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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. r andr are frequency dispersive.

2. r andr are negative over a similar frequency band.

3. The tryad (E, H, k) 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 r over a certainfrequency band. The model for the permittivity is:

r = 1 2ep

2 +ie. (1)

Periodic arrangement of rings (split-rings) realizes a resonant r modeled as

r = 1 F 2mp

2 2mo+im, (2)

where F is the fractional area of the unit cell occupied by the interior of the split-ring

(F 0. Therefore:

k2 =2=1

c(1 F) (

2 2ep)(2 2b )2 20

. (4)

Upon identifying the regions where r and r change signs, we can immediately get the

relation fork:

1

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2 Section 2. Argument on n

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3

2.2 1D wave equation

For the sake of simplification, let us work with a 1D problem. The wave equation

2E(r) +k2E(r) =iJ(r) , (9)

is rewritten with

E(r) = z E(x) , (10a)

J(r) = z j0(x x0) , (10b)

to yield2

x2E(x) +k2E(x) =ij0(x x0) . (11)

The solution to this equation is

E(x) = eik|xx0| , (12)

where needs to be determined. From Eq. (12), we write:

1. First derivative:E(x)

x =ik

x2|x x0| eik|xx0| . (13)

2. Second derivative:

2E(x)

x2 =ik

2

x2|x x0| eik|xx0| +(k2)(

x2|x x0|)2 eik|xx0|

=k2 eik|xx0| + 2ik(x x0) . (14)

Therefore:

2E(x)

x2 +k2E(x) =2ik(x x0) = 2i k0n (x x0) . (15)

Comparing Eq. (11) to Eq. (15), we get

=j02k0n

=j002

rn

, (16)

so that finally the solution is:

E(x) =j002

rn

eik|xx0| . (17)

If we now compute the power supplied by the current Jto the volume:

P=12

V

E J dV = 0j20

4

rn

>0 . (18)

The source must, on average, do positive work on the field. Yet, in LH regime, we have

r

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4 Section 3. Dispersion relations

Finally, we can also write the Efield as:

E(x, t) =2

j0ei(k0n|xx0|t) . (19)

Thus, plane waves appear to propagate from and + to the source, seemingly runningbackward 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

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5

p = 20e9 rad/s

mp = 20e9 rad/s

mo = 5e9 rad/s

e= m = 0

2

0

2

x 106

2

0

2

x 106

0

2

4

x 1010

kx

kz

[rad/s]

k surface

0 1 2 3 4

x 1010

0

0.5

1

1.5

2

2.5

3x 10

6

[rad/s]

k

Dispersion relation

0 1 2 3 4

x 1010

50

0

50

[rad/s]

r

Relative permittivity

0 1 2 3 4

x 1010

200

100

0

100

200

300

[rad/s]

r

Relative permeability

3 2 1 0 1 2 3

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

10 k surface (Gamma= 0)

kz

[rad/s]

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6 3.1 Lossless case (e= m = 0), mp = p

Other cases follow.

p = 30e9 rad/s

mp = 20e9 rad/s

mo = 5e9 rad/s

e= m = 0 rad/s

20

2

x 106

2

0

2

x 106

0

2

4

x 1010

kx

kz

[rad/s]

k surface

0 1 2 3 4

x 1010

0

1

2

3

4x 106

[rad/s]

k

Dispersion relation

0 1 2 3 4

x 1010

50

0

50

[rad/s]

r

Relative permittivity

0 1 2 3 4

x 1010

200

100

0

100

200

300

[rad/s]

r

Relative permeability

4 3 2 1 0 1 2 3 4

x 10

6

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

10 k surface (Gamma= 0)

kz

[rad/s]

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7

p = 30e9 rad/s

mp = 20e9 rad/s

mo = 5e9 rad/s

e= m = 10e7 rad/s

20

2

x 106

2

0

2

x 106

0

2

4

x 1010

kx

kz

[rad/s]

k surface

0 1 2 3 4

x 1010

0

1

2

3

4x 10

6

[rad/s]

k

Dispersion relation

0 1 2 3 4

x 1010

50

0

50

[rad/s]

r

Relative permittivity

0 1 2 3 4

x 1010

200

100

0

100

200

300

[rad/s]

r

Relative permeability

4 3 2 1 0 1 2 3 4

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

10 k surface (Gamma= 100000000)

kz

[rad/s]

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8 3.1 Lossless case (e= m = 0), mp = p

p = 30e9 rad/s

mp = 20e9 rad/s

mo = 5e9 rad/s

e= m = 10e8 rad/s

2

0

2

x 106

2

0

2

x 106

0

2

4

x 1010

kx

kz

[rad/s]

k surface

0 1 2 3 4

x 1010

0

0.5

1

1.5

2

2.5x 10

6

[rad/s]

k

Dispersion relation

0 1 2 3 4

x 1010

50

0

50

[rad/s]

r

Relative permittivity

0 1 2 3 4

x 1010

40

20

0

20

40

60

[rad/s]

r

Relative permeability

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

10 k surface (Gamma= 1000000000)

kz

[rad/s]

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9

p = 30e9 rad/s

mp = 20e9 rad/s

mo = 5e9 rad/s

e= m = 10e9 rad/s

2

0

2

x 107

2

0

2

x 107

0

2

4

x 1010

kx

kz

[rad/s]

k surface

0 1 2 3 4

x 1010

0

0.5

1

1.5

2

2.5

3x 10

7

[rad/s]

k

Dispersion relation

0 1 2 3 4

x 1010

8

6

4

2

0

2

[rad/s]

r

Relative permittivity

0 1 2 3 4

x 1010

5

0

5

10

15

20

[rad/s]

r

Relative permeability

4 3 2 1 0 1 2 3 4

x 107

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

10 k surface (Gamma= 1.000000e+10)

kz

[rad/s]

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10 3.1 Lossless case (e= m = 0), mp = p

p = 30e9 rad/s

mp = 20e9 rad/s

mo = 5e9 rad/s

e= m = 10e10 rad/s

42

02

4

x 107

42

02

4

x 107

0

2

4

x 1010

kx

kz

[rad/s]

k surface

0 1 2 3 4

x 1010

0

1

2

3

4

5x 10

7

[rad/s]

k

Dispersion relation

0 1 2 3 4

x 1010

0.96

0.962

0.964

0.966

[rad/s]

r

Relative permittivity

0 1 2 3 4

x 1010

0.96

0.962

0.964

0.966

[rad/s]

r

Relative permeability

5 4 3 2 1 0 1 2 3 4 5

x 107

0

0.5

1

1.5

2

2.5

3

3.5

4x 10

10 k surface (Gamma= 1.000000e+11)

kz

[rad/s

]

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11

Plotting all the 3D curves on the same scale:

5

0

5

x 106 5

05

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4

x 1010

kz

k surface (Gamma= 0)

kx

[rad/s]

5

0

5

x 106 5

05

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4

x 1010

kz

k surface (Gamma= 10000000)

kx

[rad/s]

5

0

5

x 106 5

05

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4

x 1010

kz

k surface (Gamma= 100000000)

kx

[rad/s]

5

0

5

x 106 5

05

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4

x 1010

kz

k surface (Gamma= 1000000000)

kx

[rad/s]

5

0

5

x 106 5

05

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4

x 1010

kz

k surface (Gamma= 1.000000e+10)

kx

[rad/s]

5

0

5

x 106 5

05

x 106

0

0.5

1

1.5

2

2.5

3

3.5

4

x 1010

kz

k surface (Gamma= 1.000000e+11)

kx

[rad/s]

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12 3.1 Lossless case (e= m = 0), mp = p

For simplicity, we can study the lossy case for e = m amd mo = 0 (although we dont

really simulate the same medium, the fundamental behavior is similar, and simpler to carry out

mathematically).

The model therefore reads:

r =r =2 2p+ i

2 +i . (23)

We compute:

rr =

2 2p+i2 +i

=[2 2p+ i] [2 i]

4 +22

=2(2 2p) +22 +i2p

4 +22 .

(24a)

The real part is given by:

{rr}=2[2 (2p 2)]

4 +22 . (25)

Losses have the effect to lower the plasma frequency to

p =

2p 2 . (26)

In addition, we also see that if is very large, the plasma effect will completey dissapear

(cf. dispersion relation for = 10e10 rad/s).