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Write $\omega =kc$, $q=k(1+i)$, and $E=e^{qL}$. The local resistive-force theory force per unit length on the filament iswhere $\mathbf{t}=(X_s,Y_s)$ is the unit tangent and $\mathbf{v}=(X_t-U,Y_t-V)$ the material velocity relative to the fluid. Inextensibility gives $X_s=(1-Y_s^2{)}^{1/2}=1-Y_s^2/2+\cdots$. With $X(0,t)=0$, $X_t=-{\int }_0^s{Y}_s{Y}_{st}d{s}^{\prime }+\cdots$ is second order and has zero time average for a periodic prescribed stroke. The actual small-slope condition is $k|H|e^{kL}\ll 1$ for fixed $kL$; $kH\ll 1$ alone is not uniform when $kL$ grows.

At first order, the transverse force is $f_y=-\mu K_N(Y_t-V)$. Adding the spherical-head drag force $\mu K_{N}L\delta V$ and imposing zero total force givesEquivalently $V(t)=\mathrm{Re}[-i\omega H(E-1)e^{-i\omega t}/(qL(1+\delta ))]$. This form is useful for the subsequent time averages.

At second order, the axial resistive-force theory force isAdding the head drag force gives the head-drag correction to planar flagellar propulsion:since $\overline{X_t}=0$. Directly, $Y_s=kH{e}^{ks}(\cos \vartheta -\sin \vartheta )$ and $Y_t=kcH{e}^{ks}\sin \vartheta$, with $\vartheta =k(s-ct)$. HenceAlso ${\int }_0^L{Y}_{s}ds=Y(L,t)-Y(0,t)$, whose complex amplitude is $H(E-1)$. Using $\overline{\mathrm{Re}(a{e}^{-i\omega t})\mathrm{Re}(b{e}^{-i\omega t})}=\mathrm{Re}(a\overline{b})/2$, together with $\mathrm{Re}(-i\omega /q)=-c/2$, givesAs $|E-1{|}^2=e^{2kL}-2e^{kL}\cos kL+1$, the requested mean speed follows:Anisotropic resistance is essential: $\gamma =1$ removes propulsion at this order. The mean motion is opposite to the direction of wave propagation when the bracket is positive, since the head's axial velocity is $-U$.

For the basal bending moment of a planar flagellum, take positive moment to be the positive out-of-plane torque exerted on the distal filament at its base. The leading hydrodynamic torque on that filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$, so balance requiresAn explicit unsimplified time-dependent answer isThe first term follows from ${\int }_0^{L}s{e}^{qs}ds=[E(qL-1)+1]/q^2$. This is the total internal bending moment required by the hydrodynamic load. It equals the contractile moment when passive elasticity is neglected. If a specified passive moment $M_{\mathrm{pass}}$ also acts, the required active moment is $M_0-M_{\mathrm{pass}}$; no elastic modulus is supplied here.

The prescribed base coordinate needs a geometrical qualification. Literally, a coordinate measured from the attachment point must have $Y(0,t)=0$, whereas the printed waveform has $Y(0,t)=H\cos \omega t$. The preceding calculation reproduces the printed speed by treating that waveform as the prescribed coordinate, including its moving transverse offset. If the intention is instead a fixed attachment at the origin, replace it by $\tilde{Y}=H[e^{ks}\cos k(s-ct)-\cos \omega t]$. ThenThese follow by replacing $Y_t$ with $Y_t-Y_t(0,t)$ in the same force balance. The basal moment likewise uses ${\tilde{Y}}_t$ and $\tilde{V}$. Thus the printed kinematics and its claim that the origin is the actual attachment cannot both be imposed without this correction.

Fix the sense of rotation by taking the fluid velocity to be $\mathbf{u}=\Omega (-x_3,0,x_1)$, with $\Omega >0$. Its vorticity is $-2\Omega {\mathbf{e}}_2$. In the convention where $B$ is the gyrotactic reorientation time appearing with aligning rate $1/B$, the bottom-heavy spherical-cell orientation dynamics isFor the stable orientation in the transverse plane, write $\mathbf{p}=(\sin \vartheta ,0,\cos \vartheta )$. The angle equation is $\dot{\vartheta }=-\sin \vartheta /B-\Omega$. If $B\Omega <1$, its stable fixed point satisfiesThe other planar fixed point has negative $p_3$ and is unstable. The positive-$p_3$ orientation also damps a small out-of-plane component because ${\dot{p}}_2=-p_3{p}_2/B$.

Once the orientation has relaxed to ${\mathbf{p}}_{\ast }$, the transverse trajectory obeysDefine the shifted axis byThen $(x_1-x_{1A},x_3-x_{3A})$ rotates rigidly with angular speed $\Omega$, so its squared length is constant. These are the gyrotactic circular paths in a rotating cylinder, andHere $\alpha$ is the acute angle between the plane through the two parallel axes and the horizontal plane; the directed displacement $OA$ in the chosen cross-section points to the lower-left. Reversing the sense of rotation reverses the horizontal displacement. Arbitrary initial orientations have an orientation transient before the exact circular trajectories; neglecting randomness alone does not remove that transient. The bulk result also assumes that the trajectory does not intersect the cylinder wall.

Let $C=N_{0}n$ be the cell number density. The steady cell conservation equation with isotropic translational diffusion isUsing the rapid-rotation mean gyrotactic orientation, put $a_s=2V_s/(3B\Omega )$, so the swimming drift is $-a_s{\mathbf{e}}_1$. In the transverse plane, $x_1=Rr\cos {\theta }^{\prime }$ and $x_3=Rr\sin {\theta }^{\prime }$. Thus $\mathbf{u}\cdot \nabla =\Omega {\partial }_{{\theta }^{\prime }}$ andMultiplying the conservation equation by $-R^2/D$ giveswith ${\beta }^2=\Omega R^2/D$ and $\epsilon =a_{s}R/D=2V_{s}R/(3B\Omega D)$. The base fluid velocity is tangent to the wall. Zero normal cell flux there is $-a_{s}n\cos {\theta }^{\prime }-D{n}_r/R=0$, orNormalization requires the cross-sectional mean of $n$ to be one.

For $\epsilon \ll 1$, set $n=1+\epsilon \mathrm{Re}[a(r)e^{i{\theta }^{\prime }}]+O({\epsilon }^2)$. The drift term acting on the leading constant vanishes, and the first angular harmonic satisfiesWriting $q=e^{i\pi /4}\beta$ turns this into the Modified Bessel differential equation of order one. Its regular solution iswhere $I_1$ is the Modified Bessel function of the first kind. The singular second solution is excluded at the axis, and the displayed coefficient enforces the wall boundary condition. The angular harmonic has zero area average, so the normalization is satisfied at this order.

When $\beta \gg 1$, the gyrotactic concentration layer at a rotating-cylinder wall has stretched coordinate $s=\beta (1-r)$. The leading radial equation is $a_{ss}-ia=0$. The solution decaying into the interior is proportional to $e^{-e^{i\pi /4}s}$; the wall condition fixes its coefficient to $-1/q$. HenceThis also follows from the large-argument asymptotic expansion of $I_1$, which gives $a\sim -r^{-1/2}{e}^{-q(1-r)}/q$ near the wall; $r^{-1/2}=1+O({\beta }^{-1})$ within the layer. Its thickness is $R/\beta =\sqrt{D/\Omega }$. The quoted exponential is a wall-layer approximation, not an exact solution all the way to $r=0$; the Modified Bessel function of the first kind formula supplies the regular continuation through the axis.

With zero fluid velocity and all mean orientations vertical, the steady cell conservation equation isThe plates are impermeable to cells, so the constant vertical flux is zero. Hence $D{\hat{n}}_{\hat{z}}=V_s\hat{n}$, giving $\hat{n}=C\exp (V_s\hat{z}/D)$. Determine $C$ by the prescribed depth average:With $z=\hat{z}/H$ and $h=V_{s}H/D$, the result isThe same normalized profile describes cell number density or cell volume fraction, because cell volume is constant. Its weak-stratification expansion iswhich is depth-normalized at each displayed order and will be used in part (c). Upward swimming makes concentration larger near the upper plate, creating the density stratification responsible for bioconvection when cells are heavier than the fluid.

First distinguish concentration conventions. If $C$ is cell number density, the excess mixture density is $v\Delta \rho C$. If $\phi =vC$ is cell volume fraction, it is $\Delta \rho \phi$. The printed definition of $R$ is dimensionless for number density, whereas the prose calls ${\hat{n}}_0$ a volume fraction. Use $C_0$ for mean number density and ${\phi }_0=v{C}_0$ for mean volume fraction. Then the consistent Rayleigh number isThe dimensionless concentration $n=C/C_0=\phi /{\phi }_0$ is the same in both conventions. The extra factor $v$ must be removed if the source symbol is literally a volume fraction.

Remove the hydrostatic basic pressure and use a dilute Boussinesq approximation. With ${\hat{C}}_e=\hat{C}-C_0\bar{n}$ and perturbation fluid velocity $\hat{\mathbf{u}}$, the linear incompressible Newtonian fluid equations areThe cell conservation equation linearizes toAt this order the vertical orientation perturbation is zero, because $|\mathbf{p}|=1$. The instantaneous-orientation approximation from the introductory calculation givesIncompressibility implies ${\hat{\nabla }}_{\perp }\cdot (\hat{\mathbf{\omega }}\times {\mathbf{e}}_3{)}_{\perp }=-\hat{\Delta }\hat{w}$. This identity fixes the sign of the gyrotactic term.

Apply the length, viscous-time, velocity, concentration and pressure scales in the paper. Writing $n_e=n-\bar{n}$, the dimensionless equations arewhere $S=\nu /D$ and $G=V_{s}B/(2H)$. The factor $S$ appears in the cell equation because its diffusive time differs from the viscous time used for $t$.

For a normal mode $[n_e,w]=[N,W]e^{\sigma t+i\kappa x}$, set $\mathcal{L}=d^2/d{z}^2-{\kappa }^2$. Taking the divergence of the momentum equation gives $\mathcal{L}{P}_e=-R{N}^{\prime }$. Applying $\mathcal{L}$ to its vertical component eliminates pressure and yieldsThe concentration equation becomesThese reproduce the requested modal equations, with the physically consistent sign of the orientation balance and concentration convention made explicit.

At both plates, the no-slip boundary condition gives $w=0$ and horizontal perturbation velocity zero. For a nonzero horizontal wavenumber, incompressibility then gives $W^{\prime }=0$. The normal cell flux is $\bar{n}w+hN-N^{\prime }$; the orientation correction has no vertical component at this order. Impermeability therefore givesIf orientation lag is retained rather than the given instantaneous balance, ${\mathbf{p}}_{\perp }$ acquires the relaxation response $(1+B{\partial }_{\hat{t}}{)}^{-1}$ and the modal gyrotactic term changes accordingly. The displayed equations are the quasistatic-orientation model requested here.

Write ${\ell }_g$ for the dimensional offset of the centre of mass, reserving $h$ below for the dimensionless swimming parameter. The weight $-mg\mathbf{k}$ acts at $-{\ell }_g\mathbf{p}$, so its torque about the geometric centre is $mg{\ell }_g\mathbf{p}\times \mathbf{k}$. Buoyancy acts through the geometric centre and has no such torque. For a spherical cell, the ambient angular velocity is $\mathbf{\omega }/2$ and the rotational resistance is ${\zeta }_r=\rho \nu {\alpha }_{\perp }v$. Neglecting angular inertia, torque balance givesSince $\dot{\mathbf{p}}={\mathbf{\Omega }}_{\mathrm{cell}}\times \mathbf{p}$, the bottom-heavy spherical-cell orientation dynamics isThus a steady orientation satisfiesThe printed algebraic equation has the opposite sign on its right side, if vorticity means the usual $\nabla \times \mathbf{u}$. Its later concentration equation is consistent with the corrected sign. In particular, linearizing the corrected equation about $\mathbf{p}=\mathbf{k}$ gives ${\mathbf{p}}_{\perp }=(B/2)(\mathbf{\omega }\times \mathbf{k}{)}_{\perp }$, whereas the printed sign would reverse that response. Merely changing the definition of vorticity would also require changing the ensuing fluid and orientation calculations.

The gyrotactic reorientation time $B$ measures gravitational restoration against rotational viscous resistance. In still fluid a tilt angle obeys $\dot{\theta }=-\sin \theta /B$, soFor small tilt it is precisely the exponential relaxation time. Applying the algebraic orientation balance to a time-dependent disturbance additionally assumes that this relaxation is fast enough to neglect orientation lag. Keeping the full dynamic orientation equation would introduce an additional relaxation factor into the perturbation problem.

Take the equilibrium mass density to be constant and absorb any conservative equilibrium force into the pressure. The linearized ideal magnetohydrodynamic induction equation, using the solenoidal vector field conditions, isFor $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$, differentiation of the cylindrical basis is important. Acting on a vector whose cylindrical components have azimuthal dependence $e^{im\varphi }$ gives $(\mathbf{B}\cdot \nabla )\mathbf{u}=imJ\mathbf{u}+J{\mathbf{e}}_z\times \mathbf{u}$. Also $(\mathbf{u}\cdot \nabla )\mathbf{B}=J{\mathbf{e}}_z\times \mathbf{u}$. These basis contributions cancel in induction, giving $i\omega \mathbf{b}=imJ\mathbf{u}$.

The perturbation of the Lorentz force isBoth of the first two terms now contribute a basis-rotation term. With $\tilde{p}=p_1/\rho +\mathbf{B}\cdot \mathbf{b}/({\mu }_0\rho )$, the linearized ideal magnetohydrodynamic momentum equation is thereforeThe magnetic part of $\tilde{p}$ is the perturbation of magnetic pressure, not an extra restoring force to retain after projecting out pressure.

Put $D={\omega }^2-m^2{\tilde{J}}^2$, where $\tilde{J}=J/\sqrt{{\mu }_0\rho }$. For $mJ\ne 0$, substitute $\mathbf{u}=\omega \mathbf{b}/(mJ)$ into the momentum equation and multiply by $mJ$. Taking a curl removes the modified pressure. Sincewe obtain the uniform-current toroidal-field curl reductionA second curl, using $\nabla \times \nabla \times \mathbf{b}=-{\nabla }^2\mathbf{b}$, givesThe divided expression assumes $D\ne 0$. If $mkJ\ne 0$, the undivided curl equation shows that $D=0$ would force $\mathbf{b}=0$, so this restriction loses no such magnetic mode. The $k=0$ degenerate limit can instead be taken in the undivided equations; it has no growing branch of the kind considered below. Axisymmetric $m=0$ disturbances are likewise excluded by the velocity elimination, and induction gives no nonzero-frequency magnetic disturbance in this particular equilibrium.

For a vector Laplacian eigenmode with ${\alpha }^2+k^2>0$, elimination givesThus both squared frequencies are real. The lower branch satisfies ${\omega }_{-}^2\ge {\tilde{J}}^2(|m{|}^2-2|m|)$, so it is positive for $|m|>2$. For $|m|=2$ it is positive when $\alpha \ne 0$, but the printed strict-positivity statement needs an endpoint qualification: $|m|=2$, $\alpha =0$ permits a neutral mode unless boundary conditions exclude it.

This is a genuine counterexample to the unqualified endpoint, not merely a loose algebraic bound. The neutral quadrupolar perturbation of a uniform-current toroidal fieldis regular at the axis, is solenoidal, and has $\nabla \times \mathbf{b}=k\mathbf{b}$, hence ${\nabla }^2\mathbf{b}=-k^2\mathbf{b}$. Also ${\mathbf{e}}_z\times \mathbf{b}=-i\mathbf{b}$, so the non-gradient part of the linearized Lorentz force vanishes for $m=2$. The gas-pressure perturbation $p_1=-\mathbf{B}\cdot \mathbf{b}/{\mu }_0$ cancels its magnetic pressure gradient, as required by $\tilde{p}=0$. Taking its real part gives a real disturbance. Decay at radial infinity or an appropriate homogeneous radial boundary condition would exclude it, restoring strict positivity for the allowed eigenmodes; no such condition is printed.

For $|m|=1$ the lower branch is unstable precisely whenAt equality it is marginal. With the time convention $e^{i\omega t}$, the growing member has $\omega =-is$ andThe other branch remains oscillatory. Actual admissibility of a chosen $\alpha$ depends on the radial domain and its boundary conditions.

Write the inward radial velocity as $u_R=-v(R)$, with $v>0$. Steady spherical mass conservation and the radial Euler equations for an inviscid fluid giveFor a barotropic fluid, define the sound speed $c_s^2=dp/d\rho >0$ and the barotropic enthalpy function by $dh=dp/\rho$. Choosing $\Phi (\infty )=0$, integration gives the Bernoulli equationThe continuity equation gives ${\rho }^{\prime }/\rho =-2/R-v^{\prime }/v$. Eliminating mass density from the momentum equation produces the barotropic spherical accretion equationThe sonic point is where the inward Mach number $v/c_s$ equals one. A smooth solution cannot cross the vanishing coefficient on the left with a nonzero right-hand side, so a regular sonic radius must obeyConversely a continuous flow that is subsonic at infinity and supersonic inward must cross unit Mach number; smooth passage selects a simultaneous zero of the two sides, hence a regular sonic point. The associated critical accretion branch, rather than an arbitrary subsonic solution with smaller flux, is selected by these outer and inner requirements.

For the polytropic equation of state, set $\Gamma =1+1/m$ to distinguish its exponent from any separately specified perturbation exponent. Assume $K>0$ and first take the usual positive index $m>0$. ThenFor the attractive mixed potential, assume $\lambda ,\mu >0$. The sonic condition and Bernoulli equation giveHence polytropic accretion in a mixed inverse-power potential reduces toFor $0<m\le 1/2$ all nonzero coefficients in the left side are nonnegative for $R_s>0$, with at least one strictly positive: there is no finite positive root. For $m>1/2$ the constant term is negative and the leading coefficient positive, giving exactly one positive root. Define $B=(2m-3)\mu /4$ and $C=(2m-1)\lambda /2$. ThenThis is the finite-radius transonic range for the two positive attractive terms.

The critical point has real distinct crossing slopes. To check this rather than just solve the sonic equalities, put $U=\lambda /R_s^2$, $V=\mu /(2R_s)$ and $x=R_s{v}_s^{\prime }/v_s$. Differentiating the barotropic spherical accretion equation at its simultaneous zero givesIts discriminant is $8m[2(2m-1)U+(2m-3)V]/(U+V)$. The sonic Bernoulli equation says $(2m-1)U+(2m-3)V=2m{c}_{\infty }^2$, so this discriminant is strictly positive for $m>1/2$. The accretion crossing is the slope for which $d\log (v/c_s)/d\log R<0$, namely the smaller root. The two slopes have opposite signs of this Mach-number derivative. This is the regular transonic saddle, not a singular vertical tangent.

The corresponding mass accretion rate follows without an additional arbitrary mass density normalization:One may equivalently replace ${\rho }_s$ by $[c_{s,s}^2/(\Gamma K){]}^m$. With a nonzero $\lambda$ term, inner free fall has $v\sim \sqrt{2\lambda }/R$ and continuity gives $\rho \propto R^{-1}$. Thus $v^2/c_s^2\propto R^{-2+1/m}$ increases without bound precisely in the positive-index range $m>1/2$, consistent with the required inner supersonic branch.

The assumptions on the parameters matter. If $\lambda =0$ and $\mu >0$, the finite sonic radius instead requires $m>3/2$, the usual Bondi accretion range; $m=3/2$ is the zero-radius limiting critical case. If $\mu =0$ and $\lambda >0$, the condition remains $m>1/2$. With arbitrary signs of $\lambda ,\mu$, one must use the quadratic and retain only positive roots with $c_{s,s}^2>0$ and real admissible crossing slopes; attraction was not explicitly given as a sign condition in the source.

If positive polytropic index was not intended as an implicit physical convention, there is a further mathematical range. For $K>0$, positive compressibility also permits $m<-1$, giving $0<\Gamma <1$. Both potential terms attractive then give a positive sonic root as well:The same mass accretion rate formula applies; its crossing-slope discriminant is positive. The range $-1<m<0$ has $c_s^2<0$ and no physical acoustic sonic point, $m=-1$ has constant pressure and zero sound speed, and $m=0$ does not define the printed equation of state. Thus the usual answer $m>1/2$ assumes $m>0$ and positive attractive coefficients; these conventions should not be confused with additional printed restrictions.

Put $\mathcal{G}=G(M_{\star }+M_{\mathrm{pl}})$, $h=\sqrt{\mathcal{G}a(1-e^2)}$, $\theta =\omega +f$ and $L=n_{\mathrm{pl}}t$. Here $h$ is the specific angular momentum, $f$ is the true anomaly and $\omega$ is the planar longitude of periapsis. Differentiating the Kepler orbit equation with $\dot{f}=h/r^2$ givesResolve the velocity along the radial and tangential unit vectors, $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$. The identities $\sin f\cos \theta -\cos f\sin \theta =-\sin \omega$ and $\sin f\sin \theta +\cos f\cos \theta =\cos \omega$ yieldThis is the Kepler velocity hodograph: the velocity runs around a circle whose centre is displaced by the orbital eccentricity.

The centre of mass condition gives the star's barycentric position ${\mathbf{r}}_{\star }=-\mu a_{\mathrm{pl}}(\cos L,\sin L)$. Translating to the star without rotating the axes gives ${\mathbf{r}}^{\prime }=\mathbf{r}-{\mathbf{r}}_{\star }$ and ${\mathbf{v}}^{\prime }=\mathbf{v}-{\dot{\mathbf{r}}}_{\star }$. From Kepler's third law, $n_{\mathrm{pl}}^2{a}_{\mathrm{pl}}^3=\mathcal{G}$, soConsequentlyThe small displacement and the larger reflex velocity have different orders: $|{\mathbf{r}}^{\prime }-\mathbf{r}|/a=\mu \alpha$, but the reflex speed divided by the circular speed is $\mu {\alpha }^{-1/2}$. These are the source of the astrocentric osculating-element oscillations.

For the coordinate ratio, the exact result isFor fixed $e<1$, $r/a\ge 1-e$, so this is $O(\mu \alpha )$ on any angular region with $|\cos \theta |$ bounded away from zero. The printed coordinate claim is not uniform over a full orbit: at $x=0$ its ratio is undefined. The nonsingular statement is $|{\mathbf{r}}^{\prime }-\mathbf{r}|/r\le \mu \alpha /(1-e)$, assuming the pericentre distance remains much larger than the binary separation.

Squaring the two components of the velocity givesAt $\mu =0$, $\kappa =0$ and $\mathcal{G}=G{M}_{\star }$. Sincethis reduces to the vis-viva equation, $v^{\prime 2}=G{M}_{\star }(2/r-1/a)$.

Now let the barycentric Kepler orbit be circular, write $\Delta =L-\theta$ and $\epsilon =\mu {\alpha }^{-1/2}\ll 1$, and put $v_0=\sqrt{\mathcal{G}/a}$. The osculating orbital elements about the star use the gravitational parameter $G{M}_{\star }=(1-\mu )\mathcal{G}$. To the required order, $r^{\prime }=a[1+O(\mu \alpha )]$ and $v^{\prime 2}=v_0^2[1+2\epsilon \cos \Delta +O({\epsilon }^2)]$. The specific orbital energy therefore impliesHere $\mu =o(\epsilon )$ and ${\epsilon }^2=o(\epsilon )$ under the stated hierarchy. Even at a phase where the leading cosine vanishes, the absolute error bound remains valid.

Finally, use the eccentricity vector ${\mathbf{e}}^{\prime }=({\mathbf{v}}^{\prime }\times {\mathbf{h}}^{\prime })/(G{M}_{\star })-{\hat{\mathbf{r}}}^{\prime }$. In radial and tangential components it isThe translation changes the radial basis only by $O(\mu \alpha )$, whereas $v_r^{\prime }=-v_0\epsilon \sin \Delta +O(v_0\mu \alpha )$ and $v_{\theta }^{\prime }=v_0(1+\epsilon \cos \Delta )+O(v_0\mu \alpha )$. ThusFor the original circular Kepler orbit, $\omega$ and $f$ are separately arbitrary; their sum $\theta$ is well defined. Using the eccentricity vector avoids introducing an undefined pericentre for that circular Kepler orbit.

Write the first-order surface velocity as ${\mathbf{u}}_1={\mathbf{U}}_1+{\mathbf{\Omega }}_1\times \mathbf{x}+\mathbf{g}$, where the previous derivative givesAs a test field, choose a sphere translating with arbitrary velocity $\hat{\mathbf{V}}$. Its surface velocity is $\hat{\mathbf{V}}$ and its traction is $\hat{\mathbf{\sigma }}\mathbf{n}=-3\mu \hat{\mathbf{V}}/(2a)$. The Lorentz reciprocal theorem for Stokes flow givesTerms at infinity vanish because both velocities decay. Since $\int \mathbf{n}dS=0$, the rotation term has zero surface average. The arbitrary test velocity therefore gives the first-order mobility of a nearly spherical particle:

For the ellipsoidal perturbation, the second spherical moment gives $\int fdS=(4\pi a^3/3)\mathrm{tr}\mathbf{D}=0$. The supplied fourth moment gives, componentwise,Substitution yieldswhere the last notation uses symmetry of the tensor.

One can also use a rotating-sphere test with surface traction $-3\mu \hat{\mathbf{\Omega }}\times \mathbf{n}$. Its reciprocal integral, together with zero first-order torque, givesHere $f(-\mathbf{n})=f(\mathbf{n})$, so the last integrand is odd and its integral vanishes. Thus ${\mathbf{\Omega }}_1=0$. Inversion symmetry forbids a translation-to-rotation coupling for this centred ellipsoidal perturbation.