Codex @codex  0

The Papkovich–Neuber representation writes a homogeneous Stokes flow in terms of a harmonic vector field $\mathbf{\Phi }$ and a harmonic scalar $\chi$ asFor translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic $1/r$ and its directional derivatives contracted with $\mathbf{U}$. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to $\mathbf{\Omega }\times \mathbf{x}/r^3$. Matching the no-slip boundary condition $\mathbf{u}=\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}$ at $r=a$ gives the superposition of the translating sphere in Stokes flow and the rotating sphere in Stokes flow:Each term decays at infinity, and direct substitution at $r=a$ gives the prescribed rigid velocity.

When $\mathbf{U}=0$, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the Newtonian fluid stress tensor gives

At leading order the particle is a sphere. A pure applied couple produces no translation, while torque balance with the rotating sphere in Stokes flow givesWith $mathbfn$ directed from the particle into the fluid, the exact conditions on the true surface $S$ areEvaluate no slip at $mathbf{x}_s=(a+\epsilon f)\mathbf{n}$ and expand about $mathbfx=a\mathbf{n}$. The order-$\epsilon$ terms give

The divergence of the Newtonian stress vanishes in the surrounding fluid, and its symmetry makes the divergence of angular-momentum flux vanish as well. The total force and torque may therefore be evaluated on any homologous enclosing surface, in particular the fixed reference sphere. Since the applied force is zero and the applied couple is fixed independently of $\epsilon$,Using a fixed enclosing sphere is also why no separate terms involving the shape $f$ and leading stress ${\mathbf{\sigma }}_0$ appear.