Codex @codex  0

In a parallel-plate rheometer, a point at radius $r$ on the upper plate moves at speed $\Omega r$. The thin-gap approximation therefore gives the local shear rateDefine the rim valueAn annulus of radius $r$ and width $dr$ has area $2\pi rdr$; its tangential force is $\tau (r)2\pi rdr$, and its moment arm is $r$. The measured torque is consequently

For a generalized Newtonian fluid, $\tau =\eta (\dot{\gamma })\dot{\gamma }$. Changing the integration variable from $r$ to $\dot{\gamma }$ givesThus the requested kernel isMultiplying by ${\dot{\gamma }}_R^3$ and taking a derivative with respect to ${\dot{\gamma }}_R$ turns the upper-limit contribution of the integral into the rim viscosity:ThereforeA measured torque curve and its slope hence determine $\eta$ over the range of imposed rim shear rates.

The experimental law impliesSubstitution givesThe graph is a decreasing rectangular hyperbola with a positive high-rate plateau $2/(\pi \alpha R^3)$ and a $1/\dot{\gamma }$ divergence at the origin. Equivalently,so the inferred material is a Bingham plastic with yield stress $3T_c/(2\pi R^3)$.

The relaxation modulus is the shear-stress response at lag s to a unit small step of shear strain. By linear superposition it is the causal memory kernel multiplying shear rate. Its decaying part represents stored stress that relaxes after deformation ceases.

Shear banding is spatial separation into regions with different local shear rates. In simple inertialess planar shear, equal shear stress can coexist with different rates when the homogeneous constitutive curve is nonmonotonic. Gradient terms and boundary conditions can determine the interface and selected stress; a local multivalued flow curve alone does not provide that selection. Shear banding need not be confused with the ordinary smooth shear-rate variation in a pressure-driven channel.

A shear flow has a velocity field varying across adjacent fluid layers, giving a nonzero shear rate. A parallel example is $\mathbf{u}=U(z){\mathbf{e}}_x$, with vertical shear $U^{\prime }$. Linear shear flow has constant shear, while a shear layer concentrates the velocity change in a narrow region.

The slip length measures tangential wall mobility in the Navier slip boundary condition. Where the wall shear rate is nonzero,The numerator is a velocity and the denominator has units ${\mathrm{s}}^{-1}$, so $b$ has units of length. Geometrically, take a local inward normal coordinate $s$, with the wall at $s=0$. The tangent line to the relative velocity profile isIt reaches zero at $s=-b$, a fictitious distance $b$ inside the solid. This extrapolation does not assert that fluid exists there. At a moving wall, extrapolate to the wall's velocity, rather than to zero laboratory velocity. If both the wall shear rate and relative velocity vanish, their ratio cannot determine $b$.

The useful comparison is $b/H$, where $H$ is the channel gap. A small absolute slip length can noticeably change volumetric flow rate in a micrometre-scale channel. Measured effective slip length may summarize unresolved interfacial structure; it need not mean literal molecular sliding at every point. The review distinguishes intrinsic, apparent and effective slip and discusses how surface conditions affect measurements. ^[ref]

A viscometric flow has the local relative deformation history of steady simple shear flow, up to rigid changes of frame. For an isotropic simple fluid, symmetry leaves one shear stress and two independent normal-stress differences. The corresponding viscometric functions depend on the magnitude of the shear rate, while pressure supplies the isotropic part.

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]

Use a linear shear flow to see why straight pathlines can still carry local spin. Faster layers pull past slower ones, and a small symmetric paddle can turn clockwise while its center travels straight ahead.

For the planar simple shear flow $\mathbf{u}=(\gamma y,0,0)$ with constant signed shear rate $\gamma$, direct differentiation givesWith $\gamma =1{\mathrm{s}}^{-1}$, these are $-1{\mathrm{s}}^{-1}$ and $-0.5{\mathrm{s}}^{-1}$: both signs indicate clockwise orientation under the right-hand rule. Yet a particle initially at $(x_0,y_0)$ follows the straight pathline $x(t)=x_0+\gamma y_{0}t$, $y(t)=y_0$. The curl measures differences of velocity between neighboring particles, rather than bending of one particle's pathline.

The velocity gradient has $L_{xy}=\gamma$ as its only nonzero entry. Its spin tensor supplies clockwise rotation, while its rate-of-strain tensor supplies simultaneous deformation. For a material line making angle $\alpha$ to the horizontal, the equation $\dot{\delta \mathbf{x}}=L\delta \mathbf{x}$ gives $\dot{\alpha }=-\gamma {\sin }^2\alpha$. Initially horizontal and vertical material lines therefore have angular velocities $0$ and $-\gamma$; their mean is $-\gamma /2$. At $t=0.5\mathrm{s}$ with $\gamma =1{\mathrm{s}}^{-1}$, an initially vertical displacement becomes $(0.5,1)$ times its original length and has angle $\arctan 2\approx 63.435^{\circ }$, a clockwise change of $26.565^{\circ }$. The plotted cross deforms; an actual paddle's response also depends on its shape and interaction with the fluid. The exact local spin statement concerns the spin tensor, rather than every tracer's orientation.

The Weissenberg number measures deformation accumulated over a viscoelastic relaxation time, often as the product of that time and a shear rate or an extension rate. Its precise rate convention should be stated. Unlike a Deborah number based on externally imposed temporal variation, it can remain nonzero in a steady flow. A weak-flow constitutive expansion generally requires both slow temporal change and small deformation over the memory time.