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For a breathing displacement take ${\mathbf{u}}_i=q(t){\mathbf{r}}_i$. Every spring has equilibrium length $\sqrt{2}$ in the given length units, so its linear extension is $\sqrt{2}q$. At the vertex ${\mathbf{r}}_i$, the four adjacent equilibrium vertices sum to zero. Therefore the spring force there isIt remains radial and proportional to the original displacement at every vertex, proving that this symmetric displacement is a normal mode. Newton's second law givesThe first is angular frequency and the second is cycles per unit time. For $k$ in $\mathrm{N}{\mathrm{m}}^{-1}$ and $m$ in $\mathrm{kg}$, these have units $\mathrm{rad}{\mathrm{s}}^{-1}$ and $\mathrm{Hz}$ respectively. An energy check gives $T=3m{\dot{q}}^2$ and $V=12k{q}^2$, leading to the same ratio ${\omega }_b^2=24k/(6m)$.

The initial displacement lies in an invariant subspace of axial motions. Let the three opposite vertex pairs have positionsFor a vertex on the positive $x$ axis, the two springs towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,The matrix is $I+11^{\mathsf{T}}$. It has eigenvalue $4$ on the breathing vector $(1,1,1)$, and eigenvalue $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the axial normal modes of an octahedral spring framework without a large characteristic determinant.

Put ${\omega }_0=\sqrt{k/m}$ and decompose the initial vector asSince all initial velocities are zero, only cosine time factors occur. The complete motion to first order in the displacements iswith the vertex positions given by the first displayed formula. No translation or rotation is excited: the initial displacements are orthogonal to those zero modes and their initial velocities vanish. The two checks ${ϵ}_1={ϵ}_2$ and ${ϵ}_1+2{ϵ}_2=0$ respectively isolate pure breathing and pure traceless axial oscillation. Nonlinear corrections beyond the small-displacement approximation have been omitted.

For constant force $\mathbf{F}$ and displacement $\mathbf{s}$, the work isFor a variable force, $W={\int }_C\mathbf{F}\cdot d\mathbf{r}$. The work-energy theorem explains this choice: Newton's second law for fixed mass gives $\mathbf{F}\cdot d\mathbf{r}=m\mathbf{a}\cdot \mathbf{v}dt=d(m{v}^2/2)$. The component along the motion changes kinetic energy; a perpendicular force can change direction without doing work. Exerting a force on a stationary object likewise does no work on it.

Walking on level ground still requires muscular work. The centre of mass rises and falls during each step; the limbs gain and lose kinetic energy; and muscles perform positive work that replaces energy dissipated in impacts and tissue motion. Air resistance can add another cost. Returning to the same height makes the net change of gravitational potential energy zero, not every intermediate work contribution.

At ideal no-slip contact with an immovable road, the contact point is instantaneously stationary, so the road's force does no work through that point. The muscles convert internal chemical energy into motion. Zero net gain of height is compatible with positive muscular work and energy expenditure.

Take downward displacement as positive and gravitational acceleration $g$ as constant. In a vacuum, $mdv/dt=mg$, so with initial downward velocity $u$,For release from rest these become $v=gt$ and $s=g{t}^2/2$.

A raindrop encounters an increasing upward drag force as its speed grows. At terminal velocity, gravitational weight is balanced by drag force and the small buoyancy contribution:The drop remains in motion: zero acceleration does not imply zero velocity. In a quadratic drag force model, $F_{\mathrm{drag}}={\rho }_{\mathrm{air}}{C}_{D}A{v}^2/2$, giving $v_t=\sqrt{2(m-{\rho }_{\mathrm{air}}V)g/({\rho }_{\mathrm{air}}{C}_{D}A)}$. That formula assumes a fixed effective area and drag coefficient; sufficiently small drops or deforming large drops need an appropriate model. The question's “effective force” at settled terminal velocity is the approximately zero resultant force, not its nonzero gravitational weight.

Pascal's law concerns a pressure increment imposed on a confined fluid at rest. A water-filled vessel with small openings in differently oriented walls and a close-fitting piston illustrates it: pressing the piston produces jets in different directions, showing that pressure acts normally to each local surface. Compare ports at the same height to remove differences of hydrostatic pressure. Pressure gauges at equally high ports show the same imposed increment after the disturbance settles.

For an ideal hydraulic press with piston areas $A_1,A_2$,Conservation of fluid volume gives $A_1{s}_1=A_2{s}_2$, hence $F_1{s}_1=F_2{s}_2$: the larger output force accompanies a smaller displacement and does not create energy.

The arterial pulse is not simultaneous throughout the body. Pascal's law is an equilibrium statement, not instantaneous transmission. An arterial pulse travels through coupled blood and elastic arterial walls. Synchronized recordings at two sites show different arrival times of corresponding waveform features. For effective propagation distance $L$, pulse wave velocity is $c_p=L/\Delta t$. It differs from the transport velocity of individual blood cells.^[ref]

A bench demonstration uses an elastic liquid-filled tube, two pressure sensors at known positions, and a short pressure pulse at one end. The delayed second recording tests finite pressure wave propagation independently of the eventual static pressure balance.

In tapping-mode atomic force microscopy, the cantilever is driven near a resonance and its oscillation amplitude is monitored. Tip-surface interaction changes that amplitude. A controller compares it with a chosen setpoint and moves the vertical piezoelectric actuator to reduce the error. The recorded actuator displacement supplies the topography channel. A smaller amplitude during approach commonly triggers withdrawal and an increased amplitude during separation triggers approach, with the actual feedback sign set by the instrument's operating convention. Correct tracking requires enough response bandwidth without destabilizing the loop.

There is no external horizontal force, so momentum conservation keeps the student–punt centre of mass fixed if the system starts at rest. Let the student's walking direction be positive. If the punt's displacement relative to the water is $X$, the student advances $X+L$, where $L=3\mathrm{m}$ is measured relative to the punt. Constancy of the centre of mass givesThe punt moves $1.04\mathrm{m}$ opposite to the student's walking direction; the student's net displacement relative to the water is $3+X\simeq 1.96\mathrm{m}$. These displacements are relative to the initial state, not distances travelled relative to the punt.

Under the same ideal assumptions, merely changing the walking speed or walking back more slowly cannot produce unlimited travel. Returning the student to the original place on the punt restores the original internal mass distribution and hence the punt's original position. A finite punt bounds the displacement available through internal rearrangement. Propulsion over large distances requires an external horizontal impulse, for example a push against the bed or momentum transferred to the water. Such an interaction changes the isolated, frictionless model rather than evading centre of mass conservation.

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.

In Fourier transform spectroscopy, a scanning Michelson interferometer measures intensity against optical path difference $\ell$ rather than selecting one narrow wavelength at a time. After subtraction of the constant background, an ideal stationary source gives an interferogramA cosine Fourier transform recovers the spectral power density $S(\nu )$ after instrumental calibration. A mirror displacement $x$ changes the round-trip path by $2x$; the zero-path-difference burst aligns the scan, and a reference laser can calibrate the displacement.

Finite maximum path difference $L$ multiplies the interferogram by a window and convolves the recovered radiation spectrum with that window's Fourier transform. Thus $\delta \nu$ is of order $c/L$; precise factors depend on scan range and resolution convention. Uniform path samples of separation $\Delta \ell$ need ${\nu }_{\max }<c/(2\Delta \ell )$ to avoid aliasing without bandpass information. interferogram apodization reduces sidelobes at the price of broader lines. The broad accepted optical throughput can be advantageous; the multiplex signal-to-noise advantage depends on detector noise and is not universal in photon-noise-limited measurements. Source drift during a scan, finite beam divergence, imperfect sampling and wave phase errors require control. The finite-window reasoning is the delay-range resolution in Fourier transform spectroscopy.^[ref]

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

Take $m,k,b>0$, and define ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. Substitution of $e^{rt}$ into the unforced equation gives $r^2+2\gamma r+{\omega }_0^2=0$. The roots determine whether the return is oscillatory.

For return after a displacement in this ideal model, critical damping, case (b), is the usual comfort choice: it removes repeated bouncing while avoiding the slow relaxation mode of strong overdamping. For release from rest it gives a monotone return with the fastest asymptotic decay available without oscillation. This criterion does not prove optimal comfort for every road profile or initial velocity; actual comfort also depends on acceleration, road forcing and suspension travel.

For the forced response, set $\Omega =2\pi u/d$ and use $x_{\mathrm{ss}}=\mathrm{Re}(\hat{x}{e}^{i\Omega t})$. ThenThe general forced solution adds the appropriate homogeneous transient from (a), (b) or (c). Initial data were not specified, so its constants cannot be fixed. Positive damping removes this transient at long times. This solution uses the particular driving force in the supplied equation; it does not silently replace it by a different moving-support model.

Maximising $X$ is minimising its squared denominator $D$. Differentiation givesA nonzero displacement resonance exists only if $b^2<2mk$. Here ${\Omega }_r^2=0.175{\mathrm{s}}^{-2}$, and thereforeThe low-frequency limit is $X(0)=A$, whereas $X\to 0$ at large $\Omega$. The parameters give damping ratio $\zeta =b/(2\sqrt{mk})=0.456$, consistent with an underdamped system and the existence of this amplitude maximum.

A Michelson interferometer divides a source beam at a partially reflecting beam splitter. The two perpendicular arms terminate in mirrors whose reflected beams return to the splitter and overlap at the detector. The recombined fields must have overlapping wavefronts and compatible polarization of waves. Moving one mirror by $\Delta L$ changes the round-trip optical path difference by $2\Delta L$, so $N$ passing fringes determine $\Delta L=N\lambda /2$. Conversely a known mirror displacement measures wavelength. Scanning the delay can determine temporal optical coherence and, by Fourier transform spectroscopy, spectral structure. A compensating plate may equalize glass traversal, particularly for broadband illumination.

For equal real peak field amplitudes, the superposition isTime averaging the square over many optical cycles gives $⟨E^2⟩=2a^2{\cos }^2(\varphi /2)=a^2(1+\cos \varphi )$. With $\varphi =kx$ after absorbing a constant port-dependent phase into the zero of $x$,Here $I_0$ is the sum of the two separate-arm intensities; maximum intensity is $2I_0$ and the ideal minimum is zero. Equal amplitudes give unit interference visibility.

For a stationary source with a flat intensity spectrum, different spectral components add in time-averaged intensity. Keeping their individual two-arm interference terms and defining $\mathrm{sinc}z=\sin z/z$ givesOne must not coherently add unrelated source frequencies as if they had a fixed common phase. The sinc envelope is the Fourier transform of the rectangular spectral intensity.

Numerically, ${\nu }_0=5.00\times 10^{14}\mathrm{Hz}$ and $\Delta \nu =1.00\times 10^{13}\mathrm{Hz}$. The annotated interferogram usesIt is even in $x$, has central value 2 and background 1, and its carrier-fringe period in optical path difference is $c/{\nu }_0=0.600\mu \mathrm{m}$. The first envelope zeros are at $x=\pm c/\Delta \nu =\pm 30.0\mu \mathrm{m}$, corresponding to mirror shifts of $\pm 15.0\mu \mathrm{m}$ from equal paths. Side lobes persist beyond the zeros; a negative sinc reverses fringe phase. The intensity remains between $1-|\mathrm{sinc}|$ and $1+|\mathrm{sinc}|$, not between the signed sinc curves with an assumed fixed ordering.

Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

For a small longitudinal disturbance, let the equilibrium pressure and mass density be $p_0,{\rho }_0$, their perturbations be $p^{\prime },{\rho }^{\prime }$, and the fluid velocity be $v$. Linearizing conservation of mass and the inviscid momentum equation in one dimension givesWhen compression is adiabatic, the equation of state gives $p^{\prime }=c_s^2{\rho }^{\prime }$ with $c_s^2=(\partial p/\partial \rho {)}_s$. Differentiating and eliminating $v$ yieldsfor an ideal gas with heat-capacity ratio $\gamma =C_p/C_v$ and molar mass $M_{\mathrm{mol}}$. Compression raises temperature as well as density, explaining the factor $\gamma$ compared with the isothermal value $\sqrt{p_0/{\rho }_0}$.

A rightgoing plane sound wave has pressure and particle velocity in phase, $p^{\prime }={\rho }_0{c}_{s}v$, and mean intensity $⟨p^{\prime }v⟩={\hat{p}}^2/(2{\rho }_0{c}_s)={\rho }_0{c}_s{\hat{v}}^2/2$ for peak amplitudes. Particle displacement is a quarter-cycle out of phase with particle velocity: particles oscillate locally while the compression pattern transports energy. This continuum linear description requires wavelength much larger than the mean free path, small pressure perturbations and negligible viscous attenuation. Adiabatic propagation requires heat exchange across the compressed region to be slow on the oscillation time; strong thermal coupling would instead favour an isothermal response.

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

For propagation along $z$, a transverse wave has two perpendicular displacement components. With $\varphi =kz-\omega t$, write $u_x=A\cos \varphi$, $u_y=B\cos (\varphi +\delta )$. Their trajectory at a fixed point obeysThe general nondegenerate trajectory is an ellipse: elliptical polarization. For $\delta =0$ or $\pi$, the components remain in phase or antiphase and the trajectory collapses to a line. A single nonzero component is also linearly polarized. Equal amplitudes $A=B$ with $\delta =\pm \pi /2$ give circular polarization. Unequal quadrature amplitudes give an ellipse aligned with the chosen axes, while general phase differences rotate its axes.

For the displayed convention and $\delta =+\pi /2$, at $z=0$ the vector starts at $+x$ and moves towards $+y$ as time increases. Reversing $\delta$ reverses its rotation. Right- and left-handed labels require a stated viewing and phase convention, so rotation is defined explicitly here. A fixed ellipse is a coherent relationship between transverse components; unpolarized light instead has fluctuating orientation and phase relationships on the measurement timescale.

A linear polarizer transmits a linearly polarized beam with intensity proportional to ${\cos }^2\theta$, allowing its direction to be measured by an analyser. circular polarization transmits equal intensity through any linear analyser, but that test alone does not distinguish it from unpolarized light. A quarter-wave plate converts a suitable coherent circular state to linear polarization, allowing the combination to discriminate them. For light the electric field defines the usual polarization; the magnetic field remains perpendicular to it and to propagation. Longitudinal sound in a homogeneous gas lacks the two transverse degrees of freedom needed for these polarization states.

Let $x$ be mirror displacement from equal optical paths in a Michelson interferometer, so the optical path difference is $2x$. Different spectral components add detected light intensities; each component interferes with its own delayed copy. For equal line strengths and ideally balanced arms, writeThe trigonometric identity for the sum of cosines givesThus the sketch consists of rapid fringes within a slowly varying contrast envelope. With $\bar{\lambda }$ the mean wavelength and $|\Delta \lambda |\ll \bar{\lambda }$, the mirror-displacement fringe spacing is $\bar{\lambda }/2$. The interference visibility isThe first contrast zero is at $x=1/(4\Delta s)$; successive visibility maxima are separated byBetween successive contrast zeros the signed envelope changes sign, shifting the rapid fringes by half a cycle. Unequal line strengths leave a nonzero minimum contrast. Finite linewidth adds a longer-range coherence envelope, while arm imbalance and polarization mismatch can reduce contrast without indicating a different spectral spacing.

Put ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. The characteristic roots are $-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. For release from rest at displacement $x_0>0$, the three regimes areunderdamping produces decaying oscillations; critical damping is the repeated-root boundary; overdamping has two real decay rates. The critical and overdamped curves from rest approach equilibrium without crossing it. All three initially have zero slope and negative acceleration. A sufficiently strong initial kick can cause a single crossing even without an oscillatory mode.

For the critical oscillator with initial velocity $-v_0$, the repeated-root solution isIts exponential is positive at finite time. Therefore the bracket has a positive-time zero precisely whenEquality gives a pure positive exponential and no finite-time crossing.

For the specified numerical oscillator, ${\omega }_0=1{\mathrm{s}}^{-1}$ and $\gamma =2/\sqrt{3}{\mathrm{s}}^{-1}>1{\mathrm{s}}^{-1}$, verifying overdamping. Its two roots are $-1/\sqrt{3}$ and $-\sqrt{3}$ in inverse seconds. The initial conditions givewith $t$ in seconds. The slow component has a negative coefficient; the fast positive component initially dominates.

The values for the requested plot areThe curve crosses near $0.56\mathrm{s}$ and then returns toward zero from below. The exact expression provides a check on the graphical estimate:

Let the first transmitted complex amplitude be $T{E}_{\mathrm{in}}{e}^{ikd}$: two amplitude transmission factors each of magnitude $\sqrt{T}$ give magnitude $T$. Each extra internal round trip adds two amplitude reflections, with combined magnitude $R$, and phase $\delta$. With the mirror phase convention already incorporated in the stipulated $\delta$,Taking the squared modulus givesIt is $R$, not $R^2$, that multiplies the field after a round trip; $R^2$ would be the corresponding round-trip intensity fraction.

At resonance $\delta =2m\pi$, $B_{\max }=T^2/(1-R{)}^2$. If $h$ is the positive phase displacement to half maximum,For $R=0.80$, the stipulated high-reflectivity approximation gives $h=0.2236\mathrm{rad}$, while the exact value is $0.2241\mathrm{rad}$. The full width is twice this value. With $T=0.20$, $B_{\max }=1$, $B_{\min }=T^2/(1+R{)}^2=1/81$ and the pattern repeats every $2\pi$. Its ideal finesse is approximately $\pi \sqrt{R}/(1-R)=14.05$.

A wavelength component resonates at $d=m\lambda /2$. Within the scan interval, the individual resonances areNo neighboring orders of these lines fall within the stated interval. The observed signal is the sum of their two Airy transmissions, since different spectral components add intensities. Individual resonance positions identify the two peaks to the plotted precision; the tails cause very small shifts of the maxima of the sum.

The gap-coordinate full width isThe separation is $0.100\mu \mathrm{m}$, about $2.8$ full widths, so the two components can be resolved by the ideal etalon. The equivalent wavelength resolution near order $1000$ is about $\lambda /(m\mathcal{F})\simeq 0.071\mathrm{nm}$, smaller than their $0.2\mathrm{nm}$ separation.

The added wavelength has an order-$999$ resonance atIt lies between the preceding two peaks; its adjacent orders are outside the scan. Its input strength was not specified, so the third peak's height cannot be uniquely assigned. The original plot uses equal strength only as a declared illustration.

This is an order-identification problem: the free spectral range near $d=0.5\mathrm{mm}$ is $\Delta {\lambda }_{\mathrm{FSR}}\simeq {\lambda }^2/(2d)\simeq 1\mathrm{nm}$. A line more than one free spectral range away can appear among peaks of another order. A lower-resolution prefilter or additional dispersive spectrometer is needed to identify the order; a narrow peak alone does not identify an absolute wavelength. mirror parallelism, beam collimation and detector angular acceptance must also be controlled before attributing a measured width entirely to reflectivity.^[ref]

The reciprocal lattice consists of all wavevectors $\mathbf{G}$ whose plane-wave phase is unchanged under every direct-lattice translation $\mathbf{R}$:It expresses periodicity in wavevector space, with dimensions of inverse length.

Choose incident field convention $e^{i{\mathbf{k}}_i\cdot \mathbf{r}-i\omega t}$. The wave at the second atom acquires incident phase ${\mathbf{k}}_i\cdot {\mathbf{r}}_n$ relative to the atom at the origin. Its path to a distant detector is shorter by approximately ${\hat{\mathbf{k}}}_f\cdot {\mathbf{r}}_n$, contributing outgoing phase $-{\mathbf{k}}_f\cdot {\mathbf{r}}_n$. ThereforeNeglecting variation of detector distance across the specimen, identical atoms have a common scattering amplitude factor andThe original geometry diagram shows the incident direction, outgoing direction and direct-space displacement separately; ${\mathbf{k}}_s$ is a wavevector difference, not an extra ray path.

The primitive reciprocal vectors obeywith cyclic versions for $\mathbf{B},\mathbf{C}$. If ${\mathbf{k}}_s=h\mathbf{A}+k\mathbf{B}+l\mathbf{C}$, every lattice term has phase $2\pi$ times an integer, so all terms add constructively. Conversely, full translational reinforcement requires ${\mathbf{k}}_s\cdot \mathbf{a}$, ${\mathbf{k}}_s\cdot \mathbf{b}$ and ${\mathbf{k}}_s\cdot \mathbf{c}$ to be integer multiples of $2\pi$, hence ${\mathbf{k}}_s$ is reciprocal. These are the Laue equations. A finite crystal gives finite-width maxima rather than ideal reciprocal-space delta peaks; elastic scattering must also satisfy $|{\mathbf{k}}_i|=|{\mathbf{k}}_f|$.

For the two-dimensional Bravais lattice interpretation, with one atom per primitive cell, takeSolving the duality conditions givessoThe reciprocal primitive vectors have equal length $4\pi /(\sqrt{3}a)$ and angle $120^{\circ }$; the array itself is triangular. A different primitive basis can use a $60^{\circ }$ angle without changing the lattice. The direct primitive-cell area is $\sqrt{3}{a}^2/2$ and its reciprocal-cell area is $(2\pi {)}^2$ divided by that area.

For nonzero integer pairs, $h^2-hk+k^2$ has minimum one, attained for six shortest vectors. Therefore the first nonforward maximum has $|{\mathbf{k}}_s|=4\pi /(\sqrt{3}a)$. If the forward transmitted maximum is included literally, the smallest value is $|{\mathbf{k}}_s|=0$; the nonzero value is the first diffraction shell. Reaching that shell in elastic scattering additionally requires $4\pi /(\sqrt{3}a)\le 2|{\mathbf{k}}_i|$.