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An isotropic linear elastic solid has two independent elastic constants. For infinitesimal strain,The shear modulus $G$ governs shape change and the bulk modulus $K$ governs volume change. Young's modulus $E$ and Poisson's ratio $\nu$ are equivalent parameters:Positive strain energy requires $G,K>0$, hence $-1<\nu <1/2$; $\nu \to 1/2$ is the incompressible limit. Small-amplitude bulk waves have speeds $c_P=\sqrt{(K+4G/3)/\rho }$ and $c_S=\sqrt{G/\rho }$. Isotropy removes directional dependence, not the distinction between compression and shear. These laws apply within the elastic small-strain regime, not after yielding or fracture.

Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

Apply equal compressive pressure $p$ in all directions; here $\sigma$ denotes Poisson's ratio. The three principal normal stresses are $-p$. For an isotropic linearly elastic material, Hooke's law givesand similarly for $y,z$. At small strain, the fractional volume change is the sum of these normal strains:The bulk modulus is $B=-p/(\Delta V/V)$, giving the isotropic elastic modulus relation,The result assumes linear elasticity and fixed isotropic moduli. In the incompressible limit $\sigma \to 1/2$ with finite Young's modulus, the bulk modulus tends to infinity.

For linear elasticity of an isotropic material, the isotropic elastic modulus relation givesThe printed Young's modulus is $E=10^{-3}\mathrm{GPa}=10^6\mathrm{Pa}$, and the printed Poisson's ratio is $\nu =1.47$. Formal substitution givesWith compressive pressure positive, the linear volume response is $\Delta V/V=-p/B$. Thus a thirty per cent reduction formally requiresHere $1\mathrm{atm}=101325\mathrm{Pa}$. The negative pressure is a formal tensile load, not the requested positive isotropic compression.

The original PDF really prints $1.47$. It is incompatible with a stable ordinary isotropic elastic solid: positive bulk modulus and shear modulus require $-1<\nu <1/2$. Consequently there is no physically admissible positive-pressure answer with the stated constants. If $0.47$ was intended, the corresponding linear estimates would instead beThat is an explicit alternative, not a correction established by the source. A thirty per cent volume strain is also outside a controlled small-strain approximation; accurate rubber behaviour at that deformation needs a finite-strain constitutive law. The initial volume cancels because the specified change is a fractional one.

Under a uniform compressive pressure $p$, each normal stress is $-p$. The isotropic Hooke's law gives each longitudinal strain asThus $\Delta V/V=-3p(1-2\sigma )/E$. By the definition of bulk modulus, $B=-p/(\Delta V/V)$, so the isotropic elastic modulus relation isPositive finite Young's modulus and bulk modulus requireThe incompressible limit is $\sigma \to 1/2$ with $B\to \infty$ at finite $E$. If positivity of the shear modulus $G_s=E/[2(1+\sigma )]$ is also imposed, the full stable isotropic range is $-1<\sigma <1/2$. Negative Poisson's ratio is not ruled out; the displayed bulk-modulus relation alone supplies the upper bound.

Interpret the applied pressure as a uniform increase of external hydrostatic pressure on all six faces, and approximate the rock by an isotropic material obeying linear elasticity. Taking tensile stress as positive, ${\sigma }_x={\sigma }_y={\sigma }_z=-p$. By Hooke's law, the strain along any one axis isTo first order in the small strain, the fractional volume change is the sum of the three axial strains:Thus the bulk modulus is $K=40\mathrm{GPa}$. Using the specified Young's modulus, Poisson's ratio, $V=1{\mathrm{m}}^3$ and $p=101325\mathrm{Pa}$ givesThe sign indicates contraction. This is the change relative to the unloaded reference state; simply maintaining a specimen already equilibrated at this same pressure would produce no further change.

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.

The matrix exponential is the absolutely convergent series $e^{tA}={\sum }_{n=0}^{\infty }{t}^n{A}^n/n!$, with derivative $A{e}^{tA}$ and inverse $e^{-tA}$. If $K^T=-K$, then $(e^{tK}{)}^T{e}^{tK}=e^{-tK}{e}^{tK}=I$. Its determinant is continuous in $t$, is never zero, and starts at one, so it is one throughout: $e^{tK}\in SO(3)$. Conversely, differentiating $(e^{tK}{)}^T{e}^{tK}=I$ at zero gives $K^T+K=0$. ThusThe quantifier matters: membership at one isolated time alone would not imply skewness.

Write $P(F)=DW(F)$ and $J=\det F>0$. For any skew matrix $K$, differentiate frame indifference along $e^{tK}F$:A matrix orthogonal to every skew matrix is symmetric. Therefore $P(F)F^T$ and the Cauchy stress tensor $T=J^{-1}P(F)F^T$ are symmetric.

Let $\mathcal{C}=D^{2}W(I)$ and assume $P(I)=0$. Differentiating symmetry of $P(F)F^T$ at $I$ in direction $H$ shows $\mathcal{C}[H]$ is symmetric for every $H$. Hence $c_{ijkl}=c_{jikl}$. As a Hessian, $\mathcal{C}$ also has pair-interchange symmetry $c_{ijkl}=c_{klij}$. Combining the two gives $c_{ijkl}=c_{ijlk}$. Equivalently $\mathcal{C}$ annihilates infinitesimal rotations. This proves both requested minor symmetries and shows exactly where the stress-free assumption enters. It is stress-free linearization of an objective energy.

For the particular energy, ${\sum }_i{v}_i^2=\mathrm{tr}(F^{T}F)$ and ${\prod }_i{v}_i=J$, soAt $I$ this is $[4\alpha +h^{\prime }(1)]I$. Therefore the reference is stress-free exactly when $\boxed{h^{\prime }(1)=-4\alpha }$. To differentiate again, use $\mathrm{cof}F=J{F}^{-T}$. The determinant and inverse derivatives at $I$ givewhich also proves the supplied hint. It follows thatFor a symmetric $e$ this becomesNo additional positivity of the bulk modulus follows from the stated smoothness of $h$ alone; this calculation identifies the moduli rather than imposing extra material assumptions.

Write $A=\pi a^2$ for the cross-sectional area, and use $z$ consistently for the longitudinal coordinate: the paper's later $x$ is the same coordinate. The canonical momentum is a density per reference length,The Legendre transform gives the Hamiltonian density in its proper canonical variables:The Euler-Lagrange equation isFor a twice-differentiable wave packet $\psi =f(z-ct)$, the derivatives are $\ddot{\psi }=c^2{f}^{\prime \prime }$ and ${\psi }^{\prime \prime }=f^{\prime \prime }$. Thus the nontrivial packet solves the wave equation whenThe negative sign gives the opposite propagation direction. Here $\Pi$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $\mathcal{H}$ has units $\mathrm{J}{\mathrm{m}}^{-1}$, and $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. The result follows from the supplied axial stiffness in the Lagrangian. For a freely contracting thin elastic rod the relevant longitudinal modulus is normally Young's modulus, rather than the bulk modulus; the question's given Lagrangian fixes which coefficient is being used.

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 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.

For isotropic linear elasticity with volumetric expansion coefficient $\alpha$, the stress-free thermal strain is $e^{\mathrm{th}}=\alpha (T-T_0)I/3$. Thus the thermal stress term is $-K\alpha (T-T_0)I$, where $K=\lambda +2\mu /3$ is the bulk modulus. A uniformly heated body with its full boundary fixed has zero displacement and the displayed compressive stress. For spatially varying temperature and zero body force, the displacement instead solves $\mu \Delta u+(\lambda +\mu )\nabla \mathrm{div}u=K\alpha \nabla T$ with its boundary values; the uniform formula is not automatic. A body force per volume $+K\alpha \nabla T$ cancels this thermal forcing in the equilibrium equation, but does not by itself cancel the stress or its boundary tractions.