# Waves in composites and metamaterials/Mie theory and Bloch theorem

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

## Scattering of radiation from a sphere

Recall the sphere shown in Figure 1. We set up our coordinate system such that the origin is at the center of the sphere. The sphere has a magnetic permeability of $\mu$ and a permittivity $\epsilon$ . The medium outside the sphere has a permittivity $\epsilon _{0}$ and a permeability $\mu _{0}$ . The electric field is oriented parallel to the $x_{1}$ axis and the $x_{2}$ axis points out of the plane of the paper. Figure 1. Scattering of radiation from a sphere.

Also recall that

$\mu =\mu _{0}~;~~\epsilon =\epsilon _{r}~\epsilon _{0}=n^{2}~\epsilon _{0}$ where $\epsilon _{r}$ is the relative permittivity of the material inside the sphere and that the incident plane wave is given by

$\mathbf {E} _{i}=e^{ikx_{3}}~\mathbf {e} _{1}$ where $\mathbf {e} _{1}$ is the unit vector in the $x_{1}$ direction.

The most widely used superpotentials are the electric and magnetic Hertz vector potentials ${\boldsymbol {\Pi }}_{e}$ and ${\boldsymbol {\Pi }}_{m}$ (also known as polarization potentials).

In the last lecture we discussed the Hertz vector potentials and that the $\mathbf {E}$ and $\mathbf {H}$ fields can be expressed as

{\text{(1)}}\qquad {\begin{aligned}\mathbf {E} &={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {{\boldsymbol {\Pi }}_{e}}-\mu ~{\boldsymbol {\nabla }}\times {\frac {\partial {\boldsymbol {\Pi }}_{m}}{\partial t}}\\\mathbf {H} &={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {{\boldsymbol {\Pi }}_{m}}+\epsilon ~{\boldsymbol {\nabla }}\times {\frac {\partial {\boldsymbol {\Pi }}_{e}}{\partial t}}~.\end{aligned}} For spherically symmetric time harmonic problems, such as we find in the problem of scattering of EM waves by a sphere, we stated that an important class of Hertz vector potentials are the Debye potentials of the form

${\boldsymbol {\Pi }}_{e}=u~\mathbf {r} ~;~~{\boldsymbol {\Pi }}_{m}=v~\mathbf {r} \qquad {\text{where}}\quad \mathbf {r} \equiv (x_{1},x_{2},x_{3})~.$ Let the time harmonic fields be given by

$\mathbf {E} ={\widehat {\mathbf {E} }}~e^{-i\omega t}~;~~\mathbf {H} ={\widehat {\mathbf {H} }}~e^{-i\omega t}~;~~u={\hat {u}}~e^{-i\omega t}~;~~v={\hat {v}}~e^{-i\omega t}~.$ Plugging these into (1) and dropping the hats gives the Maxwell equations at fixed frequency:

{\begin{aligned}\mathbf {E} &={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {(u~\mathbf {r} )}+i\omega \mu ~{\boldsymbol {\nabla }}\times (v~\mathbf {r} )\\\mathbf {H} &={\boldsymbol {\nabla }}\times {\boldsymbol {\nabla }}\times {(v~\mathbf {r} )}-i\omega \epsilon ~{\boldsymbol {\nabla }}\times (u~\mathbf {r} )~.\end{aligned}} Recall that the Debye potentials satisfy the homogeneous wave equations

${\text{(2)}}\qquad (\nabla ^{2}+k^{2})u=0\qquad {\text{and}}\qquad (\nabla ^{2}+k^{2})v=0~.$ To deal with the problem of scattering by a sphere, let us split the potentials $u$ and $v$ (outside the sphere) into incident and scattered fields:

$u=u_{i}+u_{s}~;~~v=v_{i}+v_{s}$ where the subscript $i$ indicates an incident field and the subscript $s$ indicates a scattered field.

Inside the sphere, the potentials are denoted by

$u=u_{r}~;~~v=v_{r}$ where the subscript $r$ indicates a refracted + reflected field.

Let us require that these potentials satisfy wave equations of the form given in (2), i.e.,

{\begin{aligned}(\nabla ^{2}+k^{2})u_{i}&=0&\qquad {\text{and}}\qquad &(\nabla ^{2}+k^{2})v_{i}=0\\(\nabla ^{2}+k^{2})u_{s}&=0&\qquad {\text{and}}\qquad &(\nabla ^{2}+k^{2})v_{s}=0\\(\nabla ^{2}+k^{2}n^{2})u_{r}&=0&\qquad {\text{and}}\qquad &(\nabla ^{2}+k^{2}n^{2})v_{r}=0~.\end{aligned}} Since each of these satisfies a scalar wave equation, we can express each potential in terms of spherical harmonics.

In particular, the Debye potentials associated with the incident field

$\mathbf {E} _{i}=e^{ikx_{3}}~\mathbf {e} _{1}$ have the expression

{\begin{aligned}ru_{i}&={\cfrac {1}{k^{2}}}\sum _{l=1}^{\infty }{\cfrac {i^{l-1}(2l+1)}{l(l+1)}}~\psi _{l}(kr)~P_{l}^{1}(\cos \theta )~\cos \phi \\rv_{i}&={\cfrac {1}{\eta k^{2}}}\sum _{l=1}^{\infty }{\cfrac {i^{l-1}(2l+1)}{l(l+1)}}~\psi _{l}(kr)~P_{l}^{1}(\cos \theta )~\sin \phi \end{aligned}} where

$\psi _{l}(\rho )={\sqrt {\cfrac {\pi ~r}{2}}}~J_{l+1/2}(\rho )~;~~\eta ={\sqrt {\cfrac {\mu _{0}}{\epsilon _{0}}}}~.$ Here $P_{l}^{1}(x)$ are the Legendre polynomials which solve

${\cfrac {d}{dx}}\left[(1-x^{2})~{\cfrac {dP_{l}^{1}}{dx}}\right]+\left[l(l+1)-{\cfrac {1}{1-x^{2}}}\right]~P_{l}^{1}=0$ and $J_{\nu }(\rho )$ are the Bessel functions which solve

${\cfrac {d^{2}J_{\nu }}{d\rho ^{2}}}+{\cfrac {1}{\rho }}~{\cfrac {dJ_{\nu }}{d\rho }}+\left[1-{\cfrac {\nu ^{2}}{\rho ^{2}}}\right]~J_{\nu }=0~.$ The functions $\psi (\rho )$ are chosen such that

$\psi _{\nu }(\rho )={\sqrt {\cfrac {\pi ~r}{2}}}~J_{\nu }(\rho )$ is regular at the origin.

The scattered fields have a similar expansion

{\begin{aligned}ru_{s}&={\cfrac {-1}{k^{2}}}\sum _{l=1}^{\infty }{\cfrac {i^{l-1}(2l+1)}{l(l+1)}}~a_{l}~\zeta _{l}(kr)~P_{l}^{1}(\cos \theta )~\cos \phi \\rv_{s}&={\cfrac {-1}{\eta k^{2}}}\sum _{l=1}^{\infty }{\cfrac {i^{l-1}(2l+1)}{l(l+1)}}~b_{l}~\zeta _{l}(kr)~P_{l}^{1}(\cos \theta )~\sin \phi \end{aligned}} where

$\zeta _{l}(\rho )={\sqrt {\cfrac {\pi ~r}{2}}}~H_{l+1/2}^{(1)}(\rho )$ and $H_{\nu }^{(1)}(\rho )$ is one of the Hankel functions solving the same equation as the Bessel function but decaying at infinity.

Inside the sphere, the expansion of the fields takes the form

{\begin{aligned}ru_{r}&={\cfrac {1}{k^{2}n^{2}}}\sum _{l=1}^{\infty }{\cfrac {i^{l-1}(2l+1)}{l(l+1)}}~c_{l}~\psi _{l}(knr)~P_{l}^{1}(\cos \theta )~\cos \phi \\rv_{r}&={\cfrac {1}{\eta k^{2}n^{2}}}\sum _{l=1}^{\infty }{\cfrac {i^{l-1}(2l+1)}{l(l+1)}}~b_{l}~\zeta _{l}(knr)~P_{l}^{1}(\cos \theta )~\sin \phi \end{aligned}} To find the constants $a_{l},b_{l},c_{l},d_{l}$ we need to apply continuity conditions across the boundary of the sphere.

To ensure that $E_{\theta },E_{\phi },H_{\theta },H_{\phi }$ (tangential components of $\mathbf {E}$ and $\mathbf {H}$ ) are continuous across the surface of the sphere at $r=a$ , it is sufficient that

$n^{2}u~,~~{\frac {\partial }{\partial r}}(ru)~;~~{\frac {\partial }{\partial r}}(rv)$ are continuous.

Applying these conditions, we get

{\begin{aligned}a_{l}&={\cfrac {\psi _{l}(\alpha )~\psi '_{l}(\beta )-n~\psi _{l}(\beta )~\psi '_{l}(\alpha )}{\zeta _{l}(\alpha )~\psi '_{l}(\beta )-n~\psi _{l}(\beta )~\zeta '_{l}(\alpha )}}\\b_{l}&={\cfrac {n~\psi _{l}(\alpha )~\psi '_{l}(\beta )-\psi _{l}(\beta )~\psi '_{l}(\alpha )}{n~\zeta _{l}(\alpha )~\psi '_{l}(\beta )-\psi _{l}(\beta )~\zeta '_{l}(\alpha )}}\end{aligned}} where

$\alpha =k~a~;~~\beta =k~n~a~.$ The scattered field $E_{\theta }$ , $E_{\phi }$ far from the sphere are given by

{\begin{aligned}E_{\theta }&={\cfrac {i~e^{ikr}}{kr}}~S_{1}(\theta )~\cos \phi \\E_{\phi }&=-{\cfrac {i~e^{ikr}}{kr}}~S_{1}(\theta )~\sin \phi \end{aligned}} where

{\begin{aligned}S_{1}(\theta )&=\sum _{l=1}^{\infty }{\cfrac {2l+1}{l(l+1)}}\left[a_{l}~\pi _{l}(\cos \theta )+b_{l}~\tau _{l}(\cos \theta )\right]\\S_{2}(\theta )&=\sum _{l=1}^{\infty }{\cfrac {2l+1}{l(l+1)}}\left[a_{l}~\tau _{l}(\cos \theta )+b_{l}~\pi _{l}(\cos \theta )\right]\end{aligned}} where

$\pi _{l}(\cos \theta )={\cfrac {P_{l}^{1}(\cos \theta )}{\sin \theta }}~;~~\tau _{l}(\cos \theta )={\cfrac {d}{d\theta }}~P_{l}^{1}(\cos \theta )~.$ Note that the tangential components of $\mathbf {E}$ fall off as $1/r$ while the radial component falls off as $1/r^{2}$ .

## Periodic Media and Bloch's Theorem

The following discussion is based on Ashcroft76 (p. 133-139). For a more detailed mathematical treatment see Kuchment93.

Suppose that the medium is such that the permittivity $\epsilon (\mathbf {x} )$ and the permeability $\mu (\mathbf {x} )$ are periodic. Recall that, at fixed frequency, the Maxwell equations are

${\text{(3)}}\qquad {\boldsymbol {\nabla }}\cdot (\epsilon ~\mathbf {E} )=0~;~~{\boldsymbol {\nabla }}\cdot (\mu ~\mathbf {H} )=0~;~~{\boldsymbol {\nabla }}\times \mathbf {E} +i\omega \mu ~\mathbf {H} =0~;~~{\boldsymbol {\nabla }}\times \mathbf {H} -i\omega \epsilon ~\mathbf {E} =0~.$ Also recall the constitutive relations

${\text{(4)}}\qquad \mathbf {D} =\epsilon ~\mathbf {E} ~;~~\mathbf {B} =\mu ~\mathbf {H} ~.$ Plugging (4) into (3), we get

${\text{(5)}}\qquad {\boldsymbol {\nabla }}\cdot \mathbf {D} =0~;~~{\boldsymbol {\nabla }}\cdot \mathbf {B} =0~;~~i~{\boldsymbol {\nabla }}\times \left({\cfrac {\mathbf {D} }{\epsilon }}\right)=\omega ~\mathbf {B} ~;~~-i~{\boldsymbol {\nabla }}\times \left({\cfrac {\mathbf {B} }{\mu }}\right)=\omega ~\mathbf {D} ~.$ Equations (5) suggest that we should look for solutions $\mathbf {D}$ and $\mathbf {B}$ in the space of divergence-free fields such that

${\text{(6)}}\qquad {\mathcal {L}}{\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}=\omega ~{\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}$ where the operator ${\mathcal {L}}$ is given by

${\text{(7)}}\qquad {\mathcal {L}}:={\begin{bmatrix}0&-i~{\boldsymbol {\nabla }}\times \mu ^{-1}\\i~{\boldsymbol {\nabla }}\times \epsilon ^{-1}&0\end{bmatrix}}~.$ Since $\epsilon$ and $\mu$ are periodic, the operator ${\mathcal {L}}$ has the same periodicity as the medium.

Clearly, equation (6) represents an eigenvalue problem where $\omega$ is an eigenvalue of ${\mathcal {L}}$ and $[\mathbf {D} ,\mathbf {B} ]$ is the corresponding eigenvector.

Let ${\mathcal {T}}_{R}$ define a translation operator which, when acting upon a pair of the fields $[\mathbf {D} ,\mathbf {B} ]$ shifts the argument by a vector ${\boldsymbol {R}}v$ , where ${\boldsymbol {R}}v$ is taken to be a lattice vector (see Figure~2), i.e.,

${\mathcal {T}}_{R}{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}={\begin{bmatrix}\mathbf {D} (\mathbf {x} +{\boldsymbol {R}}v)\\\mathbf {B} (\mathbf {x} +{\boldsymbol {R}}v)\end{bmatrix}}~.$  Figure 2. Lattice vector in a periodic medium.

Periodicity of the medium implies that ${\mathcal {T}}_{R}$ commutes with ${\mathcal {L}}$ , i.e.,

${\mathcal {T}}_{R}~{\mathcal {L}}={\mathcal {L}}~{\mathcal {T}}_{R}~.$ Note that ${\mathcal {T}}_{R}$ , like ${\mathcal {L}}$ , maps divergence-free fields to divergence-free fields.

Now, consider the space of field pairs $\mathbf {D} ,\mathbf {B}$ which are divergence-free and which are in the null space of ${\mathcal {L}}-\omega ~{\boldsymbol {1}}$ , i.e., they satisfy

$({\mathcal {L}}-\omega ~{\boldsymbol {1}}){\begin{bmatrix}\mathbf {D} \\\mathbf {B} \end{bmatrix}}={\boldsymbol {0}}~.$ This subspace is closed under the action of ${\mathcal {T}}_{R}$ which is unitary, i.e.,

${\mathcal {T}}_{R}~{\mathcal {T}}_{R}^{T}={\mathcal {T}}_{R}~{\mathcal {T}}_{-R}={\boldsymbol {1}}~.$ Also, the translation operator commutes, i.e.,

${\mathcal {T}}_{R}~{\mathcal {T}}_{R'}={\mathcal {T}}_{R'}~{\mathcal {T}}_{R}={\mathcal {T}}_{R+R'}~.$ Therefore, any solution can be expressed in fields which are simultaneously eigenstates of all the ${\mathcal {T}}_{R}$ . These eigenstates have the property

${\mathcal {T}}_{R}{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}=c({\boldsymbol {R}}v)~{\begin{bmatrix}\mathbf {D} (\mathbf {x} )\\\mathbf {B} (\mathbf {x} )\end{bmatrix}}~.$ The Bloch condition will be discussed in the next lecture.

## Footnotes

1. This discussion is based on Ishimaru78. Please consult that text and the reference cited therein for further details.