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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.

Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number $N$ and composition, specify the constraint $X$:Here $P$ denotes pressure, also written $p$ elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure, $\mathrm{P}{\mathrm{a}}^{-1}$; its reciprocal $K_X=1/{\kappa }_X$ is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility ${\kappa }_S$ along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law $PV=N{k}_{B}T$ and the reversible ideal-gas adiabat give, at the same equilibrium state,Here $C_P$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and $\gamma$ is the heat capacity ratio. Constant $\gamma$ is required for the finite-path formula $P{V}^{\gamma }=\mathrm{constant}$. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at $100\mathrm{kPa}$, $\gamma =5/3$, so ${\kappa }_T=10^{-5}\mathrm{P}{\mathrm{a}}^{-1}$ and ${\kappa }_S=6\times 10^{-6}\mathrm{P}{\mathrm{a}}^{-1}$; the corresponding values of the bulk modulus are $100\mathrm{kPa}$ and $166.67\mathrm{kPa}$.

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also implywhere $\alpha$ is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality ${\kappa }_T={\kappa }_S$ is possible when $\alpha =0$ and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

A compressional wave is a longitudinal wave whose propagation produces changes of volume or density. In an isotropic elastic solid its displacement is parallel to the wavevector; substituting that polarization into the dynamic Navier-Cauchy equation gives the displayed speed. In a fluid without static shear stiffness the analogous speed is $\sqrt{K/\rho }$, with the appropriate compressional bulk modulus. Reflection at a rigid elastic boundary can convert part of its energy into a shear wave.

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.

For a nonrelativistic zero-temperature free electron model with fixed particle number, electron degeneracy pressure obeys $P\propto (N/V{)}^{5/3}$. Consequently its bulk modulus is $K=-V(\partial P/\partial V{)}_N=5P/3=2n{E}_F/3$. At temperatures small compared with the electron Fermi temperature, the leading isothermal compressibility is $1/K$. This electronic contribution alone omits ionic binding and interaction corrections in a real metal.

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.

For an isotropic linearly elastic solid, equal compressive principal stresses $-p$ produce three equal normal strains $-(1-2\nu )p/E$. Adding them gives the displayed volume strain, or bulk modulus $K=E/[3(1-2\nu )]$. It is a small-strain relation using the pressure change, not an arbitrary absolute pressure reference.

For a nonrelativistic three-dimensional ideal electron gas at zero temperature, the pressure is $p=2n{E}_F/5$ and the electronic bulk modulus is $B=2n{E}_F/3$. This is the electronic model contribution, not automatically the total bulk modulus of a real crystal.

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.

For fixed atom number, volume scales as the cube of the touching-sphere radius. Compressing silver to the stated gold lattice size therefore requiresa volume decrease of about $0.415\%$. If the bulk modulus is constant,Using the silver value derived above givesFor this small compression, $\Delta p\simeq -B\Delta V/V$ gives almost the same estimate. This is an additional pressure relative to the initial standard-pressure state, using the question's free-electron model modulus rather than an empirical silver modulus.

Use the integrated constant-compressibility model from c, with $\kappa p=0.0059\times 100=0.59$:ThereforeCompression reduces the spacing and moves the same allowed reflection to larger deflection, as the positive pressure derivative predicts. The input compressibility corresponds to a constant bulk modulus $1/\kappa \simeq 169.5$ GPa in this ideal model.

This is the stipulated unchanged cubic-phase, constant-compressibility extrapolation, rather than a reliable prediction for real Fe at 100 GPa. Real Fe exhibits a pressure-induced body-centered-cubic to hexagonal-close-packed transformation, and compressibility can change with pressure.^[ref] A real high-pressure diffraction interpretation would require its actual phase and equation of state. The source gives neither, so its numerical exercise is answered within its explicitly stated idealization.

Take the trace of the constitutive relation at zero stress. It gives $0=(2\mu +3\lambda )e_{kk}-3c\alpha (T-T_0)$. With the observed stress-free volume strain this requiresThis is the bulk modulus. A stable solid resists uniform compression, so $\boxed{K>0}$; equivalently the elastic energy of a nonzero pure dilation is positive. The thermal strain is isotropic, $e_{ij}^{\mathrm{th}}=\alpha (T-T_0){\delta }_{ij}/3$, because $\alpha$ here is the volumetric, not linear, thermal expansion coefficient.

Let $\beta =K\alpha$. Using $e_{ij}=(u_{i,j}+u_{j,i})/2$, stress divergence isZero body force and equilibrium therefore giveFor an applied body force per unit volume $b$, equilibrium is $\nabla \cdot \tau +b=0$, so its right-hand side becomes $\beta \nabla T-b$. Consequentlycancels the effective thermal forcing in the displacement equation. Per unit mass the force is $\beta \nabla T/\rho$. This sign follows from the divergence of the thermal stress $-\beta (T-T_0)I$, which is $-\beta \nabla T$. Cancelling that divergence is distinct from removing the stress tensor itself: boundary tractions and thermal compatibility still matter.

For spatially uniform heating with the entire boundary fixed, $u=0$ is the unique equilibrium displacement in a stable elastic solid, andThe stress is compressive hydrostatic stress for a temperature rise. The fixed boundary supplies the tractions that prevent thermal expansion. If $T$ varies spatially and no compensating body force is imposed, fixing the boundary alone does not generally force the interior displacement to vanish: one must solve the displayed forced Navier-Cauchy equation with $u=0$ on the boundary, then recover the stress. The simple uniform hydrostatic expression should not be asserted for arbitrary temperature fields. This is uniform thermal stress in a fully constrained solid.

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.

Assume homogeneous Lamé parameters, with $\mu >0$ and positive bulk modulus $K=\lambda +2\mu /3$. For spherical symmetry, write $q={\tau }_{rr}$ and $t={\tau }_{\theta \theta }={\tau }_{\varphi \varphi }$. Radial force balance is $q^{\prime }+2(q-t)/r=0$. The given linear elasticity relations implySubstitution into radial force balance givesThusThe volumetric strain is $e=u^{\prime }+2u/r$, and the trace of the stress tensor isSince $\lambda +2\mu =K+4\mu /3>0$ is constant, $e^{\prime }=0$. The radial elastic compatibility condition follows:This condition is local to a region governed by homogeneous linear elasticity; it is not a constitutive condition for a plastically deforming region.

Put $\mu ={\mu }^0$, $\lambda ={\kappa }^0-2\mu /3$, and $L=\lambda +2\mu ={\kappa }^0+4\mu /3$. Assume positive bulk modulus and shear modulus. For a unit vector $\xi$, direct contraction of the isotropic elastic stiffness tensor gives the acoustic tensor and its inverse:These are its transverse and longitudinal orthogonal projections. Symmetrizing the definition of the directional elastic strain Green operator consequently givesSubstitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as distributions. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ givesThe first identity holds as a locally integrable distribution. Since the second derivative of $|s|$ is $2\delta (s)$, taking two derivatives in the first identity and four in the second givesThus an explicit complete isotropic elastic strain Green kernel isAll derivatives here are distributional derivatives, which retain the contribution at the origin.

For the requested contraction, ${\tilde{\Gamma }}_{ijkk}={\xi }_i{\xi }_j/L$, or equivalently trace the complete kernel and use $\Delta r=2/r$. ThereforeMore explicitly, its distribution form iswhere the Cauchy principal value uses spherical exclusions about the origin. The coefficient of the Dirac delta can be checked by integrating by parts over a punctured ball: ${\partial }_i{\partial }_j(r^{-1})$ has contact term $-4\pi {\delta }_{ij}\delta /3$. Taking the trace, the principal-value part has zero trace and $\Delta (r^{-1})=-4\pi \delta$, soKeeping only the ordinary kernel away from the origin would incorrectly give zero for this double trace.

Now extend the elastic polarization by zero outside $V$. Since the shear modulus is unchanged, the stiffness contrast acts only on the spherical strain:Hence $\tau =(\kappa -{\kappa }^0)\mathrm{tr}\epsilon$ is its scalar coefficient. Relative to the same external loading, the elastic strain Green operator gives $\epsilon ={\epsilon }^0-\Gamma \ast (\tau I)$. Taking the trace and using the contact identity reduces this nonlocal equation toSolving the scalar equation gives the constant-shear bulk-modulus inclusion formulaUsing the single contraction instead of the double trace givesFinally eliminate the actual trace using the scalar formula:The reference trace ${\epsilon }_{kk}^0$ in this last formula is essential and is what the original PDF prints; the converted TeX loses its superscript. The second derivatives are interpreted distributionally, or with their principal-value and contact terms explicitly included. With ordinary regularity they give the corresponding pointwise formula at regular interior points.

Under an orthogonal change of Cartesian coordinates with $x_i^{\prime }=R_{ia}{x}_a$, a rank-$n$ Cartesian tensor transforms as $T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$, summing repeated indices. An isotropic tensor is invariant under every rotation. For an isotropic fourth-rank tensor, half-turns about the coordinate axes eliminate components unless each coordinate label occurs an even number of times. Coordinate rotations make the remaining two-pair patterns depend on three constants $c_{1122}=\alpha$, $c_{1212}=\beta$, $c_{1221}=\gamma$. Invariance after rotating the first axis to $(e_1+e_2)/\sqrt{2}$ gives $c_{1111}=\alpha +\beta +\gamma$. HenceThis is a direct proof of the rank-four isotropic decomposition, without assuming any prior interchange symmetry of the four indices. Each displayed delta pairing is itself rotation invariant.

For symmetric strain, contraction gives ${\sigma }_{ij}=\alpha e_{kk}{\delta }_{ij}+(\beta +\gamma )e_{ij}$. Comparing with the Lamé parameters yieldsWrite $p=e_{kk}/3$ and $d_{ij}=e_{ij}-p{\delta }_{ij}$, so $d_{kk}=0$. Since $e_{ij}{e}_{ij}=3p^2+d_{ij}{d}_{ij}$, the strain energy density becomesThe two terms describe volumetric and shear deformation, respectively. Thus nonnegative isotropic elastic strain energy is equivalent toNecessity also follows by taking a nonzero traceless strain and a pure dilation separately. The combination $K=\lambda +2\mu /3$ is the bulk modulus; strict inequalities give positive energy for every nonzero strain.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

stress describes internal force per area. A cut with unit normal $\hat{\mathbf{n}}$ has traction $\mathbf{t}=\mathbf{\sigma }\hat{\mathbf{n}}$; its normal component is $\hat{\mathbf{n}}\cdot \mathbf{\sigma }\hat{\mathbf{n}}$ and its remainder is shear traction. force balance constrains the stress divergence, while ordinary local angular-momentum balance makes the Cauchy stress tensor symmetric in a medium without body couples.

strain describes deformation rather than force. For small displacement gradients,The antisymmetric displacement-gradient part is a rigid rotation to first order and contributes no infinitesimal strain. Engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$, so it must not be confused with the tensor component. stress has units of pressure; strain is dimensionless.

An isotropic linear elastic solid obeysTwo independent moduli are required. The shear modulus $G$ and bulk modulus $K=\lambda +2G/3$ must be positive for stability. Young's modulus and Poisson ratio satisfyFor a tensile specimen with free side faces, ${\sigma }_{xx}=F/A$ gives ${ϵ}_{xx}={\sigma }_{xx}/E$ and ${ϵ}_{yy}={ϵ}_{zz}=-\nu {\sigma }_{xx}/E$. Pure shear gives $\tau =G\gamma$; hydrostatic loading gives $\Delta V/V=-p/K$. The incompressible limit has $\nu \to 1/2$ at finite shear stiffness, not vanishing shear resistance. Crystals can be anisotropic and need a fourth-rank elastic stiffness tensor instead of the isotropic two-constant law.

elastic deformation is recoverable on unloading; it need not remain linear at large strain. Beyond yield, irreversible rearrangements such as dislocation motion cause plastic strain, often followed by hardening. Ductile samples can neck before fracture; brittle failure has much less plastic deformation. Engineering stress uses initial area, while true stress uses the changing instantaneous area. A decreasing engineering stress during necking therefore need not mean that local true stress decreases. stress concentrations, flaws and loading geometry can determine failure before a uniform-material strength is reached. Viscoelastic materials also show rate dependence, creep and stress relaxation; their losses cannot be described by a static Hooke law alone.

A tensile test should measure force and gauge-length extension independently, correct machine compliance, align the grips, and control strain rate and temperature. Unloading distinguishes recoverable strain from a residual plastic extension. Separate shear or hydrostatic tests determine moduli that one axial measurement does not fix. Compression specimens need friction and buckling checks. Work per volume is $\int {\sigma }_{ij}d{ϵ}_{ij}$; the recoverable part is elastic energy, while a dissipative cycle produces a loss rather than creating energy. The original loading/unloading curve below is a chosen constitutive sketch, not measured material data.

For hydrostatic volume strain from Young's modulus and Poisson's ratio in isotropic linear elasticity, the bulk modulus is $K=E/[3(1-2\nu )]=5.56\mathrm{GPa}$. Hydrostatic compression therefore givesHere $p=101325\mathrm{Pa}$ is the pressure increase, and the negative sign denotes contraction. Equivalently each normal strain is $-(1-2\nu )p/E$ and the three strains add. This is the small-strain isotropic model; concrete's microstructure is represented only through the supplied elastic constants.

For a homogeneous isotropic solid with small displacement $\mathbf{u}$, linear elasticity gives ${\sigma }_{ij}={\lambda }_L(\nabla \cdot \mathbf{u}){\delta }_{ij}+2\mu e_{ij}$. Momentum balance becomesA plane wave with displacement parallel to its wavevector is longitudinal, whereas perpendicular displacement is transverse. Writing $K={\lambda }_L+2\mu /3$ for the bulk modulus and $\mu$ for the shear modulus, their bulk speeds areThere are two independent transverse polarizations. The bulk modulus supplies resistance to compression; shear rigidity supplies transverse restoring stress. An ordinary fluid has no static shear rigidity and cannot sustain these propagating bulk shear waves. A thin rod with freely contracting sides instead has longitudinal speed $\sqrt{Y/\rho }$, because its lateral boundary conditions differ from a bulk plane wave. Reflection, refraction and mode conversion occur at interfaces; a free surface also supports Rayleigh waves.