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A linear molecule has collinear atomic nuclei in its idealized molecular geometry. Both a diatomic molecule and carbon dioxide can supply linear rigid rotor examples, although their electric-dipole selection rules need not give an observable permanent-electric dipole spectrum.

Named after Fritz London, this is an attractive interaction between fluctuating electric dipoles of atoms or molecules, present even without a permanent dipole. At large separation compared with atomic size in the nonretarded regime, the dipole interaction scales as $r^{-3}$ and its second-order ground-state energy correction is negative and scales as $r^{-6}$. Differentiating the energy gives an attractive force of magnitude $6C_6/r^7$. The short-distance overlap regime and the very-long-distance retarded regime require different descriptions.

The following original model isolates amplification by inverted electric dipoles. Let $N=N_u-N_l$ be the excess number in the upper member of one equally degenerate resonant pair. Let $a$ be a slowly varying cavity amplitude with photon number $n=|a{|}^2$, and let $p$ describe the collective molecular dipole component in the amplifying phase convention. On exact resonance, stipulate the linear equationsThe dipole relaxation rate is $\Gamma$, and $g_0=|d_{ul}{\mathcal{E}}_{\mathrm{vac}}|/\hslash$ is a single-molecule coupling in the chosen mode normalization. Here $d_{ul}$ is an electric dipole moment matrix element, ${\mathcal{E}}_{\mathrm{vac}}$ is the field's zero-point amplitude, and $\hslash =h/(2\pi )$. Field overlap and dipole orientation enter this coupling. The characteristic polynomial for growth rate $\lambda$ isFor $N>0$, one root becomes positive precisely when $g_0^{2}N>\kappa \Gamma /2$. If dipole relaxation is fast enough to eliminate $p$, then $p\approx g_{0}Na/\Gamma$ andThus positive population inversion alone is insufficient: the gain rate $GN$ must exceed the passive decay rate. This reduction describes the onset near resonance; it does not predict output phase noise, actual transit dynamics or an exact historical threshold. Nikolay Basov and Aleksandr Prokhorov's molecular-generator theory likewise treats self-excitation as competition between molecular gain and resonator-plus-load losses.^[ref]

To include replenishment, suppose upper-state molecules enter at rate $F$ and both levels leave with residence rate $\gamma$. A deliberately simple ordinary differential equation model isOne downward transition removes two units of the difference $N_u-N_l$, so $\dot{N}=F-\gamma N-2GNn$. Below onset the unsaturated inversion is $N_0=F/\gamma$, givingFor fixed mode volume $V$, the equivalent number density condition is $N_0/V>\kappa /(GV)$. Raising selected beam intensity raises gain; increasing quality factor lowers the loss rate at fixed frequency. These statements hold with $G$, residence time and mode overlap fixed.

Above onset, the nonzero steady state hasThe inversion is clamped while excess supply supports output. A finite initial seed is understood: the noiseless equations leave $n=0$ exactly unchanged even above onset. Actual startup has spontaneous emission and field fluctuations.

Two routes to electric polarization are displacement and reorientation. An electric field can shift the centres of positive and negative electric charge in a neutral chemical atom or molecule, inducing an electric dipole moment. A polar molecule already has an electric dipole moment; a field biases its orientation. At finite temperature, random molecular motion opposes complete alignment, so a nonzero average need not mean that every electric dipole points the same way.^[ref]

For a single species with number per unit volume $n$, the electric polarization isThe brackets denote the local average electric dipole moment, rather than its magnitude. Opposite orientations can cancel while individual electric dipole moments remain nonzero.^[ref] This averaging explains how an initially unpolarized material can respond strongly without acquiring net electric charge.

An outgoing radiation pulse carries a finite amount of electromagnetic energy through successively larger enclosing spheres. In vacuum, far from a localized source and without absorption or additional reflections, let $P_{\mathrm{emit}}(t)$ be the emitted power. The retarded time at radius $r$ is $s=t-r/c$. Define ${\overline{S}}_r$ as the angular average of the radial Poynting vector over the sphere, not a cycle average:Here $\overline{u}$ is the angularly averaged energy density in the leading radiation zone approximation. Multiplying by shell area removes geometric dilution: the energy per unit radial thickness translates outward without losing its shape. These angular averages do not assume that an actual electric dipole emits equally in all directions. Nor do they specify an exact spherically symmetric transverse solution of the Maxwell equations.

For an explicitly stipulated model, emitAfter emission ends, $t>\tau$, the pulse lies between $r=c(t-\tau )$ and $r=ct$, provided this shell is in the radiation zone. Its radial thickness remains $c\tau$. Substituting $s=t-r/c$ in the shell integral $\int 4\pi r^2\overline{u}dr$ gives $U=P_0\tau /2$ at every such time. The same substitution in the time integral of $4\pi r^2{\overline{S}}_r$ at any enclosing sphere gives that same total crossing energy. This is conservation of energy in two different views.

Along any fixed direction with nonzero outgoing power, compare the same part of the pulse at two radii, hence at different arrival times. The leading electric field amplitude falls as $1/r$, while its squared wave intensity falls as $1/r^2$. At $2r$ the field is one half and the energy flux one quarter; the signal arrives $r/c$ later. A small fixed area receiver therefore intercepts less energy at greater distance even though every enclosing sphere receives the same total pulse energy. The inverse-square law expresses spreading over area, rather than destruction of energy. Close to the source, the neglected near and far field terms prevent extrapolating these formulas down to $r=0$.

In the idealized isolated-carbon chemical atom interpretation, both beams carry approximately $2.68\mathrm{eV}$ per incident particle, close to the ${}^1{S}_0$ threshold. However equal incident energies do not imply equal excitation channels.

A single optical photon cannot strongly excite any of the listed even-parity $2p^2$ energy levels from the even-parity ${}^3{P}_0$ ground state through an electric dipole transition. In particular ${}^3{P}_0\to {}^1{S}_0$ also violates $\Delta S=0$ and the $J=0\leftrightarrow 0$ restriction. The lowest strong opposite-parity excitations lie well above $2.68\mathrm{eV}$. Thus one expects no strong resonant absorption or corresponding strong atomic fluorescence from these ground-state chemical atoms at this optical wavelength; weak higher-order processes and scattering are separate effects.

Incident electrons can undergo inelastic scattering and collisional excitation. Their orbital angular momentum and exchange with bound electrons allow transfers excluded for a single electric dipole photon. The ${}^1{D}_2$ energy level at $1.264\mathrm{eV}$ and the low fine structure excitations are energetically accessible. One therefore expects scattered electrons with discrete energy losses, possible weak forbidden spectral lines, and collisional or surface relaxation. For example ${}^1{D}_2\to {}^3{P}_2$ has wavelength approximately $10^7/(10192-43.4)=985\mathrm{nm}$; spin mixing supplies the weak spin-forbidden radiative channel.

There is a numerical threshold caveat. Literally,Neither quoted beam reaches ${}^1{S}_0$ from the exact ground state if these are taken as exact monoenergetic values. The electron value rounded to two decimal places could represent an actual energy above threshold; if so, ${}^1{S}_0$ is collisionally accessible and its decay to ${}^1{D}_2$ produces the weak electric quadrupole transition at $10^7/11456\simeq 873\mathrm{nm}$, followed by lower-level relaxation. These emissions require populations and sufficiently low collisional quenching of a metastable line.

An actual solid carbon foil has bands, collective excitations and surface losses, so it need not reproduce the isolated-chemical atom spectrum or selection rules. Its optical and electron-induced emission cannot be predicted quantitatively from the five atomic energy levels alone. This is the limitation of the foil wording in the source.

Use the sign $s=\pm 1$ to label a point charge $sq$ at $s\mathbf{p}/(2q)$, with velocity $s{\mathbf{p}}^{\prime }/(2q)$. All quantities on the right below are evaluated at $t_0=t-r/c$. To first order,The denominator in each Liénard–Wiechert potential therefore expands assoIn the sum of the two electric potentials, the $q/r$ terms cancel and the finite terms add. This gives the retarded potentials of a point electric dipole:For the magnetic vector potential, the signed electric charge and signed velocity multiply to give the same sign for both contributions:Since ${\mu }_0{ϵ}_0{c}^2=1$,The ${\mathbf{p}}^{\prime }$ term in the electric potential arises from the moving-source denominator and remains finite in the point-electric dipole limit; discarding it would fail to give the full retarded electric field.

Take the ideal electric dipole limit with a fixed smooth electric dipole moment $\mathbf{p}(t)$, so that ${\mathbf{r}}_0(t)=\mathbf{p}(t)/(2q)$; its required time derivatives are bounded on the time interval under consideration. Fix $r>0$ and put $\mathbf{n}=\mathbf{r}/r$, $t_0=t-r/c$. For $s=+1$ or $-1$, the source position is $s\mathbf{p}(t)/(2q)$. A Taylor expansion of the distance givesSubstitute this into the defining equation for retarded time:Boundedness first implies $t_r^s-t_0=O(q^{-1})$. The Taylor expansion $\mathbf{p}(t_r^s)=\mathbf{p}(t_0)+O(q^{-1})$ then changes the displayed correction only by $O(q^{-2})$. HenceThe nearer positive displacement reduces the propagation distance and consequently increases the retarded time. The Big O notation here is for $q\to \infty$ at a fixed observation event away from the origin. Keeping a fixed smooth $\mathbf{p}$ is essential: bounded amplitude alone, with arbitrarily increasing time frequency as $q$ varies, would not justify this Taylor expansion.

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

Take the wire at $(a,0)$ to carry $+\lambda$ and that at $(-a,0)$ to carry $-\lambda$. Ifthen, up to one additive constant,Thus an equipotential with $k=2\pi {\epsilon }_0\varphi /\lambda$ satisfies $r_{-}/r_{+}=e^k$. For $k\ne 0$, completing the square gives the Apollonius circleIts centre is $(a\coth k,0)$ and its radius is $a/|\sinh k|$. Positive and negative values give nested circles around the positive and negative wires. The electric field is orthogonal to these circles and points from the positive wire toward the negative wire. For $\varphi =0$, the limiting equipotential is the straight line $x=0$.

Direct superposition givesIn the limit $a\to 0$ with $\lambda a=p$, the potential and field become those of a two-dimensional electric dipole:and

Sunlight is approximately unpolarized before its interaction with the environment. For Rayleigh scattering by small isotropic scatterers, the induced electric dipoles radiate only their components transverse to the viewing direction. For unpolarized incident light scattered through angle $\chi$, the perpendicular and parallel intensities are proportional to 1 and ${\cos }^2\chi$. The degree of linear polarization is thereforeAt $90^{\circ }$ single scattering is fully linearly polarized perpendicular to the scattering plane. This explains the polarized sky away from the Sun; multiple scattering and aerosols reduce the ideal contrast.

Reflection at a dielectric surface treats the two linear polarizations differently. At Brewster incidence, $\tan {\theta }_B=n_2/n_1$, the reflected p component vanishes for ideal transparent nonmagnetic media. The remaining reflected glare is s polarized, perpendicular to the incidence plane, approximately horizontal for a horizontal water surface. Away from this angle the reflection is generally partially polarized. The transmitted beam also has unequal s/p transmission.

Refraction and internal reflection in water droplets likewise have polarization-dependent Fresnel coefficients, producing strong linear polarization in rainbow light. These are generic scattering and boundary-condition mechanisms; no biological detection mechanism is required.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

Work in the dilute, single-scattering limit and add the diffuse scattered light intensities of randomly positioned particles. The incident unpolarized light has equal, mutually uncorrelated $x$ and $y$ electric components. Each particle therefore has an induced electric dipole moment along its permitted axis, with the same mean squared magnitude proportional to $|\alpha {|}^2$.

For an observer in direction $\hat{\mathbf{n}}=(\sin \psi ,0,\cos \psi )$, choose transverse unit directions ${\mathbf{e}}_{\parallel }=(\cos \psi ,0,-\sin \psi )$ in the scattering plane and ${\mathbf{e}}_{\perp }=\hat{\mathbf{y}}$ perpendicular to it. The electric dipole radiation field is proportional to the transverse projection of the electric dipole. Hence an $x$ electric dipole contributes a wave amplitude factor $\cos \psi$ along ${\mathbf{e}}_{\parallel }$, while a $y$ electric dipole contributes its full wave amplitude along ${\mathbf{e}}_{\perp }$.

Let $C>0$ collect the common incident light intensity, electric polarizability, distance and frequency factors. ThenThe two components are uncorrelated, so the coherency matrix is diagonal in this basis. The degree of polarization isA signed degree of linear polarization, defined positive perpendicular to the scattering plane, has the same expression without the absolute value. The dominant direction is $\hat{\mathbf{y}}$ when $N_2>N_1{\cos }^2\psi$, and ${\mathbf{e}}_{\parallel }$ when the inequality reverses. If the denominator vanishes there is no scattered signal and its degree of polarization is undefined. A coherent exactly forward component, including the transmitted incident beam, is not part of this diffuse-intensity calculation; random particle positions do not by themselves destroy forward optical coherence.

Now take $\hat{\mathbf{n}}=(0,\sin \psi ,\cos \psi )$, with the same angle to the incident $z$ direction. The perpendicular direction is $\hat{\mathbf{x}}$ and the in-plane transverse direction is $(0,\cos \psi ,-\sin \psi )$. The roles of the two electric dipole populations interchange: $x$ electric dipoles radiate at full strength, whereas the observed wave amplitude from a $y$ electric dipole is reduced by $\cos \psi$.

Thus $I_{\perp }=C{N}_1$ and $I_{\parallel }=C{N}_2{\cos }^2\psi$, givingThis is also obtained by the symmetry $x\leftrightarrow y$, $N_1\leftrightarrow N_2$ in part (a). For $N_1=N_2$, both planes give the familiar Rayleigh scattering degree of polarization ${\sin }^2\psi /(1+{\cos }^2\psi )$: the diffuse signal is unpolarized forward and fully linearly polarized at right angles. At a dark direction with zero denominator, no degree of polarization can be measured.

In uniform fields, a rigid neutral electric dipole with both charges moving at the selected common velocity has zero net translational force. To leading nonrelativistic order its torque is $\mathbf{\tau }=\mathbf{p}\times (\mathbf{E}+\mathbf{v}\times \mathbf{B})$, which also vanishes at the velocity-selector condition. Thus it continues undeflected and, initially nonrotating, does not start rotating in the ideal balanced fields. An unbalanced effective field could exert a torque; the dipole orientation is not otherwise specified.

At $(x,0)$ the distances from the two centres are equal:By the electrostatic shell theorem, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, soThe requested magnitude isIt jumps from zero to $k_{e}q/a^2$ at $x_1$, decreases to $k_{e}q/(4a^2)$ just before $x_2$, jumps down to $k_{e}q/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an electric dipole with electric dipole moment $\mathbf{p}=-qa\hat{\mathbf{y}}$.

At $(x,0)$ the distances from the two centres are equal:By the electrostatic shell theorem, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, soThe requested magnitude isIt jumps from zero to $k_{e}q/a^2$ at $x_1$, decreases to $k_{e}q/(4a^2)$ just before $x_2$, jumps down to $k_{e}q/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an electric dipole with electric dipole moment $\mathbf{p}=-qa\hat{\mathbf{y}}$.

The electric dipole moment is $\mathbf{p}=2aq\hat{\mathbf{z}}$. By the superposition principle, the exact electric potential isFor $a/r\ll 1$, the binomial expansion gives each reciprocal distance as $r^{-1}(1\pm (a/r)\cos \theta +O(a^2/r^2))$. The leading $1/r$ terms cancel because the total electric charge is zero. The next even terms also cancel between the two signs. ConsequentlyThis is the leading electric dipole term of the multipole expansion; it describes the far field, not the electric potential arbitrarily close to either point charge.

Apply the negative gradient to the leading far-field electric potential. In spherical coordinates,The electric field therefore has the coordinate-independent formFor the electric dipole along $z$, use $V=\alpha z/(x^2+y^2+z^2{)}^{3/2}$. Direct partial differentiation gives the vector componentsOn the positive axis the electric field points along the electric dipole moment; in the equatorial plane it points oppositely, providing a useful sign check.

First use the usual idealization of a grounded infinite conducting plane at $z=0$, with the real point charge $+q$ at $(0,0,d)$. The method of images replaces the metal, for purposes of calculating the exterior electric field, by an image charge $-q$ at $(0,0,-d)$. The exact electric potential in $z>0$ isIt vanishes on the plane, and the uniqueness theorem for the electrostatic potential identifies it with the physical exterior solution. The tangential electric field cancels on the plane, while the normal components add:The electric field inside the conductor is zero, so the electrostatic boundary condition gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Integrating the surface charge density over the facing plane yieldsThe exact image expression is essential near the charge: using the far-field electric dipole approximation at $\rho$ comparable with $d$ would be invalid.

Electrical boundary condition. The source does not specify grounding or the initial total electric charge of the slab. The boxed total is the grounded infinite-plane answer and the large-plane approximation to the charge on its nearby face. An initially neutral, finite, electrically isolated slab retains zero total electric charge; an approximately $-q$ accumulation near the real charge is then balanced by approximately $+q$ on its remote face and edges. This gives a concrete counterexample to interpreting $-q$ as the unconditional total charge of every large metal slab. The local image formula assumes the infinite-plane limit with no applied background electric field.