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For independent sliding filaments in a filled bundle, bending energies add and $B\sim NE{r}^4\sim E{r}^2{R}^2$. A coherently bonded one-filament-thick tube has $B\sim Er{R}^3$, while a bonded filled rod has $B\sim E{R}^4$. The same torsional scalings apply with the shear modulus instead of Young's modulus if individual sliding filaments are forced to undergo a common material twist. Completely free filament spin/slip does not define that torsional response: collective helical bending alone begins at fourth order in the twist rate.

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.

For positive real mass density and shear modulus, real $k$ and $\mathrm{Im}\omega >0$, the displayed derivative of $S_z=-\frac{1}{2}\mathrm{Re}(V^{\ast }T)$ follows from the two first-order shear equations. Every nontrivial solution has strictly decreasing $S_z$. Since $S_z=\frac{1}{2}\mathrm{Re}Z|V{|}^2$, positive-real impedance at a deeper point remains positive at all shallower points; negative-real impedance at a shallower point remains negative deeper down. A zero field is the necessary exception to strictness. At complex frequency this is a spectral flux diagnostic, reducing to a physical cycle average on the real axis.

The ideal rubber elasticity relation infers active strand number density from the shear modulus. If a quoted modulus is instead Young's modulus, nearly incompressible behavior gives ${\nu }_s\simeq E/(3k_{B}T)$. This is not automatically the junction density: for junction functionality $f$ and no dangling ends or loops, counting strand ends gives ${\nu }_s=f{\nu }_j/2$. Network fluctuations and inactive strands modify the modulus, so converting it into a chemical cross-link count requires a network model.

The opening of an internal cavity under tensile loading in a solid. For spherically symmetric deformation of an isotropic hyperelastic material with incompressibility, and an empty cavity without surface tension, radial static elastic equilibrium gives the displayed critical signed external pressure, where $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$ and positive $P$ denotes compression. Indeed, the principal stress formula for incompressible hyperelasticity gives $d{t}_r/d\lambda =h^{\prime }(\lambda )/(1-{\lambda }^3)$. Integration between the free inner surface and the outer surface, followed by the vanishing-initial-hole and vanishing-current-cavity limits, gives the formula. For a neo-Hookean solid with shear modulus $\mu >0$, the tensile magnitude is $-P_{\mathrm{crit}}=5\mu /2$.

A coefficient relating an appropriate stress and strain in the linear elastic regime. Young's modulus, shear modulus and bulk modulus describe different deformation modes.

A network of flexible thermally fluctuating strands stores an order-$k_{B}T$ elastic free-energy change per active strand under an order-one deformation. For active-strand number density $n$, its shear modulus is therefore of order $n{k}_{B}T$. If one active element occupies a mesh volume $m^3$, this becomes $k_{B}T/m^3$. The numerical coefficient depends on network connectivity, defects and the deformation model. A mesh size alone does not fix the modulus of a general athermal or bending-dominated network; strand stiffness is then additional input.

For a uniform traction-free sphere, first-order self-gravity is $b=-gx/a$. Static averaging gives $\overline{\sigma }=-\rho gaI/5$. Homogeneous linear elasticity then gives $\overline{e}=-\rho gaI/(15\kappa )$. The boundary strain identity identifies $\overline{e}=u(a)I/a$, proving the displayed positive radius decrease. The local strain is not uniform, and the cancellation of the shear modulus concerns only the mean contraction.

A fluid surface has no equilibrium static in-plane shear modulus, because material can rearrange after shear. This does not set its membrane tension or membrane bending modulus to zero. Time-dependent deformation can also produce viscous in-plane shear stress; absence of a static elastic coefficient is not absence of dissipation.

Fix a positive shear modulus $\mu$ and let the other Lamé parameter tend to infinity in $S_{\lambda }=\lambda \mathrm{tr}(e_{\lambda })I+2\mu e_{\lambda }$. If strain and stress converge to finite limits and differentiation commutes with the limit, taking the trace shows $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$. Thus $\mathrm{div}u=0$ and $S=-pI+2\mu e$ for $p=-\lim \lambda \mathrm{tr}{e}_{\lambda }$. Force balance becomes the displayed Stokes equation, with an elastic shear modulus in place of fluid dynamic viscosity. The vanishing volumetric strain is the linearization of elastic incompressibility $\det Dy=1$.

Under uniform compressive pressure $p$, isotropic Hooke's law gives each normal strain as $-p(1-2\nu )/E$. The volume strain is three times this, so the bulk modulus is $B=E/[3(1-2\nu )]$, where $E$ is Young's modulus and $\nu$ is Poisson's ratio. Positivity of $E$ and finite $B$ gives $\nu <1/2$. Positive shear modulus $G_s=E/[2(1+\nu )]$ adds the lower bound $\nu >-1$. Negative Poisson's ratio is allowed; the bulk relation alone does not supply the lower bound.

A logarithmic measure of earthquake size based on seismic moment. A commonly used SI relation is $M_w=\frac{2}{3}[{\log }_{10}(M_0/(\mathrm{N}\mathrm{m}))-9.1]$. Because seismic moment integrates fault slip over the fault plane and includes the shear modulus, a peak slip value cannot replace the area integral.

For an incompressible neo-Hookean solid, a convenient strain energy density is $W=\mu (\mathrm{tr}C-3)/2$, where $C$ is the right Cauchy-Green deformation tensor. An additive constant in $W$ does not affect the stress. The shear modulus $\mu$ fixes the small-strain stiffness.

When deformation is stopped, the rate-of-strain tensor and velocity gradient vanish, leaving $\mathsf{T}+\tau {\partial }_t\mathsf{T}=0$. HenceThe positive constant $\tau$ is the viscoelastic relaxation time. For slow weak deformation, all the transport and quadratic terms are smaller than $\mathsf{T}$, so $\mathsf{T}\simeq 2\mu \mathsf{E}$. Thus $\mu >0$ is the zero-rate dynamic viscosity, measured in stress times time. The corresponding short-time shear modulus of the Maxwell fluid response is $G_{\mathrm{el}}=\mu /\tau$. The source's $\alpha$ has units of time, since $\alpha \mathsf{TE}$ must have units of stress.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

Choose zero energy in the unstrained state. The required strain energy density isIts variation is $\delta W=(\lambda \mathrm{tr}(e)I+2\mu e):\delta e$, giving the prescribed stress as the tensor gradient. This variational notation handles the symmetric off-diagonal components without a spurious factor of two.

Decompose $e=e^{\mathrm{dev}}+(\mathrm{tr}e)I/3$ into orthogonal deviatoric and spherical parts. ThenThe two squares vanish together only for $e=0$. Thus $\mu >0$ and $3\lambda +2\mu >0$ are sufficient for positive definiteness. They are necessary by testing a nonzero trace-free strain and a pure dilation separately. Equivalently both shear modulus $\mu$ and bulk modulus $K=\lambda +2\mu /3$ must be positive.

For antiplane shear, the displacement field is $\mathbf{u}=(0,0,w(x,y,t))$, independent of $z$. Its divergence is zero. Consequently the only possibly nonzero components of the stress tensor are the shear stressesThe $x$ and $y$ components of the Navier-Cauchy equation vanish identically. Its $z$ component gives $\rho w_{tt}={\partial }_x(\mu w_x)+{\partial }_y(\mu w_y)$. Each material has constant shear modulus and mass density, so the required wave equation and wave speeds are

At a perfectly bonded elastic interface, both the displacement field and the traction are continuous. Using the same normal ${\mathbf{e}}_y$ on both sides, the only nonzero traction component is ${\tau }_{zy}$. Thus the interface boundary conditions areContinuity of “stress” here means continuity of traction; it does not require equality of the full stress tensors. The rigid plate supplies the remaining boundary condition $w_1(x,0,t)=0$.

Substitution of the proposed normal mode into the wave equation gives the ordinary differential equationsA nonzero exterior displacement field has exponential decay at infinity only when $c<c_2$. Withits wave amplitude is $f_2(y)=D{e}^{-\ell (y-h)}$. At $c=c_2$ the exterior wave amplitude is affine and cannot decay; at $c>c_2$ it is oscillatory and cannot decay. Setting the exterior wave amplitude identically zero would impose both $f_1(h)=0$ and $f_1^{\prime }(h)=0$, forcing the entire normal mode to vanish.

We also need $c>c_1$. Indeed, if $c<c_1$, the rigid boundary condition gives $f_1(y)=A\sinh (qy)$ with $q=k\sqrt{1-c^2/c_1^2}>0$. At the interface,which contradicts continuity of traction because both values of the shear modulus are positive. At $c=c_1$, the layer wave amplitude is $Ay$ and the same contradiction follows from $f_1^{\prime }(h)/f_1(h)=1/h>0$.

Therefore $c_1<c<c_2$. Define the positive transverse wavenumberThe rigid boundary condition now gives $f_1(y)=A\sin (my)$. The interface boundary conditions giveNeither $\sin (mh)$ nor $\cos (mh)$ can vanish for a nontrivial normal mode with $m,\ell >0$. Dividing therefore gives the rigid-backed shear-wave dispersion relation:Conversely, positive $m,\ell$ satisfying these relations generate a nonzero normal mode through the wave amplitudes above. Its physical displacement field is the real part of the complex expression.

At height $z$ above the free lower end, a uniform vertical filament carries the weight of length $z$. With cross-sectional area $A$,The maximum stress is at the top and is independent of diameter. The diameter cancels because both mass and load-bearing area scale with $A$. Neglecting self-heating gradients, evaporation and initial fabrication stress, this top point sets the first creep-strain limit.

For a clearly labelled external reference estimate, take $\rho =1.93\times 10^4\mathrm{kg}{\mathrm{m}}^{-3}$. Then ${\sigma }_{\max }=1.89\times 10^4\mathrm{Pa}=0.0189\mathrm{MPa}$; the area is $3.14\times 10^{-8}{\mathrm{m}}^2$ and the total suspended mass is about $6.06\times 10^{-5}\mathrm{kg}$. The density is independently tabulated, not invented as a supplied 2016 value.^[ref]

Use the original map's top temperature axis at $2500^{\circ }\mathrm{C}$ and its right-hand tensile-stress axis, so no uncertain shear-stress conversion or additional shear modulus is required. The point lies in C, between the $10^{-9}$ and $10^{-8}{\mathrm{s}}^{-1}$ contours, nearer $10^{-8}$. A logarithmic interpolation of the scanned contours gives about $6\times 10^{-9}{\mathrm{s}}^{-1}$, with only graphical precision. Thusor of order three months in this steady-rate, reference-density map estimate. Reading only the bracketing contours gives the appropriately broad interval $5\times 10^6$ to $5\times 10^7\mathrm{s}$; the graph does not support an exact service-life number.

For the exact prescribed input, the answer remainsThis estimate uses the cited reference density and the grain-size assumptions of the supplied map. Thermal expansion lowers density, and an evolved area or tertiary material creep changes the rate; a steady initial-rate estimate is not a guaranteed lifetime. In particular, a different filament grain size could require a different map. Other failure mechanisms could shorten useful life.

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.