Dispertion relations in Left-Handed Materials
Massachusetts Institute of Technology
6.635 lecture notes
1
Introduction
We know already the following properties of LH media:
1. ²r and µr are frequency dispersive.
2. ²r and µr are negative over a similar frequency band.
3. The tryad (Ē, H̄, k̄) is left-handed.
4. The index of refraction is negative.
From the past lectures, we know that these materials can be realized by a succession of wires
and rods:
• Periodic arrangement of rods: realizes a plasma medium with negative ² r over a certain
frequency band. The model for the permittivity is:
²r = 1 −
2
ωep
.
ω 2 + iγe ω
(1)
• Periodic arrangement of rings (split-rings) realizes a resonant µ r modeled as
µr = 1 −
2
F ωmp
,
2 + iγ ω
ω 2 − ωmo
m
(2)
where F is the fractional area of the unit cell occupied by the interior of the split-ring
(F < 1).
In the lossless case (γe = γm = 0), we can rewrite these two relations as:
2
ω 2 − ωep
,
ω2
ω 2 − ωb2
ω 2 − ω02 − F ω 2
µr =
=
(1
−
F
)
,
ω 2 − ω02
ω 2 − ω02
²r =
(3a)
(3b)
√
where ωb = ω0 / 1 − F > ω0 . Therefore:
2 )(ω 2 − ω 2 )
(ω 2 − ωep
1
b
k 2 = ω 2 ²µ = (1 − F )
.
2
2
c
ω − ω0
(4)
Upon identifying the regions where ²r and µr change signs, we can immediately get the
relation for k:
1
2
Section 2. Argument on n < 0
ω
ωb
ω0
ωp
²r
−
−
−
+
µr
+
−
+
+
k2
−
+
−
+
The region ω ∈ [ω0 , ωb ], which also corresponds to ²r < 0 and µr < 0, corresponds to
positive, which means k real. Therefore, there is propagation in this band, but not in the
adjacent ones.
k2
It may still not be clear that k is negative, even if we write
q
√
√
k = ω 2 ²µ = ω 2 ²20 µ20 ²r µr = k0 n .
(5)
A demonstration of the fact that n is negative follows.
2
2.1
Argument on n < 0
Complex Poynting theorem
We shall first recall the derivation of the complex Poynting theorem and the signification of the
various terms.
We start from Maxwell’s curl equation
∇ × Ē = iω B̄ ,
∇ × H̄ = − iω D̄ + J¯ .
(6a)
(6b)
Upon multiplying Eq. (6a) by H̄ ? and substracting the complex conjugate of Eq. (6b) multuplied by Ē we get:
H̄ ? · ∇ × Ē−Ē · ∇ × H̄ ? = ∇ · (Ē × H̄ ? )
= iω B̄ · H̄ ? − iω D̄? · Ē − J¯? · Ē
= iω[B̄ · H̄ ? − Ē · D̄? ] − Ē · J¯? .
(7)
Upon rewriting, we get:
−Ē · J¯? = ∇ · (Ē × H̄ ? ) + iω[Ē · D̄? − B̄ · H̄ ? ] .
(8)
On the right-hand side of the equation, the first terms corresponds to the divergence of
Poyting power, which is therefore positive. The second term relates to the complex EM energy,
and is therefore also positive. Consequently, the left-hand side term must also be positive, and
actually corresponds to the power supplied by J¯ to the volume.
We shall use this result hereafter.
3
2.2
1D wave equation
For the sake of simplification, let us work with a 1D problem. The wave equation
¯ ,
∇2 Ē(r̄) + k 2 Ē(r̄) = −iωµJ(r̄)
(9)
is rewritten with
Ē(r̄) = ẑ E(x) ,
¯ = ẑ j0 δ(x − x0 ) ,
J(r̄)
(10a)
(10b)
to yield
∂2
E(x) + k 2 E(x) = −iωµj0 δ(x − x0 ) .
∂x2
The solution to this equation is
E(x) = α eik|x−x0 | ,
(11)
(12)
where α needs to be determined. From Eq. (12), we write:
1. First derivative:
∂E(x)
∂
= αik
|x − x0 | eik|x−x0 | .
∂x
∂x2
(13)
2. Second derivative:
∂ 2 E(x)
∂2
∂
=αik
|x − x0 | eik|x−x0 | + α(−k 2 )( 2 |x − x0 |)2 eik|x−x0 |
2
2
∂x
∂x
∂x
2 ik|x−x0 |
= −αk e
+ 2iαkδ(x − x0 ) .
(14)
Therefore:
∂ 2 E(x)
+ k 2 E(x) =2iαkδ(x − x0 ) = 2iα k0 n δ(x − x0 ) .
∂x2
(15)
Comparing Eq. (11) to Eq. (15), we get
α=−
j 0 η 0 µr
ωµj0
=−
,
2k0 n
2 n
(16)
so that finally the solution is:
E(x) = −
j0 η0 µr ik|x−x0 |
e
.
2 n
(17)
If we now compute the power supplied by the current J¯ to the volume:
1
P =−
2
Z
V
η0 j02 µr
Ē · J¯? dV =
> 0.
4 n
(18)
The source must, on average, do positive work on the field. Yet, in LH regime, we have
µr < 0 so that we must have n < 0 as well.
4
Section 3. Dispersion relations
Finally, we can also write the Ē field as:
η
E(x, t) = − j0 ei(k0 n|x−x0 |−ωt) .
2
(19)
Thus, plane waves appear to propagate from −∞ and +∞ to the source, seemingly running
backward in time. Yet, the work done on the field is positive so clearly the energy propagates
outward from the source.
3
Dispersion relations
At this point, we know that n < 0 and k < 0. The difference between phase and group velocity
can be directly seen on the dispersion relation diagram.
ω
,
kz
µ
¶
∂kz −1
vg =
.
∂ω
vφ =
(20a)
(20b)
√
• Free-space: k = ω ²µ where ² = cte and µ = cte.
• Metamaterial:
Let us take the following models
ωp2
,
ω 2 + iγe ω
2 − ω2
ωmp
mo
,
µr =1 − 2
2 + iγ ω
ω − ωmo
m
²r =1 −
3.1
(21a)
(21b)
Lossless case (γe = γm = 0), ωmp = ωp
We rewrite
ω 2 − ωp2
,
ω2
2
ω 2 − ωmp
,
µr = 2
2
ω − ωmo
²r =
and plot the relations with
(22a)
(22b)
(22c)
5
ωp
ωmp
ωmo
γe = γ m
=
=
=
=
20e9 rad/s
20e9 rad/s
5e9 rad/s
0
−6
k surface
10
3
x 10
Dispersion relation
2.5
4
2
2
k
ω [rad/s]
x 10
1.5
1
0
2
2
0
−6
x 10
k
0.5
0
−2
−2
z
k
−6
x 10
0
x
0
Relative permittivity
50
1
2
3
2
3
4
10
ω [rad/s]
x 10
Relative permeability
300
200
r
r
100
µ
ε
0
0
−100
−50
0
1
2
ω [rad/s]
3
10
4
−200
4
0
1
10
x 10
ω [rad/s]
4
10
x 10
k surface (Gamma= 0)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0
−3
−2
−1
0
kz
1
2
3
−6
x 10
6
3.1
Lossless case (γe = γm = 0), ωmp = ωp
Other cases follow.
ωp
ωmp
ωmo
γe = γ m
=
=
=
=
30e9 rad/s
20e9 rad/s
5e9 rad/s
0 rad/s
−6
k surface
10
4
x 10
x 10
Dispersion relation
4
2
k
ω [rad/s]
3
0
2
1
2
0
−6
x 10
k
−2
−2
z
2
0
k
−6
x 10
0
x
0
1
Relative permittivity
50
2
3
2
3
4
10
ω [rad/s]
x 10
Relative permeability
300
200
µr
εr
100
0
0
−100
−50
0
1
2
ω [rad/s]
3
0
1
x 10
10
4
−200
4
10
ω [rad/s]
4
10
x 10
k surface (Gamma= 0)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0
−4
−3
−2
−1
0
k
z
1
2
3
4
−6
x 10
7
ωp
ωmp
ωmo
γe = γ m
=
=
=
=
30e9 rad/s
20e9 rad/s
5e9 rad/s
10e7 rad/s
Dispersion relation
−6
k surface
10
4
x 10
x 10
4
2
k
ω [rad/s]
3
0
2
1
2
0
−6
x 10
kz
−2
−2
2
0
k
−6
x 10
0
x
0
1
Relative permittivity
50
2
3
2
3
4
10
ω [rad/s]
x 10
Relative permeability
300
200
µr
εr
100
0
0
−100
−50
0
1
2
ω [rad/s]
10
4
3
−200
4
0
1
10
x 10
ω [rad/s]
4
10
x 10
k surface (Gamma= 100000000)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0
−4
−3
−2
−1
0
k
z
1
2
3
4
−6
x 10
8
3.1
ωp
ωmp
ωmo
γe = γ m
=
=
=
=
30e9 rad/s
20e9 rad/s
5e9 rad/s
10e8 rad/s
Dispersion relation
−6
k surface
10
2.5
x 10
4
x 10
2
1.5
2
k
ω [rad/s]
Lossless case (γe = γm = 0), ωmp = ωp
1
0
2
0.5
2
0
−6
x 10
0
−2
kz
−2
k
−6
x 10
0
x
0
Relative permittivity
50
1
2
3
2
3
4
10
ω [rad/s]
x 10
Relative permeability
60
40
µr
εr
20
0
0
−20
−50
0
1
2
ω [rad/s]
10
4
3
−40
4
0
1
10
x 10
ω [rad/s]
4
10
x 10
k surface (Gamma= 1000000000)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0
−2.5
−2
−1.5
−1
−0.5
0
k
z
0.5
1
1.5
2
2.5
−6
x 10
9
ωp
ωmp
ωmo
γe = γ m
=
=
=
=
30e9 rad/s
20e9 rad/s
5e9 rad/s
10e9 rad/s
−7
k surface
10
3
x 10
Dispersion relation
2.5
4
2
2
k
ω [rad/s]
x 10
1.5
1
0
2
0.5
2
0
−7
x 10
0
−2
kz
−2
−7
x 10
0
kx
0
1
Relative permittivity
20
0
15
−2
10
3
2
3
4
µ
εr
r
2
2
10
ω [rad/s]
x 10
Relative permeability
−4
5
−6
0
−8
0
1
2
ω [rad/s]
10
4
3
−5
4
0
1
10
x 10
ω [rad/s]
4
10
x 10
k surface (Gamma= 1.000000e+10)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0
−4
−3
−2
−1
0
k
z
1
2
3
4
−7
x 10
10
3.1
ωp
ωmp
ωmo
γe = γ m
=
=
=
=
30e9 rad/s
20e9 rad/s
5e9 rad/s
10e10 rad/s
−7
k surface
10
5
x 10
x 10
Dispersion relation
4
3
2
k
ω [rad/s]
4
2
0
4
1
2
0
−7
x 10
−2
kz
−4
−4
−2
0
4
2
−7
x 10
0
kx
0
Relative permittivity
0.966
0.964
0.964
εr
µr
0.966
0.962
0.96
1
2
3
2
3
4
10
ω [rad/s]
x 10
Relative permeability
0.962
0
1
2
ω [rad/s]
10
4
Lossless case (γe = γm = 0), ωmp = ωp
3
0.96
4
0
1
10
x 10
ω [rad/s]
4
10
x 10
k surface (Gamma= 1.000000e+11)
x 10
3.5
3
ω [rad/s]
2.5
2
1.5
1
0.5
0
−5
−4
−3
−2
−1
0
k
z
1
2
3
4
5
−7
x 10
11
Plotting all the 3D curves on the same scale:
10
x 10
4
3.5
3.5
3
3
2.5
2.5
ω [rad/s]
ω [rad/s]
k surface (Gamma= 10000000)
10
k surface (Gamma= 0)
x 10
4
2
2
1.5
1.5
1
1
0.5
0.5
0
−5
0
−5
0
0
5
−6
x 10
5
0
−5
x
5
−6
x 10
−6
3.5
3
3
2.5
2.5
ω [rad/s]
ω [rad/s]
3.5
2
!#"$&%'#(*)+,
k surface (Gamma= 1000000000)
10
x 10
4
1.5
2
1.5
1
1
0.5
0.5
0
−5
0
−5
0
0
5
−6
x 10
5
0
−5
!#"#1&%2'(*)+,
.34
k surface (Gamma= 1.000000e+10)
10
x 10
3.5
3.5
3
2.5
2.5
ω [rad/s]
3
2
1.5
!#"5&%'#(*)+,
k surface (Gamma= 1.000000e+11)
10
x 10
4
x 10
kz
kx
4
5
0
−5
−6
x 10
kz
.-/0
5
−6
x 10
−6
kx
ω [rad/s]
z
k surface (Gamma= 100000000)
10
x 10
x 10
k
kx
4
5
0
−5
−6
x 10
kz
k
2
1.5
1
1
0.5
0.5
0
−5
0
−5
0
0
5
−6
x 10
5
0
−5
kx
.6*7
5
−6
−6
kz
!#"8&%2'(*)+,
x 10
x 10
5
0
−5
−6
x 10
kz
kx
90
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12
3.1
Lossless case (γe = γm = 0), ωmp = ωp
For simplicity, we can study the lossy case for γe = γm amd ωmo = 0 (although we don’t
really simulate the same medium, the fundamental behavior is similar, and simpler to carry out
mathematically).
The model therefore reads:
ω 2 − ωp2 + iγω
.
²r = µ r =
ω 2 + iγω
(23)
We compute:
ω 2 − ωp2 + iγω
√
² r µr =
ω 2 + iγω
[ω 2 − ωp2 + iγω] [ω 2 − iγω]
=
ω4 + γ 2 ω2
2
2
ω (ω − ωp2 ) + γ 2 ω 2 + iγωωp2
=
.
ω4 + γ 2 ω2
(24a)
The real part is given by:
ω 2 [ω 2 − (ωp2 − γ 2 )]
√
<{ ²r µr } =
.
ω4 + γ 2 ω2
Losses have the effect to lower the plasma frequency to
q
ωp0 = ωp2 − γ 2 .
(25)
(26)
In addition, we also see that if γ is very large, the plasma effect will completey dissapear
(cf. dispersion relation for γ = 10e10 rad/s).