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An active swimming beam has a total internal bending moment $G$ containing passive elasticity and muscular actuation. Neglecting cross-sectional rotary inertia, slice angular momentum balance identifies the internal transverse force with $G_x$. Slice linear momentum balance then gives $G_{xx}=m_b{h}_{tt}-F_{\mathrm{body}}$. In Lighthill elongated-body theory, $F_{\mathrm{body}}=-\mathcal{D}(m\mathcal{D}h)$, where $\mathcal{D}={\partial }_t+U{\partial }_x$. The body and its hydrodynamic added mass therefore contribute with the same sign to the acceleration coefficient. The positive quantity $\mathcal{D}(m\mathcal{D}h)$ is the reaction on the fluid.

With $r=x-\bar{x}$, lateral relative velocity $w=({\partial }_t+U{\partial }_x)h$, $Q=\int mwdx$ and $P=\int rmwdx$, the Lighthill elongated-body theory force on the fluid is $F_z=({\partial }_t+U{\partial }_x)(mw)$. Integration by parts gives $\int F_{z}dx=\dot{Q}+U[mw{]}_0^L$ and $\int r{F}_{z}dx=\dot{P}+U[rmw{]}_0^L-UQ$. The endpoint terms are momentum fluxes and must not be discarded when a tail has nonzero added mass. If both endpoint added masses vanish, total lateral body-plus-fluid momentum is conserved; a finite-mass body cannot then finish with zero relative crossflow and a new nonzero transverse velocity.

The Lighthill elongated-body theory is a reactive, nearly inviscid flow description of a long slender swimmer, appropriate when the longitudinal Reynolds number is large but its lateral displacement and slope are small. To leading order, crossflow past each section is the two-dimensional potential flow generated by lateral motion of that section. Longitudinal variation is slow on the cross-sectional scale, and axial speed $U$ is prescribed at this order. With $x$ increasing from nose to tail, fluid passes sections in increasing $x$, and the linearized relative lateral velocity isThe section's added mass per unit length $m(x)$ is defined by its crossflow kinetic energy $\frac{1}{2}m(x)w^2$. Equivalently, if $\Phi$ is the velocity potential for unit lateral section motion, then $m=\rho \int |{\nabla }_{\perp }\Phi {|}^{2}dA$ over the exterior crossflow plane. It has units of mass per length and depends on cross-sectional shape and density. For a circular section of radius $a$, $\Phi =-a^2\cos \phi /r$ gives $m=\rho \pi a^2$. It is not the body's mass per unit length.

The transverse fluid impulse attached to a section is $m(x)w$. A control segment has both changing impulse and an impulse flux $Umw$ through its ends. Taking their sum per unit length gives the reaction on the fluid:The force on the fish is $-F_z$. In particular, $m$ is inside the convective derivative; dropping its spatial derivative or the endpoint fluxes is not justified merely because the fish is slender.

The recoil correction in elongated-body theory adds unknown whole-body translation and yaw to a prescribed active bending shape:They are determined by total lateral linear momentum and yaw angular momentum balance, rather than by imposing that the deforming swimmer remains on its original axis. Body inertia and hydrodynamic added inertia have the same sign. If the translation variable is required to be the literal centre of mass, the active deformation must first have its mass-weighted mean removed: the reference displacement $Z_r$ is otherwise not the centre of mass displacement during the stroke. The corresponding deformation contribution to angular momentum must also be retained when it is significant.

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.

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.

The normal axial force on an element isThe prescribed tangential force gives $d{F}_{T,x}=K_T(c-q{X}_s{)}^{3/2}{X}_s{s}^{1/2}ds$. Zero mean axial force therefore gives the nonlinear resistive-force balance for a travelling filamentThe absolute values are present in the original PDF. In a periodically beating, constant-mean-speed model, phase-average the integrals where necessary; the leading approximation below is phase independent. Local force laws neglect nonlocal hydrodynamic interactions and assume nearly steady local response.

For the sinusoidal displacement, let $\epsilon =\beta k\ll 1$. Then $Y_s=\epsilon \cos ks$ and $X_s=(1-Y_s^2{)}^{1/2}=1+O({\epsilon }^2)$. Over one wavelength,Replace $X_s$ by one in the leading tangential term and balance the forces. After multiplication by $3/2$,This replacement also requires $c{\epsilon }^2\ll c-V+U$ for a uniformly small relative error inside the fractional power. If the tangential slip is itself $O(c{\epsilon }^2)$, its variation along the body must be retained. Periodicity gives $V/c=⟨X_s⟩=1-{\epsilon }^2/4+O({\epsilon }^4)$, so small geometric slope alone does not guarantee that additional slip condition.

For Lighthill elongated-body theory, put $h(x,t)\simeq \beta \sin [k(x-Vt)]$ and $w_{\perp }=h_t+U{h}_x$. The transverse fluid kinetic energy per unit length is $M_A{w}_{\perp }^2/2$. In the usual tail-flux approximation, with negligible leading-end contribution, the lateral work flux is $M_{A}U⟨h_t{w}_{\perp }⟩$, whereas kinetic energy carried into the wake is $M_{A}U⟨w_{\perp }^2⟩/2$. Their difference is the useful propulsive power $TU$. ConsequentlyBalancing this against the drag of the printed tangential model, $D=(2/3)K_T{L}^{3/2}(c-V+U{)}^{3/2}$, givesFor a genuine local laminar skin-friction estimate, write $dD=C_T{u}_t^{3/2}{s}^{-1/2}ds$, where $C_T$ has the appropriate different dimensions. Its integrated drag is $2C_T{u}_t^{3/2}\sqrt{L}$, and the corresponding elongated-body balance is instead $M_A{\beta }^2{k}^2(V^2-U^2)/8=C_T\sqrt{L}(c-V+U{)}^{3/2}$. The printed $L^{3/2}$ expression is therefore a result of its stated force model, not of that standard boundary-layer law.

The two thrust models describe different mechanisms: cross-flow resistive-force theory uses dissipative local drag, while elongated-body theory uses reactive acceleration of added mass. Both suppress details of the wake, body-body interactions, finite slenderness, skin-friction unsteadiness, separation and possible turbulent transition. The tail-only reactive result also requires a negligible leading-end flux, usually supplied by a nose taper or small leading-end motion; a blunt constant-cross-section body waving equally at both ends cannot automatically omit its leading-end contribution. Speed fluctuations and recoil may matter, and a planar constant-amplitude waveform is an idealization of a real sea snake.

The reaction sign and the meaning of recoil translation matter here. In Lighthill elongated-body theory, put $\mathcal{D}={\partial }_t+U{\partial }_x$. The transverse momentum imparted to the fluid per unit length is $m\mathcal{D}h$, so the lateral force on the fish is $-\mathcal{D}(m\mathcal{D}h)$. Thus body mass and positive added mass enter effective inertia with the same sign. The periodic trailing-edge thrust, to quadratic order in slope, isThe reaction sign and this small-amplitude thrust formula can also be checked in the primary derivation of the elongated-body force, equations (2.6), (3.2) and the discussion after (3.5). For the long-body limit below, the recoiling leading edge contributes only at relative order ${\alpha }^{-2}$.

The requested recoil and thrust formulas cannot all hold for the literal fish described. In particular, the printed thrust would remain $-m{U}^2/4$ when $h_1=0$, although a straight, unperturbed fish has $h_t=h_x=0$ and zero reactive thrust. Its negative term is missing an amplitude-squared slope factor. The stated effective translational coefficient $M^{\prime }-1$ also has the wrong sign for positive added mass, and the literal centre of mass translation cannot be equated to the translation of an arbitrarily chosen undeformed body reference line. The PDF defines $I^{\prime }=I/[m(\alpha L{)}^3]$; the converted TeX incorrectly uses the square.

Here are the force balance and angular momentum calculations for a consistent interpretation of the stated uniform fish. They give a corrected recoil solution, as well as a direct test of the printed equations. Defineand let $R$ be actual centre of mass displacement. The prescribed deformation has mean $a(t)=h_{1}f(t)/[2(\alpha +1)]$. Its mass-weighted mean must be removed in the recoil correction in elongated-body theory:Indeed $\int hdx=\ell R$, since $\int \xi dx=0$ and $\int h_{0}dx=\ell a$. Introduce the two geometric integralsFor uniform mass per unit length, $I={\mu }_{b}J$ at leading small-amplitude order. Retaining the symbol $I$ makes the rotational momentum balance transparent.

The needed integrals of transverse acceleration and convective derivatives areThe second spatial derivative includes the hinge's slope jump, as a distributional derivative; discarding it would incorrectly remove the $U^2$ terms. Newton's lateral equation is $M\ddot{R}=-m\int {\mathcal{D}}^{2}hdx$. The angular momentum of the deforming fish is $I\dot{\theta }+{\mu }_{b}C\dot{f}$, so its time derivative equals $-m\int \xi {\mathcal{D}}^{2}hdx$. Consequently the consistent uniform-added-mass recoil equations areThese follow directly from the local Lighthill elongated-body theory model with the physical reaction sign. An alternative reference-line translation replaces $R$ by $R-a$, and changes the shape forcing, but cannot change $M+m\ell$ to a difference of two positive inertias. Leading-edge details of a uniform-depth idealization can require an end model; they do not alter this reaction-sign test or the zero-stroke counterexample.

For $f=\cos \omega t$, write $R=\mathrm{Re}(r{e}^{i\omega t})$ and $\theta =\mathrm{Re}(\tau e^{i\omega t})$, choosing the periodic response and omitting free rigid-motion constants. The two balances give the explicit recoil amplitudesAt the tail the displacement and slope amplitudes are $H_L=h_1+r+\ell \tau /2-h_1/[2(\alpha +1)]$ and $S_L=h_1/L+\tau$. Thus a corrected, directly calculable mean thrust iswithin the trailing-edge approximation, with any retained leading-edge contribution treated consistently. This expression has the correct dimensions and vanishes quadratically with stroke amplitude.

A particularly simple independent check is the limit $U=0$. For the uniform fish, $I={\mu }_{b}J$, the balances give $\ddot{R}=0$ and $\theta =-Cf/J$ for the periodic response. The tail displacement amplitude is then exactlyHence, to the requested relative order,The leading-edge contribution is only $O({\alpha }^{-2})$ here. This recoil reduction contradicts the printed positive correction, for example for $M^{\prime }=2$, $I^{\prime }=1/6$. It also shows explicitly why treating the prescribed shape's moving mass as stationary would give a different model.

For completeness, the positive correction quoted in the question has a simple conditional origin if its two displayed recoil equations are accepted as formal equations rather than derived physical balances. Put $J^{\prime }=I^{\prime }+1/6$ and keep $U/(\omega L)$ fixed as $\alpha \to \infty$, away from $M^{\prime }=1$. Adding those equations and then using the first gives their leading periodic responseUnder the additional reference-line interpretation $h=h_0+R+\xi \theta$, the in-phase tail amplitude is $h_1[1+1/(2\alpha (M^{\prime }-1))+1/(8\alpha J^{\prime })]$. Squaring it in the trailing-edge thrust gives the printed frequency correction. Even this formal route, however, gives the amplitude-consistent resultnot a term $-m{U}^2/4$ independent of $h_1$. The quadrature part of $R$ first changes the squared displacement at relative order ${\alpha }^{-2}$, while recoil changes the tail slope at that order as well. This conditional calculation explains the intended algebra without presenting the flawed starting equations as a physical theorem.

Reactive thrust is positive when the tail's lateral-motion term exceeds its streamwise slope penalty; for an unrecoiling hinged tail this requires $\omega L>U$. A real steady swimming speed additionally balances the thrust against resistive drag force, absent from this inertial idealization. Recoil can alter either term and is not universally thrust-enhancing. A pole at $M^{\prime }=1$ is not a physical negative-added-mass resonance: the genuine combined inertias remain positive.