1 Reality of Electric Susceptibility and Electric Permittivity

1 Reality of Electric Susceptibility and Electric Permittivity

In this question we want to prove that and In general we have : P~ =_oχ ~E

χis a 3 ×3 matrix, where

χ=





χxx χxy χxz

χ_yx χ_yy χ_yz χzx χzy χzz



,

χ is a tensor, and this is the case when we have e.g. Anisotropic material. Also the electric permittivity is a tensor, where = 1+χ. In this problem here, we assume χ and scalar quantities. The complex notation of the electric eld of a plane wave is

E~ =E~_oe^{i(~k·~r−ωt)} ,

and because of the linearity of Maxwell's equation, we can write any E~ as a superposition of plane waves. The same applies also toP~ as long as the medium is linear.

Then we have the wave equation

∇ ×~ ∇ ×~ E~ =−1 c²

∂²E~

∂t²

Let's go into the Fourier space, such that∂t=-iω and∇~ = -i~k, so

−~k×~k×E~ = ω² c²E ,~

this is the wave equation in Fourier space. We add the dielectric function:

−~k×~k×E~ = ω² c² ~E Let's now take the complex conjugate of this equation, so

E~^∗ =E~_o^∗e^{−i(~k·~r−ωt)}=E~_o^∗e^{i(−~k·~r+ωt)},

and we assumeE~_o^∗ is real. Now we have the same equation of E~, but with−~k and −ω. Thus, the two equations are:

1)C.C :−~k×~k×E~^∗ = ω²

c^2∗(k, ω)E~^∗ 2)(−k, ω) =−~k×~k×E~^∗ = ω²

c²(−k,−ω)E~^∗, from that we conclude that

^∗(k, ω) =(−k,−ω) consequently,

χ^∗(k, ω) =χ(−k,−ω)

We see here that the conclusion above comes from the invariance of Maxwell's equation, not from the polarisation.

2 Medium with a loss or gain

From the following equations

k=k⁰+ik⁰⁰=β+iα 2 χ=χ⁰+χ⁰⁰, we need to show that for

χ⁰⁰<< χ⁰ :α'β χ n²

As we know, the relationship between the electric permittivity and susceptibility is= 1 +χ. Sinceω² = ^k2c2, then

k= ω c

√= ω c

p1 +χ k⁰+ik⁰⁰= ω

c

p1 +χ⁰+χ⁰⁰ β+iα

2 = ω c

p1 +χ⁰+χ⁰⁰ The latter equation can be formulated as

(a+ib)² =a²−b²+ 2iab from that we get

β²−α² 4 = ω²

c²(1 +χ⁰) αβ= ω²

c²χ⁰⁰ Nowχ⁰⁰<< χ⁰

p1 +χ⁰+χ⁰⁰ =p 1 +χ⁰

s

1 +i χ⁰⁰ 1 +χ⁰, by applying Taylor's expansion, we get

'p

1 +χ⁰(1 +i χ⁰⁰ 2(1 +χ⁰)) The real part of the refractive index is

q

1 +χ⁰=√ R=n β+iα

2 ' ω

cn(1 +iχ⁰⁰ 2n²) β= ω

cn α= ω

cnχ⁰⁰ n²

this gives rise to have

α 'βχ⁰⁰ n²

This equation relates the absorption coecient to the wavenumberβ.

3 Group Velocity in a Metal

Drude Model states that the relative permittivity (ω) = 1− ^ω_ω^2p₂, where ωp is the plasma frequency.

Phase velocity: v_p = ω k = c

√ and group velocity v_g = dω dk 1

v_g = dk dω = d

dω(ω c

√)

=

√ c +ω

c d dω

√ 1

vg

=

√ c +ω

c 1 2√

d dω d

dω= d

dω(1−ω_p²

ω²) =−ω_p² d dω

1 ω² d

dω= 2ω² 1

ω³ = 2ω_p² ω²

1 ω

= (1−(ω)) 2 ω 1

vg

=

√ c +ω

c 1 2√

(1−(ω)) 2 ω 1

v_g × 1 v_p =

√ c (

√ c +1

c

√1

(1−(ω)))

=

c² + 1

c² (1−

) = 1 c² vgvp =c² , wherec²= _µ¹

oo

4 EM Wave Polarizations

As shown in the gure below, we have two crossed polarizers, and we want to prove that no light can pass through. E~ before the 1^st polarizer:

E~ = (E_x,0xˆ+E_y,0y)eˆ ^i(kz−ωt) ,

whereE_x,0 andE_y,0 are complex quantities. We assume the 1^st polarizer is parallel toxˆ, which means only the eld component parallel toxˆ will propagate.

Figure 1: Light wave propagating through the rst polarizer After 1^st polarizer:

E~ =Ex,0xeˆ ^i(kz−ωt), which means theyˆcomponent will not propagate through.

The 2^nd polarizer is along yˆ, and given that the light passed through the 1^st was along xˆ. Then, the light will not pass through the 2^nd polarizer. Hence, no transmission through the cross polarizers, irrespective to the polarization ofE~ before them.

When we place a 3^rd polarizer between the crossed polarizers, as in the gure below, the situation will change: After the 1^st polarizers

Figure 2: A third polarizer inserted between the crossed polarizers E~ =Ex,0xeˆ ^i(kz−ωt).

Now, let's see the propagation of light through the 3^rd polarizer, as claried below. After the 2^nd polarizer, the eld is

E~ =Ex,0cosθθeˆ ^i(kz−ωt).

Now, we should consider the propagation through the last polarizer, shown below. The trans- mitted eld is

E~ =E_x,0cosθsinθθˆˆye^i(kz−ωt) E~ = Ex,0

2 sin 2θˆye^i(kz−ωt)

Figure 3: The eld component propagated through the middle polarizer

Figure 4: Field components crossed the last polarizer (3^rd)

So actually inserting the third polarizer between the crossed ones, has led some light to prop- agate through the crossed polarizers. We see now, the transmission is a function ofθ and the maximum transmission occurs atθ=45^◦, as well as losing 50%in terms of the eld amplitude, which correspondingly means losing around 75%of the light intensity.