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For a periodic inextensible filament with arclength period $\Lambda$ and projected axial period $\lambda$, material points slide tangentially through the wave frame at speed $Q$. An axial pattern speed $V$ relative to mean filament motion gives $Q\lambda /\Lambda =V$. Confusing the two speeds changes the swimming-to-wave-speed ratio by $\lambda /\Lambda$.

For a periodic inextensible filament, let $q=\mathbf{t}\cdot \mathbf{i}$, $\alpha =⟨q⟩=V/c$, $\beta =⟨q^2⟩$ and $\gamma ={\kappa }_T/{\kappa }_N$. The material velocity is $(V-U)\mathbf{i}-c\mathbf{t}$. Applying the linear local drag tensor ${\kappa }_T\mathbf{tt}+{\kappa }_N(I-\mathbf{tt})$ gives thrust $T={\kappa }_{N}L\{V(1-\gamma )(1-\beta )-U[1-(1-\gamma )\beta ]\}$. Equating this to head drag $\delta {\kappa }_{N}LU$ gives the displayed speed ratio. Whole spatial periods, or period averages, are needed for $c\int qds=VL$. Cauchy-Schwarz gives ${\alpha }^2\le \beta$, with equality only for constant axial tangent projection almost everywhere.

In resistive-force theory, the flagellum is slender and moves at a small Reynolds number, so local viscous resistance is linear in its velocity and fluid and body inertia are neglected. Normal and tangential resistance coefficients are treated as positive constants, usually ${\kappa }_N>{\kappa }_T$. The approximation neglects distant hydrodynamic interactions along the filament, head-tail interactions, nearby boundaries and self-contact. The centreline is an inextensible filament with prescribed shape. Its local resistance tensor isSince the filament velocity relative to distant fluid is $-\mathbf{w}$, the fluid force on a filament element is $K\mathbf{w}ds$. The equal and opposite force on the fluid has total axial component $T$. Projecting onto $\mathbf{i}$ givesThis establishes the thrust expression with its force-on-fluid convention.

For a uniform travelling wave, write the moving centreline as $\mathbf{r}(s,t)={\mathbf{r}}_{\mathrm{wave}}(s-ct)+(V-U)t\mathbf{i}$, where $s$ is material arc length and $|{\mathbf{r}}_{\mathrm{wave}}^{\prime }|=1$. Differentiating at fixed material label gives ${\mathbf{r}}_t=(V-U)\mathbf{i}-c\mathbf{t}$. HenceLet $q=\mathbf{t}\cdot \mathbf{i}=X_s$, $\alpha =⟨q⟩$ and $\beta =⟨q^2⟩$, where these brackets are contour-length averages. The axial wavelength divided by contour wavelength is $\alpha =V/c$. For a filament comprising whole wave periods, or after the usual period averaging, $c{\int }_0^{L}qds=VL$. This averaging is needed to avoid end-phase corrections for an arbitrary fraction of a wavelength. Substitution of $\mathbf{w}$ into the thrust expression now givesFor $\gamma ={\kappa }_T/{\kappa }_N$ and $r=1-\gamma$, this is $T/({\kappa }_{N}L)=Vr(1-\beta )-U(1-r\beta )$. The swimming body is force-free, so the reaction thrust on the head-tail assembly balances its head drag. Thus $T=\delta {\kappa }_{N}LU$, and linear resistive-force propulsion of a travelling filament givesWe henceforth assume $0<\gamma <1$, $\delta \ge 0$ and $0<\beta <1$. Isotropic resistance, $\gamma =1$, gives no propulsion in this model.

The flagellar contribution to viscous dissipation is the positive quadratic form $P_f=\int \mathbf{w}\cdot K\mathbf{w}ds$. Expanding it, and using $c⟨q⟩=V$, givesThere is also head dissipation $\delta {\kappa }_{N}L{U}^2$. Equivalently, the work of the flagellar deformation relative to the translating head is $P_f+UT$; the force balance makes $UT$ precisely the head dissipation. Consequently, with $\alpha =V/c$,This derives the rate of working, rather than merely inferring it from the thrust.

The Cauchy-Schwarz inequality gives ${\alpha }^2=⟨q{⟩}^2\le ⟨q^2⟩=\beta$. Equality requires $q$ to be constant almost everywhere. For a planar periodic filament this is an ideal sawtooth with alternating transverse tangent and fixed axial tangent, or a limit with negligibly rounded corners. It is not an identity for a general sinusoidal wave, and the ideal calculation omits the elastic cost of sharp corners.

For the stipulated equality ${\alpha }^2=\beta$, put $d=1+\delta$ and eliminate $V$ at fixed $U$ using $V/U=(d-r\beta )/[r(1-\beta )]$. Direct simplification yieldsThe derivative isWriting $A=r({\delta }^2-\gamma )$ and $B=\gamma d^2$, the numerator is $A{\beta }^2+2B\beta -B$. It is negative at zero and equals $(\delta +\gamma {)}^2>0$ at one. Its derivative is positive throughout $(0,1)$: this is immediate if $A\ge 0$; if $A<0$, its minimum derivative is $2(A+B)=2(\delta +\gamma {)}^2>0$. Since $F$ diverges at both endpoints, the unique stationary point is the global fixed-speed energy optimum for a planar flagellum:This form includes the case $A=0$, when ${\beta }_{\ast }=1/2$, without a removable division by zero. With no head drag it reduces to ${\beta }_{\ast }=1/(1+\sqrt{\gamma })$.

The printed final optimality formula does not follow from the preceding thrust and power equations. For a concrete check, $\gamma =1/2$, $\delta =0$ gives $F=(2-\beta )/[\beta (1-\beta )]$. The derived minimum is at $\beta =2-\sqrt{2}$. The printed formula instead gives ${\beta }^2=\sqrt{6}-2$, hence $\beta \approx 0.67044$, where $F^{\prime }\ne 0$ and the expenditure is greater. Thus the thrust and working formulas can both be established as above, but the last printed conclusion must be replaced by the derived optimum.

Use the arc length $s$ and the unit tangent $\mathbf{t}={\partial }_s\mathbf{r}$ of an inextensible filament. In the high-force state choose its positive longitudinal branch. The constraint givesThe extension is $z={\int }_0^L{t}_{z}ds=L-\frac{1}{2}{\int }_0^L|{\mathbf{t}}_{\perp }{|}^{2}ds+\cdots$. The curvature of a space curve obeys ${\kappa }^2=|{\partial }_s\mathbf{t}{|}^2$. Since ${\partial }_s{t}_z=-{\mathbf{t}}_{\perp }\cdot {\partial }_s{\mathbf{t}}_{\perp }+\cdots$ is second order in small fluctuations, its squared contribution is fourth order. Thus the worm-like chain energy through quadratic order isPositive force suppresses transverse orientation fluctuations, while the filament bending modulus penalizes their spatial variation.

To apply the equipartition theorem without missing either component or a mode-counting factor, expand each component in a real orthonormal basis of eigenfunctions of $-{\partial }_s^2$, $t_{\alpha }(s)={\sum }_q{a}_{\alpha q}{w}_q(s)$, with $\int w_q{w}_{q^{\prime }}ds={\delta }_{q{q}^{\prime }}$ and $\alpha =x,y$. The quadratic energy is $\frac{1}{2}{\sum }_{\alpha ,q}(A{q}^2+f)a_{\alpha q}^2$. Each real quadratic degree of freedom contributes $k_{B}T/2$, soFor a chain long compared with the force-dependent correlation length, the interior is translation invariant and the mode sum per length becomes $\int dq/(2\pi )$. Equivalently, for the Fourier transform ${\tilde{t}}_{\alpha }(q)=\int e^{-iqs}{t}_{\alpha }(s)ds$,The negative and positive wavenumbers here represent the real-field mode count consistently, rather than two independent complex modes. Summing both transverse components gives the high-force worm-like chain elasticitywhere the integral follows by setting $q=\sqrt{f/A}u$ and using ${\int }_{-\infty }^{\infty }(1+u^2{)}^{-1}du=\pi$. ThereforeHere $L_p$ is the three-dimensional zero-force persistence length. The result requires $f{L}_p\gg k_{B}T$, negligible backbone stretching, and a long-chain bulk approximation. It does not describe the eventual bond-stretching regime. For a continuum polymer description, the force-dependent length should also exceed the microscopic cutoff.

For comparison, a three-dimensional freely jointed chain has contour length $L=Nb$ and independent link orientations. At force $f$, a link of polar angle $\vartheta$ has Boltzmann factor $e^{x\cos \vartheta }$ with $x=fb/(k_{B}T)$. Its angular canonical partition function isThe total canonical partition function is $Z_1^N$. Differentiating with respect to force gives the force-extension of a three-dimensional freely jointed chain:ConsequentlyThe extension deficits scale as $f^{-1}$ and $f^{-1/2}$ respectively. Inverting them gives force divergences proportional to the first and second inverse powers of the deficit. Freely jointed link rotations cost no bending energy; the worm-like chain instead has a continuum of coupled bending modes whose contributing wavelengths change with force. Setting a long-wavelength coarse-grained link length $b=2L_p$ can match weak-force statistics, but does not make these high-force laws identical.

For the transverse tangent correlation of a stretched worm-like chain, use a separation $r$ consistently: the source writes $C(y)$ but integrates a separation denoted $r$. In the translation-invariant bulk the spatial average is redundant andTo evaluate the integral without an assumed decay law, let $G(r)$ be the inverse Fourier transform of $(A{q}^2+f{)}^{-1}$. Then $(-A{\partial }_r^2+f)G=\delta (r)$. Away from zero, decay at both infinities requires $G=B{e}^{-|r|/\xi }$ with $\xi =\sqrt{A/f}$. Integrating across zero yields $-A[G^{\prime }(0^{+})-G^{\prime }(0^{-})]=1$, so $B=1/(2\sqrt{Af})$. ThusAt $r=0$ this agrees with the two-component equipartition theorem calculation. The correlation length decreases with force and is distinct from the zero-force persistence length. The full tangent correlation has a nonzero aligned background; the exponentially decaying quantity here is specifically the transverse correlation.

For a finite open chain, the source's integral samples $s+r$ beyond $L$ and supplies no endpoint tangent conditions. An exact finite-length correlation is therefore not uniquely specified. The result above is the intended bulk form when $L\gg \xi$. As one explicit finite convention, periodically extending the tangent fluctuations gives the periodic finite-length tangent correlation of a stretched worm-like chainwhere $d_L(r)$ is the distance from zero around the periodic interval. Then $C_L(0)=(k_{B}T/\sqrt{Af})\coth [L/(2\xi )]$, and the extension formula acquires this same finite-length factor. The bulk law is recovered as $L/\xi \to \infty$. This makes the endpoint assumption explicit rather than claiming an unspecified finite-chain correlation is exactly translationally invariant.

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.

Take $h$ positive downwards and write $B$ for the filament bending modulus. Under the material-modulus interpretation of the printed $A$ as Young's modulus, the circular cross-section gives the second moment of area and bending modulusIndeed a small curvature $\kappa$ gives axial strain $-y\kappa$, so integrating $A{y}^2{\kappa }^2/2$ over the section gives bending energy $B{\kappa }^2/2$ per unit length. Keeping this distinction is essential to the powers of radius in the answer.

After allowing for buoyancy, the effective downward force per unit length isThe filament loses potential energy as it moves down; the displaced fluid's hydrostatic contribution is included through the density difference. For an inextensible filament with arc-length coordinate $s$, the geometric bending and gravitational energy is ${\int }_0^L[B\kappa (s{)}^2/2-wh(s)]ds$, up to a constant. For the small-slope Monge representation this reduces toThe upward-positive convention reverses the sign of $h$ and the gravitational term together. There is no stretching contribution at this order when bending a slender filament without imposed axial tension. The wording does not explicitly distinguish a material elastic modulus from a flexural modulus: if $A$ instead denotes the filament bending modulus in the intended convention, set $B=A$ throughout; then the $\pi a^4/4$ factor is already included in $A$.

The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

For a straight, unloaded inextensible filament with small transverse displacement, resistive-force theory gives viscous force density $-c_{\perp }{y}_t$. Variation of the bending energy gives $-B{y}_{xxxx}$, while the induced filament tension is higher order. Their instantaneous balance yields $c_{\perp }{y}_t=-B{y}_{xxxx}$. It is an overdamped fourth-order diffusion equation: short-wavelength bends relax much faster than long ones. Wiggins and Goldstein's flexive-propulsion paper develops the elastic-wave mechanism for low-Reynolds-number propulsion.