Lighthill elongated-body theory

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.

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.

A deforming free swimmer must satisfy total linear and angular momentum balances. The unknown lateral translation and yaw are its recoil correction. If the reference translation is the true centre of mass, the prescribed deformation must first have its mass-weighted mean removed. Otherwise the reference translation differs from actual centre of mass displacement. The body mass and its hydrodynamic added mass contribute with the same sign to its effective inertial coefficient.

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.

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.

For $G_{xx}=R$, zero bending moment and transverse force at the leading end give $G(x)={\int }_0^x(x-s)R(s)ds$. The two trailing-end free-end bending boundary conditions hold exactly when the displayed two integrals vanish. They express total transverse force and torque balance. Adding unknown rigid translation and yaw to a prescribed deformation supplies the two degrees of freedom of the swimming recoil correction. Its principal inertial matrix is the weighted moment matrix of $1$ and $x-x_c$ with weight $m_b+m$, hence is positive definite for a nondegenerate positive mass distribution.