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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.

An infinitely long circular cylinder of radius $a$ translating across its axis through otherwise stationary, inviscid incompressible fluid has added mass $m_a=\rho \pi a^2$ per unit length. The decaying, circulation-free velocity potential is $\varphi =-U{a}^2\cos \theta /r$. In moving centre coordinates, the fixed-laboratory derivative is ${\varphi }_t=-\dot{U}{a}^2\cos \theta /r-U^2{a}^2\cos (2\theta )/r^2$. The Unsteady Bernoulli equation gives $p-p_{\infty }=\rho a^2\dot{U}\cos \theta /r+\rho U^2{a}^2\cos (2\theta )/r^2-\rho U^2{a}^4/(2r^4)$. Integrating $-pa\cos \theta$ around $r=a$ yields $F_x=-\rho \pi a^2\dot{U}$; the transverse force vanishes. Taking a derivative at fixed centre coordinates would omit the necessary convective correction.

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.

In the small-amplitude slender-body theory approximation, an undulating swimmer accelerates an added mass of fluid per unit length $m(x)$. With $\mathcal{D}={\partial }_t+U{\partial }_x$ and transverse displacement $h$, the lateral force on the body is $-\mathcal{D}(m\mathcal{D}h)$. Its reaction on the fluid has the opposite sign. For a periodic stroke and the usual negligible leading-edge added-mass contribution, the mean trailing-edge thrust is $m⟨h_t^2-U^2{h}_x^2⟩/2$. This is a reactive inertial model, distinct from viscous resistive-force theory.

Fluid added mass increases inertia but need not alter the same restoring force. A submerged spring's constant buoyancy shifts equilibrium, whereas a pendulum's buoyancy reduces its restoring torque. Thus an unbounded-fluid sphere has spring frequency factor $\sqrt{\rho /(\rho +{\rho }_f/2)}$ and downward-pendulum factor $\sqrt{(\rho -{\rho }_f)/(\rho +{\rho }_f/2)}$. The latter stable configuration requires $\rho >{\rho }_f$.

On the translating circular cylinder $r=a$, the preceding velocity potential and streamfunction giveUsing the boundary integral for kinetic energy of planar potential flow,ThereforeAs a direct check, $|\mathbf{u}{|}^2=U^2{a}^4/r^4$, and integration over $r>a$ gives the same result.

The fluid kinetic energy has the same dependence on body speed as the kinetic energy of an extra mass $M$ carried by the body. Thus $M$ is its added mass, here equal to the mass of displaced fluid per unit cylinder length. For a cylinder of physical mass per length $m$, the combined kinetic energy is $\frac{1}{2}(m+M)U^2$. Accelerating the body requires accelerating the surrounding fluid, producing the inertial reaction $-M\dot{U}$ along the direction of motion. This is the added mass of a circular cylinder.

Write $r=x-\bar{x}$, $A(t)={\int }_0^t\alpha (s)ds$, and use the usual reduced rigid-recoil representation $h=Z+r\theta -\alpha r^2$, with zero initial recoil and a smooth stroke returning $\alpha$ and $\dot{\alpha }$ to zero. For this representation,Define the fluid impulse and its first moment byThe force law from part (i) gives, by integration by parts,These are the endpoint momentum flux in elongated-body recoil. PutThe endpoint quantities are needed in addition to the integrals $M_j$ unless further end-shape assumptions are made. Lateral force and yaw torque balance areExpanding gives the requested simultaneous second-order equations, with the retained endpoint fluxes:Substituting $\ddot{Z}=\ddot{\gamma }-U\dot{\theta }$ and integrating once from the initially straight, unrecoiling state yieldsThese are first-order coupled equations for $\gamma ,\theta$ with known forcing.

The claimed mass-independent final angle requires a qualification. LetAfter the stroke, $\alpha =\dot{\alpha }=0$ and $A=A_T={\int }_0^T\alpha dt$. If the post-stroke dynamics are stable, the first-order system relaxes toLet $C=B_0(B_2-M_1)-B_1(B_1-M_0)$. Solving this two-by-two system givesThis is quadratic-bend turning with finite body mass. At the limiting state $\dot{\theta }=\dot{\gamma }=0$, so $\dot{Z}=-U{\theta }_{\infty }$. The forward velocity is $-U$ in the chosen $x$ direction, and its lateral component has precisely the body's slope: the fish glides parallel to itself. The angle is generally reached after a relaxation transient, not necessarily immediately at $t=T$.

The advertised result is obtained in the thin-body approximation that neglects $M_b$ relative to fluid added mass, or in the special geometry $B_1=M_0$:For example, take negligible nose added mass and nonzero tail added mass $m_T$, so $B_j=m_T(L-\bar{x}{)}^j$. In the zero-body-mass approximation, the post-stroke equations for $(\gamma ,\theta -2U{A}_T)$ are homogeneous. The inertia matrix is positive definite by Cauchy-Schwarz inequality, andThe two eigenvalues of ${\mathsf{M}}^{-1}\mathsf{K}$ therefore have positive real parts: their sum and product are positive, whether they are real or conjugate. For $U>0$ the transient decays, proving parallel gliding and the displayed turn in this approximation.

With finite $M_b$, the angle is not a consequence of the printed force law for arbitrary $m(x)$. A concrete stable counterexample uses $L=1$, $\bar{x}=1/2$, and $m(x)=m_{\ast }{x}^2$. Then $M_0=m_{\ast }/3$, $M_1=m_{\ast }/12$, $B_0=m_{\ast }$, $B_1=m_{\ast }/2$, $B_2=m_{\ast }/4$, and $C=m_{\ast }^2/12$. Hencewhich differs from $2U{A}_T$ for positive body mass. This example has positive determinant and positive matrix trace of the relaxation matrix. Another immediate check is that if both endpoint added masses vanish, integrating the force law conserves total lateral momentum. A finite-mass fish cannot then end with $w=0$ and a nonzero new transverse velocity. Thus endpoint and inertia assumptions cannot silently be omitted.

If $Z$ is literally the centre of mass position of a deforming body, use a mass-centered deformation rather than the raw quadratic. For body mass density ${\mu }_b(x)$, its mean is $-\alpha J_2/M_b$, with $J_2=\int {\mu }_b{r}^{2}dx$; in the slender approximation $J_2=I_b$. The reference translation above is then $Z_r=Z+\alpha J_2/M_b$. Likewise the active angular momentum contains $-\dot{\alpha }{J}_3$, where $J_3=\int {\mu }_b{r}^{3}dx$, which vanishes for symmetric mass distribution. These corrections modify derivative forcing during the stroke. Since the shape and its time derivative return to zero, they do not remove the finite-body-mass qualification of the final impulse relation.

The principal physical limitations are small slopes and small total turning angle, prescribed approximately constant forward speed, slowly varying slender geometry, and neglected viscosity, separation and detailed three-dimensional wake interactions. Fast large-amplitude C-starts need more than this linear theory. Neglecting body mass is most reasonable for a laterally thin body whose crossflow added mass greatly exceeds its displaced mass. Stable relaxation also requires an appropriate tail/end geometry; it is not guaranteed for every added-mass profile.

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.

Let $V=4\pi b^3/3$ and $B=Vg({\rho }_0-\rho )/{\rho }_{\mathrm{ref}}$. The four balances describe, respectively, fluid entrainment through the spherical surface at normal speed $\alpha u$, the increase of vertical momentum due to buoyancy, conservation of total buoyancy in a uniform ambient, and motion of the thermal centre. Entrained ambient fluid initially has no vertical momentum, so the momentum equation includes the inertia needed to accelerate it. This model neglects added mass and drag; these effects would alter the numerical similarity coefficients.

With $|\rho -{\rho }_0|\ll {\rho }_0$, the Boussinesq approximation replaces inertial density by ${\rho }_0$ while retaining the density difference in buoyancy. Take ${\rho }_{\mathrm{ref}}={\rho }_0$ to this accuracy. ThenFor a point source, $b(0)=z(0)=0$ and the initial integrated impulse is zero. Therefore $b=\alpha z$ and $Vu=B_{0}t$. Integrating $b^3\dot{b}=3\alpha B_{0}t/(4\pi )$ yieldsThe mass density follows from conserved buoyancy:Thus the radius and rise height grow as $t^{1/2}$, the rise speed decays as $t^{-1/2}$, and the density deficit decays as $t^{-3/2}$. In height coordinates, put $K=4\pi {\alpha }^3/3$ and $C=\sqrt{B_0/(2K)}$; then $V=K{z}^3$ and $u=C/z$.

If a reference density different from the inertial ambient density is retained, $d(Vu)/dt=({\rho }_{\mathrm{ref}}/{\rho }_0)B_0$: replace $B_0$ in the kinematic formulas by $({\rho }_{\mathrm{ref}}/{\rho }_0)B_0$, while retaining the original $B_0$ in the density-deficit formula. This keeps reference-density normalization separate from the physical acceleration.

The point-source spherical thermal similarity is singular at $t=0$ and cannot satisfy the small-density-difference condition arbitrarily close to its formal origin. A finite-radius, zero-impulse regularization has $b=b_s+\alpha z$ and $b^4=b_s^4+3\alpha B_0{t}^2/(2\pi )$, with $u=B_{0}t/V$. This displays the required initial data rather than attributing the singularity to a physical infinite initial speed.

Choose the internal transverse force $S(x,t)$ so that the net internal transverse force on a slice is $[S(x+dx,t)-S(x,t)]$. Its linear momentum balance is thenThe coordinate $x$ labels body slices, so their lateral material acceleration is $h_{tt}$; the fluid transport operator $\mathcal{D}$ must not be substituted for this body acceleration.

Let $G$ be the total internal bending moment, including both passive elastic and active muscular contributions. With the compatible moment orientation, the net couple on the slice is $(G_x-S)dx$. The cross-sectional rotary inertia is smaller than the translational inertia by the square of the transverse-to-longitudinal length ratio, and is neglected in this slender-beam approximation. Slice angular momentum balance therefore gives $S=G_x$. Eliminating $S$ from the slice linear momentum balance provesThis equation is the balance law for an active bending beam for a swimming fish; a constitutive relation for the passive bending moment or the muscle actuation is not needed to derive it. With the force-on-body convention established above it becomesAs a check, constant $m$ and rigid lateral acceleration give the effective inertia $m_b+m$. Combining the printed positive force in part (i) with the minus sign in the beam equation would instead give $m_b-m$. That is the same sign defect, rather than a negative hydrodynamic added mass. If the positive quantity in part (i) is called $F_y$ as a force on the fluid, the beam equation must contain $+F_y$.

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.

Stable stratification is a constraint on motion, not an isolated parcel spring. A light-over-heavy density profile has $N^2=-(g/{\rho }_0){\bar{\rho }}_z>0$. An adiabatically displaced parcel has buoyancy anomaly $b=-N^2{\xi }_z$ relative to its surroundings. If one neglects the pressure response and movement of surrounding fluid, the vertical equation would be ${\ddot{\xi }}_z=-N^2{\xi }_z$. A finite spherical blob cannot satisfy that isolated-oscillator picture: incompressible flow requires surrounding fluid motion, and the pressure perturbation communicates its acceleration to the rest of the fluid.

This can be shown directly with the linear Boussinesq equations. For a uniform background, take a Fourier wavevector $({\mathbf{k}}_h,m)$. Pressure removes the component of the buoyancy force parallel to this wavevector. Taking the divergence of momentum to determine pressure and projecting the vertical component givessoThis pressure projection in a displaced stratified blob produces frequencies less than $N$ except for wavevectors with $m=0$. A localized spherical perturbation contains a range of directions and therefore a range of frequencies. It deforms and emits internal gravity waves, rather than remaining one sphere oscillating coherently at $N$. The added mass of a rigid sphere is a useful inertia analogy, but its numerical correction does not define an exact oscillation frequency for a deformable material blob in stratified fluid.

Internal waves and energy transport. The restoring force supports anisotropic internal gravity waves: their frequency depends on the direction, not just the length, of the wavevector. For a two-dimensional positive-frequency branch, $\omega =Nk/\sqrt{k^2+m^2}$ with $k>0$. Differentiation givesThe vertical group and phase velocities have opposite signs. Wave energy therefore travels along a direction different from the direction of phase propagation, a point essential in selecting radiation conditions for wave-generating obstacles.

For constant $N$, the linear perturbation energy density per reference mass isDot momentum with velocity and multiply buoyancy evolution by $b/N^2$; the $bw$ terms cancel, leaving ${\partial }_t\mathcal{E}+\nabla \cdot (\Pi \mathbf{u})=0$. The second term is the available potential energy stored by displacement against the stable density profile. In an ideal fluid, that energy is exchanged with kinetic energy or transported away, not irreversibly dissipated by buoyancy diffusion.

Steady flows, layers and topography. Stable stratification penalizes vertical displacement and favors flows with small vertical motion when the stratified Froude number $U/(NH)$ is small. The natural vertical displacement scale associated with kinetic energy $U^2$ is of order $U/N$; obstacles requiring much larger displacements can cause blocking and diversion instead of simple overpassing. In the ideal model fluid retains its buoyancy label, so motion tends to follow density surfaces. This inhibits vertical mixing but does not forbid horizontal rearrangement or turbulence powered by a shear flow.

A uniform current crossing topography generates stationary internal waves when its intrinsic frequency lies in the internal-wave range. For horizontal wavenumber $k$, stationarity gives $m^2=N^2/U^2-k^2$. Height-dependent wind and stratification replace this by the Scorer parameter relation derived in Question 1. A decrease of that parameter can produce a turning level, evanescence and vertical trapping, explaining lee-wave trains confined near a lower boundary. The exact channel superpositions in Question 1 also show that nonlinearity does not automatically destroy every wave solution.

Stability of the background density profile does not imply stability of every flow built on it. Sufficiently large waves can overturn density surfaces, and shear can supply energy to disturbances. Ideal advection can create fine structure, but irreversible homogenization of buoyancy in a real fluid additionally involves diffusion or dissipation. A finite-amplitude ideal solution is consequently distinct from a guaranteed stable or physically realizable state.

How rotation changes the picture. With the vertical Coriolis parameter $f$, the corresponding uniform-background inertia–gravity dispersion becomesWhen $N>|f|$, the propagating branch lies between $|f|$ and $N$. Rotation also allows a slow balanced sector: horizontal pressure and Coriolis forces balance, vertical pressure and buoyancy balance, and the circulation can be described through potential-vorticity inversion. Small Rossby number and small stratified Froude number lead to the quasi-geostrophic equations rather than to the fast internal-wave dynamics. Vertical shear is then tied to horizontal buoyancy gradients through thermal-wind balance.

In this balanced sector, a planetary or topographic PV gradient supports Rossby waves, whose restoring mechanism is conservation of PV under displacement across that gradient. Stable stratification couples horizontal levels through the ${\partial }_z[(f_0^2/N^2){\psi }_z]$ term, and the mountain-wave calculation in Question 3 illustrates its vertical propagation filter. Thus stratification can support rapid buoyancy-restored waves, constrain slowly evolving balanced circulation, and shape interactions with boundaries; which description applies depends on the velocity, length, time and rotation scales.

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.

Let the fluid approach a uniform, quiescent state $({\rho }_0,p_0,c_0)$ at infinity. Use the outward unit normal vector $N=\nabla f/|\nabla f|$, pointing from the object into the fluid. Write $H=H(f)$ for the Heaviside step function, $q=\rho -{\rho }_0$, and ${\delta }_S=\delta (f)|\nabla f|$. With the viscous stress tensor ${\tau }_{ij}$, defineThus $T_{ij}$ is the Lighthill stress tensor, and $L_i$ is the surface loading exerted on the fluid. The total force on the object has the opposite sign. Extend $q$ by zero inside the object. Since the boundary is fixed and impermeable, $u\cdot N=0$, and the distributional mass conservation and momentum conservation equations areFor example, the surface contribution to the momentum equation is $(\rho u_i{u}_j+P_{ij})N_j{\delta }_S=L_i{\delta }_S$. Differentiating the first equation in time and subtracting the divergence of the second gives the fixed-body Ffowcs Williams-Hawkings equation:There is no acoustic thickness noise: a fixed impermeable boundary injects no mass. Apply the retarded acoustic Green function $\delta (t-R/c_0)/(4\pi c_0^{2}R)$, where $R=|x-y|$. The supplied surface-delta identity converts the loading term to a surface integral. For an exterior observer, the required integral equation isThis acoustic analogy is an exact rearrangement of the fluid equations; source approximations enter later. In the distant linear acoustic region, $p^{\prime }=c_0^{2}q$.