Numerical errors can be avoided if the number of operations that have to be performed numerically is reduced. This can be achieved by choosing a coordinate system and basis functions that are suited to the geometry of the problem, and by evaluating integrals analytically when possible. In this section, we present an example of such prior considerations that is vital for the later evaluation of MIET experiments, namely the total energy flux of a dipole emitter close to a metal layer.
In section 2.4.5, we determined the angular distribution of radiation d2S/dΩ2(θ, φ) of a dipole in a stratified environment, i.e. in a system of several planar layers consisting of different materials. If at least one of these materials is absorbing (e.g. a metal), the energy emitted in the far-field R
(d2S/dΩ2)dΩ2 is smaller than the total energy emitted by the dipole Stot. The latter can be obtained by integrating the time-averaged Poynting vector over a surface that encloses the dipole but no other energy sources or sinks. Since the interaction of the dipole field with planar interfaces is most easily described using plane waves, integrating over a spherical surface as we did for spherical geometries in section 2.4.6 seems ill-advised. Instead, we form a closed integration surface by integrating over two infinite planes, one above and one below the emitter, but both in the same medium as the emitter5. Provided that the medium containing the dipole has a real refractive index, which is a reasonable assumption for all our experiments, the sum of the two integrals then yields Stot.
We start by considering the simple case that the dipole is situated in a medium with refractive index n1 which occupies the whole upper halfspace z > 0, while arbitrary layers make up the lower halfspace z < 0. Then, the interface between the dipole emitter’s medium and the adjoining material at z = 0 is a good choice for the lower of the two integration surfaces. Denoting the z-position of the dipole as z0, we saw in section 2.3.2 that the field atz = 0 is given by: These two fields cause a total flux through the plane z = 0 that is given by:
S− = c
5 Mathematically speaking, two parallel planes intersect at infinity. From a physical point of view, we can simply say that all energy except the negligible fraction being emitted atexactly θ=π/2 will cross one of the two planes sooner or later.
= cn31k6v
The naive implementation of this equation would require the numerical evaluation of five integrals, entailing a high risk of numerical errors. However, in this formula, we recognize the definition of Dirac’s delta-distribution:
δ2(q−q0) = 1 (2π)2
Z
d2ρexp[iρ·(q−q0)]. (3.31) Taking the integral over ρ thus produces the termδ2(q−q0), which, when integrating over the whole q0-plane, is zero everywhere except at q0 = q. Taking into account that q0 =k1sinθ10(cosφ0eˆx+ sinφ0eˆy), the integral over q0-space is given by q0dq0dφ0 = effect on the amplitude, as expected for evanescent waves. We split the integral into propagating and evanescent waves in order to take the real part because the integration measure dθ1 is real for real θ1 and imaginary for the pathθ1 = π/2 +i∞ →π/2. Since we assumed that no other materials are present above the dipole, there is also no energy sink, and we can choose an arbitrary z > z0 for the upper integration surface. It then
3
(c) phase accumulated due to reflections (a) geometry
Figure 3.6: Multiple reflections in a stratified sample. (a) Sample geometry: The dipole (green arrow) is situated at z= z0 in a medium with refractive index n1 that extends from z= 0 to z =d. The interaction with the layers above and below is described by the effective reflection coefficients ru and rd, which were calculated in section 3.2. (b) Unit vectors of the polarization of p- and s-waves for downwards (-) and upwards (+) traveling waves. (c) Due to multiple reflections between the top and bottom interface of the dipole’s medium, a plane wave changes its amplitude and accumulates a phase shift. When we sum these waves, we do so at fixed r, thus the phase is only acquired due to propagation alongz. The shift along the perpendicular direction is only shown for clear labeling of the waves. The expressions on the right give the phase and amplitude for downwards traveling waves at the marked positions, assuming that the amplitude was one and the phase zero at the position of the dipole.
makes sense to simply use the far-field ADR from equation (2.170) and integrate it over θ1 ∈[0, π/2].
The situation gets slightly more complex if other materials are present above the dipole, too. Then, a plane wave that hits the lower interface gets partially reflected, reaches the upper interface, gets partially reflected again and so on. In order to take this into account, we consider the geometry in figure 3.6(a): The dipole’s medium with refractive indexn1 fills the space 0≤z ≤d. Above and below, there are arbitrary other materials.
The effective Fresnel reflection coefficients for the upper and lower interfaces at z =d and z = 0 are ru,p/s andrd,p/s, respectively, calculated according to section 3.2. With these multiple reflections, it is easier to define θ1 as in figure 3.6(c)6 and to take the different propagation directions into account via the sign of w1 =k1cosθ1.
The expressions on the right side of figure 3.6(c) correspond to the amplitude and phase that accumulate due to the multiple reflections. The electric field at z = 0 is then composed of four different types of waves, those that have been originally emitted into the upper or lower halfspace by the dipole (blue or red in the figure), which are each either travelling in positive or negative z-direction at z = 0. For p-waves, the former
6 That is, withθ1∈[0, π/2] for propagating waves andθ1∈π/2 +i[0,∞] for evanescent waves.
category determines the initial amplitude according to ˆe±p1·p, while the latter category gives the polarization as ˆe±p1 (see figure 3.6(b)). For s-waves, all four types of waves have the same initial amplitude and polarization. The phase and final amplitude of each wave type at z = 0 is found by summing up all single plane waves belonging to the same category: where we used the fact that each sum is a geometric series and thereby eliminated another source of numerical errors. Thus, when going back to expression (3.28) for the electric field at the interface, all we have to do is replace the single plane waves by these sums:
The total flux through the plane z = 0, S−, can then be derived as before. Since the reflection coefficients do not depend on φ, the φ-integration can be carried out analytically.
We give the explicit result for two special cases, a vertical dipole with pk eˆz, and a horizontal dipole with pkeˆx:
These formulas were used, for example, to generate the curves in figure 2.18b that show Stot for a dipole in water close to a silver-coated glass cover slip. All calculations of MIET curves were done with these equations, too, exploiting the fact that for a dipole at an angle θ relative to the z-axis (see the end of section 2.5.3),
S−(θ) = cos2θ·S−,⊥+ sin2θ·S−,k . (3.37) As a quick consistency check, also consider the free space case. Then, all reflection coefficients are zero, and out of the four wave types only cdd 6= 0. This leads to S−=cn1kv4p2/6 =S0/2, which is the correct result.
Due to the symmetry of the situation, a reflection of the whole geometry converts the flux S+ towards the upper layers into S−. Thus, S+ can be obtained by exchanging all material parameters in the formulas above.