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Calculation of material properties and ray tracing in transformation media

D. Schurig, J. B. Pendry, and D. R. Smith

Open Access Open Access

Abstract

Complex and interesting electromagnetic behavior can be found in spaces with non-flat topology. When considering the properties of an electromagnetic medium under an arbitrary coordinate transformation an alternative interpretation presents itself. The transformed material property tensors may be interpreted as a different set of material properties in a flat, Cartesian space. We describe the calculation of these material properties for coordinate transformations that describe spaces with spherical or cylindrical holes in them. The resulting material properties can then implement invisibility cloaks in flat space. We also describe a method for performing geometric ray tracing in these materials which are both inhomogeneous and anisotropic in their electric permittivity and magnetic permeability.

©2006 Optical Society of America

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Figures (3)
Fig. 1.
Fig. 1. The thick blue line shows the path of the same ray in (A) the original Cartesian space, and under two different interpretations of the electromagnetic equations, (B) the topological interpretation and (C) the materials interpretation. The position vector x is shown in both the original and transformed spaces, and the length of the vector where the transformed components are interpreted as Cartesian components is shown in (C).

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Fig. 2.
Fig. 2. Rays traversing a spherical cloak. The transformation media that comprises the cloak lies between the two spheres.

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Fig. 3.
Fig. 3. Rays traversing a cylindrical cloak at an oblique angle. The transformation media that comprises the cloak lies in an annular region between the cylinders.

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Equations (58)

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F α β , μ + F β μ , α + F μ α , β = 0
G , α α β = J β
( F α β ) = ( 0 E 1 E 2 E 3 − E 1 0 − c B 3 c B 2 − E 2 c B 3 0 − c B 1 − E 3 − c B 2 c B 1 0 )
( G α β ) = ( 0 − c D 1 − c D 2 − c D 3 c D 1 0 − H 3 H 2 c D 2 H 3 0 − H 1 c D 3 − H 2 H 1 0 )
( J β ) = ( cρ J 1 J 2 J 3 )
G α β = 1 2 C α β μ ν F μ ν
C α ′ β ′ μ ′ ν ′ = ∣ det ( Λ α α ′ ) ∣ − 1 Λ α α ′ Λ β β ′ Λ μ μ ′ Λ ν ν ′ C α β μ ν
Λ α α ′ = ∂ x α ′ ∂ x α
ε i ′ j ′ = ∣ det ( Λ i i ′ ) ∣ − 1 Λ i i ′ Λ j j ′ ε ij
μ i ′ j ′ = ∣ det ( Λ i i ′ ) ∣ − 1 Λ i i ′ Λ j j ′ ε ij
ε i ′ j ′ = ∣ det ( g i ′ j ′ ) ∣ − 1 ⁄ 2 g i ′ j ′ ε
μ i ′ j ′ = ∣ det ( g i ′ j ′ ) ∣ − 1 ⁄ 2 g i ′ j ′ μ
g i ′ j ′ = Λ k i ′ Λ l j ′ δ kl
r = ( x i x j δ ij ) 1 ⁄ 2 = ( x i ′ x j ′ g i ′ j ′ ) 1 ⁄ 2
r ′ = ( x i ′ x j ′ δ i ′ j ′ ) 1 ⁄ 2
r ′ = b − a b r + a
x i ′ r ′ = x i r δ i i ′
x i ′ = b − a b x i δ i i ′ + a x i r δ i i ′
∂ ∂ x j x i r = − x i x k δ kj r 3 + 1 r δ j i
Λ j i ′ = r ′ r δ j i ′ − a x i x k δ j i ′ δ kj r 3
( Λ j i ′ ) = ( r ′ r − a x 2 r 3 − axy r 3 − axz r 3 − ayx r 3 r ′ r − ay 2 r 3 − ayz r 3 − azx r 3 − azy r 3 r ′ r − az 2 r 3 )
( x i ) = ( r , 0 , 0 )
det ( Λ j i ′ ) = r ′ − a r ( r ′ r ) 2
ε i ′ j ′ = μ i ′ j ′ = b b − a [ δ i ′ j ′ − 2 ar ′ − a 2 r ′ 4 x i ′ x j ′ ]
ε = μ = b b − a ( I − 2 ar − a 2 r 4 r ⊗ r )
det ( ε ) = det ( μ ) = ( b b − a ) 3 ( r − a r ) 2
Z ij = δ 3 i δ 3 j
T ij = δ 1 i δ 1 j + δ 2 i δ 2 j
ρ i = T j i x j
( Λ j i ′ ) = ( ρ ′ ρ − ax 2 ρ 3 − axy ρ 3 0 − ayx ρ 3 ρ ′ ρ − ay 2 ρ 3 0 0 0 1 )
Λ j i ′ = ρ ′ ρ T j i ′ − a ρ i ρ k δ i i ′ δ kj ρ 3 + Z j i ′
( x i ) = ( ρ i ) = ( ρ , 0 , 0 )
det ( Λ j i ′ ) = ρ − a ρ ρ ′ ρ
ε = μ = ρ ρ − a T − 2 aρ − a 2 ρ 3 ( ρ − a ) ρ ⊗ ρ + ( b b − a ) 2 ρ − a ρ Z
det ( ε ) = det ( μ ) = ( b b − a ) 2 ρ − a ρ
∇ × E = − ∂ B ∂ t ∇ × H = ∂ D ∂ t
E = E 0 e i ( k 0 k · x − ωt ) H = 1 η 0 H 0 e i ( k 0 k · x − ω t )
D = ε 0 ε E B = μ 0 μ H
k × E 0 − μ H 0 = 0 k × H 0 + ε E 0 = 0
k × ( μ − 1 ( k × E 0 ) ) + ε E 0 = 0
K ik ≡ ε ijk k j
( K μ − 1 K + ε ) E 0 = 0
det ( K μ − 1 K + ε ) = 0
det ( K n − 1 K + n ) = 1 det ( n ) ( knk − det ( n ) ) 2
H = f ( x ) ( knk − det ( n ) )
d x d τ = ∂ H ∂ k
d k d τ = − ∂ H ∂ x
( k 1 − k 2 ) × n = 0
H ( k 2 ) = 0
∂ H ∂ k · n > 0
H = 1 2 b − a b ( knk − det ( n ) )
∂ H ∂ k = k − 2 ar − a 2 r 4 ( x · k ) x
∂ H ∂ x = − 2 ar − a 2 r 4 ( x · k ) k + 3 ar − 2 a 2 r 6 ( x · k ) 2 x − ( b b − a ) 2 ( ar − a 2 r 4 ) x
H = 1 2 ρ − a ρ ( knk − det ( n ) )
H = 1 2 kTk − 1 2 2 aρ − a 2 ρ 4 ( ρ · k ) 2 + 1 2 [ b ( ρ − a ) ρ ( b − a ) ] 2 ( kZk − 1 )
∂ ρ ∂ x = T
∂ H ∂ k = Tk − 2 aρ − a 2 ρ 4 ( ρ · k ) ρ + [ b ( ρ − a ) ρ ( b − a ) ] 2 Zk
∂ H ∂ x = 3 aρ − 2 a 2 ρ 6 ( ρ · k ) 2 ρ − 2 aρ − a 2 ρ 4 ( ρ · k ) Tk + ( b b − a ) 2 aρ − a 2 ρ 4 ( kZk − 1 ) ρ
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