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.

In the oscillating rigid rod in resistive-force theory, $\sin \theta \cos \theta =\theta +O({ϵ}^3)$ and $\dot{x}=-ϵa\omega \sin \omega t$. Sincethe mean transverse force from a rocking rod, on the fluid in the tilt convention of part (a), isThe parallel and perpendicular drag coefficients of a slender filament are per unit length here. Their difference is essential: isotropic local drag gives zero transverse force. The hydrodynamic force on the rod is the negative of the displayed answer.

For the usual anisotropy $c_{\perp }>c_{\parallel }$ and $a>0$, the maximum positive mean transverse force from a rocking rod occurs atThe maximum magnitude has either quadrature phase $\varphi =\pm \pi /2$. If positive tilt is defined towards $-x$, the positive-force optimum is instead $+\pi /2$. At our optimum $\theta =ϵ\sin \omega t$ is opposite in sign to $\dot{x}$: the rod tilts so that the weaker tangential drag and stronger normal drag combine into a positive $y$ force on both strokes.

For $\varphi =0$ or $\pi$, orientation is a single-valued function of the centre displacement. The configuration retraces its path under time reversal, and kinematic reversibility of Stokes flow makes the force impulse cancel. Equivalently, $⟨\theta \dot{x}⟩=0$. The enclosed loop in the displacement-orientation plane is largest in quadrature and vanishes for a reciprocal stroke, illustrating the principle behind the scallop theorem.

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.

A Stokeslet is the flow produced by a point force. With $C=1/(8\pi \mu )$, its tensor is $G_{ij}=C({\delta }_{ij}/r+r_i{r}_j/r^3)$. A separated pair of equal and opposite forces has no force monopole. Expanding their difference in the separation gives a directional derivative of this tensor. For an axial pair, choose the sign of the strength so thatThe trace part of $\mathbf{ee}$ does not affect the flow because ${\partial }_j{G}_{ij}=0$. The remaining symmetric force-dipole contribution is the stresslet, decaying as $r^{-2}$ instead of the Stokeslet's $r^{-1}$. A force-free swimming cell has thrust balanced by drag, so this is typically its leading far field. With $S>0$ it is a pusher microswimmer: axial outflow and transverse inflow. It is distinct from the antisymmetric torque singularity, the rotlet.

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.