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Two equal point masses can move around the same fixed circle of radius $a$ in diametrically opposite positions. Their stationary centre of mass is the circle's center. A separation $2a$ gives gravitational force $G{m}^2/(4a^2)$ on each mass, and Newton's second law equates it to $ma{\omega }^2$. The common angular speed is therefore fixed by the displayed relation. Opposite tangential velocities produce the same sense of angular rotation.

The Lighthill acoustic analogy for a moving body has a volume acoustic quadrupole, a surface acoustic monopole from acoustic thickness noise, and a surface acoustic dipole from acoustic loading noise. For an impermeable moving surface, with outward normal $n$, normal surface speed $v_n$, and fluid and body normal velocities equal, the surface mass coefficient is ${\rho }_0{v}_n$ and the surface loading is the force exerted on the fluid. Schematically, the density-source equation isThe density perturbation is understood with the chosen interior extension. Permeable-surface versions have additional mass and momentum flux terms. Source approximations must distinguish local thickness radiation from cancellation of its compact net-volume contribution.

With drag ratio $\rho =c_{\parallel }/c_{\perp }$, squared-tangent average $\beta ={\Lambda }^{-1}\int x_s^{2}ds$, and axial wave speed $V$, resistive-force theory gives the displayed axial swimming ratio. Pure axial free swimming additionally requires vanishing cross-tangent average, as in reflection-symmetric waveforms; otherwise a transverse force or translation is present.

For a smooth passing encounter of a force-driven sphere with a fixed sphere, isotropy and reflection symmetry imply $\mathbf{U}=[m_{\perp }(R)I+(m_{\parallel }(R)-m_{\perp }(R))\mathbf{nn}]\mathbf{F}$. Along a force axis $X$, $U_X$ is even and the transverse velocity is odd in $X$. Uniqueness then makes the trajectory fore-aft symmetric and restores the incoming impact parameter downstream. This consequence of kinematic reversibility of Stokes flow remains valid with lubrication resistance, provided the particles remain smooth, noncontacting and subject only to the stated forces. Surface roughness and contact can invalidate these assumptions.

For a translating potential energy profile with equal values at the two ends and linear drag force $-\zeta \dot{x}$, a finite passing trajectory satisfies $v\Delta x=\int {\dot{x}}^{2}dt\ge 0$. Equivalently, since $\int Fdy=0$, $\Delta x=\int u^2/[v(v-u)]dy$. Compact support of the force alone is insufficient. At high speed the displacement is $v^{-2}\int (F/\zeta {)}^{2}dy+O(v^{-3})$.

For inner radius $a$, mean gap $\Delta \ll a$, fixed eccentricity parameter $|\alpha |<1$, gap $h=\Delta (1+\alpha \sin \theta )$, and inner angular velocity $\Omega$, the constant circumferential flux isIt follows by integrating the lubrication theory relation $p_s=6\mu \Omega a/h^2-12\mu q/h^3$ over a full period. The leading holding force is horizontal:Integration by parts computes this force without a closed-form pressure. The fluid couples about the separate cylinder axes obey $C_i+C_o=\alpha \Delta F_x$; moments about a common axis balance exactly at this order.

A freely translating inner axle with prescribed positive angular velocity $\Omega$ and downward load $F>0$ has full-film journal bearing force balance equationswhere $R_r,R_t$ are the squeeze resistance of an eccentric journal bearing. Rotation supplies tangential fluid force $R_t\Omega r/2$. The equilibrium lies horizontally to the right and satisfies $F=R_t\Omega r/2$. Linearization there has imaginary eigenvalues: the prescribed rotating axle continuously supplies energy, so viscous dissipation does not imply settling. For an initially concentric axle, putting $e=r/\Delta$ and $K=6\pi \mu \Omega a^3/{\Delta }^2$ gives the closed orbitThis follows by eliminating time and integrating $x_e+3ex/[2(1-e^2)]=\Omega {\Delta }^2{R}_{0}e/[2F(1-e^2{)}^{3/2}]$. It applies to the leading inertia-free full-film model; adding cavitation, constraints or finite-length leakage changes the prediction.

Gravitational softening regularizes a simulation particle's short-distance force, for example by replacing $-Gm/r$ by $-Gm/\sqrt{r^2+{ϵ}^2}$. It suppresses artificial close encounters and effectively raises the diffusion cutoff to order $\max (ϵ,Gm/u^2)$, but alters the resolved force below $ϵ$. Particle number, softening and duration must be chosen together; a point-particle relaxation estimate alone does not specify a simulation's accuracy.

A fixed sphere in a slowly varying incident Stokes flow requires an external force $\mathbf{F}=-6\pi \mu a{\mathbf{u}}_{\infty }$ and external couple $\mathbf{G}=-4\pi \mu a^3{\mathbf{\omega }}_{\infty }$ to leading order. These follow from Faxén translation law and Faxén rotation law; the incident Stokeslet supplies ${\mathbf{u}}_{\infty }$ and ${\mathbf{\omega }}_{\infty }$. Finite-radius Laplacian terms enter the translational force at the next nonzero order. The signs here refer to applied holding loads, not hydrodynamic loads.

For locally bed-tangent drag force and an upslope angle $a$, a spherical-grain force balance gives ${\tau }_{\mathrm{th}}(a)={\tau }_{\mathrm{th},0}(\cos a+\sin a/{\mu }_s)$. The small-slope term is $\delta {\tau }_{\mathrm{th}}=({\tau }_{\mathrm{th},0}/{\mu }_s){\eta }_x$. A horizontal drag force instead changes the normal contact force and gives a different slope correction.

If Euler's number were $a/b$, choose $n\ge \max (b,2)$ and multiply its factorial-series tail by $n!$. This produces an integer, but the tail is positive and smaller than ${\sum }_{j\ge 1}(n+1{)}^{-j}=1/n<1$. The contradiction proves that the number is an irrational number. The argument illustrates how a rapidly convergent rational series can force a hypothetical rational remainder into an impossible integer interval.

Linear drag opposes the velocity relative to the surrounding medium with magnitude proportional to its speed, where $\alpha >0$. Its power is ${\mathbf{F}}_d\cdot \mathbf{v}=-\alpha |\mathbf{v}{|}^2$, so it dissipates kinetic energy. Under a constant external force, the resulting linear ordinary differential equation relaxes exponentially on the time scale $m/\alpha$.

The dimensional linear friction coefficient relates drag force and velocity, $F_d=-\zeta v$, and has units of force divided by speed. It differs from the dimensionless aerodynamic drag coefficient. For an isolated sphere in Stokes flow, the Stokes drag law gives $\zeta =6\pi \eta a$.

A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

Lubrication theory predicts large viscous resistance when fluid must be squeezed through a narrow gap. For normal approach of smooth curved surfaces with curvature scale $a$ and minimum separation $h\ll a$, the resisting force scales as $\mu a^{2}U/h$. Under a bounded normal force this gives $|\dot{h}|=O(h)$, so ideal smooth surfaces do not make contact in finite time. Tangential and rolling motions have different resistance laws; the normal squeeze law should not be applied to them.