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A corotational Jeffreys fluid obeys $\stackrel{\circ }{S}+S/\tau =2{\mu }_0\stackrel{\circ }{D}+2{\mu }_{1}D/\tau$, with deviatoric stress $S$, rate-of-strain tensor $D$, Jaumann derivative and relaxation time $\tau >0$. Put $A=S-2{\mu }_{0}D$. In the corotating-frame representation of a Jaumann derivative, ${\tilde{A}}^{\prime }+\tilde{A}/\tau =2({\mu }_1-{\mu }_0)\tilde{D}/\tau$. An integrating factor gives an exponentially weighted history transported between the rotating frames. The infinite-past formula selects histories with $e^{t/\tau }\tilde{A}(t)\to 0$ as $t\to -\infty$; finite initial data also contribute an exponentially decaying homogeneous term.

A diamond anvil cell compresses a sample between opposed diamond tips. The flat diamond anvil culets load a metal gasket whose hole holds the specimen and a pressure-transmitting medium. Transparent diamond admits excitation and returning fluorescence, enabling an internal ruby pressure sensor.^[ref]

The functions in the drawing answer different questions. The drive applies force, the gasket confines the sample, the medium transmits stress, and the ruby pressure scale assigns a local pressure from an optical signal. A large drive reading cannot establish either the sample's pressure or the absence of deviatoric stress. Optical access also permits repeated readings without unloading the sample: a change during a long measurement can be detected rather than silently folded into a material-property curve.

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 constitutive law for ice creep relates deviatoric stress ${\tau }_{ij}$ and strain rate ${\dot{ϵ}}_{ij}$ through an effective stress ${\tau }_e$ and a temperature-dependent rate factor $\mathcal{A}$. The commonly used exponent is near three. Homogeneity of degree $n$ explains the combination $u/N^n$ in pressure-dependent basal sliding laws when geometry and rate-factor scales are fixed.

With tension taken positive, hydrostatic stress is the stress tensor $\mathbf{\sigma }=-p\mathbf{I}$, where $\mathbf{I}$ is the identity matrix. Every plane has the same compressive normal stress, and no tangential traction. More generally,where $\mathbf{s}$ is deviatoric stress. Two samples can share the same mean pressure while experiencing different deviatoric stresses. A stationary liquid cannot sustain equilibrium shear stress; a solid transmitting medium can support it. Consequently, a load divided by surface area does not by itself establish hydrostatic sample conditions. A transmitting medium's state at the actual pressure and temperature, and contact between sample and walls, matter.

Bridgman's Figure 4 compares mild steel fractured in tension at atmospheric pressure with steel stretched further in a liquid at $25000\mathrm{kgf}\mathrm{c}{\mathrm{m}}^{-2}$.^[ref] The imposed tension remains a directional load: the experiment combines confinement with deformation, rather than measuring a hydrostatic stress alone. Its comparison motivates recording the loading path and the surrounding medium as well as pressure; it does not establish a universal fracture threshold for all steels.

A single optical shift represents pressure only to the extent that the calibration's mechanical and thermal conditions apply. Under hydrostatic stress, all directions have the same compressive normal stress. Under deviatoric stress, differently oriented ruby grains can respond differently; spatially varying pressure can also superpose shifted lines. These mechanisms can produce different optical signals at equal mean pressure.^[ref]

The 2001 account describes scanning local line positions and measuring line width over an extended region. The former resolves spatial differences; the latter supplies a rapid warning.^[ref]

Neither a broadened line nor a narrow one uniquely determines the stress tensor. Instrument resolution, temperature variation and unresolved overlapping signals can also affect width. A small observed region can have a narrow line while differing from another region. Compare locations, reference traces and instrument settings before treating width as evidence for a particular mechanical explanation. These are proposed diagnostic checks, not additional historical controls attributed to the 1975 experiment.

Temperature is another independent input. A useful local expansion is $\Delta \lambda \simeq s_{p}p+s_T(T-T_0)$: one measured shift cannot determine both unknowns. Measure or control temperature, or supply a second independently calibrated observable. The coefficients need not remain constant across a large range. Even when a pressure sensitivity transfers over some temperatures, the reference line, line width and practical readout quality can change.

The historical high-temperature account reports degrading pressure measurement quality above about $300{}^{\circ }\mathrm{C}$.^[ref]

Sliding and cavities. Regelation is melting under increased pressure and refreezing when the pressure is released. For ice near its melting point, the Clapeyron equation gives a lower melting temperature at greater pressure. Against an obstacle, ice melts on the high-pressure upstream face, the water passes around it, and it refreezes on the lower-pressure downstream face. Transport of the released latent heat back towards the melting face sustains the process. Together with viscous deformation around obstacles, this permits basal sliding; regelation is especially effective for small obstacles, for which heat conduction distances are short.

Subglacial cavitation occurs where the contact pressure predicted behind an obstacle falls below the water pressure: the ice separates from the bed and a water-filled cavity opens. The effective pressure in subglacial drainage isthe ice overburden pressure minus water pressure. Smaller $N$ facilitates larger cavities and reduces the resisting ice–bed contact. The exponent $n$ comes from Glen's flow law, with strain rate proportional to the $n$th power of deviatoric stress, commonly $n\simeq 3$. Its homogeneity makes $N^n$ the natural velocity scale, yielding $\tau /N=f(u/N^n)$ after including the fixed bed and ice-rheology scales.

Without cavities, material creep gives a drag law with $f(x)\propto x^{1/n}$. Cavities lower the resistance relative to that extrapolation. On a bounded-slope bed the drag is bounded in proportion to $N$; depending on bed geometry, the curve may level off or pass through a maximum and weaken as contact diminishes. The illustrated weakening curve is a qualitative possibility, not a universal formula for all beds.

Channels versus distributed drainage. In Röthlisberger channel drainage, frictional dissipation of flowing water supplies heat to melt the channel walls, while material creep of ice closes them under effective pressure. An area balance has the schematic formwhere $\Psi$ is hydraulic-potential drop per length, $L_f$ is latent heat per mass, and the positive constant $\mathcal{A}$ incorporates ice material creep and geometry. Water-flow resistance relates the flux to channel size and potential gradient. Balancing melting against closure produces the stated increasing relation $N_R(Q)$.

A channel gaining flux can maintain lower water pressure, since larger $N$ means smaller $p_w$. It draws still more water from neighboring routes, melts wider, and causes competing small conduits to lose flux and close. This positive feedback concentrates flow into an arterial network.

A linked cavity drainage system consists of water-filled lee-side cavities connected by narrow passages. Sliding over bed obstacles opens cavities; ice material creep tends to close them. The given decreasing relation $N_K(Q)$ means that carrying more water requires higher water pressure. A route gaining flux then has a higher pressure and diverts water towards neighboring routes. This negative redistribution feedback supports many connected routes and a distributed system.

Coexistence instability. At a coexistence point let $N_R(Q_R)=N_K(Q_K)$ and hold $Q=Q_R+Q_K$ fixed. A loss $\Delta Q_R=-\epsilon$ from the channel gives $\Delta Q_K=\epsilon$. The changes in the two equilibrium effective pressures areConsequentlyIf the bracket is positive, the cavity system's common effective pressure exceeds what the reduced-flux channel can support against material creep closure. The channel shrinks, loses further flux, and closes down. Thus the channel-cavity drainage instability condition is

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

The spin tensor $W$ is skew, and $\dot{Q}{Q}^T=W$ means $\dot{Q}=WQ$ and ${\dot{Q}}^T=-Q^{T}W$. Differentiate all factors to obtain the corotating-frame representation of a Jaumann derivative:Thus the Jaumann derivative is the ordinary derivative of the tensor components in this rotating frame.

Put $S={\sigma }^d$, $\Delta \mu ={\mu }_1-{\mu }_0$, $A=S-2{\mu }_{0}D$ and $\tilde{A}=Q^{T}AQ$, $\tilde{D}=Q^{T}DQ$. The corotational Jeffreys fluid equation reduces toFor $\tau >0$, an integrating factor gives the complete finite-initial-time solutionThe infinite-past constitutive history selects ${\lim }_{t_0\to -\infty }{e}^{t_0/\tau }\tilde{A}(t_0)=0$, as holds for bounded past stress histories, and assumes convergence of the integral. Taking that limit and rotating back provesA general initial stress would retain the displayed homogeneous term. Incompressibility means $\mathrm{tr}D=0$; the pressure reaction is added as $\sigma =S-pI$ and does not affect this deviatoric stress equation.

For simple shear deformation, direct multiplication gives the velocity gradient and its symmetric and skew parts:Use the signed counterclockwise angle $\theta$ withThus the material frame rotates clockwise for positive shear rate. This angle convention is necessary for the signs of the sine terms. Set $\delta =\theta (t)-\theta (s)$. With $H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $R_{\delta }=Q(t)Q^T(s)$, multiplication yieldsSubstitution in the history formula provesThe suppressed third row and column are zero for the selected infinite-past history.

For constant shear rate $g=\dot{\gamma }$, $2\delta =-g(t-s)$. Put $u=t-s\ge 0$. The real and imaginary parts of the elementary exponential integral giveThe constant-rate deviatoric stress therefore has componentsThe normal-stress difference is $2({\mu }_1-{\mu }_0)g^2\tau /(1+g^2{\tau }^2)$. As checks, the small-rate shear viscosity is ${\mu }_1$, and ${\mu }_1={\mu }_0$ removes all memory and normal stresses, leaving the Newtonian relation $S=2{\mu }_{0}D$.

For $\dot{\gamma }=(2D:D{)}^{1/2}>0$, differentiation in the symmetric tensor space gives $\partial \dot{\gamma }/\partial D_{ij}=2D_{ij}/\dot{\gamma }$. Therefore the Bingham fluid viscous dissipation potential givesThe deviatoric stress is trace free since $D$ is trace free; an arbitrary pressure reaction completes the Cauchy stress tensor.

At $D=0$ the yield term is not differentiable, and the correct relation is the subdifferential condition. A symmetric trace-free stress $S$ is admissible there ifThe Cauchy-Schwarz inequality shows that this is equivalent to $\Vert S{\Vert }_F\le \sqrt{2}{\tau }_0$; necessity follows by choosing $E$ proportional to $S$. Hence the no-strain-rate restriction isThe quadratic term has zero derivative at zero and does not change this restriction. This is the unyielded part of the Bingham plastic law.

For flow along the plates, write $y=x_2$, $v=(u(y),0,0)$ and take $G=-p_{,1}>0$ as the pressure-drop magnitude. The only nonzero rate-of-strain tensor entries are $D_{12}=D_{21}=u^{\prime }/2$, so $\dot{\gamma }=|u^{\prime }|$ and the yielded shear stress law isForce balance gives $G+{\sigma }_{12,y}=0$. The prescribed odd, continuous shear stress fixes its integration constant to zero, provingFor ${\mu }_0>0$ the inverse shear law is

The original PDF also calls $u$ odd. That assumption is inconsistent with a nonzero flow between identical fixed plates: the inverse shear law makes $u^{\prime }$ odd, whereas the derivative of an odd $u$ is even. Both can hold only if $u^{\prime }=0$, and the no-slip wall values then give $u=0$. Thus with the literal odd-velocity requirement, there is no admissible flowing solution when $Gh>{\tau }_0$. The standard pressure-driven velocity is even; replacing that erroneous parity yields the intended profile below.

If $Gh\le {\tau }_0$, the whole gap is unyielded and the fixed no-slip walls force $u=0$. Otherwise let $y_0={\tau }_0/G<h$. The plug flow of a yield-stress fluid occupies $|y|\le y_0$. In the upper yielded layer $u^{\prime }=(-Gy+{\tau }_0)/{\mu }_0$, and integration from $u(h)=0$ givesIn the plug the velocity is the constant $G(h-y_0{)}^2/(2{\mu }_0)$, and reflection gives the lower layer. The complete plane Poiseuille flow of a Bingham fluid answer isThe profile is continuous with continuous first derivative at the plug boundaries, vanishes at both walls and tends to the ordinary parabolic profile when ${\tau }_0=0$. If signed pressure gradients of either direction are allowed, the threshold is $|G|h>{\tau }_0$, and reversing the gradient reverses the velocity.

Let the flow direction be $x$, the gradient direction $y$, and the vorticity direction $z$, with simple shear flow $u=(sy,0,0)$ and signed shear rate $s=\dot{\gamma }$. Use the gradient convention established in part (a):At steady homogeneous shear the ordinary time derivative and advection vanish. Put $X={\sigma }_{xx}$, $Y={\sigma }_{yy}$ and $S={\sigma }_{xy}$. The corotational Maxwell fluid law then givesThe $zz$ equation gives ${\sigma }_{zz}=0$, and the unforced $xz,yz$ equations force those components to zero. Solving the three displayed equations gives the steady shear of a corotational Maxwell fluid:The deviatoric stress is indeed traceless. Its isotropic pressure contribution is irrelevant to either shear viscosity or a normal-stress difference.

Consequently the steady shear viscosity isFor $s>0$, $d{\eta }_{\mathrm{sh}}/ds=-2\mu {\lambda }^{2}s/(1+{\lambda }^2{s}^2{)}^2<0$ when $\mu ,\lambda >0$; it decreases with $|s|$ and tends to zero at large rate. This proves shear thinning directly, without confusing a decreasing viscosity with a necessarily decreasing shear stress.

With the standard definitions $N_1={\sigma }_{xx}-{\sigma }_{yy}$ and $N_2={\sigma }_{yy}-{\sigma }_{zz}$,At zero shear both vanish; at small shear they are quadratic, consistently with part (b). The signs and the fact $|N_2|<N_1$ agree qualitatively with many polymeric-fluid experiments. However the fixed ratio $|N_2|/N_1=1/2$ often overestimates the measured second difference and cannot capture variation of that ratio with material and shear rate. For example, measurements on polyethylene show a smaller, rate-dependent relative second normal-stress difference. Thus the qualitative normal-stress signs are reasonable, but the predicted relative magnitude is not generally quantitatively satisfactory; these signs are not a universal statement about every non-Newtonian fluid.

A resistance pressure gauge is a pressure sensor whose electrical resistance is calibrated against pressure. Even a simple local response needs both pressure and temperature. For an original illustrative model definewhere $R_0$ is the reference electrical resistance at $p=0$, $T=T_0$, and zero history offset. The derivatives $\alpha$ and $\beta$ are local sensitivities with units $\mathrm{P}{\mathrm{a}}^{-1}$ and ${\mathrm{K}}^{-1}$, respectively; $h$ is a dimensionless history offset. Constant sensitivities are an approximation over a chosen range, not material constants valid at every pressure.

Choose synthetic coefficients $\alpha =0.020\mathrm{GP}{\mathrm{a}}^{-1}$ and $\beta =2.0\times 10^{-4}{\mathrm{K}}^{-1}$, with $T-T_0=10\mathrm{K}$ and $h=0$. A measured ratio $R/R_0=1.022$ then gives $p=(0.022-0.002)/0.020=1.00\mathrm{GPa}$. Ignoring the temperature change gives $1.10\mathrm{GPa}$, a $0.10\mathrm{GPa}$ error. This model is deliberately stipulated; these numbers are not measured manganin coefficients or Bridgman's data.

If two histories differ by $\Delta h=0.0004$, their outputs at the same pressure differ by $0.0004R_0$; interpreting both with a single history-free curve gives an apparent pressure separation $\Delta h/\alpha =0.020\mathrm{GPa}$. The half separation is $0.010\mathrm{GPa}$. Reversing the drive can additionally cause mechanical friction hysteresis, independently of this sensor model. Record both the electrical calibration history and the loading direction.

For independent small standard measurement uncertainties $u_y,u_T,u_h$ and exact stipulated $\alpha ,\beta$, first-order propagation givesFor $u_y=10^{-4}$, $u_T=0.5\mathrm{K}$, and $u_h=10^{-4}$, this is $u_p=0.00866\mathrm{GPa}$. Real calibration also includes coefficient measurement uncertainties, correlations, reference-point measurement uncertainty, and any unresolved nonhydrostatic loading. A sensor calibrates its local environment; it cannot by itself certify the absence of deviatoric stress throughout the sample.

Construct a synthetic local calibration to show the calculation independently of the historical data. Choose reference pairs $(\Delta {\lambda }_i,p_i)=(1.00,2.69),(2.00,5.42),(3.00,8.09)$, with shifts in $\mathrm{nm}$ and pressures in $\mathrm{GPa}$. The reference zero is already subtracted. Treat the shifts as exact for this fit and give the three reference pressures equal weights. Minimizing the ordinary least squares objectiveand setting its derivative to zero givesThe regression residuals are $(-0.01,0.02,-0.01)\mathrm{GPa}$. Their weighted moment ${\sum }_i\Delta {\lambda }_i(p_i-a\Delta {\lambda }_i)$ vanishes, as required by the least-squares normal equations. An unknown reference offset requires fitting an intercept instead; forcing it to zero can bias the slope. If shift errors matter, this elementary fit is insufficient.

Use the local measurement modelHere $p$ denotes pressure relative to the reference, $a$ is calibrated sensitivity, and $k_T$ is a local wavelength sensitivity to temperature at fixed pressure. Choose $a=2.70\mathrm{GPa}\mathrm{n}{\mathrm{m}}^{-1}$, $k_T=0.010\mathrm{nm}{\mathrm{K}}^{-1}$ and a reading $\lambda -{\lambda }_0=2.00\mathrm{nm}$ at $T=T_0$. The inferred pressure is $5.40\mathrm{GPa}$. These coefficients and readings are stipulated, not measured ruby constants. With the same pressure, a $10\mathrm{K}$ increase in temperature would add $0.100\mathrm{nm}$ to the shift in this model; neglecting it would overestimate pressure by $0.270\mathrm{GPa}$.

Assign independent standard uncertainties $u_a=0.020\mathrm{GPa}\mathrm{n}{\mathrm{m}}^{-1}$, $u_{\lambda }=u_0=0.010\mathrm{nm}$ and $u_T=0.50\mathrm{K}$ for the temperature difference. Here $u_a$ is an independently assigned calibration uncertainty, not an estimate from the regression residuals above; treat $k_T$ as exact in this example. First-order propagation of uncertainty follows by differentiating the measurement model:The four standard contributions are $0.0400,0.0270,0.0270,0.0135\mathrm{GPa}$, giving $u_p=0.0569\mathrm{GPa}$. This is a standard uncertainty, not automatically a $95\%$ interval. Correlated reference and loaded wavelength readings require the corresponding covariance term; common wavelength-axis drift may cancel in their difference. Uncertain $k_T$, nonzero $T-T_0$, model inadequacy and unresolved deviatoric stress require a broader budget. Taking more optical readings reduces some random errors while leaving a common reference-scale error.

For simple shear flow $u=(\dot{\gamma }y,0,0)$, put $s=\dot{\gamma }$ and $\ell =\lambda s$. The steady deviatoric stress isThis follows by setting ${\omega }_{xy}=-s$, ${\omega }_{yx}=s$ in the corotational Maxwell fluid equation: ${\sigma }_{xx}=\lambda s{\sigma }_{xy}$, ${\sigma }_{yy}=-\lambda s{\sigma }_{xy}$, and ${\sigma }_{xy}+\lambda s({\sigma }_{xx}-{\sigma }_{yy})/2=\mu s$. Hence the shear viscosity is $\mu /(1+{\lambda }^2{s}^2)$, which exhibits shear thinning, while $N_1=2\mu \lambda s^2/(1+{\lambda }^2{s}^2)$ and $N_2=-N_1/2$. These differences are independent of the isotropic pressure.

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]

The Von Mises yield criterion depends only on the second invariant of deviatoric stress. With shear yield stress $k$, its surface is ${\sigma }^{\prime }:{\sigma }^{\prime }=2k^2$; the uniaxial yield stress is $\sqrt{3}k$. An associated flow rule gives plastic strain rate proportional to ${\sigma }^{\prime }$, up to a positive normalization of the plastic multiplier.