Scattering by an infinite cylinder: two polarizations

A cylinder is the two-dimensional cousin of a sphere, and it keeps a feature the sphere loses by symmetry: it tells the two polarizations of the incident light apart. Rayleigh solved the problem for a dielectric cylinder at normal incidence in 1918; Bohren and Huffman's chapter 8 is the standard modern account and the source of the notation below. Everything on this page is for an infinitely long circular cylinder of radius $a$ in a host of index $n_h$, lit by a plane wave of vacuum wavelength $\lambda$. The size parameter and relative index are as for a sphere,

Cylindrical waves and the two cases

In cylindrical coordinates $(\rho, \varphi, z)$ the vector Helmholtz equation separates into modes $e^{i m\varphi}\, e^{i k_z z}$ whose radial dependence obeys Bessel's equation (the Foundations table). The regular solution is $J_m(q\rho)$ and the outgoing one $H^{(1)}_m(q\rho)$, with $q = \sqrt{k^2 - k_z^2}$ the radial wavenumber. At normal incidence, $k_z = 0$, $q = k$, and the problem splits into two independent cases, named after the field component along the axis:

• Case I: the incident electric field is parallel to the cylinder axis (TM with respect to the axis; the calculator's E along the axis). The axial component $E_z$ is the scalar field; $\mathbf{H}$ lies in the cross section.
• Case II: the incident magnetic field is parallel to the axis (TE; H along the axis). Now $H_z$ is the scalar, and $\mathbf{E}$ lies in the cross section.

For each order $m$ the two boundary conditions, continuity of the tangential $\mathbf{E}$ and $\mathbf{H}$ at $\rho = a$, fix one scattering coefficient per case, exactly as for the transfer matrix of any stratified geometry. Bohren and Huffman's results for a non-magnetic cylinder are

the relative index $m$ in the two expressions sitting where the sphere's $a_\ell$ and $b_\ell$ have it, swapped between the cases. The resonances are the near-zeros of the denominators: for a dielectric they are the whispering-gallery or leaky modes of the rod, which make a silicon nanowire of a given diameter absorb one colour preferentially (Cao et al. 2009); for a metal they are the surface plasmons of the wire.

Efficiencies per unit length

An infinite cylinder has no cross-sectional area to normalise with, so one works per unit length. The scattering and extinction widths $C_\mathrm{sca}$, $C_\mathrm{ext}$ (a length) divided by the geometric width $2a$ give the efficiencies

and the same with $a_m^{\mathrm{II}}$ for case II; $Q_\mathrm{abs} = Q_\mathrm{ext} - Q_\mathrm{sca}$, and unpolarized light averages the two cases. The orders $\pm m$ give equal contributions, hence the factor 2. The series stops a few orders beyond $x$, as Mie's does. Bohren and Huffman plot instead the scattering per unit volume, $C_\mathrm{sca}/V = 4 Q_\mathrm{sca}/(\pi d)$ with $d = 2a$, which tends to the geometric-optics asymptote $8/(\pi d)$ as $Q \to 2$: the quantity a chemist measuring turbidity cares about.

import numpy as np
import pystratify as ps
from pystratify.docs import material  # n + ik of a refractiveindex.info page

lam = (400 + 4 * np.arange(151)) * 1e-3  # wavelength, µm
N = np.column_stack([np.full(lam.shape, (1.55+0j)), np.full(lam.shape, (1+0j))])  # n + ik: core, then the host
radii = [0.25]  # outer radii, µm
q_ext = {pol: np.array([ps.cross_widths(ps.solve_cylinder(radii, N[i], wl), pol)['efficiency_extinction']
                        for i, wl in enumerate(lam)]) for pol in ('unpolarized', 'axial-electric', 'axial-magnetic')}

The two polarizations differ most strongly at a few hundred nanometres of radius, where case I rises first and case II lags: a resonance of case I at a given order $m$ needs the electric field to fit along the axis, that of case II needs the magnetic field to, and for $m > 1$ these fall at different sizes. At large $x$ both tend to 2 with a decaying ripple, and the polarization dependence fades.

The thin wire

For $x \ll 1$ the two cases behave very differently, and the difference is the whole story of metal nanowires. In case I the field along the axis induces a current along the wire with no charges at any surface; the response is proportional to $(m^2 - 1)$ and the scattering efficiency grows like $x^3$, with no resonance. In case II the field across the wire polarises it like a two-dimensional dipole, and the leading coefficient is

The denominator is the two-dimensional analogue of the sphere's $\varepsilon + 2\varepsilon_h$: a metal wire in vacuum resonates where $\varepsilon = -1$, which for a Drude metal is $\omega = \omega_p/\sqrt{2}$ rather than the sphere's $\omega_p/\sqrt{3}$. Embedding the wire in a medium of permittivity $\varepsilon_h$ moves the resonance to $\varepsilon = -\varepsilon_h$. The wires and fibres section follows this resonance down to a wire one nanometre thick, where the local description reaches its limit.

Coated cylinders and oblique incidence

A cylinder with shells is solved by the same $2\times 2$ recursion as a multilayered sphere: one transfer matrix per interface and per order $m$, multiplied inward from the host. PyStratify's cylinder solver does this for any number of shells, with the amplitudes carried in logarithmic form so that a thick absorbing shell or a high order never overflows. At oblique incidence $k_z = k\cos\zeta$ is no longer zero: $q$ shrinks, every Bessel argument changes, and, except for $m = 0$, the two polarizations couple at each interface, so the transfer matrix becomes $4\times 4$ per order and a case-I wave scatters partly into case-II light. The calculator's incidence angle for cylinders is this $\zeta$ (90° is normal incidence); the hybrid modes it then finds are the ones that guide light along an optical fibre, which is the subject of the wires and fibres section.

How much can one channel scatter? Superscattering

Each order $m$ is a separate channel, and for a lossless cylinder its coefficient can at most reach the value that makes $|b_m|^2$ (or $|a_m|^2$) equal to 1: the contribution of a single channel to the scattering width is bounded by $2\lambda/\pi$ (counting $\pm m$ together). A subwavelength object that resonates in one channel can therefore never scatter more than this, no matter how strong the resonance; it is the two-dimensional analogue of the single-channel limit of a sphere. Ruan and Fan noticed that nothing stops several channels from resonating at the same wavelength if the structure has enough free parameters, and designed a three-layer plasmonic rod whose $m = 1, 2, 3, 4$ resonances coincide, so that the scattering width exceeds the single-channel bound several times over: superscattering. The design is sensitive by construction, because four resonances have to be tuned onto each other.