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A journal bearing supports a rotating axle inside a surrounding cylindrical sleeve with a thin lubricating film. An eccentric gap generates lubrication pressure as rotation transports fluid into and out of narrower regions. A full-film journal bearing model imposes periodic pressure around the entire circumference; a model with cavitation needs additional free-boundary conditions. The two models can predict different loads and dynamics.

Let $\delta \ll a$ be the minimum gap as the spheres approach, and $w=-\dot{\delta }>0$ their normal relative approach speed. For smooth equal-radius spheres the gap has local form $\delta +O(s^2/a)$, so the thin-gap region has lateral scale $\ell \sim (a\delta {)}^{1/2}$. Incompressibility drives a radial escape speed of order $w\ell /\delta$. The lubrication pressure therefore scales asThe available gravitational forcing remains finite. Thus lubrication resistance gives $w=O(W\delta /(\mu a^2))$: the normal approach becomes proportionally slower as the gap closes. In particular, for a bounded coefficient $C$, $\dot{\delta }\ge -C\delta$, so $\delta (t)\ge \delta (t_1)e^{-C(t-t_1)}>0$ at every finite time. Equivalently the estimated time to reach zero contains ${\int }_0^{{\delta }_1}d\delta /\delta$, which diverges. This is how lubrication prevents finite-time collision of smooth spheres.

The small nonzero initial lateral offset breaks the exactly collinear configuration; the spheres can deflect sideways and pass with a positive gap rather than touch. The condition $R>2a$ leaves space for a side-by-side passage. Smooth surfaces, continuum hydrodynamics and the no-slip boundary condition are essential to the noncontact conclusion; the argument is not a model of surface roughness or molecular contact.

Take $p$ as excess pressure relative to the fluid far from the gap, and let ${\dot{h}}_0<0$ for approach. The lower surface of the sphere is $h_0+a-\sqrt{a^2-r^2}$ above the plane. In the dominant region $r=O(\sqrt{a{h}_0})\ll a$, this is the parabolic lubrication gapIts slope is $O(\sqrt{h_0/a})\ll 1$, justifying lubrication theory. Both rigid surfaces have zero radial velocity, so the no-slip boundary condition and radial Stokes equation give the radial volume flux per unit circumferenceAxisymmetric conservation of mass gives $h_t+r^{-1}(rq{)}_r=0$. Symmetry at $r=0$ therefore yields $q=-r{\dot{h}}_0/2$, hence $p_r=6\mu {\dot{h}}_{0}r/h^3$. Matching to $p\to 0$ outside the pressure-bearing region givesExtending the parabolic profile to infinity in this integral is a leading-order matching device; the main pressure and force come from $r=O(\sqrt{a{h}_0})$.

The radial scale $\sqrt{a\hat{h}}$ balances the two contributions to the gap. The time scale $\hat{h}/\hat{v}$ is the time for a typical gap change, and the lubrication pressure scale $\mu a\hat{v}/{\hat{h}}^2$ follows either from the formula above or from $p\sim \mu \hat{v}{\ell }^2/{\hat{h}}^3$ with ${\ell }^2=a\hat{h}$. The elastic surface's upward displacement is $-\eta p$, so the actual gap is increased by $+\eta p$. Thus the dimensionless gap in weakly compliant sphere-plane squeeze flow isA dot now means differentiation with respect to $T$. The dimensionless Reynolds lubrication equation isEven without fluid inertia, the elastic surface has a pressure-dependent normal velocity, so ${\mathcal{H}}_T$ includes $\epsilon P_T$; omitting it would wrongly discard the acceleration term below.

Write $d=H_0+R^2/2$ and abbreviate $H=H_0$, $v={\dot{H}}_0$, $b={\ddot{H}}_0$ during the calculation. The rigid term isAt first order the Reynolds equation givesBecause $RdR=dd$, one integration givesSet $\xi =R^2/(2H)$, so $d=H(1+\xi )$. Dividing and changing variables yieldsThe first three terms are derivatives of $I_3,I_4,I_5$, respectively. Integrating from infinity to enforce $P_1\to 0$ givesThe rational expansion of $I_n$ follows by differentiating $\log [\xi /(1+\xi )]+{\sum }_{j=1}^{n-1}[j(1+\xi {)}^j{]}^{-1}$, obtaining $1/[\xi (1+\xi {)}^n]$, and fixing its value to zero at infinity.

Near $\xi =0$ every $I_n$ has the same singular term $\log \xi$. Its coefficient must vanish to keep the pressure regular, soEquivalently, the integrated radial flux above must vanish at $R=0$; otherwise $P_{1R}$ would have a $1/R$ singularity. Using $I_4=I_3+[3(1+\xi {)}^3{]}^{-1}$ and $I_5=I_4+[4(1+\xi {)}^4{]}^{-1}$ cancels all logarithms and gives the simplified regular expression

The upward hydrodynamic force has scale $\mu a^2\hat{v}/\hat{h}$. Its leading contribution is pressure integrated over horizontal projected area; vertical shear forces are smaller in the lubrication limit. Thus $F=2\pi {\int }_0^{\infty }(P_0+\epsilon P_1+\cdots )RdR$. Use $RdR=H_{0}d\xi$ and ${\int }_0^{\infty }(1+\xi {)}^{-n}d\xi =1/(n-1)$. The two contributions areTherefore

For weight-controlled sedimentation, let the fixed effective load be $F=6\pi A$, $A>0$, and neglect particle inertia as well as fluid inertia. At leading order ${\dot{H}}_0=-A{H}_0$ and ${\ddot{H}}_0=A^2{H}_0$. The first-order bracket evaluated on this trajectory is $-(7/5)A^2/H_0^2$. Solving the force balance perturbatively therefore givesHence weak compliance initially increases the sedimentation speed in the regime where the perturbation is valid. The squeeze pressure indents the coating away from the sphere, widens the flow passage and lowers resistance. This is a load-controlled conclusion on the slow sedimenting branch, not a claim about an arbitrary externally imposed acceleration at startup.

The load-controlled breakdown of weak squeeze-flow compliance occurs when the relative correction $O(\epsilon A/H_0^2)$ becomes order one:The central rigid pressure has size $P_0(0)\sim A/H_0$, so the relative elastic gap change is also $\epsilon P_0(0)/H_0\sim \epsilon A/H_0^2$. Thus breakdown means that elastic deflection is comparable with the nominal undeformed gap. The regular expansion about a rigid plane then ceases to be uniform and the coupled elastic-lubrication problem must be retained. It does not by itself imply contact, fluid inertia or failure of the thin-gap geometry.

Uniform normal injection of speed $V$ through the lower plate of a stationary axisymmetric gap of thickness $h$ and radius $R$ producesWith ambient pressure at the rim, the lubrication pressure is $p-p_0=3\mu V(R^2-r^2)/h^3$, and its integral supports the load $3\pi \mu V{R}^4/(2h^3)$.

In an equal-density, no-slip lubrication approximation, repulsive disjoining pressure $A/(6\pi d^3)$ sustains a premelting film of thickness $d$ beneath a rigid disk of radius $a$. With imposed radial undercooling $\Delta T(1-r^2/a^2)$, local thermomolecular pressure is $P_T=\rho L\Delta T(1-r^2/a^2)/T_m$, and $d^3=A/(6\pi P_T)$. A constant vertical lifting rate $U$ requires inward volume flux $q_r=-Ur/2$. Integrating the resulting lubrication pressure and disk force balance gives the displayed rate, where $W$ is the apparent load, $\mu$ is liquid dynamic viscosity, $L$ is specific latent heat, and $T_m$ is the absolute bulk melting temperature. Positive lifting requires $W<\pi a^2\rho L\Delta T/(2T_m)$. For fixed positive load, the rate rises from zero at that threshold towards $A/(6\pi \mu a^2)$ as undercooling increases. That plateau is an asymptotic limit within this constitutive model, not a claim that continuum-film physics persists at arbitrarily low temperatures.

Incompressible mass conservation in a moving gap givesCombined with the local Couette-Poiseuille flux, this is the Reynolds lubrication equation.

For a thin gap $h$ with tangential wall velocities ${\mathbf{v}}_0,{\mathbf{v}}_h$, lubrication theory and the no-slip boundary condition give the integrated volume fluxConservation of mass gives $h_t+{\nabla }_{\parallel }\cdot \mathbf{q}=0$. This is the Reynolds equation; it determines the lubrication pressure from wall motion and gap geometry.

For an inner sphere of radius $a$, outer radius $a+\Delta$, and signed offset $\alpha \Delta$ along a fixed axis, the leading lubrication pressure generated by offset speed $V=\Delta \dot{\alpha }$ is $p=3\mu a^{2}V/[\alpha {\Delta }^3(1-\alpha \cos \theta {)}^2]+\text{constant}$. Its force on the inner sphere is $-\mathcal{R}V$ along that axis. Remove the spatially constant divergent term before taking $\alpha \to 0$; the resistance then tends to $8\pi \mu a^4/{\Delta }^3$. As $\alpha \to 1$, it is $6\pi \mu a^4/[{\Delta }^3(1-\alpha )]$ to leading order.