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Take the difference $w$ of two solutions with the same body force, initial displacement and initial velocity. It satisfies the homogeneous linear elasticity equations. With $e=e(w)$ the infinitesimal strain tensor, define the bulk energyThe usual stability assumptions make the elastic quadratic form nonnegative and $\rho >0$. Multiplying momentum balance by $w_t$, using integration by parts and using symmetry of the stress tensor givesAll interior terms cancel: the strain term is exactly the time derivative of elastic energy.

For the restoring boundary law, $t_w=kw$. The prescribed support displacement cancels in the difference. Since $k$ is a time-independent symmetric matrix,Hence the elastic energy uniqueness with restoring boundary springs identity isThe boundary contribution is nonnegative because $k$ is negative semidefinite. The initial difference and its velocity vanish, so the conserved energy is zero. Its kinetic term implies $w_t=0$ everywhere; the zero initial displacement then gives $w=0$. This proves uniqueness without needing to exclude rigid displacements separately.

Physically, $-k$ is a nonnegative boundary spring-stiffness matrix, and the prescribed vector $U$ is the moving support position. The traction restores the displacement towards that support; null directions of $k$ have no spring force. The interpretation and conserved-energy proof use a fixed spring matrix, as indicated by the given boundary law.

On the mixed part $S_1$, write $t=\sigma n$. The condition isThus the tangential traction and normal displacement are prescribed. The normal traction is a constraint reaction; tangential displacement is free. Only the tangential component of $F$ matters. On $S_2$, the entire displacement is prescribed.

For the difference of two such solutions, $w\cdot n=0$ and $(I-n{n}^{\mathsf{T}})t_w=0$ on $S_1$. Consequently $w_t$ is tangential and $t_w$ normal, so $w_t\cdot t_w=0$. On $S_2$, $w_t=0$. The boundary power vanishes everywhere, giving $d{E}_{\mathrm{bulk}}/dt=0$. The same zero-initial-energy argument gives $w=0$. Therefore the mixed boundary problem also has at most one solution.

Use the local momentum equation in the current volume:Symmetry of the Cauchy stress tensor and the product rule giveApply the divergence theorem and use $t_j={\sigma }_{jk}{n}_k$. The mean stress and strain boundary identities follow:For infinitesimal strain tensor, $e_{ij}=(u_{i,j}+u_{j,i})/2$, and a second application of the divergence theorem gives

To first order, evaluate self-gravity in the undeformed uniform sphere. The shell theorem says that only the enclosed mass $M(r)=4\pi \rho r^3/3$ contributes to the inward field. ConsequentlyThe field vanishes at the centre and increases linearly to its surface value.

The requested radius change concerns the final traction-free static elastic equilibrium, not the acceleration immediately after gravity is switched on. In equilibrium the boundary traction and acceleration vanish. Spherical symmetry givesThe homogeneous linear elasticity law commutes with volume averaging:Thus $\overline{e}=e_{0}I$ with $e_0=-\rho ga/(15\kappa )$. This conclusion is about the mean strain, not a claim that the local strain is uniform.

Let the boundary displacement be $u(a)=u_{a}n$. Since ${\int }_S{n}_i{n}_{j}dS=(4\pi a^2/3){\delta }_{ij}$, the boundary identity gives $\overline{e}=(u_a/a)I$. Therefore the equilibrium contraction of a self-gravitating elastic sphere isThe shear modulus cancels because the mean deformation is purely volumetric.

The full radial field provides a useful check without assuming uniform deformation. For $u=f(r)e_r$, momentum balance isRegularity at the centre gives $f=Ar+B{r}^3$, withfrom ${\sigma }_{rr}(a)=0$. Substitution reproduces $f(a)=-\rho g{a}^2/(15\kappa )$. During the transient, the acceleration term in the mean-stress identity remains; gravitational feedback from the small density and shape changes is higher order in the equilibrium calculation.

Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.

For a free surface at $x_3=0$ with solid occupying $x_3>0$, let $(q_0,a_0)$ be a unit-amplitude incident upgoing mode. The reflected field consists of the three downgoing propagating or depth-decaying modes:Frequency and horizontal slowness are common to all modes, by translation invariance of the surface. Zero surface traction imposes the three equationsHence the free-surface anisotropic reflection amplitudes arewhenever $B$ is invertible. This specifies mode conversion as well as reflection. If $B$ is singular, a nonzero outgoing homogeneous surface mode exists; these are surface-wave poles, including Rayleigh waves in the isotropic case. At such parameters one must use the appropriate limiting radiation solution and check solvability instead of assuming a unique finite inverse.

Use the Unscaled Papkovich–Neuber representationwith harmonic vector and scalar potentials. For a torque-driven sphere chooseIts components are harmonic for $r>a$, and both $\mathbf{x}\cdot \mathbf{\Phi }$ and $\nabla \cdot \mathbf{\Phi }$ vanish. HenceThe last equality follows from the no-slip boundary condition at $r=a$. To check that $\mathbf{G}$ is indeed the applied couple, the traction on the sphere, with $\mathbf{n}=\mathbf{x}/a$ directed into the fluid, is ${\mathbf{\sigma }}_G\mathbf{n}=-3\mu {\mathbf{\Omega }}_G\times \mathbf{n}$. Its hydrodynamic torque isHere $\int \mathbf{nn}dS=(4\pi a^2/3)I$. The applied couple balances this resisting torque. The resulting field is the rotlet of a rotating sphere in Stokes flow.

For the Lorentz reciprocal theorem for Stokes flow, let ${\mathbf{n}}_D$ point out of the fluid domain $D$ and take the same viscosity in both flows. With $\nabla \cdot {\mathbf{\sigma }}_i+{\mathbf{f}}_i=0$, the theorem isIt follows by taking the divergence of the boundary integrand: incompressibility removes pressure-work terms, the two Newtonian strain contractions are equal, and the remaining stress divergences give the displayed body-force terms. This convention matters: on the inner boundary of an exterior domain, ${\mathbf{n}}_D=-\mathbf{n}$.

Measure pressure relative to its hydrostatic value and let $\Delta P$ be the lower-side excess pressure minus the upper-side excess pressure. The long unobstructed parts of the tube act as Poiseuille resistances in series. Their lengths add to $L$ at leading order because $l\ll L$; entrance and exterior resistances are smaller for $L\gg a$. ConsequentlyHere $Q$ is downward laboratory-frame volume flux.

Take $z$ upwards in the particle frame and let $s$ measure distance from the particle surface to the tube wall. In this frame the particle is stationary and the wall moves upwards at speed $U$. In the thin, slowly varying gap, pressure is uniform across $s$ and the leading axial Stokes equation is $\mu w_{ss}=p_z$. No slip givesThe gap's upward relative flux is constant, while conservation of displaced volume gives $2\pi aq=\pi a^{2}U-Q$. Thus $p_z=6\mu U/h^2-12\mu q/h^3$, and integration from bottom to top yieldsCurvature corrections to the gap circumference are relative order $h/a$. Rapid end regions do not contribute to the leading singular neck resistance.

For the force use a cylindrical fluid control volume containing the particle, with full-area end caps just beyond it and the particle surface as an inner boundary. The upward pressure force on the caps is $\pi a^2\Delta P$. The outer wall exerts axial traction $\mu w_s(h)$ on the fluid, whereas the particle's traction on the fluid integrates to minus its upward hydrodynamic resistance $F$. The zero-inertia momentum balance is thereforeThe velocity profile gives $w_s(h)=4U/h-6q/h^2$, provingUsing the outer-wall traction is essential; directly using the particle-side shear and a flat-end pressure force would omit the pressure on the sloping particle surface. This is control-volume drag formula for a sphere in a tube.

The four dimensionless relations areSolving them givesWithin the preceding lubrication and tube-resistance approximations, the force isSince $h/a\ll 1$, $I_3^{\ast }\gg I_2^{\ast }\gg I_1^{\ast }$. Dropping the smaller additive $-2I_2^{\ast }+4I_1^{\ast }/3$ terms in the numerator yields the requested leading formulaThe Cauchy-Schwarz inequality gives $(I_2^{\ast }{)}^2\le I_1^{\ast }{I}_3^{\ast }$, so the retained $\delta$ term is positive. The term multiplied by $\delta$ must be retained: it can compete with $I_3^{\ast }$ even though the unmultiplied lower integrals cannot.

For a sphere of radius $a(1-ϵ)$, the neck has $h=aϵ+z^2/(2a)$ at leading order. With $z=a\sqrt{2ϵ}t$, the integrals reduce to $\sqrt{2}{ϵ}^{1/2-n}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-n}dt$. For $n=1,2,3$ these last integrals are $\pi ,\pi /2,3\pi /8$, giving the full two-sided integrals. PutThe printed sphere integrals are half these values: they correspond to integrating over one side of the neck rather than the stated interval spanning both sides. The substitution above establishes the radius-deficit normalization of sphere-tube lubrication integrals. Using the two-sided values, the force becomesso the three answers areThese are the corrected sphere-tube drag regimes with a radius-deficit gap. If the supplied half-integrals are treated as stipulated data instead, the same algebra gives $3\pi {ϵ}^{-5/2}/(8\sqrt{2})$, $1/\delta$ and $\pi \sqrt{2}{ϵ}^{-1/2}/3$ in the three regimes. Thus the first and third numerical drag coefficients double when the geometrically correct integrals are used; the regime boundaries and dominant flow patterns are unchanged.

In the first regime, $Q^{\ast }\sim \delta c_3{ϵ}^{-5/2}\ll 1$, $q^{\ast }\sim 1/4$, and $\Delta P^{\ast }\sim c_3{ϵ}^{-5/2}\sim F^{\ast }$. Almost no displaced fluid escapes through the long tube; it bypasses upwards through the thin annulus. The large pressure-driven bypass resistance sets the drag, with shear smaller.

In the second regime, $Q^{\ast }\sim 1$, $\Delta P^{\ast }\sim F^{\ast }\sim 1/\delta$, andAlmost all displaced volume travels through the unobstructed tube, whose pressure resistance dominates the force. Both terms in this small gap flux are worth displaying: their crossover at $\delta \sim {ϵ}^{3/2}$ changes the balance between pressure-driven and wall-driven gap transport but does not create a fourth leading force regime.

In the third regime, $1-Q^{\ast }\sim 4ϵ/3$, $q^{\ast }\sim ϵ/3$, and $\Delta P^{\ast }\sim 1/\delta \ll F^{\ast }$. Tube throughflow still carries almost all the displaced volume, but the dominant resistance is local annular lubrication rather than the net tube pressure drop. The relative gap flux is now of order $Uh$, so wall-driven and pressure-driven components both matter locally. The formulas describe hydrodynamic resistance after subtracting hydrostatic buoyancy; terminal settling would balance it against the particle's excess weight.

Write $⟨q⟩=|\Omega {|}^{-1}{\int }_{\Omega }qdx$, $E=⟨\epsilon ⟩$ and $S=⟨\sigma ⟩$. Use the infinitesimal strain tensor, symmetric stress, zero body force, and the outward unit normal $n$. The divergence theorem applied separately to the two displacement derivatives givesFor static elastic equilibrium, ${\sigma }_{ik,k}=0$, and therefore ${\partial }_k({\sigma }_{ik}{x}_j)={\sigma }_{ij}$. With traction $t_i={\sigma }_{ik}{n}_k$, the divergence theorem now givesThe right-hand side is automatically symmetric because the volume stress tensor is symmetric; it can also be written as the integral of $(t_i{x}_j+t_j{x}_i)/2$.

To prove the work identity, put $v=u-Ex$. The constant symmetric tensor $S$ is equilibrated, so integration by parts givesHere the two cross terms each equal $|\Omega |S:E$, and the last constant term restores one of them. Division by the volume proves the required identity. In particular, affine boundary displacement makes $v=0$ on the boundary and yields the Hill-Mandel energy identity. The zero-body force hypothesis matters: otherwise the first integration by parts has an additional volume-force term.

At a perfectly bonded elastic interface without an interfacial force, displacement and traction are continuous:Tangential derivatives of the common displacement trace are also continuous. Thus the jump of a piecewise constant displacement gradient tensor annihilates every tangential vector, and has the form $\alpha \otimes n$. Taking its symmetric part shows that the strain jump is $B\alpha$, whereThe adjoint formula follows from $\tau :B\alpha =(\tau n)\cdot \alpha$ for symmetric stress tensors. Tangential-tangential strain components are therefore common to all layers. Conversely, such gradient jumps can be realized by continuous piecewise affine displacement in the layered geometry.

Take positive elastic stiffness tensors with their usual minor and major symmetries, and a positive comparison elastic stiffness tensor $C^0$. A fictitious layer can have the same tangential displacement derivatives and the same traction as the physical layers. Hence ${\epsilon }^r={\epsilon }^0+B{\alpha }^r$. The common traction condition giveswhere $K^r=B^T{C}^{r}B$ is the positive acoustic tensor, with components $K_{ik}^r=C_{ijkl}^r{n}_j{n}_l$. Solving for ${\alpha }^r$ givesExpanding the two occurrences of $B$ gives exactly the symmetrization over $(ij)$ and $(kl)$ in the directional elastic strain Green operator specified in the paper.

For any two layers, the inverse identity for their acoustic tensors proves the required resolvent relation without interchanging noncommuting tensors:With $D_r=C^r-C^0$, this identity also gives ${\tilde{\Gamma }}^0{D}_r{\tilde{\Gamma }}^r={\tilde{\Gamma }}^0-{\tilde{\Gamma }}^r$. ThereforeThese finite-dimensional operators are inverses. Define $A_r=[I+{\tilde{\Gamma }}^0(C^r-C^0){]}^{-1}$. Then ${\epsilon }^r=A_r{\epsilon }^0$, and averaging strains and stresses givesEliminating the fictitious strain yieldsAll products have the displayed order; the elastic stiffness tensors need not commute.

It remains to justify the inverse and comparison independence. Consider two compatible equilibrated layer solutions with the same mean strain. Their difference has zero common tangential-tangential strain, so $\delta {\epsilon }^r=B{\beta }^r$, with ${\sum }_r{c}_r{\beta }^r=0$ because $B$ is injective and the mean difference is zero. Their traction difference is one common vector $g$. ThereforePositive elastic energy forces every $\delta {\epsilon }^r$ to vanish. In particular, if $⟨A⟩{\epsilon }^0=0$, all physical layer strains vanish. Their common tangential components then force ${\epsilon }^0=B{\beta }^0$, and their zero traction gives $K^0{\beta }^0=0$. Hence ${\epsilon }^0=0$, proving that $⟨A⟩$ is invertible. Finally, every positive choice of $C^0$ constructs the unique layer solution for the same $E$. Its averaged stress is unique, so the displayed effective stiffness is independent of the fictitious comparison material. This is the exact layered constitutive response, rather than a claim that piecewise constant fields meet arbitrary finite-body boundary data.

Let $u$ be the actual displacement field, let $u^{\prime }$ be an admissible trial displacement, and set $v=u^{\prime }-u$. On the displacement-prescribed part of the boundary, $v=0$. With actual stress $\sigma =D_{\epsilon }W(\epsilon )$, the supporting-plane inequality for the convex strain energy density givesBoth $\sigma$ and ${\sigma }_0$ satisfy static elastic equilibrium with the same body force, so $\nabla \cdot (\sigma -{\sigma }_0)=0$. On the traction-prescribed boundary their tractions agree. Since ${\epsilon }^{\prime }-\epsilon =\mathrm{sym}\nabla v$, integration by parts givesSubtracting the prescribed external-work representative proves the minimum potential energy principle in elasticity:Strict convexity gives uniqueness of the minimizing strain, although rigid displacements can remain when the boundary conditions allow them.

For the comparison bound, put $F(e,x)=W(e,x)-W_0(e)$ and use the lower conjugate, with its infimum sign convention,Its defining inequality is $F_{\ast }(\tau ,x)\le \tau :e-F(e,x)$, hence $W(e,x)\le W_0(e)+\tau :e-F_{\ast }(\tau ,x)$ for every $e$. Integrate this pointwise majorant for each compatible strain, then take the infimum over the same admissible set. This provesNo convexity of $F$ is used for this inequality. If $F_{\ast }(\tau ,x)=-\infty$, it yields only the extended-real upper bound $+\infty$. For example, $W_0=0$ and $W(e)=|e{|}^2/2$ give $F_{\ast }(\tau )=-\infty$ for every finite $\tau$. A useful finite bound requires a suitable comparison energy; choosing a sufficiently stiff quadratic comparison makes the usual excess energies concave. The derivative used below additionally presupposes finite differentiable duals, or an appropriate supergradient formulation.

Now normalize $|\Omega |=1$, take positive constant $C^0$, and write $F_r(e)=W_r(e)-e:C^{0}e/2$. For phasewise constant elastic polarization $\tau ={\sum }_r{\chi }_r{\tau }^r$, its lower conjugate contribution is constant within each phase. The remaining minimization isWrite the minimizing strain as $\hat{e}=E+\eta$, where $\eta$ comes from a zero-boundary displacement field. The static elastic equilibrium equation is $\nabla \cdot (C^0\eta +\tau )=0$, so the supplied elastic strain Green operator gives $\eta =-\Gamma \tau$. Its zero mean removes the cross term with the constant $E$. Using its reciprocity and projection property $\Gamma C^0\Gamma =\Gamma$, we obtainHere the brackets in the last two terms denote the spatial integral and tensor contraction, not just a phase average.

Let $c_r={\int }_{\Omega }{\chi }_r$. The zero output mean and reciprocity of $\Gamma$ imply that it annihilates constant elastic polarizations on either side. Thus subtracting $c_r$ and $c_s$ from the phase indicators does not change their bilinear integral. Define the fourth-order block tensorsConsequently $A_{rs}=A_{sr}^T$, ${\sum }_s{A}_{rs}=0$, andThese identities follow from reciprocity and ${\sum }_s{\chi }_s={\sum }_s{c}_s=1$; no phase independence assumption is needed. The phase polarization comparison bound, valid for every polarization with finite dual terms, is thereforeIn particular, it can be optimized over the phase elastic polarizations without changing the upper-bound direction.

To derive the final displayed form, impose the stationarity equations in the question. For $c_r>0$, put $e^r=D{F}_{r\ast }({\tau }^r)$; these are dual strains, initially distinct from the comparison solution's phase averages. Reciprocity of $A$ givesThus stationarity is exactlyThe comparison solution has phase meanso the dual strains become its phase means at stationarity. Multiplying the stationarity equations by ${\tau }^r$ and summing gives ${\sum }_{r,s}{\tau }^r:A_{rs}{\tau }^s={\sum }_r{c}_r{\tau }^r:(E-e^r)$.

For clarity, the concave biconjugate here uses another infimum: $F_{r\ast \ast }(e)={\inf }_s\{s:e-F_{r\ast }(s)\}$. The lower conjugate $F_{r\ast }$ is concave, since it is an infimum of affine functions. At a differentiable finite point, its supporting-plane inequality shows that $s={\tau }^r$ minimizes this last expression when $e=e^r$. HenceSubstitute both identities into $\mathcal{B}$ to obtainThis is the requested final bound with the accompanying phase equations. It remains valid at any well-defined stationary choice; asserting existence of such a differentiable choice for arbitrary $W_0$ would require extra hypotheses. Without stationarity, the earlier $\mathcal{B}(E,\tau )$ is the correct general bound. In the concave closed excess-energy case $F_{r\ast \ast }=F_r$; for general excess energies replacing the concave biconjugate by $F_r$ without proof would reverse the needed majorant logic.

An elastic deformation is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous stress is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the displacement field $u_i$ gives the infinitesimal strain tensor $e_{ij}=(u_{i,j}+u_{j,i})/2$. The Cauchy stress tensor is defined by traction $t_i={\sigma }_{ij}{n}_j$ on a plane of unit normal $\mathbf{n}$; absence of body couples makes ${\sigma }_{ij}={\sigma }_{ji}$.

Let $\mathcal{E}(e)$ be the recoverable elastic energy per unit mass. The local mechanical power per unit volume isbecause the symmetric Cauchy stress tensor annihilates the antisymmetric part of the velocity gradient. Conservation of energy during the deformation gives $\rho \dot{\mathcal{E}}={\sigma }_{ij}{\dot{e}}_{ij}$. Since $\dot{\mathcal{E}}=(\partial \mathcal{E}/\partial e_{ij}){\dot{e}}_{ij}$ for arbitrary symmetric strain rates,Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.

Take a stress-free reference state of mass density ${\rho }_0$. The Taylor series of the elastic energy starts quadratically:Using $\rho ={\rho }_0+O(|e|)$ gives ${\sigma }_{ij}=c_{ijkl}{e}_{kl}+O(|e{|}^2)$, the relation of linear elasticity. If the reference state has initial stress, the first-order expansion also retains that constant stress. Symmetry of stress and strain gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of elastic energy gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the elastic stiffness tensor is a symmetric map on a six-dimensional strain space: there are $6\cdot 7/2=21$ independent coefficients.