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A rigid body undergoes translation when every point has the same displacement at each time, so its orientation stays fixed. For a freely moving system of particles coupled only by internal springs, the common displacement is a zero-frequency normal mode: internal spring lengths and potential energy are unchanged. Its mode coordinate is $A+Bt$, representing constant centre-of-mass velocity, rather than an oscillation at nonzero frequency.

An RLC circuit combines electrical resistance $R$, inductance $L$, and capacitance $C$. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. Their exchange can produce an oscillation, while Joule heating removes energy. A series RLC circuit gives a particularly clear example: changing $R$ turns ringing into a direct return to equilibrium, and changing $L$ or $C$ tunes a resonance.

These circuits connect two useful viewpoints. In time, they describe how stored energy is released after switching. In frequency, they select a range of signals: the resonant series RLC response explains radio tuning and simple filters. The same mathematics describes a damped harmonic oscillator, so insight about mechanical ringing transfers directly to an electrical circuit. ^[ref]; ^[ref].

A circularly driven spin becomes time-independent after a rotation $R=e^{-i\omega t{S}_z/\hslash }$. The transformed Hamiltonian operator is $R^{†}HR-i\hslash R^{†}\dot{R}$, so the longitudinal frequency is shifted by $-\omega$. This is an exact change of frame for a circular drive. The rotating-wave approximation is a separate approximation needed when a drive also contains a counterrotating component.

A rotating ideal half-wave plate flips circular polarization. The spin change has magnitude $2\hslash$ per photon, so its mechanical energy exchange produces a frequency shift of magnitude $2|\Omega |$. For Jones convention $H(\theta )=R(\theta )\mathrm{diag}(1,-1)R(-\theta )$ and $Q=\mathrm{diag}(1,i)$, the combination $QH(\theta )Q=e^{i2\theta {\sigma }_x}$. Acting on $(1,1{)}^T/\sqrt{2}$ gives $e^{i2\theta }$; with time dependence $e^{-i\omega t}$ and $\theta =\Omega t$, the output frequency is $\omega -2\Omega$. The opposite input eigenpolarization reverses the shift.

A planar quadratic saddle $m{\omega }^2(x^2-y^2)/2$ rotated at angular speed $\Omega$ becomes linearly confining for the displayed strict inequality. Its rotating-frame mode frequencies obey ${ϵ}_{\pm }^2={\Omega }^2\pm {\omega }^2$. Equality has a zero-frequency secular mode and is not robust bounded-motion stability. This is a rotating potential model; a rolling ball on an actual surface has additional constraints.

A rigid pendulum tilted from the upward vertical and driven at fixed azimuthal angular velocity $\Omega$ has effective potential $\mu {\theta }^2/2+mga\cos \theta -m{a}^2{\Omega }^2{\sin }^2\theta /2$. With $\mu =2mga$, its upright equilibrium loses linear stability when ${\Omega }^2$ passes $g/a$. At equality the leading restoring torque is cubic and the potential has a positive quartic minimum. This gives a stable quartic oscillator with zero infinitesimal harmonic frequency, rather than a freely drifting neutral state.

In this article the rotational constant is a frequency,Its dimensions are ${\mathrm{s}}^{-1}$, since the Planck constant has dimensions $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and a moment of inertia has dimensions $\mathrm{kg}{\mathrm{m}}^2$. Some authors instead call $hB={\hslash }^2/(2I)$ the energy-valued rotational constant; this is the convention in the earlier synthetic molecular beam electric resonance diagram. A third convention uses the spectroscopic wavenumber $\tilde{B}=B/c$, with $c$ the speed of light. Always check units before using $\nu =2B(J+1)$.

If $B$ is known and the two masses are specified, the rigid two-atom model gives the bond length $r=\sqrt{h/(8{\pi }^2\mu B)}$. This is why rotational spectroscopy can constrain molecular geometry. A real fitted constant can depend on molecular vibration, so this inference alone does not establish an equilibrium bond length.

A frequency change associated with rotation and exchange of angular momentum. If an optical element rotates at angular velocity $\Omega$ about a chosen axis and gains angular momentum $\Delta L$ from a photon, its mechanical energy gain is $\Omega \Delta L$ in the prescribed slow-rotation model. Energy conservation then gives photon frequency change $\Delta \omega =-\Omega \Delta L/\hslash$. The rotating half-wave plate frequency shift is a spin-polarization example.

For a rigid heteronuclear diatomic rotor, $E_J={\hslash }^{2}J(J+1)/(2I)$ and electric-dipole absorption has $\Delta J=+1$. Therefore the $J^{\prime }\to J^{\prime }-1$ line frequency is $h{J}^{\prime }/(4{\pi }^{2}I)$ and adjacent absorption lines are spaced by $h/(4{\pi }^{2}I)$. With reduced mass $\mu$, the bond-length estimate is $r=\sqrt{I/\mu }$. The conversion assumes frequency rather than angular frequency; centrifugal distortion makes real high-J spacings imperfectly uniform.

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.

A rotational spectral line assignment associates a measured spectral line with a molecule, an internal quantum state, and an initial and final set of quantum numbers. For the spin-free linear rigid rotor, absorption is labelled $J\to J+1$. A regularly spaced group suggests a candidate rotational constant, but the lowest line inside a scanned band need not be $0\to 1$. Several molecules can contribute to the same spectrum, and an apparent missing member can lie outside the band or below the detection threshold.

First determine line centers and their uncertainties from the instrument's readout, accounting for resolved Doppler shifts or hyperfine structure rather than treating every peak as a separate rotational transition. Then propose integer assignments, fit a rotational Hamiltonian, and predict additional spectral lines. For a weakly distorted linear molecule within one vibrational quantum state, the centrifugal distortion model gives the useful diagnosticHere $B,D,{\nu }_m$ are all frequencies in $\mathrm{Hz}$. Correct assignments make the second expression a straight line against $m^2$, with intercept $B$ and slope $-2D$. This follows by division of the transition formula; no additional model of molecular geometry is required. A direct weighted least squares fit of ${\nu }_m$ to $2Bm-4D{m}^3$ uses the measured uncertainties. If instead the transformed values are fitted, their uncertainties must also be divided by $2m$.^[ref]

For an original example, stipulate $B=5.000000\mathrm{GHz}$ and $D=2.000\mathrm{kHz}$, and suppose a scan covers $38$–$72\mathrm{GHz}$. The model places four consecutive transitions at $39.999488$, $49.999000$, $59.998272$, and $69.997256\mathrm{GHz}$. Their assignments are $m=4,5,6,7$, hence $J=3,4,5,6$. An extra stipulated line at $53.2\mathrm{GHz}$ is unrelated to this series. All these frequencies are calculated example inputs, not measured OCS data.

At $m=4$ and $m=7$, the transformed values are $4999.936$ and $4999.804\mathrm{MHz}$. Their straight line has slope $(4999.804-4999.936)/(49-16)=-0.004\mathrm{MHz}$, recovering $D=0.002\mathrm{MHz}$ and $B=5000\mathrm{MHz}$. Both intermediate transitions satisfy the same line exactly. Merely relabelling the first visible transition $0\to 1$ fails this diagnostic. The model predicts a withheld $7\to 8$ transition at $79.995904\mathrm{GHz}$: an additional measurement there would test the assignment rather than merely refit the original four lines.

An assignment earns confidence when several independently observed spectral lines, including lines outside the fitted interval, agree within uncertainty and expected strong accessible lines are present. Inspect the residual of an inverse problem for each line against $J$: systematic curvature can signal inadequate centrifugal distortion, whereas isolated discrepancies suggest a blend or wrong assignment. Comparing a known sample, a blank, and an isotope shift of a linear rotor supplies different checks. These are general inference controls; the actual reported controls of the NIST trace-gas FTMW validation experiment are described separately. One isolated match or one fitted rotational constant alone cannot certify molecular identity.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

For a stationary rotating apparatus driven by a source with optical coherence and proper frequency ${\nu }_0$, define ${\lambda }_0=c/{\nu }_0$. The source wave phase advances by $2\pi {\nu }_0\Delta \tau$ between the emission times of the two contributions that meet at one readout spacetime event. Define the signed phase lag $\Delta \varphi$ of the $+$ contribution and the corresponding displacement $N$ in units of one interference fringe byThe sign of a photographed spatial displacement also depends on the imposed fringe orientation; this definition fixes the theoretical sign through the delayed beam. In an equal-amplitude output, a representative light intensity is $I(x)=I_{\max }[1+\cos (Kx-\Delta \varphi )]/2$, where $K$ sets the chosen spatial interference fringe spacing $2\pi /K$. Moving $\Delta \varphi$ by $2\pi$ shifts the pattern by one fringe. A fixed output exactly at a bright or dark fringe has zero first-order intensity slope, so a known phase bias or a spatial fringe readout is useful for determining a small signed delay.

Choose a square of side $0.30\mathrm{m}$, a rectangle $0.40\mathrm{m}$ by $0.20\mathrm{m}$, ${\lambda }_0=633\mathrm{nm}$ and $\Omega =+10\mathrm{rad}{\mathrm{s}}^{-1}$, with angular velocity parallel to the positive normal vectors. Both perimeters are $1.20\mathrm{m}$, but the areas are $0.090{\mathrm{m}}^2$ and $0.080{\mathrm{m}}^2$. Using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the leading predictions areThese are original model choices and predictions, not measured points. The square has $9/8$ times the response despite the equal path lengths. Reversing from $+\Omega$ to $-\Omega$ gives $N(-\Omega )-N(+\Omega )=-2N(+\Omega )$: the displacement magnitude between opposite rotations is twice the displacement magnitude from rest. Confusing these two comparisons introduces a factor-of-two error.

At uniform sampling times $t_n=n/f_s$, adding an integer multiple of the sampling rate to a sinusoid's frequency changes its complex phase by an integer multiple of $2\pi n$. The samples therefore agree. Real cosines also identify the two signs of frequency, giving the displayed representative below the Nyquist frequency. A chosen nonzero sampling-time origin changes the equivalent phase, rather than the alias frequency.

A Schwarzschild static observer at a circular orbit measures speed $v_{\mathrm{loc}}=r(d\varphi /dt)/\sqrt{1-2\mu /r}$. Combining the circular geodesic frequency with this clock rate gives the displayed formula. The observer-relative Lorentz factor is $\gamma =\sqrt{(r-2\mu )/(r-3\mu )}$, and $rd\varphi /d\tau =\gamma v_{\mathrm{loc}}$. Thus the local speed is $c/2$ at the Schwarzschild innermost stable circular orbit. Coordinate and locally measured speeds are distinct.

The Schwarzschild coordinate rotation frequency at the marginally stable circular orbit, measured per unit time of a stationary clock at infinity. It scales inversely with black-hole mass. A finite-radius static clock adds a gravitational time-dilation factor; a quasi-periodic brightness signal need not directly equal this orbital frequency.

With $\vartheta =\omega t+\delta$, a weak cubic potential produces a nonzero mean position and a second harmonic at first order. There is no first-order frequency shift. Higher-order frequency changes can accumulate phase over sufficiently long times, so the first-order expression is not a uniform all-time approximation.

Second-harmonic generation converts an incident optical frequency $\nu$ into $2\nu$ through a material's nonlinear response. In the photon description of an ideal lossless conversion, two input photons supply one output photon:Here $h$ is the Planck constant. The relation states conservation of energy; the material mediates the interaction. It does not say that each input photon independently splits its wavelength in half.

For vacuum wavelengths, ${\lambda }_{0,\omega }=c/\nu$ and ${\lambda }_{0,2\omega }=c/(2\nu )$, where $c$ is the speed of light. HenceThe second relation uses the refractive index for each propagating mode and frequency. Within a medium with optical dispersion, the wavelength need not halve. For an invented example, ${\lambda }_{0,\omega }=800\mathrm{nm}$, $n_{\omega }=1.5$, and $n_{2\omega }=1.6$ give $400\mathrm{nm}$ in vacuum, but in the material the two wavelengths are $533.33\mathrm{nm}$ and $250\mathrm{nm}$; their ratio is $0.46875$.

The difference between the harmonic's wave number and that of the quadratic driving electric polarization. It has units of inverse length. For collinear modes, $\Delta k=2\omega (n_{2\omega }-n_{\omega })/c$, so matching frequencies alone does not ensure matching spatial wave phases.

Two fundamental harmonics of frequency one multiply into only a constant and second harmonics. The first-order solvability condition in the method of multiple scales therefore has no amplitude-changing resonant coupling. Their first-order particular corrections can mix back into the fundamental at second order, selecting the possible scale $t=O({ϵ}^{-2})$. Detuning may suppress the resulting resonance, so identifying this scale alone does not prove instability for every initial state.