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A small conducting sphere of radius $a$ has electric polarizability $\alpha =4\pi {ϵ}_0{a}^3=a^3/k$. In the field $E\simeq k{Q}_1/R^2$ of a much more strongly charged distant source, its leading induced contribution is $-\alpha E^2/2$, in addition to $k{Q}_1{Q}_2/R$ for its fixed net charge. Taking $-dU/dR$ gives $F_R\simeq k{Q}_1{Q}_2/R^2-2k{Q}_1^2{a}^3/R^5$. The approximation requires $a/R\ll 1$ and control of the source's finite-size corrections; the threshold is not an exact finite-sphere result.

A neutral isotropic atom with static electric polarizability $\alpha >0$ in the field of a distant point charge $Q$ has induced dipole $\mathbf{p}=\alpha \mathbf{E}$ and energy $U=-\alpha E^2/2$. Since $E=|Q|/(4\pi {ϵ}_0{r}^2)$,The negative radial force is attractive for either sign of $Q$. Reversing $Q$ reverses the induced dipole but not the energy or leading force. The factor $1/2$ accounts for the work of inducing the dipole; using the permanent-dipole energy $-\mathbf{p}\cdot \mathbf{E}$ without that work would double the answer. The approximation assumes weak fields, a slowly varying field across the atom and negligible electronic overlap.

Electric polarizability relates an induced electric dipole moment to an applied electric field. In a weak-field model with linearity, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$. A tensor response allows different electric field directions to induce electric dipole moments in different directions. A response that changes with molecular rotation enables rotational Raman spectroscopy.

Electronegativity describes a chemical atom's tendency to attract electron density towards itself in a chemical bond. It helps answer a useful question: when two different chemical atoms share electrons, which end of the chemical bond tends to be richer in electron density? For hydrogen chloride, the answer is chlorine: the covalent bond is a polar covalent bond, conventionally written ${\mathrm{H}}^{\delta +}-{\mathrm{Cl}}^{\delta -}$. The symbols indicate partial atomic charges, not isolated ions carrying whole elementary electric charges.^[ref]

Electronegativity is a comparison scheme, not the number of electrons on a chemical atom. A partial atomic charge depends on how the electron density is partitioned; formal charge instead follows an integer allocation in a Lewis structure. Neither is numerically equal to electronegativity. Nor is electronegativity identical to electron affinity, which concerns adding an electron to an isolated species. Different electronegativity scales use different measured inputs and conventions; compare numbers only after specifying the scale.^[ref]

For a concrete comparison, the Royal Society of Chemistry lists Pauling electronegativities ${\chi }_{\mathrm{P}}(\mathrm{H})=2.20$ and ${\chi }_{\mathrm{P}}(\mathrm{Cl})=3.16$. Their difference is $0.96$, so the electronegativity rule predicts bond polarization towards chlorine in hydrogen chloride. An isolated molecular hydrogen molecule has identical ends and zero electronegativity difference. These examples explain a direction of unequal sharing, not a quantitative partial atomic charge.^[ref][ref]

A larger electronegativity difference often suggests stronger bond polarization, but there is no universal numerical boundary where a covalent bond suddenly becomes ionic bonding. Tables of cutoffs are classification aids with exceptions. Hydrogen fluoride, for example, is described by a polar covalent bond even when its electronegativity difference exceeds a textbook's nominal ionic threshold. The actual electron density, physical setting, and chemical substance's structure matter.^[ref]

The distinction from ionization energy is equally useful: an ionization energy measures the cost of removing an electron from a specified isolated species, whereas electronegativity compares attraction for electron density in a bonded environment. Neither the removal cost alone nor the attachment release alone supplies a unique bonded-atom electronegativity. A scale combines selected measured inputs with a model and a convention.^[ref]

For an original relative comparison, stipulate fictitious bonded sites $A,B,C$ on one chosen dimensionless electronegativity scale:The electronegativity heuristic predicts bond polarization towards $B$ in $A-B$, towards $C$ in $B-C$, and towards $C$ in $A-C$. Shifting ${\chi }_0$ leaves every comparison unchanged. Although the $A-C$ difference is twice the $A-B$ difference, this does not make its partial atomic charge or bond dipole twice as large: those require additional information about electron density and bond length. These symbols describe a stipulated ordering, not a table of measured chemical elements.

Keep the chemical context attached to that ordering. A chemical atom in a molecule has neighbours and an electron configuration; an isolated-atom energy scale and a bonded-atom comparison are different constructions. The chemical environment dependence of electronegativity explains why a table is a useful approximation rather than an immutable property of every site. Furthermore, an applied electric field can induce bond polarization even between identical chemical atoms, through electric polarizability; equal tabulated electronegativity does not forbid that response.

For a nondegenerate parity-symmetric ground state in an external electric field, write the perturbation as $V=-E\hat{d}$. The first-order state correction givesThe positive coefficient $\alpha$ is the electric polarizability along the applied direction. Meanwhile second-order nondegenerate perturbation theory gives $\delta E_0^{(2)}=-\alpha E^2/2$. Thus the induced electric dipole moment satisfies $d=-2\delta E_0^{(2)}/E$, not $-\delta E_0^{(2)}/E$. The Hellmann–Feynman theorem independently gives $d=-\partial E_0/\partial E$.

The Stark effect changes atomic energy levels in a static electric field. For $\mathbf{E}=\mathcal{E}\hat{\mathbf{z}}$, the electric-dipole interaction is $H^{\prime }=-d_z\mathcal{E}=e\mathcal{E}z$ for an electron, with $e>0$. A nondegenerate state of definite parity has $⟨z⟩=0$, so its leading quadratic Stark effect isThe ground state shift is negative, conventionally $-\alpha {\mathcal{E}}^2/2$ with positive electric polarizability.^[ref]

Degenerate opposite-parity states require degenerate perturbation theory. In ideal Coulomb hydrogen, the $2s,2p_0$ subspace has an off-diagonal matrix element of magnitude $3e{a}_{\mathrm{B}}\mathcal{E}$, where $a_{\mathrm{B}}$ is the Bohr radius. Its mixed states shift byThis Linear Stark effect in the first excited hydrogen eigenspace assumes fine-structure and Lamb splittings are negligible compared with the mixing. In an isolated parity eigenstate with such degeneracies resolved, sufficiently weak fields restore quadratic behaviour. Strong fields require further mixing and can ionize the atom.^[ref]

The electron's electric dipole moment is $\mathbf{d}=-e⟨x⟩\hat{\mathbf{x}}$. It vanishes in the unperturbed ground energy eigenstate. The first correction to the wavefunction is linear in the electric field and has the opposite parity; its interference with the unperturbed state therefore gives a nonzero position expectation value of first order.

In detail, the normalized ground quantum state, to the order needed, isThe normalization correction starts at second order and does not affect the linear term below. The two cross terms giveThe absence of an $E^2$ term follows from the parity operator: $⟨x⟩$ changes sign under field reversal. Comparing with the second-order nondegenerate perturbation theory energy sum yieldsThus the induced electric dipole moment points along the electric field, with positive electric polarizability. The induced dipole moment and quadratic energy shift obey $\delta {ϵ}_0^{(2)}=-{\alpha }_{\mathrm{pol}}{E}^2/2$. This factor of one half accounts for the energy needed to induce the dipole. As a check, the Hellmann–Feynman theorem gives $d_x=-\partial {ϵ}_0/\partial E$, with the same result.

At a large separation, the two like net charges give the familiar leading repulsion $k{Q}_1{Q}_2/R^2$, where $k=(4\pi {ϵ}_0{)}^{-1}$ and $R$ is centre separation. Each conductor also changes the other's surface charge distribution through electrostatic induction. The facing and remote surface charge densities need not remain equal, and an external insulating suspension prevents net charge transfer without preventing redistribution within either sphere.

Like net charges do not guarantee repulsion for arbitrary conducting spheres at every separation. Sufficiently unequal charges can give an attractive resultant after polarization. This does not contradict Coulomb's law: the law still applies to the distributed charges, while the point-charge approximation has lost the relevant induced contribution.

A useful controlled example takes a small conducting sphere of radius $a$ and fixed positive charge $Q_2$ in the nearly uniform field of a much more strongly charged distant sphere. For $a\ll R$, the small sphere has electric polarizability $\alpha =4\pi {ϵ}_0{a}^3$ and induced dipole moment $p=\alpha E$. Its leading interaction energy isDifferentiation gives the outward radial forceThe second term is attractive. With $\eta =Q_2/Q_1>0$, this leading model gives $F_R<0$ when $\eta <2(a/R{)}^3$. One can choose a sufficiently small positive $\eta$ while retaining $a/R\ll 1$, so the example does not require a neutral or earthed sphere. It is an asymptotic model, not an exact finite-sphere threshold.

An exact analysis confirms that close-range attraction is possible for like-charged conducting spheres and identifies a special contact-charge ratio for which the force remains repulsive.^[ref] This later analysis supplies a precise qualification to the elementary expectation; the source question gives neither a charge ratio nor a separation regime that would justify asserting universal repulsion.

For $\alpha >0$, the given model predicts attraction proportional to $r^{-5}$ in force, independent of the ion's charge sign; the energy is proportional to $-r^{-4}$ and vanishes at large separation. A larger electric polarizability gives a stronger interaction.

This is induction by a charged object. Ordinary nonretarded London interaction between two small neutral polarizable objects instead has an $R^{-6}$ energy dependence; geometry-integrated colloidal dispersion laws can have other powers. Calling the specified $r^{-4}$ law a generic dispersion force would conceal its charge-dependent origin.

Attraction can favor association or adsorption. A stable suspension normally requires an opposing free-energy barrier, such as double-layer or steric repulsion, relative to Brownian motion. Salt, surface coatings and solvent quality alter that balance. However, one ion–particle pair potential alone does not establish the phase behavior or aggregation rate of a many-particle suspension; concentrations, diffusion, surface chemistry and the remaining interactions also matter.

Write $C_4=\alpha Q^2/[2(4\pi {ϵ}_0{ϵ}_r{)}^2]$. Differentiating the stipulated potential givesFor positive $\alpha$, the force points toward smaller separation for either sign of $Q$.

Its physical origin is charge-induced dipole attraction, rather than London dispersion forces. A point-charge field induces $\mathbf{p}=\alpha \mathbf{E}$ and the energy cost of inducing that moment gives $u=-\alpha E^2/2$. The charge changes the electric dipole moment; it does not, within this linear model, change the intrinsic coefficient $\alpha$. For a conducting sphere in a uniform medium, $\alpha =4\pi {ϵ}_0{ϵ}_r{a}^3$; other dielectric contrasts give another electric polarizability.

There is a distance inconsistency in the stated model: the point-charge field uses center-to-center separation $R$, whereas the question calls $r$ a surface gap. For a point ion and sphere, $R=a+r$. A controlled far-field dipole model therefore uses $u\simeq -C_4/(a+r{)}^4$, with $R\gg a$. The boxed derivative is the literal result for the given mathematical $u(r)$; it is not a justification for extending that dipole formula to contact.

Spherical-particle frequency scaling. For a dielectric sphere of radius $a\ll \lambda$, the quasistatic electric polarizability isThe electric dipole radiation formula and $I_0={ϵ}_{0}c|E_0{|}^2/2$ giveThus Rayleigh scattering scales as ${\omega }^4$, or ${\lambda }^{-4}$, when the polarizability varies slowly with frequency. Material resonances and absorption can change this simple scaling. For the incident $z$ polarization, the dipole is along $z$: intensity is proportional to ${\sin }^2\chi$, where $\chi$ is the angle from $z$. There is no radiation on the dipole axis and a maximum in the $xy$ plane. In a direction $\hat{\mathbf{n}}$ the scattered electric field is along $\hat{\mathbf{z}}-(\hat{\mathbf{z}}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}$, hence linearly polarized wherever nonzero.^[ref]

Needle model used in parts (a)–(d). Let its axis be $\mathbf{u}=(u_x,u_y,u_z)=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. A very slender, high-permittivity needle has dominant longitudinal response, so use the rank-one needle polarizability $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, $s=|\mathbf{v}{|}^2=1-u_y^2$ andFor a unit incident Jones vector $\mathbf{e}$, the common far-field relation is ${\mathbf{E}}_s\propto \mathbf{v}(\mathbf{u}\cdot \mathbf{e})$, and $I_s=K{I}_{0}s|\mathbf{u}\cdot \mathbf{e}{|}^2$. Here $I_s$ is irradiance at distance $R$; multiplying by $R^2$ gives power per solid angle. No absolute value of ${\alpha }_{\parallel }$ is given or needed for the requested relative intensities.

Diffuse random cloud, with the incident polarization of (a). The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_x{u}_z$ and $u_z^2$. The isotropic orientation fourth moment gives the needle-scattering analyzer momentsFor example, $⟨u_z^4⟩=\frac{1}{2}{\int }_{-1}^1{\mu }^{4}d\mu =1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2⟨u_x^2{u}_z^2⟩=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,The cross correlation vanishes by reflection symmetry. The degree of linear polarization is $(I_z-I_x)/(I_z+I_x)=\boxed{1/2}$: random anisotropic scatterers produce partially depolarized light, even though every individual needle produces a single linear polarization.

Continuous homogeneous dielectric. One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $⟨\mathbf{u}{\mathbf{u}}^{\mathsf{T}}⟩=\mathbf{I}/3$, so the averaged response is isotropic and its induced polarization is along $z$. Its coherent radiation toward $y$, when allowed by the sample geometry, is therefore purely $z$ polarized, with $I_x=0$. The independent-needle prefactor is no longer a quantitative dense-medium model; local-field and collective effects determine the effective permittivity.

There is also coherent continuum phase matching: a uniform bulk driven by a plane wave has no arbitrary outgoing $90^{\circ }$ plane-wave component. In an ideal infinite homogeneous medium there is no diffuse side scattering at all; propagation, refraction and reflection are determined by the medium and its interfaces. A finite sample's coherent side signal depends on its shape and boundaries. Thus the old $1:3$ diffuse ratio is replaced by a coherent isotropic-medium response, not by a universal nonzero $N^2$ side-scattering intensity.

Put ${\omega }_0^2=k/m$, where $k$ is the binding spring constant rather than an optical wavenumber. The steady complex amplitude solution of the forced Lorentz oscillator model isThe acceleration complex amplitude is $-{\omega }^2{\mathbf{x}}_0$. HenceThis is the scattering cross-section of a harmonically bound charge. The displacement and induced electric dipole moment are parallel to the incident linear polarization, and integrating their dipole radiation over all directions gives the total cross section above.

For $\omega \ll {\omega }_0$, binding dominates andrecovering Rayleigh scattering with static electric polarizability $q^2/k$. For $\omega \gg {\omega }_0$, inertia dominates: ${\mathbf{x}}_0\simeq -q{\mathbf{E}}_0/(m{\omega }^2)$, so its second time derivative becomes frequency independent andThis is a classical Thomson scattering regime only while $\hslash \omega \ll m{c}^2$. Exactly at $\omega ={\omega }_0$, an undamped driven oscillator has no finite steady-state amplitude; the divergent formula must not be used. Finite damping and radiation reaction regulate the physical resonant response. Coherent contributions from many charges in an extended medium can also require a structure calculation rather than multiplying this single-charge result by a particle count.

In Rayleigh scattering, a small bound system develops an induced electric dipole moment $\mathbf{p}(t)=\mathrm{Re}[\alpha (\omega ){\mathbf{E}}_0{e}^{-i\omega t}]$. Well below its binding resonances, the electric polarizability is approximately its static value $\alpha (0)$, independent of frequency. The mean radiated power and incident light intensity areDividing gives the total scattering cross sectionThis Rayleigh scattering wavelength dependence comes from the two time derivatives of an oscillating electric dipole moment, whose radiated field is proportional to ${\omega }^2{p}_0$. It requires the same scatterer with approximately constant low-frequency electric polarizability; resonances and strong optical dispersion invalidate the simple fourth-power law.

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.

The given static induced surface charge and the dipole integral from part (b) giveHere $ϵ$ is the dimensionless relative permittivity, as required by the stated $ϵ\pm 1,2$ expression; ${\alpha }_E$ is the SI electric polarizability. Take ${\mathbf{E}}_{\mathrm{inc}}(t)=E_0\cos \omega t\hat{\mathbf{e}}$ and $k=\omega /c=2\pi /\lambda$. In the Rayleigh scattering regime the field is approximately uniform across the sphere. Its induced dipole radiates mean powerThe incident time-averaged light intensity in vacuum is $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$. Dividing power by intensity gives the total cross section,The units are ${\mathrm{m}}^2$; the comparison is with the geometric projected area $\pi R^2$, not with a length:The inequality holds for an ordinary nonabsorbing dielectric, with finite contrast factor and $kR\ll 1$. It explains why a sufficiently small transparent sphere scatters weakly and why its scattered power has the characteristic ${\lambda }^{-4}$ dependence. Near a material resonance where $ϵ+2$ is small, this simple inequality and static-polarizability approximation need reconsideration, including loss and radiative corrections. For the stated low-frequency ordinary dielectric the supplied polarizability is the intended approximation.

The electric field is approximately radial in the long interior, but near an electrode end its lines spread into a fringing field of a capacitor. The electric field magnitude falls on moving outward through that end region. The dielectric's induced electric dipoles experience a net force toward the stronger field, pulling the material into the capacitor. For a weakly perturbing polarizable element this is expressed by $\mathbf{F}=\frac{1}{2}\alpha \nabla E^2$, with positive electric polarizability $\alpha$.

In the sketch the field lines run from the positive electrode to the negative one, curving outward near their ends. The material is attracted toward the interior where the field is stronger. A perfectly uniform interior field by itself would not give a longitudinal dipole-gradient force; the end region is essential to the local explanation. Neglecting its small contribution to the total capacitance does not justify omitting the finite insertion force that it transmits. The energy argument of part (d) gives that force without needing the detailed fringe solution.

In electrostatics, Gauss's law isThe surface normal points outward. Spherical symmetry makes the field radial and constant in magnitude on a concentric Gaussian surface.

For a uniformly charged sphere, enclosed charge is $4\pi {\rho }_0{r}^3/3$ inside and $Q=4\pi {\rho }_0{R}^3/3$ outside. Thus the radial field isSet potential zero at infinity. Outside, $V=Q/(4\pi {ϵ}_{0}r)$. Inside, integrate $dV/dr=-E_r$ from the known boundary value, obtainingThe field rises linearly from zero to its maximum at $R$, then falls as $r^{-2}$. The potential falls quadratically from ${\rho }_0{R}^2/(2{ϵ}_0)$ to its surface value and then as $r^{-1}$. Both are continuous at the surface; there is no separate sheet of surface charge.

For the electric field of a hydrogen ground-state electron cloud, its enclosed charge isWriting $u=2s/a_0$ and using $\int u^2{e}^{-u}du=-e^{-u}(u^2+2u+2)$ givesThe total electron-cloud charge tends to $-e$, as required. Dividing by $4\pi {ϵ}_0{r}^2$ yieldsThe expression is negative and therefore points inward. Its apparently singular terms cancel at small radius, leaving $E_{e,r}\simeq -er/(3\pi {ϵ}_0{a}_0^3)$.

A distant external charge polarizes the neutral atom. Its electric field displaces the positive and negative charge distributions relative to one another, giving an induced dipole $\mathbf{p}=\alpha {\mathbf{E}}_{\mathrm{ext}}$ with static electric polarizability $\alpha >0$. The field is stronger on the nearer side, so the opposite-charge attraction exceeds the like-charge repulsion. This is charge-induced dipole attraction, not a net monopole charge on the atom.

Quantitatively, for external charge $Q=\pm e$ at separation $r$, the polarization energy is $U=-\alpha E_{\mathrm{ext}}^2/2$, henceHere negative $F_r$ means decreasing separation. Reversing the external charge reverses the induced dipole but leaves this leading long-range force attractive and of the same magnitude. The factor one half accounts for the energy cost of inducing the dipole.

The supplied charge density describes the unperturbed atom. Adding its electron field to the nucleus field leaves $E_{\mathrm{atom}}=e{e}^{-2r/a_0}(r^{-2}+2/(a_{0}r)+2/a_0^2)/(4\pi {ϵ}_0)$, an exponentially small residual because the cloud has an infinite tail. A completely frozen density would give only that tiny sign-dependent force. Allowing polarization is essential to obtain the dominant attractive power-law force for either sign of external charge when $r\gg a_0$.

An electric dipole is a localized pair of opposite electric charges, or more generally a neutral electric charge distribution with nonzero first electric moment. For charges $-q$ and $+q$ separated by displacement $\mathbf{d}$ directed from negative to positive, the electric dipole moment isA neutral molecule can have a permanent electric dipole moment when its positive and negative charge centres do not coincide. An applied electric field can also produce an induced electric dipole moment by distorting an initially nonpolar electric charge distribution.

At distance $r\gg |\mathbf{d}|$ in vacuum, expand the difference of the two point-charge potentials to first order in the separation. The leading electric potential isThe monopole term cancels because the net electric charge is zero. Taking $\mathbf{E}=-\nabla \Phi$ gives the electric dipole fieldOn the dipole axis its magnitude is $2p/(4\pi {\epsilon }_0{r}^3)$ and its direction is along $\mathbf{p}$; in the equatorial plane the magnitude is $p/(4\pi {\epsilon }_0{r}^3)$ and the field points opposite to $\mathbf{p}$. The $r^{-3}$ decay is faster than the $r^{-2}$ field of an isolated point charge.

In a uniform external electric field, the opposite forces on the two charges cancel in their resultant but form a couple. Its torque isFor a fixed permanent electric dipole moment, the interaction potential energy isDifferentiating with respect to orientation recovers the torque. Parallel alignment minimizes $U$ and is stable against a small rotation; antiparallel alignment maximizes $U$ and is unstable. Rotating a fixed dipole slowly from parallel to antiparallel requires external work $2pE$.

A nonuniform electrostatic electric field can also exert a net force. Expanding the two charge forces gives $\mathbf{F}\simeq (\mathbf{p}\cdot \nabla )\mathbf{E}$, which equals $\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed dipole in a curl-free electrostatic field. Thus a suitably aligned dipole is attracted toward stronger field. An induced electric dipole moment requires care in computing energy: when $\mathbf{p}=\alpha \mathbf{E}$ for constant isotropic electric polarizability, building the dipole costs energy and the effective interaction is $U=-\alpha E^2/2$, rather than substituting directly into the fixed-dipole formula. These field, torque and energy effects explain molecular orientation and electric polarization.

Choose the $e^{-i\omega t}$ convention. The steady displacement in the Lorentz oscillator model has complex amplitudeThe induced electric dipole moment is $p_{\omega }=q{u}_{\omega }$, so the microscopic electric polarizability is $\alpha (\omega )=q^2/[m({\omega }_T^2-{\omega }^2-i\gamma \omega )]$. For $N$ identical oscillators per unit volume, $P_{\omega }=N{p}_{\omega }={ϵ}_0{\chi }_{\omega }{E}_{\omega }$. ConsequentlyHere the dimensionless quantity called polarizability in the paper is the electric susceptibility; the single-electron polarizability $\alpha$ has different units. The sign of the electron charge disappears because both the driving force and dipole moment contain $q$.

Multiplying by the complex conjugate of the denominator givesFor positive damping and positive frequency, ${ϵ}_r^{\prime \prime }\ge 0$, as required for passive wave absorption in this Fourier convention. At zero frequency, ${ϵ}_r^{\prime }=1+{\omega }_p^2/{\omega }_T^2$ and ${ϵ}_r^{\prime \prime }=0$. Near a weakly damped resonance the imaginary part has a narrow positive peak, while the real part rises on the low-frequency side and then falls steeply through ${ϵ}_r^{\prime }=1$ at ${\omega }_T$. It can become negative above resonance for sufficiently strong oscillator strength and weak damping. At large frequency, ${ϵ}_r^{\prime }=1-{\omega }_p^2/{\omega }^2+O({\omega }^{-4})$ and ${ϵ}_r^{\prime \prime }\sim {\omega }_p^2\gamma /{\omega }^3$: both approach the vacuum values. The absorption maximum is close to, but need not be exactly at, ${\omega }_T$ for finite damping.

Rotational Raman spectroscopy observes a frequency change of scattered light associated with molecular rotation. The coupling uses the induced electric dipole moment, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$, where $\mathbf{\alpha }$ is the electric polarizability tensor. For a linear molecule, the useful anisotropy is ${\alpha }_{\parallel }-{\alpha }_{\perp }$: the response differs for an electric field parallel or perpendicular to the molecular axis. A permanent electric dipole moment is not required.^[ref]

For the spin-free linear rigid rotor in one internal quantum state, the anisotropic tensor depends quadratically on the molecular direction. Its angular part has even parity and degree two in spherical harmonics. The triangle inequality for coupled angular momenta permits $|\Delta J|\le 2$, and even parity removes odd $\Delta J$, leaving $\Delta J=0,\pm 2$. The $\Delta J=0$ contribution is unshifted light; shifted rotational lines have $\Delta J=\pm 2$. Consequently the rigid-model Stokes shifts, in ordinary frequency units, areSuccessive allowed shifts are separated by $4B$, rather than the $2B$ absorption spacing of pure rotational electric-dipole selection rules. In spectroscopic wavenumber units replace $B$ by $\tilde{B}=B/c$ and $\Delta \nu$ by $\Delta \tilde{\nu }=\Delta \nu /c$, using the speed of light in $\mathrm{cm}{\mathrm{s}}^{-1}$ for $\mathrm{c}{\mathrm{m}}^{-1}$.

A homonuclear diatomic molecule such as molecular hydrogen has no permanent electric dipole moment in its symmetric internal state, so it lacks the ordinary electric-dipole microwave ladder. Its anisotropic electric polarizability can nevertheless give rotational Raman spectroscopy lines. Nuclear-spin symmetry can change the populations of even and odd $J$ levels, so the spin-free population example above must not be reused unchanged for molecular hydrogen. The method of coupling, as well as the energy eigenvalues, determines what the experiment can see.