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The zeroth law of turbulence states that at fixed outer velocity $u$ and integral scale of turbulence $\ell$, the mean viscous dissipation per unit mass approaches a nonzero finite value as the Reynolds number $\mathrm{Re}=u\ell /\nu$ becomes large. In an established forward energy cascade, this gives $ϵ\sim C_{ϵ}{u}^3/\ell$, with $C_{ϵ}$ of order one. The small-scale velocity gradients become large enough to offset decreasing kinematic viscosity; this is the turbulent dissipation anomaly.

At the Kolmogorov microscales, the local Reynolds number is of order one, $v\eta /\nu \sim 1$, and the viscous dissipation is $ϵ\sim \nu (v/\eta {)}^2$. Solving these balances givesSubstituting the outer-scale estimate, and omitting order-one constants, givesThe associated eddy turnover time is $\eta /v=(\nu /ϵ{)}^{1/2}$.

Write $\mathbf{\tau }$ for the Kirchhoff stress tensor and $S$ for the second Piola-Kirchhoff stress tensor. Since $\mathbf{\tau }=FS{F}^T$ and $\dot{F}=LF$ for the velocity gradient,Therefore the upper-convected rate as a stress push-forward is

For the simple shear flow, put $\gamma =f_{,2}$ and $g=\dot{\gamma }$. The velocity gradient and rate-of-strain tensor areTheir upper-convected derivative givesDenote the symmetric extra stress by $E={\sigma }^d$, and the relaxation time by ${\tau }_r$ to distinguish it from $\mathbf{\tau }$. The Oldroyd-B model in the question becomesFor the steady constitutive response, $E_{22}=E_{33}=E_{13}=E_{23}=0$, $E_{12}=\mu g$ and $E_{11}=2(\mu -{\mu }_r){\tau }_r{g}^2$. HenceA constant imposed shear rate can still have startup transients; these values describe the steady response.

For circular Couette flow $v(r)e_{\theta }$, the velocity gradient in the orthonormal polar basis $(e_r,e_{\theta })$ isFor completeness, in fixed Cartesian coordinates let $\omega (r)=v(r)/r$ and $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Then $v=\omega Jx$ and $L=\omega J+({\omega }^{\prime }/r)(Jx)\otimes x$. Along a particle, the polar basis rotates at angular speed $\omega$. Its contribution to the material derivative cancels the rigid-rotation part of $L$ in the upper-convected derivative. The rotating-frame reduction of circular viscoelastic shear therefore leavesThe flow direction is now $e_{\theta }$ and the gradient direction is $e_r$. With $g=v^{\prime }-v/r$, the same simple-shear constitutive calculation gives

Angular force balance implies $r^2\mu g$ is constant. Integrating $v^{\prime }-v/r\propto r^{-2}$ gives $v=Ar+B/r$. With $v(a)=V_a$ and $v(b)=0$,The radial force balance is $-p^{\prime }-2(\mu -{\mu }_r){\tau }_r{g}^2/r=0$. Thus the inertialess Oldroyd-B circular Couette pressure isThe constant $p_0$ is fixed by a pressure or normal-traction boundary condition.

For Cartesian components, ${\partial }_j{x}_i={\delta }_{ij}$, the Kronecker delta. Differentiating $r^2=x_i{x}_i$ givesPut $s=k_{\ell }{x}_{\ell }$. Using the product rule and ${\partial }_j{r}^{-q}=-q{x}_j{r}^{-q-2}$ gives the velocity gradientThis is the rate of strain and vorticity of a Stokeslet, apart from the overall physical normalization. Its symmetric and antisymmetric parts of a matrix areHere $E_{ji}=E_{ij}$ and $W_{ji}=-W_{ij}$, and $d=E+W$. Taking the trace gives the divergenceFor the curl, use the Levi-Civita symbol. Because $d_{kj}={\partial }_j{u}_k$, the symmetric part makes no contribution, andThus $\nabla \times \mathbf{u}=2\mathbf{k}\times \mathbf{x}/r^3$. All derivative identities hold away from the singular origin.

Take a steady parallel velocity $(u(y),0)$ and interpret the specified gradient as $p_x=G>0$. The horizontal Navier-Stokes equation reduces to $\mu u^{\prime \prime }=G$: the pressure force opposes positive $x$ motion. The no-slip boundary condition imposes $u(0)=0$, $u(h)=U$. Integrating twice givesThe velocity gradient is $u^{\prime }=G(y-h/2)/\mu +U/h$, an increasing function of $y$. It is negative somewhere in the channel exactly when its lower-wall limit is negative:Under this condition the negative-gradient interval is $0<y<h/2-\mu U/(Gh)$.

The volume flux per unit span isThus the zero-flux Couette-Poiseuille flow requiresThe profile starts at zero, reaches $u=-U/3$ at $y=h/3$, crosses zero at $y=2h/3$, and reaches $U$ at the moving wall. Negative and positive flux contributions cancel, even though the local velocity is nonzero.

An objective time derivative transforms as a physical tensor under a time-dependent rigid change of observer. If $x^{\ast }=Q(t)x+c(t)$, with $Q$ orthogonal and $A^{\ast }=QA{Q}^T$, objectivity requiresThis means the constitutive equation makes the same physical prediction in translating and rotating frames, rather than requiring individual tensor components to be invariant.

The printed plus commutator uses the derivative-index-first velocity gradient, $G_{ij}={\partial }_i{u}_j$. This is the transpose of the component-index-first convention $L_{ij}={\partial }_j{u}_i$ often used for the velocity gradient. In this notation $\omega =G-G^T$, and the rate is the Jaumann derivativeTo check the covariance of the Jaumann derivative, let $R=\dot{Q}{Q}^T$. Direct transformation givesThe observer-rotation terms cancel in the displayed corotational rate. Equivalently, in the $L$ convention the physical spin tensor is $W=(L-L^T)/2=-\omega /2$, and the same rate is $D_{t}A-WA+AW$. One must transpose the gradient and reverse the commutator together; using $L$ in the printed plus-sign expression would not be objective.

Another example is the upper-convected derivative, written consistently in the $L$ convention asThe extra terms in $L^{\ast }=QL{Q}^T+R$ cancel those in $D_t^{\ast }{A}^{\ast }$, so this also transforms covariantly. In contrast, the uncorrected material derivative $D_{t}A$ of a second-rank tensor is not objective: its transformation contains $R{A}^{\ast }-A^{\ast }R$. For instance a constant anisotropic tensor in a stationary fluid has zero material derivative in its original frame, but its components have a nonzero commutator derivative in a rotating observer's frame. The ordinary partial time derivative is likewise not an objective tensor rate.

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ are the symmetric rate-of-strain tensors. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.

In components, with derivatives ${\partial }_j{u}_i$, use the product rule to compute the divergence of the cross-work flux:The pressure term vanishes by incompressibility; the antisymmetric part of the velocity gradient has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. ThusIntegrate over $V$ and apply the divergence theorem. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the Lorentz reciprocal theorem for Stokes flow:This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.

Use the paper's velocity gradient convention $A_{ij}={\partial }_i{v}_j$, so $\Omega =(A-A^T)/2$ and $D=(A+A^T)/2$. In a rigid motion there is rotation but no rate-of-strain tensor: $D=0$. A polar order parameter attached to the material must rotate exactly with it. In this convention its material derivative is $-\Omega \mathbf{p}$, and its corotational derivative of a polar vector must vanish.

Thus the compensating term is $+\Omega \mathbf{p}$ with coefficient exactly one. This follows from an objective time derivative and is independent of material properties. By contrast, the response to $D\mathbf{p}$ describes deformation and flow alignment of a polar order parameter; its coefficient $\xi$ can depend on the material. The transposed velocity gradient convention is essential to the sign of $\Omega$.

The polar molecular field isFor a spatially uniform polar order parameter, $\mathbf{h}=(a+b{p}^2)\mathbf{p}$, so it is colinear with $\mathbf{p}$ and its magnitude is $|\mathbf{h}|=|a+b{p}^2|p$. The scalar factor can be negative; colinearity does not necessarily mean parallel orientation.

For the simple shear flow, the paper's velocity gradient convention givesSubstitute these matrices into $\dot{\mathbf{p}}=-\Omega \mathbf{p}+\xi D\mathbf{p}-\Gamma \mathbf{h}$. Projection along and perpendicular to the polar order parameter yieldsAssume $g\ne 0$, $\Gamma >0$, $b>0$ and $p>0$. The flow-alignment angle in planar shear must satisfyFor $\xi =-1$, use the cosine equation: $\theta =\pi /2$ modulo $\pi$, and the displayed tangent is infinite. The radial equation giveswhere the sign distinguishes the angular branches, and only positive right-hand sides are ordered solutions. Linearizing the angular equation gives $\delta \dot{\theta }=-g\xi \sin 2\theta \delta \theta$. For $|\xi |>1$, the stable angular branch therefore hasIts radial relaxation eigenvalue is $-2\Gamma b{p}^2<0$. At $|\xi |=1$ the angular linearization is marginal and $p^2=-a/b$ requires $a<0$. If $g=0$, the shear restriction disappears and any orientation is allowed for $p^2=-a/b>0$. The printed request for dependence only on $a,b,\Gamma ,g$ omits $\xi$: in general the magnitude necessarily depends on it.

Write the velocity gradient aswhere $\mathbf{\Omega }$ is the spin tensor. Expanding the upper-convected derivative in the structure equation givesFor $\xi =0$, the first four terms reproduce the upper-convected derivative. Settingtherefore givesWith polymeric stress ${\mathbf{\tau }}_p=b_1\mathbf{\alpha }$, this is the Oldroyd-B model. For $c_1>0$, its relaxation time and polymer viscosity areIndeed the total stress obeys

For $\xi =2$, the coefficient of
$\dot{\mathbf{\gamma }}\mathbf{\alpha }+\mathbf{\alpha }\dot{\mathbf{\gamma }}$
is $+1/2$, so the objective derivative becomes the lower-convected derivative. Hencerecovers the Oldroyd-A model, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1{c}_2=0$ give the degenerate Newtonian limit.

Now both plates at $z=0$ and $z=H$ are stationary and both obey the Navier slip boundary condition with the same slip length $b$. Drive steady flow of a Newtonian fluid in incompressible flow by the constant pressure gradient $G=-dp/dx>0$, with constant dynamic viscosity $\mu >0$ and no other streamwise body force. The planar Poiseuille flow profile and its volumetric flow rate per unit width areThe full volumetric flow rate for channel width $W$ is $Q=Wq$ when side-wall effects are negligible. Thus $q$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, while $Q$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-1}$. Denote the value at $b=0$ by $q_0=G{H}^3/(12\mu )$.

The two wall checks expose the normal signs:Both Navier slip boundary conditions therefore hold exactly. The two walls exert backward tangential tractions of magnitude $GH/2$ on the fluid. Increasing $b$ at fixed $G$ adds a uniform velocity offset, leaving the velocity gradient and wall shear stress unchanged. This differs from Couette flow with a slipping lower wall, where upper-wall velocity is fixed and shear stress decreases.

For $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $G=10^3\mathrm{Pa}{\mathrm{m}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, the wall velocity is $10\mu \mathrm{m}{\mathrm{s}}^{-1}$, the centre velocity is $22.5\mu \mathrm{m}{\mathrm{s}}^{-1}$, and $q=1.83\times 10^{-10}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. The enhancement is $q/q_0=2.2$. Equivalently, a prescribed volumetric flow rate needs only $1/2.2$ of the original pressure gradient. This makes slip length useful for estimating pumping requirements in small channels.

At fixed $H$, $G>0$ and $\mu$, $b\to 0$ recovers ordinary planar Poiseuille flow. As $b\to \infty$, the velocity and $q$ grow without bound in this model, although the varying part of the profile stays the same. At exactly zero tangential wall traction, no steady solution can balance this nonzero pressure gradient: $u^{\prime }(0)=u^{\prime }(H)=0$ conflicts with $\mu [u^{\prime }(H)-u^{\prime }(0)]=-GH$. Unlimited enhancement is consequently not a prediction for a real channel; finite wall friction, finite driving capacity and the validity of the assumed steady flow must be assessed. At fixed finite $q$ instead, $G\to 0$ and the profile approaches uniform velocity $q/H$.

A short derivation integrates the reduced Navier-Stokes equation $\mu u^{\prime \prime }=-G$ twice. Equal slip lengths and the opposite inward unit normals give $u=G[z(H-z)+bH]/(2\mu )$. Its integral over the gap supplies the enhancement factor $1+6b/H$.

Using the positive polar molecular field $h_i=\delta F/\delta p_i$ and velocity gradient ${\partial }_i{v}_j$, the reversible order-parameter stress consists of a distortion contribution and rotational and alignment contributions:For a local free-energy density, a representative is ${\Sigma }_{ij}^{(1)}=(f-p_k{h}_k){\delta }_{ij}-f_{{\partial }_i{p}_k}{\partial }_j{p}_k$. The Euler-Lagrange equation verifies its divergence. Integration by parts gives $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j$ for an incompressible displacement without surface work; the corresponding mechanical power has the opposite sign. Pressure contributions may be reassigned in an incompressible flow.

Using the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$, the Rivlin-Ericksen tensors are defined by the displayed recurrence. In particular $A_1$ is twice the rate-of-strain tensor. These tensors transform objectively under time-dependent rigid observer changes. With this convention the upper-convected derivative of $A_1$ is $A_2-2A_1^2$, an identity useful in the slow-flow expansion of a second-order fluid.

For azimuthal velocity $v(r)e_{\theta }$, the velocity gradient in the rotating polar basis is $\left(\begin{array}{cc}0& -v/r\\ v^{\prime }& 0\end{array}\right)$. The basis rotates along a particle at angular speed $v/r$. Accounting for that rotation in the material derivative replaces $L$ by $L-(v/r)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ in the upper-convected derivative. Its remaining shear rate is $v^{\prime }-v/r$, so the local viscoelastic normal-stress difference follows the same constitutive algebra as simple shear flow.

Although the local viscous dissipation is $2\nu S_{ij}{S}_{ij}$, increasingly intense velocity gradients can compensate for vanishing kinematic viscosity. A Burgers vortex illustrates this mechanism: its radius shrinks while its excess integrated viscous dissipation remains finite. Such a local model does not by itself prove the zeroth law of turbulence for a whole flow.

Differentiate the Kirchhoff stress tensor relation $\mathbf{\tau }=FS{F}^T$, using $\dot{F}=LF$ for the velocity gradient. Subtracting the two deformation terms gives the displayed upper-convected derivative. This removes the transport generated solely by deformation of a fixed material stress tensor.

The velocity divergence is the trace of the velocity gradient. It is the local fractional rate of volume expansion of a moving material element.