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The conventional electron plasma frequency isUse $n_e=10^{25}{\mathrm{m}}^{-3}$, $e=1.602\times 10^{-19}\mathrm{C}$, $m_e=9.109\times 10^{-31}\mathrm{kg}$ and ${\epsilon }_0=8.854\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$. ThenThe equal proton density ensures charge neutrality; it does not double the electron density in this formula. Protons have their own plasma frequency,If both cold species are allowed to move in the relative charge-separation oscillation, its squared frequency is ${\omega }_{pe}^2+{\omega }_{pi}^2$. This gives $1.7843\times 10^{14}\mathrm{rad}{\mathrm{s}}^{-1}$, only about $0.027\%$ above the electron-only answer, because $m_p\gg m_e$.

Two interface effects. For $n_2<n_1$, Snell's law gives $\sin {\theta }_t=(n_1/n_2)\sin {\theta }_i$. If $\sin {\theta }_i>n_2/n_1$, no propagating transmitted ray exists: there is total internal reflection, with an evanescent wave on the far side. An optical fibre uses this to guide light.

For ordinary isotropic nonmagnetic media, $\tan {\theta }_i=n_2/n_1$ is the Brewster angle. The reflected p-polarized component, whose electric field lies in the incidence plane, vanishes; the reflected s-polarized component generally remains. Reflection polarizers and polarized glare reduction exploit this selectivity.

The plasma response. Neglect collisions, thermal-pressure corrections and background magnetic field. Let the wave vary as $e^{-i\omega t}$. An electron displacement obeysIts dipole moment is $-e{\mathbf{x}}_e$, so the electron electric polarization is ${\mathbf{P}}_e=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$. Ions of charge $Ze$ give a further contribution ${\mathbf{P}}_i=-nZ{e}^2\mathbf{E}/(m_i{\omega }^2)$, smaller by $Z{m}_e/m_i$. Thus heavy ions move negligibly at these electron-wave frequencies. Using $\mathbf{D}={\epsilon }_0\mathbf{E}+\mathbf{P}$ gives the cold collisionless plasma dielectric response,For this model ${\mu }_r\simeq 1$, so $n_{\mathrm{refr}}^2={\epsilon }_r$ and $c^2{k}^2={\omega }^2-{\omega }_p^2$ for the transverse electromagnetic wave.

The vertical cutoff and density. Below the plasma frequency, the squared refractive index is negative: the wave is an evanescent wave instead of propagating through a layer hundreds of kilometres thick. The observed $5\mathrm{MHz}$ cutoff therefore identifies $f_p=5.0\times 10^6\mathrm{Hz}$, with ${\omega }_p=2\pi f_p$. ConsequentlyThis uses ${\epsilon }_0=8.854\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $m_e=9.109\times 10^{-31}\mathrm{kg}$ and $e=1.602\times 10^{-19}\mathrm{C}$. The density is also $3.10\times 10^5\mathrm{c}{\mathrm{m}}^{-3}$.

The transmission elevation. At $f=8\mathrm{MHz}$, the ionosphere has refractive indexIf $\alpha$ is elevation above horizontal, incidence at the layer is ${\theta }_i=\pi /2-\alpha$. Avoiding total internal reflection requires $\sin {\theta }_i=\cos \alpha <n_i$. The limiting elevation is thereforeStrictly, it must point above this limiting angle to carry upward power through the finite layer: at equality the transmitted wave is grazing. With a suitable satellite position, the stated orbital altitude imposes no additional angular constraint in this flat-layer approximation.

Polarization-selective ground reflection. The ray's incidence plane is vertical. Horizontal electric polarization is s polarization; the other transverse polarization lies in the vertical incidence plane and is p polarization. Reflection of s but not p identifies the Brewster angle, not total internal reflection, which would reflect both in this lossless isotropic model. ThusFor a reflecting height $h=200\mathrm{km}$, the horizontal range consists of two equal legs. The Brewster reflection from an ionospheric interface givesAt this angle ${\theta }_B$ is below the critical incidence angle, so a transmitted ray also exists; the detected horizontal component is an ordinary partial reflection. The s-polarized reflected power fraction is approximately $[(1-n_i^2)/(1+n_i^2){]}^2\simeq 0.059$, whereas the ideal p-polarized reflected fraction is zero. These results use the question's sharp planar interface model and exclude upper-interface reflections.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

The mobile electrons in a metal screen an incident optical field. In the weak-damping Drude model, $\epsilon (\omega )\simeq {\epsilon }_{\infty }-{\omega }_p^2/{\omega }^2$. Below the screened plasma frequency, the real dielectric response is negative, the wave is evanescent inside the metal, and a clean surface is highly reflective. Above the plasma edge, the response can become positive and propagating waves are allowed, explaining the qualitative transmission window at sufficiently high, often ultraviolet, frequencies.

This is a free-carrier argument, not a statement that every metal is transparent throughout the ultraviolet. Interband absorption, damping and sample thickness can still produce strong absorption above the plasma edge.

In the Lorentz oscillator model, each participating bound electron is displaced by an applied electric field against a restoring force. The model replaces complicated interband transitions by a driven damped harmonic oscillator. If $u$ is the electron displacement along the field and the electron charge is $-e$, its equation of motion isHere ${\omega }_0$ describes binding and $\gamma >0$ describes loss of coherent motion. For $E(t)=\mathrm{Re}(E_0{e}^{-i\omega t})$, the complex amplitude isThe induced electric dipole moment is $-eu$, so the electric polarization amplitude for number density $n$ isUsing $D={ϵ}_{0}E+P={ϵ}_0{ϵ}_{r}E$ givesThe paper's $ϵ$ is this dimensionless relative permittivity, rather than the absolute permittivity ${ϵ}_0{ϵ}_r$. Independent classes of bound electrons contribute additive susceptibilities, usually with different resonance frequencies and optical oscillator strengths.

Writing $D_{\omega }=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$ separates the response intoAt zero frequency the real part is $1+{\omega }_p^2/{\omega }_0^2$ and the imaginary part vanishes. Near resonance the imaginary part has a positive photon absorption peak and the real part changes rapidly, passing through $1$ at ${\omega }_0$. Above resonance the real part can become negative when damping is sufficiently weak; at very high frequency it approaches $1$ from below while the imaginary part tends to zero. Finite damping removes the undamped divergence.

For a nonmagnetic material take the passive square-root branch $N=\eta +i\kappa =\sqrt{{ϵ}_r}$, with $\kappa \ge 0$. A plane wave then has an intensity attenuation coefficient $\alpha =2\omega \kappa /c$. Thus the imaginary permittivity expresses dissipation, whereas the actual photon absorption coefficient is determined by the imaginary refractive index. The normal-incidence power reflectance isA large refractive-index mismatch produces wave reflection even without loss. In the undamped model ${ϵ}_r<0$ between ${\omega }_0$ and $\sqrt{{\omega }_0^2+{\omega }_p^2}$, so the field is evanescent and wave reflection is total in a semi-infinite sample. With damping the response includes both wave reflection and photon absorption; a peak in the imaginary permittivity should not simply be identified with a reflectance peak.

To infer the short-wavelength background from $R=0.35$, assume weak photon absorption and take the physical semiconductor branch $\eta >1$. ThensoThis is an estimate of the valence-electron background in a frequency window above the doped-carrier response and below the effective bound-electron resonance. One reflectance value alone does not determine both parts of a complex dielectric constant. Even for real ${ϵ}_r$, a reciprocal-index branch gives the same $R$; the ordinary background index greater than unity selects the branch used here.

Using the supplied conventional cubic lattice parameter, the participating valence-electron number density isUsing $m=m_e$ for the stipulated free-electron comparison givesAn angular frequency “in eV” means its quantum energy $\hslash \omega$, not $h\omega$. No doped-carrier effective mass should be inserted into this valence-electron count.

The low-frequency limit of the bound-electron contribution obeysConsequentlyThis effective resonance is much larger than the quoted $0.2\mathrm{eV}$ band gap. There is no contradiction: the single oscillator has been assigned the strength of all sixteen participating valence electrons per conventional cell, whereas the band gap is only the lowest interband excitation threshold. A real spectrum distributes optical oscillator strength over many transitions. Equating the oscillator resonance to the lowest threshold would misrepresent the assumed background response. Here “low frequency” means low relative to ${\omega }_0$; the static response of the doped conducting sample also contains its free carriers.

The additional mobile carriers in doped indium antimonide contribute a Drude frequency response, obtained by setting their restoring frequency to zero. At infrared frequencies well below the effective valence resonance, the sum is approximatelyAt the shorter wavelengths the carrier term is small and $R$ approaches the background value. As wavelength increases, the negative free-carrier contribution lowers the real permittivity towards unity. For weak damping, refractive-index matching to vacuum gives the Drude-Lorentz reflectivity minimum atThis minimum is near ${ϵ}_r^{\prime }=1$, not its zero. At the slightly lower screened plasma frequency ${\omega }_{pc}/\sqrt{{ϵ}_b}$, the real permittivity crosses zero; at longer wavelengths it becomes negative and the material becomes strongly reflective. Finite damping prevents a perfectly vanishing minimum and rounds the edge. At still lower frequency the large conducting response also drives $R$ towards unity.

For fixed carrier effective mass and background permittivity, ${\omega }_{pc}\propto \sqrt{n_c}$ and ${\lambda }_{\min }\propto n_c^{-1/2}$. The observed shift of the minimum towards shorter wavelength as doping increases is therefore the expected free-carrier trend. The figure is an original model illustration, not a digitization or quantitative fit of the supplied curves; the carrier effective mass, damping and their density dependence have not been specified.

Use a representative conduction-electron density $n\sim 10^{29}{\mathrm{m}}^{-3}$, roughly one mobile electron per atom in a dense metal, and the bare electron mass for this estimate. The plasma frequency is thenThis is an order-of-magnitude answer because no particular metal or carrier density is specified. It corresponds to an ordinary frequency $f_p\simeq 2.8\times 10^{15}\mathrm{Hz}$, excitation energy $\hslash {\omega }_p\simeq 12\mathrm{eV}$, and free-space wavelength near $1.1\times 10^{-7}\mathrm{m}$.

The physical oscillation is collective motion of the electron gas against the positive ionic background. Displacing the electrons relative to the ions produces a restoring electric field. In the simplest uniform-displacement picture, $\ddot{x}+{\omega }_p^{2}x=0$: the frequency increases with the density of mobile electric charge and decreases with carrier inertia. Quantizing this longitudinal collective oscillation gives a plasmon.

The same scale controls the ideal collisionless Drude model optical response. Its cold collisionless plasma dielectric response is ${ϵ}_r(\omega )=1-{\omega }_p^2/{\omega }^2$. Below ${\omega }_p$, the negative relative permittivity makes a transverse electromagnetic wave evanescent, producing strong reflection; above it, the ideal model admits propagation. Real-metal collisions, interband photon absorption and background polarization modify this simple optical edge. Thus ${\omega }_p$ is a collective charge-response scale, not a single-electron scattering rate or a sharp transparency threshold for every real metal.

Use physical fields proportional to the real part of $e^{-i\omega t}$. The Drude frequency response isThus the sketch is an even Lorentzian line shape, with maximum ${\sigma }_0^2$ at zero frequency, half that value at $|\omega |={\tau }^{-1}$, and an ${\omega }^{-2}$ tail. The positive-frequency half is shown in the preceding figure. Squaring the modulus differs from plotting either $|\sigma |$ or the real part of $\sigma$.

For a single carrier species, the plasma frequency and electrical conductivity satisfyThereforeThe corresponding relaxation time is $\tau \simeq 3.57\times 10^{-14}\mathrm{s}$. This is an angular frequency; the ordinary frequency is $\boxed{f_{1/2}={\omega }_{1/2}/(2\pi )\simeq 4.46\times 10^{12}\mathrm{Hz}}$. The units of electrical conductivity used in the calculation are ${\mathrm{\Omega }}^{-1}{\mathrm{m}}^{-1}$, consistently with the SI plasma-frequency relation.

In a simple metal, conduction electrons can move collectively in response to an incident electromagnetic wave. The Drude model gives, with time dependence $e^{-i\omega t}$,In the low-loss range $\gamma \ll \omega <{\omega }_p$, the relative permittivity is approximately $1-{\omega }_p^2/{\omega }^2<0$. The wavevector inside is predominantly imaginary, so the field is evanescent and penetrates only a short distance. Surface currents reradiate the field; little energy enters a good conductor. The normal-incidence Fresnel equations giveMany ordinary metals have their plasma frequency in the ultraviolet, placing visible light below this reflection edge. For $\omega >{\omega }_p$ the free-carrier contribution permits propagation; at $\omega \gg {\omega }_p$, ${ϵ}_r\to 1$ and $R\sim {\omega }_p^4/(16{\omega }^4)\to 0$. The electrons' inertial response becomes too small to screen the rapidly oscillating field.

Ultraviolet transparency is a simple-metal model prediction, not a universal property of real metals. Electronic transitions between bands and bound-electron absorption can remain strong, shift the plasma edge and prevent transparency; transmission also depends on thickness and attenuation. The robust explanation is the change from a reflective negative-permittivity range to a possible propagating range above the plasma edge.

At normal incidence from vacuum, the Fresnel equations give the reflectanceA good metal has $R$ close to one below its plasma frequency. At very low frequency the electromagnetic field penetrates only a skin depth; the local Drude limit gives $1-R\simeq \sqrt{8{ϵ}_0\omega /{\sigma }_0}$. In the collisionless negative-permittivity range the transmitted field is evanescent and reflection remains strong. Near the plasma edge the reflectance drops sharply; above it ${ϵ}_r^{\prime }>0$ and propagating waves become possible. In the ideal free-electron high-frequency limit,Finite damping rounds the plasma edge, and interband absorption changes the detailed response of actual metals.

Use quasineutrality, $n_e=Z{n}_i$. Multiplying the supplied collision time by the electron plasma frequency and using the preceding definition givesThus ${\tau }_{ei}\gg 2\pi /{\omega }_p$ requires $N_D\gg (2\pi /9.6)Z\ln \Lambda$; for ordinary $Z$ and $\ln \Lambda \sim 1$–$30$, this is a large plasma parameter. Many particles then collectively screen a charge, fluctuations in a Debye sphere are relatively small, and mean-field electrostatics descriptions become meaningful. A large plasma parameter is also associated with weak Coulomb coupling, well-separated screening and close-collision scales, and collective plasma oscillations surviving many oscillations before collisionally relaxing. Roughly, relative counting fluctuations scale as $N_D^{-1/2}$ and the ratio of interaction to thermal energy decreases as $N_D$ increases. Small $N_D$ signals that strong correlations can invalidate the simple weakly coupled plasma description.

The electron Debye length is the scale over which a weak electrostatic disturbance is screened. Take the ions as an immobile neutralizing background and the electrons as an isothermal Boltzmann distribution. For a small potential $\varphi$,Poisson's equation then becomes ${\nabla }^2\varphi =\varphi /{\lambda }_D^2$ away from the perturbing source, withA test charge has a screened potential proportional to $e^{-r/{\lambda }_D}/r$. For fully static screening with mobile ions, inverse squared Debye lengths from all species add; the displayed expression is the electron scale used below.

The electron plasma frequency is the natural collective oscillation frequency when electrons are displaced from a nearly fixed ion background. Displace a uniform electron fluid by $\xi$ along a slab normal. The charge separation produces the restoring electric force $-n_e{e}^2\xi /{ϵ}_0$ on an electron. Hence $m_e\ddot{\xi }=-n_e{e}^2\xi /{ϵ}_0$ andIt sets a response time $1/{\omega }_p$ and the cold unmagnetized electromagnetic cutoff frequency. These collective effects arise from long-range electric fields, even when individual binary collisions are infrequent.

Use collisional inverse bremsstrahlung in a weakly coupled, unmagnetized plasma. The specified density is $n_e=10^{26}{\mathrm{m}}^{-3}$, and $k_{B}T=1000\mathrm{eV}$. The electron Debye length is $2.35\times 10^{-8}\mathrm{m}$, the plasma frequency is $5.64\times 10^{14}{\mathrm{s}}^{-1}$, and the laser angular frequency is $\omega =2\pi c/\lambda =1.79\times 10^{15}{\mathrm{s}}^{-1}$. Thus $X={\omega }_p^2/{\omega }^2\simeq 0.099<1$: the plasma is underdense and a propagating wave exists.

Estimate $b_{\min }$ as the larger of the classical Coulomb-deflection length $Z{e}^2/(4\pi {ϵ}_0{k}_{B}T)$ and the electron quantum length $\hslash /\sqrt{m_e{k}_{B}T}$. Both are about $9\times 10^{-12}\mathrm{m}$ here, giving $\ln {\Lambda }_C\simeq 8$. With the usual thermal collision convention,The $Z$ factor follows from $n_i{Z}^2=n_{e}Z$.^[ref]

The weakly collisional Drude model gives ${ϵ}_r=1-{\omega }_p^2/[\omega (\omega +i{\nu }_{ei})]$ for fields proportional to $e^{-i\omega t}$. Expanding at ${\nu }_{ei}\ll \omega$ yields a real refractive index $n_r=\sqrt{1-X}\simeq 0.949$ and intensity attenuation coefficientConsequently the intensity e-folding absorption length isi.e. of order a centimetre. Taking a Coulomb logarithm between about 5 and 10 gives roughly $5$–$10\mathrm{mm}$. The electric-field amplitude decay length is twice this intensity length. This estimate assumes a uniform slab thick enough to define an absorption length and neglects gradients, nonlinear quiver heating and other absorption mechanisms.

For carrier number density $n$, charge $q$ and effective mass $m$, the plasma frequency is ${\omega }_p^2=n{q}^2/({ϵ}_{0}m)$. Combining the second London equation with the magnetostatic Ampère law and Gauss's law for magnetism gives ${\nabla }^2\mathbf{B}=({\omega }_p^2/c^2)\mathbf{B}$. Therefore the decaying planar solution has length ${\lambda }_L=c/{\omega }_p=\sqrt{m/({\mu }_{0}n{q}^2)}$.

A driver excites charge-density oscillations in a plasma whose longitudinal electric field transfers energy to a trailing witness beam. Density fixes a characteristic plasma frequency and wavelength, while driver shape and loading determine the actual field and transfer efficiency.

A radio wave traversing a dilute cold plasma has a group velocity that depends on its frequency. In the ideal unmagnetized limit, the dispersion relation is ${\omega }^2=c^2{k}^2+{\omega }_p^2$, where ${\omega }_p$ is the angular plasma frequency. For $\omega \gg {\omega }_p$, differentiate to obtain $v_g=c\sqrt{1-{\omega }_p^2/{\omega }^2}$, then expand $1/v_g$ to first order. The excess travel time over a vacuum path isHere $f=\omega /(2\pi )$ is ordinary frequency, $e$ is the magnitude of the elementary charge, $m_e$ is the electron mass, ${ϵ}_0$ is the vacuum permittivity, $n_e$ is electron number density, and $c$ is the speed of light. This approximation neglects magnetic corrections, collisions and multipath scattering; its positive sign means lower frequencies arrive later.

With immobile ions, negligible collisions and electron density $n_e$, the plasma frequency is ${\omega }_p^2=n_e{e}^2/(m_e{ϵ}_0)$. Combining inertial electron current with vacuum displacement current gives ${\omega }^2={\omega }_p^2+c^2{k}^2$. Transverse fields are evanescent below ${\omega }_p$ and propagate above it.

Xiaoge Liu is the first-listed author of the 2014 study of band nonparabolicity and the optical properties of indium tin oxide. The Stanford laboratory's alumni record also identifies Liu as a former doctoral student.^[ref][ref]

The physical issue can be expressed through the Drude model relationwhere $n$ is carrier density and $m^{\ast }$ the relevant effective mass. If $m^{\ast }$ were fixed, doubling $n$ would double ${\omega }_p^2$. A density-dependent mass changes that prediction. Combining electrical and optical measurements can therefore test a band model rather than treating every shift in plasma frequency as a direct measure of carrier density alone.