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For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

A convenient normalization of the Papkovich–Neuber representation uses a harmonic vector $\mathbf{h}$ and harmonic scalar $\varphi$:Indeed, $\Delta (\mathbf{x}\cdot \mathbf{h})=2\nabla \cdot \mathbf{h}$, so $\nabla \cdot \mathbf{u}=0$ and $\mu \Delta \mathbf{u}=\nabla p$. This is a rescaling of the usual harmonic-potential representation of Stokes flow.

Measure $\mathbf{x}$ from the sphere center, set $r=|\mathbf{x}|$ and $\mathbf{n}=\mathbf{x}/r$. A translational vector harmonic $\mathbf{U}/r$ provides the decaying $1/r$ force field, while the scalar dipole $(\mathbf{U}\cdot \mathbf{x})/r^3$ adjusts the surface velocity without changing that leading far field. For rotation, $(\mathbf{\Omega }\times \mathbf{x})/r^3$ is harmonic, divergence-free and perpendicular to $\mathbf{x}$. Thus tryThe translational velocity is $\frac{A}{2r}(I+\mathbf{nn})\mathbf{U}-\frac{C}{2r^3}(I-3\mathbf{nn})\mathbf{U}$. Matching its independent tangential and radial components at $r=a$ gives $A=3a/2$, $C=-a^3/2$; matching rotation gives $B=a^3$. ConsequentlyThis superposes the translating sphere in Stokes flow and the rotating sphere in Stokes flow. It is exactly $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at the sphere and tends to zero at infinity; the additive ambient pressure is set to zero.

The minimum dissipation theorem fixes the boundary velocities and the far-field velocity: among admissible incompressible velocity fields, the Stokes flow minimizesFor a competitor $\mathbf{u}+\mathbf{v}$ with zero boundary data for $\mathbf{v}$, integration by parts eliminates the cross term and leaves $\mathcal{D}[\mathbf{u}+\mathbf{v}]-\mathcal{D}[\mathbf{u}]=2\mu \int \mathbf{e}(\mathbf{v}):\mathbf{e}(\mathbf{v})dV\ge 0$.

Extend the actual two-sphere velocity rigidly through sphere 2. It is an admissible field in the one-sphere exterior, with the same translation and rotation of sphere 1. It is continuous across the filled boundary, incompressible, and adds zero strain dissipation inside. Applying the theorem to the exact isolated-sphere flow therefore givesWith all applied couples and the second force zero, the boundary-work identity is ${\mathcal{D}}_{\mathrm{pair}}={\mathbf{F}}_1\cdot {\mathbf{U}}_1$. Thus $6\pi \mu a|{\mathbf{U}}_1{|}^2\le {\mathbf{F}}_1\cdot {\mathbf{U}}_1\le |{\mathbf{F}}_1||{\mathbf{U}}_1|$, which provesThis fixed-force comparison of minimum viscous dissipation uses a comparison at fixed actual velocity first; directly comparing different-force or different-velocity solutions would not justify the result.

If ${\mathbf{G}}_1\ne 0$, the power is instead ${\mathbf{F}}_1\cdot {\mathbf{U}}_1+{\mathbf{G}}_1\cdot {\mathbf{\Omega }}_1$, so the preceding estimate no longer bounds the force contribution alone. The two-sphere hydrodynamic mobility matrix generally has a nonzero self translation-rotation coupling: the freely moving second sphere reflects the first sphere's torque field. In the planar geometry this coupling produces a velocity perpendicular to $\mathbf{R}$; by choosing the sign and magnitude of the couple when the force has a component in that direction, its contribution to ${\mathbf{F}}_1\cdot {\mathbf{U}}_1$ can exceed the isolated force-only value. There is no universal force-only inequality with an additional applied couple. Special geometries can eliminate the coupling, but do not restore a general theorem.

Write $\mathbf{n}=\mathbf{R}/R$. At fixed geometry, Linearity of Stokes flow makes the translational velocities linear in the couples. With zero forces and couples perpendicular to $\mathbf{R}$, isotropy and reflection symmetry allow only terms proportional to ${\mathbf{G}}_j\times \mathbf{n}$ in each ${\mathbf{U}}_i$. There is no longitudinal term: reflect in the plane containing $\mathbf{R}$ and the common torque axis, taking account of the axial-vector transformation of torque, and use linearity under torque reversal. HenceThe direction of $\mathbf{R}$ can still change; constant separation does not mean stationary centers.

When the couples are equal, a rotation through $\pi$ about their common axis exchanges the identical spheres and reverses the in-plane translational velocities while leaving the axial angular velocities unchanged. Uniqueness of Stokes flow then givesThe midpoint is stationary and the pair can orbit it. The body spin need not equal the angular rate of that orbit. These symmetry statements hold at arbitrary noncontact separation, not just in a distant rotlet approximation.

The rotlet is the point-torque fundamental solution of Stokes flow. Its torque strength $\mathbf{R}$ is an axial vector, and its circulating velocity decays as $r^{-2}$. A freely swimming cell with no external torque is torque-free. The motor applies opposite internal torques to its body and flagellar apparatus, so the net torque monopole satisfiesIndividual body and flagellar torques need not vanish. A spatially separated pair can therefore leave a rotlet dipole even though there is no far-field rotlet monopole.