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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 the polar decomposition of an invertible real matrix, start with $C=F^{T}F$. For nonzero $y$, $y^{T}Cy=|Fy{|}^2>0$, since $\det F>0$ makes $F$ invertible. The spectral theorem for real symmetric matrices gives an orthonormal basis $n_a$ with $C{n}_a={\lambda }_a^2{n}_a$ and ${\lambda }_a>0$. DefineThen $U$ is a positive-definite symmetric matrix, $U^2=C$, andbecause $\det U=\sqrt{\det (F^{T}F)}=\det F$. Thus $R$ is an orthogonal matrix with positive determinant. Set $V=RU{R}^T$; it too is a positive-definite symmetric matrix, and $VR=RU=F$. Moreover $V^2=F{F}^T$. This provesThe positive square root is unique: any positive symmetric square root of $C$ commutes with $C$, preserves its eigenspaces and has the positive scalar square root on each of them. Hence $U$, then $R$ and $V$, are unique. These are the right stretch tensor, rotation and left stretch tensor of the polar decomposition in continuum mechanics.

A standard spectral strain measure isTake $f\in C^1(0,\infty )$, $f(1)=0$ and $f^{\prime }(1)=1$, with $f$ strictly increasing; the customary stronger condition $f^{\prime }(\lambda )>0$ gives a regular one-to-one parameterization of all positive principal stretches. The zero condition makes a rigid rotation strain free, monotonicity distinguishes extension from contraction and prevents nonunit stretch from being strain free, and the derivative normalization gives the ordinary infinitesimal strain. Indeed, for $F=I+H$,Under a superposed spatial rotation $F\mapsto QF$, $F^{T}F$ and hence $E^f$ remain unchanged, giving material frame indifference. Choosing $f(\lambda )=\lambda -1$ satisfies all these conditions and gives the Biot strain tensor

The meaning of work-conjugate stress and strain here is that a symmetric material tensor $T^f$ satisfiesfor every admissible rate, with both powers measured per reference volume. To keep the nominal-stress index convention explicit, write $M_{iI}=P_{Ii}$, so $M=P^T$ and the power is $M:\dot{F}$. Differentiating $F=RU$ gives $\dot{F}=R(\Omega U+\dot{U})$, where $\Omega =R^T\dot{R}$ is skew. Conservation of angular momentum gives symmetry of $M{F}^T$, or equivalently of $R^{T}MU$. ThereforeThe first term vanishes because a symmetric tensor has zero contraction with a skew tensor. Since ${\dot{E}}^{(1)}=\dot{U}$, the symmetric Biot stress isIf the nominal stress is instead stored spatial-index first as $M$, the same answer is $\mathrm{sym}(R^{T}M)$. The symmetrization is essential: $PR$ need not itself be symmetric.

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.