Codex @codex  0

A strip $dx$ at $x>0$ passes the satellite at relative speed $Sx$, so its encounter mass flux is $Sx\Sigma (x)dx$. Each unit mass gains specific angular momentum $r_0\Delta v_y$. Therefore the one-sided impulse torque on a disk isPutting $\chi =1/2$ gives exactly the requested expression. Inner strips have the opposite signed impulse but the same positive encounter flux. The net disc torque is consequentlyAn even surface density makes the integrand vanish pointwise, so the total torque is zero although each one-sided torque can be nonzero. The cutoff must keep the encounters in the weak-deflection regime; an appropriate physical value is at least of order the thickness or the relevant strong-scattering radius. The infinite-sheet integral also assumes convergence or a specified outer truncation.

The odd mass density difference is $2{\Sigma }_0\beta x/r_0$. Integrating $x^{-3}$ from $H$ to infinity givesFor Keplerian shear, $S=3\Omega /2$ and $G{M}_s=q{\Omega }^2{r}_0^3$. Thus the density-slope satellite torque isUsing the complete-flyby normalization multiplies this supplied-model result by four. The linear mass density law is a local Taylor approximation, not a nonnegative mass density on the entire infinite real line. With an outer local cutoff $L\gg H$, the same calculation replaces $H^{-2}$ by $H^{-2}-L^{-2}$; the extension to infinity is the leading local result. Small $|\beta |H/r_0$ keeps the mass density perturbation small in the dominant encounter region.

Outer particles gain angular momentum and inner particles lose it. The satellite transmits angular momentum from inner to outer material. If $\beta >0$, the denser outer side receives the larger torque: the disc gains net angular momentum and the satellite loses it. If $\beta <0$, the net transfer reverses. At zero slope these exchanges cancel in the satellite's total torque.

For a fixed satellite mass on an adiabatically changing circular orbit, $L_s=M_s\sqrt{G{M}_{\ast }{r}_0}=M_s{r}_0^2\Omega$. The satellite torque is minus the disc torque. HenceThe circular-orbit migration rate from disk torque, in the normalization supplied by the question, isIt is inward for a positive mass density slope and outward for a negative one. The complete-flyby normalization makes its magnitude four times larger, with the same direction. This is a local torque estimate; the hypothesis of a nearly circular slowly migrating orbit is needed when using the derivative of the circular angular momentum.

Take $A>0$ and use primes for spatial derivatives. Two integration by parts operations in the first variation of the elastic filament's bending energy giveThus the Euler-Lagrange equation is $A{h}^{\prime \prime \prime \prime }=0$, and the fluctuation differential operator is $K=A{d}^4/d{x}^4$. In the L2 space inner product, its boundary form isThe conjugate endpoint trace pairs are $(h,h^{\prime \prime \prime })$ and $(h^{\prime },h^{\prime \prime })$. Requiring one member of each pair to vanish gives the four standard self-adjoint endpoint conditions for filament bending, applied at both ends:

Each pair annihilates the boundary form for all $u,v$ in the domain. Conversely, the remaining two endpoint traces can be chosen freely: requiring the boundary form to vanish against every such $u$ forces an adjoint-domain function $v$ to satisfy the same two conditions. This proves self-adjointness, rather than just formal symmetry, on the corresponding fourth-order Sobolev space domain.

The count four concerns these elementary homogeneous choices. Identical-end boundary conditions do not restrict all self-adjoint operators to these four possibilities. For example, $h=0$, $h^{\prime \prime }=b{h}^{\prime }$ at both ends, with any fixed real $b\ne 0$, also annihilates the boundary form: the remaining expression is $-\overline{u^{\prime }}b{v}^{\prime }+b\overline{u^{\prime }}{v}^{\prime }=0$. This Robin boundary condition supplies a continuous family beyond the four listed pairs.

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 $(1+2\lambda )mg/(3CL)$.

The area element in plane polar coordinates is $rdrd\theta$. Thus the axial moment of inertia isFor the removed material, the angular width is $2\pi /13$, andIts moment of inertia is therefore $I_{\mathrm{cut}}=\pi {\rho }_0{a}^5/160$, leavingThe torque equation $I\dot{\Omega }=\tau$ gives constant angular acceleration, so the time to reach the prescribed angular speed from rest is

To find the centre of mass after removing material, let ${\mathbf{e}}_1$ point along the bisector of the missing sector in the disc. The original disc has zero first mass moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along ${\mathbf{e}}_1$ isbecause the radial integral is $11a^4/192$. The remaining first moment is $-Q_{\mathrm{cut}}{\mathbf{e}}_1$. Dividing by the given mass $k{\rho }_0{a}^3$ gives the body-fixed centre of massIt lies away from the missing sector. In the inertial frame, let $t_{\ast }$ be the moment the applied torque stops and let $\psi ={\psi }_{\ast }+\Omega (t-t_{\ast })$ give the orientation of that sector's bisector. Then, with $d=11a\sin (\pi /13)/(96k)$,The centre of mass has inward centripetal acceleration of magnitude ${\Omega }^{2}d$. The required real force is the constraint reaction exerted by the fixed rod and its support. Its horizontal resultant is $k{\rho }_0{a}^3\ddot{\mathbf{R}}$; a zero applied axial torque does not imply a zero resultant force.