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Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

The exact gap between the plane and the circle isIn the narrow region relevant to lubrication theory,so changes occur over the streamwise scale $x=O(\sqrt{a{h}_0})$; below use $x=\sqrt{2a{h}_0}X$.

Choose the positive $x$ direction to be the direction of the cylinder surface velocity $U=a\Omega$ in the gap, and write $\mu =\rho \nu$ for the dynamic viscosity. The lubrication velocity profile and volume flux per unit axial length areSince $Q$ is constant and the pressure returns to the same ambient value as $x\to \pm \infty$, ${\int }_{-\infty }^{\infty }{p}_{x}dx=0$. ThereforeUsing $h=h_0(1+X^2)$ and the supplied integrals givesin the direction of the cylinder surface motion.

The shear stress exerted by the fluid on the cylinder is opposite toEliminating $p_x$ and inserting $Q=2U{h}_0/3$ giveswhere the sign is relative to $U$. Sincethe resisting torque per unit axial length is

Finally, the ratio of streamwise fluid inertia $U^2/\sqrt{a{h}_0}$ to transverse viscous acceleration $\nu U/h_0^2$ is of orderThe stated restriction makes this lubrication-scale Reynolds number small, justifying neglect of inertia.

Reflection in $y=0$ preserves the prescribed translational velocity and axial angular velocity. A force transforms as a polar vector, whereas a torque transforms as an axial vector; consequently $F_y=G_x=G_z=0$. Reflection in $x=0$ reverses both imposed motions but preserves the normal force component. The linearity of Stokes flow then gives $F_z=0$. Thus only $F_x$ and $G_y$ can be nonzero.

There is a sign inconsistency in the printed question. In right-handed coordinates, $\mathbf{\Omega }=\Omega {\mathbf{e}}_y$ gives a rotational velocity $-\Omega a{\mathbf{e}}_x$ at the sphere's lowest point. The printed shear and force formulas instead use the opposite rotational sense. Below, let $\omega$ be the right-handed component along $+{\mathbf{e}}_y$; the paper's displayed formulas are recovered by setting $\omega =-\Omega$.

In the translating frame, the parabolic gap and leading boundary velocities areThe no-slip boundary condition and lubrication theory giveThe gap is stationary in this frame, so ${\nabla }_{\parallel }\cdot \mathbf{q}=0$. The Reynolds lubrication equation is thereforeSince $h_x=x/a$ and ${\nabla }_{\parallel }^{2}h=2/a$,Differentiating the velocity profile gives the leading shear stressesThe first expression becomes the printed formula after $\omega =-\Omega$.

For the logarithmic lubrication resistance of a sphere near a wall, write $\ell =\log (1/ϵ)$. The logarithmic region is $a\sqrt{ϵ}\ll r\ll a$, where $h\sim r^2/(2a)$. ConsequentlyThese follow from $\int r/hdr=a\ell +O(a)$ and $\int r^3/h^{2}dr=2a^2\ell +O(a^2)$, with a fixed small outer cutoff. The general logarithmic radial numerator is $r^{2n-1}$ over $h^n$; the numerator in the printed integration hint appears to have a typographical error.

The force exerted on the fluid through the upper gap boundary includes pressure acting on its slope:The pressure cancels the $x^2/h^2$ term, givingPressure exerts no torque about the centre of a sphere because its traction is radial. To logarithmic order, the shear has lever arm $-a{\mathbf{e}}_z$, soThe other components vanish by parity: $G_x$ involves an integral of the $xy$ term, and $G_z$ has an integrand odd in $y$. The combined hydrodynamic resistance matrix isIts equal off-diagonal coefficients are required by the Lorentz reciprocal theorem for Stokes flow. It is a positive-definite matrix, as required by viscous dissipation.

In the rotational convention of the printed force formula, set $\Omega =-\omega$ and measure the scalar couple along $-{\mathbf{e}}_y$, so $G=-G_y$. ThenFor sedimentation under the buoyancy-adjusted weight $F$, a uniform sphere is torque-free. Thus $\omega a=U/4$, and the force balance givesThe rotational scalar in the paper's displayed-force convention is $\Omega =-U/(4a)$. The sphere rolls in the right-handed $+{\mathbf{e}}_y$ sense but still slips: its lowest point has laboratory velocity $U-a\omega =3U/4$. The signs and coefficients of the right-handed resistance agree with the rigid-wall terms of Bertin et al., equations (4.4)–(4.5), after reversing the forces and torques there from fluid-on-sphere to sphere-on-fluid.

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark-matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where $I=\mathrm{diag}(I_1,I_2,I_3)$, for example,Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $D\propto a$, and ${\bar{\rho }}_m\propto a^{-3}$. The initial $I$ and $T$ are time independent, soThis is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

Let the heavier block move downward, so no slip gives linear acceleration $a\dot{\omega }$. Newton's equations and the pulley torque equation areEliminating the tensions gives the first-order equationWith $\omega (0)=0$, its solution is

For a thin slice of the solid of revolution, $dM=\rho \pi b(x{)}^{2}dx$ and its axial moment of inertia is $dI=\frac{1}{2}b(x{)}^{2}dM$. Consequently

The torque about the origin isSince $\dot{L}=\tau$, the angular momentum $L=mr\times \dot{r}$ is conserved. The force is the negative gradient of the inverse-square potentialso the conserved energy isDirectly, $\dot{E}=\dot{r}\cdot (m\ddot{r}+kmr/r^3)=0$.

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega |$; it precesses in the positive axial sense when $(I_3-I){\omega }_3>0$ and in the opposite sense when this product is negative.

The work law means that friction exerts a constant opposing torque of magnitude $W$. Until the cone stops,Its initial angular speed is ${\omega }_0=\sqrt{2K_0/I}$, and uniform angular deceleration givesTherefore

The uniform rod is invariant under inversion through its centre. This centrosymmetry makes the translation-rotation blocks of its generalized hydrodynamic resistance matrix vanish: a polar translation velocity changes sign under inversion, whereas an axial angular velocity does not, so an invariant linear coupling between them is impossible.

Gravity acts through the centre of mass and supplies no torque. The remaining rotational resistance is positive definite, so zero applied torque implies zero angular velocity. Thus an initially inclined rod keeps its orientation while settling.

Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.

In a parallel-plate rheometer, a point at radius $r$ on the upper plate moves at speed $\Omega r$. The thin-gap approximation therefore gives the local shear rateDefine the rim valueAn annulus of radius $r$ and width $dr$ has area $2\pi rdr$; its tangential force is $\tau (r)2\pi rdr$, and its moment arm is $r$. The measured torque is consequently

For a generalized Newtonian fluid, $\tau =\eta (\dot{\gamma })\dot{\gamma }$. Changing the integration variable from $r$ to $\dot{\gamma }$ givesThus the requested kernel isMultiplying by ${\dot{\gamma }}_R^3$ and taking a derivative with respect to ${\dot{\gamma }}_R$ turns the upper-limit contribution of the integral into the rim viscosity:ThereforeA measured torque curve and its slope hence determine $\eta$ over the range of imposed rim shear rates.

The experimental law impliesSubstitution givesThe graph is a decreasing rectangular hyperbola with a positive high-rate plateau $2/(\pi \alpha R^3)$ and a $1/\dot{\gamma }$ divergence at the origin. Equivalently,so the inferred material is a Bingham plastic with yield stress $3T_c/(2\pi R^3)$.

Pressure acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder isSubstitution of the pressure gradient givesA torque-free cylinder requires ${\int }_{-\infty }^{\infty }{\tau }_{c}dx=0$. Also,Using $Q=2h_0(V+a\Omega )/3$, the torque condition reduces toHence

The force is central force, so its torque about the star vanishes. The specific angular momentum is therefore constant:Taking the scalar product of $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$ with $\dot{\mathbf{r}}$ givesThus conservation of energy gives the second constant

At exact conjunction the planet's force is radial and has no instantaneous torque, but the approach and departure do not cancel when conjunction is displaced from apoapsis. Write the conjunction longitude as ${\lambda }_c={\varpi }_d+\pi +\delta$. At conjunction,The resonant part of the given evolution law has ${\dot{a}}_d=-8e_d{a}_{d}D\sin \varphi$. Thus a conjunction after apoapsis, $\delta >0$, gives $\sin \varphi <0$ and transfers angular momentum to the dust, increasing $a_d$. A conjunction before apoapsis removes angular momentum, while one exactly at apoapsis has zero secular transfer by symmetry.

Poynting–Robertson drag removes angular momentum, so a trapped grain requires positive resonant torque from the planet. Part (c) shows that conjunction must occur after the dust has passed apoapsis. In the planet's rotating frame, the four-lobed pattern is therefore rotated so that the relevant outer lobe lags the fixed planet in the direction of orbital motion. Equivalently,with the stable phase displaced beyond $\varphi =\pi$ rather than sitting at the torque-free value $\varphi =\pi$.

For two perpendicular rods of length $2L$ joined at one endpoint, the slender-body force density gives translation resistance $CL\mathrm{diag}(3,3,4)$, rotation resistance $C{L}^3\mathrm{diag}(8/3,8/3,16/3)$, and translation–rotation block $C{L}^2\left(\begin{array}{ccc}0& 0& -2\\ 0& 0& 2\\ 2& -2& 0\end{array}\right)$. The opposite block is its transpose. Here $G$ is the torque about the joint and $F$ the force on the fluid. Integrating along each rod proves the entries and makes the symmetry and positivity of a rigid-body resistance matrix explicit.