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Characterization of Left-Handed Materials

Massachusetts Institute of Technology 6.635 lecture notes

Introduction
1. How are they realized? 2. Why the denomination Left-Handed? 3. What are their properties? 4. Does it really work?

It has already been shown (see previous classes) that rings, or split-rings, can realize a negative permeability ( < 0) over a certain frequency band. In addition to this, we need to realize a negative permittivity ( < 0). It has also been shown (see previous classes) that: lossless:
metal 2 p , 2

=1

where p =

ne2 0 me

(n: electron density, e: electron charge, me : eective mass of electrons). lossy:

metal 2 p =1 . ( + i )

A typical transmission curve looks like shown in Fig. 1.

T 1 no transmission ( < 0)

PSfrag replacements

transmission ( > 0)

Section 2. Why left-handed?

With these characteristics, < 0 has been realized already at infrared frequencies (where metals behave like plasmas). Problem: how to realize it at GHz frequencies? Solution: by reducing n, the electron density. One way of doing this is to conne the electrons in space. This can be achieved by an array of rods for example, as shown in Fig. 2.

a
PSfrag replacements

Figure 2: Array of rods conning the electrons in space.

Note: it is important that the wires are thin, so as to reduce the radiation interaction and allow penetration into the structure. Eect of the wires: to reduce n to ne = n r2 . a2 (1)

Finally, note also that the rods have to be parallel to the electric eld. This, plus the (known already) fact that rings have to be perpendicular to the magnetic eld, gives an idea on how to realize physically LH metamaterials (see Fig. 3).

Why left-handed?

At this point, we have a metamaterial which can realize < 0, < 0 . We shall now see what does it imply on the electromagnetic elds. Let us write Maxwells curl equations for plane wave solutions and time harmonic notations: (2)

E ( ( k r) = H r) , H ( ( k r) = E r) .

(3a) (3b)

3
0.5 mm 5 mm

z
1.3 cm

s
FRONT BACK

(a)
1 mm

y x s

(b)

Figure 3: A realization for LH material.

) forms a right-handed system. H, k In standard materials, Eq. (3) implies that the tryad (E, However, under Eq. (2), we will have: ) form a left-handed (LH) tryad. H, k (E, However, the time average Poynting power is still ( <S r) >= 1 ( ( {E r) H r)} 2 (4)

and remains in the same direction so that we have the set up shown in Fig. 4. Characteristics: is in the phase velocity direction. k phase velocity and energy ux are in opposite directions.

Properties of LH media

Some know characteristics are: Reversed Doppler eect (track the phase), Reversed Cerenkov radiation (cf. 6.632),

3.1

Reversed Snells law

k
PSfrag replacements

H
) and Poynting power ), magnetic eld (H ), wave-vector (k Figure 4: Electric eld (E (S ) in an LH medium.

Negative index of refraction. This last item is very signicant, and we shall spend some time discussing it. The index of refraction of a medium is dened as n=
r

r ,

(5)

r ( ) r ( ) .

(6) < 0 and < 0),

For those frequencies inside the left-handed band (i.e. in the band where we can write: ( ) < 0 ( ) < 0 Eventually, we write n from Eq. (6): n= | ( )( )| ei = | ( )( )| . ( ) = | ( )| ei ,

(7a) (7b)

( ) = |( )| ei ,

(8)

3.1

Reversed Snells law

An important consequence of this fact is the reversal of Snells law. Ray diagram:

diagram for an LH medium: k

PSfrag replacements kx

1 k

2 k

kx S

k1z

k2z

3.2

Energy

Traditionally, the energy is given by W = E 2 + H 2 . (9)

What happens if < 0 and < 0? Is W < 0? Actually no, but this direct conclusion from Eq. (9) shows that this equation is not valid as is. In fact, these materials have to be modeled by frequency dispersive permittivity and permeability. In that case, the relation of Eq. (9) becomes (from Poyntings theorem): W = and we must have: ( ) > 0, ( ) > 0. (11a) (11b) ( ) 2 ( ) 2 E + H (10)

3.3

Properties of an LHM slab

When LH materials are studies as bulk materials, two models are commonly used for the permittivity/permeability: 1. Drude model:
2 ep , ( + ie ) 2 mp , ( + im )

=1

(12a)

r = 1

(12b)

which is schematically represented in Fig. 5.

Real(eps/eps0) 10
0.02 Imag(eps/eps0)

0.015

0.01

10
0.005

20

0.005

30
0.01

40 wp = 100 GHz wp = 266.5 GHz wp = 500 GHz wp = 1000 GHz 50 0 0.2 0.4 0.6 0.8 1 w/w0 1.2 1.4 1.6 1.8 2
0.015 wp = 100 GHz wp = 266.5 GHz wp = 500 GHz wp = 1000 GHz 0 0.2 0.4 0.6 0.8 1 w/w0 1.2 1.4 1.6 1.8 2

0.02

(a)

( ).

(b)

( ).

Figure 5: Permittivity for various values of ep in the Drude model (f0 = 30 GHz, ep = p ).

2. Resonant model:
2 2 ep eo , 2 + i 2 eo e 2 2 mp mo , 2 2 mo + im

=1

(13a)

r = 1

(13b)

where (em)o are the electric/magnetic resonant frequencies and (em)p are the electric/magnetic plasma frequencies. An illustration of this model is given in Fig. 6.

3.3

Imag(eps/eps0) 1 wp = 266.5 GHz wp = 500 GHz

80

0.8

60

0.6

40

0.4

20

0.2

20

0.2

40

0.4

60

0.6

80

0.8

100

1
0 0.2 0.4 0.6 0.8 1 w/w0 1.2 1.4 1.6 1.8 2

0.2

0.4

0.6

0.8

1 w/w0

1.2

1.4

1.6

1.8

(a)

( ).

(b)

( ).

Figure 6: Permittivity for various values of ep in the resonant model of Eq. (13a) (f0 = 30 GHz, ep = p , 0 = 100GHz ).

PSfrag replacements

3.3

= r = 1 (working at the right frequency).

1. From simple ray diagrams, using the reversed Snells law: PSfrag replacements x #0 #1 #2

S I1

I2 z

d We see that is the source is close enough to the slab (distance< d), the slab will produce two images, one inside the slab and one outside. The distance from source to the second image is S I2 = 2d . 2. Rigorous calculation: Let us consider a TE wave impinging on this slab. We write, for a single interface: 0 = E y E0 eikx xit [eikz z + reikz z ] , 1 = E y E0 eikx xit teikz z . (14a) (14b)

We need to match the boundary conditions and, for simplicity, we set the boundary to be at z = 0. We get: eld: Tangential E eld (H = Tangential H
1 i

eikz z + reikz z = teikz z . ): E 0 kz ikz z te . 1 k z

eikz z reikz z =

Upon solving, we get the reection/transmission coecient from free-space to the medium: 21 kz , 1 k z + 0 k z 1 k z 0 k z r= . 1 k z + 0 k z t= (15a) (15b)

In a similar way, the reection/transmission coecients from the medium to free-space are: 20 kz , 1 k z + 0 k z 0 k z 1 k z r = . 1 k z + 0 k z t = (16a) (16b)

In order to obtain the eld inside the slab, we shall compute the transmission coecient as:

T = teikz d t + teikz d r eikz d r eikz d t + tt (r )2 e5ikz d +

= tt eikz d
n=0

(r )n e2inkz d =

tt eikz d . 1 r 2 e2ikz d

(17)

r

= r = 1: T =

(18)

(19)

2 k 2 k 2 k 2 k

Evanescent waves (k < k ) kz = i kz = i

2 k2 k 2 k2 k

kz = kz

kz = k z

Performing the calculation, we get: Propagating waves (kz = kz ): T = Evanescent waves (kz = kz ): T =
2 eikz d 4kz = eikz d . 2 4kz e2ikz d 2 eikz d 4kz = eikz d . 2 4kz

10

Therefore, for all waves, we get: T = eikz d . (20)

Conclusions: Evanescent waves are amplied by the medium. Propagating waves are backward waves. Taking an innite amount of those creates a source. Two images can be formed, like shown in the ray diagram of the previous subsection.

Does it really work?

The theoretical predictions have been veried in experiments. To date, essentially two experiments have been done: 1. Prism, 2. Gaussian beam.

4.1

Prism experiment

LHM

4.2

Gaussian beam experiment

First publication was theoretical. Later experiments followed, but still need to be improved.

11

RHM

LHM

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