= Logarithmic lubrication resistance of a sphere near a wall
A sphere of radius $a$ at gap $a\epsilon$ 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 $h=a\epsilon+r^2/(2a)$. 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> gives
$$
p=\frac{6\mu}{5}(U+a\omega)\frac{x}{h^2}.
$$
Integrating the upper-boundary <traction>, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid <force> and <torque>
$$
\begin{pmatrix}F_x\\G_y/a\end{pmatrix}\sim\frac{4\pi\mu a}{5}\log(1/\epsilon)\begin{pmatrix}4&-1\\-1&4\end{pmatrix}\begin{pmatrix}U\\a\omega\end{pmatrix}.
$$
The 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 $3\pi\mu aU\log(1/\epsilon)$. Compare the rigid-wall terms of https://vincent-bertin.github.io/Papers/09_Bertin2022JFM.pdf[Bertin et al., equations (4.4)–(4.5)]; those forces act on the sphere and have the opposite sign.
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