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One mass per cell. If $u_j$ is the displacement from equilibrium, the two springs give $m{\ddot{u}}_j=\alpha (u_{j+1}+u_{j-1}-2u_j)$. Substituting $u_j=U{e}^{i(qja-\omega t)}$ yieldsThe first Brillouin zone is $-\pi /a\le q\le \pi /a$; adding $2\pi /a$ leaves the sampled displacements unchanged. The branch is linear near zero, with sound speed $a\sqrt{\alpha /m}$, and reaches $2\sqrt{\alpha /m}$ at either zone edge.

Two masses per cell. Take a cell of length $2a$, with light displacement $u_j=U{e}^{i(2jaq-\omega t)}$ and heavy displacement $v_j=V{e}^{i[(2j+1)aq-\omega t]}$. The coupled equations reduce toA nonzero displacement requires determinant zero:Solving this quadratic in ${\omega }^2$ givesNow the first zone is $|q|\le \pi /(2a)$. The minus branch is acoustic: as $q\to 0$, both sublattices move together and ${\omega }_{-}\to 0$. Its sound speed is $a\sqrt{2\alpha /(M+m)}$. The plus branch is optical: at $q=0$ the atoms move oppositely with $mU+MV=0$, giving finite frequency $\sqrt{2\alpha (1/m+1/M)}$. This terminology describes relative motion; optical activity itself also requires coupling to the electromagnetic field.

diatomic-chain zone-edge polarizations and dimensionality. At the edge, $\cos qa=0$, so the heavy and light sublattices decouple: the acoustic edge frequency is $\sqrt{2\alpha /M}$ and the optical edge frequency is $\sqrt{2\alpha /m}$. HenceThe acoustic edge mode moves only the heavy sublattice; the optical edge mode only the light one. If $M=m$, the gap closes and the two branches are the zone-folded monatomic branch.

A three-dimensional crystal with two atoms per primitive cell has six branches: three acoustic and three optical. For propagation along a suitable symmetry direction these resolve into one longitudinal and two transverse branches of each type. The wavevector and Brillouin zone are three-dimensional; anisotropic force constants determine the dispersions, and polar crystals can have longitudinal–transverse optical splitting. A universal six-branch curve cannot be specified from masses alone.

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.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

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:

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

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.

Differentiating the expectation value and using ${\partial }_t|\psi ⟩=-i\hat{H}|\psi ⟩/\hslash$ and ${\partial }_t⟨\psi |=i⟨\psi |\hat{H}/\hslash$ givesThe second term means explicit operator time dependence, distinct from the state's evolution.

The commutator product rule follows by adding and subtracting $\hat{B}\hat{A}\hat{C}$:For the mathematical induction step, let $\hat{D}=[\hat{A},\hat{B}]$ and use $[\hat{A},\hat{D}]=0$. The case $n=1$ is immediate. If $[{\hat{A}}^m,\hat{B}]=m\hat{D}{\hat{A}}^{m-1}$, thenThus the claimed identity holds for every positive integer. Commutation of $\hat{B}$ with $\hat{D}$ is not needed for this particular step. This is the central-commutator differentiation rule. Termwise application to a convergent power series, on a common domain where the operations are justified, gives

For position and momentum, $[\hat{x},\hat{p}]=i\hslash \hat{I}$. Hence $[{\hat{p}}^2,\hat{x}]=-2i\hslash \hat{p}$, $[V(\hat{x}),\hat{x}]=0$ and $[V(\hat{x}),\hat{p}]=i\hslash V^{\prime }(\hat{x})$. Substitution gives Ehrenfest theorem:The mean momentum changes by the mean force. It does not follow for a general nonlinear Potential energy that $⟨V^{\prime }(\hat{x})⟩=V^{\prime }(⟨\hat{x}⟩)$: a closed classical trajectory for the mean position requires an additional condition, such as a quadratic Potential energy or a sufficiently localized wave packet.

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

The effective mass follows from the curvature of the electronic band structure. Differentiating twice givesAt either first Brillouin zone boundary $k=\pm \pi /a$,The boundary is a band maximum, with negative curvature and zero group velocity. In the semiclassical single-band description, an applied force changes $\hslash k$; the resulting acceleration is opposite the force near that maximum. This is a band-response coefficient, not a negative microscopic electron rest mass. A nearly filled band is often described more conveniently through positively charged electron holes near its maximum.

Take the electron's signed charge to be $e=-q$, where $q>0$ is the elementary-charge magnitude. Conventional electric current flows along $+x$, while the mean electron drift velocity is negative. With $\mathbf{B}=B_z\hat{\mathbf{z}}$, the magnetic Lorentz force separates charge across the width until a transverse electric field balances it. The accumulated edge charge then sustains a Hall voltage without a transverse steady current. In a one-carrier model, the Hall coefficient identifies the sign and number density of carriers; real multiband materials require a more careful interpretation.

The acceleration term contains the effective mass describing the local band response. The relaxation term removes drift momentum on the mean momentum-relaxation time $\tau$. The remaining two terms are the electric and magnetic Lorentz force on signed charge $e$; the magnetic force is perpendicular to velocity and does no instantaneous work. The relaxation-time approximation replaces individual scattering events by an averaged drift equation.

Let $n$ be the electron number density, distinct from the negative charge $e$. In transverse open circuit $⟨\mathbf{v}⟩=(v_x,0,0)$ and $j_x=ne{v}_x=I_x/(wt)$. The steady component equations givehenceConsequentlyThere is no sign ambiguity in this field-defined $R_H$ once axes and carrier charge are fixed.

A measured voltage ratio additionally needs probe polarities, which the geometric bar diagram does not label. To reproduce the positive ratio specified for electrons, define the longitudinal voltage drop $V_x=\phi (0)-\phi (l)=E_{x}l$ and the transverse rise $V_y=\phi (w)-\phi (0)=-E_{y}w$. ThenReversing either voltage's probes reverses the measured ratio, without changing the physical carrier density. With both voltages instead defined as drops along their positive axes, the algebraic ratio would have the opposite sign. The figure records the convention used here.

For the density, geometry gives $\rho =Rwt/l=Rt/(l/w)$ and $|R_H|=t|V_y|/(I_x|B_z|)=tR|V_y/V_x|/|B_z|$. ThusandThe mobility is $\mu =|R_H|/\rho =0.0130{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. With $m^{\ast }=m_e$,

A monovalent copper model assigns roughly one delocalized conduction electron per atom to a broad, approximately free-electron-like band. Using $m_e$ is therefore a sensible first estimate for its transport scale. Monovalence alone does not prove that every band curvature or orbit has exactly the free mass: periodic-potential mixing, the nonspherical Fermi surface and direction-dependent scattering limit the model. The calculated $\tau$ is conditional on the stipulated effective-mass approximation.

Experimentally, reverse both current and magnetic field and extract the part of the transverse voltage odd in each. This rejects fixed contact offsets and much longitudinal pickup; use four-terminal longitudinal sensing, measure dimensions independently and check heating before assigning material parameters. These controls are implemented in a reusable Hall/conductivity teaching experiment, although that experiment uses germanium rather than the copper sample here.^[ref]

The characteristic impedance of a stretched string is $Z=T/c=\rho c$, since its wave speed obeys $c^2=T/\rho$. Put the endpoint at $x=0$ and use displacement complex amplitudes $A$ incident and $B$ reflected:At the massless end, the string's transverse force $-T{\Psi }_x$ balances the damper's force $-\alpha {\Psi }_t$. Thus $T{\Psi }_x=-\alpha {\Psi }_t$, or $Z(A-B)=\alpha (A+B)$. This viscous termination of a stretched string hasHalf the incident mean power is reflected when $r=\pm 1/\sqrt{2}$, giving two positive solutions:They are approximately $0.1716\rho c$ and $5.828\rho c$, in $\mathrm{kg}{\mathrm{s}}^{-1}$. The smaller coefficient produces no reflected displacement-phase reversal; the larger produces a $\pi$ reversal. Both dissipate the remaining half of the incident power. Choosing the impedance-matched value $\alpha =\rho c$ would instead absorb all of it.

To find the displacement peak, minimize the squared denominator:The nonzero minimum givesFor stronger damping there is no positive-frequency displacement maximum. When $\omega \gg {\omega }_0$ and $b\ll m\omega$,The inertia then dominates the restoring and damping forces.

Use the $e^{i\omega t}$ convention for the harmonic force and displacement complex amplitudes. Substitution in the forced harmonic oscillator equation givesIts magnitude and phase relative to the force areAt zero frequency, $|R|=1/k$ and $\varphi =0$. At ${\omega }_0=\sqrt{k/m}$, $\varphi =-\pi /2$; at high frequency the phase approaches $-\pi$. For light viscous damping, the quality factor $Q=m{\omega }_0/b$ is large and there is a tall narrow displacement peak close to ${\omega }_0$.

A band effective mass describes the acceleration of an electronic wavepacket in a periodic lattice through the local band curvature. It incorporates the periodic lattice potential into a useful transport parameter without changing the bare electron mass. For a Bloch state in an isolated band,Differentiation givesIn one dimension this becomes $\boxed{m^{\ast }={\hslash }^2/(d^2ϵ/d{k}^2)}$. A locally parabolic isotropic band minimum has $ϵ\simeq {ϵ}_c+{\hslash }^2|\mathbf{k}-{\mathbf{k}}_c{|}^2/(2m^{\ast })$ with positive $m^{\ast }$. A flatter band has a larger effective mass. At a band maximum the electron curvature is negative, so the electron accelerates oppositely to the applied force. Describing the missing electron as a positive-charge electron hole with positive mass is often simpler.

In the scalar relaxation-time approximation, a conduction electron in an electric field reaches ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }$. ThusFor a crystal with different directional curvatures, conductivity is a tensor and involves the occupied states and their scattering. Contributions from electrons and holes add: $\sigma =e(n{\mu }_e+p{\mu }_h)$ in an isotropic two-carrier description. A filled band contributes no ordinary longitudinal current because opposite group velocities cancel; a partially filled band or holes permit a current response.

Effective mass therefore connects measured electrical transport to electronic band structure, but carrier density and momentum relaxation time matter as much as curvature. Ionised impurities and phonons alter $\tau$. A smaller mass does not guarantee a higher conductivity if carrier population or scattering changes. Nor are all reported masses identical: density-of-states, cyclotron and conductivity masses weight anisotropic bands differently. The simple semiclassical formula assumes an isolated band, slowly varying weak fields and negligible interband transitions; near degeneracies or strong fields a single constant scalar mass is inadequate.

For the steady response of a forced harmonic oscillator, write $x=\mathrm{Re}(A{e}^{i\omega t})$ and use the same complex amplitude convention for the driving force. Substitution givesThus the response magnitude and displacement phase relative to the force areThe phase goes from $0$ at low frequency, through $-\pi /2$ at ${\omega }_0$, towards $-\pi$ at high frequency. The static magnitude is $1/(m{\omega }_0^2)=1/k$, while the high-frequency magnitude is approximately $1/(m{\omega }^2)$.

Define the quality factor by $Q={\omega }_0/\gamma$. If $Q>1/\sqrt{2}$, the displacement peak occurs at ${\omega }_r=\sqrt{{\omega }_0^2-{\gamma }^2/2}$; for $Q\le 1/\sqrt{2}$ the magnitude decreases monotonically from its static value. For $Q\gg 1$ there is a tall narrow peak close to ${\omega }_0$, of height approximately $Q/k$ and width approximately $\gamma$. At very small $Q$ the resonant peak disappears and the response is broad. The plotted functions are normalized as $k|R|$, at fixed $m,k$.

The symbol $\rho$ here is linear mass density, with units $\mathrm{kg}{\mathrm{m}}^{-1}$. A short element $[x,x+\delta x]$ has transverse forcebecause small slopes give $\sin \theta \simeq \tan \theta ={\Psi }_x$. Its mass is $\rho \delta x$. Applying Newton's second law, dividing by $\delta x$, and taking the limit gives the wave equation on a string,The derivation assumes constant positive tension, small transverse displacements and slopes, a flexible string with negligible bending stiffness, and negligible damping. Variations of tension due to the motion enter beyond the linear approximation.

A characteristic wave impedance relates an effort-like quantity to its conjugate flow-like quantity in a single travelling wave: for a mechanical wave these are force and velocity, and their product is power. The sign depends on the propagation and force conventions. Use the transverse force transmitted towards increasing $x$, $F=-T{\Psi }_x$. A rightgoing travelling wave $\Psi =f(x-vt)$ has transverse velocity $u={\Psi }_t=-v{\Psi }_x$, soThis is the positive characteristic impedance of a stretched string. A leftgoing component has $F/u=-Z$ with the same force orientation. Its units differ from pressure-based acoustic impedance because it uses force, rather than force per area.

For a rightgoing harmonic travelling wave in the lossless string, its force and transverse-velocity complex amplitudes satisfy $F=Zu$ with real $Z=\sqrt{T\rho }$. Thus the supplied mean-power expression giveswhere $A$ is the displacement complex amplitude and $u=-i\omega A$ for convention $e^{i(kx-\omega t)}$. The same result follows directly from instantaneous string power $P=-T{\Psi }_x{\Psi }_t$ averaged over a cycle. The amplitudes here are peak amplitudes; with root mean square force or velocity the factor $1/2$ is absent. If reflections return to the driver, its net mean input is still $\mathrm{Re}(F{u}^{\ast })/2$, but its input mechanical impedance need not equal the travelling-wave value $Z$.

At an ordinary point or a massless join, the string is continuous, soWith no concentrated transverse force or point mass, transverse force balance requiresFor the common tension in this problem, the slope $\partial \Psi /\partial x$ is therefore continuous as well. A jump in linear mass density changes wave speed, but introduces no finite inertial force at a zero-length join. A point mass, external support force, or unequal tensions would require modified boundary conditions.