Waves in composites and metamaterials/Bloch waves and the quasistatic limit

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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.

Bloch Theorem

In the previous lecture we showed that Maxwell's equations at fixed frequency can be formulated in terms of the fields 𝐃 and 𝐁 as [1]

(1)𝐃=0;𝐁=0;i×(𝐃ϵ)=ω𝐁;i×(𝐁μ)=ω𝐃.

Equations (1) suggest that we should look for solutions 𝐃 and 𝐁 in the space of divergence-free fields such that

(2)β„’[𝐃𝐁]=ω[𝐃𝐁]or(β„’ω1)[𝐃𝐁]=0

where the operator β„’ is given by

(3)β„’:=[0i×μ1i×ϵ10].

If the medium is such that the permittivity ϵ(𝐱) and the permeability μ(𝐱) are periodic, i.e.,

ϵ(𝐱)=ϵ(𝐱+𝐑);μ(𝐱)=μ(𝐱+𝐑)

where 𝐑 is a lattice vector (see Figure 1) then the operator β„’ has the same periodicity as the medium.

File:Lec16Fig1.jpg
Figure 1. Lattice vector in a periodic medium.

Also recall the translation operator 𝒯R defined as

𝒯R[𝐃(𝐱)𝐁(𝐱)]=[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)].

Periodicity of the medium implies that 𝒯R commutes with β„’, i.e.,

𝒯Rβ„’=ℒ𝒯R.

[2] The translation operator is unitary, i.e.,

𝒯R𝒯RT=𝒯R𝒯R=1.

This means that the adjoint operator 𝒯RT is equal to the inverse operator 𝒯R.

The translation operator also commutes, i.e.,

𝒯R𝒯R=𝒯R𝒯R=𝒯R+R.

[3] Also, since β„’ and 𝒯R commute, the operators β„’ω1 and 𝒯R must also commute. This implies that

𝒯R(β„’ω1)[𝐃(𝐱)𝐁(𝐱)]=0=(β„’ω1)𝒯R[𝐃(𝐱)𝐁(𝐱)]=(β„’ω1)[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)].

Hence the eigenstates of β„’ω1 and the eigenstates of 𝒯R lie in the same space. Therefore, any solution can be expressed in fields which are simultaneously eigenstates of all the 𝒯R, i.e., these eigenstates have the property

(4)[𝒯Rc(𝐑)1][𝐃(𝐱)𝐁(𝐱)]=0.

Since 𝒯R𝒯R=𝒯R+R, we have

c(𝐑)c(𝐑)=c(𝐑+𝐑).

So it suffices to know c(𝐑) when 𝐑=𝐚1,𝐚2,𝐚3 where the 𝐚i's are the primitive vectors of the lattice, i.e.,

ϵ(𝐱+𝐚j)=ϵ(𝐱);μ(𝐱+𝐚j)=μ(𝐱).

Let us assume that

c(𝐚j)=e2πiαjj=1,2,3

for a suitable choice of αj.

Then for any lattice vector

𝐑=n1𝐚1+n2𝐚2+n3𝐚3

we have

c(𝐑)=c(n1𝐚1+n2𝐚2+n3𝐚3)=c(n1𝐚1)c(n2𝐚2)c(n3𝐚3)=c(i=1n1𝐚1)c(i=1n2𝐚2)c(i=1n3𝐚3)=[c(𝐚1)]n1[c(𝐚2)]n2[c(𝐚3)]n3=e2πiα1n1e2πiα2n2e2πiα3n3=e2πi(α1n1+α2n2+α3n3)

or,

c(𝐑)=e2πi(α1n1+α2n2+α3n3).

Define a vector

𝐀:=α1𝐛1+α2𝐛2+α3𝐛3

where the vectors 𝐛i are the reciprocal lattice vectors satisfying

𝐛i𝐚j=2πδij.

Then,

𝐀𝐑=(α1𝐛1+α2𝐛2+α3𝐛3)(n1𝐚1+n2𝐚2+n3𝐚3)=α1n12π+α2n22π+α3n32π

or,

𝐀𝐑=2π(α1n1+α2n2+α3n3).

Therefore, we have

c(𝐑)=ei𝐀𝐑.

Plugging this expression into (4), we get

[𝒯Rei𝐀𝐑1][𝐃(𝐱)𝐁(𝐱)]=0

or,

(5)[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)]=ei𝐀𝐑[𝐃(𝐱)𝐁(𝐱)].

Equation (5) is called the Bloch condition.

In summary, the solutions to the electromagnetic equations in a periodic medium can be expressed in Bloch waves where each Bloch wave is a time harmonic solution to the electromagnetic equations which in addition satisfies the Bloch condition for all lattice vectors 𝐑 and for some appropriate choice of 𝐀.

Note that for any vector 𝐱, the Bloch condition implies that

ei𝐀(𝐱+𝐑)[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)]=ei𝐀𝐑i𝐀𝐑i𝐀𝐱[𝐃(𝐱)𝐁(𝐱)]=ei𝐀𝐱[𝐃(𝐱)𝐁(𝐱)].

Therefore the quantity

ei𝐀𝐱[𝐃(𝐱)𝐁(𝐱)]

is periodic.

Quasistatic Limit

Let us now consider the solution of Maxwell's equation in periodic media in the quasistatic limit. [4] Consider the periodic medium shown in Figure 2. The lattice spacing is η.

File:Lec17Fig1.jpg
Figure 2. Periodic medium with x and y spaces.

Define

ϵη(𝐱):=ϵ(𝐱η)=ϵ(𝐲);μη(𝐱):=μ(𝐱η)=μ(𝐲).

These are periodic functions, i.e.,

ϵ(𝐲+𝐚i)=ϵ(𝐲);μ(𝐲+𝐚i)=μ(𝐲)

where 𝐚i are the primitive lattice vectors. We may also write these periodicity conditions as

ϵη(𝐱+η𝐚i)=ϵη(𝐱);μη(𝐱+η𝐚i)=μη(𝐱).

Similarly, define

𝐃η(𝐱):=𝐃(𝐲);𝐁η(𝐱):=𝐁(𝐲);𝐄η(𝐱):=𝐄(𝐲);𝐇η(𝐱):=𝐇(𝐲).

Then Maxwell's equations can be written as

(6)𝐃η=0;𝐁η=0;×𝐄ηiω𝐁η=0;×𝐇η+iω𝐃η=0.

Let us look for Bloch wave solutions of the form

𝐄η(𝐱)=ei𝐀𝐱𝐞η(𝐱);𝐃η(𝐱)=ei𝐀𝐱𝐝η(𝐱);𝐇η(𝐱)=ei𝐀𝐱𝐑η(𝐱);𝐁η(𝐱)=ei𝐀𝐱𝐛η(𝐱)

where 𝐞η,𝐝η,𝐑η,𝐛η have the same periodicity as ϵ and μ, i.e.,

𝐞η(𝐱+η𝐚i)=𝐞(𝐱);𝐝η(𝐱+η𝐚i)=𝐝(𝐱);𝐑η(𝐱+η𝐚i)=𝐑(𝐱);𝐛η(𝐱+η𝐚i)=𝐛(𝐱).

From the constitutive relations, we get

𝐝η(𝐱)=ϵη(𝐱)𝐞η(𝐱);𝐛η(𝐱)=μη(𝐱)𝐑η(𝐱).

Recall that, for periodic media, Maxwell's equations may be expressed as

(β„’ω1)[𝐃𝐁]=0.

Here ω is an eigenvalue of β„’. However, β„’ depends on ω via ϵ(ω) and μ(ω). {\bf Bloch wave solutions do not exists unless ω takes one of a discrete set of values.}

Let these discrete values be

ω=ωηj(𝐀)

where the superscript j labels the solution branches.

Let us see what the Bloch wave solutions reduce to as η0. Following standard multiple scale analysis, let us assume that the periodic complex fields have the expansions

(7)𝐞η(𝐱)=𝐞0(𝐲)+η𝐚1(𝐲)+η2𝐚2(𝐲)+𝐝η(𝐱)=𝐝0(𝐲)+η𝐝1(𝐲)+η2𝐝2(𝐲)+𝐑η(𝐱)=𝐑0(𝐲)+η𝐑1(𝐲)+η2𝐑2(𝐲)+𝐛η(𝐱)=𝐛0(𝐲)+η𝐛1(𝐲)+η2𝐛2(𝐲)+

Let us also assume that the dependence of ω on η and 𝐀 has an expansion of the form

(8)ω=ωηj(𝐀)=ω0j(𝐲)+ηω1j(𝐲)+η2ω2j(𝐲)+

Plugging (8) and (7) into (6) gives

(9)(eiη𝐀𝐲[𝐝0(𝐲)+η𝐝1(𝐲)+])=0(eiη𝐀𝐲[𝐛0(𝐲)+η𝐛1(𝐲)+])=0×(eiη𝐀𝐲[𝐞0(𝐲)+η𝐞1(𝐲)+])i[ω0j+ηω1j+][𝐛0(𝐲)+η𝐛1(𝐲)+]=0×(eiη𝐀𝐲[𝐑0(𝐲)+η𝐑1(𝐲)+])+i[ω0j+ηω1j+][𝐝0(𝐲)+η𝐝1(𝐲)+]=0.

Define

y:=(y1,y2,y3).

Then, for a vector field 𝐯(𝐲), using the chain rule we get

(10)𝐯(𝐲)=1ηy𝐯(𝐲);×𝐯(𝐲)=1ηy×𝐯(𝐲).

Using definitions (10) in (9) and collecting terms of order 1/η gives

(11)y𝐝0(𝐲)=0y𝐛0(𝐲)=0y×𝐞0(𝐲)=0y×𝐑0(𝐲)=0

These are the solutions in the quasistatic limit. Also, from the constitutive equations

𝐝0(𝐲)=ϵ(𝐲)𝐞0(𝐲);𝐛0(𝐲)=μ(𝐲)𝐑0(𝐲).

Similarly, collecting terms of order 1 from the expanded Maxwell's equations (9) we get

(12)i𝐀𝐝0(𝐲)+y𝐝1(𝐲)=0i𝐀𝐛0(𝐲)+y𝐛1(𝐲)=0i𝐀×𝐞0(𝐲)+y×𝐞1(𝐲)iω0j𝐛0(𝐲)=0i𝐀×𝐑0(𝐲)+y×𝐑1(𝐲)+iω0j𝐝0(𝐲)=0.

Since 𝐝1(𝐲),𝐛1(𝐲),𝐞1(𝐲),𝐑1(𝐲) are periodic, this implies that

(13)y𝐝1(𝐲)=0y𝐛1(𝐲)=0y×𝐞1(𝐲)=0y×𝐑1(𝐲)=0.

where is the volume average over the unit cell. So a necessary condition that equations (12) have a solution is that

(14)i𝐀𝐝0(𝐲)=0i𝐀𝐛0(𝐲)=0i𝐀×𝐞0(𝐲)iω0j𝐛0(𝐲)=0i𝐀×𝐑0(𝐲)+iω0j𝐝0(𝐲)=0.

Note that the second pair of (14) implies the first pair.

Footnotes

  1. ↑ The following discussion is based on Ashcroft76 (p. 133-139).
  2. ↑ We can see that the two operators commute by working out the operations. Thus,
    𝒯Rβ„’[𝐃𝐁]=𝒯R[i×[μ1(𝐱)𝐁(𝐱)]i×[ϵ1(𝐱)𝐃(𝐱)]]=[i×[μ1(𝐱+𝐑)𝐁(𝐱+𝐑)]i×[ϵ1(𝐱+𝐑)𝐃(𝐱+𝐑)]]=β„’[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)]=ℒ𝒯R[𝐃𝐁].
  3. ↑ We can see that the translation operator commutes by working out the operations. Thus,
    𝒯R𝒯R[𝐃(𝐱)𝐁(𝐱)]=𝒯R[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)]=[𝐃(𝐱+𝐑+𝐑)𝐁(𝐱+𝐑+𝐑)]=𝒯R+R[𝐃(𝐱)𝐁(𝐱)]=𝒯R[𝐃(𝐱+𝐑)𝐁(𝐱+𝐑)]=𝒯R𝒯R[𝐃(𝐱)𝐁(𝐱)].
  4. ↑ The following discussion is based on Milton02

References

  • N. W. Ashcroft and N. D. Mermin. Solid State Physics. Saunders, New York, 1976.
  • G. W. Milton. Theory of Composites. Cambridge University Press, New York, 2002.