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A symmetric thin viscous sheet with shear-free faces has leading longitudinal velocity independent of thickness coordinate $z$, and incompressibility gives $w=-z{u}_x$. Young–Laplace equation and the Newtonian fluid stress tensor give ${\sigma }_{zz}=-p_a+\gamma h_{xx}/2$ and ${\sigma }_{xx}=-p_a+\gamma h_{xx}/2+4\mu u_x$. The effective longitudinal tension per unit transverse width is $\mathcal{T}=4\mu h{u}_x+\gamma h{h}_{xx}/2-\gamma h_x^2/4+2\gamma$. The last two terms include the horizontal projections of the two surface tension pulls. Force balance gives ${\mathcal{T}}_x=0$, yielding the displayed longitudinal equation; the face kinematic conditions give the displayed conservation of mass equation.

Differentiating the Stokeslet gives its rate-of-strain tensor and vorticity, displayed above for $r=|\mathbf{x}|>0$. The pressure is $p=\mathbf{F}\cdot \mathbf{x}/(4\pi r^3)$. Integrating the Newtonian fluid stress tensor over a sphere enclosing the origin gives traction resultant $-\mathbf{F}$, fixing the point-force normalization.

For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

For Stokes flow without body forces, $\nabla \cdot \mathbf{\sigma }=0$. Symmetry of the Newtonian fluid stress tensor also gives $\nabla \cdot (\mathbf{x}\times \mathbf{\sigma })=0$, interpreted componentwise as the angular-momentum flux. The divergence theorem makes the resultant force and torque the same on any two homologous enclosing surfaces lying in a nonsingular fluid region. In a boundary perturbation of a nearly spherical particle, evaluating each perturbation field on a fixed enclosing surface avoids spurious geometric force terms: the geometric terms from changing the surface normal and area cancel those from evaluating the base stress at the displaced surface.

For $\Phi =U+\alpha aU/r$ and $\chi =\beta a^{3}U\cdot \nabla (1/r)$ in the Unscaled Papkovich–Neuber representation, the Newtonian fluid stress tensor gives the displayed traction at $r=a$. The $\alpha$ contribution is the Stokeslet stress; the $\beta$ contribution is twice the viscosity times the Hessian of its harmonic scalar potential. The prefactor is $6\mu /a$, not $12\mu /a$. The rigid translating-sphere limit is a direct normalization check.

A spherical inviscid bubble moving slowly through a viscous liquid has a dipolar surfactant perturbation. The associated Marangoni stress reduces the tangential velocity from its clean-bubble value towards zero. In the linear regime, with ${\mathbf{u}}_s=A(I-\mathbf{nn})\mathbf{U}$, the consistent Newtonian fluid stress tensor gives $A=-1/[2(1+2\lambda )]$.