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The contact traction on an oriented material cut is assumed to depend locally on the point, time and unit normal, with no independent dependence on the cut's curvature or global geometry. Together with local momentum balance and regularity sufficient for shrinking-volume arguments, this implies linear normal dependence through a stress tensor. Referential traction is measured per reference area and yields the first Piola-Kirchhoff stress tensor; spatial traction is measured per current area and yields the Cauchy stress tensor. These are different representations and the nominal tensor is not generally symmetric.

Consider an incompressible neo-Hookean prism with energy $\alpha (|F{|}^2-3)$, zero nominal end traction normal to $e_3$, and equal nominal lateral tractions $TN$ on side faces whose normals enforce both in-plane principal stresses. Within positive diagonal homogeneous deformations the first Piola-Kirchhoff stress tensor has entries $t_i=2\alpha v_i-p/v_i$. Equilibrium makes $p$ constant; end freedom gives $p=2\alpha v_3^2>0$.

The function $2\alpha v-p/v$ is strictly increasing on positive $v$, so equal lateral stresses require $v_1=v_2$. The determinant constraint then gives $v_3=v^{-2}$ and the displayed load relation. Its derivative is $2\alpha (1+5v^{-6})>0$, with limits minus infinity and infinity at the two ends of the positive axis. Thus there is exactly one positive diagonal equilibrium for every signed load. This uniqueness is confined to the homogeneous diagonal class and does not rule out other modes or nonhomogeneous equilibria under a general dead-loading problem.

A hyperelastic material derives its elastic stress from a strain energy density. In reference-volume units its first Piola-Kirchhoff stress tensor is the derivative of that density with respect to the deformation gradient.

Here the convention places the material index first: $P_{Ii}$. Its stress power per reference volume is $P_{Ii}{\dot{F}}_{iI}$. It is the transpose of the first Piola-Kirchhoff stress tensor; some texts instead call that first tensor itself nominal stress.

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

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.

Cauchy's stress hypothesis says that traction across a cut depends locally on its point, time and oriented normal, rather than on the global shape of the cut. Here all areas and normals are referential: $E$ is a fixed material region in reference coordinates, ${\rho }_R$ its mass density per reference volume, $v$ the current particle velocity, $b$ body force per unit mass, and $s_R$ force per unit reference area. The time derivative on the left is total material momentum rate; the two right terms are contact and body forces.

A shrinking pillbox gives $s_R(-N)=-s_R(N)$. In a shrinking tetrahedron the volume contributions scale as its linear size cubed, whereas face forces scale as size squared. Their leading balance expresses traction on an arbitrary face as the normal-component-weighted sum of tractions on three coordinate faces. Defining those three tractions as tensor columns gives $\boxed{s_R(N)=T_{R}N}$. Applying the divergence theorem to momentum balance and localizing over arbitrary $E$ then yieldsThe referential tensor is the first Piola-Kirchhoff stress and need not be symmetric; it should not be silently identified with the spatial Cauchy stress.

Incompressibility is $\boxed{\det F=1}$. For a homogeneous hyperelastic material the referential constitutive relation becomeswhere the undetermined scalar pressure $p$ enforces the constraint. Homogeneity removes explicit reference-position dependence from $W$, not the possible spatial variation of that pressure.

For the neo-Hookean energy, $DW=2\alpha F$. For the positive diagonal ansatz the nominal principal stresses are $t_i=2\alpha v_i-p/v_i$. Since $F$ is constant, equilibrium requires $\mathrm{cof}F\nabla p=0$, so $p$ is constant. Stress-free endfaces give $\boxed{p=2\alpha v_3^2>0}$. The side normals include $(0,-1,0)$ and $\pm (\sqrt{3}/2,1/2,0)$, so their prescribed tractions force $\boxed{t_1=t_2=T}$.

For fixed positive $p$, the scalar function $2\alpha v-p/v$ is strictly increasing on $v>0$. Thus equality of the two nominal stresses implies $v_1=v_2=v$. Incompressibility gives $v_3=v^{-2}$, and substitution givesThe load function has derivative $2\alpha (1+5v^{-6})>0$, tends to minus infinity as $v\downarrow 0$ and to infinity as $v\to \infty$. It therefore determines exactly one positive $v$ for every specified load, with $v\ge 1$ for outward tensile $T\ge 0$. The resulting constant stress satisfies equilibrium and every prescribed face traction, proving existence and uniqueness within the requested positive diagonal class. It does not claim uniqueness among all possible nonhomogeneous deformations. This is the equibiaxial dead loading of a neo-Hookean prism.

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.