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The angular momentum associated with motion in a degenerate bending normal mode of a linear molecule. A one-quantum bend has $l=\pm 1$, and the total rotational quantum number obeys $J\ge |l|$.

The harmonic-oscillator entropy contribution for one vibrational coordinate, with $x={\theta }_{\mathrm{vib}}/T$. Contributions from independent normal modes add, with one term for each component of a degenerate mode. A temperature-independent zero-point energy does not change this entropy.

Excitation of one quantum of a normal mode from the vibrational ground state. For an infrared-active mode in the linear-dipole harmonic approximation, this is the allowed absorption underlying the usual fundamental band. A vibrational overtone excites two or more quanta from the ground state.

Coupling between electronic and nuclear vibrational motion. Dependence of an electronic electric dipole transition moment on a normal mode coordinate can give intensity to a transition whose purely electronic matrix element vanishes at the equilibrium geometry.

Across a jump of $U^{\prime }$ in a constant-density parallel flow with continuous $U$, the normal mode vertical velocity and pressure are continuous. The linearized momentum equation gives $\hat{p}=\rho [(U-c){\hat{w}}^{\prime }-U^{\prime }\hat{w}]/(ik)$, yielding the displayed conditions. Equivalently, away from $U=c$, $(U-c)[{\hat{w}}^{\prime }]=[U^{\prime }]\hat{w}$. Derivative continuity is generally incorrect when the background vorticity jumps. This is the constant-density, continuous-velocity specialization of jump conditions for stratified inviscid shear flow.

An ideal uniform, flexible string under constant tension $T>0$ has transverse displacement $y(x,t)$ satisfyingwhere $\mu >0$ is linear mass density, meaning mass per unit length of the straight reference string. The approximation assumes small slope $|y_x|\ll 1$, transverse motion, negligible bending stiffness, negligible damping, and no distributed external force. Constant tension and uniform linear mass density make $c$ constant. Some texts use $\rho$ for this same linear mass density, giving $c=\sqrt{T/\rho }$; here $\rho$ must have units $\mathrm{kg}{\mathrm{m}}^{-1}$, rather than units of volume mass density. ^[ref]

The units provide a useful check: $T/\mu$ has units $(\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2})/(\mathrm{kg}{\mathrm{m}}^{-1})={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For $T=100\mathrm{N}$ and $\mu =0.010\mathrm{kg}{\mathrm{m}}^{-1}$, the wave speed is $100\mathrm{m}{\mathrm{s}}^{-1}$, so a pulse crosses a one-metre string in $0.010\mathrm{s}$. Quadrupling tension doubles wave speed; quadrupling linear mass density halves it. These square-root laws explain how material and loading set the time scale of the vibration.

For a length-$L$ string with fixed endpoints, the homogeneous Dirichlet boundary conditions are $y(0,t)=y(L,t)=0$. Its nonzero normal modes haveThus $cn\pi /L$ is an angular frequency, in radians per second, whereas $cn/(2L)$ is the frequency in hertz. The wavelength is $2L/n$, and each node of a standing wave lies at $x=jL/n$ for some $j=0,\dots ,n$. For the numerical string above with $L=1\mathrm{m}$, the first three frequencies are $50$, $100$ and $150\mathrm{Hz}$, with angular frequencies $100\pi$, $200\pi$ and $300\pi {\mathrm{s}}^{-1}$. ^[ref]

Each normal mode is a standing wave, but it also has a travelling wave description:At a fixed endpoint an incident displacement pulse reflects with reversed sign, allowing incident and reflected displacements to cancel at the endpoint. Periodic reflection for a fixed-end string implements both endpoints through odd, $2L$-periodic extensions of the initial data. This is the bridge between moving pulses and the stationary Fourier sine basis. ^[ref]

A short derivation comes from Newton's second law. The transverse force on a small segment is approximately $T[y_x(x+\delta x,t)-y_x(x,t)]$, and its mass is $\mu \delta x$. Dividing the force balance by $\delta x$ and taking the limit gives $\mu y_{tt}=T{y}_{xx}$. Separation of variables then gives ${\varphi }^{\prime \prime }+k^2\varphi =0$; the fixed-end boundary conditions select $k=n\pi /L$, while the temporal equation is a harmonic oscillator equation with $\omega =ck$.

A waveguide confines a wave transversely while allowing propagation along a longitudinal direction. Material properties and boundary conditions select transverse normal modes, whose longitudinal dispersion relations may have cutoffs even when the bulk material is nondispersive. Examples include an optical fiber, a hollow electromagnetic waveguide, and a membrane waveguide. Guiding reduces geometric spreading along the chosen route but does not eliminate absorption, scattering or leakage.

The waveguide cutoff frequency separates oscillatory axial propagation from exponential decay for a particular normal mode. For the vacuum-filled rectangular electromagnetic waveguide, the dispersion relation isFor a transverse electric mode of a rectangular waveguide, $m,n$ are nonnegative integers but not both zero. For a mode with $B_z=0$ and nonzero $E_z$, both indices must be positive. The lowest cutoff for $a>b$ is consequently ${\omega }_{10}=\pi c/a$, or ordinary frequency $f_c=c/(2a)$.

A square cross-section instead gives equal ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$ cutoffs. The absence of a transverse electromagnetic mode is specific to this hollow pipe, rather than all electromagnetic waveguides.^[ref]

For $\omega >{\omega }_{mn}$, choose real $\beta >0$ for propagation toward $+z$. For $0<\omega <{\omega }_{mn}$, a field decaying into a semi-infinite guide hasA finite section can contain both growing and decaying axial solutions set by its two ends; the single decaying solution is not a complete finite-section transmission calculation. At exact cutoff $\beta =0$, the two axial solutions become a constant and a linear function of $z$: there is no finite axial wavelength for a traveling sinusoid.

In the chosen $a=30\mathrm{mm}$ guide at $f=0.8f_c=3.99723277\mathrm{GHz}$, the ${\mathrm{TE}}_{10}$ branch has $\alpha =20\pi {\mathrm{m}}^{-1}$. Its amplitude decay length is $1/\alpha \simeq 15.9155\mathrm{mm}$; after $50\mathrm{mm}$, its normalized wave amplitude is $e^{-\pi }\simeq 0.0432139$. This is geometric evanescence even with ideal walls, not Joule heating. A single decaying branch has zero cycle-averaged axial Poynting vector: substituting $\beta =i\alpha$ makes $B_x$ a quarter-cycle out of phase with $E_y$. Coupling across a finite section requires the complete boundary-value solution.

For the free-slip plane layer, onset selects the smallest Rayleigh number across all admissible stationary and oscillatory normal modes. The first vertical mode wins because increasing vertical index raises viscous cost and reduces oscillatory admissibility. Oscillatory minimization replaces the right-hand side of the stationary wave-number equation by $P^2\mathrm{Ta}{\pi }^2/(1+P{)}^2$. For the oscillatory minimum also require positive angular frequency squared. Where both selected curves are admissible, at rapid rotation the ratio of the minimized thresholds is $R_{o,c}/R_{s,c}\sim 2P^{4/3}/(1+P{)}^{1/3}$. Equality gives $8P^4-P-1=0$, with positive root about $0.67660$; below it sufficiently rapid rotation can select oscillatory primary onset.

Positive steady solutions of the balanced thermocapillary thin-film equation with zero volume flux satisfy $H_{XX}+\ln H=0$ after choosing the thickness scale. Their first integral has potential $V(H)=H(\ln H-1)$, with minimum $V(1)=-1$ and limiting value $V(0^{+})=0$. At $E=-1$ the film is uniform; for $-1<E<0$ it is positive and periodic. At $E=0$ a limiting drop has peak $e$, half-width $\sqrt{\pi e}$, zero limiting contact angle and divergent curvature at zero thickness. Small periodic profiles have period $2\pi$ and approach the neutral, rather than fastest-growing, normal mode of the thermocapillary film instability.