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Use the inverse Lorentz transformationAlong the particle's worldline $d{x}^{\prime }=u^{\prime }d{t}^{\prime }$, so the relativistic velocity-addition formula isFor subluminal collinear velocities, put $u/c=\tanh {\phi }_u$, $u^{\prime }/c=\tanh {\phi }_{u^{\prime }}$ and $v/c=\tanh {\phi }_v$. The hyperbolic tangent addition formula, obtained by dividing the supplied sine/cosine addition formulas, givesBecause hyperbolic tangent is injective on the real line,Thus rapidity is additive even though velocity is not.

Each positive increment in the instantaneous rest frame adds the same rapidity $\alpha =\mathrm{artanh}(1/2)=\frac{1}{2}\log 3$. Starting from zero rapidity, iterated collinear boosts with equal rapidity giveEvery finite number of increments leaves $u_n<c$; the speed of light is approached as $n\to \infty$.

Choose the positive-exponential branches from part (i) and set their additive constants to zero. First take $0<b<1$, and defineThen $\tan ({\varphi }_b/4)=e^{\gamma (x-vt)}$ and $\tan ({\varphi }_a/4)=e^{-\gamma (x+vt)}$. The tangent subtraction formula givesConsequently the allowed Sine-Gordon superposition formula produces the smooth fieldThis is the negative of the Sine-Gordon two-kink solution, and hence a two-antikink configuration. The auxiliary seeds have opposite topological charges, but their charges cannot simply be added to infer the charge of the nonlinear two-step Bäcklund transformation. Indeed, the displayed final field tends to $2\pi$ at the left spatial end and $-2\pi$ at the right, so its total topological charge is $-2$.

Let $T=|t|$ become large. Near the right transition, $x=vT+O(1)$, the tangent argument has the asymptotic formso the local field is $-4\arctan e^{\gamma (x-vT)+\log v}$, a single antikink. Near the left transition, the local field is $4\arctan e^{-\gamma (x+vT)+\log v}+o(1)$, again a decreasing antikink. The resulting asymptotic center lines areThus two incoming antikinks with topological charges $(-1,-1)$ and velocities $(+v,-v)$ separate again with exactly the same topological charges and velocities. There is no radiative tail in these asymptotic profiles. Labeling the outgoing objects by their preserved rapidities makes this elastic soliton scattering; labeling the left and right lumps instead describes reflection with exchanged velocities.

For the right-moving soliton, its incoming intercept is ${\gamma }^{-1}\log v$ and its outgoing intercept is $-{\gamma }^{-1}\log v$. The spatial shifts are therefore $\Delta x_{+}=-2{\gamma }^{-1}\log v$ and $\Delta x_{-}=2{\gamma }^{-1}\log v$. Define the soliton time delay as the change in arrival time at a fixed distant spatial point relative to continuation of the incoming straight line, so $\Delta t=-\Delta x/v_{\mathrm{particle}}$. Both objects have the same signed soliton time delay, which is an advance:This is the Sine-Gordon two-kink time advance. In physical coordinates $T_{\mathrm{phys}}=t/m$, the time shift is $2\log v/(m\gamma v)$. The explicit intercepts fix the sign convention unambiguously.

The remaining real parameter choices are covered without changing the calculation. For any $b\ne 0$ with $b^2\ne 1$, put $v_b=(1-b^2)/(1+b^2)$, $u=|v_b|$, $\gamma =(|b|+|b{|}^{-1})/2$ and $ϵ=\mathrm{sgn}(b{v}_b)$. The same choice of zero additive constants givesEach scattered object's topological charge is $-ϵ$, the velocities are $\pm u$, and the signed soliton time delay is $2\log u/(\gamma u)$. If $b=\pm 1$, the superposition coefficient vanishes and this representative is the vacuum; there is no pair of separated moving solitons and no scattering delay to assign. Thus the scattering conclusion requires the nondegenerate case $0<u<1$.

With zero seed, both Bäcklund transformation equations concern ${\varphi }_a/2$. On a nonconstant branch, separation of variables uses $d\log |\tan ({\varphi }_a/4)|=d{\varphi }_a/[2\sin ({\varphi }_a/2)]$, givingFor $C>0$, absorb its magnitude into an additive constant $c=\log C$ in the exponent. A convenient smooth representative isThe constant-phase condition for this traveling profile determines its velocity:The profile is a traveling soliton with velocity $v_a=(1-a^2)/(1+a^2)$, strictly between $-1$ and $1$. The scalar-field vacua on the two sides differ by $2\pi$. With the topological charge convention $Q=[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$, this branch has $Q=\mathrm{sgn}a$: it is a Sine-Gordon kink for $a>0$ and an antikink for $a<0$. Its width is proportional to ${\gamma }_a^{-1}$, and its derivative decays exponentially away from its center, giving a finite-energy field configuration.

For completeness, the classical rest mass in this paper's coupling convention is $8m/\beta$, as derived in Question 2; a Lorentz boost gives energy $8m{\gamma }_a/\beta$. This localized, topologically protected traveling field is the required classical field-theory soliton. A negative $C$ reverses the field and hence the topological charge; $C=0$ gives the vacuum rather than a soliton. Vacuum shifts by $4\pi$ can be made without changing the displayed Bäcklund transformation equations.

Let $k=|k_x|$, $c_s^2=(dP/d\Sigma {)}_0$ and ${\kappa }^2=2\Omega (2\Omega -S)$. Linearize the barotropic closure of a razor-thin disk and shearing sheet about the given state. Write the radial and azimuthal velocity perturbations as $u$ and $v$. Axisymmetry removes advection by the background shear, givingThe softened potential is ${\Phi }_1=-2\pi G{\Sigma }_1{e}^{-kϵ}/k$. Eliminating the velocity components for the compressive branch gives the softened Toomre dispersion relationThe Coriolis/shear combination supplies the radial epicyclic restoring term, pressure supplies the short-wave term, and self-gravity lowers the squared frequency. The full three-variable determinant also has a stationary axisymmetric geostrophic mode, with $u=0$ and azimuthal flow balancing the pressure-plus-gravity gradient. The displayed relation describes the density-wave pair; eliminating by division by $\omega$ must not silently deny that stationary mode.

The conventional complete-flyby impulse approximation uses a straight incoming path $x=x_0$, $y=-S{x}_{0}t$, with $S>0$. Integrate the satellite's transverse acceleration over the whole encounter. For $x_0>0$,The hint's integral is one over a half encounter, and two over the full encounter. The leading longitudinal impulse vanishes by oddness. In weak elastic scattering, conservation of the relative speed $V=S|x_0|$ gives the next-order change along the original velocity as $-\Delta v_x^2/(2V)$. With the original outer flow along negative $y$, this meansThe same signed expression holds for inner particles and has the opposite sign there. This is the complete-flyby gravitational impulse. Its weak-deflection requirement is $G{M}_s/(S^2|x_0{|}^3)\ll 1$. Rotation and tidal dynamics retained throughout the encounter give a more detailed response; this impulse model does not claim to solve the full Hill scattering problem exactly.

The printed coefficient $1/2$ is four times smaller than this complete-flyby value. It is obtained if the single-sided transverse impulse $G{M}_s/(S{x}_0^2)$ is inserted into the same quadratic longitudinal estimate, leaving out the other half. Thus the scaling and sign agree, but that numerical coefficient is not derived by the usual full-encounter prescription. To keep the subsequent requested formulas unambiguous, write the satellite impulse normalizationThe following parts are derived for general $\chi$ and specialize to the supplied value. The complete-flyby one-sided torque coefficient $8/27$ is also the normalization used in the primary coplanar impulse calculation summarized by Chametla and collaborators.

The leading Stokes drift comes from evaluating the oscillatory velocity at the displaced particle position. Taylor expansion givesHere $u_x=-ak\omega e^{kz}\sin \vartheta$ and $u_z=ak\omega e^{kz}\cos \vartheta$. Inserting the initial-position displacements from the preceding solution yieldsIn particular the instantaneous difference is zero at $t=0$, as it must be. The unaveraged constant equality in the PDF is incompatible with its initial labels. Averaging over a period $\mathcal{T}=2\pi /\omega$ removes the oscillatory term:Equivalently, using mean parcel labels and the purely oscillatory displacements gives $a^{2}k\omega e^{2kz}({\sin }^2\vartheta +{\cos }^2\vartheta )$ at this order. This recovers the intended constant result. The period-mean drift is in the wave-propagation direction and decreases as $e^{2kz}$. The corresponding second-order vertical difference is $a^{2}k\omega e^{2kz}\sin (\omega t)$ for initial labels and has zero mean. The resulting horizontal displacement per cycle is $u_s\mathcal{T}$; closed first-order circles do not imply zero second-order transport.

For an incompressible flow, write the Newtonian fluid stress tensor as $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$ is the rate-of-strain tensor. The Stokes equation gives ${\partial }_j{\sigma }_{ij}=0$. Symmetry of the Cauchy stress tensor therefore givesIntegrating and applying the divergence theorem expresses the viscous dissipation as boundary power:At a moving rigid body, the surface velocity is $U+\Omega \times (x-x_c)$. Its boundary-power contribution is $U\cdot F+\Omega \cdot G$, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power givesApply the Lorentz reciprocal theorem to $(u_0,{\sigma }_0)$ and $(u^{\prime },{\sigma }^{\prime })$ in the fluid outside the inclusion. On the outer boundary $u^{\prime }=0$, soWriting $n_p=-n$ for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor $E$ is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of $E$ eliminates its torque. To derive the disturbance, put $r=|x|$, $Q=x\cdot Ex$ and seekThe incompressible flow condition and Stokes equation reduce toA decaying family satisfying these equations is $f=A{r}^{-5}$, $g=B{r}^{-5}-5A{r}^{-7}/2$, $j=2B{r}^{-5}$. The no-slip boundary condition requires $f(a)=-1$ and $g(a)=0$, so $A=-a^5$ and $B=-5a^3/2$. ConsequentlyThe total velocity is zero at $r=a$, and the disturbance decays as $r^{-2}$. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere $u^{\prime }=-u_0=-aE{n}_p$. Contracting the supplied Newtonian fluid stress tensor with $n_p$ makes its two terms proportional to $(n_p\cdot E{n}_p)n_p$ cancel, leaving $\sigma n_p=5\mu E{n}_p$. Hence the extra dissipation due to a rigid inclusion isHere ${\int }_{S^2}{n}_j{n}_{k}d\Omega =4\pi {\delta }_{jk}/3$. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.

If $N_v$ denotes the number of spheres per unit volume, $N_v(4\pi a^3/3)=\varphi$. ThusAdd this to the particle-free viscous dissipation density $2\mu E:E$. Comparing with $2{\mu }_{\mathrm{eff}}E:E$ gives the Einstein viscosity formula for a dilute suspension:to first order in $\varphi$. The hydrodynamic interactions neglected here contribute beyond that dilute order.

For two Stokes flows $(u^{(1)},{\sigma }^{(1)})$ and $(u^{(2)},{\sigma }^{(2)})$ of the same dynamic viscosity in the same region, with no volume force, the Lorentz reciprocal theorem statesIndeed, the divergence of their cross-work difference isThe pressure terms vanish by incompressible flow, and the derivatives of the Cauchy stress tensors vanish by the Stokes equation. The divergence theorem proves the result.

Use the convention that $F,G$ are the force and torque exerted by the body on the fluid. By Linearity of Stokes flow, $(F,G)=\mathsf{R}(U,\Omega )$. Applying the Lorentz reciprocal theorem to two independent rigid motions gives $q_1^T\mathsf{R}{q}_2=q_2^T\mathsf{R}{q}_1$, so the hydrodynamic resistance matrix is a symmetric matrix. The boundary-power identity givesThis is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The no-slip boundary condition would then force $q=0$. Therefore $\mathsf{R}$ is a positive-definite matrix. Reversing to the fluid-on-body force changes the sign of the force law, not the positive resistance coefficients.

For the two rods, take the torque about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\le s\le 2L$, and its slender-body force density isOn the $y$-rod it isIntegrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int sds=2L^2$ and $\int s^{2}ds=8L^3/3$. A convenient dimensionally uniform statement of the complete right-angle two-rod resistance matrix isIn physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $C{L}^2$, and rotation–rotation entries as $C{L}^3$. The planar block has inverseThis verifies the printed hydrodynamic mobility matrix with its third velocity component $L{\Omega }_z$; the TeX aid's ${\Omega }_x$ is an OCR error.

Take laboratory vertical velocity positive upwards. The body axes are $e_1=(\cos \theta ,\sin \theta )$ and $e_2=(-\sin \theta ,\cos \theta )$. In quasistatic sedimentation the force and torque exerted on the fluid equal the gravitational resultants on the body:The out-of-plane block is unforced, so its positive hydrodynamic resistance matrix gives $U_z={\Omega }_x={\Omega }_y=0$. Substitution in the planar hydrodynamic mobility matrix givesandFor $0\le \lambda <1$, this ordinary differential equation has positive right side on $0\le \theta <\pi /4$, with a stable zero at $\pi /4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda >1$ the initial angular velocity is negative, and the stable equilibrium reached from zero is $-3\pi /4$. The heavier-end body turns clockwise through $3\pi /4$; it does not settle at $\pi /4$. More explicitly, with $K=(1-\lambda )mg/(4C{L}^2)$,where the continuous branch has $\theta -\pi /4\in (-\pi ,0)$. At $\lambda =1$, $\theta$ remains zero.

Transforming the translational velocity back to laboratory axes givesFor $\lambda \ne 1$, eliminate time between $U_h$ and the angular velocity:At either limiting orientation $\sin \theta =\cos \theta$. Thus both cases have the same net horizontal displacement,For $\lambda <1$ the drift is monotonically rightwards. For $\lambda >1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt{2})$ at $\theta =-\pi /4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt{2}-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda =1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda \to 1$ is consequently singular.

For the requested sedimentation drift of a weighted two-rod body sketch, a full parametric trajectory follows by also integrating $U_v/\dot{\theta }$. Put $\psi =\theta -\pi /4$ and set the initial height to zero:It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda <1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda >1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda )mg/(3CL)$.