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For the neo-Hookean solid, the axial Cauchy stress tensor under the unloaded lateral boundary conditions isWith $K,\mu >0$, the spring balance is equivalent, after multiplication by $\alpha >0$, to the cubic equationFor a fixed $\gamma \ge 1$, write its left side as $h(\alpha )$. It obeys $h(0)=-\mu <0$ and tends to $+\infty$ as $\alpha \to \infty$. MoreoverThus $h$ first decreases from a negative value and then increases strictly, crossing zero exactly once on $\alpha >0$. For $\gamma >1$,so the unique equilibrium has $1/\gamma <\alpha <1$. The tissue is elastically compressed, while its actual length $\alpha \gamma$ exceeds one and compresses the spring.

To determine the large-growth limit without presuming a power law, solve for $\gamma$:This is a strictly decreasing function of $\alpha >0$, since its derivative isIt tends to $+\infty$ only as $\alpha \to 0^{+}$. Consequently unlimited growth forces $\alpha \to 0^{+}$, andMore precisely, the dominant balance is $K\gamma {\alpha }^3\sim \mu$, givingThe surrounding spring softens the divergence relative to rigid confinement but does not bound the compressive stress. Both axial extension and lateral expansion help accommodate growth:whose product is the total volume $\gamma$, as required by elastic incompressibility and the imposed growth tensor.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

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 the local momentum equation in the current volume:Symmetry of the Cauchy stress tensor and the product rule giveApply the divergence theorem and use $t_j={\sigma }_{jk}{n}_k$. The mean stress and strain boundary identities follow:For infinitesimal strain tensor, $e_{ij}=(u_{i,j}+u_{j,i})/2$, and a second application of the divergence theorem gives

To first order, evaluate self-gravity in the undeformed uniform sphere. The shell theorem says that only the enclosed mass $M(r)=4\pi \rho r^3/3$ contributes to the inward field. ConsequentlyThe field vanishes at the centre and increases linearly to its surface value.

The requested radius change concerns the final traction-free static elastic equilibrium, not the acceleration immediately after gravity is switched on. In equilibrium the boundary traction and acceleration vanish. Spherical symmetry givesThe homogeneous linear elasticity law commutes with volume averaging:Thus $\overline{e}=e_{0}I$ with $e_0=-\rho ga/(15\kappa )$. This conclusion is about the mean strain, not a claim that the local strain is uniform.

Let the boundary displacement be $u(a)=u_{a}n$. Since ${\int }_S{n}_i{n}_{j}dS=(4\pi a^2/3){\delta }_{ij}$, the boundary identity gives $\overline{e}=(u_a/a)I$. Therefore the equilibrium contraction of a self-gravitating elastic sphere isThe shear modulus cancels because the mean deformation is purely volumetric.

The full radial field provides a useful check without assuming uniform deformation. For $u=f(r)e_r$, momentum balance isRegularity at the centre gives $f=Ar+B{r}^3$, withfrom ${\sigma }_{rr}(a)=0$. Substitution reproduces $f(a)=-\rho g{a}^2/(15\kappa )$. During the transient, the acceleration term in the mean-stress identity remains; gravitational feedback from the small density and shape changes is higher order in the equilibrium calculation.

An elastic deformation is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous stress is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the displacement field $u_i$ gives the infinitesimal strain tensor $e_{ij}=(u_{i,j}+u_{j,i})/2$. The Cauchy stress tensor is defined by traction $t_i={\sigma }_{ij}{n}_j$ on a plane of unit normal $\mathbf{n}$; absence of body couples makes ${\sigma }_{ij}={\sigma }_{ji}$.

Let $\mathcal{E}(e)$ be the recoverable elastic energy per unit mass. The local mechanical power per unit volume isbecause the symmetric Cauchy stress tensor annihilates the antisymmetric part of the velocity gradient. Conservation of energy during the deformation gives $\rho \dot{\mathcal{E}}={\sigma }_{ij}{\dot{e}}_{ij}$. Since $\dot{\mathcal{E}}=(\partial \mathcal{E}/\partial e_{ij}){\dot{e}}_{ij}$ for arbitrary symmetric strain rates,Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.

Take a stress-free reference state of mass density ${\rho }_0$. The Taylor series of the elastic energy starts quadratically:Using $\rho ={\rho }_0+O(|e|)$ gives ${\sigma }_{ij}=c_{ijkl}{e}_{kl}+O(|e{|}^2)$, the relation of linear elasticity. If the reference state has initial stress, the first-order expansion also retains that constant stress. Symmetry of stress and strain gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of elastic energy gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the elastic stiffness tensor is a symmetric map on a six-dimensional strain space: there are $6\cdot 7/2=21$ independent coefficients.

For twofold elastic symmetry, choose the symmetry axis as $x_3$. Rotation through $\pi$ is $Q=\mathrm{diag}(-1,-1,1)$, and transforms the infinitesimal strain tensor into $Qe{Q}^T$. Components $e_{11},e_{22},e_{33},e_{12}$ are even and $e_{23},e_{13}$ are odd. Invariance of elastic energy excludes every coupling between these parity classes. In the engineering strain ordering $(e_{11},e_{22},e_{33},2e_{23},2e_{13},2e_{12})$, with corresponding stress ordering $({\sigma }_{11},{\sigma }_{22},{\sigma }_{33},{\sigma }_{23},{\sigma }_{13},{\sigma }_{12})$, the symmetric stiffness matrix has the formIts symmetric even block has ten independent entries and its symmetric odd block has three. Hence a twofold axis leaves $13$ independent coefficients. No invariance under other rotation angles has been assumed.

Let $c=\cos (\alpha {\xi }_3)$, $s=\sin (\alpha {\xi }_3)$ and $r^2=x_1^2+x_2^2=({\xi }_1^2+{\xi }_2^2)/\lambda$. Differentiating the deformation givesFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$. The constrained nominal stress tensor is therefore $N_{\alpha i}=\mu A_{i\alpha }+q(A^{-1}{)}_{\alpha i}$. The multiplier cannot be set arbitrarily to a constant; equilibrium and the traction-free curved surface determine it.

Push the nominal stress to the Cauchy stress tensor: since $J=1$, $\sigma =AN=\mu A{A}^T+qI$. In the current cylindrical orthonormal basis, the left Cauchy-Green deformation tensor has entriesThus ${\sigma }_{rr}=\mu /\lambda +q$, ${\sigma }_{\vartheta \vartheta }=\mu /\lambda +\mu {\alpha }^2{r}^2+q$, and ${\sigma }_{\vartheta 3}=\mu \alpha \lambda r$. Static radial equilibrium without body force givesThe current outer radius is $a=R/\sqrt{\lambda }$. Setting ${\sigma }_{rr}(a)=0$ integrates the reaction field:The other curved-surface traction components vanish because ${\sigma }_{r\vartheta }={\sigma }_{r3}=0$.

On the reference end ${\xi }_3=H$, the outward reference normal is $e_3$, and the nominal traction has components $t_i=N_{3i}$. The third row of $A^{-1}$ is $(0,0,{\lambda }^{-1})$. Hence the nominal end traction of a twisted neo-Hookean cylinder isThe current $x_1,x_2$ here are evaluated at ${\xi }_3=H$. These components are per unit reference area, not per unit current area.

For the moment, use current lever arms and reference-area traction. Its axial component isThe axial traction gives no axial moment. This torque also agrees with the reference-energy derivative at fixed axial stretch: the twist-dependent energy per reference length is $\mu {\alpha }^2\pi R^4/(4\lambda )$. It is important that $\alpha$ is twist per reference length, not twist per current length.

Let two observers be related by ${\mathbf{x}}^{\ast }=\mathbf{c}(t)+Q(t)\mathbf{x}$, with Q a proper rotation matrix. A spatial second-rank tensor has material frame indifference if its values at corresponding events satisfy $\boxed{J^{\ast }=QJ{Q}^T}$. This defines an objective second-rank tensor, and allows the observer's rotation to depend on time.

For Cauchy stress tensor, consider a material surface element with unit normal n and contact traction $\mathbf{t}=\sigma \mathbf{n}$. A rigid observer change preserves its area and rotates both its normal and actual contact force: ${\mathbf{n}}^{\ast }=Q\mathbf{n}$, ${\mathbf{t}}^{\ast }=Q\mathbf{t}$. Thus ${\sigma }^{\ast }Q\mathbf{n}=Q\sigma \mathbf{n}$ for every n, proving ${\sigma }^{\ast }=Q\sigma Q^T$. Fictitious body forces of an accelerated observer do not alter this transformation of physical contact traction.

To keep the vorticity convention explicit, write $L_{ij}={\partial }_j{v}_i$, $E=(L+L^T)/2$ and $\Omega =(L^T-L)/2$, the negative of the usual component-first spin tensor. Equivalently, with the derivative-index-first $\nabla v$ used in the paper, $\Omega =[\nabla v-(\nabla v{)}^T]/2$. Put $R=\dot{Q}{Q}^T$, which is antisymmetric. The transformed velocity is ${\mathbf{v}}^{\ast }=\dot{\mathbf{c}}+\dot{Q}\mathbf{x}+Q\mathbf{v}$, and direct differentiation givesThe additional observer spin cancels from the symmetric part, proving that the rate-of-strain tensor is objective.

Along the same material particle, differentiating $J^{\ast }=QJ{Q}^T$ givesThe two extra terms cancel those from ${\Omega }^{\ast }{J}^{\ast }-J^{\ast }{\Omega }^{\ast }$. Moreover $E^{\ast }{J}^{\ast }+J^{\ast }{E}^{\ast }=Q(EJ+JE)Q^T$. Hence the complete objective time derivative satisfiesThis proves covariance for every constant scalar $\alpha$; changing the spin convention without changing the commutator sign would not. It is a Gordon--Schowalter derivative with the paper's slip-parameter convention.

For the Johnson-Segalman fluid, first use the standard uniaxial extensional flow $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y/2,-\dot{ϵ}z/2)$, with $\dot{ϵ}>0$. Here $\Omega =0$ and E is diagonal. The steady conformation equation giveswith vanishing off-diagonal components on the stable steady branch. The tensile normal-stress difference is $\alpha G_0(A_{xx}-A_{yy})+3{\mu }_0\dot{ϵ}$. Dividing by the extension rate gives the Johnson-Segalman extensional response:The conformation tensor evolution in each principal direction is ${\dot{A}}_{ii}+(1/\tau -2\alpha E_{ii})A_{ii}=1/\tau$. Therefore a positive, attracting steady conformation requires both displayed denominators positive. For $\alpha >0$ the first pole is at $\dot{ϵ}=1/(2\alpha \tau )$; for $\alpha <0$ it is at $\dot{ϵ}=1/(|\alpha |\tau )$ in the transverse components. At a pole the corresponding component grows linearly in time, and beyond it generically grows exponentially. The model has a finite-rate extensional catastrophe; the algebraic viscosity cannot be continued as a physical steady response past the first pole. At zero rate, ${\eta }_E=3({\mu }_0+{\alpha }^2{G}_0\tau )$, giving Trouton ratio three. For $\alpha =0$ the polymer contribution to stress disappears and the response is simply Newtonian.

The question does not specify extension geometry. If a planar convention $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y,0)$ is intended instead, the same diagonal calculation gives $({\sigma }_{xx}-{\sigma }_{yy})/\dot{ϵ}=4{\mu }_0+4{\alpha }^2{G}_0\tau /[1-4{\alpha }^2{\tau }^2{\dot{ϵ}}^2]$, valid on the stable branch $|2\alpha \tau \dot{ϵ}|<1$. The preceding boxed formula uses the standard uniaxial definition of extensional viscosity.

For simple shear flow $v_x=\dot{\gamma }y$, write $A_{xx}=X$, $A_{yy}=Y$, $A_{xy}=A_{yx}=Z$, and $A_{zz}=1$. The xx, yy and xy components of the steady constitutive equation giveSubstituting the first two relations into the third yields $Z=\alpha \tau \dot{\gamma }/D$, with $D=1+(1-{\alpha }^2){\tau }^2{\dot{\gamma }}^2$. Define the viscometric functions by ${\sigma }_{xy}=\mu \dot{\gamma }$, $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, using x for flow, y for the shear gradient and z for vorticity. The Johnson-Segalman steady viscometric functions are thereforeThe solvent creates no normal-stress differences in this flow. For nonzero $\alpha$, the fluid is shear thinning with positive first and negative second normal-stress differences.

To classify the stress curves, set $x=\tau \sqrt{1-{\alpha }^2}\dot{\gamma }$ and $b={\mu }_0/({\alpha }^2{G}_0\tau )$. The scaled shear stress is $S=x[b+(1+x^2{)}^{-1}]$, andThe second term has minimum $-1/8$ at $x^2=3$. Thus the shear stress is monotone for $b\ge 1/8$ and has a decreasing segment for $\boxed{0<b<1/8}$. At equality it has a horizontal tangent but no decreasing interval. Below the threshold, its maximum and minimum occur atwith the smaller root giving the maximum. Negative-rate curves are obtained by odd reflection.

For small b, polymer stress dominates the low-rate response, while the solvent restores a positive high-rate slope. The Johnson-Segalman negative-slope shear threshold signals a constitutive instability of homogeneous shear: different local shear rates can carry the same shear stress, permitting shear banding. A stress plateau or a selected interface position cannot be inferred from the local curve alone; spatial regularization and boundary/history information matter. This limitation is illustrated in a primary diffusive Johnson-Segalman study. If $\alpha =0$, the scaling by b is inapplicable and the stress is just ${\mu }_0\dot{\gamma }$.

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 $F$ be an invertible, orientation-preserving deformation gradient, with $J=\det F>0$. To prove Nanson's formula, take two reference surface tangent vectors $a,b$. Their current images are $Fa,Fb$, and their oriented area vectors are proportional to $a\times b$ and $(Fa)\times (Fb)$. For every vector $c$, the scalar triple product givesThus $F^T[(Fa)\times (Fb)]=J(a\times b)$, andThis proof also identifies the area transformation as the cofactor of $F$, rather than a transformation by $F$ itself.

Use the paper's material-index-first nominal stress tensor $P_{Ii}$. Equality of force on the same material surface in its two configurations gives $P_{Ii}d{S}_{0I}={\sigma }_{ji}d{S}_j$. Substitute the area transformation to obtainHere $\tau$ is the Kirchhoff stress tensor. With symmetric Cauchy stress tensor, the matrix forms are $P=J{F}^{-1}\sigma$ and $\tau =FP$. This $P$ is the transpose of the first Piola-Kirchhoff stress tensor; keeping the index convention prevents a spurious transpose later.

For the velocity gradient $L=\dot{F}{F}^{-1}$, let $D=(L+L^T)/2$ be the rate-of-strain tensor. Then the reference-volume mechanical power isThe last step uses stress symmetry, as supplied by angular-momentum balance in an ordinary continuum without couple stresses: a symmetric tensor has zero contraction with the skew part of $L$.

By work-conjugate stress and strain, a pair $(T,E)$ must satisfy $T:\dot{E}=\tau :D$ for every deformation rate, with both sides measured per reference volume. For the Green-Lagrange strain tensor, set $C=F^{T}F$ and calculateCyclically rearranging the contraction shows $T^{(2)}:{\dot{E}}^{(2)}=(F{T}^{(2)}{F}^T):D$. Hence the symmetric conjugate isthe second Piola-Kirchhoff stress tensor.

The inverse right Cauchy-Green strain is $E^{(-2)}=(I-C^{-1})/2$, because $C^{-1}=F^{-1}{F}^{-T}$. Differentiating $C{C}^{-1}=I$ givesConsequently $T^{(-2)}:{\dot{E}}^{(-2)}=(F^{-T}{T}^{(-2)}{F}^{-1}):D$, andThe two stress measures have a geometric interpretation in the convected coordinate net. Its basis vectors are $g_I=F{e}_I$ and its reciprocal basis is $g^I=F^{-T}{e}_I$. The entries of $T^{(2)}$ are the contravariant components in $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$, whereas $T_{IJ}^{(-2)}=g_I\cdot \tau g_J$ are the covariant components, equivalently $\tau =T_{IJ}^{(-2)}{g}^I\otimes g^J$. They describe the same spatial Kirchhoff stress tensor in a deforming, generally nonorthonormal coordinate net.