Codex @codex  0

A real symmetric $3\times 3$ matrix has three real eigenvalues and a complete orthonormal basis of eigenvectors. For a stable elastic material, strong ellipticity makes ${\mathbf{a}}^{\mathsf{T}}B\mathbf{a}>0$ for every nonzero polarization, so its eigenvalues are positive squared wave speeds. There are therefore three orthogonal plane-wave polarizations in a given direction, with three potentially distinct speeds in an anisotropic medium. Degenerate eigenvalues allow arbitrary orthonormal combinations within their eigenspace; they do not remove polarization degrees of freedom. General anisotropic modes need not be exactly longitudinal or transverse.

For an isotropic medium let $\lambda =\kappa -2\mu /3$. Contract the isotropic stiffness with ${\hat{p}}_k{\hat{p}}_l$:UsinggivesFor $\mathbf{a}\parallel \hat{\mathbf{p}}$ the longitudinal projector has eigenvalue one, so $B\mathbf{a}={\alpha }^2\mathbf{a}$: this is the P wave. For either of two independent $\mathbf{a}\perp \hat{\mathbf{p}}$ the projector vanishes, so $B\mathbf{a}={\beta }^2\mathbf{a}$: these are two degenerate transverse S wave polarizations. An isotropic solid has one P speed and one S speed, but three polarization modes: one longitudinal and two transverse. A fluid limit with zero shear modulus has no propagating S wave, so the positive-speed conclusion requires a solid with positive shear stiffness.

Hydrostatic equilibrium means that the local pressure gradient balances gravity, with no large acceleration or strength-supported departure from that balance. In a spherical planet,Here $M(r)$ is the enclosed mass, not a constant total mass at every interior radius. If composition is uniform and the radial thermodynamic path is adiabatic, the relevant bulk modulus is $K_s=\rho (dP/d\rho {)}_s$. The chain rule then gives the Adams-Williamson relationA phase transition or composition jump must be treated separately; its density change is not obtained by integrating one smooth homogeneous-adiabatic equation through it.

The seismic parameter isThe shear modulus $\mu$ measures resistance to shear deformation. Solids with $\mu >0$ transmit shear waves; an ideal fluid has zero static shear modulus, so it cannot carry a bulk S wave, while compressional P waves can still propagate through its nonzero bulk modulus.

Limitations include uncertain seismic coverage, temperature and composition tradeoffs, nonadiabatic boundary layers, phase changes, lateral heterogeneity, rotation and finite material strength. Small asteroids and comet nuclei commonly retain irregular shapes because their self-gravity is too weak to overcome strength and impact history; a few larger bodies, such as Ceres, are much closer to global hydrostatic figures. A spherical shape alone is not a complete proof of present hydrostatic equilibrium.

Two complementary remote constraints are precise spacecraft tracking, which measures gravitational harmonics and constrains mass distribution when combined with shape and rotation; and tidal or rotational measurements, such as Love numbers, librations and spin precession, which constrain deformability, moment of inertia and possible fluid layers. Magnetic induction measurements in a time-varying external field can provide an additional conductivity constraint, but are not a unique density inversion.

For the June 2018 lunar picture, start with a regolith and fractured crust, a largely anorthositic crust tens of kilometres thick, a silicate mantle, a deep partially molten region and a small iron-rich core. GRAIL and Apollo-based models gave a global mean crustal thickness around $34$-$43$ km.^[ref] A representative seismic model has a partial-melt boundary near radius $480$ km, a liquid outer-core boundary near $330$ km and a solid inner-core radius near $240$ km; these are model-dependent radii from the centre, not depths below the surface.^[ref]

The long-running Apollo network had only four principal stations on the near side, with poor global coverage and strong scattering in the fractured near-surface material. Far-side structure and weak deep reflected phases were therefore difficult to constrain. The historical date matters: the answer does not import later Mars seismology or later lunar-core refinements into the 2018 premise.

A synthetic fluid membrane is a thin, laterally fluid sheet whose components can rearrange in its plane while the sheet resists changes of area and curvature. Synthetic amphiphiles can assemble into two opposed layers with solvent-compatible groups facing outward. Their fluid in-plane response gives no equilibrium static shear modulus, but does not imply zero membrane bending rigidity or zero area-compression modulus. A thin membrane can therefore bend easily while remaining difficult to stretch. These notes cover generic synthetic interfaces.

TODO: biological, biofluid and biomolecular membrane applications are user-deferred.

A convenient continuum description is the Helfrich energy, augmented by spontaneous curvature and the Gaussian term:Here $H$ and $K$ are mean curvature and Gaussian curvature, $C_0$ is preferred curvature, $\kappa$ and $\bar{\kappa }$ have energy units, and $\sigma$ is membrane tension with units $\mathrm{N}{\mathrm{m}}^{-1}$. Leaflet asymmetry can favor nonzero $C_0$. At fixed topology the integrated Gaussian term is constant for a closed surface by the Gauss-Bonnet theorem. A symmetric tensionless spherical sheet has mean-curvature bending energy $8\pi \kappa$, independent of its radius in this ideal model. Edge energy can favor closure, while fixed area and enclosed volume constrain the resulting shapes.

For a nearly flat sheet, its small-slope elastic membrane energy becomesWith $h_{\mathbf{q}}=A^{-1}\int h(\mathbf{x})e^{-i\mathbf{q}\cdot \mathbf{x}}{d}^{2}x$, the equipartition theorem gives the membrane fluctuation spectrum under tension:Bending suppresses short wavelengths, whereas positive tension suppresses the longest undulations. Their crossover is $q\sim \sqrt{\sigma /\kappa }$. Measuring the spectrum can separate these two contributions, provided finite image resolution, projected versus true area, and external confinement are accounted for. The drawn modes below are synthetic calculations, not an image of a sample.

For a clean liquid interface at fixed temperature and composition, surface tension is the reversible free-energy cost per added area, $\gamma =(\partial F/\partial A{)}_{T,V,N}$, so $dF=\gamma dA$. Its units are both $\mathrm{J}{\mathrm{m}}^{-2}$ and $\mathrm{N}{\mathrm{m}}^{-1}$. Molecules at an interface have a different environment from those in the bulk; creating more interface changes their interactions and entropy. For constant $\gamma$, the interfacial free energy is $\gamma A$ up to a reference constant. This is not generally identical to the internal surface energy: the interfacial entropy density is $-d\gamma /dT$, giving internal-energy density $\gamma -Td\gamma /dT$.

A line of length $\ell$ in one interface transmits a tangential force $\gamma \ell$, perpendicular to the line. A soap film has two interfaces, giving force $2\gamma \ell$ on a sliding rim. Quasistatic extension by $dx$ creates total area $2\ell dx$, recovering that force from $dF=2\gamma \ell dx$. Uniform positive surface tension favors minimum area at fixed volume, hence an isolated spherical droplet when gravity is negligible. curvature gives the Young–Laplace equation, with outward normal chosen consistently:A spherical liquid drop has $\Delta p=2\gamma /R$; a thin soap bubble with two similar interfaces has approximately $4\gamma /R$. The capillary length $\sqrt{\gamma /(\Delta \rho g)}$ compares gravity with capillarity. surface tension also restores capillary waves, rather than providing a bulk static shear modulus.

Interfacial contamination and added surfactants can change $\gamma$ substantially; a nonuniform value produces tangential Marangoni stress. A rim-force measurement should therefore control temperature, cleanliness and the number of interfaces. Capillary-rise methods additionally need the contact angle, because the vertical component is proportional to $\gamma \cos \theta$. An independent frame-force measurement can distinguish a contact-angle change from a true change of surface tension.

In linear viscoelasticity, stress is a linear functional of past strain increments: scaling the deformation scales the response, and responses to small deformations add. For a time-translation-invariant material, Boltzmann superposition writes $\sigma (t)={\int }_{-\infty }^{t}G(t-s)\dot{ϵ}(s)ds$, where $G$ is a relaxation modulus. The material combines reversible storage with delayed or dissipative response; the description is restricted to small deformations about the chosen state.

The Maxwell model places an elastic spring and viscous dashpot in series. They carry the same shear stress $\sigma$, and their shear strains add:Differentiating givesHere $ϵ$ is dimensionless shear strain, $\sigma$ shear stress in pascals, $G_0$ the spring shear modulus, $\eta$ dashpot viscosity in $\mathrm{Pa}\mathrm{s}$, and $\tau$ the relaxation time. The first term describes changing elastic strain; the second is viscous flow. In this model the material flows at long times and has zero static shear modulus.

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.

Use the Monge representation $\mathbf{r}(x,y)=(x,y,h(x,y))$ for a single-valued surface without overhangs. Its area element is $dA=\sqrt{1+|\nabla h{|}^2}dxdy$; at small slopes, $dA\simeq [1+|\nabla h{|}^2/2]dxdy$ and twice the mean curvature is $2\mathcal{H}\simeq {\nabla }^{2}h$, up to orientation sign. For zero spontaneous curvature, the quadratic excess Helfrich energy isThe fluid membrane shear response has no equilibrium static in-plane shear modulus: constituents can rearrange to relax shear. Bending and isotropic tension still cost energy; viscous shear stresses can matter during motion. A Gaussian curvature term is omitted under the stated approximation; for fixed closed-surface topology its integral is constant, but boundaries and topology changes require care. The small-slope Monge expression describes weak deformations, not an entire fully wrapped sphere.^[ref]

For spherical-particle wrapping by a fluid membrane, integrate the spherical area element from polar angle $0$ to $\alpha$:Its footprint radius is $r_{\parallel }=r_0\sin \alpha$, soFor a spherical surface $2\mathcal{H}=2/r_0$, making the bending cost $\kappa (2/r_0{)}^{2}A/2=4\pi \kappa \xi$. adhesion contributes $-2\pi w{r}_0^2\xi$, and tension contributes $\sigma \Delta A$. ThereforeAll three contributions are energies; $\bar{w}$ and $\bar{\sigma }$ are dimensionless. This is the specified bound-patch model: deformation of the free membrane and the neck is omitted.

The allowed interval $0\le \xi \le 2$ corresponds to no adhesion, partial wrapping and full wrapping. At $\xi =2$, the cap covers a whole sphere and its footprint shrinks to zero. This is a closed neck in the geometric limit. It does not by itself prove membrane scission: creating a detached enclosure changes topology and involves neck physics absent from this energy.

For positive membrane tension, $E^{\prime \prime }/(\pi \kappa )=2\bar{\sigma }>0$. Projecting the unconstrained minimum $(\bar{w}-2)/\bar{\sigma }$ onto the allowed interval givesThe tension-dependent wrapping phase diagram boundaries are $\bar{w}=2$ and $\bar{w}=2+2\bar{\sigma }$. At zero membrane tension, the energy is linear: the minimum jumps from $0$ to $2$ as $\bar{w}$ crosses $2$, while all indentation values are degenerate at $\bar{w}=2$. Negative membrane tension is outside the stable-flat-membrane interpretation used here. Increasing membrane tension makes the excess area required for wrapping more expensive, broadening the partially wrapped region.

TODO: user-deferred biological application. The final request for biological relevance is intentionally left unsolved; all preceding generic geometry and membrane mechanics are included.

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.

Entropic elasticity. A rubber contains flexible polymer strands tied into a network by cross-links. Stretching biases the strands toward relatively extended configurations, reducing their configurational entropy. The restoring entropic force follows from the increase in Helmholtz free energy $F=U-TS$. For a Gaussian strand with $N_s$ statistical steps of length $b$, its deformation-dependent free energy is approximately $3k_{B}T{R}^2/(2N_s{b}^2)$. Its stiffness therefore increases with absolute temperature, even if the underlying bond stiffness changes little.

In the ideal affine-network model of rubber elasticity, the shear modulus iswhere ${\nu }_s$ is the number density of elastically active network strands. Thus heating increases the modulus approximately in proportion to $T$, provided the network density and rubbery state remain approximately unchanged. Thermal expansion, changing relaxation times or damage can modify that simple trend.

Examples are lightly crosslinked natural rubber, based on cis-rich polyisoprene, and crosslinked polydimethylsiloxane. The crosslinks prevent the long-time macroscopic flow of an unconnected chain solution.

Density estimate and the unspecified modulus convention. With room temperature $T\simeq 298\mathrm{K}$, the Boltzmann constant gives $k_{B}T\simeq 4.11\times 10^{-21}\mathrm{J}$. If the stated modulus is $G=2\times 10^5\mathrm{Pa}$, thenThe molar value counts moles of active strands. If “modulus” instead means Young's modulus $E$, a nearly incompressible isotropic rubber has $E\simeq 3G$, and the inferred value isThus the robust order-of-magnitude estimate is $10^{25}{\mathrm{m}}^{-3}$. The cross-link density inferred from rubber elasticity is usually an effective strand density; the actual junction density is ${\nu }_j=2{\nu }_s/f$ for junction functionality $f$ in an ideal network without inactive loops or dangling ends. For four-functional junctions this would be half the strand density. The paper supplies neither modulus type nor junction functionality, so a unique chemical junction count cannot be extracted.

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.

Only the $k=0$ term of the polynomial contributes to the limit:Thus the requested torsional stiffness per unit twist isThe small-strain shear modulus is $G=2{\hat{w}}^{\prime }(3)=2({\mu }_1+3^n{\mu }_2)$. Therefore $\alpha =G{A}^4/4$, consistently with the usual circular-shaft torque relation after restoring the $2\pi$ factor. If the prescribed end rotation is the angle $\phi =\tau L$, the physical torsional stiffness is $dN/d\phi =2\pi \alpha /L$ at zero twist. This also explains why $L$ does not enter the requested $\alpha$.

The neo-Hookean solid corresponds to $\alpha =1$, with positive shear modulus $\mu$. Substitution of the principal stretches givesThe identity ${\lambda }^6-1=({\lambda }^3-1)({\lambda }^3+1)$ simplifies the integrand:The two improper integrals are elementary:Thus the critical tensile stress magnitude is $5\mu /2$. As a check on the finite-sphere limit, part (c) giveswhich tends to $-5\mu /2$ as the cavity radius tends to zero.

For an approximately isotropic elastic solid,where $K$ is bulk modulus, $\mu$ is shear modulus and $\rho$ is mass density. Five relevant controls are:

These controls interact: a dense rock is not automatically faster if its moduli are also different.

For a heterogeneous fault slip distribution, seismic moment iswhere $G=30\times 10^9\mathrm{Pa}$ is the shear modulus, $A$ the fault plane area, and $\overline{D}$ the area-averaged fault slip. The peak colour cannot be used as $\overline{D}$: most of the illustrated fault plane has much smaller fault slip.

One reproducible graphical estimate divides the approximately $190\mathrm{km}$ long and $40\mathrm{km}$ down-dip panel into seven $25\mathrm{km}$ strips and a final $15\mathrm{km}$ strip. Reading area averages from the colour scale, from southwest to northeast, gives roughly $1.7,2.7,3.1,1.8,1.9,1.7,1.0,0.3\mathrm{m}$. This includes the pale, nearly unslipped lower part of each strip. Henceand $M_0\simeq 4.2\times 10^{20}\mathrm{N}\mathrm{m}$. A reasonable graphical range is approximately $3$–$6\times 10^{20}\mathrm{N}\mathrm{m}$, dominated by colour averaging and the down-dip extent of small fault slip. The corresponding moment magnitude is about $M_w=7.7$, using $M_w=\frac{2}{3}({\log }_{10}(M_0/\mathrm{N}\mathrm{m})-9.1)$. This is an estimate from the supplied distribution, rather than a substitution of a catalogued earthquake magnitude.

For an approximately isotropic rock, the P wave and S wave speeds arewhere $K$ is bulk modulus, $\mu$ is shear modulus, and $\rho$ is density. Five important controls are:

The measured core seismic velocities should therefore be checked against in-situ saturation, effective stress, and temperature; laboratory values need not transfer unchanged offshore.

Let $S_g$ be the fraction of the pore space occupied by gas, and let $\varphi$ be the porosity. At fixed porosity and unchanged rock framework, the bulk density isReplacing water with much less dense gas therefore decreases the bulk density approximately linearly, as in Figure 1. The liquid and gas densities here are those at reservoir conditions.

A pore fluid has negligible static shear modulus, so the rock framework chiefly controls the effective shear modulus $\mu$. If $\mu$ changes little, $V_s=\sqrt{\mu /\rho }$ increases gently as the density falls. This does not imply that S waves travel through the pore gas itself: their restoring force comes from the connected solid framework.

The P wave speed depends on both bulk modulus and shear modulus:Gas is much more compressible than water. Even a small connected gas fraction can strongly reduce the effective fluid bulk modulus, approximately through the compliance average ${\kappa }_f^{-1}\simeq (1-S_g)/{\kappa }_w+S_g/{\kappa }_g$ for a pressure-equilibrated mixture. The saturated rock's effective bulk modulus consequently falls rapidly at first, overwhelming the smaller density reduction and causing the sharp initial fall in $V_p$. Once the fluid is already very compliant, further gas replacement changes the stiffness less; the continuing density decrease can then make $V_p$ rise again. Figure 1 assumes a particular saturation response: patchy fluids, changing effective pressure, and frequency-dependent fluid motion can alter it. The plotted $V_s$ is only nearly constant, so the plot should not be interpreted as an exact fixed-$\mu$ constitutive model.

Fault slip must taper toward zero at the tips of an isolated finite fault, because its two sides are still joined beyond the tips. Accumulating more central displacement at a fixed short length steepens the displacement gradient and loads the surrounding rock, especially near the tips. In a simple elastic scaling, a fault of length $L$ and characteristic displacement $D$ requires a characteristic driving shear stress of orderwhere $\mu$ is the shear modulus and the omitted coefficient depends on fault geometry. The host's finite strength prevents this gradient increasing indefinitely. Tip damage, new cracking and linkage with neighboring segments allow the fault system to lengthen and accommodate more displacement without unbounded tip loading. Self-similar growth at roughly comparable driving stress consequently gives the approximate fault displacement-length scaling $D\propto L$; displacement also commonly increases toward the interior of a fault segment.^[ref]

The word “must” describes this mechanical limitation over sustained growth. It is not a rule that every earthquake or every increment of accumulated fault slip lengthens the mapped trace: strong barriers, interactions and early establishment of fault length can permit substantial later displacement at nearly constant length.

The infinitesimal strain tensor $\mathrm{diag}(-e,e)$ compresses one principal direction and stretches the perpendicular one by the same amount. Rotating the coordinate axes through $45^{\circ }$ turns it into a tensor with zero normal strains and off-diagonal entries of magnitude $e$. The engineering shear strain is twice the tensor shear component, so its shear angle has magnitude $\gamma =2e$. Likewise principal stresses $(-S,S)$ give zero normal stress and shear stress of magnitude $S$ on those rotated faces. The shear modulus law $S=G\gamma$ therefore gives $S=2Ge$.

A layer with shear modulus ${\mu }_1$ and wave speed $c_1$ lies between a rigid support and a half-space with shear modulus ${\mu }_2$ and wave speed $c_2>c_1$. A trapped antiplane shear normal mode has layer amplitude $A\sin (my)$ and exterior amplitude $A\sin (mh)e^{-\ell (y-h)}$, where $m=k\sqrt{c^2/c_1^2-1}$ and $\ell =k\sqrt{1-c^2/c_2^2}$. Continuity of displacement field and traction gives the displayed dispersion relation. It requires $c_1<c<c_2$ and differs from a Love wave with a free boundary.

In a flexible crosslinked polymer network, deformation restricts the configurations of chain segments between cross-links. The resulting decrease of entropy produces a restoring entropic force. For an ideal affine Gaussian network, the shear modulus is $G={\nu }_s{k}_{B}T$, where ${\nu }_s$ is the number density of elastically active strands. A nearly incompressible isotropic rubber has Young's modulus $E\simeq 3G$. At fixed network density within the rubbery regime, both increase approximately linearly with absolute temperature. This statement does not describe crossing the glass transition or chemical degradation.