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The bulk modulus relates hydrostatic compressive pressure to fractional volume compression. In isotropic three-dimensional linear elasticity, the spherical part of the stress tensor is $\kappa \mathrm{tr}(\epsilon )I$, so uniform dilation has strain energy density $\kappa (\mathrm{tr}\epsilon {)}^2/2$. Positive bulk modulus and positive shear modulus make the isotropic elastic stiffness tensor positive on nonzero symmetric strains.

When only the bulk modulus differs from an isotropic matrix, the elastic polarization is spherical: $\tau =(\kappa -{\kappa }^0)\mathrm{tr}(\epsilon )I$. The double trace of the isotropic elastic strain Green kernel makes the trace equation local, giving the displayed formula. Individual strain components still depend on a nonlocal potential of the spherical polarization, so the complete infinitesimal strain tensor is not generally a local multiple of its reference value.

A constitutive equation relates continuum fields such as stress, strain, and the rate-of-strain tensor to characterize a material beyond universal balance laws.

Deformation changes a body's shape or size. Strain quantifies relative changes in lengths and angles, whereas a rigid translation or rotation alone produces no strain. The continuum approximation describes these changes using displacement fields and their spatial variation.

The effective elastic energy is the minimum averaged strain energy density under prescribed affine boundary displacement. For differentiable stable elastic response without body force, varying the prescribed mean strain gives $D{W}^{\mathrm{eff}}(E)=⟨\sigma ⟩$. The result follows by integration by parts, because the interior variation with zero boundary data does no work against an equilibrated stress tensor.

An elastic laminate has planar perfectly bonded layers with a common normal. Its piecewise constant strains differ by $\mathrm{sym}(a\otimes n)$, while its normal traction is common to every layer. For positive elastic stiffness tensors, these conditions and the mean strain uniquely determine the layer strains: the difference of two solutions has zero mean and zero total elastic work, and positivity forces it to vanish.

The elastic polarization is the excess stress relative to a homogeneous comparison elastic stiffness tensor. In heterogeneous linear elasticity, $\tau =(C-C^0)\epsilon$. The comparison static elastic equilibrium equation gives the integral equation $\epsilon ={\epsilon }^0-\Gamma \tau$, where ${\epsilon }^0$ is the reference strain under the same external loading.

An elastic stiffness tensor maps a small displacement gradient tensor to the stress tensor. Minor symmetries in each index pair ensure symmetric stress and dependence only on symmetric strain; major symmetry $c_{ijpq}=c_{pqij}$ follows from an elastic energy and underlies reciprocity. Stability requires positive elastic energy on nonzero strains. In an isotropic material, $c_{ijpq}=\lambda {\delta }_{ij}{\delta }_{pq}+\mu ({\delta }_{ip}{\delta }_{jq}+{\delta }_{iq}{\delta }_{jp})$, with Lamé parameters $\lambda ,\mu$.

The two minor symmetries of the elastic stiffness tensor make it a map between six-dimensional spaces of symmetric strain and stress. An elastic energy gives major symmetry, so the matrix of this map is symmetric and has 21 independent entries. For a symmetry operation acting on strains with parity spaces of dimensions $m$ and $6-m$, invariance eliminates the mixed block and leaves $m(m+1)/2+(6-m)(7-m)/2$ independent coefficients.

The infinitesimal strain tensor is the symmetric part of a displacement gradient tensor. It is the linear approximation to a finite strain measure for small displacement gradients. An infinitesimal rigid rotation contributes no strain, so the elastic energy depends on the symmetric part rather than the entire gradient.

For a three-dimensional isotropic comparison material, the contraction of its elastic strain Green operator is the displayed distributional derivative. Its double trace is ${\Gamma }_{iikk}=\delta /L$, because $\Delta |x{|}^{-1}=-4\pi \delta$. Away from zero the contraction is $({\delta }_{ij}|x{|}^2-3x_i{x}_j)/(4\pi L|x{|}^5)$; the full distribution also includes ${\delta }_{ij}\delta /(3L)$. This contact term must not be discarded when computing hydrostatic strain.

Let a convex strain energy density have derivative equal to the actual stress tensor. Under mixed displacement and traction data, any equilibrated stress ${\sigma }_0$ with the prescribed body force and boundary tractions represents the external work. For any compatible trial strain ${\epsilon }^{\prime }$, the convex supporting-plane inequality gives $W({\epsilon }^{\prime })-W(\epsilon )\ge \sigma :({\epsilon }^{\prime }-\epsilon )$. The integral of $(\sigma -{\sigma }_0):({\epsilon }^{\prime }-\epsilon )$ vanishes by integration by parts. Thus the actual strain minimizes $\int (W-{\sigma }_0:\epsilon )$.

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.