# Waves in composites and metamaterials/Perfect lenses and negative density materials

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.

## Perfect Lenses

Recall the interface separating media with permittivities $\epsilon _{1},\epsilon _{2}$ and permeabilities $\mu _{1},\mu _{2}$ that "behaves like a mirror" (see Figure 1).
* Figure 1. "Reflection" at an interface.
For the situation shown in the figure, on the left hand side (LHS) of the interface, let $\mathbf {E} _{L}$ and $\mathbf {H} _{L}$ solve Maxwell's equations
    ${\boldsymbol {\nabla }}\times \mathbf {E} _{L}+i\omega ~\mu _{1}~\mathbf {H} _{L}={\boldsymbol {0}}~;~~{\boldsymbol {\nabla }}\times \mathbf {H} _{L}-i\omega ~\epsilon _{1}~\mathbf {E} _{L}={\boldsymbol {0}}~.$
Let the solutions on the LHS be of the form
    $\mathbf {E} _{L}(\mathbf {x} )=[E_{1}(\mathbf {x} ),E_{2}(\mathbf {x} ),E_{3}(\mathbf {x} )]~~{\text{and}}~~\mathbf {H} _{L}(\mathbf {x} )=[H_{1}(\mathbf {x} ),H_{2}(\mathbf {x} ),H_{3}(\mathbf {x} )]~.$
Also, on the right hand side, let
    ${\boldsymbol {\nabla }}\times \mathbf {E} _{L}=[F_{1}(\mathbf {x} ),F_{2}(\mathbf {x} ),F_{3}(\mathbf {x} )]~~{\text{and}}~~{\boldsymbol {\nabla }}\times \mathbf {H} _{L}=[I_{1}(\mathbf {x} ),I_{2}(\mathbf {x} ),I_{3}(\mathbf {x} )]~.$
On the right hand side of the interface, let $\mathbf {E} _{R}$ and $\mathbf {H} _{R}$ solve the Maxwell equations
    ${\boldsymbol {\nabla }}\times \mathbf {E} _{R}+i\omega ~\mu _{2}~\mathbf {H} _{R}={\boldsymbol {0}}~;~~{\boldsymbol {\nabla }}\times \mathbf {H} _{R}-i\omega ~\epsilon _{2}~\mathbf {E} _{R}={\boldsymbol {0}}~.$
If the interface acts as a mirror such that the right hand side (RHS) of the interface has reflected fields, then
    ${\begin{aligned}\mathbf {E} _{R}(\mathbf {x} )&=[-E_{1}(-x_{1},x_{2},x_{3}),E_{2}(-x_{1},x_{2},x_{3}),E_{3}(-x_{1},x_{2},x_{3})]~~{\text{and}}~~\\\mathbf {H} _{R}(\mathbf {x} )&=[-H_{1}(-x_{1},x_{2},x_{3}),H_{2}(-x_{1},x_{2},x_{3}),H_{3}(-x_{1},x_{2},x_{3})]~.\end{aligned}}$
Then, to the right of the interface, we have
    ${\begin{aligned}{\boldsymbol {\nabla }}\times \mathbf {E} _{R}=&\left\{\left[{\frac {\partial }{\partial x_{2}}}[E_{3}(-x_{1},x_{2},x_{3})]-{\frac {\partial }{\partial x_{3}}}[E_{2}(-x_{1},x_{2},x_{3})]\right],\right.\\&\left[-{\frac {\partial }{\partial x_{3}}}[E_{1}(-x_{1},x_{2},x_{3})]+{\frac {\partial }{\partial x_{1}}}[E_{3}(-x_{1},x_{2},x_{3})]\right],\\&\left.\left[-{\frac {\partial }{\partial x_{1}}}[E_{2}(-x_{1},x_{2},x_{3})]+{\frac {\partial }{\partial x_{2}}}[E_{1}(-x_{1},x_{2},x_{3})]\right]\right\}\end{aligned}}$
or,
    ${\boldsymbol {\nabla }}\times \mathbf {E} _{R}=[F_{1}(-x_{1},x_{2},x_{3}),-F_{2}(-x_{1},x_{2},x_{3}),-F_{3}(x_{1},x_{2},x_{3})]~.$
Similarly, on the RHS, we have
    ${\boldsymbol {\nabla }}\times \mathbf {H} _{R}=[I_{1}(-x_{1},x_{2},x_{3}),-I_{2}(-x_{1},x_{2},x_{3}),-I_{3}(x_{1},x_{2},x_{3})]~.$
For continuity of the fields at the interface, we must have
    $\mu _{2}=-\mu _{1}\qquad {\text{and}}\qquad \epsilon _{2}=-\epsilon _{1}~.$
This implies that negative permeability and permittivity have the bizzare property of reflecting the fields.

### A series of interfaces

Let us consider a slab of one material immersed in another medium. Let the permittivity and permeability of the surrounding medium be $\epsilon =\mu =1$ (normalized with respect to the values for free space). Let the normalized permittivity and permeability of the slab be $\epsilon =\mu =-1$. Hence, the first interface between the medium and the slab acts as a mirror in that it reflects the electric field $\mathbf {E}$. The second interface also acts as a mirror and reflects the field $\mathbf {E}$ to the original orientation (see Figure 2).
* Figure 2.The effect of a slab of material with $\epsilon =\mu =-1$ on the field $\mathbf {E}$.
If the source is located at a distance $d_{0}$ from the first interface, and the slab has a thickness $d$, we have
    $\mathbf {E} (x_{1}+2d,x_{2},x_{3})=\mathbf {E} (x_{1},x_{2},x_{3})\qquad {\text{for}}\quad -d<x_{1}<0~.$
Therefore, the effect of the slab is just a translation. The same is true for the $\mathbf {H}$, $\mathbf {D}$, and $\mathbf {B}$ fields.
Let $\mathbf {E} _{0}$ be the field which solves the electromagnetic problem with the slab removed for a given source. Let us now insert a negative index slab in the field. The effect of the slab is that the fields appear to move to the right of the slab, i.e., to the right of the slab it appears as if all the fields have been moved a distance $2d$. In other words, it appears that the source has been moved a distance $2d$ to the right (see Figure 3).
* Figure 3. The effect of a negative index slab is to make the source appear to be at the right of the slab.
This implies that the slab works as a "perfect lens" in the sense that the image to the right of the slab is not diffraction limited. This observation was first made by Pendry [Pendry2000] and was a surprising result because most lenses were though to be diffraction limited.
Consider for instance the ordinary lens shown in Figure 4(a). From geometric optics we expect the rays from the source to be focussed at a point. However, if we consider the wave nature of electromagnetic radiation, several Fourier components of the wave are superimposed at the focal point and the maximum resolution of the image can never be greater than $\lambda /2$ where $\lambda$ is the wavelength.
* Figure 4. Focussing due to an ordinary lens versus a "perfect" lens.
On the other hand, the lensing effect with a slab of negative index material is expected to lead to a point source being exactly represented at the focal point. This idea dates back to Veselago [Veselago1968]. From Figure 4(b) we observe that there will appear to be sources inside the lens and at the focal point when a negative index slab is used as a lens. However, a point source leads to a singularity in the Maxwell equations and there should be no singularities where there no physical point sources. This is a paradox.
The paradox can be resolved by observing that, in fact, a solution does not exist to the time harmonic equations if $\epsilon =\mu =-1$ in the slab. However, if we let
    $\epsilon =-1+i\delta \qquad {\text{and}}\qquad \mu =-1+i\delta$
and let $\delta \rightarrow 0$, then we do have a solution. This is equivalent to assuming that there some loss in the material due to the electrical conductivity of the material.
For $d_{0}>d/2$ (see Figure 5), the fields blow up to infinity within a strip of width $2(d-d_{0})$ starting from the focal point within the slab to the focal point outside the slab. They develop more and more oscillations (in space), i.e., at finer and finer length scales. In the remaining regions, the field converge to Pendry's solution. Therefore, the image looks like a point source only on one side of the lens if $\delta \neq 0$. However, in the limit that $\delta \rightarrow 0$, the image also looks like a point source.
* Figure 5. Behavior of fields around and inside a "perfect" lens.
If we look at the wave vectors of the electromagnetic waves, then the from the reflected direction of the wave vector inside the lens it appears that light travels backwards inside a negative refractive index lens. But one has to remember that it is the wave crests that are travelling backwards and transport of energy is in the direction propagation of the EM waves (and therefore it is not useful to think of the direction of the wave vector as the direction of the Poynting vector). Note that the phase velocity is negative if the refractive index of the material is negative. Figure 6 illustrates this point.
* Figure 6. Direction of the wave vector and the wave packets inside a negative refractive index lens.

## Negative, Complex Anisotropic Density Materials

An important question in elasticity is how can we get materials with negative moduli. The answer is "through resonance". In this section we approach the problem by exploring the difference between "dynamic" density and "static" density. The static density is defined as the mass per unit volume whereas the dynamic density is defined as the inertial density that appears in Newton's law $F=m~a$. Is the dynamic density the same as the static density?
This question was first addressed by Sheng and coworkers [Sheng2003]. This work was extended by Liu et al. [Liu2005] and the mathematical analysis by Avila et al. [Avila2005]. More recently, several models have been explored by Milton and Willis [Milton2007]. The following discussion is based on one of the models presented by Milton and Willis.
Consider a rigid bar with $n$ voids each of width $d$ as shown in Figure 7. Each cavity contains a spherical ball of mass $m$ and radius $r$. The ball is attached to the walls of the cavity by springs with sprint constant $K$ (the spring constant may be complex valued to allow for materials with viscous damping). The spring attached to the left wall of the cavity exerts a force $f_{1}(t)$ on the wall while the spring attached to the right wall exerts a force of $f_{2}(t)$ on the wall. A force $F(t)$ is applied on the left side of the rigid bar. Our aim is to find the response of the bar as a function of time.
* Figure 7. A rigid bar containing $n$ voids. Each void contains a spherical ball that is attached to the bar by springs.[p]
Let us assume that all quantities depend harmonically on time. Let us also assume that a one-dimensional approximation of the problem is adequate. Then the forces acting on the bar are given by
    ${\text{(1)}}\qquad F(t)={\text{Re}}({\widehat {F}}~e^{-i\omega t})~;~~f_{1}(t)={\text{Re}}({\widehat {f}}_{1}~e^{-i\omega t})~;~~f_{2}(t)={\text{Re}}({\widehat {f}}_{2}~e^{-i\omega t})$
where the amplitudes ${\widehat {F}}$, ${\widehat {f}}_{1}$, and ${\widehat {f}}_{2}$ are generally complex.
From Newton's second law,
    ${\text{(2)}}\qquad F(t)={\cfrac {d}{dt}}[P(t)]$
where $P(t)$ is the linear momentum which is of the harmonic form
    ${\text{(3)}}\qquad P(t)={\text{Re}}({\widehat {P}}~e^{-i\omega t})~.$
Therefore, plugging equations (1) and (3) into equation (2), we get
    ${\text{Re}}({\widehat {F}}~e^{-i\omega t})={\text{Re}}(-i\omega ~{\widehat {P}}~e^{-i\omega t})$
or
    ${{\widehat {F}}=-i\omega ~{\widehat {P}}~.}$
Let the time-dependent position of the left side of each cavity be given by
    $X(t)=X_{0}+U(t)=X_{0}+{\text{Re}}({\widehat {U}}~e^{-i\omega t})$
where $X_{0}$ is the initial position and ${\widehat {U}}$ is the complex valued displacement of the bar.
Similarly, the position of each ball is assumed to be given by
    ${\text{(4)}}\qquad x(t)=X(t)+{\cfrac {d}{2}}+u(t)=X_{0}+{\cfrac {d}{2}}+[U(t)+u(t)]=X_{0}+{\cfrac {d}{2}}+{\text{Re}}({\widehat {u}}~e^{-i\omega t})$
where ${\widehat {u}}$ is the complex valued displacement of each ball.
Then the velocity of the rigid bar is
    ${\cfrac {dX}{dt}}={\text{Re}}(-i\omega ~{\widehat {U}}~e^{-i\omega t})={\text{Re}}({\widehat {V}}~e^{-i\omega t})~.$
Assume that the rigid bar has mass $M_{0}$. Therefore, the linear momentum of the rigid bar is
    $P_{\text{bar}}(t)=M_{0}~{\cfrac {dX}{dt}}=M_{0}~{\text{Re}}({\widehat {V}}~e^{-i\omega t})~.$
Similarly, the velocity of each sphere is
    ${\cfrac {dx}{dt}}={\text{Re}}(-i\omega ~{\widehat {u}}~e^{-i\omega t})={\text{Re}}({\widehat {v}}~e^{-i\omega t})~.$
If there are $n$ balls, the total linear momentum of the balls is
    $P_{\text{ball}}(t)=n~m~{\cfrac {dx}{dt}}=n~m~{\text{Re}}({\widehat {v}}~e^{-i\omega t})~.$
Therefore, the total linear momentum of the system is
    $P(t)=M_{0}~{\text{Re}}({\widehat {V}}~e^{-i\omega t})+n~m~{\text{Re}}({\widehat {v}}~e^{-i\omega t})={\text{Re}}({\widehat {P}}~e^{-i\omega t})~.$
From Newton's second law, the applied force equals the rate of change of linear momentum. Therefore,
    $F(t)={\cfrac {d}{dt}}[P(t)]=M_{0}~{\text{Re}}(-i\omega ~{\widehat {V}}~e^{-i\omega t})+n~m~{\text{Re}}(-i\omega ~{\widehat {v}}~e^{-i\omega t})={\text{Re}}(-i\omega ~{\widehat {P}}~e^{-i\omega t})~.$
Therefore,
    ${\text{(5)}}\qquad {{\widehat {P}}=M_{0}~{\widehat {V}}+n~m~{\widehat {v}}~.}$
But ${\widehat {v}}$ is unobservable since it is in the hidden part of the bar and we need to relate ${\widehat {P}}$ directly to the observable velocity ${\widehat {V}}$.
Let us now consider the free-body diagram of each spring inside a cavity (see Figure 8).
* Figure 8. Free-body diagrams for the spring-mass system and for each spring.
Hooke's law for each spring implies that (note that $u(t)$ is positive in the positive $x$ direction)
    $-f_{1}(t)=K~u(t)=f_{2}(t)$
where $K$ is the complex spring constant.
Recall (from equation 4) that the displacement of the spring is given by
    $u(t)={\text{Re}}({\widehat {u}}~e^{-i\omega t})-{\text{Re}}({\widehat {U}}~e^{-i\omega t})~.$
Using the assumed harmonic forms of $f_{1}(t)$, $f_{2}(t)$ and $u(t)$, we then have
    ${\text{Re}}({\widehat {f}}_{1}~e^{-i\omega t})=-{\text{Re}}({\widehat {f}}_{2}~e^{-i\omega t})=K~\left[{\text{Re}}({\widehat {U}}~e^{-i\omega t})-{\text{Re}}({\widehat {u}}~e^{-i\omega t})\right]$
or
    ${\text{(6)}}\qquad {{\widehat {f}}_{1}=-{\widehat {f}}_{2}=K~({\widehat {U}}-{\widehat {u}})~.}$
Next, considering the free body diagram of the sping-mass system (see Figure 8), the balance of linear momentum for the spring-mass system implies that
    $f_{1}(t)-f_{2}(t)=m~{\cfrac {d^{2}x}{dt^{2}}}=m~{\text{Re}}(-\omega ^{2}~{\widehat {u}}~e^{-i\omega t})~.$
Therefore, substituting in the harmonic forms of $f_{2}(t)$ and $f_{1}(t)$, we get
    ${\text{(7)}}\qquad {{\widehat {f}}_{1}-{\widehat {f}}_{2}=-m~\omega ^{2}~{\widehat {u}}~.}$
From equations (6) and (7), we have
    ${\widehat {f}}_{1}-{\widehat {f}}_{2}=-m~\omega ^{2}~{\widehat {u}}=2~K~({\widehat {U}}-{\widehat {u}})$
or,
    $2~K~{\widehat {U}}=(2~K-m~\omega ^{2})~{\widehat {u}}\qquad \implies \qquad {\widehat {u}}={\cfrac {2~K}{2~K-m~\omega ^{2}}}~{\widehat {U}}~.$
Now ${\widehat {V}}=-i\omega {\widehat {U}}$ and ${\widehat {v}}=-i\omega {\widehat {u}}$. Hence,
    ${\text{(8)}}\qquad {{\widehat {v}}={\cfrac {2~K}{2~K-m~\omega ^{2}}}~{\widehat {V}}~.}$
Plugging equation (8) into equation (5), we get
    ${{\widehat {P}}=\left(M_{0}+{\cfrac {2~K~n~m}{2~K-m~\omega ^{2}}}\right)~{\widehat {V}}=M~{\widehat {V}}}$
where $M$ is the effective mass. Clearly, the effective mass depends on the frequency $\omega$ and is different from the static mass.
A normalized plot of the effective mass versus the frequency in shown in Figure 9.
* Figure 9. The effective mass of the bar as a function of the frequency. $\omega _{r}$ is the resonant frequency.
At $\omega =0$, the effective mass is equal to the rest mass $M_{0}+n~m$. The resonant frequency is given by
    $\omega _{r}={\sqrt {\cfrac {2~K}{m}}}~.$
Close the resonant frequency, the effective mass can either take high positive values or negative values.