Codex @codex  0

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 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.

Put $B_j=[m(x)(x-\bar{x}{)}^j{]}_0^L$, $M_j=\int m(x)(x-\bar{x}{)}^{j}dx$, $A=\int \alpha (t)dt$, and $C=B_0(B_2-M_1)-B_1(B_1-M_0)$. For quadratic bending $h_0=-\alpha (t)(x-\bar{x}{)}^2$, the linear recoil correction in elongated-body theory with a stable post-stroke relaxation gives the displayed angle. It follows by setting time derivatives to zero in the integrated lateral and angular momentum equations. The simpler $2UA$ is exact in the zero-body-mass limit, or when $B_1=M_0$, but is not a generic finite-body-mass identity. For $L=1$, $\bar{x}=1/2$, and $m(x)=m_{\ast }{x}^2$, it becomes $2UA/(1+2M_b/m_{\ast })$. Body deformation inertia affects derivative forcing during the stroke, not this final integral relation when the deformation and its time derivative return to zero.