A Negative Refractive Index Has to Be Built: Resonance, Bandwidth and Loss — Epoche C2
Textbooks print refractive indices in tables: water 1.33, crown glass 1.52, diamond 2.42. The format invites an inference — that the index is a property of a substance, and that some compound might have an index of $-1$. It is wrong, and the reason predicts the field's two chronic features: a narrow band and unavoidable loss. The index is not a primitive quantity For a plane wave in a uniform medium, Maxwell's equations give a phase velocity $c/\sqrt{\varepsilon\mu}$, so $n^{2}=\varepsilon\mu$. Here $\varepsilon$ is the relative permittivity — how strongly the medium's charges polarise in an electric field — and $\mu$ the relative permeability — how strongly its currents magnetise in a magnetic field. In Gaussian units, used throughout as in Veselago (1968) and Landau and Lifshitz, both are dimensionless and equal 1 in vacuum. The index is derived: whatever one wants from $n$ must be arranged first in $\varepsilon$ and $\mu$. That splits the problem in two, and the split is what the catalogue hides. If only one of the pair is negative, then $\varepsilon\mu < 0$, so $n^{2} < 0$ and $n$ is purely imaginary: the wave decays instead of propagating, which is why a metal below its plasma frequency is a mirror. If both are negative, the wave propagates, but the root to take is the negative one. Veselago (1968) fixed the sign from the curl equations rather than by fiat, and the argument is worth having in front of one because everything downstream depends on it. For a plane wave with dependence $\exp[i(\mathbf{k}\cdot\mathbf{r}-\omega t)]$ the two curl equations become $$\mathbf{k}\times\mathbf{E} = \frac{\omega}{c}\,\mu\mathbf{H}, \qquad \mathbf{k}\times\mathbf{H} = -\frac{\omega}{c}\,\varepsilon\mathbf{E}.$$ With $\varepsilon$ and $\mu$ both positive, these say that $\mathbf{E}$, $\mathbf{H}$ and $\mathbf{k}$ form a right-handed triad in that order. Flip the sign of both, and the two right-hand sides reverse together: the triad becomes left-handed, which is the origin of the name left-handed media. The Poynting vector $\mathbf{S}=(c/4\pi)\,\mathbf{E}\times\mathbf{H}$, however, is defined without reference to $\varepsilon$ or $\mu$ and is unaffected. So $\mathbf{S}$ and $\mathbf{k}$ point in opposite directions: energy flows one way and the phase advances the other. Since $n$ is defined by the phase, its sign must be taken negative: $$n(\omega) = -\sqrt{\varepsilon(\omega)\,\mu(\omega)}, \qquad \varepsilon(\omega) < 0,\ \ \mu(\omega) < 0.$$ In a real, lossy medium the choice of branch is not made by inspection but by passivity: $\varepsilon$ and $\mu$ are complex, $n$ is complex, and the branch is the one for which the wave decays in the direction the energy travels. Where the real parts of $\varepsilon$ and $\mu$ are both negative and the losses are small, that prescription reproduces the negative root above. Nature supplies one half and refuses the other The electric half is free. Any metal below its plasma frequency has $\varepsilon < 0$, and for silver the free-electron plasma energy inferred from the optical constants is about 9 eV. That is the unscreened value; polarisation of the filled d bands lowers the frequency at which the real part of $\varepsilon$ actually crosses zero, but for silver the crossing still falls in the near ultraviolet, so $\varepsilon < 0$ across the whole visible range. The magnetic half is the obstacle, and Landau and Lifshitz show why it is not a gap in the periodic table waiting to be filled. Magnetisation at optical frequencies would have to come from orbital currents. The magnetic energy of a current loop is smaller than the electric energy of the same charges by a factor of order $(v/c)^{2}$, where $v$ is the orbital speed of a bound electron — the magnetic field of a moving charge is smaller than its electric field by $v/c$, and the energy is quadratic in the field. For an outer electron $v/c$ is of order the fine-structure constant, so the suppression is $(1/137)^{2}=5.3\times10^{-5}$. Above the ionic and spin resonances, $\mu$ therefore departs from 1 by at most about $10^{-4}$. Nothing in that estimate refers to a particular element: the suppression is kinematic, not chemical, and no compound repairs it. This is why Landau and Lifshitz argue that at optical frequencies $\mu$ ceases to have independent meaning and should simply be set to 1, all the medium's response being folded into $\varepsilon$. The mechanisms that do give a large permeability — ferromagnetic and antiferromagnetic resonance — are driven by internal exchange and anisotropy fields rather than by the applied field, and their frequencies are set by those fields. They stop near $10^{11}$ Hz. Visible light at $\lambda_{0}=6.0\times10^{-5}$ cm has $f=c/\lambda_{0}=(3\times10^{10})/(6.0\times10^{-5})=5\times10^{14}$ Hz: a ratio of $5\times10^{3}$, between three and four orders of magnitude. There is no natural magnetic response anywhere near the optical band, and the gap is not the sort a better material closes. The composite builds each response separately Since neither response can be ordered from a chemist, both are built. The two constructions are independent, which is the point: each sub-lattice is designed to control one of $\varepsilon$ and $\mu$ while leaving the other alone. Pendry, Holden, Stewart and Youngs (1996) showed that a lattice of thin metal wires of radius $r$ on a square spacing $a$ behaves as a dilute electron gas with the Drude permittivity $\varepsilon(\omega)=1-\omega_{p}^{2}/[\omega(\omega+i\gamma)]$, where $\omega_{p}$ is the plasma frequency and $\gamma$ the damping rate, and that $\omega_{p}$ can be dragged down into the gigahertz range. Two effects do the dragging, and it is worth seeing why the second is the larger. Diluting the metal reduces the carrier density to $n_{\mathrm{eff}}=n\pi r^{2}/a^{2}$, the fraction of the cell cross-section that carries current. Separately, a current $I$ in a wire sets up a circulating magnetic field of magnitude $2I/(c\rho)$ at di