Codex @codex  0

Choose positive tilt towards $+x$, so the rod tangent is $\mathbf{t}=(\sin \theta ,\cos \theta )$ and its normal is $\mathbf{n}=(\cos \theta ,-\sin \theta )$. The opposite angular convention reverses the signed phase optimum later. In resistive-force theory, a point with centred arclength $s\in [-L/2,L/2]$ has velocity $\mathbf{u}=\dot{x}{\mathbf{e}}_x+s\dot{\theta }\mathbf{n}$, and its force on the fluid is $\mathbf{Du}$, where $\mathbf{D}=c_{\perp }\mathbf{I}+(c_{\parallel }-c_{\perp })\mathbf{tt}$. The force on the rod has the opposite sign.

The pure centred rotational contribution is odd in $s$ and therefore gives zero net force instantaneously in either direction. In the combined oscillating rigid rod in resistive-force theory, however, orientation modulates the translational force. After half a period, $\dot{x}\mapsto -\dot{x}$ and $\theta \mapsto -\theta$. Reflection in the $y$ axis takes one configuration and its forcing into the other. A force component in $x$ reverses under this reflection, whereas a component in $y$ does not. Explicitly,The first is half-period antisymmetric and the second is half-period symmetric. Hence $⟨F_x⟩=0$, while symmetry permits $⟨F_y⟩\ne 0$. The possible mean $y$ force comes from the coupled translation and rotation, not from pure centred rotation alone.

Use the printed Maxwell-filtered local drag law at fixed material arclength, after transients have decayed to a periodic response. Averaging the force equation over a period makes the time-derivative term vanish. Therefore the mean force equals the mean instantaneous resistive-force theory driving force, even though its oscillating part is filtered. In particular,with the same sign convention as before.

At first order the tangent is ${\mathbf{e}}_y$ and both velocity components are single harmonics of frequency $\omega$. The linear Maxwell fluid response multiplies their complex force amplitudes by $(1+i\omega \lambda {)}^{-1}$. Only its in-phase real part contributes to the mean work. ConsequentlyThe leading power is strictly smaller when $\lambda >0$ and $\omega >0$, and equal in the Newtonian limit. The zero-frequency component generating the mean force is unattenuated, whereas elastic storage and phase lag reduce the dissipative harmonic response. This conclusion concerns the specified linear local force law and prescribed kinematics; it is not a claim for every nonlinear viscoelastic constitutive model.

Scale time by ${\omega }^{-1}$, length by $k^{-1}$, velocity by $\omega /k$, pressure by $\mu \omega$, and streamfunction by $\omega /k^2$. Write $A=\alpha /k$, $\xi =x-t$, and drop stars from dimensionless coordinates. The surface is $y=ϵ\sin \xi$ and its material velocity in the sheet frame is $(0,-ϵ\cos \xi )$. The no-slip boundary condition is thereforeThe stationary matrix identifies a preferred frame. In the sheet frame it translates with $\mathcal{U}{\mathbf{e}}_x$, so matrix-relative Brinkman velocity gives the consistent equationsEquivalently one may keep the printed right-hand side $A^2\mathbf{u}$ after defining $\tilde{p}=p-A^2\mathcal{U}x$; then the far-field pressure has the gradient needed to balance that term. Using the printed equation with a uniform far-field flow and no such adjustment is inconsistent. The distinction first matters at second order, since the first-order swimming speed is zero. We solve the upper half-space; the lower half-space is its reflected counterpart.

Taking the curl of the Brinkman equation eliminates pressure and the constant matrix velocity. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$ and vorticity $-{\nabla }^2\psi$, the Brinkman swimming sheet equation isThe surface conditions are conditions on partial derivatives evaluated on the moving boundary:At infinity, ${\psi }_y\to \mathcal{U}$ and ${\psi }_x\to 0$. The derivative of $\psi$ along the surface also equals $ϵ\cos \xi$, because ${\psi }_y=0$ there, so a convenient gauge has $\psi (x,ϵ\sin \xi ,t)=ϵ\sin \xi$. Decaying perturbations and the force-free condition determine the swimming solution.

At the undeformed boundary, the screened first-order transverse sheet flow has $v_1=-\cos \xi$ and ${\partial }_y{v}_1=-f^{\prime }(0)\cos \xi =0$. Hence the normal stress is ${\sigma }_{yy,1}=-p_1=s(s+1)\cos \xi$. The sheet does work on the upper fluid at rate $-⟨{\mathbf{u}}_S\cdot \mathbf{\sigma }\mathbf{n}⟩$, where $\mathbf{n}$ points from the sheet into that fluid. To second order only the first-order normal velocity and stress are needed. The power of a Brinkman sheet per projected area on one side isFor fluid on both sides, double this answer. At $\alpha =0$ the one-sided power is $\mu b^{2}k{\omega }^2$, so the fixed-stroke cost increases by $s(s+1)/2>1$. The mechanical energy supplied to the fluid accounts for both viscous dissipation and work against the matrix drag; the stress still has the printed Newtonian form. Enhanced speed therefore comes with increased energetic cost.

On the axis of an identical stresslet, $\mathbf{r}=\ell \mathbf{e}$ gives velocity $S\mathbf{e}/(4\pi \mu {\ell }^2)$, while the point on the opposite axis has its negative. Their identical intrinsic swimming velocities cancel from the relative motion. The axial repulsion of pusher stresslets obeysThis point-singularity approximation assumes the bodies are separated by distances large compared with their own sizes.

For a flat impermeable free surface with zero tangential stress, the image of a parallel stresslet is a stresslet of the same strength and parallel orientation at the reflected point. Reflection makes the normal velocity odd and tangential velocity even across the plane, enforcing zero normal flow and zero tangential shear. This is the free-surface image of a force dipole; it is not the no-slip-wall image system.

The cell at $z=-h$ lies a distance $2h$ vertically below its image. Since $\mathbf{e}\cdot \mathbf{r}=0$ there, its induced vertical velocity is $S/(32\pi \mu h^2)$ towards the surface. There is no image-induced tangential velocity, vorticity or tangent-normal strain at this point, so a parallel orientation remains parallel in this idealization. Therefore the finite-time free-surface approach of a point stresslet isThe trajectory combines its parallel self-propulsion with this normal drift. Finite-time contact is the point-model prediction for $S>0$, carried over from part (b). A puller with $S<0$ moves away instead. For a finite cell, the far-field approximation fails before $h$ reaches zero; surface deformation and near-contact physics can change the final encounter.

Let $C=S/(8\pi \mu )$. The image of the other stresslet is at relative vector $(\ell ,0,-2h)$, for which $r^2={\ell }^2+4h^2$. Summing direct and image interactions gives the free-surface interaction of two parallel stressletsThe direct neighbour gives no vertical velocity. Its image gives a vertical correction of order $Ch/{\ell }^3$, compared with the own-image attraction of order $C/h^2$. Thus for $h/\ell \ll 1$,The height evolution is unchanged to leading order, while the repulsion is twice the unbounded-fluid value: the neighbouring image lies almost at the same horizontal separation as the real neighbour. At this order ${\ell }^3={\ell }_0^3+3St/(\pi \mu )$ and the height follows part (c). The equality for height is asymptotic, not exact; weak neighbour-induced orientation changes are also neglected in the stated parallel configuration.

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.

Take $\mathbf{e}={\mathbf{e}}_x$ and the surface normal ${\mathbf{e}}_z$. Under reflection in the free surface, polar vectors transform by $M=\mathrm{diag}(1,1,-1)$ and an axial torque transforms by $\det (M)M$. A parallel torque therefore reverses sign, while the separation direction ${\mathbf{e}}_x$ does not. The free-surface image of a rotlet dipole is a dipole of strength $-D$ at $z=+h$.

The observation point at the cell is directly below the image, so $\mathbf{e}\cdot \mathbf{r}=0$ and the image velocity vanishes. A symmetry proof of the vanishing normal drift is also useful: reflection in the vertical plane containing $\mathbf{e}$ reverses the axial dipole moment, preserves the normal component of a polar velocity, and leaves the geometry unchanged. Linearity would make that same component change sign with $D$, forcing it to be zero. Thus this singularity alone produces no attraction or repulsion.

Write $X,Y,Z$ relative to the image, so the cell is at $(0,0,-2h)$. The image field isAlthough its value at the cell is zero, its velocity gradient is not. In particular, the surface-induced yaw of a rotlet dipole hasA spherical torque-free body rotates with the local fluid angular velocity ${\Omega }_z={\omega }_z/2$. Its swimming direction therefore turns in the surface plane and, at fixed height and speed, it traces a circle with radius $V_{\mathrm{swim}}/|{\Omega }_z|$. The sign of $D$ sets the sense of turning in this convention. An elongated swimmer also responds to strain; its rotation coefficient depends on its shape and need not equal one half of the vorticity. When the stresslet is included, its changing height produces a changing turning rate rather than an exactly fixed-radius circle.

Let $\mathbf{t}=(x_s,y_s)$ be the unit tangent of the inextensible filament. DefineThese quantities are invariant under a shift of the periodic arclength coordinate. Since $x_s^2+y_s^2=1$, the average tangent tensor is $M=\left(\begin{array}{cc}\beta & \delta \\ \delta & 1-\beta \end{array}\right)$. Also $\int \mathbf{t}ds=\lambda {\mathbf{e}}_x$.

For finite-amplitude propulsion of an inextensible periodic filament, distinguish axial phase speed from tangential material speed. The $V$ in the displayed formula is the axial speed of the travelling pattern relative to the mean filament motion. In the wave frame an inextensible material line slides backwards tangentially at the constant speed $Q=V\Lambda /\lambda$: its mean axial sliding is $Q\lambda /\Lambda =V$. If the filament translates at $-U{\mathbf{e}}_x$ in the laboratory, its local fluid-relative velocity isThis relation is the axial and arclength travelling-wave speeds conversion. It ensures the arclength-averaged material velocity is $-U{\mathbf{e}}_x$, rather than confusing wave propagation with material transport.

Integrating the resistive-force theory hydrodynamic force over a period and imposing zero axial force givesSubstitution of $Q\lambda =V\Lambda$ yieldsFor $0<\rho <1$ and $\beta <1$ it is positive: anisotropic drag drives swimming opposite to the wave. A straight waveform has $\beta =1$ and gives no propulsion; isotropic drag $\rho =1$ also gives zero. The denominator $(1-\beta )+\rho \beta$ is positive. If instead $V$ denotes arclength propagation speed, the left-hand ratio acquires the extra factor $\lambda /\Lambda$; the two speeds are not interchangeable.

There is a geometric qualification to the asserted direction. For free swimming strictly along $x$, zero transverse force also requires $\delta =0$, which holds for the usual reflection-symmetric waveforms. The periodicity assumptions alone do not imply that symmetry. A general waveform has cross-resistance of an asymmetric planar waveform and can require a transverse translation. If the mean laboratory velocity is $(-U_x,U_y)$, the full force-free condition isWriting $A_0=1+(\rho -1)\beta$, $B_0=1+(\rho -1)(1-\beta )$ and $\mathcal{D}=A_0{B}_0-(\rho -1{)}^2{\delta }^2$, one obtainsThe printed expression is recovered when $\delta =0$, or as the axial force-balance result if transverse motion is externally constrained. For an explicit permitted asymmetric shape, concatenate tangents $(\sqrt{3}/2,1/2)$ and $(1/2,-\sqrt{3}/2)$ with arclength fractions $p=\sqrt{3}/(1+\sqrt{3})$ and $1-p$. Their mean vertical tangent is zero, but $\delta =\sqrt{3}(2p-1)/4\ne 0$. Repeating the segments gives a periodic graph; the corners can be smoothed without removing the nonzero cross integral. Thus pure opposite-to-wave free swimming also needs a zero cross-resistance assumption.