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A dilute suspension of identical rigid spheres raises the dynamic viscosity by a fraction $5\varphi /2$, where $\varphi$ is the particle volume fraction. At imposed trace-free rate-of-strain tensor $E$, a sphere of radius $a$ adds viscous dissipation $20\pi \mu a^3(E:E)/3$. Multiplication by the number density $3\varphi /(4\pi a^3)$ gives $5\mu \varphi E:E$, which added to $2\mu E:E$ proves the formula. This is the leading noninteracting-sphere term; hydrodynamic interactions enter at higher concentration.

In the free-draining approximation, each separated component of a body experiences its isolated viscous drag, and hydrodynamic interactions between components are neglected. For widely separated spheres of radii $a_i$ in an unbounded Newtonian fluid of dynamic viscosity $\mu$, Stokes law gives ${\mathbf{f}}_i=-{\zeta }_i{\mathbf{v}}_i$, with ${\zeta }_i=6\pi \mu a_i$. This is a leading approximation in sphere size divided by separation. Any connecting rod must also have negligible drag; a small diameter alone does not guarantee this for an arbitrarily long rod.

Represent a dilute polymer by two beads connected by a spring obeying Hooke's law, with stiffness $H>0$ and bead drag coefficient $\zeta >0$. Neglect bead inertia and hydrodynamic interactions. For component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the configuration Fokker-Planck probability current isThe factor two comes from relative motion of both beads. At rest its normalized probability density is ${\psi }_{\mathrm{eq}}=(H/(2\pi k_{B}T){)}^{3/2}{e}^{-H{R}^2/(2k_{B}T)}$. The conformation tensor $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ has dimensions of length squared; $(H/(k_{B}T))C$ is its dimensionless version. The Hookean dumbbell in planar extension is a particular imposed flow in this model.

With a consistent force convention, the mobility matrix maps applied generalized forces to rigid-body velocities; it is the inverse of the hydrodynamic resistance matrix. For a sphere with force on the fluid $\mathbf{F}$, its isolated mobility is $1/(6\pi \mu a)$. Coupling between bodies represents hydrodynamic interactions.

The Oseen tensor is the unbounded three-dimensional incompressible Stokes flow Green tensor relating a point force to the velocity it produces: $u_i={\mathsf{G}}_{ij}{F}_j$. At separation $r$, its longitudinal mobility is twice its transverse mobility. It generates the leading far-field hydrodynamic interaction between externally forced particles. The singular value at $r=0$ is not a particle's self mobility; a no-slip sphere instead has self mobility $1/(6\pi \eta a)$. Walls and near-contact lubrication require different or corrected tensors.

Use the deterministic Hookean dumbbell model, with a zero-rest-length spring and negligible connector drag. For $\mathbf{R}={\mathbf{r}}_1-{\mathbf{r}}_2$, its elastic forces on the two beads areHere $H>0$ is the spring constant. No independent thermal forcing from Brownian motion or bead inertia is included. The unperturbed linear velocity field is $\mathbf{Lr}$, with $\mathrm{tr}\mathbf{L}=0$ by incompressibility.

The fluid velocity used in each bead's drag law is the imposed flow plus the disturbance from the other bead, with its own singular field omitted. At leading order in $a/R$, the evenness of the Oseen tensor, $\mathsf{G}(-\mathbf{R})=\mathsf{G}(\mathbf{R})$, givesThese are the pair hydrodynamic mobility matrix relations. Their signs follow from balancing each spring force with $-\zeta ({\dot{\mathbf{r}}}_i-{\mathbf{u}}_{\mathrm{ext},i})$, the fluid drag on that bead.

Adding the equations cancels both the direct spring slips and the cross-induced velocities, since ${\mathbf{F}}_1+{\mathbf{F}}_2=0$. For the midpoint ${\mathbf{r}}_c=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$,The midpoint therefore follows the undisturbed linear velocity field. Subtracting the equations instead gives the hydrodynamic interaction of a Hookean dumbbell:The dimensionless interaction matrix isIts transverse and longitudinal eigenvalues are $3a/(4R)$ and $3a/(2R)$, respectively. Since the spring force is longitudinal, the separation equation can also be writtenThe relative correction to the free-draining spring relaxation rate is therefore $3a/(2R)$, explicitly proportional to $a/R$. Entrainment by the other bead reduces their relative approach for given opposing spring forces.

This is a far-field hydrodynamic interaction, valid for $R\gg a$; it is not a near-contact law and omits higher reflections and lubrication. The printed target also fixes the spring's rest length implicitly. For an ordinary spring obeying Hooke's law with nonzero rest length $R_0$, the elastic force would be $-H(R-R_0)\hat{\mathbf{R}}$, and every spring term above would acquire the factor $1-R_0/R$. The displayed result uses the zero-rest-length spring convention appropriate to the ideal Hookean dumbbell model.

Take $z$ positive upward, let sphere A's centre be $Z(t){\mathbf{e}}_z$, and write its downward velocity as $-U{\mathbf{e}}_z$, with $\dot{Z}=-U$. At B's centre the displacement from A is $\mathbf{r}=\ell {\mathbf{e}}_x-Z{\mathbf{e}}_z$. The vorticity of translating-sphere Stokes flow therefore givesB is held in translation but has zero applied torque. By Faxén's rotational law, it rotates with half the incident vorticity:Integrating the rotation induced by a passing Stokes sphere along the complete path cancels $U$:Its magnitude is $3a/(2\ell )$ radians, with direction about the negative $y$ axis by the right-hand rule. All instantaneous rotations have this same axis, so the integrated angular velocity gives the net angle without a noncommuting-rotation correction. The result is independent of B's radius in the prescribed incident-flow calculation.

The inequality $\ell >a+b$ ensures no overlap. It does not by itself make mutual hydrodynamic interactions negligible: the stated flow is the isolated-A solution, and a held finite B also disturbs A's flow. Thus the displayed angle is the result for that supplied incident flow, or the leading one-way approximation to separated bodies. A fully coupled two-sphere calculation uses the method of reflections for Stokes flow or an exact pair resistance solution.

Use the free-draining approximation: neglect the hydrodynamic interaction between the widely separated spheres and neglect the connector's viscous drag. Stokes law gives ${\mathbf{f}}_i=-6\pi \mu a_i\mathbf{U}$ for each sphere. Their resultant isThis is the leading two-sphere approximation, not an exact finite-separation hydrodynamic resistance matrix. Writing $d=|{\mathbf{r}}_2-{\mathbf{r}}_1|$ and $\mathbf{n}=({\mathbf{r}}_2-{\mathbf{r}}_1)/d$, the first Stokeslet reflection gives, if the connector remains negligible,The negative correction expresses reduced drag from fluid entrainment by the other sphere. A very thin but arbitrarily long rod need not have negligible total drag: its typical transverse drag scales as $\mu d/\log (d/b)$ for rod radius $b$. Thus the simple boxed result uses the intended negligibly dragging connector limit as well as $a_i/d\ll 1$; the two printed size inequalities alone do not specify a uniform limit in which rod drag vanishes.

Hold the two-sphere object fixed in a uniform stream $\mathbf{V}$. In the free-draining approximation, its drag forces are ${\mathbf{f}}_i={\zeta }_i\mathbf{V}$, with ${\zeta }_i=6\pi \mu a_i$. The hydrodynamic torque about ${\mathbf{r}}_c$ isTo make it zero for every incoming-stream direction, choose the hydrodynamic centre of a two-sphere dumbbell:It lies closer to the larger sphere, whose larger drag requires a shorter lever arm. Uniform translation does not spin an isolated sphere, so no additional isolated-sphere spin torque enters this leading approximation.

Finite-separation hydrodynamic interactions and connector drag can change the displayed location; the calculation has the same approximation as part c.

Put $r=|\mathbf{x}|$ and $s=\mathbf{F}\cdot \mathbf{x}$. In the Unscaled Papkovich–Neuber representation, take the harmonic potentials $\mathbf{\Phi }=-\mathbf{F}/(8\pi \mu r)$ and $\chi =0$. Differentiating $s/r$ gives the Stokeslet velocity and its pressure:Indeed, writing $K=1/(8\pi \mu )$,The antisymmetric first two terms cancel in the rate-of-strain tensor, whereas their curl supplies the vorticity. ThusThe rate-of-strain tensor has zero trace, as required by incompressibility. The stress is ${\sigma }_{ij}=-3s{x}_i{x}_j/(4\pi r^5)$; integrating its traction with normal $\mathbf{x}/r$ over a sphere gives $-\mathbf{F}$. This fixes the force normalization of the Stokeslet.

For a sphere in a uniform straining Stokes flow, write the ambient velocity as $\mathbf{Ex}$, where $\mathbf{E}$ is symmetric and trace-free, and set $q=\mathbf{x}\cdot \mathbf{Ex}$. Symmetry gives zero translation and rotation of the centred force-free, torque-free sphere. Seek a decaying disturbance using the Unscaled Papkovich–Neuber representation withThe components of $\mathbf{Ex}/r^3$ are derivatives of $1/r$, and $q/r^5$ is a trace-free harmonic quadrupole, so both potentials satisfy Laplace's equation outside the sphere. Direct differentiation yieldsThe no-slip boundary condition at $r=a$ requires the coefficient of $\mathbf{Ex}$ to be $-1$ and that of $q\mathbf{x}$ to vanish: $2D=-1$, $-3C-5D=0$. Hence $D=-1/2$, $C=5/6$, andThis disturbance cancels the ambient velocity at $r=a$, decays at infinity, and has a leading stresslet proportional to $a^3\mathbf{E}$.

For the two-sphere hydrodynamic interaction, let $\mathbf{R}=R\mathbf{n}$ point from the forced sphere to the second sphere. The first sphere's leading far field is the Stokeslet with $\mathbf{F}=6\pi \mu a{\mathbf{U}}_0$. Its incident rate-of-strain tensor at the second sphere isThe force-free sphere translates with the local uniform incident flow, and the torque-free sphere rotates with half the incident vorticity. Those parts cause no leading Stokeslet or rotlet disturbance. The remaining incident strain produces the stresslet just calculated. At displacement $-\mathbf{R}$ from the second sphere its velocity isThe first sphere translates in this returned incident flow. Finite-size sampling and higher incident multipoles enter only beyond the retained reflection order. Thus the stresslet reflection between two force-free and forced spheres givesThe factor $|{\mathbf{U}}_0|$ makes explicit the velocity scale implicit in the printed remainder.

The resulting input power, equal to the viscous dissipation, isIt decreases at fixed force. The Minimum-dissipation theorem for Stokes flow instead compares admissible fields at fixed first-sphere velocity: extending a two-sphere field rigidly into the second sphere gives an admissible one-sphere field, so the extra rigid inclusion can only increase the minimum resistance. Allowing the second sphere to choose its translation and rotation minimizes over its rigid motions, producing zero force and couple there. An increase of this effective hydrodynamic resistance matrix decreases its inverse hydrodynamic mobility matrix, and hence decreases fixed-force power. The negative mobility correction is therefore consistent with minimum dissipation.

If the second sphere is prevented from rotating, its angular velocity relative to the incident fluid becomesUse the rotating sphere in Stokes flow at $-\mathbf{R}$. The additional reflected rotlet givesAdding this rotation-clamped reflection between two spheres to the strain reflection,The holding couple supplies no power because the held sphere's angular velocity is zero.

For a Newtonian fluid, write $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation without body force gives ${\partial }_j{\sigma }_{ij}=0$. Since the stress tensor is symmetric, its contraction with the antisymmetric part of the velocity gradient is zero. ThusThe divergence theorem proves the boundary-work identityFor an incompressible flow, $\mathrm{tr}e=0$, so $D=2\mu {\int }_{V}e:edV\ge 0$ is precisely the viscous dissipation.

Apply this identity to the particle-free volume and then to the fluid outside the added particle. On the rigid particle, the velocity has the form $V+\Omega \times x$. Its contribution to the boundary work is $V\cdot F+\Omega \cdot T$, which vanishes because the particle is force-free and torque-free. The outer velocity is unchanged. Subtracting the two boundary-work identities therefore givesTo move this expression to the particle surface, apply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the punctured fluid domain. For completeness, its local proof isOn the outer boundary $u^{\prime }=0$. Let $N=-n$ be the particle-outward normal, opposite to the fluid-outward normal there. Integrating the preceding divergence identity givesThis expression includes the removal of the original fluid inside the particle; that removal must not be counted separately again.

For a sphere in a uniform straining Stokes flow, $E$ is symmetric and trace free, and $u_0=Ex$. Symmetry gives zero rigid translation and rotation. Put $H=x\cdot Ex$ and $r=|x|$. Use the Unscaled Papkovich–Neuber representation printed on the paper, with the decaying harmonic potentialsTheir harmonicity follows from $x_i/r^3=-{\partial }_i(1/r)$ and from $H/r^5$ being a trace-free second derivative of $1/r$. Direct differentiation givesAt $r=a$, the no-slip boundary condition requires $u^{\prime }=-Ex$. Matching the independent $Ex$ and $Hx$ terms fixes $B=-1/2$ and $C=5/6$. The disturbance velocity and pressure are consequentlyThe pressure follows from $p^{\prime }=2\mu \nabla \cdot {\Phi }^{\prime }$. The disturbance decays, the total velocity vanishes at the sphere, and the harmonic representation verifies the Stokes equation. The boundary traction has zero total force by odd symmetry, and zero total torque because $E$ is symmetric, confirming the assumed force-free and torque-free motion.

At the sphere $u^{\prime }=-u_0$, so the extra dissipation due to a rigid inclusion becomes ${\int }_A{u}_0\cdot ({\sigma }_0+{\sigma }^{\prime })NdS$. Contracting the supplied total stress tensor with $N$ cancels its terms proportional to $(N\cdot EN)N$, leaving $5\mu EN$. ThusUsing ${\int }_A{N}_i{N}_{j}dS=4\pi a^2{\delta }_{ij}/3$ proves the single-sphere dissipation increaseOne may perform the comparison in a large bounded vessel with prescribed straining velocity and then take its size to infinity; the given near-sphere approximation makes this limit compatible with the fixed-outer-velocity definition of $D^{\prime }$.

The sphere number density is $n_s=\varphi /(4\pi a^3/3)=3\varphi /(4\pi a^3)$. Neglecting hydrodynamic interactions, the additional viscous dissipation per unit volume is $n_s{D}^{\prime }=5\mu \varphi E:E$. Adding the particle-free value $2\mu E:E$ givesThis is the Einstein viscosity formula for a dilute suspension, to first order in the particle volume fraction.

For an incompressible flow, write the Newtonian fluid stress tensor as $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation gives ${\partial }_j{\sigma }_{ij}=0$. Symmetry of the Cauchy stress tensor therefore givesIntegrating and applying the divergence theorem expresses the viscous dissipation as boundary power:At a moving rigid body, the surface velocity is $U+\Omega \times (x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega \cdot G$, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power givesApply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the fluid outside the inclusion. On the outer boundary $u^{\prime }=0$, soWriting $n_p=-n$ for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor $E$ is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of $E$ eliminates its torque. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seekThe incompressible flow condition and Stokes equation reduce toA decaying family satisfying these equations is $f=A{r}^{-5}$, $g=B{r}^{-5}-5A{r}^{-7}/2$, $j=2B{r}^{-5}$. The no-slip boundary condition requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. ConsequentlyThe total velocity is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere $u^{\prime }=-u_0=-aE{n}_p$. Contracting the supplied Newtonian fluid stress tensor with $n_p$ makes its two terms proportional to $(n_p\cdot E{n}_p)n_p$ cancel, leaving $\sigma n_p=5\mu E{n}_p$. Hence the extra dissipation due to a rigid inclusion isHere ${\int }_{S^2}{n}_j{n}_{k}d\Omega =4\pi {\delta }_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.

If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\varphi$. ThusAdd this to the particle-free viscous dissipation density $2\mu E:E$. Comparing with $2{\mu }_{\mathrm{eff}}E:E$ gives the Einstein viscosity formula for a dilute suspension:to first order in $\varphi$. The hydrodynamic interactions neglected here contribute beyond that dilute order.

In resistive-force theory (RFT), the hydrodynamic force per unit arclength exerted by the fluid on a slender filament isHere $\mathbf{t}$ is the unit tangent and $\mathbf{u}$ the local velocity relative to the background fluid. The parallel and perpendicular drag coefficients of a slender filament are positive and generally satisfy ${\xi }_{\perp }>{\xi }_{\parallel }$; their leading logarithmic ratio is about two. This local approximation assumes a small filament radius compared with length and curvature scales, negligible fluid inertia, and a Newtonian fluid. It represents drag by the local tangent direction and neglects nonlocal hydrodynamic interactions between separated filament segments. Boundaries, close approaches and end corrections may require slender-body theory or a more complete flow calculation. In this problem the background fluid is at rest, and the drag coefficients are taken uniform along the filament.

The rigid-body velocity in a deforming swimmer frame is the sum of translation, rotation and material deformation. With $\mathbf{r}=x{\mathbf{e}}_x+ϵg{\mathbf{e}}_y$,so, in instantaneous swimmer-frame components,The reference-frame conditions $g(0,t)=g_x(0,t)=0$ attach the origin and orientation to the filament's base and base tangent. They prevent arbitrary shape translations or tilts from being absorbed into the definition of $g$.

The exact tangent and arclength element areSince $U,V,\Omega =O(ϵ)$, the local velocity is $O(ϵ)$. Changing the drag tensor by its $O(ϵ)$ tangent correction therefore changes $\mathbf{f}$ only at $O({ϵ}^2)$. The arclength correction is smaller still at this order. Also $-ϵ\Omega g=O({ϵ}^2)$. Thus $\mathbf{t}={\mathbf{e}}_x$ is sufficient at order $ϵ$. This assumes the small-slope expansion is uniform, $ϵ|g_x|\ll 1$.

Put $G_0(t)=⟨g_t{⟩}_x$ and $G_1(t)=⟨x{g}_t{⟩}_x$, with $⟨h{⟩}_x=L^{-1}{\int }_0^{L}hdx$. The first-order local force isIts total hydrodynamic forces areTaking the moment about the swimmer-frame origin, $M_z=\int (x{f}_y-ϵg{f}_x)ds$, the second term is beyond first order. HenceFor force-free and torque-free motion, the three leading coefficients vanish. The longitudinal force immediately gives $U_1=0$. The two transverse equations areTheir determinant is $L^2/12>0$. Solving yields the first-order free swimming of a planar filament:The resistive-force theory coefficient cancels because both equations use the same transverse drag. Another interpretation is the least-squares projection of filament deformation velocity: $-V_1-{\Omega }_{1}x$ is the best affine approximation to $g_t$ on $[0,L]$. Zero transverse force and torque mean that the residual is orthogonal to $1$ and $x$.

If $g$ is sufficiently differentiable and periodic with period $T$, each $g_t(x,t)$ has zero temporal mean. More explicitly,Both bracketed quantities return to their initial values over a period. Thus the first-order periodic transverse swimming velocity satisfiesThese results concern first order and swimmer-frame components. Changes of orientation can affect laboratory displacements at second order; higher-order net swimming is not ruled out.

Use the same resistive-force theory convention as before: the force density is the fluid's force on the filament. Let $\phi$ be the usual cylindrical azimuth. Choose the orientation of a left-handed helix so that, as $z$ increases, $\phi$ decreases. With $s=\sin \theta$, $c=\cos \theta$,Reversing the orientation of the tangent leaves the local drag tensor unchanged. Define $k={\xi }_{\perp }-{\xi }_{\parallel }$. From resistive-force theory, the relevant force-density components areThese scalar axial and azimuthal components are uniform along the helix. Integration over its contour length $\ell$, and use of the axial torque density $a{f}_{\phi }$, give the axial resistance matrix of a slender helix:Thus $B=C$, as also required by the Lorentz reciprocal theorem. The signs here belong to $(F,L{)}^T=-\left(\begin{array}{cc}A& B\\ B& D\end{array}\right)(U,\Omega {)}^T$, with $F,L$ forces and moments on the helix. The handedness reversal of helical hydrodynamic resistance reverses $B,C$ but leaves $A,D$ unchanged. If the positive rotation direction or handedness convention is reversed, the coupling sign reverses with it.

As a check on physical admissibility, the determinant of helical resistance in resistive-force theory isTogether with $A,D>0$, this makes the hydrodynamic resistance matrix positive definite and the viscous power loss $A{U}^2+2BU\Omega +D{\Omega }^2$ positive. Isotropic local drag would have $B=0$, so rotating a helix would not propel it in this approximation.

For the pair, use a local additive resistive-force theory model: both helices have the same drag coefficients, and inter-helix hydrodynamic interactions and unresolved end effects are omitted. The right-handed helix has resistance entries $(nA,-nB,-nB,nD)$. Define the signed motor motion byThe printed relative rotation specifies a magnitude rather than which helix rotates relative to which. Choosing ${\Omega }_2-{\Omega }_1=\omega$ instead reverses all signed speeds below.

The common translation speed $U$ and zero total force and torque obeyInternal motor forces and torques cancel in these totals. IntroduceSolving the three linear equations yields the opposite-handed counterrotating helical swimmer:For $n>0$, ${\xi }_{\perp }>{\xi }_{\parallel }$ and $\omega >0$, $B>0$, so $U<0$, ${\Omega }_1>0$ and ${\Omega }_2<0$. The opposite-handed helices counterrotate but contribute thrust in the same axial direction. The formula satisfies the specified relative rotation without identifying either laboratory rotation rate with the motor rate.

At $n=0$, the second helix supplies no hydrodynamic resistance. The remaining helix must have $F=L=0$, and invertibility of its hydrodynamic resistance matrix forces $U={\Omega }_1=0$. Formally ${\Omega }_2=-\omega$ is the rotation of a zero-resistance motor shaft or vanishing second rotor; there is no physical finite second helix to propel or to provide a reaction torque. This is the vanishing reaction rotor in a helical swimmer.

At $n=1$, equal-length opposite-handed helices haveThe translation-generated axial torques cancel between the two helices, and equal counterrotation balances the rotational torques. Their propulsive forces add, rather than cancel, because handedness and rotation both reverse. These are equal-length opposite-handed helices.

For $n\to \infty$, the large reaction helix limit isThe increasingly long second helix nearly anchors the whole assembly: its very small translation and rotation suffice to balance the finite force and torque generated by the first helix. The first helix rotates at almost the full motor rate, but moving the large resistive second helix makes the common translation tend to zero. These are idealized limits of the additive local model, rather than a resolution of short-helix end effects or infinitely extended interacting filaments.

In the imposed free-draining polymer friction model, the drag coefficients of the $N$ monomers add, giving ${\zeta }_1=N(6\pi \eta a)$ and $D_1=k_{B}T/(6\pi \eta aN)$. Taking the same three-dimensional centre-of-mass RMS convention as part (ii),The polymer is a one-dimensional connected contour, but that alone does not restrict its centre of mass to one spatial axis. If the intended coordinate is instead literal one-dimensional curvilinear motion along an entanglement tube, its RMS contour displacement is $\sqrt{2D_{1}t}=\sqrt{k_{B}Tt/(3\pi \eta aN)}$. That is a different coordinate and should not be compared with a three-dimensional displacement without stating the conversion/model.

Additive local drag is the free-draining Rouse model approximation. It neglects long-range solvent-mediated hydrodynamic interactions and is not a complete theory of an entangled chain. In a tube/reptation description, curvilinear diffusion can scale as $N^{-1}$ while long-time spatial centre-of-mass diffusion scales as $N^{-2}$ after the chain renews its tube. Conversely a dilute unentangled coil can exhibit Zimm model hydrodynamic coupling. Monomer motion, centre-of-mass motion and contour sliding therefore cannot all be identified with the same displacement law.

In a dilute solution, solvent-mediated hydrodynamic interactions couple distant chain segments. The Zimm model describes the resulting cooperative dynamics: an intermediate-time segment mean-square displacement scales as $t^{2/3}$, while whole-chain translation has $D_Z\sim k_{B}T/({\eta }_{s}R)$. Once internal modes relax, each tagged segment follows the chain's ordinary long-time diffusion, with mean-square displacement asymptotic to $6D_{Z}t$.

In a concentrated solution or polymer melt, hydrodynamic coupling is screened. The unentangled Rouse model has an intermediate $t^{1/2}$ segment regime and whole-chain diffusivity proportional to $N^{-1}$. Long chains with polymer entanglement are additionally confined by neighbouring chains: reptation gives a slower intermediate $t^{1/4}$ regime and an asymptotic chain diffusivity scaling approximately as $N^{-2}$. The time windows depend on segment, entanglement and terminal relaxation times, so these powers do not all describe the same time interval.

After sufficiently long times both dilute and concentrated chains undergo ordinary translational diffusion. The main long-time difference is the coefficient and its chain-length dependence, with entanglement strongly reducing mobility when the chains are long enough; concentration alone does not guarantee an entangled regime.

In a sufficiently uniform flow the Stokes force is $F=\gamma v$ toward trap 2. Its work over separation $d$ lowers the second effective well by $Fd$. Equal effective free energies require $\gamma vd=k_{B}T\log 2$. Henceat 298 K. The temperature was not specified, so the numerical result scales as $T/298\mathrm{K}$. This approximation treats a weak uniform drag as a constant tilt and neglects changes in barrier shape and hydrodynamic interactions with boundaries.

A moving or externally forced colloid disturbs the surrounding fluid, changing the velocity of other colloids even without direct contact. These hydrodynamic interactions are long-ranged at small Reynolds number. In an unbounded incompressible Newtonian liquid, the Stokes flow response to a point force is the Stokeslet:Its Oseen tensor isThis is a mobility tensor, with units velocity per force, rather than a stress tensor. At separation $r$ the response parallel to the line of centres is $1/(4\pi \eta r)$; the perpendicular response is $1/(8\pi \eta r)$. Flow is not isotropic even though the solvent is. The disturbance decays as $1/r$, so treating all drag forces as independent can miss correlated motion.