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For the material-first nominal stress tensor, the displayed local conservation of energy law balances mechanical work, heat supply and outgoing heat flux. Integrating over a fixed reference material volume and using the divergence theorem gives its integral form. The full total-energy balance includes kinetic energy and external mechanical work; momentum conservation reduces it to this internal-energy balance.

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.

The factor $A$ can encode an internal relaxed configuration, while $F_{\ast }=F{A}^{-1}$ determines the instantaneous elastic energy. Differentiating $F_{\ast }$ at fixed $F$ or fixed $A$ produces the corresponding nominal stress tensor and thermodynamic force conjugate to an internal variable. This is a kinematic decomposition, not an assertion that each factor separately derives from a global deformation map.

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 $\sigma$.

Use the material-first nominal stress tensor convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}{\dot{A}}_{i\alpha }$. At a regular constraint point, $F_{,A}\ne 0$, all admissible deformation rates satisfy $F_{,A_{i\alpha }}{\dot{A}}_{i\alpha }=0$. Equality of mechanical power and the derivative of the strain energy density givesfor every vector in this eight-dimensional tangent hyperplane. Its orthogonal complement is the span of $F_{,A}$. Consequently the constraint reaction in hyperelastic stress isThe scalar Lagrange multiplier $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular incompressibility constraint by $(\det A-1{)}^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.

For incompressibility use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ givesThe first expression is the cofactor entry; contracting it with $A_{j\alpha }$ gives $J{\delta }_{ij}$, proving the inverse formula. On $J=1$ this yieldsWith this plus-sign convention $q$ is the negative of the usual pressure multiplier.

For an equibiaxially stretched sheet, incompressibility requires the principal stretches $(\lambda ,\lambda ,{\lambda }^{-2})$. The printed sheet expression's ${\lambda }^{-1/2}$ is inconsistent with this constraint and must be ${\lambda }^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to principal stretch ${\lambda }_j$, evaluated on this constrained path. With no applied thickness-direction traction,Isotropy gives $W_1=W_2$ at equal in-plane stretches. HenceThe equibiaxial nominal tension of an incompressible sheet is thereforeThe two equal in-plane nominal tensions do work $2N_{11}d\lambda$ per reference volume; the thickness-direction nominal stress does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a neo-Hookean solid gives $N=\mu (\lambda -{\lambda }^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu /4$ at that state.

For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2{h}_0$. Its two tangential stretches are $\lambda =a/a_0$, while its thickness stretch is ${\lambda }^{-2}$ and its current thickness is $h=h_0{\lambda }^{-2}$. Define $\hat{W}(\lambda )=W(\lambda ,\lambda ,{\lambda }^{-2})$. Its total elastic energy at leading thin-shell order isDuring a quasistatic increment $da$, the enclosed volume changes by $d\mathcal{V}=4\pi a^{2}da$, and $d\lambda =da/a_0$. Equating energy change to work of the internal excess pressure givesThis is the inflation pressure of a thin incompressible elastic balloon. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll 1$. It also agrees with membrane force balance: $p=2h{\sigma }_t/a$, ${\sigma }_t=\lambda N$, and $N={\hat{W}}^{\prime }/2$. Here pressure is force per current area, while $N$ is force per reference area; confusing them would lose a factor of $\lambda$.

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 $F$ 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.

For fixed nominal boundary tractions and dead body forces, use the total potentialThe dead-load terms are linear in the deformation map. Let $x(\epsilon )$ be a constraint-preserving perturbation path through equilibrium, with $H=\delta F$ and $K={\delta }^{2}F$. Differentiating $\varphi (F)=0$ twice gives ${\varphi }_{,F}:H=0$ and ${\varphi }_{,F}:K=-{\varphi }_{,FF}[H,H]$. Equilibrium has nominal stress tensor $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$. Its virtual-work relation, applied to ${\delta }^{2}x$, gives the otherwise remaining first-derivative contribution ${\Pi }^{\prime }[{\delta }^{2}x]=-\int q{\varphi }_{,F}:KdX$. Therefore the constrained second variation isFor a time-dependent linearized perturbation $y$, the incremental reaction stress contains $\delta q{\varphi }_{,F}$, but its power vanishes because ${\varphi }_{,F}:\nabla \dot{y}=0$. Multiply the incremental momentum equation by $\dot{y}$, integrate by parts, and use zero incremental dead-load traction or the prescribed displacement condition. One obtainsPositive $\mathcal{Q}$ is therefore the standard constrained second-variation stability in elasticity criterion: perturbations cannot grow in this quadratic energy norm. A statement in a specified displacement norm also uses the usual coercivity and boundary conditions removing uncontrolled rigid-motion modes; translations with $\delta F=0$ are not tested by the displayed condition. This is a small-perturbation energy argument, not a claim of global nonlinear stability.

For the cube use incompressibility in the form $\varphi =\det F-1$. A neo-Hookean solid has $W_{,F}=\mu F$, and hence the spatial-index-first transpose of nominal stress is $\mu F+q{F}^{-T}$. On the diagonal family, equality to $TI$ givesThe first relation immediately gives $\boxed{q=T\lambda -\mu {\lambda }^2}$. Eliminating $q$ and factoring yieldsThus $\boxed{\lambda =1\text{or}T/\mu =\lambda +{\lambda }^{-1/2}}$.

Let $t=T/\mu$ and $h(\lambda )=\lambda +{\lambda }^{-1/2}$. Its derivative $h^{\prime }=1-\frac{1}{2}{\lambda }^{-3/2}$ vanishes only at ${\lambda }_c=2^{-2/3}$, where $h^{\prime \prime }>0$ and $h_c=32^{-2/3}$. Also $h\to \infty$ at both ends of $(0,\infty )$. Therefore $h(\lambda )=t$ has two distinct positive roots for $t>h_c$, one repeated root at $h_c$, and none below. There is an exception to the printed assertion that both roots differ from one: at $t=2$, one root is exactly $\lambda =1$. Indeed, with $x=\sqrt{\lambda }$, the equation becomes $(x-1)(x^2+x-1)=0$, so the only other positive stretch is $\lambda =(3-\sqrt{5})/2$.

Evaluate the stability form directly on the prescribed uniform perturbationIt satisfies $\mathrm{tr}(F^{-1}H)=0$. The determinant Hessian identity isMeanwhile $W_{,FF}[H,H]=\mu (1+\frac{1}{2}{\lambda }^{-3})(\delta \lambda {)}^2$. For the unit reference cube,On the trivial branch, $q=T-\mu$ andThus the undeformed state is stable in this perturbation family for $t<2$, unstable for $t>2$, and quadratically neutral at $t=2$.

On a nontrivial branch, substitute $t=h(\lambda )$ and put $z={\lambda }^{-3/2}$:The smaller root always has $\lambda <{\lambda }_c$, hence $z>2$, and is stable. For $h_c<t<2$, the larger root lies between ${\lambda }_c$ and one, so $1<z<2$ and it is unstable; the trivial branch is the second stable equilibrium. For $t>2$, the larger root lies above one and has $0<z<1$, so it is stable; the trivial branch is now unstable. Whenever there are three distinct equilibria in this family, precisely two are stable and one unstable against the specified uniform perturbations. This does not establish stability to every nonuniform or symmetry-breaking perturbation.

The homogeneous tensile bifurcation of a neo-Hookean cube has a nontrivial-branch fold at $t=32^{-2/3}$ and a meeting with the undeformed branch at $\boxed{t=2}$. The latter is the bifurcation from the undeformed state. Stating both values distinguishes this branch intersection from the earlier creation of the two nontrivial solutions.

Use the material-first nominal stress tensor convention $P_{Ii}$, with $F_{iI}$ the deformation gradient. For any fixed reference material subvolume $V_0$ with outward unit normal $N$, conservation of energy isHere $b$ is body force per mass, $r$ is heat supply per mass and $q$ is nominal heat flux. Using momentum conservation to remove the kinetic energy balance gives the integral Lagrangian internal energy balance:The integral Lagrangian entropy inequality iswhere $\theta >0$ is temperature and $q/\theta$ is the thermodynamic entropy flux.

Since the subvolume is arbitrary, the divergence theorem givesExpand $\dot{U}=U_{,F_{iI}}{\dot{F}}_{iI}+U_{,\eta }\dot{\eta }+U_{,{\xi }_r}{\dot{\xi }}_r$ and eliminate the heat supply between these relations. The Clausius-Duhem inequality becomesUnder the usual local constitutive independence assumptions, the reversible mechanical and thermal variations can have either sign. The Coleman-Noll procedure therefore imposesDefine the thermodynamic force conjugate to an internal variable by $f_r=-{\rho }_0{U}_{,{\xi }_r}$. The energy balance and the remaining entropy constraint now give, respectively,

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.

Using $a=A$ and the end-face shear stress from part (c), the printed torque normalization givesThe binomial theorem expands the remaining integral into powers of $r$:Integrating each term gives an expression valid also at $\tau =0$:For example, $n=1$ gives $M=({\mu }_1+3{\mu }_2)A^4\tau /2+{\mu }_2{A}^6{\tau }^3/3$. The torque is an odd polynomial in the twist per unit reference length, as expected when reversing the twist reverses the end-face shear stress.

For a physical end-face torque $N$, integration over the azimuth normally givesThe original PDF explicitly defines $M$ without this $2\pi$, so the boxed answer retains its convention. Since $\mathbf{T}$ is the Cauchy stress tensor, its shear component here is spatial $T_{\theta z}$; the printed capital $Z$ in $T_{\theta Z}$ does not require a change to a nominal stress tensor.

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.

The second tensor has two material indices and is symmetric when the Cauchy stress tensor is symmetric. The displayed relation uses the material-first nominal stress tensor convention. In particular, $PF$ is generally not this tensor. These conventions agree with A. F. Bower's definitions of stress measures.

The admissible rates obey ${\varphi }_{,F}:\dot{F}=h^{\prime }(\theta )\dot{\theta }$. A reaction multiplier $q$ in the reversible nominal stress tensor therefore has an accompanying specific entropy contribution:Their reaction contributions cancel in the Clausius-Duhem inequality on admissible rates. Omitting the entropy term would generally violate the constraint's power balance.