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Work in the laboratory frame, with the fluid at rest at infinity. SetAt $r=a$, $f=1$ and $g=0$, so the no-slip boundary condition is $\mathbf{u}=\mathbf{U}$ on the translating sphere. As $r\to \infty$, $\mathbf{u}=O(r^{-1})\to 0$ and the gauge-fixed pressure tends to zero.

It is also possible to check the exterior Stokes equations directly. Differentiating the radial functions givesandThus the Newtonian fluid stress tensor $\sigma =-pI+\mu (\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)$ satisfies $\nabla \cdot \sigma =0$ throughout $r>a$.

Let $\mathbf{n}$ point away from the common centre on each concentric sphere. Applying the divergence theorem to the intervening fluid shell, whose inner normal is $-\mathbf{n}$, yieldsThis is the force flux at infinity in Stokes flow. It is legitimate even though the stress tensor tends to zero pointwise, because the area of the large sphere grows as $R^2$.

For an explicit evaluation, put $A=3a/4$ and $\mathbf{n}=\mathbf{x}/R$. The leading terms are $\mathbf{u}=A[\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/R$ and $p=2\mu A(\mathbf{U}\cdot \mathbf{n})/R^2$. Their velocity gradient isAfter symmetrizing and subtracting the pressure contribution,By rotational symmetry, $\int n_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$. HenceThis is Stokes's law, with the force on the sphere opposite its velocity. For a component calculation with $\mathbf{U}=U{\mathbf{e}}_z$, the same integral is $F_z=-(9\mu aU/2)\int {\cos }^2\theta d\Omega$.

Use the force flux at infinity in Stokes flow. With an outward normal from a large sphere, the force on the body is $\mathbf{F}={\int }_{S_R}\sigma \mathbf{n}dS$. Let $C=3a/4$. The leading far-field velocity and pressure areTheir rate-of-strain tensor isIn the Newtonian fluid stress tensor ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}$, the isotropic pieces cancel. Consequently,The second spherical moment integral $\int n_i{n}_{j}d\Omega =(4\pi /3){\delta }_{ij}$ follows from rotational symmetry and ${\sum }_i{n}_i^2=1$. Since $dS=R^{2}d\Omega$, taking $R\to \infty$ gives the Stokes drag law:The omitted dipole contributes a surface integral of order $R^{-2}$ and vanishes in this limit. The sign refers to the fluid's drag on the moving sphere.

The stress tensor of the incompressible Newtonian fluid is $\mathbf{\sigma }=-pI+2\mu e$, where $e=(\nabla \mathbf{u}+\nabla {\mathbf{u}}^T)/2$ is the rate-of-strain tensor. The homogeneous Stokes equations give $\nabla \cdot \mathbf{\sigma }=0$. Apply the divergence theorem to the fluid between the body and a surrounding sphere $S_R$. If ${\mathbf{n}}_B$ points from the body into the fluid, the fluid-domain normal on the inner boundary is $-{\mathbf{n}}_B$. HenceThis is the force flux at infinity in Stokes flow. A nonzero resultant force must be supported by a stress of order $R^{-2}$ on a surface of area of order $R^2$. More precisely, the leading exterior solution is a Stokeslet with force on the fluid $\mathbf{f}=-\mathbf{F}$:Thus $\mathbf{u}=O(R^{-1})$, $p-p_{\infty }=O(R^{-2})$ and $e=O(R^{-2})$. These are leading generic orders for a translating body with nonzero drag; the integral alone does not determine every angular component, and force-free disturbances may decay faster.