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A planar freezing front can push a spherical particle steadily when repulsive sphere-plane dispersion force balances lubrication resistance. The thin parabolic gap gives hydraulic drag $6\pi \mu V{a}^2/d$, and balancing it against $Aa/(6d^2)$ gives the displayed speed. This balance assumes a nonretarded interaction, small gap and negligible inertial and other particle forces.

Take an attractive nonretarded molecular potential $U(R)=-\beta /R^6$, with molecular number densities $n_1,n_2$. Its force magnitude is $6\beta /R^7$. For one molecule at height $z$ above a half-space, integration over cylindrical radius and depth givesIts downward force is $\pi n_2\beta /(2z^4)$. A sphere's horizontal cross-section a distance $y$ above its bottom has area $\pi (2ay-y^2)$, so its total attractive force isFor $d_0\ll a$ the near-bottom region dominates. Replacing its area by $2\pi ay$ and extending the upper limit givesThis derives the sphere-plane dispersion force from the pairwise Van der Waals force. Hereafter $A>0$ denotes the magnitude of the effective repulsive interaction across water, rather than the signed attractive Hamaker constant. Its planar pressure magnitude is $\Pi (d)=A/(6\pi d^3)$.

For a nearly planar freezing front, the spherical gap is $d(r)=d_0+r^2/(2a)$. In the particle frame the incoming normal motion has speed $V$. Axisymmetric mass conservation gives the outward gap flux per circumference $q_r=Vr/2$. The lubrication flux law therefore givesIntegrating its axial force givesSteady particle pushing by a freezing front requires $F_{\mu }=Aa/(6d_0^2)$, and henceThe viscous drag is not the isolated-sphere Stokes drag: its divergence as the gap closes is essential.

For the deformed interface let $c=\cos {\theta }_0$ and measure $\theta$ from the sphere's downward pole, as in the original geometry. The height below the remote melting isotherm is $a(\cos \theta -c)$. Under the imposed temperature gradient, the undercooling is $Ga(\cos \theta -c)$. With capillary depression neglected, the local phase-equilibrium pressure isHere $T_m$ is the absolute melting temperature; a constant reference ice density is used in the thermomolecular relation. ThusThe numerical value of $K$ will cancel from the speed relation.

In the thin spherical film, incompressibility and the incoming normal velocity $V\cos \theta$ giveSymmetry supplies zero pole flux. The tangential Couette contribution is smaller by $d/a$ and is negligible at this order. Applying the pressure-driven flux law along arclength $a\theta$ giveswhere pressure is matched to the outer liquid at ${\theta }_0$.

A cap area element is $2\pi a^2\sin \theta d\theta$ and its vertical normal component is $\cos \theta$. Put $y=\cos \theta$. The upward molecular and downward hydraulic forces are thereforeBalancing these forces establishes premelted-film particle pushing:The factorizations are $(1-c{)}^2(c+2)$ and $(1-c{)}^3(c+3)$. They confirm positivity on the physical cap range and provide a numerically better-conditioned shallow-cap expression. Since $h=a(1-c)$, the shallow-cap limit within $d\ll h$ is $V\sim A/(12\pi \mu ah)$; it is not a limit in which the film can be identified with the flat-front gap $d_0$.

The local film formula diverges right at the rim ${\theta }_0$. Accordingly the thin-film calculation is an outer approximation away from a small matching fringe, not a claim of a uniformly thin film at its contact with bulk water. Both integrated stress densities vanish at that rim; in the thin-film limit the fringe contributes only a higher-order correction. Surface-tension curvature, altered thermal gradients or appreciable density-change fluxes would require a different balance.