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An upward drag force $F_D\ge 0$ changes the variable-mass system equation toFor an ordinary drag force increasing with positive speed, the new curve lies below the no-drag curve. It still starts with slope $g$ if $F_D=0$ at rest. The dashed curve illustrates quadratic drag on a sphere with constant radial growth, using $F_D=\frac{1}{2}{C}_D{\rho }_{\mathrm{air}}\pi r^2{v}^2$ with constant drag coefficient $C_D$ and air mass density ${\rho }_{\mathrm{air}}$. The equation for these dimensionless quantities isThe figure uses $\eta =2$ solely to show the shape; it is not an additional datum of the paper.

There is a limitation to the usual terminal velocity sketch here. The examiner's commentary describes a curve approaching terminal velocity, but the paper maintains unbounded radius growth and specifies no drag law. A horizontal limit is justified if one additionally treats the radius as approximately fixed, or otherwise assumes a drag-to-mass ratio with a fixed limiting balance. With the growing radius and constant-$C_D$ quadratic drag just written, the late balance instead givesThis curve flattens in the sense that its slope tends to zero, but its speed remains unbounded. Thus a lower, more strongly flattening curve is warranted; a finite terminal speed requires an additional assumption. The no-growth approximation gives the familiar finite value $v_T=\sqrt{8\rho gr/(3C_D{\rho }_{\mathrm{air}})}$ for a specified fixed radius.^[ref]

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

For a cylinder of radius $R_c$ in unbounded two-dimensional incompressible potential flow, a velocity potential is $\phi =u(r+R_c^2/r)\cos \theta$. It givesOn the surface $v_r=0$. The points nearest the banks are the sides perpendicular to the stream, $\theta =\pi /2,3\pi /2$, hence the speed is $2u$ at each. This is the inviscid outer-flow speed; a real no-slip boundary layer has zero velocity at the material surface. The small-radius/wide-river assumption neglects bank effects.

For an elastic eccentric impact of a thin disc, take upward velocity as positive and let $J$ be the short upward impulse. The disc's moment of inertia about a horizontal diameter is $I=M{a}^2/4$. With impact offset $d=a/2$,The contact point has upward velocity $v_C+d\omega$. Unit restitution against the fixed post requires that velocity to be $+u$, soSince $I=M{d}^2$, $J=Mu$ and the centre's instantaneous speed is zero. The disc rotates with $\omega =2u/a$; its rotational kinetic energy $I{\omega }^2/2=M{u}^2/2$ equals the entire initial energy. Gravity's impulse during the impact is neglected. Using the polar-axis moment of inertia would describe the wrong rotation axis.

The vacuum wave equation gives ${\omega }^2=c^2[k_g^2+(\pi /a{)}^2]$. The cutoff frequency is $f_c=c/(2a)=1.499\mathrm{GHz}$. In a lossless empty guide, energy travels at group velocity,The larger phase velocity $\omega /k_g=c^2/v_E$ is not the speed of energy transport.

Oceanic magnetic anomalies. Newly formed oceanic crust contains magnetic minerals. As it cools through their magnetic blocking temperatures, it acquires thermoremanent magnetization recording the ambient field polarity. Continuing seafloor spreading carries that record away from the ridge. Geomagnetic reversals therefore generate approximately paired marine magnetic stripes parallel to the ridge, with older crust farther from its axis. A magnetometer measures the superposed field above this magnetized material; subtracting a reference field reveals magnetic anomalies. The anomaly sign and shape depend on magnetization direction, latitude, source depth and geometry, so a positive measured anomaly is not universally identical to normal rock polarity.^[ref]

Matching a sequence of stripes to an independently dated reversal chronology gives plate ages and spreading histories. A stripe of width $w$ formed over interval $\Delta t$ implies half-spreading rate $v=w/\Delta t$ under steady symmetric spreading; full separation speed is $2v$. Unequal widths on opposite flanks can reflect asymmetric spreading or later deformation. Alteration, remagnetization and overlapping magnetic sources complicate inversion. The sequence correlation is stronger evidence than an isolated stripe or a field-strength peak.

Rayleigh-Bénard convection. Consider a horizontal fluid layer of thickness $d$, hotter below by $\Delta T>0$. The conductive state is motionless with a linear temperature gradient. A rising warm parcel is buoyant, but viscosity resists its motion and thermal diffusion erases its temperature anomaly. In the Boussinesq approximation, density variations are retained in buoyancy while the flow is otherwise incompressible. The competing effects are measured by the Rayleigh number and Prandtl number,For stress-free impermeable, fixed-temperature boundaries, linear stationary disturbances with horizontal wavenumber $q/d$ giveMinimizing gives $q_c=\pi /\sqrt{2}$ and the critical Rayleigh number for an isothermal layer, ${\mathrm{Ra}}_c=27{\pi }^4/4\simeq 657.5$. Two rigid no-slip, fixed-temperature boundaries instead give approximately $1708$; the critical value is not independent of boundary conditions.^[ref]

Above onset, buoyancy can sustain overturning rolls and heat transfer exceeds pure conduction, with Nusselt number $\mathrm{Nu}>1$. Nonlinear advection eventually limits the disturbance; larger driving can produce time-dependent or turbulent states. The ideal Boussinesq problem does not require hexagonal cells: planform selection depends on boundaries and non-Boussinesq effects. Convection driven by surface tension at a free surface is a distinct mechanism. An adverse temperature gradient supplies driving, but overcoming diffusion and viscosity requires a finite threshold.

The phase velocity in Figure 1 is measured along the Earth's surface. Spherical geometry makes the local horizontal phase speed at radius $r$ equal to $cr/R_E$. In the ray approximation, squared radial shear wavenumber isThe spherical surface-wave evanescence threshold immediately below the discontinuity is thereforeEquality is the grazing limit. Using $5.08\mathrm{km}{\mathrm{s}}^{-1}$ directly would omit the requested spherical correction. Immediately above the discontinuity the corresponding boundary is $4.87(6371/5961)\simeq 5.20\mathrm{km}{\mathrm{s}}^{-1}$. Hence the interval $5.20<c<5.43\mathrm{km}{\mathrm{s}}^{-1}$ specifically describes propagation just above the jump and evanescence just below it. Slower modes can already be evanescent above $410\mathrm{km}$.

Intersecting the plotted branches with $5.43\mathrm{km}{\mathrm{s}}^{-1}$ gives the following approximate upper periods. Below each upper period, that branch is evanescent immediately beneath the discontinuity:

For these overtones the displayed ranges thus start near $10\mathrm{s}$ and end at their respective crossings. Any still higher curve already above the threshold at the lower plotted period has no evanescent interval within the displayed range. Readings of the closely spaced higher curves are only accurate to about a second, and the plot does not establish their continuation to unshown periods. Nor does the local inequality prove that a mode remains evanescent at every greater depth in an unspecified Earth model.

Use the flat two-layer model, with thickness $h=30\mathrm{km}$, crustal shear speed ${\beta }_1=3.5\mathrm{km}{\mathrm{s}}^{-1}$ and mantle speed ${\beta }_2=4.5\mathrm{km}{\mathrm{s}}^{-1}$. For SH-wave travel times in a crustal waveguide, unfolding each reflected path gives all reflection orders at once:with $x$ in kilometres and times in seconds. The direct wave begins at the origin; reflected orders have intercepts $60m/3.5=17.14m\mathrm{s}$ and approach the direct-wave slope at large distance.

The critical angle, measured from the vertical, is $i_c=\arcsin (3.5/4.5)\simeq 51.1^{\circ }$. Associated critical refraction and its multiples giveEach head-wave line is tangent to its reflection curve at the critical distance. In particular the primary head wave exists before it becomes the first arrival; it overtakes the direct wave only around $170\mathrm{km}$.

The slope of a reflected travel-time curve is $p=d{t}_m/dx=\sin i/{\beta }_1$. For $x<x_{c,m}$, $p<1/{\beta }_2$, so the mantle supports outgoing shear radiation and successive reverberations leak energy. For $x>x_{c,m}$, $1/{\beta }_2<p<1/{\beta }_1$: the mantle field is evanescent and ideal interface reflection is total. These leaking and trapped SH-wave reverberations explain the annotations in the original travel-time drawing below. A reflected ray is not itself a separate Love mode: constructive interference of the trapped contributions, including reflection phase, selects the normal modes.

For a horizontally propagating mode with ${\beta }_1<c<{\beta }_2$, define vertical wavenumbersThe free upper surface and welded interface impose the Love-wave dispersion conditionwhere ${\mu }_j={\rho }_j{\beta }_j^2$ are shear moduli. Each phase-velocity branch falls towards $3.5\mathrm{km}{\mathrm{s}}^{-1}$ with increasing frequency. Higher branches join $4.5\mathrm{km}{\mathrm{s}}^{-1}$ at their low-frequency cutoffs. At cutoff $s\to 0$, soThus the first three overtones have maximum periods about $10.77$, $5.39$ and $3.59\mathrm{s}$. The fundamental mode has $n=0$ and no positive-frequency cutoff in this flat layer-over-half-space model: it approaches $c={\beta }_2$ as $f\to 0$, with ever-deeper penetration. The phase-velocity sketch uses an explicitly illustrative equal-density choice, because the question gives speeds but no density or rigidity ratio. That missing ratio changes detailed branch curvature, not the cutoff values or the requested branch topology.

For the actual finite spherical Earth, the very-long-period continuation must instead be described by discrete toroidal normal modes, not a flat infinite waveguide at arbitrary wavelength. The flat-model answer therefore does not assert a continuum of arbitrarily low-frequency global nonrigid oscillations. A global lowest oscillation frequency cannot be computed from the supplied crust-and-half-space data alone.^[ref]

A dense array does not correct an incorrect velocity model. The faster southwest block permits shorter travel times than an assumed uniform medium, with refraction favouring paths through fast rock. First arrivals may include refracted or seismic head wave paths rather than the assumed direct straight paths. Fitting those arrivals with one constant speed can make an earthquake on the fault appear closer to southwest stations and farther from northeast stations.

The sign follows even from a two-station linearization. Put southwest and northeast stations at $x=-L,+L$, and the true event at $x=0$ with depth $h$. In a constant-speed model,A negative observed differential residual therefore produces $\delta x<0$, an apparent southwest displacement. A common origin-time adjustment cannot remove this directional residual. The stated kilometre-scale offset is a model bias, not proof that the fault dips towards the apparent epicentres. Its exact magnitude cannot be predicted without the actual velocities and station geometry; the question supplies that magnitude as an observation.

The relevant contrast is independently documented at Parkfield: higher velocities lie southwest of the fault, with lower velocities northeast.^[ref] Account for the lateral structure and correct phases, then relocate and test whether the bias remains.

Subtract the background hydrostatic pressure. A relief amplitude $h$ produces a restoring pressure perturbation $p^{\prime }\sim \rho gh$ and horizontal gradient $\rho gh/\lambda$. Incompressibility makes the disturbance penetrate a depth of order its horizontal scale. For sinusoidal viscous rebound, with relaxation speed $v\sim h/\tau$, the viscous force density is $\eta v/{\lambda }^2$. Balancing givesThis is a small-amplitude viscous half-space estimate; numerical factors from the sinusoidal geometry are absent from a scaling argument.

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]

At the south pole $\mathbf{r}=-R\hat{z}$. Choose plate C as the instantaneous reference frame. Since $R=4\times 10^9$ mm, the two components of relative angular speed give 20 and 48 mm/year. The plate velocities in the fixed tangent-plane axes areThe plate velocity triangle therefore has sidesThe physical boundary rays point along $+x$ for AB, $+y$ for AC and $(12,-5)$ for BC. Plate A occupies the quadrant between the first two rays, B the narrow wedge between AB and BC, and C the remainder. The stated overriding relations make AB and AC subduction trenches, while BC remains a transform fault: this is a trench-trench-transform triple junction. Put the subduction teeth on B's side of AB and A's side of AC, identifying B above A and A above C.

For kinematic stability there must be a single junction velocity $\mathbf{U}$ compatible with all three boundary-normal motions. Use the standard idealization that each trench's normal velocity is that of its overriding plate, with no independent rollback. AB is parallel to $x$, so $U_y=v_{B,y}=20$. AC is parallel to $y$, so $U_x=v_{A,x}=-48$. The transform requires $\mathbf{U}-{\mathbf{v}}_C$ to be parallel to $(12,-5)$, which is true for this intersection. HenceThe southern triple junction is instantaneously kinematically stable. It is stationary relative to B, migrates along AB relative to A, and migrates along BC relative to C. This compatibility result does not promise long-term stability if plate rotations or trench rollback change. Notice that the three relative velocities close even though the overriding order is not a statement that all three plates move toward one common stationary point.

At the north pole $\mathbf{r}=+R\hat{z}$, so the velocities reverse:The triangle still has side lengths 20, 48 and 52 mm/year. Across AB, A moves toward positive $y$ relative to B; across AC, A moves toward positive $x$ relative to C. Both boundaries therefore open and are spreading ridges, with full opening speeds 20 and 48 mm/year. Together with BC they form a ridge-ridge-transform triple junction.

In the symmetric normal-spreading model, a ridge's normal velocity is the mean normal velocity of its two plates. AB gives $U_y=(0-20)/2=-10$, and AC gives $U_x=(48+0)/2=24$. Their intersection lies on the BC transform velocity line:This northern junction is also instantaneously kinematically stable. The diagrams use the same fixed geocentric tangent axes at both poles; reversing velocities is not the same operation as reflecting a map drawn from an exterior observer's viewpoint.

Not every ridge-ridge-transform geometry works. For pure normal spreading, each ridge velocity line is the perpendicular bisector of the corresponding side of the plate velocity triangle. Their intersection is the triangle's circumcenter. A transform junction must also lie on the third side's line. A nondegenerate triangle's circumcenter lies there only at that side's midpoint, so the opposite angle is a right angle. Equivalently, $({\mathbf{v}}_B-{\mathbf{v}}_A)\cdot ({\mathbf{v}}_C-{\mathbf{v}}_A)=0$. The two ridge directions are then perpendicular, as they are here. RRF junctions are conditionally stable, with the right-angle ridge geometry in this symmetric pure-normal model; they are not all stable. Oblique or asymmetric accretion changes the velocity-line assumptions and must be specified separately.

First-motion source mechanisms. Locate an earthquake from arrival times and use a seismic velocity model to calculate each station's source takeoff direction. Measure the initial direct P wave polarity after checking instrument sign, component orientation, timing and phase identification. An upward trace at an instrument is not automatically a source compression without the propagation and component convention. Plot the inferred source radiation signs on a focal sphere, commonly using a lower-hemisphere equal-area projection, and seek nodal planes separating positive and negative observations.

For a shear dislocation with orthogonal unit slip and fault-normal vectors $\mathbf{s},\mathbf{n}$, the double-couple fault source has $M=M_0(\mathbf{s}{\mathbf{n}}^{\mathsf{T}}+\mathbf{n}{\mathbf{s}}^{\mathsf{T}})$. Its far-field P radiation sign is that ofThe two zeros are orthogonal nodal planes; the four quadrants alternate between outward and inward initial radiation. Their orientations and quadrant signs determine candidate strike, dip and rake. Normal, reverse and strike-slip mechanisms give different projected patterns. The earthquake pressure and tension axes are the negative and positive moment-tensor principal directions, respectively; they do not by themselves measure the complete regional stress tensor.

There is a fundamental fault-plane ambiguity: exchanging $\mathbf{s}$ and $\mathbf{n}$ leaves the radiation tensor unchanged. First motions therefore give a fault plane and an auxiliary plane without saying which actually slipped. Use mapped rupture, aftershock geometry or independent geodetic displacement to select the physical fault where possible. Polarity signs alone do not measure scalar seismic moment, source duration or rupture extent. A poorly sampled focal sphere or many near-nodal observations can admit several mechanisms rather than one defensible beachball.

Propagate uncertainty in event depth, velocity gradients and takeoff angles, allow for mispicked or reversed polarities, and report the range of mechanisms consistent with the observations. A primary method did this by testing alternative locations and velocity models and validated its uncertainty estimates using noisy synthetic data and clusters of events with similar waveforms.^[ref] Broadband waveforms and amplitude ratios add information, especially for non-double-couple sources, but they are additional measurements rather than a way to extract amplitude information from a sign-only data set.

Radial density structure. In an isotropic adiabatic material, the seismic parameter isFor a chemically homogeneous region whose background stratification is adiabatic, $d\rho /dP=\rho /K_S=1/\Phi$. Combine this with spherical hydrostatic balance and mass conservation to obtain the Adams-Williamson equationsThe inward density increase follows compressibility under the increasing pressure. A specified starting density and measured $\Phi (r)$ allow integration within a homogeneous adiabatic layer. Seismic velocities alone, however, do not determine that density normalization, since they contain ratios of moduli to density. Chemical boundaries, phase transitions, nonadiabatic thermal gradients and anisotropy require additional modeling; a composition jump is not obtained by blindly integrating the smooth relation across it.

Use travel times and surface waves to constrain elastic velocities, reflected/converted phases to locate interfaces, and normal-mode frequencies to constrain density as well as elastic structure through their sensitivity kernels. Impose independently known total mass and moment of inertia:Because these weight depth differently, they help distinguish admissible profiles, but two integrals alone do not specify an arbitrary radial function. Laboratory equations of state and mineral physics restrict composition, compressibility and density jumps. pressure and enclosed mass are matched across interfaces, while density may jump. In a liquid region $\beta =0$ and $\Phi ={\alpha }^2$, so the shear-wave absence constrains phase state without itself providing a direct density measurement.

The Preliminary Reference Earth Model is a concrete joint inversion of seismic travel times, normal modes, mass and inertia, including density and attenuation along with elastic structure. The original analysis combined these complementary data rather than deriving density uniquely from P wave speed.^[ref] A defensible radial density model combines hydrostatic/thermodynamic relations with seismic and global integral constraints, and states where homogeneity and adiabaticity fail.

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.

Use geometric rays for body-wave arrivals and a traction-free upper surface. A compact isotropic explosion source has moment tensor proportional to ${\delta }_{ij}$ and produces direct P body radiation in a homogeneous isotropic medium. Thus the basic near-offset arrival is the direct upper-layer P wave, with time $X/{\alpha }_1$. A pure explosion is not a shear fault: it does not automatically emit a direct S wave of comparable strength. Interfaces can nevertheless convert its P energy into SV motion, and the free surface can support Rayleigh waves.

Downward P energy reflects from the interface to give a PP reflection returning to the receiver; for a horizontal interface its ray is symmetric and its time is $\sqrt{X^2+4H^2}/{\alpha }_1$. Other transmitted P energy enters the faster half-space. At $\sin i_c={\alpha }_1/{\alpha }_2$, a critical ray travels along the interface and feeds an upgoing seismic head wave, often labeled Pn in this simplified crustal model. Its ideal travel-time branch isIt requires enough offset for the two inclined upper-layer legs, $X\ge 2H\tan i_c$. Once $X$ exceeds the larger first-arrival crossover distance $X_c=2H\sqrt{({\alpha }_2+{\alpha }_1)/({\alpha }_2-{\alpha }_1)}$, this branch arrives before the direct upper-layer wave because it spends most of the path in faster material. A transmitted ray descending into an otherwise uniform infinite half-space does not return as a separate reflection without an additional reflector.

At oblique incidence, the bonded-interface conditions couple P and SV polarizations. A downgoing P ray can return as a converted SV reflection in the upper layer, with common horizontal slowness $p=\sin i_P/{\alpha }_1=\sin i_S/{\beta }_1$. There are transmitted P/SV branches and, after further interaction with the free surface or interface, mixed-mode reflections and layer reverberations. Conversion coefficients depend on density and impedance as well as wave speeds, and some vanish at normal incidence. The supplied inequalities do not order ${\beta }_2$ against ${\alpha }_1$, so one cannot universally draw a P-incident critical head ray travelling at the lower-layer S speed: that direct conversion construction requires ${\beta }_2>{\alpha }_1$. A shear head-wave phase can be present if the appropriate SV energy and ray conditions are available, but is not an inevitable direct explosion arrival from the stated data alone.

At later times a shallow explosion can excite dispersive Rayleigh-wave trains, with coupled P-SV motion decaying with depth. Short wavelengths mainly sample the slow upper layer, whereas long wavelengths penetrate the lower half-space; group and phase speeds need not be equal. Multiple reflected/converted body waves can overlap this coda. The shear-speed contrast also permits Love-wave modes structurally, but SH motion is decoupled from P-SV in this flat isotropic model. An ideal isotropic explosion does not excite Love waves here; appreciable SH/Love energy would require source asymmetry, scattering or a broken symmetry of the model.

Exact amplitudes, resolved arrival separation and attenuation cannot be calculated without densities, source spectrum, damping and instrument response. The qualitative robust features are a near-offset direct P arrival, reflected and converted phases, a faster-lower-layer P head wave that can become first at long range, reverberations and P-SV surface-wave energy. The ray diagram uses declared illustrative speed ratios solely to show the geometry; they are not additional supplied material constants.

Use the shock frame, with upstream and downstream fluid velocities $u_1=u$ and $u_2=u/4$ pointing in the same direction. These are the Strong-shock Rankine-Hugoniot conditions for a nonrelativistic monatomic gas with adiabatic index $5/3$. The two scattering media approach one another at relative speed $U=u_1-u_2=3u/4$, to first order in $u/c$. An energetic electron repeatedly crosses the shock wave, scatters from magnetic turbulence carried by the local plasma, and returns. Each scattering approximately conserves its energy in the local fluid rest frame while changing its direction; comparing successive fluid frames produces a systematic energy and momentum gain. The energy comes from the converging bulk flow, through the moving magnetic field irregularities and their associated electric fields. This is diffusive shock acceleration, a first-order Fermi acceleration process.

Let $\eta$ be the positive cosine of the crossing angle with the shock wave's normal. The Lorentz transformation between fluid frames gives, for either crossing direction with the corresponding sign of the normal reversed,For an almost isotropic population, the number crossing a unit area is weighted by its normal velocity. Thus the normalized crossing distribution is $2\eta d\eta$, rather than a uniform distribution in $\eta$, andFor an ultrarelativistic electron, $E\simeq pc$ and its speed $v\simeq c$, soMore generally $dE/dp=v$ gives $⟨\Delta p⟩/p=2U/(3v)$, while a nonrelativistic kinetic energy has $⟨\Delta K⟩/K=4U/(3v)$. A complete upstream–downstream–upstream cycle contains two crossings and has ultrarelativistic fractional energy gain $u/c$.

The initially cold electrons need an injection stage: shock wave heating and small-scale plasma processes must first produce particles with speed much larger than $u$ that can return across the front. The $v\simeq c$ result is a description of that accelerated population; a cold electron cannot simply be assigned $E\simeq pc$ at its first encounter.

Interpret the terminal infall speed as the free-fall speed from a radius much larger than the neutron star radius $R_{\ast }=12\mathrm{km}$. In the Newtonian estimate,For fully ionized hydrogen with electrons and protons at a common temperature, there are two particles per atom and $\mu =(m_p+m_e)/2\simeq m_p/2$. The result of e then yieldsThis is the thermal energy scale $k_B{T}_d$ per particle. In the same classical monatomic gas approximation, the mean random kinetic energy per particle is $3k_B{T}_d/2\simeq 59\mathrm{MeV}$. Before the shock wave, the bulk kinetic energy per proton is $m_p{v}_u^2/2=G{M}_{\ast }{m}_p/R_{\ast }\simeq 208\mathrm{MeV}$. Distinguishing these quantities avoids an ambiguity in “particle energy”. If only the ions support the downstream pressure because electrons have not equilibrated, $\mu \simeq m_p$ instead gives $k_B{T}_i\simeq 78\mathrm{MeV}$; that is a different, two-temperature assumption.

These are estimates within the question's Newtonian, $\gamma =5/3$ model. Its self-consistency is poor here: $2G{M}_{\ast }/(R_{\ast }{c}^2)\simeq 0.443$, and $39\mathrm{MeV}$ is much larger than the electron rest energy $0.511\mathrm{MeV}$. Relativistic electrons, electron–positron pair production, radiation, unequal species temperatures, and general relativity can alter the equation of state and shock wave structure. In particular, the numerical $39\mathrm{MeV}$ must not be presented as a precision surface temperature prediction.

In the rest frame, conservation of energy would require $E_{\gamma }=m{c}^2$, whereas conservation of momentum would require the photon momentum to vanish. These contradict $|{\mathbf{p}}_{\gamma }|=E_{\gamma }/c$. More generally, four-momentum conservation would equate the initial massive four-momentum to a null four-momentum, although their Minkowski norms differ:An isolated massive particle cannot decay into one photon in any inertial frame. Directly, a laboratory speed $u<c$ gives $E=\gamma m{c}^2$ and $|\mathbf{p}|=\gamma mu$; matching a photon would require $E=c|\mathbf{p}|$, hence $u=c$. Recoil against another object could change the kinematics, but would introduce an additional participant.

Write the four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$ and four-acceleration $A^{\mu }=d{U}^{\mu }/d\tau$. Since $U_{\mu }{A}^{\mu }=0$, spatial orthogonality $\mathbf{U}\cdot \mathbf{A}=0$ implies $U^0{A}^0=0$. Thus $A^0=cd\gamma /d\tau =0$ and the speed is constant. If “acceleration” is read as the laboratory three-acceleration instead, $\gamma \mathbf{v}\cdot d\mathbf{v}/dt=0$ immediately gives the same conclusion.

For four-acceleration in uniform circular motion, the speed is $r|\omega |<c$, so $\gamma =(1-r^2{\omega }^2/c^2{)}^{-1/2}$ is constant. Differentiating the four-velocity using $d/d\tau =\gamma d/dt$ givesThe proper acceleration is $\sqrt{-A_{\mu }{A}^{\mu }}={\gamma }^{2}r{\omega }^2$. The vanishing time component reflects constant energy; the spatial components describe the changing momentum direction.