In optics, the Ewald–Oseen extinction theorem, sometimes referred to as just the extinction theorem, is a theorem that underlies the common understanding of scattering (as well as refraction, reflection, and diffraction). It is named after Paul Peter Ewald and Carl Wilhelm Oseen, who proved the theorem in crystalline and isotropic media, respectively, in 1916 and 1915. Originally, the theorem applied to scattering by isotropic dielectric objects in free space. The scope of the theorem was greatly extended to encompass a wide variety of bianisotropic media.
Overview An important part of optical physics theory is starting with microscopic physics—the behavior of atoms and electrons—and using it to derive the familiar, macroscopic, laws of optics. In particular, there is a derivation of how the refractive index works and where it comes from, starting from microscopic physics. The Ewald–Oseen extinction theorem is one part of that derivation (as is the Lorentz–Lorenz equation etc.). When light traveling in vacuum enters a transparent medium like glass, the light slows down, as described by the index of refraction. Although this fact is famous and familiar, it is actually quite strange and surprising when you think about it microscopically. After all, according to the superposition principle, the light in the glass is a superposition of:
The original light wave, and The light waves emitted by oscillating electrons in the glass. (Light is an oscillating electromagnetic field that pushes electrons back and forth, emitting dipole radiation.) Individually, each of these waves travels at the speed of light in vacuum, not at the (slower) speed of light in glass. Yet when the waves are added up, they surprisingly create only a wave that travels at the slower speed. The Ewald–Oseen extinction theorem says that the light emitted by the atoms has a component traveling at the speed of light in vacuum, which exactly cancels out ("extinguishes") the original light wave. Additionally, the light emitted by the atoms has a component which looks like a wave traveling at the slower speed of light in glass. Altogether, the only wave in the glass is the slow wave, consistent with what we expect from basic optics. A more complete description can be found in Classical Optics and its Applications, by Masud Mansuripur. A proof of the classical theorem can be found in Principles of Optics, by Born and Wolf, and that of its extension has been presented by Akhlesh Lakhtakia.
Derivation from Maxwell's equations
Introduction When an electromagnetic wave enters a dielectric medium, it excites (resonates) the material's electrons whether they are free or bound, setting them into a vibratory state with the same frequency as the wave. These electrons will in turn radiate their own electromagnetic fields as a result of their oscillation (EM fields of oscillating charges). Due to the linearity of Maxwell equations, one expects the total field at any point in space to be the sum of the original field and the field produced by oscillating electrons. This result is, however, counterintuitive to the practical wave one observes in the dielectric moving at a speed of c/n, where n is the medium index of refraction. The Ewald–Oseen extinction theorem seek to address the disconnect by demonstrating how the superposition of these two waves reproduces the familiar result of a wave that moves at a speed of c/n.
Derivation The following is a derivation based on a work by Ballenegger and Weber. Let's consider a simplified situation in which a monochromatic electromagnetic wave is normally incident on a medium filling half the space in the region z>0 as shown in Figure 1.
The electric field at a point in space is the sum of the electric fields due to all the various sources. In our case, we separate the fields in two categories based on their generating sources. We denote the incident field
E v a c {\displaystyle \mathbf {E} _{\mathrm {vac} }}
and the sum of the fields generated by the oscillating electrons in the medium
E r a d ( z , t ) . {\displaystyle \mathbf {E} _{\mathrm {rad} }(z,t).}
The total field at any point z in space is then given by the superposition of the two contributions,
E ( z , t ) = E v a c ( z , t ) + E r a d ( z , t ) . {\displaystyle \mathbf {E} (z,t)=\mathbf {E} _{\mathrm {vac} }(z,t)+\mathbf {E} _{\mathrm {rad} }(z,t).}
To match what we already observe, E v a c {\displaystyle \mathbf {E} _{\mathrm {vac} }} has this form. However, we already know that inside the medium, z>0, we will only observe what we call the transmitted E-field E T {\displaystyle \mathbf {E} _{\mathrm {T} }} which travels through the material at speed c/n. Therefore, in this formalism,
E r a d ( z , t ) = − E v a c ( z , t ) + E T ( z , t ) {\displaystyle \mathbf {E} _{\mathrm {rad} }(z,t)=-\mathbf {E} _{\mathrm {vac} }(z,t)+\mathbf {E} _{T}(z,t)}
This to say that the radiated field cancels out the incident field and creates a transmitted field traveling within the medium at speed c/n. Using the same logic, outside the medium the radiated field produces the effect of a reflected field E R {\displaystyle \mathbf {E} _{R}} traveling at speed c in the opposite direction to the incident field.
E r a d ( z , t ) = − E v a c ( z , t ) − E R ( z , t ) {\displaystyle \mathbf {E} _{\mathrm {rad} }(z,t)=-\mathbf {E} _{\mathrm {vac} }(z,t)-\mathbf {E} _{R}(z,t)}
assume that the wavelength is much larger than the average separation of atoms so that the medium can be considered continuous. We use the usual macroscopic E and B fields and take the medium to be nonmagnetic and neutral so that Maxwell's equations read
∇ ⋅ E = 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 J + ϵ 0 μ 0 ∂ E ∂ t {\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &=0\\\nabla \cdot \mathbf {B} &=0\\\nabla \times \mathbf {E} &=-{\frac {\partial \mathbf {B} }{\partial t}}\\\nabla \times \mathbf {B} &={\boldsymbol {\mu }}_{0}\mathbf {J} +\epsilon _{0}{\boldsymbol {\mu }}_{0}{\frac {\partial \mathbf {E} }{\partial t}}\end{aligned}}}
both the total electric and magnetic fields
E = E v a c + E r a d , B = B v a c + B r a d {\displaystyle \mathbf {E} =\mathbf {E} _{\mathrm {vac} }+\mathbf {E} _{\mathrm {rad} },\quad \mathbf {B} =\mathbf {B} _{\mathrm {vac} }+\mathbf {B} _{\mathrm {rad} }}
the set of Maxwell equations inside the dielectric
∇ ⋅ E r a d = 0 ∇ ⋅ B r a d = 0 ∇ × E r a d = − ∂ B r a d / ∂ t ∇ × B r a d = μ 0 J + ϵ 0 μ 0 ∂ E r a d / ∂ t {\displaystyle {\begin{array}{l}{\nabla \cdot \mathbf {E} _{\mathrm {rad} }=0}\\{\nabla \cdot \mathbf {B} _{\mathrm {rad} }=0}\\{\nabla \times \mathbf {E} _{\mathrm {rad} }=-\partial \mathbf {B} _{\mathrm {rad} }/\partial t}\\{\nabla \times \mathbf {B} _{\mathrm {rad} }=\mu _{0}\mathbf {J} +\epsilon _{0}\mu _{0}\partial \mathbf {E} _{\mathrm {rad} }/\partial t}\end{array}}}
where J {\displaystyle \mathbf {J} } includes the true and polarization current induced in the material by the outside electric field. We assume a linear relationship between the current and the electric field, hence
J = σ ( E v a c + E r a d ) {\displaystyle \mathbf {J} ={\sigma }\left(\mathbf {E} _{\mathrm {vac} }+\mathbf {E} _{\mathrm {rad} }\right)}
The set of Maxwell equations outside the dielectric has no current density term
∇ ⋅ E v a c = 0 ∇ ⋅ B v a c = 0 ∇ × E v a c = − ∂ B v a c / ∂ t ∇ × B v a c = ϵ 0 μ 0 ∂ E v a c / ∂ t {\displaystyle {\begin{array}{l}{\nabla \cdot \mathbf {E} _{\mathrm {vac} }=0}\\{\nabla \cdot \mathbf {B} _{\mathrm {vac} }=0}\\{\nabla \times \mathbf {E} _{\mathrm {vac} }=-\partial \mathbf {B} _{\mathrm {vac} }/\partial t}\\{\nabla \times \mathbf {B} _{\mathrm {vac} }=\epsilon _{0}\mu _{0}\partial \mathbf {E} _{\mathrm {vac} }/\partial t}\end{array}}}
The two sets of Maxwell equations are coupled since the vacuum electric field appears in the current density term. For a monochromatic wave at normal incidence, the vacuum electric field has the form
E v a c ( z , t ) = E v a c exp [ i ( k z − ω t ) ] , {\displaystyle \mathbf {E} _{\mathrm {vac} }(z,t)=\mathbf {E} _{\mathrm {vac} }\exp[i(kz-\omega t)],}
with k = ω / c {\displaystyle k=\omega /{c}} . Now to solve for E r a d {\displaystyle \mathbf {E} _{\mathrm {rad} }} , we take the curl of the third equation in the first set of Maxwell equation and combine it with the fourth.
∇ × ( ∇ × E r a d ) = − ∂ ∂ t ( ∇ × B r a d ) ∇ × ( ∇ × E r a d ) = − ∂ ∂ t ( μ 0 J + ϵ 0 μ 0 ∂ E r a d ∂ t ) {\displaystyle {\begin{aligned}\nabla \times (\nabla \times \mathbf {E} _{\mathrm {rad} })&=-{\frac {\partial }{\partial t}}(\nabla \times \mathbf {B} _{\mathrm {rad} })\\[1ex]\nabla \times (\nabla \times \mathbf {E} _{\mathrm {rad} })&=-{\frac {\partial }{\partial t}}\left(\mu _{0}\mathbf {J} +\epsilon _{0}\mu _{0}{\frac {\partial \mathbf {E} _{\mathrm {rad} }}{\partial t}}\right)\end{aligned}}}
We simplify the double curl in a couple of steps using Einstein summation.
∇ × ( ∇ × E ) i = ϵ i j k ϵ k l m ∂ j ∂ l E m = ( δ i l δ j m − δ i m δ j l ) ∂ j ∂ l E m = ∂ i ( ∂ j E j ) − ∂ j ∂ j E i {\displaystyle {\begin{aligned}\nabla \times (\nabla \times \mathbf {E} )_{i}&=\epsilon _{ijk}\epsilon _{klm}\partial _{j}\partial _{l}E_{m}\\&=(\delta _{il}\delta _{j\mathrm {m} }-\delta _{i\mathrm {m} }\delta _{jl})\partial _{j}\partial _{l}E_{m}\\&=\partial _{i}(\partial _{j}E_{j})-\partial _{j}\partial _{j}E_{i}\end{aligned}}}
Hence we obtain,
∇ × ( ∇ × E r a d ) = ∇ ( ∇ ⋅ E r a d ) − ∇ 2 E r a d {\displaystyle \nabla \times (\nabla \times \mathbf {E} _{rad})=\nabla (\nabla \cdot \mathbf {E} _{rad})-\nabla ^{2}\mathbf {E} _{rad}}
Then substituting J {\displaystyle \mathbf {J} } by σ ( E v a c + E r a d ) {\displaystyle {\sigma }\left(\mathbf {E} _{\mathrm {vac} }+\mathbf {E} _{\mathrm {rad} }\right)} , using the fact that ∇ ⋅ E r a d = 0 {\displaystyle \nabla \cdot \mathbf {E} _{\mathrm {rad} }=0} we obtain,
∇ 2 E r a d = ∂ ∂ t ( μ 0 σ E v a c + μ 0 σ E r a d + ϵ 0 μ 0 ∂ E r a d / ∂ t ) {\displaystyle \nabla ^{2}\mathbf {E} _{\mathrm {rad} }={\frac {\partial }{\partial t}}(\mu _{0}{\sigma }\mathbf {E} _{\mathrm {vac} }+\mu _{0}{\sigma }\mathbf {E} _{\mathrm {rad} }+\epsilon _{0}\mu _{0}\partial \mathbf {E} _{\mathrm {rad} }/\partial t)}
Realizing that all the fields have the same time dependence exp ( − i ω t ) {\displaystyle \exp(-i\omega t)} , the time derivatives are straightforward and we obtain the following inhomogeneous wave equation
∇ 2 E r a d + μ 0 ω 2 ( ϵ 0 + i σ / ω ) E r a d = − i μ 0 ω σ E v a c ( z ) {\displaystyle \nabla ^{2}\mathbf {E} _{\mathrm {rad} }+\mu _{0}\omega ^{2}\left(\epsilon _{0}+i\sigma /\omega \right)\mathbf {E} _{\mathrm {rad} }=-i\mu _{0}\omega \sigma \mathbf {E} _{\mathrm {vac} }(z)}
with particular solution
E r a d P = − E v a c ( z ) {\displaystyle \mathbf {E} _{\mathrm {rad} }^{P}=-\mathbf {E} _{\mathrm {vac} }(z)}
For the complete solution, we add to the particular solution the general solution of the homogeneous equation which is a superposition of plane waves traveling in arbitrary directions
( E r a d c ) i = ∫ g i ( θ , ϕ ) exp ( i k ′ ⋅ r ) d Ω {\displaystyle \left(\mathbf {E} _{\mathrm {rad} }^{c}\right)_{i}=\int g_{i}({\boldsymbol {\theta }},{\boldsymbol {\phi }})\exp \left(i\mathbf {k} '\cdot \mathbf {r} \right)d\Omega }
where k ′ {\displaystyle k'} is found from the homogeneous equation to be
k ′ 2 = μ 0 ϵ 0 ω 2 ( 1 + i σ ϵ 0 ω ) {\displaystyle k^{\prime 2}=\mu _{0}\epsilon _{0}\omega ^{2}\left(1+i{\frac {\sigma }{\epsilon _{0}\omega }}\right)}
Note that we have taken the solution as a coherent superposition of plane waves. Because of symmetry, we expect the fields to be the same in a plane perpendicular to the z {\displaystyle z} axis. Hence k ′ ⋅ a = 0 , {\displaystyle \mathbf {k} '\cdot \mathbf {a} =0,} where a {\displaystyle \mathbf {a} } is a displacement perpendicular to z {\displaystyle z} . Since there are no boundaries in the region z > 0 {\displaystyle z>0} , we expect a wave traveling to the right. The solution to the homogeneous equation becomes,
E r a d c = E T exp ( i k ′ z ) {\displaystyle \mathbf {E} _{\mathrm {rad} }^{c}=\mathbf {E} _{T}\exp \left(ik'z\right)}
Adding this to the particular solution, we get the radiated wave inside the medium ( z > 0 {\displaystyle z>0} )
E r a d = − E v a c ( z ) + E T exp ( i k ′ z ) {\displaystyle \mathbf {E} _{\mathrm {rad} }=-\mathbf {E} _{\mathrm {vac} }(z)+\mathbf {E} _{T}\exp \left(ik'z\right)}
The total field at any position z {\displaystyle z} is the sum of the incident and radiated fields at that position. Adding the two components inside the medium, we get the total field
E ( z ) = E T exp ( i k ′ z ) , z > 0 {\displaystyle \mathrm {E} (z)=\mathrm {E} _{T}\exp \left(ik'z\right),\qquad z>0}
This wave travels inside the dielectric at speed c / n , {\displaystyle c/n,}
n = c k ′ / ω = 1 + i σ ϵ 0 ω {\displaystyle n=ck'/\omega ={\sqrt {1+i{\frac {\sigma }{\epsilon _{0}\omega }}}}}
We can simplify the above n {\displaystyle n} to a familiar form of the index of refraction of a linear isotropic dielectric. To do so, we remember that in a linear dielectric an applied electric field E {\displaystyle \mathbf {E} } induces a polarization P {\displaystyle \mathbf {P} } proportional to the electric field P = ϵ 0 χ e E {\displaystyle \mathbf {P} =\epsilon _{0}\chi _{e}\mathbf {E} } . When the electric field changes, the induced charges move and produces a current density given by ∂ P / ∂ t {\displaystyle \partial \mathbf {P}