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Let $N$ be circumferential tension. A small arc subtending $d\theta$ has inward resultant $Nd\theta$ and mass $\rho Ard\theta$. Its centripetal acceleration gives $N=\rho A{\omega }^2{r}^2$. Circumferential stress is $N/A$ and the strain is $\Delta (2\pi r)/(2\pi r)=\Delta r/r$. Thus the centrifugal strain of a thin elastic hoop isUse the unstrained radius and cross-section at this order. The approximation requires $\rho {\omega }^2{r}^2/E\ll 1$, a slender hoop and linear elastic response; the area cancels.

A small displacement $\mathbf{u}$ in a homogeneous isotropic elastic solid obeys linear elasticity. With Lamé constants $\lambda ,\mu$ and mass density $\rho$, momentum balance givesA longitudinal plane wave has displacement parallel to its propagation direction and changes local volume. A transverse wave has perpendicular displacement and produces shear. Substitution givesAn ideal fluid has no static shear modulus and cannot support this bulk transverse elastic wave. A thin rod with free lateral surfaces instead supports an extensional long-wavelength mode with speed $\sqrt{E/\rho }$; its transverse contraction changes the effective stiffness, so it should not be identified with the unbounded-solid longitudinal mode.

Stress and displacement boundary conditions produce reflection, transmission, mode conversion and guided waves. Viscoelasticity and scattering can attenuate waves; finite geometry can make guided modes dispersive even when the bulk elastic constants are frequency independent. Wave speed depends on both material stiffness and the allowed deformation.

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.

At distance $y$ from the neutral axis, bending strain has magnitude $y/R$ and the corresponding stress is $Ey/R$. Its moment is thereforeThe second moment of area is about the neutral bending axis and has units of length to the fourth power. This assumes a slender linear-elastic beam, small strains and plane sections remaining plane, not necessarily small deflections.

For horizontal compression $F>0$, moment balance across a short arc gives $dB/ds=-Fdy/ds=-F\sin \theta$. Since $B=EI{\theta }^{\prime }$, the exact-angle Euler elastica equation and its first integral areThe first integral follows by multiplying by ${\theta }^{\prime }$ and integrating; no small-angle replacement has been made.

For tension set $F=-T$. The straight far-field ends give $C=T$, so on the increasing-angle tensile elastica loop branchChoosing $s=0$ at $\theta =\pi$ givesThe terminal angles are $0$ and $2\pi$, the same physical tangent direction. At the top, ${\theta }^{\prime }(0)=2\alpha$ and the bending moment is $2EI\alpha$. The horizontal tensile force has lever arm $h$, soIntegrating $(x^{\prime },y^{\prime })=(\cos \theta ,\sin \theta )$ independently gives $x=s-2\tanh (\alpha s)/\alpha$, $y=2\mathrm{sech}(\alpha s)/\alpha$, confirming the height. This ideal loop neglects self-contact and thickness at its crossing.

Let $A$ be the initial plan area and take compression normal to the sheet. Smooth plates exert no lateral traction, so the uniaxial stress is ${\sigma }_z=-F/A$ and the two lateral strains are ${\epsilon }_x={\epsilon }_y=\nu F/(EA)$. To first order in strain,This area expansion under uniaxial compression is the increase of the sheet's plan area, equivalently the area of either broad face. Counting the sum of both broad faces would give twice this change; edge area would require thickness information. The cancellation of $A$ is a small-strain linear-elastic result, not a statement that arbitrarily large compressions need no geometry data.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

First estimate the excess pressure as weight divided by contact area:For the equilibrium estimate, use the Clapeyron equation for melting. The specific volume change on melting is $\Delta v=1/{\rho }_w-1/{\rho }_i$, and the latent heat of fusion $L$ gives $\Delta s=L/T_m$. Thus the common-pressure melting-point slope isTake $T_m\simeq 273.15\mathrm{K}$ and the usual rounded water density ${\rho }_w\simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$, in addition to the supplied ${\rho }_i=920\mathrm{kg}{\mathrm{m}}^{-3}$ and $L=3.3\times 10^5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. ThensoThe sign is negative because ice contracts on melting. This is a linear equilibrium estimate treating the load as a common local pressure on coexisting ice and water. The actual contact stress is nonuniform, and frictional heating, deformation and drainage also affect skating; the estimate is not a complete account of the lubricating water film.

Constitutive law and Poisson ratio. Use positive stress for tension. In an isotropic linearly elastic medium, Hooke's law givesThe analogous equations follow by cyclic permutation. Under uniaxial tension in direction 1, $e_1={\tau }_1/E$ and $e_2=e_3=-\sigma {\tau }_1/E$. Thus Poisson's ratio $\sigma =-e_2/e_1$ measures transverse contraction relative to longitudinal extension when the transverse stresses are zero. It is dimensionless; it is not the ratio of transverse stress to longitudinal stress.

Radial and hoop strains. Let a material point originally at radius $r$ move to $r+R(r)$, with small axisymmetric displacement. Two neighbouring radii separated by $dr$ move apart by $dr+dR$, so the radial fractional extension is $dR/dr$. A circle's circumference changes from $2\pi r$ to $2\pi (r+R)$, so its fractional extension is $R/r$. ConsequentlyThe absence of longitudinal stress is a plane stress condition, not a claim of zero longitudinal strain.

Equilibrium of an annular wedge. Consider a slice of unit axial length, radial thickness $dr$ and small angular width $d\varphi$. The difference of the outward and inward radial face forces is $(r{\tau }_r{)}^{\prime }drd\varphi$ to leading order. The two tangential face tensions have inward radial resultant ${\tau }_{\varphi }drd\varphi$: each contributes ${\tau }_{\varphi }dr\sin (d\varphi /2)$. With no body force, their balance gives the radial equilibrium of a thick elastic cylinder,This hoop contribution is why the radial stress need not be spatially constant in a thick wall.

Displacement equation. With ${\tau }_z=0$, the two Hooke's law equations readInvert them:Substitute in ${\tau }_{\varphi }=(r{\tau }_r{)}^{\prime }$. After cancelling the common factor, the right-hand side is $r{R}^{\prime \prime }+(1+\sigma )R^{\prime }$, soThe cancellation explains why this displacement equation does not depend on the value of Poisson's ratio.

Power laws and wall pressure. Try $R\propto r^n$. The indicial equation is $n(n-1)+n-1=n^2-1=0$, soThe resulting Lamé stress distribution in a pressurized cylindrical wall can be writtenThe pressure boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. Solving them yieldsHence the positive compressive radial pressure in the wall isIt equals $p_1$ at $a$ and $p_0$ at $b$. This directional radial pressure is not an isotropic fluid pressure: radial and tangential stresses differ. The associated displacement is $R=[(1-\sigma )Pr+(1+\sigma )Q/r]/E$. Also $e_z=-\sigma ({\tau }_r+{\tau }_{\varphi })/E=-2\sigma P/E$, generally nonzero even though ${\tau }_z=0$.

Maximum absolute tangential stress. For $p_1>p_0$, $Q>0$ and ${\tau }_{\varphi }^{\prime }=-2Q/r^3<0$. Therefore the maximum of its absolute value must be at one of the two surfaces. Keeping the specified external pressure gives the fully general answerFor the usual high internal pressure with negligible external pressure, the largest magnitude is the tensile stress on the inner wall:For finite $p_0$, the inner-wall expression is $[p_1(a^2+b^2)-2p_0{b}^2]/(b^2-a^2)$, but its absolute value is not always the larger endpoint value. For example, nearly equal internal and external pressures can leave the outer surface more compressive. With $p_1=p_0=p$, both stresses correctly reduce to $-p$ everywhere. These results use a homogeneous linear elastic wall without axial loading; a closed-ended pipe would require a different longitudinal-stress condition.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

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 a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

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.

Assume a slender homogeneous elastic beam obeying linear elasticity, with plane sections remaining plane and no resultant longitudinal force. A length element of its neutral axis subtending a small angle $d\varphi$ has length $Rd\varphi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $zd\varphi$, giving longitudinal strain $\epsilon =z/R$ with signs chosen consistently for tensile and compressed fibres. By Hooke's law, its longitudinal stress is $\sigma =E\epsilon =Ez/R$.

The torque of these longitudinal forces about the neutral axis is thereforeHere $E$ is Young's modulus and $I$ the second moment of area. The displayed relation uses a signed radius and bending moment; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative bending moments in the later parts, take the deflection $y$ positive downward and the signed curvature as $1/R\simeq y^{\prime \prime }$. Then the small-slope Euler-Bernoulli beam equation reads $EI{y}^{\prime \prime }=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.

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.

The depth-averaged added horizontal compressive stress is $\bar{\sigma }=F/a$. Resolving its traction on a plane dipping at angle $\delta$ gives the fault shear stress from a plate-integrated force,Keeping the previous topographic correction gives about $13.8\mathrm{MPa}$. This is the mean additional shear from ridge push, assuming the force is transmitted through thickness $a$ and resolving the nonisotropic stress increment. An isotropic lithostatic pressure adds normal traction but no shear. The force integral alone does not specify the local stress at each point on the fault or establish failure.

Earthquake location. The hypocentre is rupture initiation; its surface projection is the epicentre. Record calibrated arrival times at stations of known coordinates, identify P wave and S wave phases, and predict their travel times through a stated velocity model. Earthquake location by arrival times fitsfor the three source coordinates and origin time $t_0$. Linearization supplies residual equations $\delta t_i=\delta t_0+\nabla T_i\cdot \delta \mathbf{x}$. A grid search or iterative weighted least squares inversion can use pick uncertainties, while station geometry controls whether those parameters are well constrained. Four independent arrival constraints are the minimum for four unknowns, not a guarantee of a reliable solution.

In a homogeneous medium, an arrival-time difference between an S wave and a P wave cancels origin time and estimates source-to-station distance,Several distances intersect to locate the event; real layered or three-dimensional Earth structure requires curved-ray travel times. Nearby stations and diverse takeoff angles help resolve depth. Clock errors, phase misidentification, poor azimuthal coverage and incorrect velocities introduce uncertainty or systematic bias. Closely spaced earthquakes can be relocated precisely relative to each other because common-path errors partly cancel, even when their absolute locations remain uncertain. Report a location uncertainty and model assumptions, not merely a best-fit point.

H. F. Reid's earthquake-cycle model. A fault can remain locked while long-term tectonic motion slowly deforms the surrounding rocks. Elastic strain energy and shear stress accumulate. When resistance is overcome, rapid fault slip releases part of that stored energy as seismic waves, frictional heat and other irreversible work. The adjoining rocks partly recover their elastic distortion while retaining a permanent displacement discontinuity across the fault: this is elastic rebound. Surface markers initially crossing the fault become bent during loading and offset after slip.^[ref]

A simple sketch is a slow rise of stress followed by a rapid stress drop, then renewed loading; the plate displacement continues throughout. Geodetic deformation, observed fault offsets and earthquake stress changes can test the picture. Actual cycles need not repeat with identical slips or times: creep, afterslip, stress transfer and heterogeneous friction complicate the idealization. The release need not remove all pre-existing stress. Elastic rebound explains storage and release; it is not a deterministic earthquake clock.

Let the seismic moment tensor of a double-couple seismic source be $M=M_0(\mathbf{s}\otimes \mathbf{n}+\mathbf{n}\otimes \mathbf{s})$ for perpendicular unit slip and normal vectors. Its T axis is $\mathbf{T}=(\mathbf{s}+\mathbf{n})/\sqrt{2}$, with eigenvalue $+M_0$, and its P axis is $\mathbf{P}=(\mathbf{s}-\mathbf{n})/\sqrt{2}$, with eigenvalue $-M_0$. They bisect the fault-plane and auxiliary-plane quadrants.

These earthquake pressure and tension axes describe the source deformation. They are not generally the principal axes of the ambient stress tensor. Friction, inherited fault orientation and local stress changes can rotate their relationship to that tensor; a stress inversion requires further assumptions and multiple mechanisms.

Elastic rebound. Tectonic motion loads a fault whose shallow portion is locked. Rock on either side deforms and stores elastic strain energy while slip is inhibited. When failure permits rapid slip, part of the stored energy becomes seismic radiation and part is dissipated in friction and damage. The adjacent rock partly relaxes, while a permanent displacement remains across the fault. Geodetic strain accumulation and coseismic offsets support this loading–release picture.^[ref]

The model is useful for brittle crustal faulting but is not a prediction of precise recurrence, rupture size or frictional evolution. Aseismic creep can release deformation without a large sudden event. Viscoelastic flow, afterslip, fluid-pressure changes and interacting fault segments complicate a simple locked-fault cycle. Deep-focus earthquakes and non-shear source components require additional physics; not every earthquake source is adequately described by a shallow brittle double couple. Rebound also need not release all the pre-existing stress.

A thin negative-gradient velocity zone. In horizontal layering, use the seismic ray parameter $p=\sin i/v$, the conserved horizontal slowness. A ray turns when $pv=1$. Downward passage through decreasing speed makes $pv$ smaller, so a surface ray cannot turn in the negative-gradient interior. Rays either turn above the zone or penetrate it and turn below where speed has recovered sufficiently. This produces missing turning depths and a seismic shadow range from a low-velocity zone in the ideal ray model.

Retain all branches usingThe travel-time slope is $dT/dx=p$. Just below the critical value that permits passage beyond the upper speed maximum, the ray makes a finite extra excursion through the low-velocity region. Range can jump, and the deep branch can fold back before resuming its normal trend, giving multiple arrivals outside the shadow. A first-arrival curve alone hides some of these branches. Define the seismic intercept time byThe $x$–$p$ representation displays the gap and fold; the $\tau$–$p$ representation has separate shallow and deep branches and a branch jump at the critical ray parameter for the piecewise-linear cusp model. Do not join them by invented turning rays. Diffraction and finite-frequency energy soften a purely geometrical shadow.

The figures use an explicit illustrative profile, not measured Earth parameters: speed rises from 1 to 2 over depths 0 to 1, falls to 1.8 by depth 1.2, then rises with its original gradient. The asthenosphere beneath oceanic lithosphere is an important setting for a low-velocity zone. A major P-wave speed reduction also occurs at the boundary between the mantle and the outer core of the Earth; S waves do not propagate through the liquid outer core of the Earth.^[ref]

New oceanic lithosphere is hot and mechanically weak. As it cools, the depth over which rock can sustain elastic stress on the loading timescale generally increases, so effective elastic thickness tends to increase with plate age at the time of loading. It is this mechanical state when the load is emplaced, together with subsequent relaxation, that matters; the present age alone is insufficient for an ancient load.

The half-space cooling model illustrates the thermal dependence. With age $\tau$, surface temperature $T_s$, initial interior temperature $T_m$ and constant thermal diffusivity $\kappa$,A chosen temperature $T_{\ast }$ lies atIf the base of the mechanically strong layer approximately follows such an isotherm, $T_e$ initially follows a similar age trend. Cooling models constrained by ocean-floor depth and heat flow provide the thermal motivation; their fitted thermal thickness must not be mistaken for measured elastic thickness.^[ref]

In practice $T_e$ is an effective bending thickness rather than the depth of one universal temperature. Stress-dependent yielding, grain size, water content, composition, loading duration and reheating can alter it. Finite-plate cooling and basal heat input also limit indefinite square-root growth at old ages. Nor should $T_e$ be identified with either the thickness of the crust or the full seismically defined lithosphere.