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For an orientation-preserving deformation gradient $F$, oriented area transforms by its cofactor matrix: $(Fa)\times (Fb)=(\det F)F^{-T}(a\times b)$. Taking the dot product with $Fc$ proves this from the determinant identity for the scalar triple product. Applying it to tangent vectors of a surface proves Nanson's formula. Equating current and reference tractions then gives the material-first nominal stress tensor $P=J{F}^{-1}\sigma$ for symmetric Cauchy stress tensor $\sigma$.

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.

Using $J=1$ and the preceding first Piola-Kirchhoff stress tensor,Its eigenvalues are $\mu (1+\sqrt{5})/2$, $\mu (1-\sqrt{5})/2$, and zero. Because it is symmetric, its largest traction magnitude on a unit spatial normal is the largest absolute eigenvalue. Put $\phi =(1+\sqrt{5})/2$. The positive eigenvalue has the strictly largest modulus, with normalThe question asks for the plane in the reference configuration, so this spatial normal must be pulled back. Nanson's formula gives $N$ parallel to $F^{T}n$, andTherefore the required plane and maximum areEither sign of its unit normal describes the same plane. Maximizing over reference normals is equivalent to maximizing over spatial normals because the map $N\mapsto F^{-T}N/|F^{-T}N|$ is a bijection of unit normals.

We first derive the stress and energy facts used in both parts. The Cauchy traction $s=Tn$ is force per unit area in the deformed configuration. The first Piola-Kirchhoff traction $s_R=T_{R}N$ is force per unit reference area. Their force vectors are measured in the same spatial coordinates, but the area measures and the normals used to specify the surface are different.

Parametrize a reference surface by $x=x(u,v)$. Its oriented area vector is $(x_u\times x_v)dudv$. Under deformation, the two tangents become $F{x}_u,F{x}_v$. The cross-product identitytherefore proves Nanson's formula $nda=J{F}^{-T}NdA$. Equality of the physical force on corresponding surface patches now givesSince this holds for every reference normal,The first Piola-Kirchhoff stress tensor is generally not symmetric, whereas the Cauchy stress tensor is symmetric in the absence of couple stresses.

For the stated energy on orientation-preserving gradients, differentiation givesA stress-free state satisfies $F{F}^T=J^{-1}I$. Taking determinants yields $J^2=J^{-3}$, so $J=1$ and $F{F}^T=I$: precisely $F\in SO(3)$. These states are also the global energy minimizers. If $v_i>0$ are the singular values, the arithmetic-geometric mean inequality gives $|F{|}^2\ge 3J^{2/3}$, with equality only when the singular values are equal. The function $3J^{2/3}/2+J^{-1}$ decreases up to $J=1$ and increases thereafter, with minimum $5/2$. Thus $W\ge 5\mu /2$, with equality precisely at rotations. A constant shift of the energy can set this minimum to zero without changing the stress.

For the given simple shear,For a unit normal $N$, $|T_{R}N{|}^2={\mu }^2(N_1^2+N_2^2)={\mu }^2(1-N_3^2)$. Hence every reference plane whose normal lies in the $e_1,e_2$ plane is maximizing:There is a family of maximizing planes, rather than one distinguished normal.

The first Piola-Kirchhoff stress tensor $T_R$ maps a reference normal to force per reference area. The Cauchy stress tensor $T$ maps a current normal to force per current area. Nanson's formula gives $\mathbf{n}da=J{F}^{-T}\mathbf{N}dA$. Equating the actual surface force in both descriptions gives $T_R=JT{F}^{-T}$, and thereforeFor incompressibility $\boxed{J=1}$. The strain-energy derivative only fixes the constitutive stress up to a constraint reaction in hyperelastic stress: $T_R=DW-p\mathrm{cof}F$, hence $T=DW{F}^T-pI$. The scalar $p$ is a Lagrange multiplier determined by equilibrium and boundary tractions, rather than by the isochoric energy alone.

In principal stretch axes, the extra principal stress is $v_i\partial W/\partial v_i=2\alpha v_i^2-2\beta v_i^{-2}$, up to that arbitrary spherical reaction. With $B=F{F}^T$, its eigenvalues are $v_i^2$, so isotropy yieldsFor the prescribed simple shear,The $x_3$ faces retain normal $\pm {\mathbf{e}}_3$, and zero traction there gives $p=2(\alpha -\beta )$. ThusThis constant stress satisfies the interior equilibrium equation, and the remaining face forces maintain the deformation.

To compute the angle, the material face $x_1=1$ becomes the spatial plane $y_1-\gamma y_2=1$. Its outward unit normal and an in-plane unit tangent areResolve the current traction $\mathbf{t}=T\mathbf{n}$ into these directions:Since $N<0$ for nonzero $\gamma$, the angle lies above $\pi /2$, andUsing $\arctan z\sim z$ at zero givesUsing the undeformed normal ${\mathbf{e}}_1$ would give the wrong coefficient. The transformed unit normal is essential to this simple-shear traction angle for a Mooney-Rivlin solid.

The surface element relation is Nanson's formula. The force on a corresponding pair of current and reference surface elements must agree:Therefore the first Piola-Kirchhoff stress tensor isIt measures force per reference area, with a current force direction and a reference normal direction.

In components, differentiate $T_{iA}=S_{ij}(y(x))(\mathrm{cof}F{)}_{jA}$. The Piola identity says ${\partial }_A(\mathrm{cof}Dy{)}_{jA}=0$. One proof writes the cofactor matrix as $\frac{1}{2}{\epsilon }_{jkl}{\epsilon }_{ABC}{y}_{k,B}{y}_{l,C}$; differentiating in $A$ produces symmetric second derivatives contracted against antisymmetric epsilon symbols, hence zero. The chain rule consequently givessince $F_{kA}(\mathrm{cof}F{)}_{jA}=J{\delta }_{kj}$. Thus, with the coordinate composition made explicit,The distinction between reference divergence and current divergence is essential to this identity.