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The Lighthill elongated-body theory approximation treats each cross-section of a slender swimmer as a locally two-dimensional potential flow problem. The transverse dimension is small compared with the length and the scale over which cross-sections vary. The displacement and slope are small enough to linearize the impermeability condition about a straight body. The Reynolds number is sufficiently large for lateral loading to be dominated by fluid inertia; viscosity, axial friction and higher-order end corrections are neglected at this order. This reactive approximation must be distinguished from the viscous resistive-force theory used for a flagellum.

In coordinates translating with the swimmer, the undisturbed fluid passes in the positive $x$ direction at speed $U$. Applying the impermeability condition to the moving centre plane gives the transverse fluid velocityThe local added mass $m(x)$ is defined so that the transverse fluid hydrodynamic impulse per unit axial length is $P_y=m(x)v_y$. A fixed body-coordinate interval receives an impulse flux $U{P}_y$ at its leading end and loses the corresponding flux at its trailing end. Its rate of change of impulse plus the net outward impulse flux is thereforeThis is the transverse force exerted by the body on the fluid. By Newton's third law, the transverse force exerted by the fluid on the body has the opposite sign:The distinction is essential here: the printed positive expression is labelled as the force on the fish, so that label and expression are inconsistent. For example, take a uniform cross-section and a rigid lateral displacement $h=H(t)$. The swimmer must accelerate fluid of added mass $m$, which reacts on it with force $-m{H}^{\prime \prime }$, not $+m{H}^{\prime \prime }$. Thus the positive expression is recovered exactly with the force-on-fluid convention; the physically consistent force on the fish is the negative expression. In the following parts, $F_y$ denotes $F_{\mathrm{body}}$ so that the beam balance uses a genuine external force on the fish.

By the inverse identity established in the root solution, the composite $T$ is a symmetric relation precisely whenThus condition (i) implies condition (iii). Conversely, if $S\circ R\subseteq R\circ S$, take inverse relations. Since $R,S$ are symmetric, this gives the reverse inclusion $R\circ S\subseteq S\circ R$. The two composites are equal, making $T$ symmetric. Therefore (i) and (iii) are equivalent, and either one forces commutation.

Let ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi },{\mathbf{e}}_z$ be the cylindrical unit vectors. The surface has an upward normal vector proportional to ${\mathbf{e}}_z-h^{\prime }(r){\mathbf{e}}_r$. If $N_z$ is the vertical component of the normal force, its radial component is $-h^{\prime }(r)N_z$. In the stated approximation, vertical acceleration and the vertical driving-force component are neglected, so vertical balance gives $N_z\simeq Mg$. The horizontal contact force is consequently $-Mg{h}^{\prime }(r){\mathbf{e}}_r$. The projected velocity and its speed areThe driving force is parallel to this velocity, with components $F\dot{r}/u$ and $Fr\dot{\varphi }/u$. Use the acceleration in polar coordinates and Newton's second law to obtainThese equations apply where $u\ne 0$; at rest the direction of an exactly velocity-aligned drive is not determined.

For a prescribed path, write $r^{\prime }=dr/d\varphi$, $r^{\prime \prime }=d^{2}r/d{\varphi }^2$. The chain rule gives $\dot{r}=r^{\prime }\dot{\varphi }$ and $\ddot{r}=r^{\prime \prime }{\dot{\varphi }}^2+r^{\prime }\ddot{\varphi }$. Subtract $r^{\prime }/r$ times the azimuthal equation from the radial equation. The driving force and $\ddot{\varphi }$ both cancel:Thus the constraint for tangentially driven motion on an axisymmetric surface isThe denominator must give a nonnegative squared speed, and the calculation uses $r>0$. At the polar origin a limiting Cartesian description is appropriate.

For the paraboloid and Archimedean spiral, putThen, for $\varphi >0$,The branch tending to $\varphi =\infty$ has positive angular velocity, henceSince $u=ϵ\dot{\varphi }\sqrt{1+{\varphi }^2}$, solve the azimuthal equation for the required force:This is positive for $\varphi >0$, consistent with a forward drive. Taking the limit gives

These are the large-time limits of the printed projected equations. Their mechanical approximation has a precise range limitation: on the exact surface, $z=h(r)$ gives $\dot{z}=(2r/\ell )\dot{r}$, which along this solution grows like $2{ϵ}^2\alpha \varphi /\ell$. At fixed positive $ϵ$, it is not small as $\varphi \to \infty$. Thus the requested limits are formal limits of that reduced model; neglect of vertical motion is not a uniform approximation to the full three-dimensional motion all the way to infinity.

The fluid is outside the body, so its normal on the inner boundary is $-n^{\prime }$. Let $t=\sigma n^{\prime }$ be traction on the body's surface from the fluid. For an exterior point and a flow vanishing at infinity, the boundary representation readsFor $r=|y|\gg a$, expand the single layer through its first moment and the double layer through its zeroth moment:The next single-layer and double-layer terms are both $O(r^{-3})$ at fixed body size and boundary data. Differentiating $J$ givesHence the dipole terms combine intoThe antisymmetric part obeys $(\Sigma -{\Sigma }^T)y=G\times y$ for $G_j=-{ϵ}_{j\alpha \beta }{\Sigma }_{\alpha \beta }$. For the symmetric trace-free tensor $S$, $y\cdot My=y\cdot Sy+(\mathrm{Tr}M)r^2/3$. Also $\mathrm{Tr}\Sigma -\mathrm{Tr}M=2\mu \mathrm{Tr}U=2\mu Q$. Substitution givesThis derives the Stokes far-field multipoles of a deforming body. $F$ is the force exerted by the body on the fluid, the negative of the integrated fluid traction. $G$ is its torque on the fluid: $G=-\int x\times tdS$. Their kernels are the Stokeslet and rotlet. $Q=\int u\cdot n^{\prime }dS$ is the rate of increase of the body's volume, by the Reynolds transport theorem. It gives a radial source flow with volume flux $Q$. A volume-preserving body has $Q=0$. The trace-free tensor $S$ gives the stresslet, which can remain nonzero for a force-free, torque-free swimmer.

Use the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$. The Stokes equations for a point force on the fluid are $\nabla \cdot u=0$ and $\nabla \cdot \sigma +F\delta (x)=0$. Put $r=|x|$. Away from the origin, a harmonic vector potential proportional to $F/r$ in the Unscaled Papkovich–Neuber representation gives the rotationally covariant decaying solution. Specifically, take $\Phi =-F/(8\pi \mu r)$ and $\chi =0$. Differentiation yieldsThis satisfies the homogeneous Stokes equations for $r>0$. Its stress isTo verify the point-force normalization, on a sphere $r=\epsilon$ the traction is $\sigma n=-3(F\cdot n)n/(4\pi {\epsilon }^2)$. Since ${\int }_{S^2}{n}_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$, its surface integral is $-F$. Thus integrating momentum balance over the small ball gives exactly the applied force $F$, fixing the normalization of the Stokeslet. Its radial velocity is $(F\cdot n)/(4\pi \mu r)$, soThere is no net volume or mass source at the singularity. The kernels have the parity $J(-x)=J(x)$ and $K(-x)=-K(x)$.

For two Stokes flows with the same viscosity and body forces $f^{(1)},f^{(2)}$, the Lorentz reciprocal theorem including body forces isIndeed, apply the divergence theorem to $u_i^{(1)}{\sigma }_{ij}^{(2)}-u_i^{(2)}{\sigma }_{ij}^{(1)}$. The cross-gradient terms cancel: incompressibility removes each pressure contraction and the remaining terms are both $2\mu e^{(1)}:e^{(2)}$. The two stress divergences give the body-force terms shown.

Choose flow 1 to be the desired body-force-free field and flow 2 to be a Stokeslet with arbitrary force vector $q$ at $y$. For $y\in V$, the body-force term is $-q\cdot u(y)$. Rearrange the reciprocal identity and use the parity of $J,K$; equality for every $q$ givesFor $y$ outside the volume there is no point force in $V$, so the same two integrals sum to zero. To obtain the smooth boundary value, exclude a small hemisphere about $y$, use continuity of $u$ to replace it by $u(y)$ on that hemisphere, and take the radius to zero. A full small sphere contributes $u(y)$; a hemisphere contributes half of it. Equivalently, the local solid-angle factor is $2\pi /(4\pi )$. The result is the boundary integral representation of Stokes flow with right side $\frac{1}{2}u(y)$ on a smooth boundary, and with the double layer interpreted by its boundary principal value. A nonsmooth corner would have a different solid-angle coefficient.

For body-force-free Stokes fields with the same viscosity,To prove this Lorentz reciprocal theorem, integrate the divergence of the difference of the two cross-work fluxes. Both stress divergences are zero; incompressibility and symmetry of the Newtonian stress reduce each cross-gradient contraction to $2\mu e^{(1)}:e^{(2)}$. These are equal, so the divergence integrates to zero. The argument applies to an exterior domain by first cutting it off and taking the outer radius to infinity; the decaying velocity and stress make the outer boundary contribution vanish.

Define the generalized force as force and torque exerted by the rigid body on the fluid. With a resting far field and no-slip boundary conditions, $u=U+\Omega \times x$ on the body, and linearity givesFor two such motions the reciprocal theorem becomesSince the six-component velocities can be chosen independently, $\boxed{\mathsf{R}={\mathsf{R}}^T}$. The ordinary work identity givesfor every nonzero rigid-body motion. If the integral were zero, $e=0$ would make the connected exterior flow an infinitesimal rigid motion. Its decay at infinity requires that motion to be zero, which is incompatible with a nonzero no-slip body velocity. Thus $\mathsf{R}$ is positive definite. Fluid-on-body forces have the opposite sign; their map is $-\mathsf{R}$, not a positive-definite resistance matrix.

Write $s_{\varphi }=\sin \varphi$ and $c_{\varphi }=\cos \varphi$. The helix has $|dX/d\theta |=b\sqrt{1+{\tan }^2\varphi }=b\sec \varphi$, soIts unit tangent is $X^{\prime }=c_{\varphi }{e}_{\theta }+s_{\varphi }{e}_z$, where $e_{\theta }=(-\sin \theta ,\cos \theta ,0)$. The given slender-body force density is the force per arc length exerted on the fluid. For pure axial translation, $V=U{e}_z$, it givesFor pure rotation, $V=b\Omega e_{\theta }$, it givesThe axial torque per unit arc length is $b{f}_{\theta }$. Integrating along the wire gives the axial resistance matrix of a slender helixThus translation gives $(F_z,G_z)=(\mathcal{A}U,\mathcal{B}U)$, and rotation gives $(\mathcal{B}\Omega ,\mathcal{D}\Omega )$. These are the requested axial components; the axial swimmer approximation does not require proving statements about the other components. The equality of the two off-diagonal coefficients checks reciprocity, andFor the organism, retain the specified entrainment approximation: the head moves with the local large-scale fluid translation, and its angular velocity $\Omega -\omega$ differs from the local fluid rotation $\Omega$ by $-\omega$. Put $D_0=8\pi \mu a^3$. Its torque on the fluid is therefore $-D_0\omega$, and its force is neglected. Neutral buoyancy and absence of external torque giveSolving,This is the entrained-head approximation for a helical microswimmer, which differs from treating the head as an isolated sphere in otherwise stationary fluid. The flagellum's angular speed is also $\Omega =D_0\omega (1+c_{\varphi }^2)/(CL{b}^2)$.

At fixed $b,L,a,\omega ,ϵ$, the speed is maximal at $\varphi =\pi /4$ and tends to zero at either limiting pitch. At fixed other parameters it decreases as $1/L$: the longer flagellum has more resistance at the fixed torque supplied in this model. Holding the number of turns fixed instead would make $L$ change with pitch, so that is a different comparison.

The motor exerts equal and opposite torques of magnitude $D_0\omega$ on flagellum and head, whose relative angular speed is $\omega$. Therefore its rate of working isEquivalently, the work delivered to the fluid is $G_{\mathrm{flag}}\Omega +G_{\mathrm{head}}(\Omega -\omega )=D_0\omega \Omega -D_0\omega (\Omega -\omega )$. Translation contributes no net work because the total force vanishes. This explains the independence from $U$ and $\Omega$, without assigning an isolated-fluid drag law to the entrained head.

The material derivative follows surfactant carried by the moving interface. The term $-C{\nabla }_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n){\nabla }_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s{\nabla }_s^{2}C$ represents surface diffusion, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per surface area, as in surfactant transport with exchange relaxation.

In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C^{\prime }$ with uniform $C_0$ and regard both $u$ and $C^{\prime }$ as first order in the rising speed. Advection of $C^{\prime }$ and its product with surface dilation are second order. Hence the linear steady equation isSpherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A{I}_{s}U$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.

On $r=a$, differentiating $n=x/r$ givesEach component of $n$ is a degree-one spherical harmonic, soOne may also obtain the coefficient from $n\cdot n=1$: its surface Laplacian gives $n\cdot {\nabla }_s^{2}n=-|{\nabla }_{s}n{|}^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_{s}U=U-n(U\cdot n)$ then yieldsThe candidate $C^{\prime }=BU\cdot n$ therefore solves the linear transport equation ifIts spherical mean is zero. This is the dipolar surfactant distribution. Assuming $k{a}^2+2D_s>0$, its consistency condition isDiffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface Péclet number, fast exchange, or sufficiently immobilized surface velocity. These same conditions justify neglecting the perturbation's advection. Small capillary number and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.

Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general interfacial stress balance with variable surface tension isFor $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part is $-{\nabla }_s\gamma ={\gamma }_{1}B{I}_{s}U/a$. Thus the requested form isThis positive parameter measures the strength of the restoring Marangoni stress relative to viscous stress, including the relaxation of concentration by diffusion and exchange.

The far velocity $-U$ is produced by the constant harmonic vector potential $\Phi =U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi =\beta a^{3}U\cdot \nabla (1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The Unscaled Papkovich–Neuber representation gives, at arbitrary radius,The velocity at $r=a$ agrees with the printed expression. However, the original PDF's traction formula has a factor-of-two error. Direct differentiation with $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ givesnot the printed prefactor $12\mu /a$. This is traction of translating-sphere Papkovich–Neuber potentials. The $\alpha$ term follows immediately from the Stokeslet stress, and the scalar-potential part gives $6\mu \beta [U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.

With this consistent traction, no penetration requires $\beta -\alpha =1/2$, tangential velocity requires $A=-(1+\alpha +\beta )$, and tangential stress balance requires $\beta =A\lambda$. Solving,For comparison, taking the printed $12\mu /a$ literally would instead giveThose solve the supplied algebraic pair but not the Newtonian stress calculated from the supplied potentials. The factor-six result is the physically consistent translating surfactant-coated bubble solution.

As $\lambda \to 0$, $A\to -1/2$ and $\beta \to 0$: the clean bubble has a mobile interface and zero tangential traction. As $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$ and $\beta \to -1/4$: the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere. The limiting tangential traction is $-3\mu I_{s}U/(2a)$, a useful check on the correct normalization.

The normal dynamic traction isapart from an arbitrary constant pressure. In the normal stress balance on a translating bubble, its degree-one part is balanced by the buoyant hydrostatic pressure difference and by the variation $-2{\gamma }_1{C}^{\prime }/a$ of normal capillary stress. A uniform internal gas pressure balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta \rho$ is liquid minus bubble mass density, the degree-one balance can be writtenIt fixes $U=\Delta \rho g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this force balance.

Let $ϵ=1-\alpha \ll 1$, and use local distance $\rho =a\theta$ from the closest pole. The gap isHorizontal rotation has nonzero wall speed $V_s\sim |{\Omega }^{\prime }|a$ at this pole. In the broad region with $h=O(\Delta )$, shear is $O(\mu |{\Omega }^{\prime }|a/\Delta )$, the area is $O(a^2)$ and the lever arm is $O(a)$. Its torque is therefore $O(\mu |{\Omega }^{\prime }|a^4/\Delta )$. The smallest gap patch has radius $O(a\sqrt{ϵ})$ and area $O(a^2ϵ)$; its shear is $O(\mu |{\Omega }^{\prime }|a/(ϵ\Delta ))$. It gives the same bounded torque scale.

These are the two estimates requested in the paper. They omit an intermediate annulus, so the printed uniform $O(\mu |{\Omega }^{\prime }|a^4/\Delta )$ assertion is not correct as $ϵ\to 0$. For $a\sqrt{ϵ}\ll \rho \ll a$, $h\sim \Delta {\rho }^2/a^2$. An annular shear-torque contribution scales asIntegrating between the inner and outer scales produces $\log (1/ϵ)$.

A viscous dissipation lower bound makes this conclusion independent of possible signs of local tractions. A fluid velocity changing by $V_s$ between two no-slip walls dissipates at least $\mu V_s^2/h$ per unit area, by the Cauchy-Schwarz inequality in the cross-gap coordinate. The lubrication velocity profile gives more explicitlywith no cross term. The second term is nonnegative. Thus the annulus entails a logarithmic divergence in total power and hence in $|G^{\prime }|$, since the centre is held fixed and all boundary work is $G^{\prime }{\Omega }^{\prime }$.

The horizontal rotation in a nearly touching spherical shell calculation also gives the corrected force scale. Choose local $x$ along $v_s$ and put $V_s=a{\Omega }^{\prime }$ with its signed value. At leading order the stationary-gap Reynolds lubrication equation isFor $h=h_0+(x^2+y^2)/(2R)$ it has the particular pressurewhich satisfies the equation by direct differentiation. The pressure itself gives no torque about the centre of an exactly spherical inner body, because its normal is radial. It does contribute to tangential shear through the pressure gradient.

For any fixed local outer cutoff, the logarithmically divergent integrals areThe resisting tangential traction is $\mu V_s/h+(h/2)p_x$. Integration by parts gives torque magnitude $a\mu |V_s|(I_1+3I_2/(5R))$ to logarithmic order. Consequently, at leading order in $\Delta /a$,The strict corrected order is $\Theta ((\mu |{\Omega }^{\prime }|a^4/\Delta )\log (1/ϵ))$; the paper's scale is the bounded contribution of each of its two nominated regions, not the full asymptotic answer.

For the horizontal force, pressure dominates shear by a factor $a/\Delta$. Its horizontal projection on the inner sphere has normal component $n_x\simeq x/a$, giving $|F_p|\sim (6\mu |V_s|/(5a))I_2$. Hence the required holding-force magnitude isThe broad region and smallest patch each supply $O(\mu |{\Omega }^{\prime }|a^4/{\Delta }^2)$; the intermediate annulus again supplies the extra logarithm. The shear force is smaller by $\Delta /a$. These expressions assume the completely filled lubrication film of the problem.

The vertical holding force is zero. The geometry is invariant under a half-turn about the vertical axis. That operation reverses the horizontal angular velocity but leaves a vertical force component unchanged. Linearity would reverse that force component too, forcing it to vanish. Equivalently, the induced pressure has the first azimuthal harmonic, whose vertical-force integral over azimuth is zero. The horizontal holding force is the negative of the fluid's horizontal force on the body.

Set $V=\Delta \dot{\alpha }$, positive downwards. Then $h_t=-V\cos \theta$. The leading axisymmetric lubrication flux is $q_{\theta }=-h^3{p}_{\theta }/(12\mu a)$. Wall Couette flux from the translating sphere is smaller by $\Delta /a$ than the squeeze flux and does not affect this leading pressure. The spherical Reynolds lubrication equation isRegularity at the pole gives $q_{\theta }=(aV/2)\sin \theta$, soThe apparent singularity at $\alpha =0$ is a constant-pressure gauge term: subtract $3\mu a^{2}V/(\alpha {\Delta }^3)$ before taking the limit. The remaining pressure tends to $6\mu a^{2}V\cos \theta /{\Delta }^3$.

To leading lubrication order the force is the pressure integral. Its downward component on the inner sphere isThe original PDF's first integration hint is incorrect: the rational term needs $1-b^2$, not $(1-b{)}^2$. From the antiderivative $1/(1-t)+\log (1-t)$,Thus the squeeze resistance of eccentric nested spheres givesIt opposes the velocity for either sign of $\alpha$. The continuous concentric limit is $F_z=-8\pi \mu a^{4}V/{\Delta }^3$, since the bracket is $4{\alpha }^3/3+O({\alpha }^5)$. As $\alpha \to 1$, its leading magnitude is $6\pi \mu a^4|V|/[{\Delta }^3(1-\alpha )]$.

Let the tangential velocity scale be $U$, gap scale $H$ and along-gap length $L\gg H$. The tangential Stokes equation balances $P/L$ with $\mu U/H^2$, while wall shear is $\tau \sim \mu U/H$. ThereforeThis is lubrication pressure dominates shear stress; pressure force projections may nevertheless be small in a particular direction or may give no torque on a spherical surface.

Write the signed centre displacement along the downward axis as $d=\alpha \Delta$. A point $an$ on the inner sphere has distance $|an+d{e}_z|=a+d\cos \theta +O({\Delta }^2/a)$ from the outer centre. The leading normal gap to radius $a+\Delta$ is consequentlyThe local-curvature corrections are smaller by $\Delta /a$. This approximation also follows by solving the outer-sphere intersection along the ray $n$ from the inner centre and expanding its positive root.

Let $q(t)=\mathbf{r}(t)\cdot \mathbf{B}$. Since the field is constant, dotting the equation of motion with $\mathbf{B}$ eliminates the magnetic term:This is a harmonic oscillator with frequency $\omega =\sqrt{k/m}$. Initially $q(0)={\mathbf{r}}_0\cdot \mathbf{B}$ and $\dot{q}(0)=0$, because the particle starts at rest. ThereforeThe magnetic force modifies the transverse motion but not this projection. The formula also holds trivially when $\mathbf{B}=0$.

Write $\mathbf{v}=\dot{\mathbf{r}}$. The original PDF contains the cross product $\dot{\mathbf{r}}\times \mathbf{r}$; the TeX aid drops its velocity dot. Since $\mathbf{v}\times \mathbf{v}=0$, differentiation and the equation of motion giveThe radial restoring force contributes no torque. The vector triple-product identity yieldsMeanwhileThe derivatives cancel, provingThis is the magnetic axial angular-momentum invariant. Its first term is the negative of the ordinary angular momentum projected on $\mathbf{B}$; retaining that sign is essential.

For constant particle masses define the centre of mass by $M\mathbf{X}={\sum }_i{m}_i{\mathbf{r}}_i$, with $M={\sum }_i{m}_i$. Summing Newton's second law over all particles givesEach pair of internal forces cancels by Newton's third law. HenceOnly the action-reaction condition is needed for this translation law; centrality of the internal forces is needed for the angular momentum result in part (ii).

Let $r=|{\mathbf{r}}_1-{\mathbf{r}}_2|$ and $\hat{\mathbf{r}}=({\mathbf{r}}_1-{\mathbf{r}}_2)/r$. A differentiable central potential energy gives the internal forcesThere are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore givesSince $\dot{r}=\hat{\mathbf{r}}\cdot ({\mathbf{v}}_1-{\mathbf{v}}_2)$, the last expression is $-U^{\prime }(r)\dot{r}=-dU/dt$. ThusThis conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

Take $x$ tailward, so forward swimming has laboratory velocity $-U\mathbf{i}$. For an inextensible travelling waveform, material motion in the swimmer frame is $V\mathbf{i}-c\mathbf{t}$: a material point passes through the wave along its tangent while the wave pattern translates at $V$. Thus the relative velocity of otherwise stationary fluid isPut $q=V-U$ and choose $\mathbf{t}=(X_s,Y_s)$, $\mathbf{n}=(-Y_s,X_s)$, with $X_s^2+Y_s^2=1$. ThenOn the forward-swimming branch $0<U<V<c$ and $X_s>0$, the tangential slip is positive. A law for reversed tangential slip would also need its sign, not just its absolute magnitude.

The quadratic normal law represents inertial cross-flow drag: dynamic pressure scales as fluid density times $w_n^2$, while $|w_n|w_n$ chooses the force direction. Its coefficient contains the cross-sectional size and an empirical drag coefficient. The $|w_t{|}^{3/2}$ exponent is characteristic of an attached laminar boundary layer: its thickness scales as $\delta \sim (\nu s/|w_t|{)}^{1/2}$ and its shear as $\mu |w_t|/\delta$. However, this gives local shear proportional to $|w_t{|}^{3/2}{s}^{-1/2}$, not $s^{1/2}$. The positive power of $s$ is genuinely printed in this paper and repeated in its target formulas. Below it is treated as the prescribed phenomenological weight, rather than misidentified as standard local laminar skin-friction scaling along a slender swimmer.