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Let $\Delta \rho ={\rho }_p-{\rho }_f>0$. A spherical grain has submerged weightNeglect lift and contact torque, and use a sliding static friction model with normal reaction $R=G$. The inertial quadratic drag iswhere the last equality fixes the convention $\tau ={\rho }_f{U}^2$. Equivalently $U$ is the bed shear velocity and $C_D$ is an effective drag coefficient referred to it. A literal grain-level flow speed can differ from the shear velocity; that conversion must then be absorbed into $C_D$.

Downstream sliding begins when $F_D>{\mu }_{s}R$. Defining the Shields parameter by $\Theta =\tau /(\Delta \rho gd)$, the threshold force balance isThe grain moves downstream above this threshold in the stated sliding model. Real grain motion can instead involve lift, rolling, irregular contacts, or viscous drag; the printed constant belongs to the particular inertial-drag convention and force balance above.

The normal contact force and resisting static friction balance the drag force and submerged weight at impending motion. The right panel uses locally bed-tangent drag force.

By Newton's second law, force is mass times acceleration. Hence its dimensions are

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

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.

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

Substitute $y=y_0+y_F$ into the energy functional and integrate the resulting polynomial:The term linear in $F$ cancels. HenceAs required, $E_0$ is independent of the force and $\mathcal{E}$ is a quadratic function of it.

Taking the derivative of the force-dependent energy givesThus the derivative of the additional minimized internal energy with respect to the applied force equals the resulting endpoint displacement. This is the endpoint force derivative of clamped-free beam energy.

The physical dimension of a quantity records its powers of chosen base dimensions, independently of units. For example, velocity has dimension $L{T}^{-1}$ and force has dimension $ML{T}^{-2}$. Dimensional analysis compares these expressions in physical equations.

A harmonic line force localized at one point of a two-dimensional sheet gives $\hat{\eta }=F/D$ and $\Delta =\gamma D$ in the Fourier transform. The outgoing upper pressure is the displayed amplitude. Dividing by $c_0^2$ converts it to acoustic density perturbation. The method of steepest descent gives cylindrical sound from the saddle point, while crossed zeros of the dispersion relation give guided or leaky modal residues. Retaining $k_0=\omega /c_0$ is essential for dimensionally consistent spreading and prefactors.

A polar vector transforms as ${\mathbf{v}}^{\prime }=Q\mathbf{v}$ under an orthogonal matrix $Q$, including an orientation-reversing transformation. Displacement, velocity and force are polar vectors. An axial vector instead transforms as ${\mathbf{a}}^{\prime }=(\det Q)Q\mathbf{a}$.

The leading uniform stellar radiation pressure is a differential acceleration relative to a planet, of scale $\beta G{M}_{\star }/a_p^2$. Relative to planetocentric gravity at dust radius $a$, its magnitude isThus lying inside a Hill sphere alone does not guarantee small radiative perturbations. Approximately circular tightly bound dust requires this ratio sufficiently small, as well as weak tides; the uniform radiative force can otherwise excite orbital eccentricity or unbind grains.

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.

The congruence is soluble exactly for $p=2$ or $p\equiv 1(\mathrm{mod}4)$. For odd $p$, Fermat's little theorem applied to a prospective root forces $(p-1)/2$ even. Conversely, when $p\equiv 1(\mathrm{mod}4)$, pair opposite factors in $(p-1)!$ and use the Wilson theorem to obtain $(((p-1)/2)!{)}^2\equiv -1(\mathrm{mod}p)$.

For a two-mode tensorial squirmer flow of radius $a$ and a passive equal sphere held at $X=Re$, with $a/R\ll 1$, let $s=e\cdot Be$. Faxén's first law gives holding force $F_h=-9\pi \mu a^{3}se/R^2+O(\mu a^4|A|/R^3+\mu a^5\Vert B\Vert /R^4)$. Its reflected Stokeslet changes swimming velocity by $-9a^{3}se/(4R^3)$. Since this leading force is radial, its vorticity at the swimmer vanishes. The next holding force, generated by the potential dipole, yields rotation $a^{4}A\times X/(4R^6)$. Terms from the passive sphere's finite radius, reflected stresslet and rotlet are needed at higher order. These formulas assume fixed slip and no externally applied swimmer force or torque.

The streaming velocity is the local mean stellar velocity obtained by averaging with the galactic distribution function. In an axisymmetric disk it usually denotes $⟨v_{\varphi }⟩$. It differs from the circular speed, which is set by the force field, and from the velocity dispersion, which measures fluctuations about the mean.

For a grain of mass density ${\rho }_p$ and volume $\mathcal{V}$ in a fluid of mass density ${\rho }_f$, its effective downward force is $({\rho }_p-{\rho }_f)g\mathcal{V}$, after subtracting buoyancy from weight.

Traction is the force per unit area transmitted across an oriented surface. For unit normal vector $\mathbf{n}$, the stress tensor gives $\mathbf{t}=\mathbf{\sigma }\mathbf{n}$. Its normal component is normal stress, and its tangential component is shear stress.