Why Metals Shine: Free Electrons, Not Smooth Surfaces — Epoche B2
The problem: a plausible but false explanation Ask why a metal mirror shines, and most people answer: because its surface is extremely smooth. The explanation sounds reasonable, yet it fails a simple test. A polished block of glass is just as smooth as polished silver, but it transmits light instead of throwing it back. Roughness only decides whether the reflected light is directed (a mirror image, because parallel rays bounce off parallel) or diffuse (a dull grey sheen, because a rough surface sends parallel rays off in scrambled directions); it cannot explain why nearly all the light returns at all. Glass, for its part, transmits because its electrons are locked to their atoms: to absorb a photon, an electron would have to be lifted across an energy gap of roughly 8–9 electron-volts, while visible photons carry only 1.8–3.1, so the light passes through with nothing able to take its energy. The real question is therefore: what property of a metal forbids light from travelling through it? The free-electron picture The answer lies in the conduction electrons. In a metal, each atom donates one or more of its outer electrons to a shared "gas" of mobile charges — this is the same fact that makes metals conduct electricity. The simplest serious model of this gas is due to Paul Drude (1900): treat each conduction electron as a classical particle of charge $-e$ and mass $m$, free except that from time to time it collides with something and loses its momentum, on average once every collision time $\tau$. Now shine light on it. Light is an oscillating electromagnetic field, and a wave of angular frequency $\omega$ (angular frequency is just $2\pi$ times the ordinary frequency in cycles per second) exerts an oscillating force $-eE$ on each electron. Newton's second law for one electron, with the collisions entering as a friction-like drag $-mv/\tau$, reads $m\,\dot{v} = -eE - mv/\tau$. Solving this for a field oscillating as $e^{-i\omega t}$ gives the electron's displacement, and multiplying by the charge $-e$ and the number of electrons per unit volume $n$ gives the polarisation — the density of induced electric dipoles, i.e. how much the charge cloud shifts in response to the field. The standard bookkeeping quantity for this response is the dielectric function $\varepsilon(\omega)$: the factor by which the medium's shifted charges modify the electric field of a wave of frequency $\omega$, with $\varepsilon = 1$ meaning "responds like vacuum". Carrying the algebra through yields the Drude result $$\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2 + i\gamma\omega},$$ where $\gamma = 1/\tau$ is the collision rate. In the limit $\omega\tau \gg 1$ — the wave oscillates many times between collisions, so the drag term is negligible; this holds well for good metals at visible frequencies, where $\omega \sim 10^{15}\,\mathrm{s^{-1}}$ while $1/\tau \sim 10^{13}\text{–}10^{14}\,\mathrm{s^{-1}}$ — this reduces to $$\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}, \qquad \omega_p^2 = \frac{n e^2}{\varepsilon_0 m},$$ where $n$ is the electron density, $e$ the electron charge, $m$ its mass and $\varepsilon_0$ the vacuum permittivity. These are the SI forms; Ashcroft and Mermin work in Gaussian units, where the same result reads $\omega_p^2 = 4\pi n e^2/m$ and $\varepsilon_0$ does not appear at all. The combination $\omega_p$, the plasma frequency, is not an arbitrary lump of constants: it is the natural oscillation frequency of the electron gas itself. Displace the whole electron sea a small distance $x$ relative to the fixed positive ions, and sheets of uncompensated charge appear at the surfaces, producing a restoring field $E = nex/\varepsilon_0$. Newton's law then gives $m\ddot{x} = -ne^2x/\varepsilon_0$ — the equation of a mass on a spring, oscillating at exactly $\omega_p$. The dielectric function above simply compares the driving light to this internal spring: drive faster than the spring ($\omega > \omega_p$) and the electrons cannot keep up; drive slower ($\omega The solution: a negative dielectric function Put in the numbers. For typical metals $n \sim 10^{28}\text{–}10^{29}\,\mathrm{m^{-3}}$; silver, with one conduction electron per atom, has $n = 5.86\times10^{28}\,\mathrm{m^{-3}}$ (Ashcroft & Mermin, Table 1.1). Then $\omega_p^2 = ne^2/(\varepsilon_0 m) = (5.86\times10^{28} \times 2.57\times10^{-38})/(8.85\times10^{-12} \times 9.11\times10^{-31}) \approx 1.9\times10^{32}\,\mathrm{s^{-2}}$, so $\omega_p \approx 1.4\times10^{16}\,\mathrm{s^{-1}}$ — a photon energy near $9\,\mathrm{eV}$, or a vacuum wavelength around $140\,\mathrm{nm}$, deep in the ultraviolet. (In real silver, absorption by the more tightly bound d-electrons pulls the observed edge down to about $3.9\,\mathrm{eV}$, roughly $320\,\mathrm{nm}$ — still comfortably beyond the violet end of vision.) For all visible light, then, $\omega Why does a negative $\varepsilon$ forbid propagation? The refractive index $n_{\text{opt}}$ — the factor by which light slows in a medium, so that a wave travels as $e^{i(n_{\text{opt}}\omega z/c - \omega t)}$ — satisfies $n_{\text{opt}}^2 = \varepsilon$ in a non-magnetic material. A negative $\varepsilon$ makes $n_{\text{opt}}$ purely imaginary, say $n_{\text{opt}} = i\kappa$. Substitute that into the wave and the oscillating factor $e^{i n_{\text{opt}}\omega z/c}$ becomes $e^{-\kappa\omega z/c}$: not a wave that travels and repeats, but an amplitude that only dies away with depth $z$. Such a field is called evanescent . Its decay length, the skin depth, is $\delta \sim c/\omega_p$ once $\omega \ll \omega_p$; with the silver value above, $\delta \approx (3\times10^8)/(1.4\times10^{16}) \approx 20\,\mathrm{nm}$ — a few tens of nanometres, a hundredth of the wavelength. Light does enter the metal, but only as this thin exponential tail. Now follow the energy. In the ideal collisionless limit the metal absorbs nothing, the wave transmits nothing, and energy conservation leaves only one exit: back ou