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A viscoelastic material stores part of the deformation energy and dissipates part of it. In linear viscoelasticity, a small imposed shear strain produces a stress determined by its history, $\tau (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. A shear step probes stress relaxation; a stress step probes creep compliance. A purely elastic solid has persistent stress after a held strain, while a Newtonian fluid relaxes that stress and creeps continuously under constant stress.

For a Maxwell model, a spring of shear modulus $G_0$ and a dashpot of viscosity $\eta$ are in series:For the convention $\gamma \propto e^{i\omega t}$,The storage modulus dominates at short times or high frequencies; the loss modulus dominates at long times or low frequencies. The energy dissipated per cycle per volume is $\pi G^{\prime \prime }{\gamma }_0^2$. A Kelvin-Voigt model places spring and dashpot in parallel and instead gives bounded creep $J(t)=(1-e^{-G_{0}t/\eta })/G_0$: a simple illustration of a viscoelastic solid. Real materials can require a spectrum of relaxation times rather than one element.

A small-amplitude oscillatory rheometer measures stress amplitude and phase relative to imposed strain, giving $G^{\prime }=|{\tau }_0/{\gamma }_0|\cos \delta$ and $G^{\prime \prime }=|{\tau }_0/{\gamma }_0|\sin \delta$. First sweep strain amplitude to identify the linear range; control temperature, sample ageing, wall slip and instrument inertia. Frequency-dependent phase lag alone is insufficient if these apparatus effects are unresolved.

The independent fluid-mechanical relation iswhere $\rho$ is fluid mass density, $\eta$ dynamic viscosity and $\nu$ kinematic viscosity. For a generic rigid sphere, the convention must specify whether $L$ is its diameter or radius; their Reynolds numbers differ by a factor of two. The ratio compares advective inertia $\rho U^2/L$ with viscous stress gradients $\eta U/L^2$. Small Reynolds number supports a creeping-flow approximation when unsteady fluid inertia is also negligible. No organism-specific size, speed or propulsion calculation is made here.

A viscoelastic material stores part of the work of deformation and dissipates part. In the linear regime its stress depends on the strain history, not just the instantaneous strain or shear strain rate. For simple shear, a causal relaxation modulus $G(t)$ gives $\sigma (t)={\int }_{-\infty }^{t}G(t-t^{\prime })\dot{\gamma }(t^{\prime })d{t}^{\prime }$. A sudden strain then probes stress relaxation, while a constant imposed stress probes material creep; these are different experiments.

For steady sinusoidal strain use the convention $\gamma (t)={\gamma }_0\cos \omega t$ and $e^{i\omega t}$ complex amplitudes. The complex viscoelastic modulus isso $\sigma (t)={\gamma }_0[G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t]$. The storage modulus $G^{\prime }$ is the in-phase elastic component. The loss modulus $G^{\prime \prime }$ multiplies the component in phase with shear strain rate; stress leads strain by $\delta =atan2(G^{\prime \prime },G^{\prime })$. The loss tangent is $\tan \delta =G^{\prime \prime }/G^{\prime }$ when $G^{\prime }>0$.

Multiplying stress by shear strain rate and averaging gives $⟨P⟩/V=\frac{1}{2}\omega G^{\prime \prime }{\gamma }_0^2$. Hence the dissipated energy per oscillatory shear cycle isThe elastic contribution returns its work over the cycle. A passive material has nonnegative dissipation, which fixes the sign of the loss term in the chosen Fourier convention. This complex viscoelastic modulus is a response function, not the mathematical absolute value of a complex number.

A Maxwell model places a spring of shear modulus $G$ in series with a dashpot of viscosity $\eta$, so the strains add at common stress:The Maxwell fluid dynamic modulus isAt $\omega \tau \ll 1$ the response is predominantly viscous, $G^{\prime \prime }\simeq \eta \omega$ and $G^{\prime }\simeq G(\omega \tau {)}^2$. At $\omega \tau \gg 1$ it approaches the elastic plateau $G^{\prime }=G$. The two moduli cross, and $G^{\prime \prime }$ reaches its maximum, at $\omega \tau =1$. A Kelvin-Voigt model instead has the spring and dashpot in parallel, giving $G^{\ast }=G+i\eta \omega$ and no fluid-like zero-frequency terminal flow. Real materials often need a spectrum of relaxation times rather than one Maxwell time.

Oscillatory rheometry measures stress amplitude and phase while imposing a small strain. An oscillatory-shear amplitude sweep first identifies a linear regime; an oscillatory-shear frequency sweep then measures $G^{\prime }$ and $G^{\prime \prime }$. Repeat at controlled temperature, suppress evaporation, check wall slip and sample aging, and correct instrument inertia at high frequency. Frequency-dependent structure and a nonlinear response cannot be inferred reliably by fitting one spring and dashpot to one point.

The requested model is the standard linear solid model: a spring of modulus $E_0>0$ in parallel with a Maxwell branch consisting of a spring $E_1>0$ in series with a dashpot of viscosity $\eta >0$. Both parallel branches have the same strain; their stresses add. In the Maxwell branch the spring and dashpot have the same stress and their strains add. The equilibrium spring prevents indefinite flow at long times.

For harmonic strain $ϵ(t)=\mathrm{Re}[{ϵ}_0{e}^{i\omega t}]$, define the dynamic modulus by $\tilde{\sigma }=E^{\ast }(\omega )\widetilde{ϵ}$. The Maxwell branch has compliance $1/E_1+1/(i\omega \eta )$. Defining $\tau =\eta /E_1$ therefore givesThe same mechanical model can represent a shear modulus if the appropriate shear stresses and strains are used consistently.

Writing $E^{\ast }=E^{\prime }+i{E}^{\prime \prime }$, its storage modulus and loss modulus areWith this time convention the physical stress is ${ϵ}_0[E^{\prime }\cos \omega t-E^{\prime \prime }\sin \omega t]$ and leads the strain by the loss angle. The dissipated energy per unit volume per cycle is $\pi E^{\prime \prime }{ϵ}_0^2\ge 0$. Thus $E^{\prime }$ describes elastic storage and $E^{\prime \prime }$ describes dissipation.

At low frequency, $\omega \tau \ll 1$, the dashpot flows during a cycle and relaxes the Maxwell spring, leaving $E^{\prime }\to E_0$, $E^{\prime \prime }\sim E_1\omega \tau$. At high frequency, $\omega \tau \gg 1$, the dashpot hardly deforms within a cycle, so both springs carry elastic load: $E^{\prime }\to E_0+E_1$, while $E^{\prime \prime }\sim E_1/(\omega \tau )$. The loss modulus is maximal at $\omega \tau =1$, with peak value $E_1/2$.

The equivalent constitutive equation isAfter a sudden strain ${ϵ}_s$ is imposed and held, the stress relaxes as $\sigma (t)={ϵ}_s[E_0+E_1{e}^{-t/\tau }]$. It falls from the instantaneous modulus $E_0+E_1$ to the nonzero relaxed modulus $E_0$. This single-time relaxation is the simplest improvement over a purely elastic solid or a Maxwell fluid with zero long-time modulus.

The loss tangent is $\tan \delta =E^{\prime \prime }/E^{\prime }$. Its maximum need not coincide with the loss-modulus maximum: the peak frequencies in a standard linear solid give $\omega \tau =\sqrt{E_0/(E_0+E_1)}$ for the loss tangent. The model captures one viscoelastic crossover between two finite elastic plateaux, with dissipation concentrated near the relaxation frequency.

Oscillatory response. A material with viscoelasticity stores some deformation energy and dissipates some over a loading cycle. In small-amplitude oscillatory shear, represent the strain by $\gamma (t)=\mathrm{Re}({\gamma }_0{e}^{i\omega t})$. The shear-stress amplitude is related to the strain amplitude by the complex viscoelastic modulusHere $G_1=G^{\prime }$ is the storage modulus, the component in phase with strain, and $G_2=G^{\prime \prime }$ is the loss modulus, the dissipative quadrature component. Both have units of pressure. The stress leads the strain by an angle $\delta$, withFor this $e^{i\omega t}$ convention, a passive material has $G_2\ge 0$ at positive frequency. Integrating $\tau \dot{\gamma }$ over one period gives dissipated energy per unit volume $\pi {\gamma }_0^2{G}_2$, and the average input power per unit volume is $\omega {\gamma }_0^2{G}_2/2$. Thus the loss tangent compares the dissipative and elastic parts of the response. The frequency dependence reflects which internal relaxation processes can follow the imposed deformation.

Torsional measurement. Use $\rho$ for radial position inside the rod, reserving $r$ for its outer radius. A small angular displacement $\theta$ of one end relative to the fixed other end gives shear strainIn harmonic amplitude notation, the shear stress is $G^{\ast }\rho \Theta /\ell$. Its restoring torque is obtained by multiplying by the lever arm and integrating over the circular section:The factor $J={\int }_A{\rho }^{2}dA=\pi r^4/2$ is the polar second moment of area; it is not the disc's mass moment of inertia $I$. The torsional spring constant becomes the complex stiffness $K^{\ast }=J{G}^{\ast }/\ell$. Angular Newton's second law therefore gives the paper's compact representationIts precise interpretation is a torsional measurement of a complex shear modulus. At a specified frequency, with complex angular amplitude $\Theta$, the driven response obeysSetting the external torque to zero supplies the free-mode condition, whose damped frequencies are generally complex. A dissipative free oscillator does not execute a sustained nonzero oscillation at a real frequency without driving. In real time, a general frequency-dependent complex viscoelastic modulus represents a causal memory response, rather than a complex instantaneous force on a real displacement. The derivation assumes a uniform slender circular rod, small twist, negligible rod inertia relative to the disc, and no additional bending, slip or external damping.

Frequency regimes and sketches. A permanently crosslinked elastomer has a low-frequency rubbery plateau $G_R>0$. Slow chain rearrangements can follow the strain, so the network is relatively compliant but still elastic. At high frequency the relevant segmental motions are frozen, giving a much larger unrelaxed or glassy stiffness $G_U$. Between these regimes the motions lag behind the loading and dissipate strongly.

A standard linear solid model supplies a consistent illustrative sketch of this process. Put $x=\omega {\tau }_r$, where ${\tau }_r$ is the stress viscoelastic relaxation time, and $\Delta G=G_U-G_R>0$. ThensoAt very low frequency, $G_1\to G_R$, $G_2\sim \Delta Gx$ and $\tan \delta \to 0$. Through the relaxation region $G_1$ rises, $G_2$ has a maximum and $\tan \delta$ has a loss peak. At high frequency, $G_1\to G_U$, $G_2\sim \Delta G/x$ and $\tan \delta \to 0$ again. Real polymers usually have a spectrum of relaxation times, so these features are broader and can have further structure.

The two kinds of loss peak need not coincide. The peak frequencies in a standard linear solid areThe first follows by differentiating $x/(1+x^2)$; the second by differentiating $x/(G_R+G_U{x}^2)$. This is why the sketch marks the peaks separately. If the polymer is an uncrosslinked melt rather than a permanent network, there is an additional ultimate low-frequency terminal-flow regime, with $G_1\propto {\omega }^2$ and $G_2\propto \omega$; a permanent elastomer does not have that zero-frequency fluid limit.

Interstitial atoms in a metal. A suitable mechanism is the Snoek effect, a form of anelasticity. An interstitial defect distorts its local crystal structure anisotropically. Different site orientations become unequally favorable under an applied stress, so hopping atoms redistribute towards the preferred orientations and supply additional strain. The hopping time is thermally activated, approximately ${\tau }_a={\tau }_0{e}^{E_a/(k_{B}T)}$, with activation energy $E_a$.

At low frequency, $\omega {\tau }_a\ll 1$, the populations nearly equilibrate during each cycle. The defect contribution to strain is nearly in phase with stress, giving a relaxed compliance and little dissipation. At intermediate frequency, $\omega {\tau }_a\sim 1$, redistribution lags behind the stress and energy dissipation is largest. At high frequency, $\omega {\tau }_a\gg 1$, the populations are effectively frozen; the extra relaxation strain cannot develop, and the response approaches the stiffer unrelaxed elastic limit. For weak single-mode relaxation, the loss tangent is approximately proportional to $\omega {\tau }_a/[1+(\omega {\tau }_a{)}^2]$. The metal remains a solid in the low-frequency regime: the extra delayed strain is recoverable, not unrestricted viscous flow.

Use a rheometer, for example a cone-and-plate rheometer or a parallel-plate rheometer, to impose a small deformation and measure the required torque, or impose a torque and measure the deformation. The apparatus geometry converts these into shear stress and strain. First check that reducing the imposed amplitude does not change the normalized response; this identifies the linear regime and avoids yielding or structural rearrangement.

Three complementary protocols characterize the material's memory. In a stress relaxation experiment impose a small step ${ϵ}_{0}H(t)$ and hold the deformation fixed. The output is the relaxation modulus $G(t)=\sigma (t)/{ϵ}_0$, revealing how stored stress decays and whether an elastic plateau remains. In a creep compliance measurement impose a small stress step ${\sigma }_{0}H(t)$ and record $J(t)=ϵ(t)/{\sigma }_0$. Removing the load also distinguishes recoverable deformation from permanent flow.

In small-amplitude oscillatory shear, impose sinusoidal strain over a range of frequencies and measure both stress amplitude and phase. With $ϵ={ϵ}_0\cos \omega t$, fitThe outputs are the storage modulus $G^{\prime }$ and loss modulus $G^{\prime \prime }$, or equivalently amplitude ratio and loss tangent $\tan \delta =G^{\prime \prime }/G^{\prime }$. These measure elastic storage and dissipative response, respectively. The step experiments give time-domain information and the oscillatory experiment gives the corresponding frequency-domain response. All comparisons must use the same deformation convention and controlled temperature.

The dynamic modulus, or complex viscoelastic modulus, is the ratio of complex stress amplitude to complex strain amplitude in the periodic steady response:This defines the sign convention; for passive response the loss modulus is nonnegative at positive frequency. A Newtonian fluid obeys $\sigma =\eta \dot{ϵ}$, while a Hooke's law solid obeys $\sigma =G_0ϵ$. HenceThe first is purely dissipative and the second purely elastic. Here $ϵ$ denotes the scalar deformation used in the problem, with any shear-versus-extension geometric factors already included in its modulus and viscosity.

Parallel elements have the same strain, while their stresses add. The spring supplies $G_0ϵ$ and the viscous element supplies ${\eta }_0\dot{ϵ}$, so the Kelvin-Voigt model has constitutive equationFor the oscillatory deformation this givesPhysical stress and strain are the real parts of these expressions. Thus the storage modulus is $G_0$, independent of frequency, and the loss modulus is $\omega {\eta }_0$. The equation is the quasistatic material law; an inertial equation for an attached mass would require additional mechanical data.

For elements in series the stress is common and the strains add. If ${\widehat{ϵ}}_K$ and ${\widehat{ϵ}}_1$ are the Kelvin element and added spring amplitudes, thenThe compliances therefore add, yieldingThis is a standard linear solid model. DefineWriting $x=\omega \tau$, its storage modulus and loss modulus areThus $G^{\prime }$ increases from the low-frequency plateau $G_L$ to the high-frequency plateau $G_H$. The loss modulus has low-frequency slope $+1$ and high-frequency slope $-1$ on a log-log plot, with maximum $\Delta G/2$ at $\omega =1/\tau$. At that frequency $G^{\prime }=(G_L+G_H)/2$. There need not be a crossing of the two curves for arbitrary spring constants.

A Kelvin-Voigt model has no finite instantaneous elastic compliance: the viscous element prevents a sudden deformation under a finite stress step. Its creep compliance isConversely an ideal strain step requires an impulsive stress,After the impulse the held-strain response is simply constant. There is no finite stress-decay interval, and the loss modulus grows without bound as $\omega \to \infty$. This fails to represent a material with both a finite instantaneous stiffness and delayed stress relaxation.

The added series spring resolves this particular defect. At the instant of loading the Kelvin element cannot move but the added spring can. The standard linear solid model consequently hasandIts instantaneous stiffness is $G_H=G_1$, its long-time stiffness is $G_L$, and it exhibits genuine stress relaxation as well as delayed creep. The creep retardation time ${\eta }_0/G_0$ differs from the stress-relaxation time $\tau ={\eta }_0/(G_0+G_1)$. These properties also follow directly from its constitutive equation,It remains an idealized solid with one relaxation mode: it cannot describe unrestricted long-time viscous flow, nonlinear response, or a broad spectrum of relaxation times.

By definition, the relaxation modulus is the stress generated by a unit step of strain from an initially relaxed state. Set $ϵ(t)=H(t)$, where $H$ is the Heaviside step function, and write $\sigma (t)=G(t)$. Integrating the linear ordinary differential equation across the step shows that neither $G$ nor $G^{\prime }$ can jump: a jump would produce a Dirac delta function or its derivative, absent from the forcing. Hence $G(0^{+})=G^{\prime }(0^{+})=0$.

For $t>0$, a particular solution is $1$. The characteristic equation factors asso $G=1+A{e}^{-t/5}+B{e}^{-t/3}$. The initial conditions give $A+B=-1$ and $A/5+B/3=0$, whenceThe time and modulus units are those implicit in the printed coefficients. A Laplace transform check gives $\hat{G}(s)=1/[s(5s+1)(3s+1)]$.

The original PDF really does put undifferentiated strain on the right-hand side. The resulting response rises from zero to one, with $G^{\prime }(t)=\frac{1}{2}(e^{-t/5}-e^{-t/3})>0$, rather than exhibiting ordinary stress relaxation. There is a stronger physical problem if this is intended as a passive material law: with the convention $e^{i\omega t}$ the stress-to-strain amplitude ratio isThe loss modulus would give negative cycle-averaged supplied power $⟨\sigma \dot{ϵ}⟩=\omega G^{\prime \prime }|{ϵ}_0{|}^2/2$, contrary to passivity in linear viscoelasticity. The boxed expression is therefore the formal relaxation modulus of the equation as printed, not a claim that this equation describes an ordinary passive viscoelasticity model. No derivative of the input has been silently inserted.

Use SI units, uniform absolute permittivity $\epsilon$ and a neutral bulk reservoir. Local equilibrium of ideal mobile ions gives $n_i=n_{i,\infty }{e}^{-z_{i}e\psi /(k_{B}T)}$. Combining their charge with the Poisson equation yieldsThe reference is $\psi =0$ in the bulk. Boundary data specify fixed charge, fixed potential or a surface reaction/charge regulation law. These are physically different problems. The mean-field theory assumes point ions, thermal equilibrium and a continuum solvent; ion-size packing, dielectric structure and strong electrostatic correlations can invalidate it.

For a symmetric monovalent electrolyte let $n_{\infty }$ count each sign and define $u=e\psi /(k_{B}T)$, ${\kappa }_D^2=2e^2{n}_{\infty }/(\epsilon k_{B}T)$. Outside fixed sources,For $|u|\ll 1$, the Debye-Hückel approximation is ${\nabla }^{2}u={\kappa }_D^{2}u$. A planar wall bounding electrolyte in $z>0$ has $\psi ={\psi }_0{e}^{-{\kappa }_{D}z}$; assuming zero field on the other side, its charge is $\sigma =\epsilon {\kappa }_D{\psi }_0$. A sphere of radius $a$ and total charge $Q$ instead givesThe factors arise from geometry and boundary conditions; planar screening cannot be substituted unchanged for a small sphere. The weak-potential condition must hold where the potential is largest, particularly at the surface.

The nonlinear planar Poisson-Boltzmann profile is also exactly integrable. Multiply $u^{\prime \prime }={\kappa }_D^2\sinh u$ by $u^{\prime }$ and use $u,u^{\prime }\to 0$ at infinity:The sign selects a profile decaying to the bulk for either sign of surface potential. Integration givesFor weak $u_0$ this reduces to the linear solution. Strong surface potentials have a bounded far-field amplitude $4\gamma$, while the near-wall ion density still grows strongly. There is no need to permit the negative ion densities produced by an invalid extrapolation of a linearized density formula.

The same first integral yields the Grahame equation and its diffuse-layer differential capacitance per unit area:These describe the diffuse layer. A compact interfacial layer adds a voltage drop and can place its capacitance in series with $C_D$; a measured electrode capacitance need not equal the ideal diffuse contribution.^[ref]

For comparison, a planar counterion-only double layer next to a negative plane has positive counterions and no salt reservoir. Its profile is algebraic:Integrating $en(z)$ over $z>0$ gives $|\sigma |$, checking charge neutrality. This problem cannot be obtained by putting a finite reservoir density into the symmetric-electrolyte answer without changing its boundary conditions.

Applications include screening around synthetic colloids, diffuse-layer electrode capacitance and forces between charged plates. In planar geometry, the bulk-subtracted normal stress is constant:At the midpoint of identical plates, symmetry sets $E=0$, so the remaining excess ionic osmotic pressure determines their repulsion. A comparison with force or capacitance measurements should vary salt and surface potential independently, test charge regulation and adsorption, and distinguish compact-layer effects from failures of diffuse screening. Nonlinear Poisson-Boltzmann theory retains finite-potential Boltzmann populations, but still omits correlations and finite-ion-volume saturation.

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.

A dead-load traction is prescribed force per unit reference area whose vector is independent of the deformation. It is imposed through $T_{R}N=t_R$ on the reference boundary. It does not rotate or rescale with the current surface as a follower load would.

Elastic incompressibility imposes $\det F=1$ and adds a scalar Lagrange multiplier to the constitutive law. With the pressure sign convention used here,The pressure field is determined by equilibrium and boundary conditions, rather than by the strain energy density alone.

For the diagonal homogeneous deformation, $F=\mathrm{diag}(v_1,v_2,v_3)$ and $v_1{v}_2{v}_3=1$. Isotropy makes $DW(F)$ diagonal here, with entries ${\Phi }_i$, soThe six face conditions are exactly $T_R{e}_i=\tau e_i$. Thus ${\Phi }_i-p/v_i=\tau$ for each $i$, orNecessity follows from the boundary conditions. Conversely these equations give a constant pressure and constant stress satisfying those conditions, while $\mathrm{Div}{T}_R=0$ because the stress is constant and there is no body force. This proves sufficiency as well.

For the particular energy, ${\Phi }_i=\mu (v_i-v_i^{-2})$. Put $S=v_1+v_2+v_3$ and $Q=v_1{v}_2+v_1{v}_3+v_2{v}_3$. Elastic incompressibility gives $v_i^{-1}=v_j{v}_k$, henceEquality for every pair is therefore equivalent to $(\mu S-\tau )(v_i-v_j)=0$. If all stretches are equal, their positive product one gives $v_1=v_2=v_3=1$, with $p=-\tau$. Otherwise $S=\tau /\mu$. Conversely any positive triple with this sum and product one solves the equations, with $p=-\mu Q$. Thus the complete solutions within the stated homogeneous diagonal ansatz are the unit triple and the stated sum-product family.

By the arithmetic-geometric mean inequality, $S\ge 3$, with equality only at the unit triple. Consequently $\tau \le 3\mu$ admits only that triple, including equality, when the two descriptions coincide.

To prove infinitely many for $s=\tau /\mu >3$, choose $v_1=t$ near one and setAt $t=1$ the discriminant is $(s-1{)}^2-4>0$; it stays positive on an interval about one, and both roots remain positive. Every such $t$ gives a different admissible triple. HenceThe degeneracy has an energy explanation: under the product constraint the strain energy density is $\mu S^2/2$, and the total potential under dead loading is $\mu S^2/2-\tau S$. It depends on the three stretches only through their sum. Its minimum over attainable $S\ge 3$ moves from $S=3$ to the whole level set $S=\tau /\mu$ when the load exceeds $3\mu$.

For a hyperelastic material $T_R(F)=DW(F)$. Material frame indifference of this first Piola-Kirchhoff law meansThe spatial force components rotate under a superposed rigid rotation, whereas the reference normal does not.

If $W(QF)=W(F)$, differentiate with respect to $F$ in direction $H$ to obtain $DW(QF):(QH)=DW(F):H$, hence the displayed stress transformation. Conversely, assume that stress transformation. For fixed $Q$, the derivative of $W(QF)-W(F)$ with respect to $F$ vanishes. The positive-determinant matrix domain is connected, as follows from polar decomposition and the connectedness of $SO(3)$ and of the positive-definite cone. Thus $W(QF)-W(F)=c(Q)$, independent of $F$. Iterating givesThe assumed lower bound on $W$ forces $c(Q)=0$: either nonzero sign would tend to minus infinity for one direction of integer $n$. This proves the required equivalence and explains the use of the lower-bound hypothesis.

In the frame-indifferent setting, isotropy means invariance under reference rotations as well: $W(FQ)=W(F)$ for $Q\in SO(3)$. The distinction matters: left rotations encode changes of observer, and right rotations encode material symmetry.

An orientation-preserving singular value decomposition is $F=R\mathrm{diag}(v_1,v_2,v_3)S$, with $R,S\in SO(3)$ and $v_i>0$. Material frame indifference and isotropy remove $R,S$, leaving a function $\Phi$ of the stretches. Permuting the stretches using signed permutation rotations does not change the value, so $\Phi$ is symmetric. Conversely a symmetric stretch function is unchanged by both left and right rotations, in particular is isotropic.

The eigenvalues of $B=F{F}^T$ are $v_i^2$. Its three principal invariants areThey determine the unordered positive eigenvalues as the roots of ${\lambda }^3-I{\lambda }^2+II\lambda -III=0$. Therefore a symmetric stretch function defines a well-defined $h(I,II,III)$ on the admissible invariant triples. Conversely a function of these invariants is a symmetric stretch function, proving the equivalence of all three descriptions.

The printed domain of $\Phi$ must be $(0,\infty {)}^3$, since it takes three arguments. Also isotropy alone as right invariance would not imply the spectral representation without material frame indifference: $\mathrm{tr}(AF{F}^T)$ for a fixed nonscalar symmetric positive-definite $A$ is right invariant but not generally objective. The equivalence here uses the frame-indifferent constitutive setting.

For the specific energy, Newton identities for the three eigenvalues gives ${\sum }_i{v}_i^6=I^3-3III+3III$, while $v_1{v}_2{v}_3=J=\sqrt{III}$. ThusTo differentiate it, useThese follow from $I=|F{|}^2$, the displayed trace formula for $II$, and $III=J^2$. The three partial derivatives of $h$ are $h_I=\alpha +3\beta I^2-3\beta II$, $h_{II}=-3\beta I$ and $h_{III}=3\beta -\gamma II{I}^{-3/2}/2$. The chain rule therefore givesFinally $-6\beta II=-3\beta (\mathrm{tr}B{)}^2+3\beta \mathrm{tr}(B^2)$, and $BF=F{F}^{T}F$. This is precisely the requested expression, with the coefficient of $F$ separated from the $BF$ and cofactor terms. The positive sign of the $6\beta J$ term comes from the $3\beta III$ contribution in the energy.

Retain the material frame indifference assumption used in part (c), in addition to material isotropy. Put $p=u_{,1}$, $q=u_{,2}$ and $s=p^2+q^2$. For this antiplane shear,The invariant identities give $I_B=I{I}_B=3+s$ and $II{I}_B=1$. Hence the restriction of the strain energy density isAntiplane energy does not determine the full stress: the restriction gives one scalar energy along antiplane gradients. It does not determine the full stress derivative with respect to every entry of $F$.

Indeed, using the isotropic hyperelastic stress invariant representation and differentiating the energy, with $h_j$ denoting the corresponding partial derivatives, givesUsing $S=J^{-1}T{F}^T$, the general objective isotropic Cauchy stress tensor isAll derivatives are evaluated at the current invariant triple. The extra terms generally remain nonzero on antiplane gradients, so the printed assertion that the full stress must be $\beta B$ does not follow.