Lecture 34
Rayleigh Scattering, Mie
Scattering
34.1 Rayleigh Scattering
Rayleigh scattering is a solution to the scattering of light by small particles. These particles
are assumed to be much smaller than wavelength of light. Then a simple solution can be found
by the method of asymptotic matching. This single scattering solution can be used to explain
a number of physical phenomena in nature. For instance, why the sky is blue, the sunset so
magnificently beautiful, how birds and insects can navigate without the help of a compass.
By the same token, it can also be used to explain why the Vikings, as a seafaring people,
could cross the Atlantic Ocean over to Iceland without the help of a magnetic compass.
339
340 Electromagnetic Field Theory
Figure 34.1: The magnificent beauty of nature can be partly explained by Rayleigh scattering
[182, 183].
When a ray of light impinges on an object, we model the incident light as a plane elec-
tromagnetic wave (see Figure 34.2). Without loss of generality, we can assume that the
electromagnetic wave is polarized in the z direction and propagating in the x direction. We
assume the particle to be a small spherical particle with permittivity εs and radius a. Essen-
tially, the particle sees a constant field as the plane wave impinges on it. In other words, the
particle feels an almost electrostatic field in the incident field.
Figure 34.2: Geometry for studying the Rayleigh scattering problem.
Rayleigh Scattering, Mie Scattering 341
34.1.1 Scattering by a Small Spherical Particle
The incident field polarizes the particle making it look like an electric dipole. Since the
incident field is time harmonic, the small electric dipole will oscillate and radiate like a
Hertzian dipole in the far field. First, we will look at the solution in the vicinity of the
scatter, namely, in the near field. Then we will motivate the form of the solution in the far
field of the scatterer. Solving a boundary value problem by looking at the solutions in two
different physical regimes, and then matching the solutions together is known as asymptotic
matching.
A Hertzian dipole can be approximated by a small current source so that
J(r) = ˆzIlδ(r) (34.1.1)
In the above, we can let the time-harmonic current I = dq/dt = jωq
Il = jωql = jωp (34.1.2)
where the dipole moment p = ql. The vector potential A due to a Hertzian dipole, after
substituting (34.1.1), is
A(r) = μ
4π
V
dr′ J(r′)
|r − r′| e−jβ|r−r′|
= ˆz μIl
4πr e−jβr (34.1.3)
Near Field
From prior knowledge, we know that the electric field is given by E = −jωA − ∇Φ. From a
dimensional analysis, the scalar potential term dominates over the vector potential term in
the near field of the scatterer. Hence, we need to derive the corresponding scalar potential.
The scalar potential Φ(r) is obtained from the Lorenz gauge that ∇ · A = −jωμεΦ.
Therefore,
Φ(r) = −1
jωμε ∇ · A = − Il
jωε4π
∂
∂z
1
r e−jβr (34.1.4)
When we are close to the dipole, by assuming that βr 1, we can use a quasi-static approx-
imation about the potential.1 Then
∂
∂z
1
r e−jβr ≈ ∂
∂z
1
r = ∂r
∂z
∂
∂r
1
r = − z
r
1
r2 (34.1.5)
or after using that z/r = cos θ,
Φ(r) ≈ ql
4πεr2 cos θ (34.1.6)
1This is the same as ignoring retardation effect.
342 Electromagnetic Field Theory
This dipole induced in the small particle is formed in response to the incident field. The
incident field can be approximated by a constant local static electric field,
Einc = ˆzEi (34.1.7)
The corresponding electrostatic potential for the incident field is then2
Φinc = −zEi (34.1.8)
so that Einc ≈ −∇Φinc = ˆzEi, as ω → 0. The scattered dipole potential from the spherical
particle in the vicinity of it is given by
Φsca = Es
a3
r2 cos θ (34.1.9)
The electrostatic boundary value problem (BVP) has been previously solved and3
Es = εs − ε
εs + 2ε Ei (34.1.10)
Using (34.1.10) in (34.1.9), and comparing with(34.1.6), one can see that the dipole moment
induced by the incident field is that
p = ql = 4πε εs − ε
εs + 2ε a3Ei (34.1.11)
Far Field
In the far field of the Hertzian dipole, we can start with
E = −jωA − ∇Φ = −jωA − 1
jωμε ∇∇ · A (34.1.12)
But when we are in the far field, A behaves like a spherical wave which in turn behaves
like a local plane wave if one goes far enough. Therefore, ∇ → −jβ = −jβˆr. Using this
approximation in (34.1.12), we arrive at
E = −jω
(
A − ββ
β2 · A
)
= −jω(A − ˆrˆr · A) = −jω(ˆθAθ + ˆφAφ) (34.1.13)
where we have used ˆr = β/β.
34.1.2 Scattering Cross Section
From (34.1.3), we see that Aφ = 0 while
Aθ = − jωμql
4πr e−jβr sin θ (34.1.14)
2It is not easier to get here from electrodynamics. One needs vector spherical harmonics [184].
3It was one of the homework problems.
Rayleigh Scattering, Mie Scattering 343
Consequently, using (34.1.11) for ql, we have in the far field that4
Eθ ∼ = −jωAθ = − ω2μql
4πr e−jβr sin θ = −ω2με
( εs − ε
εs + 2ε
) a3
r Eie−jβr sin θ (34.1.15)
Hφ ∼ =
√ ε
μ Eθ = 1
η Eθ (34.1.16)
where η = √μ/ε. The time-averaged Poynting vector is given by 〈S〉 = 1/2<e {E × H∗}.
Therefore, the total scattered power is
Ps = 1
2
π
0
r2 sin θdθ
2π
0
dφEθ H∗
φ (34.1.17)
= 1
2η β4
( εs − ε
εs + 2ε
)2 a6
r2 |Ei|2r2
( π
0
sin3 θdθ
)
2π (34.1.18)
But
π
0
sin3 θdθ = −
π
0
sin2 θd cos θ = −
π
0
(1 − cos2 θ)d cos θ
= −
−1
1
(1 − x2)dx = 4
3 (34.1.19)
Therefore
Ps = 4π
3η
( εs − ε
εs + 2εs
)2
β4a6|Ei|2 (34.1.20)
The scattering cross section is the effective area of a scatterer such that the total scattered
power is proportional to the incident power density times the scattering cross section. As
such it is defined as
Σs = Ps
1
2η |Ei|2 = 8πa2
3
( εs − ε
εs + 2ε
)2
(βa)4 (34.1.21)
In other words,
Ps = 〈Sinc〉 × Σs
It is seen that the scattering cross section grows as the fourth power of frequency since
β = ω/c. The radiated field grows as the second power because it is proportional to the
acceleration of the charges on the particle. The higher the frequency, the more the scattered
power. this mechanism can be used to explain why the sky is blue. It also can be used to
explain why sunset has a brilliant hue of red and orange. The above also explain the brilliant
glitter of gold plasmonic nano-particles as discovered by ancient Roman artisans. For gold,
4The ω2 dependence of the following function implies that the radiated electric field in the far zone is
proportional to the acceleration of the charges on the dipole.
344 Electromagnetic Field Theory
the medium resembles a plasma, and hence, we can have εs < 0, and the denominator can be
very small.
Furthermore, since the far field scattered power density of this particle is
〈S〉 = 1
2η Eθ H∗
φ ∼ sin2 θ (34.1.22)
the scattering pattern of this small particle is not isotropic. In other words, these dipoles ra-
diate predominantly in the broadside direction but not in their end-fire directions. Therefore,
insects and sailors can use this to figure out where the sun is even in a cloudy day. In fact,
it is like a rainbow: If the sun is rising or setting in the horizon, there will be a bow across
the sky where the scattered field is predominantly linearly polarized.5 Such a “sunstone” is
shown in Figure 34.3.
Figure 34.3: A sunstone can indicate the polarization of the scattered light. From that, one
can deduce where the sun is located (courtesy of Wikipedia).
34.1.3 Small Conductive Particle
The above analysis is for a small dielectric particle. The quasi-static analysis may not be
valid for when the conductivity of the particle becomes very large. For instance, for a perfect
electric conductor immersed in a time varying electromagnetic field, the magnetic field in the
long wavelength limit induces eddy current in PEC sphere. Hence, in addition to an electric
dipole component, a PEC sphere has a magnetic dipole component. The scattered field due
to a tiny PEC sphere is a linear superposition of an electric and magnetic dipole components.
5You can go through a Gedanken experiment to convince yourself of such.
Rayleigh Scattering, Mie Scattering 345
These two dipolar components have electric fields that cancel precisely at certain observation
angle. This gives rise to deep null in the bi-static radar scattering cross-section (RCS)6 of a
PEC sphere as illustrated in Figure 34.4.
Figure 34.4: RCS (radar scattering cross section) of a small PEC scatterer (courtesy of Sheng
et al. [185]).
34.2 Mie Scattering
When the size of the dipole becomes larger, quasi-static approximation is insufficient to
approximate the solution. Then one has to solve the boundary value problem in its full glory
usually called the full-wave theory or Mie theory [186, 187]. With this theory, the scattering
cross section does not grow indefinitely with frequency. For a sphere of radius a, the scattering
cross section becomes πa2 in the high-frequency limit. This physical feature of this plot is
shown in Figure 34.5, and it also explains why the sky is not purple.
6Scattering cross section in microwave range is called an RCS due to its prevalent use in radar technology.
346 Electromagnetic Field Theory
Figure 34.5: Radar cross section (RCS) calculated using Mie scattering theory [187].
34.2.1 Optical Theorem
Before we discuss Mie scattering solutions, let us discuss an amazing theorem called the
optical theorem. This theorem says that the scattering cross section of a scatterer depends
only on the forward scattering power density of the scatterer. In other words, if a plane wave
is incident on a scatterer, the scatterer will scatter the incident power in all directions. But
the total power scattered by the object is only dependent on the forward scattering power
density of the object or scatterer. This amazing theorem is called the optical theorem, and
the proof of this is given in J.D. Jackson’s book [42].
The true physical reason for this is power orthogonality. Two plane waves can interact or
exchange power with each other unless they share the same k or β vector. This happens in
orthogonal modes in waveguides [75, 188].
The scattering pattern of a scatterer for increasing frequency is shown in Figure 34.6. For
Rayleigh scattering where the wavelength is long, the scattered power is distributed isotrop-
ically save for the doughnut shape of the radiation pattern, namely, the sin2(θ) dependence.
As the frequency increases, the power is scattered increasingly in the forward direction. The
reason being that for very short wavelength, the scatterer looks like a disc to the incident
wave, casting a shadow in the forward direction. Hence, there has to be scattered field in the
forward direction to cancel the incident wave to cast this shadow.
Rayleigh Scattering, Mie Scattering 347
In a nutshell, the scattering theorem is intuitively obvious for high-frequency scattering.
The amazing part about this theorem is that it is true for all frequencies.
Figure 34.6: A particle scatters increasingly more in the forward direction as the frequency
increases.Courtesy of hyperphysics.phy-astr.gsu.edu.
34.2.2 Mie Scattering by Spherical Harmonic Expansions
As mentioned before, as the wavelength becomes shorter, we need to solve the boundary
value problem in its full glory without making any approximations. This can be done by
using separation of variables and spherical harmonic expansions that will be discussed in the
section.
The Mie scattering solution by a sphere is beyond the scope of this course.7 This problem
have to solved by the separation of variables in spherical coordinates. The separation of
variables in spherical coordinates is not the only useful for Mie scattering, it is also useful for
analyzing spherical cavity. So we will present the precursor knowledge so that you can read
further into Mie scattering theory if you need to in the future.
34.2.3 Separation of Variables in Spherical Coordinates
To this end, we look at the scalar wave equation (∇2 +β2)Ψ(r) = 0 in spherical coordinates. A
lookup table can be used to evaluate ∇·∇, or divergence of a gradient in spherical coordinates.
Hence, the Helmholtz wave equation becomes8
( 1
r2
∂
∂r r2 ∂
∂r + 1
r2 sin θ
∂
∂θ sin θ ∂
∂θ + 1
r2 sin2 θ
∂2
∂φ2 + β2
)
Ψ(r) = 0 (34.2.1)
Noting the ∂2/∂φ2 derivative, by using separation of variables technique, we assume Ψ(r) to
be
Ψ(r) = F (r, θ)ejmφ (34.2.2)
7But it is treated in J.A. Kong’s book [31] and Chapter 3 of W.C. Chew, Waves and Fields in Inhomogeneous
Media [34] and many other textbooks [42, 59, 154].
8By quirk of mathematics, it turns out that the first term on the right-hand side below can be simplified
by observing that 1
r2
∂
∂r r2 = 1
r
∂
∂r r.
348 Electromagnetic Field Theory
where ∂2
∂φ2 ejmφ = −m2ejmφ. Then (34.2.1) becomes
( 1
r2
∂
∂r r2 ∂
∂r + 1
r2 sin θ
∂
∂θ sin θ ∂
∂θ − m2
r2 sin2 θ + β2
)
F (r, θ) = 0 (34.2.3)
Again, by using the separation of variables, and letting further that
F (r, θ) = bn(βr)P m
n (cos θ) (34.2.4)
where we require that
{ 1
sin θ
d
dθ sin θ d
dθ +
[
n(n + 1) − m2
sin2 θ
]}
P m
n (cos θ) = 0 (34.2.5)
when P m
n (cos θ) is the associate Legendre polynomial. Note that (34.2.5) is an eigenvalue
problem, and |m| ≤ |n|.
Consequently, bn(kr) satisfies
[ 1
r2
d
dr r2 d
dr − n(n + 1)
r2 + β2
]
bn(βr) = 0 (34.2.6)
The above is the spherical Bessel equation where bn(βr) is either the spherical Bessel function
jn(βr), spherical Neumann function nn(βr), or the spherical Hankel functions, h(1)
n (βr) and
h(2)
n (βr). The spherical functions are related to the cylindrical functions via [34, 43] 9
bn(βr) =
√ π
2βr Bn+ 1
2 (βr) (34.2.7)
It is customary to define the spherical harmonic as
Ynm(θ, φ) =
√
2n + 1
4π
(n − m)!
(n + m)! P m
n (cos θ)ejmφ (34.2.8)
The above is normalized such that
Yn,−m(θ, φ) = (−1)mY ∗
nm(θ, φ) (34.2.9)
and that
2π
0
dφ
π
0
sin θdθY ∗
n′m′ (θ, φ)Ynm(θ, φ) = δn′nδm′m (34.2.10)
These functions are also complete10 if like Fourier series, so that
∞ ∑
n=0
n ∑
m=−n
Y ∗
nm(θ′, φ′)Ynm(θ, φ) = δ(φ − φ′
)δ(cos θ − cos θ′
) (34.2.11)
9By a quirk of nature, the spherical Bessel functions needed for 3D wave equations are in fact simpler than
cylindrical Bessel functions needed for 2D wave equation. One can say that 3D is real, but 2D is surreal.
10In a nutshell, a set of basis functions is complete in a subspace if any function in the same subspace can
be expanded as a sum of these basis functions.
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