Codex @codex  0

The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

With the right Cauchy-Green deformation tensor $C=F^{T}F$ and the second Piola-Kirchhoff stress tensor $S$, this mixed material stress is generally nonsymmetric. In the material-first nominal stress tensor convention, $PF=SC=M^T$. Thus replacing $P{F}^{-T}$ by $PF$ changes the stress measure. The relation follows from $\mathcal{P}=FS$, without any assumption of material isotropy.

A constitutive equation is frame indifferent when a superposed rigid spatial rotation changes its outputs by their tensor transformation rules. For an elastic Cauchy stress tensor, this requires $\sigma (QF)=Q\sigma (F)Q^T$. It differs from material isotropy, which rotates reference material directions.

Let $B=A^{\mathsf{T}}A$. Since $\det A>0$, $B$ is a positive-definite matrix. The spectral theorem gives a unique principal square root of a positive semidefinite matrixThen $R^{\mathsf{T}}R=U^{-1}B{U}^{-1}=I$ and $\det R=\det A/\det U=1$, so $R$ is a rotation matrix. Define $V=RU{R}^{\mathsf{T}}$. It is a symmetric matrix, is positive definite, and satisfies $V^2=A{A}^{\mathsf{T}}$. Consequently the polar decomposition in continuum mechanics isHere $U$ and $V$ are the right stretch tensor and left stretch tensor.

For the Cauchy stress tensor, use both material isotropy and material frame indifference. Their respective transformation laws areThe first law and $A=VR$ give $\sigma (A)=\sigma (V)$. Combining the two laws gives $\sigma (QV{Q}^{\mathsf{T}})=Q\sigma (V)Q^{\mathsf{T}}$. In a principal stretch basis, $V=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3)$. Each half-turn about a coordinate axis leaves $V$ unchanged; its conjugation changes the signs of the corresponding off-diagonal stress tensor entries. Those entries must therefore vanish. If two principal stretches coincide, rotations within their eigenspace additionally force the stress tensor to be scalar on that eigenspace. Thus the Cauchy stress tensor and left stretch tensor have common principal axes, including at repeated principal stretches. This proves coaxiality of isotropic elastic stress.

The precise symmetry conclusion is equivariance of isotropic principal stresses. Every permutation of the principal stretches can be realized by a signed permutation rotation matrix, sowhen the indexing is permuted consistently. Equivalently, permuting the arguments permutes the labelled principal stresses in the same way. In particular, ${\sigma }_1$ is symmetric in ${\lambda }_2,{\lambda }_3$, and analogous statements hold for ${\sigma }_2,{\sigma }_3$.

For distinct principal stretches, interpolate the three values ${\sigma }_i$ at the three points ${\lambda }_i$:Uniqueness of this interpolation implies that the scalar coefficients ${\beta }_j$ are symmetric functions of the unordered principal stretches; they may be expressed through the elementary symmetric invariants. Smooth isotropic constitutive laws have the corresponding invariant representation across repeated stretches as well. The wording about symmetric functions must be understood in this collective sense: each labelled principal stress need not be symmetric in all three arguments. For example, the objective isotropic law $\sigma =V^2$ has ${\sigma }_i={\lambda }_i^2$, which disproves that stronger interpretation.

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

An isotropic simple fluid has local stress determined by the deformation history of the material element. A viscometric flow is one whose relative deformation history is that of steady simple shear flow, up to rigid changes of frame. At a given element the shear rate is constant along its history, although it can differ between elements. This includes the locally simple-shear kinematics used in standard viscometric measurements.

Use the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$. The first two Rivlin-Ericksen tensors areIn the axes of a steady shear $v_x=sy$, the relative deformation gradient over a lag $q$ is $F=I+qK$, since $K^2=0$. The material history is determined by $s$, and its tensor symmetries exclude $xz$ and $yz$ stress components. The remaining symmetric stress consists of an isotropic part, one shear component, and two independent normal-stress differences. The relevant matrices areThus $I,A_1,A_1^2,A_2$ span the allowed stress components. Material isotropy and reversal of the shear show that their scalar coefficients depend on $|s|=(A_1:A_1/2{)}^{1/2}$. Define the viscometric functions bySolving for the matrix coefficients givesThis is an exact steady-viscometric representation, rather than an assumption that the fluid is everywhere a second-order fluid. The shear viscosity $\mu$ measures the shear resistance, while ${\psi }_1$ and ${\psi }_2$ measure the first and second normal-stress differences generated by shear. Their zero-rate values are understood as limits when these exist. An isotropic pressure changes neither normal difference.

For a flow in the lubrication approximation with gap $H$, longitudinal scale $\ell \gg H$ and speed $U$, the shear-rate scale is $U/H$. The viscous shear force per unit volume scales as $\mu U/H^2$. Quadratic viscoelastic normal stresses scale as ${\psi }_i{U}^2/H^2$; their longitudinal gradients, and the comparable mixed-component corrections, contribute forces of scale ${\psi }_i{U}^2/(H^2\ell )$. Therefore the normal-stress smallness in lubrication criterion isEquivalently, the thin-gap aspect ratio must multiply the shear-based elastic parameter to give a small quantity: $(H/\ell )|{\psi }_i|\dot{\gamma }/\mu \ll 1$. Normal stresses need not themselves be smaller than the shear stress; it is their effect on the leading force balance that matters. If the flow changes on an additional short time scale $T$, the $D_t{A}_1$ term also requires $|{\psi }_1|/(\mu T)\ll 1$ for its neglect. Material coefficients should be evaluated at the characteristic shear rate.

Now take positive cylinder rotation, measure $\theta$ counterclockwise from the rightward horizontal, and let $y=r-a$ increase outward through the film. The tangent has upward component $\cos \theta$, so the tangential gravitational acceleration is $-g\cos \theta$. To leading order in $h/a$, neglect inertia and higher-order hydrostatic or capillary pressure gradients, and write $U=\Omega a$. Tangential force balance and the shear-free free surface givehenceThe wall shear stress is $-\rho gh\cos \theta$. It opposes rotation on the uphill side and assists it on the downhill side.

For a power-law fluid, ${\sigma }_{\theta y}=k|u_y{|}^{n-1}{u}_y$. Its inversion must retain the sign:Integrating with the no-slip boundary condition $u(0)=U$ yieldsThe volume flux per unit axial length is $Q={\int }_0^{h}u(y)dy$. The integrated bracket is $[(n+1)/(2n+1)]h^{(2n+1)/n}$, so the power-law coating on a rotating cylinder hasAt $\cos \theta =0$ the gravitational term vanishes by continuity. Steady mass conservation makes $Q$ independent of $\theta$.

On the uphill half, put $D=(\rho g\cos \theta /k{)}^{1/n}$ and $m=(2n+1)/n$. The local flux is $Q(h)=Uh-(D/m)h^m$, withIt has a unique maximum atA positive coating must carry the same steady flux at every angle. The tightest local bound occurs at $\cos \theta =1$, not on the downhill side, where the flux increases monotonically with thickness. ConsequentlyFor $0<Q<Q_{\max }$ the thinner positive root on the uphill side connects through $\cos \theta =0$ to the unique downhill root, giving a periodic leading-order coating profile. At the limiting flux the two uphill roots meet at $\theta =0$; it is the maximum flux of this lubrication model, with a possible limiting corner resolved by neglected higher-order effects. For $n=1$ the check is $Q_{\max }=(2/3)U\sqrt{kU/(\rho g)}$. All these conclusions require the resulting film to remain thin enough for the lubrication approximation and for the neglected normal stresses and inertia to satisfy their scaling restrictions.

Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

Use both material isotropy and material frame indifference. The polar decomposition in continuum mechanics writes $F=VR$, where $V$ is the left stretch tensor. Isotropy removes the reference rotation: $\sigma (F)=\sigma (V)$. Choose an orthonormal eigenbasis of $V$. A half-turn about any principal axis leaves $V$ fixed but reverses appropriate off-diagonal entries of $\sigma (V)$. Frame indifference therefore makes those entries zero. If a principal stretch is repeated, rotations in its eigenspace additionally force the stress to be isotropic there. Thus the principal stresses share the stretch axes: this proves coaxiality of isotropic elastic stress with $V$ and with $B=V^2=F{F}^T$.

For simple shear, the left Cauchy-Green deformation tensor isIts principal axes make an angle $\vartheta$ with the coordinate axes satisfying $2\cot (2\vartheta )=\gamma$ when $\gamma \ne 0$. If the two principal stresses are $s_1,s_2$, rotating their diagonal matrix back givesConsequently the universal simple-shear normal-stress identity isEquivalently, it follows immediately from the off-diagonal entry of $\sigma B-B\sigma =0$. At $\gamma =0$, the undeformed material has isotropic stress, so the identity holds there too. No particular elastic constitutive equation has been used.

For simple shear, $B=\left(\begin{array}{cc}1+{\gamma }^2& \gamma \\ \gamma & 1\end{array}\right)$. The off-diagonal entry of $\sigma B-B\sigma$ is $\gamma ({\sigma }_{11}-{\sigma }_{22})-{\gamma }^2{\sigma }_{12}$. Coaxiality of isotropic elastic stress therefore proves the identity for $\gamma \ne 0$. At $\gamma =0$, material isotropy makes the stress isotropic, proving the same identity.