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With the right Cauchy-Green deformation tensor $C=F^{T}F$, the Green-Lagrange strain tensor measures the change in the material metric. Since ${\dot{E}}^{(2)}=F^{T}DF$, its work-conjugate stress and strain partner is the second Piola-Kirchhoff stress tensor $F^{-1}\tau F^{-T}$. The formula follows by cycling factors in $T:{\dot{E}}^{(2)}$, not by identifying nominal and symmetric material stress.

For $C=F^{T}F$, differentiation of $C^{-1}$ gives ${\dot{E}}^{(-2)}=F^{-1}D{F}^{-T}$. Its reference-power conjugate is $T^{(-2)}=F^T\tau F$. In a convected coordinate basis $g_I=F{e}_I$, these are the covariant components of the Kirchhoff stress tensor. The second Piola-Kirchhoff stress tensor instead gives its contravariant components, since $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$.

The Kirchhoff tensor is the Cauchy stress tensor multiplied by the local volume ratio. Its stress power is per reference volume. Pushing forward the second Piola-Kirchhoff stress tensor gives the displayed formula.

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.

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.

The specific Helmholtz free energy $\psi =U-\theta \eta$ satisfies $d\psi ={\rho }_0^{-1}{P}_{Ii}d{F}_{iI}-\eta d\theta -{\rho }_0^{-1}{f}_{r}d{\xi }_r$. For the multiplicative decomposition of the deformation gradient $F=F_{\ast }A$, adopt $P_{Ji}^{\ast }={\rho }_0\partial \psi /\partial (F_{\ast }{)}_{iJ}$, consistently with the material-first nominal stress tensor. Differentiating $F_{\ast }=F{A}^{-1}$ givesConsequentlyorThese relations follow from the chain rule; the transpose convention matters.

Differentiating the dissipation potential, with the Frobenius norm of $Q$, givesAn integrating factor yields the exponential memory from tensor relaxation:For $n>0$, the driving term is continuously zero at $Q=0$.

Taking the product $T=PF$ explicitly defined in the question, $F=F_{\ast }A$ gives $T=QA$, and hence its requested history functional isThere is a naming error in the PDF: with its $P_{Ii}$ convention, the actual second Piola-Kirchhoff stress tensor is $S=P{F}^{-T}$, while $PF=SC$ is the transpose of the Mandel stress tensor, where $C=F^{T}F$. Thus $S=QA{C}^{-1}$ if that stress measure is required. This agrees with the stress push-forward used in Question 5; the printed product is fully determined by the history above, whereas recovering $S$ also requires the deformation or an invertible elastic constitutive relation.

The multiplicative decomposition of the deformation gradient is used on histories where $A$ stays invertible. Thermodynamic admissibility additionally requires $\mathrm{tr}(Q\dot{A})=\alpha \Vert Q{\Vert }^{n+1}-\mathrm{tr}(AQ)/\tau \ge 0$: positivity of the constants alone does not ensure this for arbitrary prescribed histories.

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.

The Green-Lagrange strain tensor has eigenvalues $({\lambda }_j^2-1)/2$. Its Frobenius norm and trace are invariant under the orthogonal principal-axis change, soFor the homogeneous dilation $F=\rho I$, put $K=\mu +3\lambda /2$. Then $E=({\rho }^2-1)I/2$. Differentiating the energy through $E$ gives the second Piola-Kirchhoff stress tensor $2\mu E+\lambda \mathrm{tr}(E)I=K({\rho }^2-1)I$, and multiplication by $F$ gives the first Piola-Kirchhoff stress tensor. ThusThe normals of the reference and deformed spheres point in the same radial direction. As $\rho \downarrow 0$, the reference traction magnitude $|TN|$ tends to zero, while the current traction magnitude $|Sn|$ tends to infinity, provided $K>0$, as for physically stable Lamé parameters. The negative signs of the stress coefficients for $0<\rho <1$ indicate compression.

There is no contradiction: the current area is ${\rho }^2$ times the reference area, so the same force divided by the vanishing current area has the larger stress. Indeed $TN={\rho }^{2}Sn$. The limiting map at $\rho =0$ is not itself an admissible deformation in continuum mechanics; the statement concerns the approach through positive $\rho$. A degenerate choice $K=0$ would not have the asserted growth without bound.

These stress measures refer force or area to the reference configuration. Different index-order conventions distinguish a first Piola-Kirchhoff stress tensor from its transposed nominal stress tensor. The second Piola-Kirchhoff stress tensor has two material indices.