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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 $\tau =T_{IJ}^{(2)}{g}_I\otimes g_J$.

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.

Differentiate the pulled-back Kirchhoff stress tensor $T=F^T\tau F$, using the velocity gradient convention $L=\dot{F}{F}^{-1}$:Thus the lower-convected stress pull-back givesFor an explicit objective time derivative check, superpose $x^{\ast }=Q(t)x+c(t)$, with $Q^{T}Q=I$ and $\Omega =\dot{Q}{Q}^T$ skew. Then ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\Omega$, whereasThe additional terms in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$ are $-\Omega {\tau }^{\ast }+{\tau }^{\ast }\Omega$. They cancel the extra material-derivative terms, proving $\boxed{{\mathcal{D}}_l{\tau }^{\ast }=Q({\mathcal{D}}_l\tau )Q^T}$. The translation has no effect on a stress or velocity gradient.

For the incompressible Oldroyd-A model, avoid using the same symbol for a stress tensor and its relaxation time by referring to $\tau >0$ below only as the scalar relaxation time. Define $S={\sigma }^d-2{\mu }_{r}D$. The constitutive equation becomesBy the pull-back identity, $A=F^{T}SF$ satisfiesAn integrating factor givesPush forward by $F(t{)}^{-T}$ and $F(t{)}^{-1}$. For a convergent, fading remote-past history with the transported homogeneous term tending to zero as $t_0\to -\infty$, this proves the fading-memory representation of the Oldroyd-A model:For arbitrary finite-time initial data retain $e^{-(t-t_0)/\tau }F(t{)}^{-T}A(t_0)F(t{)}^{-1}$. Thus the printed integral-only expression incorporates an initial-memory condition; it is not an unconditional replacement for all initial-value solutions.

For time-dependent simple shear,Put $\Delta \gamma =\gamma (s)-\gamma (t)$. The relative deformation is $F(s)F(t{)}^{-1}=\left(\begin{array}{cc}1& \Delta \gamma \\ 0& 1\end{array}\right)$, so the entire transported integrand simplifies toConsequently the explicit nonpressure stresses areSymmetry gives ${\sigma }_{21}^d={\sigma }_{12}^d$, and the unshown third-direction nonpressure components vanish. The superscript $d$ here labels the paper's nonpressure constitutive stress; it need not have zero trace. Subtracting $pI$ supplies the total Cauchy stress tensor.

For constant shear rate $s_0$, use $\gamma (s)-\gamma (t)=-s_0(t-s)$ and the integrals ${\int }_0^{\infty }{e}^{-a/\tau }da=\tau$ and ${\int }_0^{\infty }a{e}^{-a/\tau }da={\tau }^2$. This gives the steady simple shear of an Oldroyd-A fluid resultand thereforeThe incompressibility pressure cancels from this normal-stress difference. With the usual $\mu \ge {\mu }_r\ge 0$, it is nonnegative, while the second normal-stress difference equals its negative.

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.

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.

A stress measure $T$ is work-conjugate to a strain measure $E$ when their contraction gives the mechanical power per specified volume for every admissible deformation rate. The reference-volume power is $P_{Ii}{\dot{F}}_{iI}=\tau :D$, with Kirchhoff stress tensor $\tau$ and rate-of-strain tensor $D$. A strain constraint can leave the conjugate stress undetermined by a reaction component that does no admissible work; uniqueness should therefore not be assumed on a restricted set of rates.