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The diagram makes $x$ the remaining distance to the wall: $x$ decreases from $L$ to $0$. With the endpoint condition $m(0)=0$, integrating the supplied mass gradient givesAssume escaping sand has the same horizontal velocity as the bucket at release. There is then no horizontal recoil: the momentum flux of the outgoing sand cancels the $\dot{m}\dot{x}$ contribution to the bucket's momentum derivative. The variable-mass system equation is thereforeSet $u=-\dot{x}\ge 0$, the speed towards the wall. Since $\ddot{x}=udu/dx$, the initial condition $u(L)=0$ givesConsequently the remaining sand has kinetic energyIn terms of $z=x/L$, $K/(TL)=-z\ln z$. Its derivative is $-\ln z-1$ and its second derivative is $-1/z<0$, givingThe magnitude of the remaining sand's momentum isIt is easier to maximize its square, $p^2/(2MTL)=-z^2\ln z$. Its derivative is $-z(2\ln z+1)$, giving the unique interior maximumBoth kinetic energy and momentum tend to zero at $x\to 0^{+}$, even though the idealized speed diverges. There is no finite-mass bucket left at the endpoint, so that divergence signals the limit of the idealization.

The leaking bucket under constant tension has $x_K=L/e<L/\sqrt{e}=x_p$: the kinetic energy maximum is closer to the wall. As the bucket loses mass, its increasing speed continues to boost $K=pu/2$ after $p$ has begun to fall. Equivalently, $p\propto z\sqrt{-\ln z}$ and $K\propto z(-\ln z)$, so kinetic energy has the stronger dependence on the growing speed. The work $T(L-x)$ is shared between the remaining sand and the sand already lost; it is not all the bucket's instantaneous kinetic energy.

Take upward velocity as positive, and let $w>0$ be the rearward rocket exhaust speed. During $dt$, the ejected parcel has mass $-dm>0$ and laboratory velocity $v-w$ to first order. The momentum change of the rocket and this parcel equals the gravitational impulse:Discarding the second-order product $dmdv$ gives the vertical rocket equationWith $\dot{m}=-\alpha$ and $m=m_0-\alpha t$, integration for free flight starting from rest at $t=0$ givesThe logarithmic term is the ideal Tsiolkovsky rocket equation contribution; $gt$ is the gravity loss. Immediate upward departure requires positive initial acceleration, orAt exact equality the initial acceleration is zero, but further burning reduces the weight and permits departure immediately thereafter. If departure is allowed to be delayed while fuel burns on the pad, the condition is instead $\alpha w>m(t)g$ at some time before burnout. With final rocket dry mass $m_f$, any powered departure before burnout requires $w>g{m}_f/\alpha$; the free-flight formula then starts at the actual departure time $t_{\ell }$ with $m(t_{\ell })$, giving $v=w\ln [m(t_{\ell })/m(t)]-g(t-t_{\ell })$.

For the specified data, the initial mass includes the payload. The rocket fuel fraction givesThe $100\mathrm{kg}$ payload is already part of the $2500\mathrm{kg}$ final mass; it is not the entire rocket dry mass and must not be added a second time. Here $m_{0}g/\alpha =196m/s$, well below the stated rocket exhaust speed, so departure is immediate. The burnout speed isThis is the speed at burnout. Subsequent motion under the assumed uniform gravity has $v(t)=v(T)-g(t-T)$.

Use $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$ for the magnetic Lorentz force on each wire segment. Let $+x$ point right, $+y$ upward and $+z$ out of the page, so $\mathbf{B}=-B\hat{\mathbf{z}}$. With clockwise electric current, the four magnetic forces areThe upper side carries electric current to the right and experiences an upward force, opposing the fall. The bottom side lies outside the magnetic field and has no magnetic force. On the right side the electric current is downward and the force is outward to the right; on the left it is upward and the force is outward to the left. Each side's in-field vertical length is $h$, so these horizontal forces cancel in the net force on the loop.

The wire's total mass is $M=4a{\rho }_m{A}_w$. Uniform wire gives each side a downward weight $Mg/4$. Internal mechanical forces at the corners transmit the unequal loads and maintain the assumed square shape; they cancel for the loop as a whole. The total external vertical force is therefore $Mg-IaB$ downward.

With $V=+20\mathrm{V}$ in the outer-minus-inner convention, $\eta =\ln 1.5=0.405465$ and the radial electric field at the orbit is $E_r=-986.52\mathrm{V}{\mathrm{m}}^{-1}$. The signed roots areUsing the proton mass and charge givesThe speeds are therefore $48.8$ and $96.7\mathrm{km}{\mathrm{s}}^{-1}$, in opposite senses of rotation. Both are nonrelativistic. The slower counterclockwise orbit has an outward magnetic force; the faster clockwise orbit has an inward magnetic force. Reversing $\mathbf{B}$ interchanges the senses, while reversing the electrode polarity changes the quadratic and may eliminate real orbits; its sign must therefore be specified.

Let $m$ be the sliding point charge's mass. Expanding the axial field of a uniformly charged annulus for $|x|\ll a,b$ givesIts electric charge is $-Q$, so its electric force points toward the centre. Newton's second law becomesTo leading order this is simple harmonic motion, withThe mass is an additional parameter, since the source supplies no numerical value. The constraint is essential: stability along the axis does not imply unconstrained three-dimensional trapping by a static electric field.

As an estimate, model the confinement by an infinite square well of width $L$ and use the stipulated electron mass. Its energy eigenvalues are $E_j=j^2{\pi }^2{\hslash }^2/(2m{L}^2)$, $j=1,2,\dots$, so the gap controlling thermal occupation of a quantum well isThus $T\ll 130\mathrm{K}$ suppresses excitation: $p_2/p_1=e^{-\Delta E/(k_{B}T)}$. For example $T\lesssim 28\mathrm{K}$ makes this ratio below one percent in this model. There is no sharp cutoff. The ground state energy itself corresponds to about $44\mathrm{K}$, but it is the excitation gap that enters the Boltzmann factor. Finite barrier height and a different effective mass would change the numerical scale; neither is supplied.

A pure stationary state of a time-independent Hamiltonian operator is $\Psi (\mathbf{r},t)=\psi (\mathbf{r})e^{-iEt/\hslash }$. Its probability density and all expectations of time-independent observables are constant. This does not require zero probability current.

For real scalar potential energy and no vector potential, substitute $i\hslash \dot{\Psi }=-({\hslash }^2/2m){\nabla }^2\Psi +V\Psi$ and its complex conjugate into $\dot{\rho }={\Psi }^{\ast }\dot{\Psi }+{\dot{\Psi }}^{\ast }\Psi$. The potential terms cancel:Thus the probability current is $\mathbf{j}=(\hslash /m)\mathrm{Im}({\Psi }^{\ast }\nabla \Psi )$. In a stationary state $\nabla \cdot \mathbf{j}=0$: flow may circulate without changing the density.

Here only the azimuthal phase contributes an imaginary logarithmic derivative. Since $\nabla \varphi =\hat{\mathbf{\varphi }}/(r\sin \theta )$,There is no radial or polar flow; the azimuthal current is independent of $\varphi$, so its divergence vanishes. The apparent coordinate singularity at $\sin \theta =0$ cancels in the final expression.

The z-component of orbital angular momentum follows from the moment of the mass current:Indeed $A^2(24a_0^5)(4/3)(2\pi )=1$. This agrees with $L_z=-i\hslash {\partial }_{\varphi }$ acting on the wavefunction.

Put $\kappa =e^2/(4\pi {ϵ}_0)$. Substitution of $\psi =R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation, followed by division by $R\Phi$ and multiplication by $r^2$, separates the angular dependence:Thus $\Phi =e^{im\varphi }/\sqrt{2\pi }$ andThe single-valued wavefunction condition, $\Phi (\varphi +2\pi )=\Phi (\varphi )$, requires $m\in \mathbb{Z}$. The regular boundary condition at the origin selects $R\sim r^{|m|}$. Accordingly use $s=|m|$ in the radial ansatz; the printed $x^m$ notation describes the $m\ge 0$ representatives.

For the two-dimensional Coulomb bound state, substitution of $R=x^s{e}^{-x/2}G$ with the stipulated radial and energy scales givesThe coefficient of $x^n$ in the power series isThe regular solution otherwise has an exponentially growing contribution at large $x$, incompatible with being normalizable. Termination at degree $n_r\ge 0$ therefore requiresThe resulting $G$ is proportional to a Generalized Laguerre polynomial.

Define $R_{\mathrm{y}}=\mu {\kappa }^2/(2{\hslash }^2)$, the hydrogenic Rydberg energy for the stipulated mass. The first three distinct two-dimensional energies are $-4R_{\mathrm{y}},-4R_{\mathrm{y}}/9,-4R_{\mathrm{y}}/25$, whereas in three dimensions they are $-R_{\mathrm{y}},-R_{\mathrm{y}}/4,-R_{\mathrm{y}}/9$. Both series approach zero from below, but the planar ground state is four times more tightly bound. Without spin the respective quantum degeneracies are $1,3,5$ and $1,4,9$; if individual degenerate states are counted, the second and third planar states already share the first excited energy. This comparison retains the supplied $1/r$ interaction; it does not replace it by the logarithmic electrostatic potential of a genuinely two-dimensional electromagnetic theory.

For the two inline operator labels, (a) $L_z$ has eigenvalue $m_{\ell }\hslash$, and (b) $L^2$ has eigenvalue $\ell (\ell +1){\hslash }^2$, with $\ell =0,1,\dots$ and $m_{\ell }=-\ell ,\dots ,\ell$. The subscript distinguishes the magnetic quantum number from the electron mass.

Put $b=a_0(\ell +1)$ and $\kappa =e^2/(4\pi {ϵ}_0)$. For the proposed radial shape,The angular-momentum term cancels the inverse-square part. The inverse-radius coefficient vanishes if ${\hslash }^2(\ell +1)/(mb)=\kappa$. ThereforeThis nodeless Coulomb radial eigenstate has principal quantum number $n=\ell +1$, not the complete set of hydrogen radial states. Since the spherical harmonic is normalized, radial normalization requiresup to an arbitrary phase.

For the hydrogen ground-state nuclear-volume probability, use $R=2a_0^{-3/2}{e}^{-r/a_0}$. With $u=r_p/a_0\ll 1$,No numerical proton radius is supplied or required. A sharp proton boundary is an approximation to a charge distribution, and the point-Coulomb potential changes inside an extended proton. Finite proton mass, relativistic and electromagnetic corrections also limit the ideal wavefunction's precision.

Classically the kinetic energy must satisfy $E-V(r)\ge 0$. For $E=-\kappa /(2a_0)$ this demands $r\le 2a_0$. The hydrogen ground-state probability beyond the classical turning point isA substantial probability lies beyond the classical turning radius; the bound-state wavefunction decays there rather than vanishing.

For a narrow packet, phase velocity is $u_p=\omega /k$ and group velocity is $u_g=d\omega /dk$. Set $b=\sigma /{\rho }_1$. The uncovered surface hasEquality occurs at $k_{\ast }^2=g/b$, also the minimum of $u_p$. Numerically,The long-wave gravity limit has $u_g/u_p\to 1/2$; the short-wave capillary limit has $u_g/u_p\to 3/2$.

A compact packet subsequently spreads because different wavenumbers have different group speeds. In the gravity regime $\omega \simeq \sqrt{gk}$, so smaller $k$ and longer waves move ahead. Individual crests move twice as fast as the envelope, appearing at its rear and disappearing at its front. In the capillary regime $\omega \simeq \sqrt{b}{k}^{3/2}$, shorter waves move ahead; the envelope overtakes its slower individual crests, which drift backward relative to it. Thus both the wavelength ordering within the spread packet and the crest motion relative to its envelope reverse.

For the sheet define $c=a{\rho }_2/{\rho }_1=8.0\times 10^{-5}\mathrm{m}$. The elastic-cover frequency ratio for water waves isAt small $k$, $F=1-ck/2+O(k^2)$: sheet inertia lowers the frequency, while bending first enters at higher order. At sufficiently large $k$, bending dominates the restoring force and sheet inertia dominates the extra mass, giving $F\sim \sqrt{B/(\sigma c)}{k}^{1/2}$. The curve initially dips below one, reaches a minimum, then rises and crosses one. This asymptote describes the supplied ideal dispersion law; arbitrarily large $k$ eventually violates a thin-sheet continuum model.

For a pinned-square surface displacement mode, $k_{nm}=\pi \sqrt{n^2+m^2}/L$ with $n,m\ge 1$. The fundamental has $k_{11}=74.05{\mathrm{m}}^{-1}$. The uncovered frequency is $4.3729\mathrm{Hz}$ and $F(k_{11})=0.997051$, soThis uses a square two-dimensional surface mode, not the one-dimensional $\pi /L$ estimate. A clamped plate or additional hydraulic boundary conditions would require a different mode calculation.

In a periodic crystal potential, atomic levels broaden into bands of allowed bloch states. Coupling between states of the same symmetry opens avoided crossings, including Bragg-coupled states in the nearly free-electron picture. Bonding and antibonding combinations provide a complementary tight-binding picture. A semiconductor has a filled valence band and an empty conduction band at zero temperature separated by a forbidden energy interval; thermal excitation, light and doping provide mobile electrons and holes. The gap depends on the crystal's periodic potential, bonding and symmetry, not on a classical collision barrier. In a direct gap, the relevant extrema coincide in wavevector.

For the stated parabolic edges, take $E_v(k)=-{\hslash }^2{k}^2/(2m_h^{\ast })$ and $E_c(k)=E_g+{\hslash }^2{k}^2/(2m_e^{\ast })$. Negligible photon momentum and no phonon make the transition nearly vertical in $E$ versus $k$. parabolic-band optical energy sharing divides the excess photon energy $0.32\mathrm{eV}$ in inverse proportion to the effective masses:Thus the initial and final electron energies, relative to the valence-band maximum, areTheir difference is $0.5000\mathrm{eV}$. The selected $|k|$ is about $9.33\times 10^8{\mathrm{m}}^{-1}$; either sign gives the same energy in this symmetric approximation.

A shallow donor supplies an electron in a screened impurity potential, so its localized level lies just below the conduction edge. A shallow acceptor binds a hole and has an electron-occupation level just above the valence edge; thermal ionization supplies mobile holes. The level positions are measured relative to the relevant band edge, and deep impurities need not obey this hydrogenic approximation.

For the supplied bound-carrier mass and unit effective ionic charge, dielectric screening and effective mass change the hydrogen binding energy toThis corresponds to $E_b/k_B\simeq 110\mathrm{K}$ and an effective Bohr radius about $6.35\mathrm{nm}$. The binding is shallow compared with the band gap and typical atomic binding. Trivalent denotes chemical valence, not an unscreened hydrogenic charge of three. An acceptor physically binds a hole; the question's stated $0.1m_e$ is used as its stipulated bound-carrier mass rather than silently replacing it by the separate band-edge $0.4m_e$.

A carrier in a locally parabolic isotropic band obeys $m^{\ast }\dot{\mathbf{v}}=q\mathbf{v}\times \mathbf{B}$, soIlluminate a high-mobility crystal with a transverse alternating electric field, sweep a calibrated static magnetic field at known microwave or far-infrared frequency, and detect resonant absorbed power. Cooling lengthens the relaxation time; resolving the resonance generally needs ${\omega }_c\tau \gg 1$. Varying field direction measures band anisotropy, while circular polarization distinguishes the two senses of rotation for opposite carrier charges. This is orbital motion rather than electron spin resonance.

The classic silicon/germanium measurements used single crystals near $4\mathrm{K}$, light modulation and circularly polarized microwaves, with directional scans to infer anisotropic carrier-energy surfaces.^[ref] Carrier generation, calibration, scattering linewidth and inhomogeneity must be separated from a genuine mass shift.

In a general band the cyclotron mass is an orbit property, $m_c={\hslash }^2(\partial A_k/\partial E)/(2\pi )$ for the appropriate field-normal orbit. It need not equal every component of the band-curvature mass tensor. For an ellipsoidal parabolic band with field along a principal $z$ direction it is $\sqrt{m_x{m}_y}$. Thus a single resonance does not determine a whole effective-mass tensor without angular information.

The mass dependence of fixed-barrier tunnelling follows by holding energy and the potential-energy barrier fixed: quantum tunnelling has the leading WKB approximationThe alpha-particle mass is about $7.3\times 10^3$ electron masses, so its decay exponent is about 85 times larger for the same $V-E$ profile. Exponential suppression therefore makes transmission much smaller. This comparison holds the potential energy and incident energy fixed: the same electrostatic potential is not the same potential-energy barrier for charges of different magnitude and sign.

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

temperature changes alter dimensions, resistances, refractive indices and offsets; gradients can generate thermoelectric voltages. Identify the dominant heat paths and the experiment's temperature coefficients before choosing insulation. Reduce convection with an enclosure or, where appropriate, vacuum; reduce conduction through supports and leads with suitable geometry and thermal anchoring; use thermal radiation shields and low-emissivity surfaces to reduce radiative exchange. vacuum alone does not remove heat conduction through wires or thermal radiation.

In first-order thermal isolation, a sample with heat capacity $C$ and thermal conductance $G$ to its surroundings obeysThermal mass and insulation suppress rapid fluctuations but only delay slow ambient drift: the zero-frequency gain remains one. Active temperature control, a stable reference environment, adequate equilibration time and an appropriate sensor location address long-term stability. Self-heating produces a steady offset $\delta T_s=\delta P/G$; increasing insulation can worsen that offset and settling time. Check temperature gradients rather than relying on a remote thermometer.

Match materials or use differential/reference measurements to reduce expansion and resistance drift; keep dissimilar-metal junctions isothermal to suppress thermoelectric offsets. Log temperature alongside the signal and deliberately vary environmental temperature to measure residual coupling, using a physically justified correction with uncertainty. Interleave calibrations and allow equilibration after changing power. Feedback must remain stable, and heater/sensor electronics should not introduce electrical interference into the measurement.

Sound compressions in air are approximately adiabatic, giving $c_s^2=(\partial p/\partial \rho {)}_s=\gamma p/\rho$. With $p=\rho RT/M$,Here dry air is approximated as a diatomic ideal gas with $\gamma =7/5$ and molar mass $M\simeq 0.029\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. The ideal-gas law alone is insufficient without this adiabatic response and mean molecular mass; humidity and composition affect a more precise inference.

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.

Confinement to length $L$ gives a momentum scale $p\sim \hslash /L$ by the Heisenberg uncertainty principle. Using the alpha-particle mass $m_{\alpha }\simeq 6.64\times 10^{-27}\mathrm{kg}$, its speed scale is $v\sim \hslash /(m_{\alpha }L)$. Counting encounters with either wall,This is an order-of-magnitude collision-rate estimate, not a classical trajectory of a stationary quantum state. Taking $\Delta x\Delta p=\hslash /2$, a different geometric interpretation of the well size, or successive returns to the same wall changes the coefficient by factors of order unity; the robust scale is $10^{20}{\mathrm{s}}^{-1}$.

Define $s_i=\sin [2\pi (t_i-T)/P+\pi ]$ and use the independent quoted Gaussian errors $\sigma =10\mathrm{m}{\mathrm{s}}^{-1}$, with period and exoplanet transit epoch fixed. The model is linear in $K,V_0$ despite its sinusoidal time dependence. Minimize ${\chi }^2={\sum }_i(V_i-K{s}_i-V_0{)}^2/{\sigma }^2$. With $N=5$ the normal equations areUsing $\sum s_i=-0.741472$ and $\sum s_i^2=2.824633$, the solution isThe residuals in observation order are $(3.67,11.01,2.74,-7.90,-9.51)\mathrm{m}{\mathrm{s}}^{-1}$. Therefore ${\chi }_{\min }^2=2.950$ for $5-2=3$ degrees of freedom, giving reduced ${\chi }^2=0.983$. The upper-tail probability is about $0.40$: the fit is consistent with the stated errors, although five points do not establish a unique physical explanation.

For known errors the covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}$. Equivalently, profiling out the fitted offset givesThis is the $\Delta {\chi }^2=1$ one-parameter uncertainty, not the curvature with $V_0$ incorrectly held fixed. Propagating it through the supplied mass relation, and treating its coefficient and $P$ as fixed,The symbol $\pm nP$ in the exoplanet transit epoch denotes periodic copies of the same exoplanet transit phase, not an uncertainty distribution to propagate. No independent error for the empirical mass coefficient was supplied.

Additional independent observations retain positive weight even with twice the error: each has one quarter of the previous per-observation weight. Their information adds to the existing information matrix, so the formal mass uncertainty decreases for informative additional orbital phases and cannot increase under this fixed-error model. It can remain unchanged in a degenerate phase configuration that supplies no further amplitude information after fitting the offset. The conclusion assumes the same correct model and independent errors, not a new fitted error rescaling or unmodelled systematic effects.

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 $2\pi$.