eigenvalues of $J$ as $e^{2{\rho }_1},e^{2{\rho }_2}$, their typical initially nondiffusive evolution isThe fluctuation scale presumes the usual central limit theorem for accumulated strain; it is not implied by incompressibility alone. During this stage the packet becomes long and thin but keeps its area and peak concentration.

In the instantaneous principal-axis basis, write $\tilde{C}$ for the rotated velocity gradient tensor. Differentiation of an eigenvalue givesThe formulas hold through a repeated eigenvalue by continuity of their sum. The compressed physical variance is $e^{2{\rho }_2}/Q$. Its contraction is arrested when it is of order $\kappa /\lambda$, the square of the Batchelor scalar microscale ${\ell }_B\sim \sqrt{\kappa /\lambda }$. If its initial length is $L\gg {\ell }_B$, the crossover occurs atThereafter ${\rho }_2$ has stationary fluctuations associated with the strain and the diffusive floor, while ${\rho }_1$ continues to grow typically at rate $\lambda$. Under sufficiently fast mixing, the central part of the distribution of ${\rho }_1$ is approximately Gaussian with mean $\lambda t+O(1)$ and variance $D_{\rho }t+o(t)$ for a strain-dependent coefficient $D_{\rho }$. This central approximation does not determine far tails, which control high concentration moments; those require a full large deviation principle. The two logarithmic widths need not be independent. The area now increases and $f$ decreases typically as $e^{-\lambda t}$. This is diffusion arrest of Gaussian packet compression.