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Put $r=|\mathbf{x}|$. In the Unscaled Papkovich–Neuber representation, choose the componentwise harmonic vector fieldBoth $\mathbf{\Omega }\times \mathbf{x}$ and $(\mathbf{\Omega }\times \mathbf{x})/r^3$ are harmonic away from the origin: the latter is a linear combination of derivatives of the harmonic function $1/r$. Also $\mathbf{x}\cdot \mathbf{\Phi }=0$, so the representation gives $\mathbf{u}=f(r)\mathbf{\Omega }\times \mathbf{x}$. The no-slip boundary condition requires $f(a)=1$ and $f(b)=0$, givingThus the Rotational Stokes flow between concentric spheres isThe divergence of $\mathbf{\Phi }$ is zero: the cross-product field itself has zero divergence, and $\nabla f$ is radial and perpendicular to $\mathbf{\Omega }\times \mathbf{x}$. Hence $p=2\mu \nabla \cdot \mathbf{\Phi }=0$ in this pressure gauge.

There is also a symmetry argument. Take the rotation axis as the polar axis. Linearity of Stokes flow makes velocity and pressure perturbations change sign when the rotation reverses. Reflection in any plane containing that axis reverses the axial rotation vector. Combined with axial symmetry, this excludes meridional velocity components and forces the pressure perturbation to be both unchanged and sign-reversed by that reflection. Equivalently a purely azimuthal axisymmetric Stokes flow has no meridional component of its vector Laplacian, so its meridional pressure gradient vanishes. Thus the pressure is constant; setting that arbitrary constant to zero gives the same result without solving the radial coefficients.

For the Newtonian fluid stress tensor, let $\mathbf{q}=\mathbf{\Omega }\times \mathbf{x}$. The symmetric part of the velocity gradient isbecause the gradient of rigid rotation is antisymmetric. Consequently the full stress tensor, with $p=0$, isFor $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, its only nonzero spherical-coordinate components are ${\sigma }_{r\varphi }={\sigma }_{\varphi r}=-3\mu B\Omega \sin \vartheta /r^3$. On the inner sphere the traction exerted by the fluid, using the sphere's outward normal $\mathbf{n}$, isIts torque isusing $\int n_i{n}_{j}d\omega =4\pi {\delta }_{ij}/3$. Therefore the applied Torque in rotational Stokes flow between concentric spheres isFor $b\gg a$, the relevant speed and length are $\Omega a$ and $a$, so negligible inertia requires the rotational Reynolds numberFor a narrow gap $d=b-a\ll a$, the viscous gradients are controlled by $d$, not $a$. A centrifugal acceleration of scale ${\Omega }^{2}a$ is then compared with viscous scale $\mu \Omega a/d^2$, rather than $\mu \Omega /a$. Their ratio is $\rho |\Omega |d^2/\mu$. This explains why the large-gap condition is not a necessary narrow-gap criterion; a thin gap greatly strengthens viscous resistance. This base-flow scaling is distinct from a criterion for stability to arbitrary short-scale disturbances.

The Minimum-dissipation theorem for Stokes flow compares a Stokes solution with all sufficiently regular incompressible trial velocities in the same fluid domain that have the same prescribed boundary velocities. With no body-force work, the dissipation functional isIf $\mathbf{v}=\mathbf{u}+\mathbf{w}$ and $\mathbf{w}=0$ on the prescribed boundary, integration by parts with $\nabla \cdot \mathbf{\sigma }(\mathbf{u})=0$ and $\nabla \cdot \mathbf{w}=0$ makes the cross term vanish. ThusThe comparison fixes velocities, rather than applied torques or forces.

Now let the particles take their actual rigid translations and rotations in the particle-containing flow. Extend the fluid velocity through each particle by its rigid velocity. No-slip boundary conditions make this extension continuous; rigid motion has zero rate-of-strain tensor and zero divergence. The extension is therefore an admissible trial field in the original particle-free annulus, and its dissipation equals that of the fluid outside the particles. Hence $D\ge D_0$.

The Stokes boundary-work identity gives $D=\mathbf{G}\cdot \mathbf{\Omega }$. The outer sphere is stationary, the maintained inner sphere has zero translational velocity, and every added particle has zero total force and torque, so none of those other boundaries contributes power. Likewise $D_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor nonzero rotation and a particle of positive volume the inequality is strict: equality in the dissipation identity would force the extension to be the unique original flow, but that original flow has nonzero strain on every open particle region, whereas a rigid interior has zero strain. This proves that force-free inclusions increase rotational resistance:No dilute-suspension approximation or assumption about the particle shapes was used.

In a slender symmetric viscous sheet, the two broad faces have zero tangential traction. A cross-thickness variation of $w$ on the thickness scale would produce leading shear $\mu w_x$, whereas $\mu u_z$ is smaller by the square of the aspect ratio. The leading transverse momentum balance and the shear-free conditions therefore give $w_x=0$: write $w=w(z,t)$.

Incompressibility gives $u_x=-w_z$. Symmetry about $x=0$ sets the integration constant to zero, soThe normal traction on either nearly vertical face matches the ambient hydrostatic pressure, giving ${\sigma }_{xx}=-p_a$ to leading order. The Newtonian fluid stress tensor then yieldsThe factor four is the planar extensional resistance, as in the planar viscous-sheet stretching equations, rather than the factor three for uniaxial extension with two contracting transverse directions.

The following slice diagram shows all axial-force contributions per unit $y$-width; the end arrows indicate positive tensile stress and reverse if that signed stress is negative.

The two end tractions have net $z$-component ${\partial }_z(h{\sigma }_{zz})\delta z$. The normal pressure on the inclined side faces contributes $p_a{h}_z\delta z$, and gravitational acceleration gives the weight $\rho gh\delta z$. Hence force balance isSubstitute the stress expression and $p_a^{\prime }={\rho }_{a}g$:soKeeping the side-pressure component is essential: dropping it would retain an incorrect dependence on the arbitrary ambient pressure level. The remaining equation from conservation of mass isIt follows either by integrating incompressibility across the moving faces or directly from the face kinematic boundary condition.

When effective gravitational acceleration vanishes, $4\mu h{w}_z$ is independent of $z$. It equals the applied excess tensile force $F(t)$, with ambient pressure removed from the end loading. Along a material element labeled by its initial position $z_0$,Integration gives uniform material thinning under planar-sheet tension:A material interval conserves its area: $h(z,t)dz=h_0(z_0)d{z}_0$. ThusWith the specified fixed material origin, integrate the latter relation to obtainThese formulas hold while the material thickness remains positive.

For the quadratic initial profile, put $q=H-\Delta >0$. If $k\ne 0$, evaluating the integral at $z_0=L_0$ givesFor $k=0$ it instead givesFor a constant positive pulling force, $\Delta =Ft/(4\mu )$ first reaches the minimum initial thickness $H$ atBoth expressions for $L$ diverge there. Write $q=F(t^{\ast }-t)/(4\mu )$. If $k\ne 0$, the inverse tangent tends to $\pi /2$ and $\Delta \to H$, givingFor $k=0$,This is finite-time extension of a quadratically thickened sheet. With nonzero $k$, the dominant extension comes from the small material region $|z_0|=O(\sqrt{q/H}/|k|)$ near the thickness minimum, where the integrand is proportional to $1/(q+H{k}^2{z}_0^2)$. Its width shrinks like $\sqrt{q}$, giving an integrated divergence $q^{-1/2}$. The fixed ends are asymptotically far away in those local coordinates, so $A$ does not depend on $L_0$. For a uniform sheet the entire material interval has denominator $q$, giving the stronger $q^{-1}$ divergence and retaining its length in $A$. The singularity is a prediction of the ideal slender-sheet model as its minimum thickness tends to zero.

For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

For a stationary solid skeleton, the Darcy velocity $\mathbf{u}$ is liquid volume transport per unit total cross-sectional area, while the pore velocity $\mathbf{v}$ is the intrinsic mean fluid velocity within the connected pore space. ThusFor an incompressible liquid in a saturated region, conservation of mass is ${\varphi }_t+\nabla \cdot \mathbf{u}=0$. With time-independent but spatially varying porosity, it becomesnot generally $\nabla \cdot \mathbf{v}=0$. Neglecting dispersion, a fluid interface is advected by the pore velocity; its mean normal speed is $V_n=\mathbf{v}\cdot \mathbf{n}=\mathbf{u}\cdot \mathbf{n}/\varphi$.

For the thin layer on the inclined plane, the gravitational acceleration components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plane. The hydrostatic approximation and constant pressure $p_a$ above the current giveWith permeability of a porous medium $k(z)$, Darcy's law therefore givesHere $u_x,u_y$ denote the components of the Darcy velocity. Integrate over depth, using the prescribed power laws:Both integrals are finite for the positive exponents in the model. Conservation of the stored fluid gives ${\mathcal{V}}_t+{\nabla }_{x,y}\cdot \mathbf{Q}=0$, orFor $0<\theta <\pi /2$, set $X=x\tan \theta$, $Y=y\tan \theta$ andSince $\cos \theta {\tan }^2\theta =\sin \theta \tan \theta$, the depth-dependent inclined porous-current equation isThe longitudinal flux carries fluid downslope, while the thickness-gradient terms spread it. The degenerate horizontal or vertical slope limits need their own coordinate scaling.

The cell conservation equation has the form ${\bar{n}}_{\bar{t}}=-\bar{\nabla }\cdot \bar{\mathbf{J}}$, withThe first term is advection by the suspension's fluid velocity, the second is active swimming at fixed speed $V_c$ in direction $\mathbf{p}$, and the last is an isotropic diffusive flux due to random swimming. Here $D$ has units length squared per time, $\bar{n}$ is cell number per volume, and the divergence measures local loss by transport. No birth or death term is included.

For a spherical cell of radius $a_c$, its excess buoyancy force is $(4\pi a_c^3/3)\Delta \rho g$. The Stokes drag law gives passive settling speed $V_{\mathrm{set}}=2a_c^2\Delta \rho g/(9\mu )$. Neglecting individual settling requires $V_{\mathrm{set}}\ll V_c$, not exact neutral buoyancy. A dilute suspension can still generate appreciable collective buoyancy because many small excess masses contribute to the bulk density.

Use the half-rate convention for the gyrotactic reorientation time $B$, required by the coefficients printed in the question:The first torque tends to point the bottom-heavy cell upward; the second rotates a sphere with half the fluid vorticity. This is a quasistatic torque balance, assuming orientation relaxation is fast enough to follow the local flow. Keeping orientation dynamics instead would put $d\mathbf{p}/d\bar{t}$ on the left. The convention with aligning rate $1/B$ uses a reorientation parameter twice this $B$. Thus the factor of two is a definition, not a physical discrepancy. Close to the upward state the quasistatic gyrotactic balance with half-rate reorientation gives ${\mathbf{p}}_{\perp }=B\bar{\mathbf{\omega }}\times {\mathbf{e}}_z$.

For a dilute Newtonian suspension with the Boussinesq approximation, the other equations areThe background-water hydrostatic pressure is absorbed in $\bar{p}$. The cell volume is $\upsilon$, density is $\rho$, cell density is $\rho +\Delta \rho$, dynamic viscosity is $\mu =\rho \nu$, and $\upsilon \bar{n}\ll 1$ is the volume fraction. Extra cell mass produces the downward buoyancy term. Constant viscosity and no explicit swimming stress are part of this closure. Rigid impermeable plates impose no-slip boundary condition and zero normal total cell flux.

At rest, the stable cell direction is upward and the steady vertical flux is $V_c\bar{n}-D{\bar{n}}_{\bar{z}}$. It is constant in depth by conservation and zero at an impermeable plate, soThe pressure gradient balances its excess weight. The mean-concentration formula follows by averaging over $0\le \bar{z}\le h$ and is continuous at $\beta =0$.

Use $\mathbf{x}=\bar{\mathbf{x}}/h$, $t=D\bar{t}/h^2$, $\mathbf{u}=h\bar{\mathbf{u}}/D$, $n=\bar{n}/{\bar{N}}_0$, and $p=h^2\bar{p}/(\rho \nu D)$. ThenwhereThe dimensionless base concentration is $n_0=e^{\beta z}$.

Linearize with the horizontal Fourier mode $e^{\sigma t+ilx+imy}$ and $k^2=l^2+m^2$. Write the first-order orientation perturbation as $\mathbf{p}={\mathbf{e}}_z+ϵ{\mathbf{p}}_1+\cdots$. Unit length gives $p_{1z}=0$, while torque balance gives ${\mathbf{p}}_{1\perp }=G{\mathbf{\omega }}_1\times {\mathbf{e}}_z$. Incompressibility impliesLet $\mathcal{L}=d^2/d{z}^2-k^2$. The perturbation of fluid advection is $n_0^{\prime }W=\beta n_{0}W$, and swimming contributes $\beta N^{\prime }+\beta n_0\nabla \cdot {\mathbf{p}}_1$. Cell conservation therefore givesorThis explains the destabilizing concentration-advection term and the gyrotactic redistribution term with their signs.

To eliminate pressure from the fluid equations, take their divergence, giving $\mathcal{L}{p}_1=-(R/\beta )N^{\prime }$. Apply $\mathcal{L}$ to the vertical equation $S{c}^{-1}\sigma W=-p_1^{\prime }+\mathcal{L}W-(R/\beta )N$. The concentration second derivatives cancel, leavingNo-slip conditions give $W=0$ and $W^{\prime }=0$ at each plate; for nonzero $k$, the latter follows from the vanishing horizontal velocity and incompressibility. Zero perturbed normal cell flux givesThe normal orientation correction is zero at this order, and the fluid velocity is zero at the plate, so neither adds another term to this concentration condition.

Now use the stated distinguished scalings $k=\beta \alpha$, $\sigma ={\beta }^2{\sigma }_2$, $G={\beta }^{-1}{G}_{-1}$, $N=N_0+\beta N_1+{\beta }^2{N}_2+\cdots$, $W=\beta W_1+{\beta }^2{W}_2+\cdots$, and $R=R_0+\beta R_1+\cdots$. Here $\alpha$ is a scaled wavenumber, unrelated to Question 2's shear rate. At order one, the concentration equation and boundary conditions give $N_0^{\prime \prime }=0$, $N_0^{\prime }=0$ at both endpoints, so $N_0$ is constant. Absorb its nonzero value into the perturbation amplitude and chooseAt order $\beta$, the fluid equation is $W_1^{(4)}=-R_0{\alpha }^2$ with $W_1=W_1^{\prime }=0$ at both ends. Its unique solution isA homogeneous difference is cubic and the four clamped conditions make it zero.

The concentration equation at order $\beta$ givesso $N_1=z-G_{-1}{W}_1+C_1$, where $C_1$ fixes the amplitude normalization. At order ${\beta }^2$ it givesThe boundary conditions expand to $N_2^{\prime }=N_1$ at each endpoint. Integrating this equation from zero to one cancels $[N_2^{\prime }{]}_0^1$ against $[N_1{]}_0^1$. Also $[W_2^{\prime }{]}_0^1=0$ andThus the required solvability condition isSince ${\int }_0^1{z}^2(1-z{)}^{2}dz=1/30$, the clamped-quartic solvability in shallow bioconvection gives

For nonzero long horizontal waves, $R_0<720$ gives decay and $R_0>720$ gives growth; the leading neutral threshold is $R_0=720$. The leading instability is stationary and independent of $G_{-1}$ and $Sc$. The exactly uniform horizontal mode $\alpha =0$ has zero growth at this order because total cell number is conserved, so it does not determine the threshold. Upward swimming produces a dense upper layer; the unstable disturbance is its buoyancy-driven overturning, or bioconvection. Higher orders are needed for wavelength selection and finite-depth threshold corrections.

With fixed material parameters and mean concentration, $R\sim h^4{V}_c\upsilon \Delta \rho g{\bar{N}}_{00}/(\rho \nu D^2)$ for small $\beta$, so a sufficiently thin chamber is stable to this branch. Formally its threshold corresponds to $h_c\sim [720\rho \nu D^2/(V_c\upsilon \Delta \rho g{\bar{N}}_{00}){]}^{1/4}$ when the weak-stratification and distinguished-gyrotaxis assumptions apply. In particular, the assumed $G=O({\beta }^{-1})$ is a specified asymptotic analysis; at fixed $B,D,V_c$ and decreasing $h$, one instead has $G=O({\beta }^{-2})$. The leading result should not be asserted unchanged in that different ordering without redoing the expansion.

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

The usual elastic secondary flow around a rotating sphere is inward towards the equator, then poleward beside the sphere and outward near both poles. It is superposed on the primary azimuthal rotating sphere in Stokes flow. Tensile normal stresses along curved azimuthal streamlines act as hoop stresses, pulling liquid towards the rotation axis. Incompressibility supplies the axial return flow. Unlike the primary rotation, this meridional circulation does not reverse when the rotation is reversed.

The direction assumes the usual polymeric sign $c={\psi }_1+2{\psi }_2>0$ and negligible inertia. A general second-order fluid has circulation proportional to $c$: a negative value reverses the arrows and $c=0$ eliminates the leading elastic circulation. Inertial secondary circulation is a different contribution, so slow rotation alone does not determine which contribution dominates. The sketch shows the elastic contribution for $c>0$:

For a quantitative check of the sketch, take sphere radius $a$, angular velocity $\Omega$ and $K=\Omega a^3$. The primary azimuthal velocity is $v_{\varphi }=K\sin \theta /r^2$, with $\theta$ measured from the rotation axis. Its only nonzero strain components in the spherical orthonormal basis are $E_{r\varphi }=E_{\varphi r}=-3K\sin \theta /(2r^3)$. The leading elastic secondary circulation around a rotating sphere has the stream functionwith $v_r=H(3{\cos }^2\theta -1)/r^2$ and $v_{\theta }=-H^{\prime }\sin \theta \cos \theta /r$. The bracket is $(1-a/r{)}^2(1+2a/r)>0$ outside the sphere. Thus $v_r<0$ at the equator and $v_r>0$ near the poles when $c>0$, and both components vanish on the sphere. The diagram uses these velocities, not independently chosen arrows.

Two other manifestations of the same elastic normal-stress differences are rod climbing, in which inward hoop stresses raise a free surface around a rotating rod, and die swell, in which relaxation of the axial stress and stored polymer deformation allows an extrudate to expand sideways after leaving the die. These are not explained by a scalar shear viscosity alone. A Newtonian extrudate can also swell through velocity-profile rearrangement, so the elastic contribution should not be identified with every instance of swelling.

The second-order fluid is a slow, weak-flow expansion of the objective constitutive history of an isotropic simple fluid. Material frame indifference requires objective tensor rates. For an incompressible flow, the leading isotropic stress is arbitrary pressure, the linear deviatoric term is $2\mu E$, and at the next order the independent contributions can be represented by $E^2$ and an objective rate of $E$. Isotropic scalar contributions are absorbed into the pressure. Their coefficients can be expressed through the two low-rate normal-stress differences, givingHere $\stackrel{△}{E}$ is the lower-convected derivative, not a division by $E$. This local expansion is appropriate when deformation and temporal change over the relevant viscoelastic relaxation time are small: both the Weissenberg number and the appropriate Deborah number must be small. It captures leading normal stresses in slow flows, but not arbitrary finite-strain history, rapid transients or strongly nonlinear shear thinning. In particular its restrictions are stronger than small-amplitude linear oscillatory response at arbitrary frequency.

For steady simple shear flow $\mathbf{v}=(\dot{\gamma }y,0,0)$, use component-first $L_{ij}={\partial }_j{v}_i$, so the paper's $\nabla v$ is $L^T$. The relevant tensors areSubstitution givesThus the first and second normal-stress differences are $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, and the shear stress is $\mu \dot{\gamma }$.