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Point-charge field and superposition. Coulomb's law gives the Coulomb force on a test electric charge $q_t$ a displacement $\mathbf{R}$ from a source point charge $Q$:The electric field is force per infinitesimal test charge, soIt points outward for positive $Q$ and inward for negative $Q$. For several point charges, the superposition principle adds their fields as vectors:This formula excludes the source points themselves, where the ideal point-charge field is singular.

Exact field components of the pair. PutThe displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$ and $(x+a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Including the signs of the charges givesReading off both components,For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.

Far-field polar components. The pair's electric dipole moment points from negative to positive charge:Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, with $\theta$ measured from that positive dipole axis. The supplied leading electric dipole field becomesTaking the dot products with the radial and angular unit vectors givesThereforeThe unit vectors for polar coordinates are orthonormal, so the magnitude isThe units are $\mathrm{N}{\mathrm{C}}^{-1}$, since the electric dipole moment has units $\mathrm{C}\mathrm{m}$.

Angular sketch and comparison. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta =0,\pi$, with value $2p/(4\pi {ϵ}_0{r}^3)$, and smallest on the perpendicular directions, $\theta =\pi /2,3\pi /2$, with value $p/(4\pi {ϵ}_0{r}^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:

The neutral charge pair has no net electric charge, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero point charge has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_{\theta }$ component.

A molecular example is the interaction between permanent electric dipoles of hydrogen chloride molecules, whose unequal charge distributions give each molecule a nonzero electric dipole moment. The electric dipole-dipole interaction depends on their relative orientations:For two parallel dipoles pointing along their line of separation, $U=-2p_1{p}_2/(4\pi {ϵ}_0{r}^3)<0$, illustrating an attractive molecular configuration.

Put the rigid lid at $z=0$, the interface at $z=-h(x,y,t)$ and ${\rho }_1={\rho }_0-\Delta \rho$. Assume hydrostatic vertical balance and a resting deep layer. Its pressure is $p_2=C-{\rho }_{0}gz$, with horizontally uniform $C$. Pressure continuity at the interface then fixes the upper-layer pressure:Hence its horizontal pressure gradient at fixed $z$ is ${\nabla }_h{p}_1=\Delta \rho g{\nabla }_{h}h$, independent of depth. Taking the horizontal momentum equation and a vertically uniform upper-layer velocity givesThe exact density denominator for upper-layer acceleration is ${\rho }_1$; the familiar ${\rho }_0$ denominator is the small-density-contrast Boussinesq approximation. The restoring pressure is weaker than for a free surface because only the density difference contributes.

The horizontal shallow-layer equations use the vertical component of planetary rotation: $f=2\Omega \sin \phi$. Thus the source's vector notation is interpreted as the projected vector $f\mathbf{k}$ after the traditional approximation. Taking the full magnitude $2\Omega$ as a midlatitude Coriolis parameter would discard the latitude dependence. Together with layer-volume conservation, the model has ${\partial }_{t}h+{\nabla }_h\cdot (h\mathbf{u})=0$.

For the linear three-reservoir carbon model, with concentrations as a column vector, the linear ordinary differential equation isA homogeneous normal mode has $\mathbf{x}=\mathbf{v}{e}^{\lambda t}$, where $J\mathbf{v}=\lambda \mathbf{v}$. For negative real eigenvalues the e-folding relaxation time is $\tau =-1/\lambda$. The row sums vanish, so ${\mathbf{v}}_3=(1,1,1{)}^T$ has ${\lambda }_3=0$ andEqual effective concentrations generate no exchange. Summing the reservoir budgets givesThus injected carbon dioxide is conserved after emissions cease: the model redistributes it but contains no permanent removal. The zero normal mode records this conserved excess mass, rather than a long but finite decay.

To check the other normal modes are damped, set $D=\mathrm{diag}(R_a,R_b,R_d)$ and $L=-DJ$. ThenThe similarity transformation $D^{1/2}J{D}^{-1/2}=-D^{-1/2}L{D}^{-1/2}$ is a symmetric negative-semidefinite matrix with a one-dimensional kernel, so the remaining two eigenvalues are strictly negative.

For a double-couple seismic source, measure P-wave first-motion polarity signs and use source-to-station takeoff angles from the location and velocity model to plot them on a focal sphere. Fit the alternating radiation-sign quadrants, optionally adding amplitude ratios and waveform information. The two nodal planes determine a focal mechanism, conventionally expressed by fault strike, fault dip and fault rake.

A more general waveform inversion fits a seismic moment tensor to several components and stations. Its double-couple part supplies candidate slip and normal vectors, but a general moment tensor should not be forced to be double-couple without justification. Polarity calibration, near-nodal observations, path effects and takeoff-angle errors limit the result.

Point-source radiation alone interchanges the fault plane and auxiliary plane without changing the double couple. Resolve this fault-plane ambiguity using mapped rupture, aftershock geometry, geodesy or finite-rupture directivity. A dense set of polarity picks does not by itself remove that fundamental ambiguity.

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.

A fault plane represents the surface across which the two sides of a fault slip. A real fault need not be perfectly planar; these angles can describe a local planar segment. Its fault strike is the azimuth, measured clockwise from north, of its horizontal intersection. The line itself has a $180^{\circ }$ ambiguity; choose the direction for which the fault dips to the right. Its fault dip, $\delta$, is the angle below horizontal measured perpendicular to that fault strike, from $0^{\circ }$ for horizontal to $90^{\circ }$ for vertical.

The fault rake, $\lambda$, is the angle of the relative slip vector within the fault plane. State the sign convention: if $\mathbf{s}$ is the chosen strike direction and $\mathbf{d}$ points down dip, useThus $\lambda =0$ is slip along $\mathbf{s}$, $+90^{\circ }$ is up-dip slip and $-90^{\circ }$ is down-dip slip. This distinguishes the direction in which the fault is tilted from the direction in which it actually moves.

The hypocentre is the point where rupture begins within the Earth. The epicentre is the point on the surface directly above it. The hypocentre need not coincide with the centroid of the final rupture. Orientation uses strike and dip; motion additionally needs rake; epicentre and hypocentre locate the rupture's start.^[ref]

For seismic source inversion, collect calibrated, consistently filtered samples from all available stations and components into a data vector $\mathbf{d}$. For each trial source position and origin time, compute the corresponding elastodynamic Green tensor in an Earth model, including the instrument response used for the data. Write the symmetric seismic moment tensor in six independent symmetric basis tensors:Each column of $\mathbf{G}$ is the synthetic seismogram for one basis tensor. An off-diagonal basis tensor contains both symmetric entries, so the double contraction and its factors of two remain consistent.

A suitable objective iswith noise covariance ${\mathbf{C}}_d$; a justified penalty can be added for regularization. At fixed $\mathbf{q}$, solve for the six entries of $\mathbf{m}$ by weighted least squares, preferably using a QR decomposition or singular value decomposition rather than explicitly inverting the matrix of the normal equations. Then search over the four nonlinear parameters using a coarse location/time grid followed by local refinement. This variable projection reduces a ten-parameter search to a four-parameter nonlinear search with an inner six-parameter linear fit. Merely having more than ten data samples does not ensure identifiability: the design matrix must have adequate rank and the network must resolve the source's radiation patterns.

Potential errors include inaccurate velocity, density, anisotropy and attenuation in the assumed Earth model; neglected finite rupture and source-time effects; timing, instrument calibration or component-orientation errors; correlated noise; uneven station coverage; and tradeoffs between depth, origin time and the moment tensor. Imposing a double-couple seismic source mechanism without evidence can hide genuine non-double-couple components. Regularization also changes the recovered source and should be reported.

Inspect residuals, singular values, recovery of synthetic sources and uncertainty under resampling or noise perturbations. Linearized covariance or a posterior distribution can quantify uncertainty within the assumed forward model, but does not automatically account for errors in that model. A good waveform fit alone is insufficient evidence that all ten parameters are uniquely recovered.

Seismic surface-wave dispersion. Rayleigh waves combine vertical and longitudinal motion in the vertical plane of propagation; Love waves have horizontal transverse motion and can be guided by a low-shear-speed surface layer. A homogeneous elastic half-space supports a Rayleigh wave with a frequency-independent speed in the ideal elastic limit, but depth-dependent Earth structure makes the surface-wave eigenfunctions and their propagation speeds depend on frequency. This structural dispersion can occur even when each constituent material is itself nondispersive.

Distinguish phase velocity $c=\omega /k$ from group velocity $U=d\omega /dk$, which controls a narrow-band envelope in the appropriate weakly attenuating limit. Longer periods generally have deeper sensitivity, while shorter periods emphasize shallower structure. In a profile whose shear speed increases with depth this often makes long-period waves faster, but low-velocity zones, anisotropy and higher modes can change that trend. Frequency dependence need not be monotonic in every Earth model.

Measure phase delays or wave-packet arrival times across known source–receiver paths for many periods, accounting for source phase, cycle ambiguity, multipathing and mode identification. Compute predicted dispersion from trial elastic profiles and compare with observations. Linearization expresses the sensitivity schematically aswhere $\alpha ,\beta$ are compressional and shear speeds. A regularized inverse problem then estimates depth structure; many crossing paths additionally provide surface-wave tomography. Joint Love wave and Rayleigh wave measurements help distinguish shear-wave polarizations and constrain radial seismic anisotropy. Finite bandwidth, depth averaging and correlated sensitivities mean that one dispersion curve does not uniquely determine a detailed Earth profile.

Seismic ray theory, the eikonal equation and Fermat's principle. In a smoothly varying isotropic medium, insert a high-frequency ansatz $u=A(\mathbf{x})e^{i\omega [T(\mathbf{x})-t]}$ into the scalar wave equation. The leading powers of frequency giveThe travel-time surfaces are wavefronts and rays are their normal trajectories, with slowness vector $\mathbf{p}=\nabla T$. The next-order equation transports amplitude, including geometrical spreading; attenuation is an additional material effect.

Fermat's principle says that the ray makes $\int ds/c$ stationary under path variations with fixed endpoints. Stationarity, rather than always the global minimum, permits multiple arrivals. Its Euler–Lagrange equation, with arclength $s$ and unit tangent $\hat{\mathbf{t}}$, isFor horizontal layering, the horizontal slowness $p_h=\sin i/c$ is conserved, giving Snell's law at interfaces and smooth refraction between them. Here $i$ is measured from vertical. A turning point satisfies $p_{h}c(z)=1$; increasing speed with depth bends a downgoing ray back upwards. Travel-time tomography uses these ray integrals to infer spatial slowness variations.

The approximation requires wavelength to be small compared with structural variation scales. Caustics, turning-point amplitudes, diffraction, scattering and finite-frequency sensitivity require wave-based corrections. Thus seismic ray theory gives an economical geometric description and useful travel times, while surface-wave dispersion retains essential interference and boundary-mode physics.

Let $\hat{x}$ point to the equatorial meridian at zero longitude, $\hat{y}$ to the equator at $90^{\circ }$ E, and $\hat{z}$ to the north pole. Use a right-hand positive Euler angular velocity vector. Relative rotations add as vectors, not as geographic coordinates:ConsequentlyHere $\lambda ={\tan }^{-1}(12/5)$ is the longitude of the positive Euler pole, and its latitude $\varphi$ is zero. The antipodal pole with the opposite angular-velocity sign is an equivalent convention; mixing a pole reversal with an unchanged sign would give the wrong motion.

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.

Write the stationary distribution as a row vector $\pi =(a,b,c,d)$. For the first transition matrix, $\pi P=\pi$ gives $a=a/2+d$, $b=a/2+b/2$, $c=b/2$ and $d=c$. Hence $b=a$ and $c=d=a/2$. Normalization yieldsBy the stated common-stationary distribution property, this is also the answer for the second transition matrix.

Any nonzero vector in ${\mathbb{F}}_2^3$ extends to a basis. Sending one such basis to another gives an element of the general linear group, so its group action on the seven nonzero vectors is a transitive group action. Counting successive independent columns givesBy the orbit-stabilizer theorem, the stabilizer $H$ of a chosen nonzero vector $v$ has order $168/7=24$.

There are three two-dimensional linear subspaces containing $v$, corresponding to the three one-dimensional linear subspaces of ${\mathbb{F}}_2^3/⟨v⟩$. The stabilizer $H$ acts transitively on them: extend a basis of either plane beginning with $v$ to a basis of the whole vector space. Thus the stabilizer of a flag $⟨v⟩<P$ has order $24/3=8$.

If $Q\le G$ has order eight, each orbit of a group action of $Q$ on the seven nonzero vectors has size a power of two. Since seven is odd, at least one orbit of a group action has size one. Thus $Q$ fixes a nonzero vector $v$. Its group action on the three planes containing $v$ likewise has a fixed plane $P$. Therefore $Q$ is contained in, and by order equals, the stabilizer of that flag. The group action of $G$ is transitive on these flags, so all such subgroups are conjugate subgroups.

Distinct flags give distinct stabilizers. In a basis adapted to a flag, its order-eight stabilizer consists ofIts only common fixed nonzero vector is $e_1$. Its only invariant plane is $⟨e_1,e_2⟩$: an invariant plane is the kernel of a linear map of a nonzero linear functional, and over ${\mathbb{F}}_2$ that linear functional must be fixed; the displayed matrices have only the fixed nonzero linear functional $e_3^{\ast }$. Hence each subgroup specifies its flag uniquely. There are seven choices of $v$ and three choices of $P$, giving

An element $\alpha$ is an algebraic element over $F$ if some nonzero polynomial in $F[X]$ vanishes at $\alpha$. A field extension is an algebraic extension if each of its elements is an algebraic element over $F$. If $[K:F]=d<\infty$, the $d+1$ vectors $1,\alpha ,\dots ,{\alpha }^d$ are linearly dependent, producing such a polynomial. Thus every finite field extension is an algebraic extension.

The converse fails for the field of algebraic numbers $\overline{\mathbb{Q}}$. It is an algebraic extension of $\mathbb{Q}$ by definition, but it contains primitive $p$th roots of unity of degree $p-1$ for arbitrarily large primes $p$, by the permitted irreducibility of the cyclotomic polynomial. The tower law would make every such degree at most $[\overline{\mathbb{Q}}:\mathbb{Q}]$ if the latter were finite, a contradiction.

Write $q=(x,X{)}^{\mathsf{T}}$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The Euler-Lagrange equations give $M\ddot{q}+Kq=0$. A normal mode is a motion $q=v\cos (\omega t+\phi )$ with fixed nonzero vector $v$; its normal frequency is $\omega >0$. Substitution gives $(K-{\omega }^{2}M)v=0$. Thus the squared normal frequencies are the assumed positive distinct roots ofFor either root, solve the associated two linear equations for the mode ratio.

For the rods, the small angles are $\theta =x/a$ and $\varphi =(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second rod's center has horizontal velocity $(\dot{x}+\dot{X})/2$ and its angular velocity is $(\dot{X}-\dot{x})/(2a)$. Its center-of-mass moment of inertia is $2m{a}^2/3$, givingAdding these kinetic energies gives $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The first center rises by $a{\theta }^2/4$, and the second by $a{\theta }^2/2+a{\varphi }^2/2$. Multiplying by the respective masses and $g$ givesTherefore the stated Lagrangian is $T-V$.

Set $\mu =2a{\omega }^2/(3g)$. The mode equation isHenceThe mode ratio is $r=X/x=(6-3\mu )/(1+\mu )$, so the ratio of the rods' angles is $\varphi /\theta =(r-1)/2=(5-4\mu )/[2(1+\mu )]$. It is positive for ${\mu }_{-}=(17-3\sqrt{21})/10$ and negative for ${\mu }_{+}=(17+3\sqrt{21})/10$. Thus the lower-frequency normal mode has the rods in phase, and the higher-frequency normal mode has them out of phase.

Let the third profit coefficient be $c_3$. For the same basis, the resource dual prices solve $2y_1+y_2=3$ and $5y_1+3y_2=c_3$, givingDual feasibility requires $y_1,y_2\ge 0$ and $3y_1+2y_2\ge 4$. These inequalities are $c_3\le 9$, $c_3\ge 15/2$ and $c_3\ge 7$, respectively. HenceThe production vector is unchanged because the resource constraints are unchanged. Equality at a boundary can allow additional optimal solutions.

In the usual linear programming sensitivity within a fixed optimal basis, retain the optimal basis $\{x_1,x_3\}$ while allowing its values to adjust. If the second resource amount is $b$, thenThe dual prices remain $(1,1)$, so optimality persists exactly while both basic variables are nonnegative:The optimal profit on that range is $33+b$. At the endpoints the basis is degenerate but remains optimal. If the wording is instead read as requiring the exact numerical production vector $(9,0,3)$ to remain unchanged and optimal, only $b=18$ does so; the standard allowable range concerns the basis and its adjusted solution.

For a scalar-valued function, the gradient is the column vectorWhen $f$ is differentiable, its total derivative is the linear functionalThus the Jacobian is the transpose of the gradient in standard Euclidean coordinates.

The $j$-th partial derivative is the vector limitwhen it exists. The total derivative is a linear map $Df(x):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe Jacobian matrix is $J_f(x)=({\partial }_j{f}_i(x){)}_{i,j}$. When the total derivative exists, all partial derivatives exist and $J_f(x)$ is its matrix in the standard bases; its $j$-th column is $Df(x)e_j$. Existence of partial derivatives alone does not guarantee total differentiability.

Write $A=(X^{T}X{)}^{-1}{X}^T$. Since $AX=I$, the estimator is $\hat{\theta }=\theta +Aϵ$. ThereforeIts expectation proves unbiasedness. It is also jointly normal, being an affine transformation of the normal error vector.

The dual prices $u,v\ge 0$ solveThe optimal tableau gives $(u,v)=c_B^T{B}^{-1}=(6,3)B^{-1}=(3/4,3/2)$. This vector is dual feasible and has objective $21$, equal to the primal value, proving optimality.

Thus an extra unit of R is worth $3/4$ while the current basis stays feasible. Increasing its supply by $\delta$ changes the basic values toFor increases, feasibility is retained exactly up to $\delta =4/3$. Therefore the firm can buy up to $4/3$ extra units at a marginal value of $3/4$ per unit. The reduced profits do not change during this right-hand-side variation.

For the proposed fourth good, the reduced profit is $8-(4u+4v)=8-9=-1$. Its profit is below the resource opportunity cost, so it is not worth producing at the current optimum. The old dual solution also satisfies the new good's inequality, certifying that introducing it cannot improve the optimum.