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A velocity field with spatially constant velocity gradient $K$. Differences of fluid velocities satisfy $\mathbf{u}(\mathbf{x}+\mathbf{R})-\mathbf{u}(\mathbf{x})=K\mathbf{R}$, even when the uniform translation $\mathbf{b}$ is nonzero. Incompressibility requires $\mathrm{tr}K=0$.

With velocity gradient $A_{ij}={\partial }_i{v}_j$, set $\Omega =(A-A^T)/2$. A vector rigidly rotating with the material has material derivative $D_t\mathbf{p}=-\Omega \mathbf{p}$, so its corotational derivative vanishes. Its unit rotational coefficient is fixed by an objective time derivative; adding $-\xi D\mathbf{p}$ describes the material's flow alignment of a polar order parameter.

Consider steady flow of a Newtonian fluid in incompressible flow with constant dynamic viscosity $\mu >0$ between parallel plates at $z=0$ and $z=H$. The bottom plate is stationary and obeys the Navier slip boundary condition with slip length $b$. The top plate moves at velocity $U>0$ and obeys the no-slip boundary condition. There is no streamwise pressure gradient or body force. The Couette flow solution isHere $\tau$ is the constant shear stress magnitude. The lower wall's Navier slip boundary condition holds exactly because $b{u}^{\prime }(0)=bU/(H+b)=u(0)$. At the top, $u(H)=U$ is exactly the prescribed no-slip boundary condition; no finite upper-wall slip length is imposed in this example.

For a numerical example, stipulate $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $U=1\mathrm{mm}{\mathrm{s}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. The bottom velocity is $0.167\mathrm{mm}{\mathrm{s}}^{-1}$ and the shear stress is $0.0833\mathrm{Pa}$, compared with $0.100\mathrm{Pa}$ when both plates obey the no-slip boundary condition. The extrapolated profile reaches zero at $z=-2\mu \mathrm{m}$.

For fixed $H$, $U$ and $\mu$, $b\to 0$ recovers $u=Uz/H$. As $b\to \infty$, $u\to U$ throughout the gap and $\tau \to 0$: the moving upper wall selects the limiting uniform velocity, while the lower wall approaches a stress-free boundary condition. The dynamic viscosity stays positive; the disappearing shear stress comes from the disappearing velocity gradient.

A short derivation uses unidirectional flow: the convective acceleration vanishes and the Navier-Stokes equation reduces to $\mu u^{\prime \prime }=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.

Use the derivative-index-first velocity gradient $G_{ij}={\partial }_i{u}_j$ and $\omega =G-G^T$. For an objective time derivative the tensor transformation $A^{\ast }=QA{Q}^T$ under $x^{\ast }=Q(t)x+c(t)$ must imply $\mathcal{D}{A}^{\ast }/\mathcal{D}t=Q(\mathcal{D}A/\mathcal{D}t)Q^T$. If $R=\dot{Q}{Q}^T$, then $G^{\ast }=QG{Q}^T-R$ andConsequently the extra terms cancel in $D_{t}A+\frac{1}{2}(\omega A-A\omega )$, proving covariance. With the component-index-first gradient $L_{ij}={\partial }_j{u}_i$, the same physical rate is $D_{t}A-WA+AW$, with spin tensor $W=(L-L^T)/2=-\omega /2$. Changing the gradient convention without changing the commutator sign destroys objectivity.

This differential model separates normalized molecular orientation $B=A/\mathrm{tr}A$ from scalar stretch $\lambda$. With component-first velocity gradient $K$, it evolves by $D_{t}A=KA+A{K}^T-(A-I)/{\tau }_1$ and $D_t\lambda =\lambda B:K-(\lambda -1)/{\tau }_2$. Its relaxed state is $A=I$, $B=I/3$, $\lambda =1$. The displayed constant-relaxation stretch law is an idealization; variants can add saturation or nonlinear relaxation.

The dissipation range contains scales at which kinematic viscosity appreciably affects velocity gradients and converts kinetic energy into heat. In a high-Reynolds-number three-dimensional cascade its characteristic length and time are the Kolmogorov microscales. Intense vortex cores can be intermittent within this range.

For a material line tangent $\mathbf{\ell }$, $D_t\mathbf{\ell }=L\mathbf{\ell }$ with velocity gradient $L_{ij}={\partial }_j{u}_i$. The product rule gives $D_t(\mathbf{\ell }{\mathbf{\ell }}^T)=L\mathbf{\ell }{\mathbf{\ell }}^T+\mathbf{\ell }{\mathbf{\ell }}^T{L}^T$. Consequently the upper-convected derivative vanishes for any transported sum of these outer products. Under the derivative-index-first convention $G=L^T$ the same formula is $D_{t}A=G^{T}A+AG$. The distinction prevents transposition errors in constitutive equations.

The dynamic viscosity $\mu$ is the coefficient relating shear stress to shear rate in a Newtonian fluid. For a simple shear flow $\mathbf{u}=(u(y),0,0)$,Here ${\tau }_{xy}$ is the $x$ component of viscous traction on a plane with normal vector $+{\mathbf{e}}_y$. The shear rate $\dot{\gamma }=du/dy$ has units ${\mathrm{s}}^{-1}$. Thus dynamic viscosity measures the stress required to sustain a given rate of deformation. It is the shear coefficient; the separate coefficient for volume change is volume viscosity. ^[ref]

For a worked Couette flow example, place a Newtonian fluid of constant dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$ between parallel plates separated by $h=0.0010\mathrm{m}$. Hold the lower plate fixed and move the upper plate at $U=0.020\mathrm{m}{\mathrm{s}}^{-1}$. Assume steady flow, no streamwise pressure gradient, and the no-slip boundary condition. ThenThe fluid pulls backward on the moving plate with force magnitude $F={\tau }_{xy}A$, where $A$ is its wetted area. For $A=0.010{\mathrm{m}}^2$ this is $0.020\mathrm{N}$. Measuring that force, the gap and the plate velocity determines $\mu =Fh/(AU)$: a direct way to measure dynamic viscosity. The linear profile and opposing wall tractions are the standard plane Couette flow solution. ^[ref]

The velocity gradient matters, rather than the velocity itself. Adding the same constant velocity to both plates and the fluid leaves $du/dy$, and hence shear stress, unchanged. In uniform flow, every layer translates together and $du/dy=0$. More generally, even a nonzero velocity gradient need not generate viscous stress: solid-body rotation has zero strain-rate tensor, as the viscous stress tensor example shows. Thin gaps amplify shear stress for a fixed plate-speed difference, making dynamic viscosity useful for predicting the force needed to move lubricated surfaces.

Flow separation occurs when near-wall fluid leaves the surface-following pattern of an attached boundary layer. An adverse pressure gradient, with pressure increasing along the external flow, decelerates the slow fluid near the wall particularly strongly. The near-wall velocity gradient can fall to zero and reverse, producing a recirculating region and a detached shear layer.

For a steady two-dimensional Newtonian fluid with no-slip boundary condition, zero wall shear stress is the classical local indicator at separation, although the full flow and geometry determine what happens next. Dye introduced near a wall or motion of surface tufts can reveal the change of flow direction; a velocity profile resolves it more quantitatively. A sudden expansion of a channel is a useful motivating geometry because fluid cannot immediately follow the receding wall. The detached region changes drag and mixing, and invalidates an attached-layer approximation continued through it.

A fluid is a material that flows under sustained shear stress and cannot maintain a static shear stress in its fluid state. Liquids and gases are the usual examples. A Newtonian fluid has deviatoric stress proportional to its rate-of-strain tensor, whereas more general fluid constitutive equations can be nonlinear or have memory. In a porous medium, the fluid occupies connected voids within a load-bearing solid framework. The important distinction is the response to a continuing tangential force: an ordinary elastic solid can settle into a fixed deformation, while an ordinary fluid continues to deform. A fluid can nevertheless be at rest when no unbalanced shear stress drives it. ^[ref]

For a worked example, place a Newtonian fluid between parallel plates separated by $h=1\mathrm{mm}$, keep the lower plate fixed and move the upper one at $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$. With no-slip boundary conditions, steady flow, no imposed pressure gradient and negligible edge effects, Couette flow has $u(y)=Uy/h$. For dynamic viscosity $\mu =0.001\mathrm{Pa}\mathrm{s}$, the shear stress magnitude isThe velocity gradient describes neighbouring layers sliding at different speeds. Doubling $U$ doubles the required shear stress; maintaining the same velocity keeps producing deformation, rather than fixing a final shape. This gives viscosity a concrete meaning: resistance to a rate of motion, not a prohibition on motion.

A homologous spherical flow has velocity proportional to radius. Its velocity gradient is $q(t)I$ and its velocity divergence is $3q(t)$.

Let $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ be the unnormalized conformation tensor, $G_{ij}={\partial }_j{u}_i$ the velocity gradient, and $M/\zeta$ a symmetric positive hydrodynamic mobility matrix independent of the individual connector. The connector Fokker-Planck equation has drift $G\mathbf{R}-(2H/\zeta )M\mathbf{R}$ and diffusion matrix $(2k_{B}T/\zeta )M$. Integrating its second moment twice by parts gives the displayed upper-convected derivative, $C^{\nabla }=D_{t}C-GC-C{G}^T$. If $M$ is a polynomial in $C$, the two matrices commute and the relaxation term simplifies to $-(4H/\zeta )M(C-k_{B}TI/H)$. No Gaussian distribution assumption is required for this moment closure.

Represent one polymer by two beads joined by an ideal linear entropic spring, and stipulate a dimensionless conformation tensor $C$ with equilibrium value $I$. The displayed model uses velocity gradient $A=\mathrm{diag}(\dot{ϵ},-\dot{ϵ},0)$ and positive polymer relaxation time $\lambda$. It omits finite chain extensibility, interactions with other chains, and feedback from the polymer onto the flow. It is a teaching model, not a fitted description of Keller's apparatus.

Define the Weissenberg number for this model as $\mathrm{Wi}=\lambda \dot{ϵ}$ and dimensionless time $u=t/\lambda$. With an initially isotropic $C=I$, the diagonal entries obeyBelow $\mathrm{Wi}=1/2$, both transverse and longitudinal responses approach positive finite steady values:For a chosen $\mathrm{Wi}=0.4$, $C_{xx}=5$ and $C_{yy}=5/9$: the longitudinal mean-square separation is five times its quiescent value, while the transverse separation contracts. A root-mean-square length grows as the square root of a variance, so the longitudinal length factor is $\sqrt{5}$, not five. The anisotropy $C_{xx}-C_{yy}$ is an orientational indicator; converting it to measured birefringence requires an additional calibrated optical relation.

At $\mathrm{Wi}=1/2$, $C_{xx}=1+u$. Above that value, the growing homogeneous solution prevents a positive finite steady state. For instance, $\mathrm{Wi}=0.6$ gives $C_{xx}=6e^{u/5}-5$. Real chains have finite contour lengths, so this growth identifies the failure of the linear spring model. It does not establish a universal threshold for arbitrary flow histories, finite chains or other definitions of polymer relaxation time. Combining the growth equation with residence time in planar extension also explains why a large instantaneous Weissenberg number alone does not guarantee a strongly stretched exiting chain.

Represent a dilute polymer by two beads connected by a spring obeying Hooke's law, with stiffness $H>0$ and bead drag coefficient $\zeta >0$. Neglect bead inertia and hydrodynamic interactions. For component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the configuration Fokker-Planck probability current isThe factor two comes from relative motion of both beads. At rest its normalized probability density is ${\psi }_{\mathrm{eq}}=(H/(2\pi k_{B}T){)}^{3/2}{e}^{-H{R}^2/(2k_{B}T)}$. The conformation tensor $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ has dimensions of length squared; $(H/(k_{B}T))C$ is its dimensionless version. The Hookean dumbbell in planar extension is a particular imposed flow in this model.

For component-first velocity gradient $L_{ij}={\partial }_j{u}_i$, ideal incompressible magnetic field stretching gives the displayed evolution of $M_{ij}=B_i{B}_j$. It has the same upper-convected derivative as a polymer conformation tensor with negligible relaxation. Finite polymer relaxation time introduces a local return toward thermal equilibrium, whereas magnetic diffusion acts through spatial gradients and preserves the solenoidal magnetic-field constraint. The analogy does not identify those dissipative mechanisms.

A material volume element contains the same fluid particles as it moves and deforms. If its edges transform with the local velocity gradient, its volume changes at a rate equal to the trace of that gradient, giving $D(\delta V)/Dt=(\nabla \cdot \mathbf{u})\delta V$. Mass conservation states $D(\rho \delta V)/Dt=0$, not that its volume is always fixed. These two identities give the material form of the continuity equation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$.

An objective spatial second-rank tensor transforms by the displayed rule under every rigid observer change $x^{\ast }=c(t)+Q(t)x$. This permits time-dependent rotations, not merely constant rotations. Cauchy stress tensor is objective because contact traction and surface normal both rotate by Q. The symmetric rate-of-strain tensor is objective because the additional observer spin in the velocity gradient is antisymmetric.

Label a trajectory by $\mathbf{a}$ and write its flow map as $\mathbf{X}(\mathbf{a},t)$, with ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. An infinitesimal material line joining neighboring labels has tangent ${\ell }_i=F_{ij}d{a}_j$, where $F_{ij}=\partial X_i/\partial a_j$ is the deformation gradient. Differentiating at fixed labels givesThis identity holds for an unnormalized material line tangent even in a compressible flow. Incompressibility also makes an advected magnetic field obey the same stretching law. A unit tangent would instead require a projection removing its change of length.

Use the source's velocity gradient convention $G_{ij}={\partial }_i{u}_j$. For a outer product $A_{ij}={\ell }_i{\ell }_j$, the product rule yieldsThe same law holds for a sum or locally transported average of such outer products. Thus the dyadic transport of material line elements motivatesThis upper-convected derivative vanishes for a symmetric tensor transported solely by stretching of material lines. It subtracts both advection and geometric deformation, allowing a constitutive equation to isolate intrinsic relaxation or production. More generally $\mathsf{A}=F{\mathsf{A}}_{\mathrm{ref}}{F}^T$ gives $\stackrel{▽}{\mathsf{A}}=F{\dot{\mathsf{A}}}_{\mathrm{ref}}{F}^T$. Under a time-dependent rigid rotation $Q$, its value transforms as $Q\stackrel{▽}{\mathsf{A}}{Q}^T$, making it an objective time derivative. This is a suitable derivative for this transport type; symmetry alone does not make it the unique possible objective time derivative.

Linearize about a relaxed state $\mathsf{T}=0$ with constant mass density ${\rho }_0$. In the unforced momentum conservation equation ${\rho }_0{D}_t\mathbf{u}=-\nabla p+\nabla \cdot \mathsf{T}$, the $x$ component of the pressure gradient is zero for a freely evolving unidirectional flow independent of $x$. A prescribed external pressure gradient would add a forcing term.

For $\mathbf{u}=u(y,t)\hat{\mathbf{x}}$, incompressibility is automatic and the only nonzero entries of the rate-of-strain tensor are $E_{xy}=E_{yx}=\frac{1}{2}{\partial }_{y}u$. Products of stress and velocity gradient are quadratic in the small amplitude. The constitutive equation consequently becomes $\mathsf{T}+\tau {\partial }_t\mathsf{T}=2\mu \mathsf{E}$ to first order. With $T=T_{xy}=T_{yx}$, its shear component and the $x$ component of momentum conservation areThe diagonal stress components have only homogeneous relaxation at this order and vanish if initially relaxed. In particular, $\alpha$ does not enter this linear response.

When deformation is stopped, the rate-of-strain tensor and velocity gradient vanish, leaving $\mathsf{T}+\tau {\partial }_t\mathsf{T}=0$. HenceThe positive constant $\tau$ is the viscoelastic relaxation time. For slow weak deformation, all the transport and quadratic terms are smaller than $\mathsf{T}$, so $\mathsf{T}\simeq 2\mu \mathsf{E}$. Thus $\mu >0$ is the zero-rate dynamic viscosity, measured in stress times time. The corresponding short-time shear modulus of the Maxwell fluid response is $G_{\mathrm{el}}=\mu /\tau$. The source's $\alpha$ has units of time, since $\alpha \mathsf{TE}$ must have units of stress.