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For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.

Subglacial till is a mixture of rock fragments and fine sediment beneath ice. A wet deformable layer can lubricate basal sliding. Its constitutive equation must be specified; Newtonian till lubrication is one idealization, rather than a claim that all subglacial till is a Newtonian fluid.

For a two-dimensional Newtonian fluid ice sheet attached to an unbuttressed ice shelf, ice-sheet flotation gives $\rho H={\rho }_{w}b$. The difference between the depth-integrated ice and ocean hydrostatic pressures is $\rho g{H}^2/2-{\rho }_{w}g{b}^2/2=\rho g^{\prime }{H}^2/2$, where $g^{\prime }=g(1-\rho /{\rho }_w)$. Shelf force balance integrates $4\mu H{u}_x-\rho g^{\prime }{H}^2/2$ to a constant; absence of a buttressing force makes that constant zero. The shallow-shelf approximation therefore gives the displayed boundary condition, or $u_x=g^{\prime }H/(8\nu )$ with kinematic viscosity $\nu =\mu /\rho$.

A uniform steady film of thickness $d$ on a plane inclined by $\theta$ above horizontal has downslope coordinate $s$ and outward normal coordinate $x$. Neglecting air shear, the pressure is $p_{\mathrm{atm}}+\rho g\cos \theta (d-x)$ and its downslope gradient vanishes. The Newtonian fluid momentum equation is $\eta v^{\prime \prime }+\rho g\sin \theta =0$. The no-slip boundary condition $v(0)=0$ and shear-free free surface $v^{\prime }(d)=0$ give the displayed parabolic profile. Integrating over thickness gives volume flux per unit width $q$. The wall shear $\rho gd\sin \theta$ balances the film's downslope weight, and the vertical-wall case is $\theta =\pi /2$.

A velocity profile records the variation of velocity across a spatial coordinate, for example $u(z)$ across a parallel viscous layer. Its derivative sets the tangential shear stress in a Newtonian fluid, while its integral gives the volume flux.

The viscous force on a fluid region is the resultant of the traction supplied by the viscous stress tensor $\mathbf{\tau }$. With $\mathbf{n}$ the outward unit normal of the region, its surface contribution is the displayed surface integral. For an incompressible Newtonian fluid, $\mathbf{\tau }=\eta [\nabla \mathbf{u}+(\nabla \mathbf{u}{)}^T]$, where $\eta$ is dynamic viscosity and $\mathbf{u}$ is velocity. A force scale is $\eta UL$ for a body of size $L$ moving at speed $U$, while the inertial force scale is $\rho U^2{L}^2$; their ratio is the Reynolds number. Total hydrodynamic drag on a body includes both viscous traction and the pressure traction induced by the flow. For example, the full Stokes drag law gives $6\pi \eta RU$ for a no-slip sphere, rather than just the viscous-traction part alone.

The viscous penetration depth is the distance over which a time-periodic shear velocity driven by an oscillating boundary decays by $1/e$ in a semi-infinite Newtonian fluid. The momentum diffusion equation gives $u=U_0{e}^{-z/\delta }\cos (\omega t-z/\delta )$. Both attenuation and phase lag are controlled by $z/\delta$. The corresponding fluid-displacement amplitude has the same spatial decay, because dividing each velocity amplitude by the common angular frequency does not change their ratio.

The viscous stress tensor is the part of the Cauchy stress tensor associated with deformation rates. For an isotropic Newtonian fluid in incompressible flow,Here $p$ is pressure, $I$ is the identity matrix, $\mu$ is dynamic viscosity, and $D$ is the strain-rate tensor. The convention is $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$; $T$ denotes transpose and $\mathrm{tr}$ denotes trace. Across a plane with unit normal vector $\mathbf{n}$, the viscous traction is ${\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}$. The tensor makes this rule work for any plane orientation. ^[ref]

In the worked Couette flow, $\mathbf{u}=(Uy/h,0,0)$, soIn particular, $D_{xy}=10{\mathrm{s}}^{-1}$ while the shear rate is $20{\mathrm{s}}^{-1}$; the factor $2$ in $2\mu D$ recovers ${\tau }_{xy}=2\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf{n}={\mathbf{e}}_y$ gives viscous traction $+2{\mathbf{e}}_x\mathrm{Pa}$; on the lower face it is $-2{\mathbf{e}}_x\mathrm{Pa}$. Constant stress can therefore transmit a force between the plates even though its divergence, and the net viscous force on each slice, vanish.

Compare planar extension $\mathbf{u}=(ax,-ay,0)$ with solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$:Extension changes lengths and produces viscous normal stresses $\pm 2\mu a$. Rotation preserves distances and gives zero viscous stress, despite its nonzero velocity gradient and vorticity $2\Omega {\mathbf{e}}_z$. For $a=\Omega =1{\mathrm{s}}^{-1}$ and $\mu =0.10\mathrm{Pa}\mathrm{s}$, extension gives ${\sigma }_{\mathrm{visc},xx}=0.20\mathrm{Pa}$ and ${\sigma }_{\mathrm{visc},yy}=-0.20\mathrm{Pa}$; rotation gives zero. A rotating fluid may still have a pressure gradient supplying its centripetal acceleration.

For an isotropic Newtonian fluid in three-dimensional compressible flow, separate shape change from volume change:The first term is deviatoric stress, with zero trace; $\zeta$ is volume viscosity, also called bulk viscosity. For uniform expansion $\mathbf{u}=b\mathbf{x}$, $D=bI$ and $\nabla \cdot \mathbf{u}=3b$, so the shear contribution vanishes and the viscous stress is $3\zeta bI$. Thus dynamic viscosity and volume viscosity describe distinct responses, and incompressible flow removes only the volume-change term. ^[ref]

For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

For a stationary flat plate in a Newtonian fluid, the streamwise shear stress exerted by the fluid on its upper face isThe sign here is positive when the fluid pulls the plate downstream; the plate exerts the opposite tangential traction on the fluid. In a Blasius boundary layer, ${\tau }_w\simeq 0.332057\rho U^2/\sqrt{{\mathrm{Re}}_x}$. A growing boundary layer thickness therefore accompanies decreasing local wall shear stress, because its near-wall velocity gradient becomes gentler. The formal divergence as $x\to 0$ is integrable, but that leading-edge limit lies outside the thin-layer hypothesis.^[ref]