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With equal masses m, the two mass-weighted Hessians areFor a uniform displacement vector $t=(1,1)$,Therefore uniform translation is not a normal mode in orientation (1) when $k_2\ne 0$: anchoring only one atom mixes translation with internal stretching. It is an exact normal mode displacement in orientation (2) because both atoms have identical restoring terms. Its frequency is nevertheless nonzero, ${\omega }_T=\sqrt{k_3/m}$; the antisymmetric stretch has ${\omega }_S=\sqrt{(2k_1+k_3)/m}$. The rigid surface breaks free translational invariance in both orientations. A vector resembling a uniform translation need not be a zero-frequency mode when the molecule is anchored.

When $|eV|$ reaches a vibrational quantum, a tunnelling electron can lose energy by exciting that vibration. Opening an inelastic channel changes $dI/dV$ at both voltage polarities. If the conductance increases in each direction, $d^{2}I/d{V}^2$ has a positive feature at positive threshold and a negative feature at the corresponding negative threshold. Approximate antisymmetry follows from approximate symmetry of the conductance; unequal tip and sample electronic structures can make amplitudes unequal.

The three energy scales are 8, 48 and 375 meV, corresponding to about 65, 387 and 3025 cm${}^{-1}$. A plausible assignment is a low-energy frustrated lateral translation or rotation, an adsorbate–metal stretch or hindered molecular mode at 48 meV, and a C–H stretching vibration at 375 meV. The terminal carbon of methyl isocyanide supplies a plausible adsorption contact, but exact mode assignment requires calculated normal modes or independent vibrational measurements. In particular the 48 meV band should not be labelled a C–N stretching vibration simply because the molecule contains that bond: its frequency is far too low for the usual internal stretching mode.

For a positive gyromagnetic ratio the magnetic moment is parallel to the spin angular momentum; for a negative ratio it is antiparallel. The Hamiltonian ${\hat{H}}_Z=-\gamma B_0{\hat{I}}_z$ gives $E_m=-\gamma \hslash B_{0}m$. Thus $m=+1/2$ is lower for the proton, whereas $m=-1/2$ is lower for nitrogen-15. Their spin precession senses are opposite for a common positive field convention; measurable absorption frequencies and energy separations remain positive.

For an exponentially correlated local field, a single-frequency illustrative relaxation rate is $T_1^{-1}=C{\tau }_c/(1+{\omega }_0^2{\tau }_c^2)$. Its derivative with respect to ${\tau }_c$ changes sign at ${\omega }_0{\tau }_c=1$. Slow motion gives $T_1\propto {\tau }_c$, and fast motion gives $T_1\propto 1/{\tau }_c$; an Arrhenius plot consequently has a minimum of $T_1$ between opposite-slope limits.

For identical spin-one-half nuclei with homonuclear dipolar relaxation, include both transition frequencies:Writing $x={\omega }_0{\tau }_c$, the maximum of the rate satisfiesLet $\ell$ be jump length, ${\tau }_j$ the total waiting time, ${\tau }_c=c{\tau }_j$, and $f_D$ the diffusion jump correlation factor. In three dimensions $D=f_D{\ell }^2/(6{\tau }_j)=f_{D}c{\ell }^2/(6{\tau }_c)$. Combining this with $D=D_0\exp [-E_D/(k_{B}T)]$ givesThe numerical constant is one for the single-frequency illustration and 0.616 for the homonuclear dipolar model. Assumptions are a single activated process, exponential field correlations, a temperature-independent prefactor and coupling amplitude, and an established relation between local correlation and translational motion. Defect creation can enter the measured activation energy; it cannot automatically be identified with a bare migration barrier. The fluctuation-spectrum approach originates in the relaxation theory of Nicolaas Bloembergen, Edward Mills Purcell and Robert Pound.^[ref]

Here $\mathbf{j}$ is electric current density, $n$ the number density of independently mobile ions, $Q$ their charge, $T$ absolute temperature and $k_B$ the Boltzmann constant. The sum runs over all nearest-neighbour destinations $j$ of site $i$, and ${\mathbf{r}}_{ij}$ is the displacement from $i$ to $j$. The applied field is $\mathcal{E}$, ${\omega }_0$ is an attempt frequency, and $\Delta E_{ij}$ is the zero-field reaction energy barrier for the indicated jump. With the normalization of the supplied equation,It has units s${}^{-1}$; it is not an equilibrium occupation probability. The current formula is the linear-response result of a small field-induced asymmetry in forward and reverse rates, requiring $|Q{\mathbf{r}}_{ij}\cdot \mathcal{E}|\ll k_{B}T$. Use its explicit factor $1/2$ consistently rather than absorbing it into a redefined jump rate.

At fixed number density, attempt frequency, nearest-neighbour distance and barrier, the conductivities have ratio ${\sigma }_{\mathrm{sc}}:{\sigma }_{\mathrm{bcc}}:{\sigma }_{\mathrm{fcc}}=1:4/3:2$. Thus the face-centred cubic array is largest. Numerically,The small value is consistent with the low supplied carrier number density and strongly activated hopping. This comparison holds $n$ fixed rather than comparing fully occupied crystal structures, whose site densities at the same neighbour distance differ.

Use Gaussian units, plane waves proportional to $e^{i\mathbf{k}\cdot \mathbf{r}-i\omega t}$, and the signed cyclotron frequency ${\Omega }_s=q_s{B}_0/(m_{s}c)$. Write ${\Omega }_e=e{B}_0/(m_{e}c)>0$ for its electron magnitude, so the signed electron value is $-{\Omega }_e$. Define the species plasma frequency by ${\omega }_{ps}^2=4\pi n_{0s}{q}_s^2/m_s$. The linearization of the cold Lorentz force equations givesSolving this linear system and using the species current density ${\mathbf{j}}_s=n_{0s}{q}_s{\mathbf{u}}_s$ gives the conductivity tensorThe magnetized cold-plasma dielectric tensor is $\mathbf{ϵ}=\mathbf{I}+(4\pi i/\omega ){\sum }_s{\mathbf{\sigma }}_s$. ThusAt the stated electron frequency scale, the ion contribution is smaller by $O(Z{m}_e/m_i)$ for $q_i=Ze$ and quasineutrality. To this accuracy,These expressions require negligible collisions and the cold ordering $k{v}_{\mathrm{th},s}\ll |\omega |$; sufficiently close to a cyclotron resonance, thermal broadening or collisions must also be retained.

Assume the stated outward-decreasing pressure, $L_p>0$, and setFor $|k_{\parallel }|<k_c$, the interchange instability hasAn outward-displaced flux tube carries higher pressure than its new surroundings. In the unfavorable magnetic field-line curvature of the Z-pinch, the resulting imbalance drives it farther outward. Parallel variation bends field lines, and magnetic tension supplies the stabilizing $k_{\parallel }^2{v}_A^2$ term. The fastest local interchange is at $k_{\parallel }=0$, with $\gamma ={\omega }_I$; this endpoint is understood as the limit of the stated Alfvénic ordering. There is no leading $k_y$ dependence once $k_y\gg k_{\parallel }$.

For $|k_{\parallel }|>k_c$, the two real frequencies are $\omega =\pm \sqrt{k_{\parallel }^2{v}_A^2-{\omega }_I^2}$: the displacement oscillates as a curvature-modified Alfvén wave. At $|k_{\parallel }|=k_c$, the leading equation is ${\ddot{\xi }}_x=0$, so the marginal solutions are a constant and a term linear in time; neglected corrections determine finer behavior. For a favorable reversed pressure gradient, $L_p<0$, this drive is stabilizing.

The leading dispersion relation is the same in ideal magnetohydrodynamics. Its adiabatic law gives $\delta p=-{\xi }_x{p}_0^{\prime }-\Gamma p_0\nabla \cdot \mathbf{\xi }$; the compressive part is smaller than the steep-gradient advection by $O(L_p/R)$ or $O(\beta )$, giving the same leading $\delta p=-{\xi }_x{p}_0^{\prime }$. Relaxing those orderings introduces compression, field-strength and field-length changes, and pressure anisotropy; the drift-kinetic equation can then introduce parallel kinetic phase mixing and Landau damping. At larger plasma beta, coupling to compressive branches matters; if pressure anisotropy develops, mirror instability and firehose instability may also become relevant. These are possibilities beyond the derived leading equation, not additional roots already established here.

The Debye model treats the crystal as an isotropic elastic continuum with independent harmonic vibrational modes. There are three acoustic polarizations, approximated by linear dispersions and a common effective speed. A cutoff is chosen to retain $3N$ modes for $N$ atoms. The model neglects anharmonic interactions and the detailed high-wavevector dispersion. In a multi-atom primitive cell, optical branches require additional treatment; copper has one atom per primitive cell even though its conventional cubic cell contains four atoms.

A frequency interval contains, per volume,Each phonon mode has Bose-Einstein distribution occupation $(e^{\hslash \omega /(k_{B}T)}-1{)}^{-1}$. The fixed zero-point energy contributes no heat capacity. Its thermal energy density isFor $T\ll {\theta }_D$, extending the upper limit to infinity gives $u_{\mathrm{ph}}=(3{\pi }^4/5)n{k}_B{T}^4/{\theta }_D^3$. ThereforeThe cubic law reflects the three-dimensional acoustic mode density proportional to ${\omega }^2$. It is a low-temperature limit, not a valid extrapolation through the classical high-temperature regime.

A phonon is one quantum of a collective harmonic normal mode of a crystal. A mode labelled by wavevector and branch has energies $E_m=\hslash \omega (m+1/2)$, with bosonic occupation statistics. The crystal momentum $\hslash \mathbf{k}$ is defined modulo a reciprocal lattice vector; it is not generally the mechanical momentum of one atom. The phonon dispersion relation assigns its frequency and polarization to the wavevector and branch.

Inelastic neutron scattering can measure this relation in a single crystal. Determine the incident neutron energy and wavevector, scatter at a chosen angle, and analyze the outgoing energy by time of flight or a spectrometer analyzer. Energy and crystal-momentum conservation giveThe signs distinguish phonon creation and annihilation with consistent momentum convention. Repeating measurements at different momentum transfers maps peaks in the dynamic scattering response into phonon branches. The measured intensity depends on occupation, scattering factors and the projection of the mode polarization along the momentum transfer; a weak or missing peak need not mean that the corresponding mode does not exist.

Let $u_p$ be the longitudinal displacement of atom $p$. Translational invariance and a symmetric harmonic pair force give the long-range harmonic-chain dispersion equationsUse a traveling normal mode $u_p=u_0{e}^{i(kpa-\omega t)}$. Dividing by its nonzero amplitude givesHencePositive pair constants give a nonnegative squared frequency and a uniform-translation zero mode. More general signed force constants require checking stability of the entire sum. The frequency is even in $k$ and periodic with $2\pi /a$, reflecting reciprocal-lattice equivalence. The force sum must converge in the model. A finite ordinary acoustic slope additionally requires convergence of $\sum n^2{C}_n$, in which case $v_s=a\sqrt{\sum n^2{C}_n/M}$. Both finite-range examples below satisfy this condition.

Write $q=ka$ and use $2\sqrt{C_1/M}$ as the frequency unit. The nearest-neighbour chain and its next-nearest-neighbour extension haveThese are each a single acoustic branch, repeated over reciprocal-lattice periods. There is no added optical branch merely because a longer-range spring is introduced. The cutoff $C_n=0$ for $|n|>1$ specifies the first chain as nearest-neighbour despite the source's repeated word in its description.

For small $|q|$, ${\omega }_1\simeq a\sqrt{C_1/M}|k|$ and ${\omega }_2\simeq a\sqrt{(C_1+4C_2)/M}|k|$. ThusThe maximum is not necessarily at the zone edge. Put $x={\sin }^2(q/2)$, so $0\le x\le 1$. The second squared dimensionless frequency isIt reaches its maximum at $x=3/4$, or $q=\pm 2\pi /3$ in the first zone. The first chain peaks at $q=\pm \pi$. ThereforeAt $q=\pm \pi$ both chains have the same frequency, but the second chain has already passed its interior maximum. The next-nearest-neighbour monatomic-chain frequency maximum is therefore crucial to a correct sketch. Periodic repetition gives the requested curves for all wavevectors.

The mass per lead atom isLet $u_n$ be its longitudinal displacement. Newton's second law gives the harmonic chain equationInsert the normal mode $u_n=u{e}^{i(kna-\omega t)}$ to obtainAt long wavelength, $\omega \simeq a\sqrt{\kappa /m}|k|$. The sound speed is the corresponding low-frequency group velocity:The relevant inertia in this chain equation is one atomic mass. The reduced mass $m/2$ belongs to the relative vibration of an isolated pair and should not be substituted for $m$ in the chain's dispersion relation.

Quantum statistics matter when thermal energy is comparable with a characteristic vibrational level spacing. The chain's natural angular frequency is ${\omega }_0=\sqrt{\kappa /m}=5.7471\times 10^{12}{\mathrm{s}}^{-1}$, givingFactors of order unity depend on which vibrational scale is used. The largest chain phonon angular frequency is $2{\omega }_0$, corresponding to $87.8\mathrm{K}$. An isolated two-atom bond has reduced mass $m/2$ and angular frequency $\sqrt{2}{\omega }_0$, corresponding to $62.1\mathrm{K}$. Thus quantum corrections become appreciable on the scale of several tens of kelvin, and all modes are classical only when $T$ is comfortably above the largest relevant vibrational scale. Below it, high-frequency phonons are progressively frozen out and equipartition fails.

For the isolated bond, the zero-temperature root-mean-square relative displacement isIt is small enough for a local bond expansion. The classical weakly anharmonic estimate in part d therefore has a useful parametric windowwhere the lower condition means comfortably above the vibrational scale, rather than a sharp threshold. The model's upper bound concerns local anharmonicity and binding, not a prediction of the melting temperature.

Let $\mathcal{R}=|r{|}^2$ be the single-mirror power reflectivity and ${\theta }_g$ the propagation angle inside the gas. Define $\lambda$ as the vacuum wavelength. Successive transmitted complex amplitudes differ by a round-trip factor $\mathcal{R}{e}^{i\delta }$, where, for real dielectric reflection coefficients whose product has positive sign,The cosine arises by comparing successive emerging rays on the same outgoing wavefront; it is not the literal zigzag distance $2d/\cos {\theta }_g$. Equivalently, the round-trip phase is $2k_{z}d$. If the source's $\theta$ is the angle in external air, Snell's law gives $n\sin {\theta }_g=\sin \theta$ andIf $\theta$ denotes the internal angle, simply set ${\theta }_g=\theta$.

For identical lossless mirrors, their power transmission is $1-\mathcal{R}$. In an energy-normalized amplitude convention this is $|t{|}^2=1-\mathcal{R}$; more general interface amplitudes require the corresponding wave impedance factors. The geometric series of transmitted amplitudes givesHenceThis is Fabry–Pérot transmission from multiple reflections.^[ref]

At normal incidence, for nondispersive $n$ and zero net reflection phase modulo $2\pi$, the resonance condition is $\delta =2\pi m$. The resonance frequencies and free spectral range areFor finite $\mathcal{R}<1$ the resonances have finite width, and transmission away from them is nonzero. Thus “only discrete frequencies” is an ideal high-reflectivity description of the selected resonances, not an exact statement about the support of the transmitted spectrum. For example, $\mathcal{R}=0.9$ at $\delta =\pi$ still gives $I_t/I_0=(0.1/1.9{)}^2\simeq 0.00277$.

At half the unit peak, $Q{\sin }^2(\delta /2)=1$. Measuring the phase displacement from a resonance yieldsThe last expression is the high-reflectivity approximation, with finesse $\pi \sqrt{\mathcal{R}}/(1-\mathcal{R})$. At $\mathcal{R}=0.9$, the full phase width is about $0.2109$ radians and $\Delta {\nu }_{\mathrm{FWHM}}/\Delta {\nu }_{\mathrm{FSR}}\simeq 0.03357$. At $\mathcal{R}=0.1$, the minimum transmission is $0.6694$, so the trace never reaches half the peak and this full width at half maximum is not defined. The exact half-maximum condition requires $\mathcal{R}\ge 3-2\sqrt{2}$.

For the absorbing Fabry–Pérot mirror phase and transmission, write $r=\sqrt{\mathcal{R}}{e}^{i{\varphi }_r}$ and let $\mathcal{T}$ be the power transmission of one mirror. Optical absorption makes $\mathcal{R}+\mathcal{T}<1$. The transmitted intensity becomesHere the phase formula is at normal incidence. The peak is reduced below unity, while a change in $\mathcal{R}$ changes the width. The complex reflection phase also moves the resonances: if ${\varphi }_r$ is approximately constant,If either $n$ or ${\varphi }_r$ depends on frequency, this is an implicit condition and the free spectral range need not be constant. No numerical shift is determined without the mirrors' optical response.

Take the wind to blow toward positive $x$, so $U_0\ge 0$. In subsonic flow, both stationary observers record the source angular frequency $\omega$ and, for the imposed density boundary data, the same acoustic density perturbation amplitude ${\rho }_0|A|$. Their spatial records distinguish the wavelengths and directions:The downstream pattern moves at $c_0+U_0$ and the upstream pattern at signed velocity $U_0-c_0$. Thus the downstream observer sees longer wavelengths while the upstream observer sees shorter ones. The interval containing multiple wavelengths lets each measure the spatial spacing and its direction of travel. The arrival delays at $\pm L$ are $L/(c_0+U_0)$ and $L/(c_0-U_0)$ respectively; after arrival their temporal traces differ by these propagation phases, not by temporal frequency.

In supersonic flow, both characteristic curves have positive slopes $dx/dt=U_0\pm c_0$. With an initially undisturbed upstream region, the observer at $-L$ sees no source-generated disturbance. The observer at $+L$ can see two downstream travelling waves, withThe faster signal arrives first, at $L/(U_0+c_0)$; the slower one arrives at $L/(U_0-c_0)$. Once both are present, their different wavenumbers can produce spatial modulation, while the time dependence at a fixed position remains harmonic with angular frequency $\omega$.

There is a qualification to the source data: the single prescribed value of ${\rho }_1(0^{+},t)$ fixes only $A_{+}+A_{-}=A$. It does not determine the two complex amplitudes separately. A second source condition, such as its downstream acoustic particle velocity with amplitude $c_0(A_{+}-A_{-})$, or a specified mass/force source model, is needed. Consequently two downstream branches are allowed; the stated density data alone do not require both to have nonzero amplitude. Extending a formally upstream-going subsonic formula into $x<0$ when $U_0>c_0$ would impose a signal arriving from upstream infinity and violate the causal source interpretation.

As $U_0\to c_0$, one laboratory phase velocity tends to zero. Its wavelength tends to zero, its wavenumber diverges, and its arrival time diverges; a finite-time experiment never attains the assumed established upstream field at fixed $L$. Finite source size, viscosity, thermal losses and eventually molecular scales invalidate this singular inviscid flow limit. With effective acoustic diffusivity ${\nu }_a$, a loss estimate is ${\nu }_a{k}^2/\omega \sim {\nu }_a\omega /(U_0-c_0{)}^2$, which diverges. Moreover the finite-amplitude change in the characteristic speed is of order $c_0|A|$, so uniform linearization requires $|U_0-c_0|\gg c_0|A|$. At exactly sonic background speed, the zero-speed branch cannot support a nonzero-$\omega$ harmonic solution. Nonlinear acoustics and losses must resolve the near-sonic regime; an unlimited spatial compression of the ideal wave is not physical.

For the radial epicyclic frequency, introduce the effective potential at fixed specific angular momentum $l$: ${\Phi }_{\mathrm{eff}}=\Phi +l^2/(2r^2)$. A circular orbit at $r_c$ satisfies ${\Phi }^{\prime }(r_c)=l^2/r_c^3=r_c{\Omega }^2(r_c)$. A small radial displacement $\xi$ obeys $\ddot{\xi }=-{\Phi }_{\mathrm{eff}}^{\prime \prime }(r_c)\xi$. Differentiate at fixed $l$ before inserting the circular-orbit relation:For a flat rotation curve, $\Omega \propto r^{-1}$, so $\kappa =\sqrt{2}\Omega$.

In a perturbation proportional to $\exp [i(\int kdr+n\varphi -\omega t)]$, $n$ is the integer azimuthal mode number, giving $|n|$ arms for a nonaxisymmetric pattern. Its angular pattern speed is ${\Omega }_p=\omega /n$ for a real frequency and $n\ne 0$; $a_0$ is the unperturbed gas sound speed, or the specified effective fluid pressure-response speed. For growing patterns the angular phase speed uses $\mathrm{Re}\omega /n$. Axisymmetric modes have $n=0$: their pattern speed is undefined, and one must use $(\omega -n\Omega {)}^2\to {\omega }^2$, not set the pattern-speed expression to zero.

Define the Toomre parameter and Toomre wavenumber byThen $q=|k|/k_T$ givesFor $a_0>0$, the minimum is negative precisely when $Q<1$. Thus this condition is necessary and sufficient for an exponentially growing axisymmetric mode within the given local razor-thin fluid model; $Q=1$ is marginal. It is not a guarantee against global or nonaxisymmetric instabilities, and the collisionless stellar-disk criterion has a different pressure-response normalization.

The Toomre unstable band and fastest growth follow by minimizing ${\omega }^2$:Here the growing branch has $\omega =i{\gamma }_{\ast }$ under the stated Fourier convention. Both signs of the radial wave number have the same growth rate. The unstable band is $q_{-}<q<q_{+}$, where $q_{\pm }=2(1\pm \sqrt{1-Q^2})/Q^2$.

For the minimum Toomre parameter in an exponential flat-rotation disk with constant $a_0$, put $x=r/r_0$:The logarithmic derivative is $d\log Q/dx=1-1/x$, so the unique minimum occurs at $r=r_0$. Both $Q\to \infty$ as $r\downarrow 0$ and $Q\to \infty$ as $r\to \infty$. Hence inner and outer regions are stable within this idealized rotation law, even when an intermediate annulus is unstable. The whole disk is strictly stable in this local axisymmetric test ifEquality makes the ring at $r_0$ marginal. The central conclusion uses the assumed flat rotation curve all the way inward; real central rotation laws need not retain that limit.

Radial epicyclic motion. At fixed specific angular momentum, use ${\Phi }_{\mathrm{eff}}=\Phi +{\ell }^2/(2r^2)$ and $\ddot{r}=-{\Phi }_{\mathrm{eff}}^{\prime }$. A circular orbit at $r_c$ obeys ${\Phi }^{\prime }(r_c)={\ell }^2/r_c^3=r_c{\Omega }^2$. Expanding $r=r_c+\xi$ gives $\ddot{\xi }+{\kappa }^2\xi =0$, whereFor ${\kappa }^2>0$ the radial epicyclic frequency describes simple harmonic motion. The angular correction follows from conservation of angular momentum, $\dot{\varphi }\simeq \Omega (1-2\xi /r_c)$, and describes an epicyclic motion around the guiding circular orbit.

Stabilizing terms. In the given local fluid dispersion relation, ${\kappa }^2$ is rotational support, important at long wavelength; $a_0^2{k}^2$ is random-motion or pressure support, important at short wavelength. The term $-2\pi G\Sigma |k|$ is destabilizing self-gravity. Although the wording refers to stellar dispersion, this algebra uses the supplied fluid approximation; a collisionless stellar treatment has a reduction factor and a different numerical normalization.^[ref]

Axisymmetric threshold. For $n=0$, set $q=|k|\ge 0$. Completing the square givesThus $Q>1$ gives local axisymmetric oscillatory stability; $Q<1$ gives a growing band, withAt $Q=1$, the minimizing mode has zero frequency and is marginal, not a strictly restoring oscillation. Local axisymmetric stability does not establish stability to every global nonaxisymmetric mode.

long-wave and short-wave branches at marginal Toomre stability. For $Q=1$ and $n\ge 1$,The long-wavelength branch, using the minus sign, requiresThese endpoints are the inner and outer Lindblad resonance conditions. The stated dispersion relation does not restrict every wave solution to that interval. Its plus-sign short-wavelength branch has $q>0$ even outside it: for example, ${\Omega }_P=\Omega +2\kappa /n$ gives $q=3\kappa /a_0$ and satisfies the equation. Additional branch selection or resonant absorption physics would be needed for an absolute confinement claim. No such condition is supplied.

Toomre stability radius of a flat-rotation disk. With $\Omega =v_0/r$, $\kappa =\sqrt{2}{v}_0/r$. Using $G=4.3009\times 10^{-3}\mathrm{pc}(\mathrm{km}{\mathrm{s}}^{-1}{)}^2{M}_{\odot }^{-1}$,The supplied local model therefore predicts stable radii below about $4.19\mathrm{kpc}$, marginality at that radius, and axisymmetric instability farther out while its assumed constant surface density and thin-disk approximation remain applicable. An infinite constant-density, flat-rotation disk is an idealization, not a global self-consistent galaxy model.

Write $U=\hat{u}$, $V=\hat{v}$, and $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$. By the Sobolev space hypotheses, $A=⟨\xi {⟩}^{s}U$ and $B=⟨\xi {⟩}^{t}V$ belong to the L2 space. Their product is an integrable function by Cauchy-Schwarz inequality. Thus $UV=⟨\xi {⟩}^{-s-t}AB$ is a locally integrable function defining a tempered distribution, with estimateThe supremum is controlled by finitely many Schwartz seminorms. Define the convolution of Sobolev distributions byIt is independent of the representatives chosen for $U,V$, and the convolution theorem shows that it agrees with ordinary convolution for Schwartz functions. More generally, approximate $u,v$ by Schwartz functions in their respective Sobolev norms; the displayed estimate shows convergence of the products in tempered distributions, so their ordinary convolutions converge to this definition. For $u,v\in L^1$ that also satisfy the stated Sobolev space hypotheses, the ordinary convolution theorem gives the same result.

There need not be any real $r$ with $u\ast v\in H^r(\mathbb{R})$. ChooseHere $1_{(0,1)}$ is an indicator function. Since ${\int }_0^1{\xi }^{-2/3}d\xi =3$ and every weight $(1+{\xi }^2{)}^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every Sobolev space. But their convolution has Fourier transform $UV={\xi }^{-2/3}{1}_{(0,1)}$, and for every real $r$,Thus $\boxed{u,v\in \bigcap _{q\in \mathbb{R}}{H}^q(\mathbb{R})\text{but}u\ast v\notin \bigcup _{r\in \mathbb{R}}{H}^r(\mathbb{R})}$. The obstruction is failure of local square integrability at zero; a weight controlling large frequencies cannot fix it.

In scanning tunneling microscopy, $I\propto e^{-2\kappa z}{\int }_{E_F}^{E_F+eV}{\rho }_s(\mathbf{r},E)dE$ under simplified constant-tip-density assumptions, with $\kappa \simeq \sqrt{2m\varphi }/\hslash$. A constant-current height map mixes actual surface height with local electronic density, barrier/work-function changes and tip structure. Changing bias can change contrast without moving atoms. Scanning tunneling spectroscopy instead records energy dependence, often through $dI/dV$; tip states, matrix elements and thermal or modulation broadening still affect its interpretation.

In atomic force microscopy, contrast comes from tip-sample forces or their gradients: short-range chemical bonding and repulsion, dispersion forces, electrostatics and sometimes magnetism or capillary forces. Contact deflection, oscillation amplitude and frequency shift are different observables. A force-gradient image can therefore differ from a geometrical-height image, and changing the tip termination can strongly change atomic contrast.

STM's exponential distance sensitivity can give excellent vertical sensitivity; usable resolution also depends on current noise, feedback, vibration and drift. Lateral contrast depends on the apex orbitals and electronic spatial extent, not merely the nominal tip radius. In AFM, broad-tip geometrical convolution broadens larger features; a sharp terminal atom and short-range forces can produce much finer atomic contrast. Cantilever stiffness, amplitude, frequency noise, force range, thermal fluctuations and feedback influence height sensitivity. In both techniques, detecting a small displacement is distinct from resolving two neighboring objects.

An example is the Cu(111) surface-state standing-wave experiment. Low-temperature STM and spatially resolved spectroscopy connect atomic steps or defects in a surface map with surrounding energy-dependent electronic interference patterns. Their period changes with bias, whereas a fixed structural corrugation would not acquire that electronic dispersion. The result shows why spectroscopic maps reveal surface-electron scattering beyond what a height image alone identifies.^[ref] The apparent fringes are not a second set of equally spaced surface atoms.