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Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

Yes: the same physical resonant cavity and its field remain resonant in every inertial description. A Lorentz transformation changes the mirror worldlines and the field together. The boundary phase matching and the constructive-interference condition are statements about the same reflection events, and the contracted wave phase is a Lorentz scalar. A boosted resonant solution therefore still satisfies the moving-mirror boundary conditions.

It would be incorrect merely to insert a Lorentz-contracted length into the formula for two stationary mirrors while leaving both propagation frequencies unchanged. For example, let the mirrors be separated by $L$ in their rest frame and let ${\omega }_{0}L/c=m\pi$, including the usual equal-mirror reflection-phase convention. For an observer boosted at $+v$ along the cavity axis, the cavity moves at $-v$ and its two travelling components have frequenciesThey need not have one common laboratory frequency. The simultaneous mirror separation is $L^{\prime }=L/\gamma$, while the forward and return flight times are $L^{\prime }/(c+v)$ and $L^{\prime }/(c-v)$. The round-trip interval between events on the same mirror is therefore $2\gamma L/c$.

Along that moving mirror the wave phase rate of either wave component is ${\omega }^{\prime }-k^{\prime }{v}_{\mathrm{mirror}}={\omega }_0/\gamma$. Its advance over the round trip isThus the resonance condition is unchanged even though the observed travelling frequencies, lengths and flight times all change. For a boost in a general direction the beam directions also undergo relativistic aberration; the event-by-event phase-invariance argument still applies. A separate stationary cavity placed in the new frame, or an external laser whose frequency is not transformed with the original setup, is a different physical system.

Away from thresholds and narrow resonances, photon-mediated annihilation gives one colour copy proportional to $Q_q^2$ relative to the muon cross-section. The accessible flavours are $u,d,s,c$, each with three colours, soBottom production is closed. QCD radiative corrections raise the leading parton-model result by a relative term beginning at ${\alpha }_s/\pi$; the requested estimate does not require a supplied value of ${\alpha }_s$.

The charged two-pion and three-pion channels can be assigned to the same spin-zero kaon by their parent mass and lifetime. For two spinless pions, total angular momentum zero forces their relative orbital angular momentum to zero, so the product of their intrinsic parities givesFor three spinless pions, use Jacobi orbital angular momenta $l,L$. Coupling them to total zero requires $l=L$, and henceThus both channels cannot conserve parity in decays of one parent with a definite parity. The weak decay interaction permits the observed coexistence.

Dalitz plots identify the three-body events, determine their internal kinematics, and allow amplitude and angular-momentum analyses through their populations and resonance structures. Together with the spin and parent-identity evidence, they support the parity assignment. They are not a standalone parity-odd observable: reversing all three momenta leaves every $s_{ij}$, and therefore every point on the plot, unchanged. A flat distribution by itself cannot establish parity violation without the channel and spin information.

Treat a nucleon as incoherent collinear spin-half quarks and antiquarks carrying fractions $x$ of its momentum. Neglect their masses and transverse motion, use a charged-current four-fermion interaction, and take $Q^2\ll m_W^2$. The subprocess energy is $\hat{s}\simeq xs$. Antineutrino scattering on a quark has the same selected-current helicity factor $(1-y{)}^2$ as in part a; crossing to an antiquark changes the allowed spin pairing and gives a flat distribution. Multiplying the elementary strength by the respective parton densities givesHere $q,\bar{q}$ are the effective weak flavour combinations with the isoscalar proton-neutron average and CKM weights absorbed into their definitions. With only light flavours and charge symmetry, the relevant up and down combinations coincide on an isoscalar target. Rescaling those density definitions requires rescaling the prefactor consistently.

The equation is a leading-order scaling result: it neglects logarithmic QCD evolution, heavy-quark thresholds, higher-twist and target-mass corrections, nuclear modifications, and weak-propagator suppression. The current produces ${\mu }^{+}$ for an incident muon antineutrino. At higher momentum transfer the weak propagator supplies a factor approximately $[m_W^2/(m_W^2+Q^2){]}^2$, and at low energy resonance and quasielastic scattering invalidate a pure DIS description.

At an electron-positron collider tuned to the charmonium resonance, select the mixed leptonic tau lepton decaysand their charge conjugates. Require exactly two well-reconstructed oppositely charged tracks, one identified electron and one identified muon, little extra visible activity, visible energy below the beam total, and significant missing momentum pointing into the instrumented region. The multiple neutrinos prevent full reconstruction of each tau lepton, but the known initial four-momentum, lepton spectra and missing energy support a statistical signal extraction. Off-resonance data constrain nonresonant electron-positron annihilation into a tau lepton pair; that continuum is present at both beam energies.^[ref]

A resonant muon pair is one possible detector background: one muon can be misidentified as an electron, while radiation escaping along the beam or a badly measured track produces apparent missing energy. Particle-identification requirements and rejection of almost back-to-back tracks near the beam momentum suppress it. This background needs detector mismeasurement; an ideal measurement would distinguish it.

Above the open-charm threshold, $\psi (3770)\to D\bar{D}$ supplies an additional resonant background. Semileptonic decays of the two D mesons can give an electron, a muon and neutrinos; unreconstructed hadrons let some events pass the selection. This on-shell D meson-pair background is kinematically forbidden at $\psi (3686)$.

The three structures are the radiative cascadesIn the quark model, the J/psi meson and $\psi (2S)$ are respectively $1{}^3{S}_1$ and $2{}^3{S}_1$ charmonium states. Thus $L=0$, $S=1$, $J=1$, and meson parity and charge conjugation gives $J^{PC}=1^{--}$. The intermediate chi c mesons are $1{}^3{P}_J$: $L=1$, $S=1$, and addition of angular momentum permits $J=0,1,2$. ConsequentlyThese are electric dipole transitions, changing $L$ by one and preserving $S$ to leading order. A singlet $1{}^1{P}_1$ state has $J^{PC}=1^{+-}$ and is not another member of these spin-triplet cascades.

The primary photon energy is $(M_{2S}^2-M_{{\chi }_J}^2)/(2M_{2S})$, about $261$, $171$ and $128\mathrm{MeV}$ for $J=0,1,2$. The secondary photon remains more energetic even after its boost into the parent rest frame. Hence $m^2(J/\psi {\gamma }_h)=M_{{\chi }_J}^2$: with this quantity on the vertical axis, each resonance gives a horizontal band. A finite decay width and detector measurement resolution broaden each band; photon ordering removes the mirrored duplication of an unordered Dalitz plot.

The faint band beside the negative-slope diagonal boundary is $\psi (2S)\to J/\psi {\pi }^0$, followed by ${\pi }^0\to \gamma \gamma$. The neutral pion is predominantly $(u\bar{u}-d\bar{d})/\sqrt{2}$, a ${}^1{S}_0$ pseudoscalar meson with $J^{PC}=0^{-+}$. This production transition violates isospin, accounting for its small rate. To locate the band, set $X=m^2(J/\psi {\gamma }_l)$ and $Y=m^2(J/\psi {\gamma }_h)$. Four-momentum conservation givesFor the neutral pion, $m_{\gamma \gamma }^2=m_{{\pi }^0}^2$ is small, putting the line just inside the $m_{\gamma \gamma }^2=0$ boundary. The allowed $J/\psi \eta$ channel, with the eta meson also $0^{-+}$, produces a parallel line further inside, at $m_{\gamma \gamma }^2=m_{\eta }^2$. The positive-slope edge $X=Y$ instead comes from ordering the two photon energies; proximity to that edge alone would not identify the two-photon state.

Fit radiatively corrected visible cross sections across the Z boson resonance. The line shape determines $M_Z$, the total decay width ${\Gamma }_Z$, and pole normalizationsThe first extraction uses the fitted total and visible partial decay widths. Assuming lepton universality, ${\Gamma }_e={\Gamma }_{\mu }={\Gamma }_{\tau }={\Gamma }_{\ell }$, an absolute leptonic pole cross section gives ${\Gamma }_{\ell }={\Gamma }_Z\sqrt{M_Z^2{\sigma }_{\ell }^0/(12\pi )}$. The hadronic-to-leptonic ratio gives $R_{\ell }={\Gamma }_{\mathrm{had}}/{\Gamma }_{\ell }$. Subtracting every visible channel then yieldsWithout lepton universality, separately fitted visible leptonic widths replace $3{\Gamma }_{\ell }$, with sufficient independent information to disentangle their products with ${\Gamma }_e$.

A second, useful peak-normalization extraction first determines the invisible-to-leptonic ratio. Becauseone obtainsThis ratio extraction does not require the measured line width. Multiplication by an independently predicted leptonic partial decay width, or by ${\Gamma }_Z/(R_{\mathrm{inv}}+R_{\ell }+3)$ from the scan, converts it to ${\Gamma }_{\mathrm{inv}}$. Alternatively dividing $R_{\mathrm{inv}}$ by the predicted single-species ratio ${\Gamma }_{\nu }/{\Gamma }_{\ell }$ gives the number of light active neutrinos. These are two standard uses of line-shape information, with different choices of theoretical input; if both use the same fitted ${\Gamma }_{\ell }$, they are algebraically equivalent and statistically correlated, not independent determinations.

Statistical uncertainties in the reference-process counts give fluctuations in the collider luminosity at individual scan points, propagating into normalization and fitted line width. They decrease with additional reference events. A common multiplicative luminosity-scale systematic largely cancels in $R_{\ell }$ and the normalized line shape, but shifts every absolute peak cross section. It therefore does not cancel from either invisible-width extraction. For a fractional upward luminosity shift $\epsilon$, ${\sigma }_{\mathrm{had}}^0$ shifts by $-\epsilon$, and at fixed $R_{\ell },M_Z,{\Gamma }_Z$,Thus overestimating luminosity increases the inferred invisible contribution. Energy-dependent systematic errors can also distort the line shape and its total decay width. The covariance matrix must retain correlations between scan points and experiments, including common theoretical uncertainty in the Bhabha scattering calculation.

Away from narrow resonances and quark thresholds, the QED annihilation cross section into each accessible quark flavour is its squared fractional electric charge times the muon-pair cross section, multiplied by the number $N_c$ of distinguishable color charges. Thus the inclusive hadronic ratio isWith $N_c=3$, the continuum predictions are $R_{uds}=2$, $R_{udsc}=10/3$ and $R_{udscb}=11/3$ as successive charm quark and bottom quark channels become accessible. The measured normalization and steps support three color charges. At leading order these hadronic events form two back-to-back jets from the quark and antiquark; quantitative comparisons include QCD corrections and avoid the nonperturbative threshold region.

The extra resolved jet is explained by $e^{+}{e}^{-}\to {\gamma }^{\ast }\to q\bar{q}g$, with gluon radiation from either the quark or antiquark. One possible color charge routing is initial $q_r{\bar{q}}_{\bar{r}}$ followed by $q_r\to q_b+g_{r\bar{b}}$; the final $q_b$, ${\bar{q}}_{\bar{r}}$ and $g_{r\bar{b}}$ combine to a net colour singlet.

The three-jet cross section contains one additional strong interaction vertex in the amplitude, so its ratio to the two-jet rate is of order ${\alpha }_s=g_s^2/(4\pi )$ times a resolution-dependent kinematic coefficient. A resolved fraction of order $15\%$ is consistent with perturbative QCD and a strong coupling constant much larger than the fine-structure constant. It is not an equality ${\alpha }_s=0.15$: the jet definition and cuts are needed. Photon radiation has a much smaller electromagnetic rate. The large hadronic third-jet rate supports gluon radiation; identifying the radiated object specifically as a colour-octet gluon also uses the QCD vertex and colour-flow description, not merely the existence of three visible jets.

The order-${ϵ}^2$ equation isUsing $\cos t\cos 2t=(\cos 3t+\cos t)/2$ and $\cos t\sin 2t=(\sin 3t+\sin t)/2$, its resonance coefficients aresecular term are removed by setting these to zero:The eigenvalues of this two-by-two system satisfyThey are real with opposite signs for $-1/12<a_2<5/12$. Generic initial amplitudes then contain an exponentially growing slow-time component. At either leading endpoint the matrix is nonzero nilpotent matrix, so suitable initial amplitudes give linear slow-time growth; higher-order detuning decides the precise corrected boundary. Outside this interval the leading eigenvalues are purely imaginary and the slow amplitudes remain oscillatory.

Substitution into $a=1+a_2{ϵ}^2+\cdots$ gives the requested leading instability interval“Can become unbounded” permits special initial amplitudes without growth. The calculation identifies the second instability tongue of a weak Mathieu oscillator and its slow amplitudes for the second Mathieu instability tongue; it is not an assertion that every solution grows throughout the closed interval.

At order one, $({\partial }_t^2+1)y_0=0$, soAt order $ϵ$,The operator sends a constant to itself and sends $\cos 2t,\sin 2t$ to minus three times themselves. HenceAn additional homogeneous $\cos t,\sin t$ contribution can be absorbed into the definitions of $A,B$. There is no resonance forcing at this order; the first amplitude evolution therefore occurs on the chosen slower scale $T={ϵ}^{2}t$.

Use the fast time $T_0=t$ and slow time $T={ϵ}^{2}t$. A possible intermediate scale $ϵt$ has no leading modulation: the absence of first-order resonance imposes independence of that scale. WriteAt first order choose the nonresonant particular solutionwith any needed fundamental homogeneous correction assigned consistently when matching the initial data. At second order,The coefficient of $e^{i{T}_0}$ on the right is $(10/3)|A{|}^{2}A-2i{A}_T$. A bounded oscillatory correction requires this resonance coefficient, and its conjugate, to vanish. The quadratic oscillator phase-modulation condition is thereforeIt resums the secular phase accumulation into a slow modulation instead of leaving it as a growing term in the correction.

With phasors proportional to $e^{i\omega t}$, the series RLC circuit hasFor fixed $V_0$ and frequency-independent positive $R$, the maximum occurs when the reactances cancel. ThusThis is the driven current resonance; a damped free-oscillation frequency is a different quantity.

A wirewound resistor uses a metal alloy chosen for stable electrical resistivity. Compared with a typical carbon composition resistor, it can offer tighter resistor tolerance, a smaller temperature coefficient of resistance, lower excess electrical noise, and greater continuous electrical dissipation when its construction provides an adequate heat path. These are design-dependent advantages, rather than guarantees for every alloy or package.

The winding has parasitic inductance and interturn capacitance. Its electrical impedance therefore ceases to be a nearly real electrical resistance at sufficiently high frequencies; an approximate model is a series circuit $R+i\omega L$ shunted by a small parasitic capacitance. A resonance can occur, making a wirewound resistor unsuitable where a nearly frequency-independent electrical resistance is required. Noninductive winding patterns reduce this problem. For comparable small-signal duties it is usually larger and more expensive. A carbon composition resistor is compact and has relatively small parasitic inductance, but commonly has poorer stability, larger resistor tolerance, and more excess electrical noise.

Put $x=\omega /{\omega }_0$. Maximizing the output wave amplitude is equivalent to minimizingIts derivative is $D^{\prime }(x)=2x[2x^2-2+Q^{-2}]$. An interior minimum at positive frequency therefore requires $Q>1/\sqrt{2}$, and occurs atIndeed $D^{\prime \prime }(x_{\mathrm{res}})=8x_{\mathrm{res}}^2>0$, and the peak value is $|H{|}_{\max }=Q/\sqrt{1-1/(4Q^2)}$. For $0<Q\le 1/\sqrt{2}$, $D$ increases for $x>0$ and the largest output is the DC value $|H(0)|=1$. At equality the putative peak has reached zero frequency; there is no distinct nonzero-frequency resonance. This threshold concerns capacitor voltage peaking: the series RLC circuit's electric current still has its maximum at ${\omega }_0$ for every positive $R$.

In the parallel LC circuit both components have the same voltage phasor. Adding their current phasors by Kirchhoff's current law gives the total electrical admittance:ThusAt the ideal parallel resonance, ${\omega }_0=1/\sqrt{LC}$, the two branch current phasors cancel. The net electrical admittance is zero and the magnitude of the complex impedance tends to infinity: the parallel combination behaves as an open circuit at that frequency. Its complex argument is undefined exactly at the pole. Finite component losses replace the infinite peak by a large finite electrical impedance. This conclusion concerns a voltage-driven sinusoidal response; an ideal resonant current drive does not have a bounded sinusoidal steady-state voltage.

The restoring, damping and electric forces on the electron give $m_e\ddot{x}=-kx-b\dot{x}-eE(t)$. Dividing by $m_e$ gives a Lorentz oscillator model withFor the $e^{i\omega t}$ convention, substitute $x=\mathrm{Re}(x_0{e}^{i\omega t})$. Then $x_0=-e{E}_0/[m_e({\omega }_0^2-{\omega }^2+i\gamma \omega )]$ andThe time-averaged dissipated power isFor positive frequency the squared term is minimized to zero only at $\omega ={\omega }_0$. Thus the power maximum is exactly at the undamped natural frequency, even though the displacement maximum is slightly shifted. Here $P_{\max }=e^2|E_0{|}^2/(2b)$. The half-power equation is $|{\omega }_0^2-{\omega }^2|=\gamma \omega$, whose two positive roots areThis is the absorbed-power bandwidth of a viscously damped oscillator, exact within the model. Near resonance the profile is approximately a Lorentzian line shape. With $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$,For this wavelength linewidth of a damped-oscillator power resonance we used ${\omega }_{+}{\omega }_{-}={\omega }_0^2$. The half-power wavelength endpoints are not exactly symmetric around ${\lambda }_{\mathrm{peak}}$, although that distinction is tiny here.

In a constant field the settled displacement is $x_s=-e{E}_{\mathrm{dc}}/k$. On sudden removal, with zero initial velocity, the free motion isThe electron oscillates about its original equilibrium, losing the previously stored spring energy. The period is $3.52\mathrm{fs}$, the amplitude decay time is $2/\gamma =0.319\mathrm{ps}$, and the cycle-averaged energy decay time is $1/\gamma =0.160\mathrm{ps}$. This is the effective classical model supplied, not a prediction of the full hydrogen line radiation spectrum.

The wave impedance is the travelling-wave field ratio $Z=E/H$; for the stated nonmagnetic media it is $Z_0/n$. A reflection coefficient is a reflected-to-incident complex amplitude ratio with an explicitly chosen field orientation. With this reflection-amplitude sign convention, using the question's formula for incidence inside silicon givesWith a common tangential electric-field direction, the usual electric-field Fresnel equations instead give $r_E=(Z_2-Z_1)/(Z_2+Z_1)=+5/9$ for this incidence. The printed formula corresponds to the opposite reflected-field sign convention, for example the magnetic amplitude convention. It gives the same reflectance and the same round-trip factor $r^2$.

Each extra internal round trip multiplies the transmitted amplitude by $r^2{e}^{i\delta }$, with $\delta =4\pi nd/\lambda$. A geometric series therefore givesThis is Fabry–Pérot transmission from multiple reflections. Assuming lossless silicon with wavelength-independent $n$, the full intensity transmittance between air regions isMaxima have $2nd/\lambda =j$, minima $2nd/\lambda =j+\frac{1}{2}$. Here $2nd=7000\mathrm{nm}$, so within the specified tuning range the only maximum is $1400\mathrm{nm}$, and the two minima are $1555.6\mathrm{nm}$ and $1272.7\mathrm{nm}$. The contrast is

Identical highly reflecting coatings increase the finesse: transmission maxima become narrower and the intensity between them falls. Ideal symmetric lossless coatings retain $T_{\max }=1$; absorption or unequal mirrors reduce it. The illustrated maxima stay fixed only because additional reflection wave phases are held unchanged. Real coating wave phases can also shift the resonances.

For $m\ddot{x}+b\dot{x}+kx=F_0\cos \omega t$, define ${\omega }_0=\sqrt{k/m}$ and $\beta =b/(2m)$. The parameter relating quality factor and viscous damping regimes isAt weak damping it also equals $2\pi$ times stored energy divided by energy lost per cycle. The stored-energy convention and frequency matter: this cycle-based interpretation should not be extended to free motions that never complete a cycle.

For $\beta <{\omega }_0$ the free motion is $e^{-\beta t}[A\cos {\omega }_{d}t+B\sin {\omega }_{d}t]$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. Thus underdamping means $Q>1/2$. High $Q$ gives many oscillations per amplitude-decay time, narrow resonance and strong frequency selectivity; a lightly loaded tuning fork, optical cavity or low-loss electrical resonator provides examples. The amplitude decay time is ${\tau }_A=1/\beta =2Q/{\omega }_0$, while the slowly averaged energy decays with ${\tau }_E=1/(2\beta )=Q/{\omega }_0$. These times differ by a factor of two.

At critical damping, $\beta ={\omega }_0$ and $Q=1/2$, with $x=(A+Bt)e^{-{\omega }_{0}t}$. There is no ringing; a simple critically damped position indicator can settle without oscillation. For $\beta >{\omega }_0$ and $Q<1/2$, there are two real negative decay rates $-\beta \pm \sqrt{{\beta }^2-{\omega }_0^2}$. Increasing very strong drag slows the long-time return, whose rate approaches ${\omega }_0^2/(2\beta )$. A strongly viscously damped mechanical indicator illustrates this regime. These classifications concern the linear homogeneous equation, not every possible nonlinear damping law.

The driven displacement amplitude isThe displacement-resonance threshold of a viscous oscillator requires $\beta <{\omega }_0/\sqrt{2}$, or $Q>1/\sqrt{2}$; it lies at $\sqrt{{\omega }_0^2-2{\beta }^2}$. Some underdamped systems therefore have no such maximum. Average absorbed power instead peaks at ${\omega }_0$ for this viscous model. Its half-power frequencies are $\sqrt{{\omega }_0^2+{\beta }^2}\pm \beta$, so their separation is $2\beta$ and $Q={\omega }_0/\Delta \omega$. For broad response this ratio remains a useful parameter but not a claim of many free oscillations. The familiar displacement-linewidth version is a high-$Q$ approximation.

A series RLC circuit has ${\omega }_0=1/\sqrt{LC}$ and $Q={\omega }_{0}L/R$ for the charge oscillation. Resistance dissipates energy, inductance provides the inertial analogue and capacitance the restoring analogue. Ringdown and a calibrated half-power scan independently estimate loss; parasitic resistance, loading and radiation change the measured value. A large $Q$ increases frequency selectivity but also lengthens settling time, so it is not universally the best instrumental choice.

A carrier in a locally parabolic isotropic band obeys $m^{\ast }\dot{\mathbf{v}}=q\mathbf{v}\times \mathbf{B}$, soIlluminate a high-mobility crystal with a transverse alternating electric field, sweep a calibrated static magnetic field at known microwave or far-infrared frequency, and detect resonant absorbed power. Cooling lengthens the relaxation time; resolving the resonance generally needs ${\omega }_c\tau \gg 1$. Varying field direction measures band anisotropy, while circular polarization distinguishes the two senses of rotation for opposite carrier charges. This is orbital motion rather than electron spin resonance.

The classic silicon/germanium measurements used single crystals near $4\mathrm{K}$, light modulation and circularly polarized microwaves, with directional scans to infer anisotropic carrier-energy surfaces.^[ref] Carrier generation, calibration, scattering linewidth and inhomogeneity must be separated from a genuine mass shift.

In a general band the cyclotron mass is an orbit property, $m_c={\hslash }^2(\partial A_k/\partial E)/(2\pi )$ for the appropriate field-normal orbit. It need not equal every component of the band-curvature mass tensor. For an ellipsoidal parabolic band with field along a principal $z$ direction it is $\sqrt{m_x{m}_y}$. Thus a single resonance does not determine a whole effective-mass tensor without angular information.