17 electromagnetic waves - paweł klimasklimas.paginas.ufsc.br/files/2017/05/chapter17-1.pdf ·...

174840

Electromagnetic waves4841

Electromagnetic waves are generated by electric charges in non-uniform motion. In next chapter4842

we shall deal with phenomenon of electromagnetic radiation. Here we shall discuss some basic4843

properties of electromagnetic waves without touching the problem of their generation. Maxwell4844

has observed that the sourceless electromagnetic field equations possess a wave solution. This4845

theoretical prediction was confirmed in experiments realized by H. Hertz in November 1886.4846

The most important Hertz’s results were published in 1888 (Ann. Phys. 34 610, Ann. Phys. 364847

769, Ann. Phys. 36 1).4848

17.1 Electromagnetic waves in non-dispersive dielectric4849

media4850

17.1.1 A strength field approach4851

The sourceless Maxwell’s equations in non-conducting continuous media have the following4852

form4853

@µHµ⌫

= 0, @↵F�� + @�F�↵ + @�F↵� = 0 (17.1.1)

where4854

(Fµ⌫) =

0BBBB@0 �E1 �E2 �E3

E1

0 �B3 B2

E2 B3

0 �B1

E3 �B2 B1

0

1CCCCA , (Hµ⌫) =

0BBBB@0 �D1 �D2 �D3

D1

0 �H3 H2

D2 H3

0 �H1

D3 �H2 H1

0

1CCCCA .

367

17. ELECTROMAGNETIC WAVES

In order to get solution one has to include the constitutive relations which provide some4855

relations between pairs (E,B) and (D,H). In the simplest case of isotropic linear media those4856

relations have the form4857

D = "E, H =

1

µB (17.1.2)

where " = const and µ = const. Equivalently, one can cast equations (17.1.1) in the form4858

�@0

D+r⇥H = 0 (17.1.3)

r ·D = 0 (17.1.4)

@0

B+r⇥E = 0 (17.1.5)

r ·B = 0 (17.1.6)

Acting with r⇥ on equations (17.1.3) and (17.1.5) we obtain4859

�"µ@0

r⇥E| {z }�@0B

+r⇥r⇥B| {z }r(r·B)�r2

B

= 0 (17.1.7)

@0

r⇥B| {z }"µ@0E

+r⇥r⇥E| {z }r(r·E)�r2

E

= 0. (17.1.8)

If one defines the d’Alembert operator4860

⇤ ⌘ "µ

c2@2t �r2 (17.1.9)

then the result can be cast in the form4861

⇤E = 0, and ⇤B = 0. (17.1.10)

Equations (17.1.10) are just wave equations, therefore the Maxwell’s theory supports the ex-4862

istence of electromagnetic waves. The characteristic velocity which appears in the d’Alembet4863

operator depends on properties of continuous media and it reads4864

v :=

cp"µ

(17.1.11)

Note that for " = 1 = µ the continuous medium is reduced to an empty space and then v = c.4865

The fact that a speed of electromagnetic waves in dielectric media is lower than the speed of4866

light in vacuum does not violate the Einstein’s postulate because the effect originates in effective4867

description of dielectric media. On quantum level photons are absorbed end emitted all the time4868

so the velocity v cannot be interpreted as a velocity of a single photon.4869

368

17.1 Electromagnetic waves in non-dispersive dielectric media

17.1.2 A potential approach4870

It follows that electromagnetic potentials Aµ ! (',A) also satisfy the wave equation. Indeed,4871

substituting expressions4872

E = �@0

A�r', B = r⇥A (17.1.12)

into first pair of Maxwell’s equations (17.1.3) and (17.1.4) we obtain4873

("µ @20

�r2

)A+r("µ @0

'+r ·A) = 0 (17.1.13)

�@0

(r ·A)�r2' = 0 (17.1.14)

Then imposing the Lorenz condition4874

"µ @0

'+r ·A = 0 (17.1.15)

one gets that the four-potential satisfies the d’Alembert equation4875

⇤Aµ= 0. (17.1.16)

The equation (17.1.16) lead to conclusion that the free electromagnetic field is represented by4876

transverse waves. In order to see it, we we split the solution of the equation (17.1.16) into two4877

components4878

Aµ ! (',A) = (',0)| {z }Aµ

1

+(0,A)| {z }Aµ

2

. (17.1.17)

From linearity they are also solutions of the equation (17.1.16). The four-potential Aµ1

leads to4879

expressions4880

E1

= �r', B1

= 0, (17.1.18)

whereas Aµ2

gives4881

E2

= �@0

A, B2

= r⇥A. (17.1.19)

It follows from (17.1.18) that r ⇥ E1

= 0 and from (17.1.19) and Lotenz condition (17.1.15)4882

for Aµ2

that r · E2

= 0. These conditions imply that the potential Aµ1

represent longitudinal14883

degrees of freedom whereas Aµ2

represent transverse2 degrees of freedom. Note that such distin-4884

guishing on longitudinal and transverse degrees of freedom in the case of time dependent fields4885

1Note that for electric field E1 = nE1(n · x, t) one gets r ⇥ E1 = (

ˆei ⇥ n)@iE1(n · x, t) =

(

ˆei ⇥ n)ni@sE1(s, t) = (n⇥ n)@sE1(s, t) = 0, where s := n · x.2Taking E2 = n0E2(n · x, t) with n0 · n = 0 one gets r ·E2 = 0.

369


is meaningless for the fields and it can be introduced only for electromagnetic potentials. The4886

longitudinal degrees of freedom can be eliminated by properly chosen gauge transformation. In4887

order to see it we consider the gauge transformation4888

'0= '� @

0

�, A0= A+r�. (17.1.20)

E =

E1z }| {[�r'] +

E2z }| {[�@

0

A]

=

E

01z }| {

[�r('� @0

�)�⇠⇠⇠r@0

�] +

E

02z }| {

[�@0

(A+r�) +⇠⇠⇠@0

r�] . (17.1.21)

The appropriate choice of �(t,x), namely4889

' = @0

�, ⇤� = 0 (17.1.22)

allows to eliminate E01

. The condition ⇤� = 0 is necessary in order to potentials ('0,A0) satisfy4890

the Lorenz condition. Note, that the choice ' = @0

� is compatible with the fact that ⇤' = 0.4891

The Lorenz condition (17.1.15) imposed on the potentials ('0,A0) reduces to the form4892

'0= 0, r ·A0

= 0 (17.1.23)

for � given by (17.1.22). It restricts a wave solutions to transverse waves. It is important to no-4893

tice that the gauge fixing (17.1.22) exist only for some free and time-dependent fields. When4894

electromagnetic field is not free then the condition (17.1.23) is not a gauge condition anymore.4895

Instead it is a kind of restriction which allows to separate out the transversal (radiation) part4896

of the electromagnetic field. The longitudinal part represents static fields and does not satisfy4897

(17.1.23). Notice that such decomposition is not invariant under Lorentz transformations.4898

17.1.3 Plane waves4899

17.1.3.1 Phase velocity4900

A special group of electromagnetic waves is given by waves characterized by constant phase4901

surfaces. Such surfaces are solutions of the equation4902

(t,x) = const. (17.1.24)

The form of the function determines geometry of the surface of constant phase. For instance4903

for a plane wave that propagates in a dielectric homogeneous medium the surface of constant4904

phase is given by equation4905

⌘ n · x� vt = const (17.1.25)

370


where n is a constant unit vector in direction of propagation of the wave. For spherical wave4906

the vector n is replaced by a radial spherical versor ˆr and for cylindrical wave by a radial4907

versor in cylindrical coordinates ˆ⇢.4908

The phase velocity vp is defined as a velocity of translocation of the phase surface. It can4909

be obtained from d = 0 which leads to the equation4910

@t dt+r · dx = 0

Dividing by |r | one can cast this equation in the form4911

vp =r |r | ·

dx

dt= � @t

|r | . (17.1.26)

It follows that the phase velocity is just a projection of dxdt on r

|r | , where dxdt is the velocity of4912

the point x belonging to the surface of constant phase and r |r | is a vector normal to this surface.4913

The phase velocity of a plane wave with the surface = const given by (17.1.25) reads4914

vp = v =

cp"µ

. (17.1.27)

17.1.3.2 Solution of wave equation in 1+1 dimensions4915

Equations ⇤E = 0 and ⇤B = 0 can be represented by a single equation ⇤� = 0 where4916

� = {E1, E2, E3, B1, B2, B3}. For the case of plane waves the surface of constant phase4917

depends in fact on a single coordinate. One can always choose one of the axes of the Cartesian4918

reference frame in direction of propagation of the wave. In such a case the function � depends4919

only on two variables (t, x). Defining two light-cone coordinates4920

x± := x± vt (17.1.28)

one gets4921

@t =@x

+

@t@+

+

@x�@t

@� = v(@+

� @�) (17.1.29)

@x =

@x+

@x@+

+

@x�@x

@� = @+

+ @� (17.1.30)

and therefore ⇤ = �4@+

@�. The equation4922

@+

@��(x+, x�) = 0 (17.1.31)

has a general solution � = �

+

(x+

) + ��(x�), then4923

�(t, x) = �

+

(x+ vt) + ��(x� vt). (17.1.32)

371


This solution describes superposition of two waves that propagate in positive ��(x � vt) and4924

negative �

+

(x+ vt) direction of axis x.4925

Let us observe that also for spherical waves one can also get an explicit form of the solution.4926

The wave equation in this case reads4927

1

v2@2t�� 1

r2@r

�r2@r�

�= 0 (17.1.33)

Substituting �(t, r) = 1

r�(t, r), one gets4928

1

r

1

v2@2t �� @2r�

�= 0. (17.1.34)

where r2@r�(t, r) = r@r�(t, r)� �(t, r). It follows that �(t, r) satisfies equation @+

@�� = 0,4929

with x± := r ± vt then the solution reads4930

�(t, r) =�+

(r + vt)

r+

��(r � vt)

r. (17.1.35)

The solution is a superposition of ingoing �+

and outgoing �� spherical wave.4931

17.1.3.3 Spectral decomposition4932

We shall consider here a plane wave which is superposition of many plane waves which differ4933

by the angular wave frequency !. The frequency of a single component enters to the solution4934

through one of the factors cos!t, sin!t, exp (�i!t). The problem of polarization of such waves4935

will be discussed in further part. In this section we shall study the case where all vectors A (or4936

E) assigned to different frequencies oscillate in one common direction. Mathematically such4937

superposition of waves can be represented by a Fourier transform.4938

Any function f(x) of class L1

i.e.4939 Z 1

�1|f(x)|dx < 1 (17.1.36)

and which satisfies a condition4940

f(x) =1

2

[f(x� 0) + f(x+ 0)] (17.1.37)

at the discontinuity points can be represented by a Fourier integral4941

f(x) =1

2⇡

Z 1

�1dk F (k)eikx. (17.1.38)

The expansion coefficients are given by the Fourier transform of f(x)4942

F (k) ⌘ F[f(x)](k) :=

Z 1

�1dx f(x)e�ikx. (17.1.39)

372


If the function f(x) depends on all spacetime coordinates x then4943

f(x) =1

(2⇡)3

ZR3

d3xF (k)eik·x, (17.1.40)

F (k) =

ZR3

d3k f(x)e�ik·x. (17.1.41)

In study of electromagnetic waves it is convenient to work with complex functions instead4944

of the real ones. However, since the electromagnetic field is a real-valued field, then one has to4945

interpret its complex version as an auxiliary field whose physical content is encoded in either4946

its real or imaginary part. The electromagnetic plane wave can be represented by a Fourier4947

decomposition on a plane monochromatic waves4948

A(t,x) =1

(2⇡)3

ZR3

d3k a(t,k)eik·x (17.1.42)

where4949

a(t,k) =

ZR3

d3xA(t,x)e�ik·x. (17.1.43)

In our approach A is a complex-valued vector function.1 An assumption that the wave is a4950

superposition of many waves with different frequencies leads to the following form of the coef-4951

ficients24952

a(t,k) = a(k)e�i!(k)t, (17.1.44)

where k := |k| is a wave number and k is called wave vector. The electromagnetic potential4953

takes the form4954

A(t,x) =1

(2⇡)3

ZR3

d3k a(k)ei(k·x�!(k)t). (17.1.45)

Let us observe that each monochromatic component must be a solution of wave equation4955 ⇣"µc2@2t �r2

⌘ei(k·x�!(k)t) = 0 ) "µ

!2

c2� k2

= 0. (17.1.46)

The last equality is called the dispersion relation and it can be written in terms of wave numberck =

p"µ!(k). The characteristic expression p

"µ is called the refraction coefficient and it isusually denoted by

n :=

p"µ.

1Note, the Fourier coefficients must obey relation a⇤(t,�k) = a(t,k) for a real-valued electromag-netic potential.

2A factor e�i!(k)t can be introduced only for complex-valued fields. For real-valued fields it must bereplaced by either cos(!(k)t) or cos(!(k)t).

373


If the x�axis is parallel to the vector k then

a(k) = a(k1) (2⇡)2�(k2)�(k3)

and consequently the Fourier integral can be written in the form4956

A(t,x) =1

2⇡

Z 1

�1dk1 a(k1)ei(k

1x�!(|k1|)t). (17.1.47)

Suppose that the coefficients a(k1) vanish outside the interval |k1 � k10

| < ✏, where " is a small4957

number. If k10

> 0 then the integral contains only contributions from k1 > 0. In such a case4958

k1 := k ⌘ |k|.1

Figure 17.1: The amplitude coefficient |a(k1)|.

4959

Expanding !(k) in Taylor series in a neighborhood of k0

4960

!(k) = !(k0

) + (k � k0

)

✓d!

dk

◆k=k0

+ . . . (17.1.48)

and defining !0

:= !(k0

) and4961

vg :=

✓d!

dk

◆k=k0

(17.1.49)

one gets4962

A(t,x) =

1

2⇡

Z k0+✏

k0�✏dk a(k)ei(kx�!0t�(k�k0)vgt) (17.1.50)

= ei(k0x�!0t) 1

2⇡

Z k0+✏

k0�✏dk a(k)ei(k�k0)(x�v

g

t)| {z }A0(t,x)

(17.1.51)

1For k10 < 0 analysis is very similar with k1 = �k.

374


The expression ei(k0x�!0t) represent a dominant frequency oscillation term. The amplitude term4963

A0

(t, x) assumes constant values on the planes x� vgt = const. This expression give a profile4964

(envelope) of a wave packet. The velocity with which the envelope moves is given by (17.1.49).4965

For this reason it is termed a group velocity.4966

When the frequency is a linear function of the wave number ! = vk, then the phase factor4967

is = kx� vkt and one can conclude that the group velocity is equal to the phase velocity4968

vg =

d!

dk= v, vp = � @t

|r | =!

k= v. (17.1.52)

Let us observe that relation ! = k vp leads to4969

vg =

d(k vp)

dk= vp + k

dvpdk

= vp � �dvpd�

(17.1.53)

where � :=

2⇡k is the length of the wave.4970

17.1.4 Monochromatic wave in homogeneous dielectrics4971

Let us consider a monochromatic electromagnetic wave with the angular frequency ! that prop-4972

agates in a homogeneous dielectric medium characterized by a constant permeabilities " and µ.4973

Such wave is described by auxiliary complex fields4974

E = E0

ei(k·x�!t), B = B0

ei(k·x�!t) (17.1.54)

where E0

and B0

are some constant complex amplitude vectors. The fields (17.1.54) are solu-4975

tions of the wave equation under condition4976

n2

!2

c2= k2. (17.1.55)

It is enough to consider k as a real vector. In order to get information about mutual orientation4977

of E, B and k one has to plug such solutions to Maxwell’s equations and analyze the resulting4978

algebraic equations. Considering that rei(k·x�!t) = ik ei(k·x�!t) one gets4979

r ·E = ik ·E, r⇥E = ik ⇥E, @tE = �i!E (17.1.56)

and similarly for B. Then4980

r ·E = 0 ) k ·E0

= 0 (17.1.57)

r ·B = 0 ) k ·B0

= 0 (17.1.58)

375


4981

r⇥B � n2

c@tE = 0 ) k ⇥B

0

+ n2

!

cE

0

= 0 (17.1.59)

r⇥E +

1

c@tB = 0 ) k ⇥E

0

� !

cB

0

= 0 (17.1.60)

Two first equations (17.1.57) and (17.1.58) imply that vectors E and B lies in the plane orthog-4982

onal to the wave vector k. Taking into account that the wave number k = n!c is given by the4983

dispersion relation (17.1.150) one gets from (17.1.59) and (17.1.60)4984

E0

= � 1

nk ⇥B

0

, B0

= n k ⇥E0

, k ⌘ k

k. (17.1.61)

A scalar product of amplitudes read4985

E0

·B0

= �[k2E0

·B0

� (k ·E0

)(k ·B0

)] = �E0

·B0

(17.1.62)

and therefore4986

E0

·B0

= 0. (17.1.63)

Then one can conclude that electric and magnetic field vectors are mutually perpendicular. The4987

amplitudes of the fields are proportional, what follows from the expression4988

E0

·E⇤0

=

1

n2

[k ⇥B0

] · [k ⇥B⇤0

] =

1

n2

[k2

(B0

·B⇤0

)� (k ·B0

)(k ·B⇤0

)]

which gives4989

|E0

|2 = 1

n2

|B0

|2. (17.1.64)

Moreover, since both vector amplitudes (17.1.61) are related by multiplication by a real vector4990

n k then it follows that the electric and magnetic field have the same phase. The square of E0

4991

is a complex number so it can be represented in the form E2

0

= |E0

|2e�2i'. The phase of this4992

number was chosen as �2i. The square of the complex magnetic field vector reads4993

(B0

)

2

= n2

(k ⇥E0

)

2

= n2|E0

|2e�2i' ) B2

0

= n2|E0

|2e�2i'.

what clearly shows that complex electric and magnetic field have the same phase. Some of the4994

above statements do not hold in conducting media where the vector k must be replaced by a4995

complex vector.4996

376


17.1.5 Polarization of electromagnetic waves4997

In order to study polarization of electromagnetic wave we choose a given point in space and4998

observe behaviour of the electric field E at this point. The magnetic field can be discarded in4999

this analysis because it not an independent field. Its orientation is determined by by an expression5000

like (17.1.61).5001

17.1.5.1 Totally polarized electromagnetic wave5002

The monochromatic wave is totally polarized because its amplitude vector E0

ia a constant5003

vector i.e. its orientation in a space remains unchanged in time. In this section we shall consider5004

such a wave. Let E = E0

ei(k·x�!t) be an complex-valued electric field vector describing the5005

electromagnetic plane wave with a single frequency. The amplitude E0

is a complex constant5006

vector and a physical electric field is given by its real part Re(E). At the fixed point the field5007

E is a time dependent function.5008

It follows from Gauss’ law that E0

· k = 0. Since E0

is a complex function then its square5009

E0

·E0

is a complex number. Following the previous section we shall parametrize this number5010

as5011

E0

·E0

= |E0

|2e�2i'. (17.1.65)

The ampliud vector E0

can be parametrized by two real vectors e1

and e2

as follows5012

E0

= (e1

+ ie2

)e�i'. (17.1.66)

The square of (17.1.66) reads E0

· E0

= (e1

2 � e2

2

+ 2ie1

· e2

)e�2i' and it is equivalent to5013

(17.1.65) for mutually perpendicular real vectors e1

and e2

. Then we have5014

e1

· e2

= 0, e1

· k = 0, e2

· k = 0. (17.1.67)

The lengths of the vectors ea := |ea|, a = 1, 2 can be expressed in terms of the amplitudes5015

and phase shift of the components of electric field in a given reference frame. Without loss of5016

generality we can choose two Cartesian versors ˆx and ˆy as vectors being parallel to the plane5017

defined by e1

and e2

. The orientation of this versors is determined by a measure aparathus5018

associated with the laboratory. A third versor ˆz is determined by ˆz =

ˆx ⇥ ˆy. The electric field5019

amplitude decomposes as follows5020

E0

= Aei↵ ˆx+B ei� ˆy (17.1.68)

377


where A, B, ↵, � are real-valued quantities. Then5021

E⇤0

·E0

= A2

+B2 (17.1.69)

E⇤0

⇥E0

= ABhei(��↵) � e�i(��↵)

iˆz = 2i AB sin(�)ˆz. (17.1.70)

where � := � � ↵. On the other side5022

E⇤0

·E0

= (e1

� ie2

) · (e1

+ ie2

) = e21

+ e22

(17.1.71)

E⇤0

⇥E0

= (e1

� ie2

)⇥ (e1

+ ie2

) = 2i e1

⇥ e2

= 2i e1

e2

ê1

⇥ ê2

. (17.1.72)

The vector ê1

⇥ ê2

is an unit vector. One can always choose both versors ê1

, ê2

in a way that5023

ê1

⇥ ê2

=

ˆz. Comparing both decompositions one gets two equations5024

e21

+ e22

= A2

+B2, e1

e2

= AB sin � (17.1.73)

which can be cast in the form5025

(e1

± e2

)

2

= A2

+B2 ± 2AB sin � (17.1.74)

The sum and difference of square roots of (17.1.74) gives5026

e1

=

1

2

hpA2

+B2

+ 2AB sin � +pA2

+B2 � 2AB sin �i

(17.1.75)

5027

e2

=

1

2

hpA2

+B2

+ 2AB sin � �pA2

+B2 � 2AB sin �i

(17.1.76)

The expression (17.1.75) and (17.1.76) gives respectively the lengths of the major and minor5028

semi-axes of the polarization ellipse. The orientation of the ellipse is given by the angle #5029

between vectors ˆx and ê1

5030

ˆx · ê1

= cos#, ˆx · ê2

= � sin# (17.1.77)

ˆy · ê1

= sin#, ˆy · ê2

= cos#. (17.1.78)

The angle # can be determined from the identity5031

Re[(E0

· ê1

)(E⇤0

· ê2

)] ⌘ 0 (17.1.79)

which can be checked immediately

[e�i'(e

1

+ ie2

) · e1

][ei'(e1

� ie2

) · e2

] = �ie1

e2

.

378


Figure 17.2

Plugging (17.1.68) to (17.1.79) one gets5032

Re[(Aei↵ ˆx · ê1

+Bei� ˆy · ê1

)(Ae�i↵ˆx · ê

2

+Be�i�ˆy · ê

2

)]

= Re[(A cos#+Bei� sin#)(�A sin#+Be�i�cos#)]

= Re[(B2 �A2

) sin# cos#+AB(e�i�cos

2 #� ei� sin2 #)]

= �1

2

(A2 �B2

) sin(2#) +AB cos � cos(2#) ⌘ 0 (17.1.80)

Then from (17.1.80) we get5033

tan(2#) =2AB

A2 �B2

cos �. (17.1.81)

Let us consider some particular cases.5034

1. The linear polarization5035

When � = {0,⇡}, cos � = ±1 then5036

e1

=

pA2

+B2, e2

= 0, tan(2#) = ± 2AB

A2 �B2

The complex electric field vector reads5037

E = (e1

+ ie2

)e�i�, � := !t� k · x+ ' (17.1.82)

what leads to the physical electric field5038

Re[E] =

pA2

+B2

cos� ê1

. (17.1.83)

379


At a given space point the electric field oscillates with the frequency ! and orientation of5039

the vector Re[E] is fixed in direction of the vector ê1

. For B = 0 the field oscillate in5040

direction ê1

=

ˆx and for A = 0 it oscillates in direction ê1

=

ˆy. For A = B the vector5041

of the electric field form angle # = ⇡/4 with x for � = 0 and # = �⇡/4 for � = ⇡.5042

2. The circular polarization5043

When � = ±⇡2

and A = B one gets5044

e1

= A, e2

= ±B ⌘ ±A, tan(2#) = undetermined. (17.1.84)

The electric field vector reads5045

E = A(ê1

± iê2

)(cos�� i sin�) (17.1.85)

and then5046

Re[E] = A[cos� ê1

± sin� ê2

]. (17.1.86)

where � := !t�k ·x+'. At a given space point the electric field vector rotates with the5047

frequency omega. The rotation is characterized by a positive helicity (anti-clockwise)5048

for � = ⇡/2 and by a negative helicity (clockwise) for � = �⇡/2. The amplitude of the5049

vector Re[E] remains constant.5050

3. The elliptical polarization5051

If non of the listed above cases is present then the electromagnetic wave has elliptic5052

polarization. The electric field vector rotates and oscillates simultaneously. Let us observe5053

that the wave with elliptic polarization5054

Re[E] = Re[(e1

+ ie2

)(cos�� i sin�)] = e1

ê1

cos�+ e2

ê2

sin� (17.1.87)

can be interpreted as a superposition of two linearly polarized waves whose planes of5055

polarization are mutually perpendicular. Each linear polarization can be decomposed on5056

a combination of two circular polarizations. Let us define the circular polarization vectors5057

ê± :=

ê1

cos�± ê2

sin�, (17.1.88)

then plugging5058

ê1

cos� =

1

2

(

ê+

+

ê�), ê2

sin� =

1

2

(

ê+

� ê�) (17.1.89)

to the formula (17.1.87) one gets5059

Re[E] = e1

ê1

cos�+ e2

ê2

sin� =

e1

+ e2

2

ê+

+

e1

� e2

2

ê�. (17.1.90)

380


It follows that the elliptically polarized electromagnetic wave can be decomposed either5060

on two linearly polarized waves whose directions of polarization are orthogonal or on two5061

circularly polarized waves with opposite helicities.5062

17.1.5.2 Partially polarized electromagnetic wave5063

In many realistic situations the electromagnetic wave is not perfectly monochromatic. Its fre-5064

quencies belong to a narrow interval �! around some frequency !. A single monochromatic5065

wave is polarized, however, superposition of such waves with different polarizations needs a5066

special treatment. At fixed space point the electric field of such a wave is of the form5067

E = E0

(t)e�i!t (17.1.91)

where the amplitude E0

(t) is slow-varying function of time. Since the vector of amplitude de-5068

scribes the polarization of the wave then the change of its orientation means that the polarization5069

of the wave changes slowly.5070

In experimental study of polarization one can measure the intensity of light beam that pass5071

through a polarizing filters in dependence on orientation of the filter. The intensity of light is a5072

quadratic function of electric field. For this reson we shall consider the quadratic functions of5073

the electric field components Ei(t). Since the auxiliary electric field is a complex field we have5074

the following possibilities5075

EiEj= Ei

0

Ej0

e�2i!t, E⇤iE⇤j= E⇤i

0

E⇤j0

e2i!t, EiE⇤j= Ei

0

E⇤j0

.

However, whats matter are not actual values of this quantities but rather theirs time average5076

values defined by the expression5077

hf(t)i := 1

T

Z T

0

dt f(t). (17.1.92)

The characteristic time scale of variation of amplitudes and the functions e±2i!t are different.5078

Then time averaging over intervals T such that the phase factor oscillates many times whereas5079

the amplitude is essentially unchanged leads to conclusion that terms which depends on dom-5080

inant frequency ! gives zero in averaging process. On the other hand, the average values5081 ⌦EiE⇤j↵

=

DEi

0

E⇤j0

Edo not vanish. It follows that properties of partially polarized electro-5082

magnetic wave are completely characterized by the tensor5083

Jij :=DEi

0

E⇤j0

E. (17.1.93)

Since the vector E0

is perpendicular to the wave vector k then Jij has only four components.5084

We choose the Cartesian coordinates x1, x2 in the plane perpendicular to the wave vector so5085

381


i, j = {1, 2} in (17.1.93). The trace of this tensor reads5086

Tr

ˆJ =

2Xi=1

Jii =⌦|E1

0

|2↵+

⌦|E2

0

|2↵=

⌦|E

0

|2↵

(17.1.94)

and it represent an intensity of the wave (density of the energy flux). This quantity is not related5087

to polarization properties of the wave and therefore one should replaced a tensor (17.1.93) by5088

another one which contains relative intensities instead of absolute ones. We define the polar-5089

ization tensor5090

⇢ij :=Jij

Tr

ˆJ=

DEi

0

E⇤j0

Eh|E

0

|2i (17.1.95)

It follows from definition of the polarization tensor that:5091

1. Tr⇢ = 1 , ⇢11

+ ⇢22

= 1,5092

2. ⇢† = ⇢ , ⇢11

, ⇢22

2 R, ⇢21

= ⇢⇤12

.5093

There are two limit cases: total polarization of the electromagnetic wave and the absence of5094

polarization. When the electromagnetic wave is totally polarized the amplitude vector E0

=5095

const what means that the time averaging drops out5096

⇢ij =Ei

0

E⇤j0

|E0

|2 . (17.1.96)

The determinant of the polarization tensor vanishes in such a case5097

det ⇢ =

1

|E0

|4⇥(E1

0

E⇤10

)(E2

0

E⇤20

)� (E1

0

E⇤20

)(E2

0

E⇤10

)

⇤= 0. (17.1.97)

When the electromagnetic wave is not polarized (natural light beam) the average intensity of5098

the electromagnetic wave is the same in all directions what gives5099 ⌦E1

0

E⇤10

↵=

⌦E2

0

E⇤20

↵=

1

2

⌦|E

0

|2↵. (17.1.98)

On the other side the components E1

0

(t) and E2

0

(t) are not correlated in absence of polarization5100

what leads to vanishing of expressions5101 ⌦E1

0

E⇤20

↵= 0 =

⌦E2

0

E⇤10

↵. (17.1.99)

Substituting this results to (17.1.95) we get the following form of the polarization tensor5102

⇢ij =1

2

�ij . (17.1.100)

382


which leads to det ⇢ =

1

4

. It follows that the determinant of the polarization tensor vanishes for5103

totally polarized electromagnetic wave and it equals to 1/4 for absence of polarization. One can5104

define the grade of polarization P 2 [0, 1]5105

det ⇢ =

1

4

(1� P 2

) (17.1.101)

where P = 0 corresponds to absence of polarization and P = 1 represent a maximal grade of5106

polarization.5107

It is convenient to decompose the polarization tensor on its symmetric Sij and anti-symmetric5108

Aij parts defined as follows5109

Sij :=1

2

(⇢ij + ⇢ji) =1

2

(⇢ij + ⇢⇤ij) 2 R (17.1.102)

Aij :=1

2

(⇢ij � ⇢ji) =1

2

(⇢ij � ⇢⇤ij) ⌘ � i

2

"ijA 2 I (17.1.103)

where A 2 R, then5110

⇢ij = Sij �i

2

"ijA (17.1.104)

Polarization tensor for totally polarized wave. Let us study the meaning of componentsof the polarization tensor in the case of totally polarized electromagnetic wave

E = E0

ei(k·x�!t) = (E1

0

ˆx+ E2

0

ˆy)eik·xe�i!t

where E1

0

= Aei↵ and E2

0

= B ei� . At given x the expression eik·x is a fixed complex number.5111

It has no influence on the tensor ⇢ij because it is an overall phase factor and therefore it does not5112

contribute to � = � � ↵. Plugging this components to (17.1.96) one gets5113

⇢ij =

1

A2

+B2

"A2 AB e�i�

AB ei� B2

#

=

1

A2

+B2

"A2 AB cos �

AB cos � B2

#| {z }

Sij

� i

2

"0 1

�1 0

#2AB

A2

+B2

sin �| {z }A

The parameter A vanishes for � = {0,⇡}. We have seen that in this case the electromagnetic5114

wave is linearly polarized. When it happens the polarization tensor is symmetric5115

⇢ij =1

A2

+B2

"A2 ±AB

±AB B2

#(17.1.105)

383


and it takes the following form for particular cases of linear polarization5116

⇢ij =

"1 0

0 0

#⇢ij =

"0 0

0 1

#⇢ij =

1

2

"1 ±1

±1 1

#(17.1.106)

corresponding respectively to polarizations along the axis x, y and the diagonal one. The circular5117

polarization is obtained for � = ±⇡2

and A = B what gives5118

⇢ij =1

2

"1 0

0 1

#� i

2

(±)

"0 1

�1 0

#. (17.1.107)

In general, the coefficient A has interpretation of degree of circular polarization. Its extremal5119

values A = �1 and A = +1 correspond to circularly polarized waves with respectively negative5120

and positive helicity.5121

17.1.5.3 Stokes parameters5122

Going back to our observation that the polarization tensor is a 2 ⇥ 2 Hermitian matrix we con-5123

clude that it can be decomposed on a set of Pauli matrices �a and the identity matrix5124

⇢ =

1

2

[I+ ⇠a�a] (17.1.108)

where the Pauli matrices have the form5125

�1

=

"0 1

1 0

#, �

2

=

"0 �i

i 0

#, �

3

=

"1 0

0 �1

#and ⇠a are termed the Stokes parameters. The determinant of the polarization tensor is related5126

to the polarization degree P (17.1.101). Explicitly5127

1

4

(1� P 2

) =

1

4

det

"1 + ⇠

3

⇠1

� i⇠2

⇠1

+ i⇠2

1� ⇠3

#=

1

4

[1� (⇠21

+ ⇠22

+ ⇠23

)]

then5128

P =

q⇠21

+ ⇠22

+ ⇠23

. (17.1.109)

It means that all states with given grade of polarization form a spherical surfaces in the space of5129

Stokes parameters. The states of maximal polarization form a sphere with the radius P = 1 and5130

the state in which polarization is absent correspond to the origin ⇠1

= ⇠2

= ⇠3

= 0.5131

In order to understand the meaning of the Stokes parameters we consider states P = 1 and5132

express ⇠a in terms of parameters A, B and �. Since5133

Tr(�a�b) = 2�ab, Tr(�a) = 0

384


then5134

⇠a = Tr(�a⇢). (17.1.110)

Plugging to this expression the polarization tensor5135

⇢ =

1

A2

+B2

"A2 AB (cos � � i sin �)

AB (cos � + i sin �) B2

#one gets5136

⇠1

=

2AB

A2

+B2

cos �, ⇠2

=

2AB

A2

+B2

sin � ⇠3

=

A2 �B2

A2

+B2

. (17.1.111)

The parameter ⇠2

= A, so it describes the grade of circular polarization. For B = 0 we have5137

⇠1

= ⇠2

= 0 and ⇠3

= +1 what correspond to polarization in direction of x-axis. Similarly, for5138

A = 0 we have ⇠1

= ⇠2

= 0 and ⇠3

= �1 one gets polarization in direction of y-axis. Then we5139

conclude that ⇠3

describes polarization in directions x and y. Finally in the case A = B and5140

� = 0 the wave is polarized in direction of a line which form the angle # = ⇡/4 with the axis5141

x. This situation correspond to values of the Stokes parameters ⇠2

= ⇠3

= 0 and ⇠1

= +1. If5142

� = 0 is replaced by � = ⇡ the parameter ⇠3

takes the value ⇠3

= �1 (polarization in direction5143

of a line which form the angle # = �⇡/4 with the axis x).5144

We have shown that for a totally polarized electromagnetic wave P = 1 it holds5145

e21

+ e22

= A2

+B2, e1

e2

= AB sin �, tan(2#) =2AB

A2 �B2

cos �.

One can express the Stokes parameters in terms of lengths of parameters that characterize the5146

ellipse of polarization5147

⇠1

⇠3

= tan(2#), ⇠2

=

2e1

e2

e21

+ e22

. (17.1.112)

The parameter ⇠2

andp⇠21

+ ⇠23

are invariant under Lorentz transformations.5148

17.1.5.4 Decomposition of partially polarized wave on polarized and unpolarized5149

components5150

Let us split the tensor Jij =

⌦EiE⇤j↵ on component J (n)

ij that represent the unpolarized elec-5151

tromagnetic wave and J (p)ij corresponding to the polarized electromagnetic wave component.5152

We have shown that for unpolarized component5153

⇢(n)ij :=

J (n)ij

J (n)=

1

2

�ij ) J (n)ij =

1

2

J (n)�ij (17.1.113)

385


Figure 17.3: The space of Stokes parameters and the meaning of points at the surface ofthe sphere with the radius P = 1.

where J (n) ⌘ Tr(

ˆJ (n)). For a polarized component the time averaging is redundant J (p)

ij =5154

Ei(p)0

E⇤j(p)0

, so the expression Jij � J (n)ij = J (p)

ij reads5155

Jij �1

2

J (n)�ij = Ei(p)0

E⇤j(p)0

. (17.1.114)

The determinant of the rhs of this expression vanishes according to (17.1.97). It leads to the5156

equation5157

det

Jij �

1

2

J (n)�ij

�= 0 (17.1.115)

386


where Jij = J⇢ij and J ⌘ Tr(

ˆJ). It constitute an equation which allows to determine the5158

intensity of non polarized component J (n)5159

det

"J⇢

11

� 1

2

J (n) J⇢12

J⇢21

J⇢22

� 1

2

J (n)

#= (J⇢

11

� 1

2

J (n))(J⇢

22

� 1

2

J (n))� J2⇢

12

⇢21

= J2

[⇢11

⇢22

� ⇢12

⇢21

]| {z }det ⇢= 1

4 (1�P 2)

+

1

4

(J (n))

2 � 1

2

J (n)J [⇢11

+ ⇢22

]| {z }Tr⇢=1

=

1

4

h(J (n)

)

2 � 2JJ (n)+ (1� P 2

)J2

i= 0 (17.1.116)

Since J (n) < J , then the physical solution reads5160

J (n)= (1� P )J (17.1.117)

The intensity of the polarized component equals to J (p)= PJ as a consequence of decompo-5161

sition J (p)ij = Jij � J (n)

ij . One can easily establish connection with the Stokes parameters. The5162

only difference is that the intensity of a polarized component is a fraction of a total intensity i.e.5163

e21

+ e22

= A2

+B2

= PJ , then5164

⇠1

⇠3

= tan(2#), ⇠2

=

2e1

e2

PJ. (17.1.118)

17.1.5.5 Decomposition of partially polarized wave on two incoherent elliptically5165

polarized waves5166

From Hermiticity of ⇢ it follows that eigenvalues �a, a = 1, 2 of this tensor are real-valued. The5167

eigenvectors n(a) of the polarization tensor are given by two complex versors n⇤(a) · n(a)= 15168

satisfying equations5169

⇢ijn(a)j = �an

(a)i . (17.1.119)

Multiplying this equation by n⇤(a)i and summing over i we get5170

�a = ⇢ijn⇤(a)i n(a)

j =

1

J

DEi

0

E⇤j0

En⇤(a)i n(a)

j =

1

J

D(Ei

0

n⇤(a)i )(E⇤j

0

n(a)j )

E=

1

J

D|Ei

0

n⇤(a)i |2

E> 0

then both eigenvalues are real-valued and positive. The eigenvalues �a can be parametrized5171

by P . Indeed, the equation5172

det[⇢� �I] = 0 , �2 � Tr⇢|{z}1

�+ det ⇢| {z }14 (1�P 2

)

= 0 (17.1.120)

387


has two solutions �1,2 =

1

2

(1± P ).5173

One can show that eigenvectors are mutually orthogonal. Multiplying the equation with5174

a = 1 by n⇤(2) and the complex conjugated equation with a = 2 by n(1)

5175 (⇢ijn

(1)

j = �1

n(1)

i /n⇤(2)i

⇢⇤ijn⇤(2)j = �

2

n⇤(2)i /n(1)

i

(17.1.121)

and subtracting the resultant equations one gets5176

(⇢ij � ⇢⇤ji)| {z }0

n(1)

j n⇤(2)i = (�

1

� �2

)n(1)

i n⇤(2)i .

Since �1

6= �2

it follows that n(1) · n⇤(2)= 0. It means that the complex eigenvectors form the5177

orthonormal set5178

n(a) · n⇤(b)= �ab. (17.1.122)

The matrix whose columns are formed by the eigenvectors is an unitary matrix5179

U :=

hn(1) n(2)

iU †

:=

"n⇤(1)

n⇤(2)

#.

It follows from this definition and the fact that n(a) are eigenvectors of ⇢ that5180

U †⇢U =

"�1

0

0 �2

#, ⇢ = U

"�1

0

0 �2

#U † (17.1.123)

The last equality reads5181

⇢ij = �1

n(1)

i n⇤(1)j + �

2

n(2)

i n⇤(2)j . (17.1.124)

A complex vector amplitude can always be chosen in the way that one of two mutually perpen-5182

dicular components is real whereas the other one is imaginary. Let e1

and e2

be two perpendic-5183

ular real vectors. We consider the following decomposition5184

n(1)

:= e1

ê1

+ ie2

ê2

(17.1.125)

then the normalization condition n⇤(1) · n(1)

= 1 leads to equation e21

+ e22

= 1. The second5185

vector n(2) is orthogonal to the first one, than substituting n⇤(2)= ↵⇤

ê1

+ �⇤ê2

one gets5186

↵⇤e1

+ i�⇤e2

= 0. The solution of the last equation which leads to normalized the second5187

vector reads ↵⇤= �ie

2

and �⇤ = e1

5188

n(2)

:= ie2

ê1

+ e1

ê2

. (17.1.126)

The vectors (17.1.125) and (17.1.126) describes two identical ellipses (the same ratio of semi-5189

axes). The major semi-axes of these ellipses form an angle ⇡/2. The polarization components5190

associated which each ellipse are mutually incoherent. It follows from the fact that in decompo-5191

sition (17.1.124) there are no cross terms.5192

388


Figure 17.4: The polarization ellipses.

17.1.6 Energy and momentum flux of electromagnetic waves5193

The energy density u and the momentum flux density S of the electromagnetic field is given by5194

expressions5195

u =

1

8⇡[E ·D +H ·B] (17.1.127)

5196

S =

c

4⇡[E ⇥H] (17.1.128)

where all fields are real-valued. We assume that the continuum medium is linear, homogeneous5197

and isotropic. It follows that D = "E and B = µH .5198

Considering the electromagnetic field as a real part of the complex auxiliary field E 2 C,5199

B 2 C5200

ReE =

1

2

(E +E⇤) ReB =

1

2

(B +B⇤) (17.1.129)

the expressions (17.1.127) and (17.1.128) are substituted by5201

u =

1

8⇡

"(ReE)

2

+

1

µ(ReB)

2

�S =

c

4⇡ µ[ReE ⇥ReB] . (17.1.130)

When fields oscillate quickly with a frequency ! the instantaneous values of u and S are less5202

relevant than theirs time average values hui and hSi. Taking a monochromatic wave in the form5203

E = E0

ei(k·x�!t) B = B0

ei(k·x�!t) (17.1.131)

389


one gets that expressions5204

⌦(ReE)

2

↵=

1

4

⇥⌦E2

↵+ 2 hE ·E⇤i+

⌦E⇤2↵⇤

=

1

2

E ·E⇤=

1

2

E0

·E⇤0⌦

(ReB)

2

↵=

1

4

⇥⌦B2

↵+ 2 hB ·B⇤i+

⌦B⇤2↵⇤

=

1

2

B ·B⇤=

1

2

B0

·B⇤0

does not contain oscillating terms because⌦E2

↵⇠

⌦e�2i!t

↵= 0,

⌦E⇤2↵ ⇠

⌦e2i!t

↵= 0.5205

Similarly the following expression5206

hReE ⇥ReBi =

1

4

[hE ⇥B⇤i+ hE⇤ ⇥Bi+ hE ⇥Bi| {z }0

+ hE⇤ ⇥B⇤i| {z }0

]

=

1

2

Re [E ⇥B⇤] =

1

2

Re [E0

⇥B⇤0

] (17.1.132)

also does not contain terms behaving as e±i!t. Then5207

hui = 1

16⇡[E ·D⇤

+H ·B⇤] (17.1.133)

5208

hSi = c

8⇡Re [E ⇥H⇤

] . (17.1.134)

We have already shown that Maxwell’s equations lead to following algebraic equations5209

E = � 1

nk ⇥B, B = n k ⇥E, |E|2 = 1

n2

|B|2 (17.1.135)

where k :=

k

|k| . Then average value of the energy density reads5210

hui =

1

16⇡

"|E|2 + 1

µ|B|2

�=

1

16⇡ µ

h"µn2

+ 1

i|B|2 = 1

8⇡µ|B|2.

Similarly one gets average value of the Poynting vector5211

hSi = c

8⇡Re

� 1

n(k ⇥B)⇥

✓1

µB⇤

◆�=

c

8⇡µnRe(|B|2)k =

c

8⇡µn|B|2k.

The last two expressions are proportional5212

hSi · khui =

c

n= v (17.1.136)

and the proportionality coefficient is equal to a velocity of propagation of electromagnetic wave.5213

390


17.1.7 Reflection and refraction of light at the surface of interface5214

In current section we shall deal with description of electromagnetic wave on the border of two5215

different homogeneous dielectrics. Such dielectrics are characterized by electric permittivities5216

"1

, "2

and magnetic permeabilities µ1

and µ2

. The refractive indices are given by n1

:=

p"1

µ1

5217

and n2

:=

p"2

µ2

. The surface of contact of two dielectric media is termed surface of interface.5218

Especially simple solution is obtained for an idealized situation i.e when the surface of interface5219

is approximated by an infinite plane. Let n be an unit vector, normal to the surface of interface5220

that point out into medium 2. As there are not free charges and free currents the fields have to5221

satisfy sourceless Maxwell’s equations5222

r ·D = 0, r⇥H � 1

c@tD = 0 (17.1.137)

r ·B = 0, r⇥E +

1

c@tB = 0 (17.1.138)

and the boundary conditions5223

n · (D2

�D1

) = 0, n⇥ (H2

�H1

) = 0 (17.1.139)

n · (B2

�B1

) = 0, n⇥ (E2

�E1

) = 0 (17.1.140)

which requires continuity of normal components Dn and Bn and continuity of tangent compo-5224

nents Ht and Et.5225

We choose a z�axis as being perpendicular to the plane and oriented in direction of n, i.e.5226

z = n. The incoming electromagnetic waves, characterized by the wave vector k0

, propagates5227

in medium 1. We shall denote by k1

the wave vector of reflected electromagnetic wave (in5228

medium 1) and by k2

the wave vector of transmitted electromagnetic wave (in medium 2). For5229

general case of oblique incidence the vectors k0

and n are not parallel therefore they define5230

plane of incidence. The angles between wave vectors k0

, k1

, k2

and the vector n are denoted5231

by ✓0

, ✓1

and ✓2

. Without loss of generality we choose versor x parallel to the incidence plane5232

(and perpendicular to z). The last versor y is defined as y = z ⇥ x.5233

17.1.7.1 Relations between the angles of incidence, reflection and refraction5234

The incident wave, reflected wave and transmitted wave are given respectively by5235

E0

= E0

0

ei(k0·r�!t), H0

=

n1

µ1

k0

⇥E0

, (17.1.141)

E1

= E0

1

ei(k1·r�!t), H1

=

n1

µ1

k1

⇥E1

, (17.1.142)

E2

= E0

2

ei(k2·r�!t), H2

=

n2

µ2

k2

⇥E2

. (17.1.143)

391


These fields have to satisfy some continuity conditions at the surface of interface z = 05236

n⇥ [E0

+E1

] = n⇥E2

n⇥ [H0

+H1

] = n⇥H2

.

The common factor e�i!t cancels out on both sides of the equations giving5237

n⇥ [E0

0

eik0·r+E0

1

eik1·r] = n⇥E0

2

eik2·r (17.1.144)

n⇥ [H0

0

eik0·r+H0

1

eik1·r] = n⇥H0

2

eik2·r (17.1.145)

The solution of these conditions cannot depend on coordinates x and y on the surface z = 0.5238

One can satisfy this condition if it holds eik0·r= eik1·r

= eik2·r, where r = x x + y y at the5239

surface of interface. Alternatively5240

k0

· r = k1

· r = k2

· r (17.1.146)

where by assumption there is no component k0y i.e. k

0

· y = 0. It follows from (17.1.146) and5241

this assumption that5242

k1

· y = 0 = k2

· y (17.1.147)

what means that vectors k1

and k2

belongs to the plane of incidence. The condition (17.1.146)5243

reduces to the following one5244

k0

· x = k1

· x = k2

· x (17.1.148)

where ka · x = ka cos (✓a + ⇡/2) = �ka sin ✓a, a = 0, 1, 2, then5245

k0

sin ✓0

= k1

sin ✓1

= k2

sin ✓2

. (17.1.149)

The dispersion relation can be written in the form5246

!

c=

k0

n1

=

k1

n1

=

k2

n2

(17.1.150)

which implies that k1

= k0

and k2

=

n2n1k0

. The first equality of (17.1.149) leads to equality of5247

the incidence and the reflection angles5248

sin ✓0

= sin ✓1

) ✓0

= ✓1

. (17.1.151)

The second equality in known as the Snell’s law5249

sin ✓0

sin ✓2

=

n2

n1

. (17.1.152)

392


The Snell law determines the angle of refraction ✓2

in dependence on the angle of incidence.5250

Note that from sin ✓2

=

n1n2

sin ✓0

it follows that there are some limitations for the solutions.5251

When n2

> n1

then there exist solution ✓2

for any ✓1

. However, for n2

< n1

there exist a5252

critical value ✓c = arcsin

n2n1

and for ✓0

> ✓c there is no solution ✓2

. The relation between the5253

incidence angle ✓0

and the refraction angle ✓2

is shown in Fig.17.5.5254

Figure 17.5: The refraction angle ✓2

= arcsin(

n1n2

sin ✓0

) in dependence on the angle ofincidence for n

1

< n2

, n1

= n2

and n1

> n2

.

17.1.7.2 Conditions for amplitudes5255

The continuity conditions for parallel components of the fields E and H read5256

n⇥ [E0

0

+E0

1

] = n⇥E0

2

(17.1.153)5257

n⇥ [H0

0

+H0

1

] = n⇥H0

2

. (17.1.154)

Since B = n k⇥E and consequently E = � 1

n k⇥B we get for E and H =

1

µB the following5258

relations5259

H =

1

Zk ⇥E, E = �Z k ⇥H (17.1.155)

where Z ⌘ µn =

qµ" is called impedance of the medium. The amplitudes of the fields satisfy5260

E = Z H . It can be seen taking a square of any of equations (17.1.155).5261

The equations (17.1.153) and (17.1.154) determine amplitudes of the electric field. The5262

generic monochromatic wave is a superposition of waves described by E0

0

perpendicular to the5263

plane of incidence and by E0

0

parallel to this plane.5264

393


17.1.7.3 Fresnel’s formulas for electric field vector perpendicular to the pane of5265

incidence5266

We shall consider E0

0

as being perpendicular to the plane of incidence. In such a case it is5267

convenient to express H by E according to (17.1.155). The continuity conditions read5268

n⇥ [E0

0

+E0

1

] = n⇥E0

2

(17.1.156)

n⇥ [k0

⇥E0

0

+ k1

⇥E0

1

] =

Z1

Z2

n⇥ [k2

⇥E0

2

]. (17.1.157)

Figure 17.6: The electric field vector perpendicular to the plane of incidence.

Since the electric field vectors are perpendicular to the plane of incidence we can choose5269

them in the form E0

a = �E0

ay, a = 0, 1, 2. It follows that the first condition (17.1.156) gives5270

�E0

1

+ E0

2

= E0

0

. (17.1.158)

Making use of orthogonality of vectors n ·E0

a = 0 and of the fact that n · k1

= �n · k0

we get5271

the second condition (17.1.157) in the form5272

Z2

(n · k0

)E0

1

+ Z1

(n · k2

)E0

2

= Z2

(n · k0

)E0

0

(17.1.159)

Multiplying (17.1.158) by Z2

(n · k0

) and adding to (17.1.159) we get5273

E0

2

=

2Z2

(n · k0

)

Z2

(n · k0

) + Z1

(n · k2

)

E0

0

(17.1.160)

394


and then5274

E0

1

=

Z2

(n · k0

)� Z1

(n · k2

)

Z2

(n · k0

) + Z1

(n · k2

)

E0

0

(17.1.161)

One can express these formulas in terms of the angles ✓0

and ✓2

. Note that5275

Z1

Z2

=

µ1

µ2

n2

n1

=

µ1

µ2

sin ✓0

sin ✓2

, (17.1.162)

where the last equality is true for all angles of incidence ✓0

if only n1

< n2

. Otherwise, it holds5276

only for ✓0

being smaller then the critical angle. When the solution ✓2

exist, all three wave5277

vectors are real-valued and have the following expansion on the Cartesian versors5278

k0

= � sin ✓0

x+ cos ✓0

z

k1

= � sin ✓0

x� cos ✓0

z

k2

= � sin ✓2

x+ cos ✓2

z.

Then the formulas take the form5279

E0

1

E0

0

=

Z2

cos ✓0

� Z1

cos ✓2

Z2

cos ✓0

+ Z1

cos ✓2

,E0

2

E0

0

=

2Z2

cos ✓0

Z2

cos ✓0

+ Z1

cos ✓2

(17.1.163)

or alternatively5280

E0

1

E0

0

=

µ2

tan ✓2

� µ1

tan ✓0

µ2

tan ✓2

+ µ1

tan ✓0

µ1=µ2=

sin(✓2

� ✓0

)

sin(✓2

+ ✓0

)

, (17.1.164)

5281

E0

2

E0

0

=

2µ2

tan ✓2

µ2

tan ✓2

+ µ1

tan ✓0

µ1=µ2=

2 cos ✓0

sin ✓2

sin(✓2

+ ✓0

)

. (17.1.165)

These formulas were derived in 1818 by Fresnel who analyzed oscillations of light in ether. The5282

result was obtained before a final formulation of Maxwell’s equations.5283

Each ratio of electric field amplitudes E0

1

/E0

0

and E0

2

/E0

0

can be studied either for n1

< n2

5284

or n1

> n2

. In Fig.17.7 we show these ratios in dependence on the angle of incidence for5285

µ1

= µ2

.5286

For n1

< n2

the ratio E0

1

/E0

0

is always negative. It means that electric field vector changes5287

its phase by ⇡ when wave reflects on the surface of interface. The amplitude of electric field5288

E2

is always less than E0

. Both ratios are related by E0

2

/E0

0

� E0

1

/E0

0

= 1 according to the5289

continuity condition of the parallel (to the surface of interface) electric field component. For a5290

normal incidence the amplitudes take values5291

E0

1

E0

0

=

µ2

n1

� µ1

n2

µ2

n1

+ µ1

n2

,E0

2

E0

0

=

2µ2

n1

µ2

n1

+ µ1

n2

. (17.1.166)

395


(a) (b)

Figure 17.7: The ratios of amplitudes of electric field perpendicular to the plane of inci-dence for µ

1

= µ2

. Note that E02

E00� E0

1E0

0= 1.

For n1

> n2

the reflected wave and incident wave have the same phase. However, the5292

refracted wave exist only for ✓0

< ✓c. This phenomenon is called total internal reflection and5293

we shall comment about it in section 17.1.7.5.5294

17.1.7.4 Fresnel formulas for electric field vector parallel to the pane of incidence5295

We shall assume H0

0

perpendicular to the plane of incidence, namely we choose H0

0

= H0

0

y. In5296

this case it is more convenient to express the electric field as E = �Zk⇥H then the continuity5297

conditions read5298

n⇥ [k0

⇥H0

0

+ k1

⇥H0

1

] =

Z2

Z1

n⇥ [k2

⇥H0

2

]. (17.1.167)

n⇥ [H0

0

+H0

1

] = n⇥H0

2

. (17.1.168)

All fields H0

a are parallel to the vector y. Taking into account that n · k1

= �n · k0

and5299

substituting the amplitude ZH = E one gets from (17.1.167)5300

(n · k0

)E0

1

+ (n · k2

)E0

2

= (n · k0

)E0

0

. (17.1.169)

The second condition (17.1.168) became5301

�Z2

E0

1

+ Z1

E0

2

= Z2

E0

0

(17.1.170)

396


Figure 17.8: The electric field vector perpendicular to the plane of incidence.

The solution of the last two equations reads5302

E0

2

=

2Z2

(n · k0

)

Z1

(n · k0

) + Z2

(n · k2

)

E0

0

(17.1.171)

E0

1

=

Z1

(n · k0

)� Z2

(n · k2

)

Z1

(n · k0

) + Z2

(n · k2

)

E0

0

(17.1.172)

For real-valued refraction angles (n2

> n1

) and for ✓0

< ✓2

in the case n2

< n1

the above5303

formulas take the form5304

E0

1

E0

0

=

Z1

cos ✓0

� Z2

cos ✓2

Z1

cos ✓0

+ Z2

cos ✓2

,E0

2

E0

0

=

2Z2

cos ✓0

Z1

cos ✓0

+ Z2

cos ✓2

. (17.1.173)

Plugging Z1Z2

=

µ1µ2

sin ✓0sin ✓2

to the first formula in (17.1.173) one gets5305

E0

1

E0

0

=

µ1

sin(2✓0

)� µ2

sin(2✓2

)

µ1

sin(2✓0

) + µ2

sin(2✓2

)

µ1=µ2=

tan(✓0

� ✓2

)

tan(✓0

+ ✓2

)

, (17.1.174)

where5306

sin(2✓0

)� sin(2✓2

)

sin(2✓0

) + sin(2✓2

)

=

2 sin

h2✓0�2✓2

2

icos

h2✓0+2✓2

2

i2 sin

h2✓0+2✓2

2

icos

h2✓0�2✓2

2

i=

tan(✓0

� ✓2

)

tan(✓0

+ ✓2

)

The second formula became5307

E0

2

E0

0

=

4µ2

cos ✓0

sin ✓2

µ1

sin(2✓0

) + µ2

sin(2✓2

)

µ1=µ2=

2 cos ✓0

sin ✓2

sin(✓0

+ ✓2

) cos(✓0

� ✓2

)

. (17.1.175)

397


The ratio of the electric field components corresponding to the incident, reflected and refracted5308

wave are shown in figure Fig.17.9, where µ1

= µ2

.5309

For n1

< n2

the ratio E0

1

/E0

0

is positive for ✓0

< ✓B and negative for ✓0

> ✓B . The angle5310

✓B is defined by condition ✓B + ✓2

= ⇡/2. According to Snell’s law5311

n1

sin ✓B = n2

sin ✓2

= n2

sin

⇣⇡2

� ✓B⌘= n

2

cos ✓B ) tan ✓B =

n2

n1

. (17.1.176)

The angle ✓B is called the Brewster angle. For the incident angle equal to the Brewster angle

(a) (b)

Figure 17.9: The amplitudes of electric field parallel to the plane of incidence for µ1

= µ2

.There is no reflected wave for the Brewster angle ✓B .

5312

the reflected electromagnetic wave vanishes. A general electromagnetic wave is a superposi-5313

tion of the wave having both components of the electric field vector - perpendicular and parallel5314

to the plane of incidence. When the wave vector of the incident wave form the Brewster angle5315

with the optic axis then the reflected wave has only a component perpendicular to the plane of5316

incidence. It means that the wave is polarized. This method is used to obtain polarized elec-5317

tromagnetic wave. The Brewster effect is associated with the fact that electromagnetic radiation5318

cannot be emitted in direction of oscillation of electric charges. The incident electromagnetic5319

wave forces electrons of the material to oscillate in direction of the vector of electric field E0

0

.5320

For the Brewster angle the refracted wave is emitted in direction perpendicular to direction of5321

propagation of the incident wave. It means that electrons oscillate along the line parallel to5322

direction in which the reflected wave would propagate. However, since there is not radiation5323

emitted in such direction consequently there is no reflected wave.5324

398


17.1.7.5 Total reflection5325

It is convenient to write the formulas that involve ✓0

and ✓2

in a way that they are true also for5326

n1

> n2

and ✓0

> ✓c. According to our previous results, all tangent components (to the surface5327

of interface) of the wave vector ka of the incident a = 0, reflected a = 1 and refracted a = 25328

wave must be equal i.e.5329

�k0

sin ✓0| {z }

k0t=k0x

= �k1

sin ✓1| {z }

k1t=k1x

= �k2

sin ✓2| {z }

k2t=k2x

where kax = ka · x = ka(ka · x) = ka cos (✓a + ⇡/2). The normal components of the incident5330

and reflected wave are opposite k1n = �k

0n. Together with the expression k22n = k2

2

� k22t =5331

k22

� k20t one gets5332

k2t = �k

0

sin ✓0

(17.1.177)

k22n = k2

2

� k20

sin

2 ✓0

= k22

"1�

✓n1

n2

◆2

sin

2 ✓0

#. (17.1.178)

These formulas are valid for any values of the angle ✓0

and the value of the ratio n1

/n2

. The5333

angle ✓2

can be derived from these formulas substituting k2t = �k

2

sin ✓2

and k2n = k

2

cos ✓2

.5334

It gives5335

n2

sin ✓2

= n1

sin ✓0

(17.1.179)

cos

2 ✓2

= 1�✓n1

n2

◆2

sin

2 ✓0

. (17.1.180)

Expression (17.1.179) is just a Snell law. The point is that the angle ✓2

may or not to have a5336

geometric meaning. In both cases it is determinable from the formula (17.1.180).5337

An important characteristic of total reflection is given by ratios of amplitudes of electric field5338

vectors. For incident and refracted waves they read5339

E0

= E0

0

ei[k0·r�!t]= E0

0

ei[(k0xx+k0zz)�!t]= E0

0

ei[k0(�x sin ✓0+z cos ✓0)�!t] (17.1.181)5340

E2

= E0

2

ei[k2·r�!t]= E0

2

ei[(k2xx+k2zz)�!t]= E0

2

ei[(�k0x sin ✓0+k2zz)�!t] (17.1.182)

where r = xx+ zz, k2x ⌘ k

2t = k0t and k

2z ⌘ k2n.5341

For n1

> n2

and ✓0

> ✓c the rhs of (17.1.180) takes negative values, then cos

2 ✓2

< 0. It5342

follows that ✓2

has no geometric meaning. Let us define5343

s2 := �k22n = k2

2

"✓n1

n2

◆2

sin

2 ✓0

� 1

#= k2

2

"✓sin ✓

0

sin ✓c

◆2

� 1

#(17.1.183)

399


(a) (b)

Figure 17.10: A total internal reflection

where n1

sin ✓c = n2

. It follows from this expression that k2z = ±is. Plugging this expression5344

to E2

one gets5345

E2

= E0

2

e�szei[�k0x sin ✓0�!t] (17.1.184)

where there term e+sz is not allowed because it reads to unlimited grows of amplitude for z ! 15346

(what is not consistent with conservation of the energy). The inverse of the expression5347

s = k2

s✓n1

n2

◆2

sin

2 ✓0

� 1 (17.1.185)

is denoted by � and it is called penetration depth5348

� :=1

s=

1

2⇡

�2r⇣

n1n2

⌘2

sin

2 ✓0

� 1

(17.1.186)

where we made use of relation between wave number and wave length k = 2⇡/�. It follows5349

from this expression and the formula (17.1.184) that refracted electromagnetic waves penetrate5350

the second medium only on depth being of the order of maginitude of the wavelength �2

. The5351

phase of the refracted wave is = �k0

x sin ✓0

� !t so the phase velocity of the refracted5352

wave reads5353

vp = � @t

|r | =!

k0

sin ✓0

=

c

n1

sin ✓0

=

c/n2

(n1

/n2

) sin ✓0

=

✓sin ✓csin ✓

0

◆v2

. (17.1.187)

where v2

= c/n2

is a phase velocity of the plane wave in the second medium. The phase5354

velocity vp is smaller than v2

because ✓c < ✓0

. The wave described by (17.1.184) is not a plane5355

400


wave because its amplitude and phase is not constant on the plane perpendicular to the vector of5356

propagation in the second medium. The phase of this wave clearly depends on properties of the5357

medium (✓c) and the angle of incidence ✓0

.5358

Substituting5359

cos ✓2

=

s1�

✓n1

n2

◆2

sin

2 ✓0

=

is

k2

(17.1.188)

into Fresnel formulas (17.1.163) and (17.1.173) one gets5360 ✓E0

1

E0

0

◆?=

Z2

cos ✓0

� iZ1

(s/k2

)

Z2

cos ✓0

+ iZ1

(s/k2

)

= e2i'? (17.1.189)✓E0

1

E0

0

◆k=

Z1

cos ✓0

� iZ2

(s/k2

)

Z1

cos ✓0

+ iZ2

(s/k2

)

= e2i'k (17.1.190)

where5361

tan'? = �Z1

(s/k2

)

Z2

cos ✓0

, tan'k = �Z2

(s/k2

)

Z1

cos ✓0

,Z1

Z2

⌘ µ1

µ2

n2

n1

. (17.1.191)

The parametrization comes from5362

1 + i⇣�Z1(s/k2)

Z2 cos ✓0

⌘1� i

⇣�Z1(s/k2)

Z2 cos ✓0

⌘=

1 + i tan'?1� i tan'?

=

cos'? + i sin'?cos'? � i sin'?

=

ei'?

e�i'? = e2i'?

and similarly for 'k. The ratio of the amplitudes has absolute value |e2i'? | = 1 and |e2i'k | = 15363

so the magnitude of reflected wave is equal to magnitude of incident wave. The only change5364

is in a phase. It follows that the energy flux of the incident wave is totally reflected. There is5365

no transfer of the energy to the second medium. For this reason the phenomenon is called total5366

reflection. The existence of penetration depth and the reflection of the whole energy do not lead5367

to contradiction if the refracted wave goes back to the first medium. This fact was confirmed5368

experimentally. Moreover, the small distance on which the refracted wave propagate parallel to5369

the surface of interface in the second medium before going back to the first medium was also5370

observed experimentally. Total reflection is responsible for formation of mirages.5371

17.1.7.6 Energy fluxes of reflected and refracted wave5372

The energy flux is represented by time average of the Poynting vector5373

hSi = c

8⇡µRe [E ⇥B⇤

] =

1

8⇡Z|E|2k =

1

8⇡Z|E0|2k (17.1.192)

401


which can be decomposed on directions parallel and perpendicular to the surface of interface5374

hSi = (hSi · n)| {z }hSi

n

n+ (hSi · t)| {z }hSi

t

t (17.1.193)

The reflection coefficient is given as the ratio5375

R :=

| hS1

i · n|| hS

0

i · n| =|E0

1

|2

|E0

0

|2(17.1.194)

where k0

· n = k1

· n because ✓1

= ✓0

. The reflection coefficient R represent a total flux of5376

energy reflected on the surface of interface. This flux can be split on contribution coming from5377

the wave having the electric field vector perpendicular and parallel to the plane of incidence. It5378

leads to definition of coefficients R? and Rk5379

R? =

Z2

cos ✓0

� Z1

cos ✓2

Z2

cos ✓0

+ Z1

cos ✓2

�2

, Rk =

Z1

cos ✓0

� Z2

cos ✓2

Z1

cos ✓0

+ Z2

cos ✓2

�2

. (17.1.195)

In the case of normal incidence ✓0

= 0 and for µ1

= µ2

the reflection coefficient has the form5380

Rn =

✓n1

� n2

n1

+ n2

◆2

. (17.1.196)

The transmission coefficient is defined as follows5381

T :=

| hS2

i · n|| hS

0

i · n| =Z1

cos ✓2

Z2

cos ✓0

|E0

2

|2

|E0

0

|2(17.1.197)

The perpendicular and parallel transmission coefficients read5382

T? =

4Z1

Z2

cos ✓0

cos ✓2

(Z2

cos ✓0

+ Z1

cos ✓2

)

2

Tk =4Z

1

Z2

cos ✓0

cos ✓2

(Z1

cos ✓0

+ Z2

cos ✓2

)

2

(17.1.198)

For normal incidence and for µ1

= µ2

one gets5383

Tn =

4n1

n2

(n2

+ n1

)

2

. (17.1.199)

Let us observe that R? + T? = 1 and similarly Rk + Tk = 1. The energy conservation requires5384

that it must hold for total coefficients5385

R+ T = 1. (17.1.200)

17.2 Electromagnetic waves in dispersive dielectric me-5386

dia5387

(Napisac ten rozdzial...........)5388

402

17.3 Electromagnetic waves in conducting media

17.3 Electromagnetic waves in conducting media5389

The constitutive relations in conducting media must include Ohm’s law which is relation be-5390

tween an electric strength field E and a current density J . We shall assume that5391

D = "E, B = µH, J = �E, ⇢ = 0 (17.3.201)

where � is electric conductivity and ⇢ stands for density of free electric charges. We shall5392

consider the homogeneous and non-dispersive medium, " = const, µ = const. The Maxwell’s5393

equations take the form5394

r ·E = 0, r⇥B � "µ

c@tE � 4⇡

c�µE = 0 (17.3.202)

r ·B = 0, r⇥E +

1

c@tB = 0. (17.3.203)

17.3.1 Relaxation time5395

Let us consider a plane electromagnetic wave in non-perfect conductor. Without loss of gen-5396

erality we can choose a Cartesian versor e3

in direction of propagation of the wave. Since the5397

electric field enters to the Maxwell’s equation by a current term one has to check what are re-5398

strictions on transversality of electromagnetic waves in such a case. For this reason we do not5399

assume that E and B are perpendicular to the wave vector. The behaviour of these fields is5400

determined by Maxwell’s equation. Let us observe that5401

r ·E = ei · (@iE) = e3

· (@3

E)

r⇥E = ei ⇥ (@iE) = e3

⇥ (@3

E)

so the Maxwell’s equations take the form5402

e3

· (@3

E) = 0, e3

· (@3

B) = 0 (17.3.204)5403

e3

⇥ (@3

B)� "µ

c@tE =

4⇡

c�µE (17.3.205)

e3

⇥ (@3

E) +

1

c@tB = 0. (17.3.206)

Taking a scalar product of (17.3.205) with e3

one gets5404

e3

· @tE = �4⇡�

"e3

·E (17.3.207)

403


Multiplying (17.3.207) by dt and summing with first equation of (17.3.204) multiplied by dx35405

we get5406

e3

·⇥@tEdt+ @

3

Edx3⇤| {z }

dE

= �4⇡�

"e3

·Edt

what gives5407

e3

·dE

dt+

4⇡�

"E

�= 0.

This is in fact equation for longitudinal component of the electric field and can be cast in the5408

form5409

dEkdt

+

1

⌧Ek = 0, ⌧ :=

"

4⇡�(17.3.208)

Ek := e3

· E and ⌧ is a constant with dimension of time and it is called relaxation time. Re-5410

peating these steps for two remainig Maxwell’s equations we get dBkdt = 0. Equation (17.3.208)5411

has solution5412

Ek(t, x3

) = Ek(0, x3

)e�t

⌧ .

It follows that in non-perfect conductors there can exist a time dependent longitudinal compo-5413

nent of the electric field which decreases with time as e�t

⌧ . The Maxwell’s equations allow only5414

for static magnetic longitudinal component Bk = const. It leads to conclusion that electromag-5415

netic waves in conductors are transversal. Any time dependent longitudinal electric component5416

vanishes exponentially with time. Note, that a longitudinal component is absent in perfect con-5417

ductors � = 1 (⌧ = 0). On the other hand in empty space � = 0 (⌧ = 1) therefore the5418

longitudinal component is static. However, a static electric field cannot be a part of a wave5419

solution so electromagnetic waves in empty space are transversal.5420

17.3.2 Dispersion relation5421

The appropriate combination of Maxwell’s equations leads to the second order equations for5422

electric and magnetic field. The steps are exactly the same as for non-conductors. Taking rota-5423

tional of Ampere’s-Maxwell equation and using other equations one can get equation for mag-5424

netic field and similarly acting with rotational on Faraday’s law one gets equation for electric5425

field. The resulting equations have the form5426

LE = 0, LB = 0 (17.3.209)

404


where L stands for a linear differential operator5427

L :=

"µ

c2@2t + 4⇡

�µ

c2@t �r2. (17.3.210)

The first order differential operator appearing in 4⇡ �µc2@t plays role of a dissipative term (sim-5428

ilarly to diffusion equation). For � ! 0 the operator L became a d’Alembert operator. The5429

fields5430

E = E0

ei(·r�!t), B = B0

ei(·r�!t) (17.3.211)

are solution of (17.3.209) if Lei(·r�!t) = 0. The last equation gives5431 �"µc2!2 � i

4⇡�µ

c2! + 2

�ei(·r�!t) = 0

which gives a dispersion relation5432

2 = "µ!2

c2

1 + i

4⇡�

"!

�. (17.3.212)

Note that rhs of (17.3.212) is a complex number. It means that is a complex-valued vector, i.e.5433

2

= 2 2 C. Let us parametrize the complex number as5434

= k + is, k, s 2 R.

Plugging this expression to (17.3.212) one gets5435

k2 � s2 = "µ!2

c2, ks = 2⇡�µ

!

c2. (17.3.213)

Then plugging s (k) from the second equation into the first one and multiplying by k2 (s2) we5436

obtain5437

k4 � "µ!2

c2k2 �

✓2⇡�µ!

c2

◆2

= 0, (17.3.214)

s4 +"µ!2

c2s2 �

✓2⇡�µ!

c2

◆2

= 0. (17.3.215)

In both cases5438

� =

✓"µ!2

c2

◆2

"1 +

✓4⇡�

"!

◆2

#(17.3.216)

then5439

k2 ="µ!2

2c2

24s1 +

✓4⇡�

"!

◆2

+ 1

35 s2 ="µ!2

2c2

24s1 +

✓4⇡�

"!

◆2

� 1

35

405


where term with �p� must be discarded in both cases because it lads to imaginary solutions5440

for k and s. Finally we obtain5441

k =

np2

⇣!c

⌘24s1 +

✓4⇡�

"!

◆2

+ 1

35 12

(17.3.217)

5442

s =np2

⇣!c

⌘24s1 +

✓4⇡�

"!

◆2

� 1

35 12

. (17.3.218)

17.3.3 Relation between amplitudes5443

Having the form of vector one can obtain an algebraic relation between E0

and B0

. Let us5444

consider an electromagnetic wave which propagate along the axis x3. Then taking the wave5445

vector in the form = (k + is)e3

one gets5446

E = E0

e�s x3ei(k x3�!t) (17.3.219)

B = B0

e�s x3ei(k x3�!t). (17.3.220)

The Faraday’s law e3

⇥ @3

E +

1

c@tB = 0, where @3

E = i(k + is)E and @tB = �i!B gives5447

B0

=

c

!(k + is) e

3

⇥E0

. (17.3.221)

where5448

= ||ei�, || =p

k2 + s2, � = arctan

⇣ sk

⌘. (17.3.222)

Plugging (17.3.217) and (17.3.218) to these formulas we obtain5449

|| = n!

c

"1 +

✓4⇡�

"!

◆2

# 14

(17.3.223)

and5450

� =

1

2

arctan

✓4⇡�

"!

◆. (17.3.224)

The second expression can be proved as follows. Let us denote5451

tan� =

s

k=

pu� 1pu+ 1

� 12

, u ⌘ 1 +

✓4⇡�

"!

◆2

(17.3.225)

406


then5452

tan(2�) =

sin(2�)

cos(2�)=

2 sin� cos�

cos

2 �� sin

2 �=

2 tan�

1� tan

2 �

=

2

sk

1��sk

�2

= 2

hpu�1pu+1

i 12

1�pu�1pu+1

= 2

hpu�1pu+1

i 12

pu+1�

pu+1p

u+1

= (

pu+ 1)

pu� 1pu+ 1

� 12

= [(

pu)2 � 1]

1/2=

pu� 1 =

4⇡�

"!.

Finally we obtain that complex electric and magnetic field amplitudes satisfy5453

B0

= n

"1 +

✓4⇡�

"!

◆2

# 14

ei� e3

⇥E0

. (17.3.226)

The factor ei� is responsible for time lag which manifests as a mutual phase shift of fields B5454

and E. Since e3

·E0

= 0 = e3

·B0

then5455

|B0

| = n

"1 +

✓4⇡�

"!

◆2

# 14

|E0

|. (17.3.227)

17.3.4 Limit cases5456

Let us analyze again the equation5457

"µ

c2@2tE + 4⇡

�µ

c2@tE �r2E = 0.

The second order time derivative term "µc2@2tE has its origin in displacement current whereas5458

the first order time derivative term 4⇡ �µc2@tE comes from the conduction current. Each of these5459

terms contribute to dispersion relation through5460

"µ

c2@2tE ! �"µ

c2!2, 4⇡

�µ

c2@tE ! �i

4⇡� µ

c2!.

The ratio of absolute values of these two terms reads5461

4⇡� µc2

!"µc2!2

=

4⇡�

"!= tan(2�). (17.3.228)

Case 4⇡�"! ⌧ 1.5462

407


In this case the effect of conduction current is small comparing with the effect of displace-5463

ment current. Expanding k in powers of 4⇡�"! one gets5464

k =

np2

⇣!c

⌘24s1 +

✓4⇡�

"!

◆2

+ 1

35 12

=

np2

⇣!c

⌘"1 +

1

2

✓4⇡�

"!

◆2

+ . . .+ 1

# 12

= n!

c

"1 +

1

4

✓4⇡�

"!

◆2

+ . . .

# 12

= n!

c

"1 +

1

2

✓2⇡�

"!

◆2

#+ . . . (17.3.229)

and similarly5465

s =

np2

⇣!c

⌘24s1 +

✓4⇡�

"!

◆2

� 1

35 12

=

np2

!

c

"1 +

1

2

✓4⇡�

"!

◆2

+ . . .+ (�1)

# 12

=

r"µ

2

!

c

1p2

4⇡�

"!+ . . . =

2⇡�

c

rµ

"+ . . . (17.3.230)

where Z :=

qµ" . Notice that the imaginary part s = Im() does not depend on the frequency5466

!. For � ⇡ 0 the phase shift is � ⇡ 0 so the fields E and B have approximately the same phase.5467

Case 4⇡�"! � 1.5468

In this limit the conduction current dominates. Such situation takes place in metals where5469

�/" ⇡ 10

18. The conduction current dominates for frequencies less than 10

17 Hz (microwaves,5470

radio-frequency, light, some range of X-ray). One gets5471

k ⇡ s ⇡ 1

c

p2⇡� µ!

then5472

|| =

pk2 + s2 ⇡ 1

c

p4⇡� µ! (17.3.231)

� = arctan

⇣ sk

⌘⇡ arctan(1) =

⇡

4

. (17.3.232)

The fields have relative phase shift � = ⇡/4 and their amplitudes satisfy5473

|B0

| ⇡ n

r4⇡�

"!|E

0

| (17.3.233)

where n takes values of the order of unity whereasq

4⇡�"! � 1 then |B

0

| � |E0

|.5474

408

17.4 Waveguides

17.3.5 Distribution of electric current in conductors5475

The displacement current @tD term is irrelevant for good conductors. Then equation5476

r2E � 4⇡�µ

c2@tE = 0

is equivalent to the following one5477

r2J � 4⇡�µ

c2@tJ = 0 (17.3.234)

where J = �E. The last equation is known as diffusion equation. Since electric field possesses5478

time dependence given by term e�i!t then the current J must contain this dependence as well5479

i.e. J = J0

(r)e�i!t. Plugging this form to last equation we obtain5480

r2J0

+ ⌧2J0

= 0, ⌧2 := i4⇡�µ!

c2(17.3.235)

or considering thatpi = 1p

2

(1 + i)5481

⌧ =

1 + i

�, � =

cp2⇡�µ!

⇡ 1

sin the limit

4⇡�

"!� 1.

In SI units � =

q2

�µ! . Let us consider a conductor in a half-space x1 � 0. The electromag-5482

netic field propagates along the axis x1 ⌘ x. For linearly polarized wave one can choose for5483

convenience the axis x3 in direction of the electric field vector, then J0

(r) = J3

0

(x)e3

. The5484

equation5485 d2

d(x)2+ ⌧2

�J3

0

(x) = 0 (17.3.236)

has a solution5486

J3

0

(x) = J3

0

(0)ei⌧ x = J3

0

(0)e�x

� eix

� . (17.3.237)

We have to reject the second one solution e�i⌧ x because it leads to non-physical behaviour in5487

conductor x � 0. The current density vanishes exponentially in the conductor. The characteristic5488

length � is called skin depth.5489

For instance, a skin depth for an electromagnetic wave with the frequency ⌫ = 4 ⇤ 109 Hz5490

which enters to silver material (��1

= 1.58 ⇤ 10

�8

⌦m) has value � =

1p⇡µ�⌫ = 10

�6m. In5491

copper (��1

= 1.68 ⇤ 10�8

⌦m) for UV light ⌫ = 10

15 Hz skin depth is � = 10

�9m.5492

409


17.4 Waveguides5493

Waveguides are formed by conducting tubes filled with some dielectric material which in gen-5494

eral posses some magnetic properties. The perpendicular cross section of the waveguide is not5495

necessary a symmetric figure.5496

The electric and magnetic field exist only inside the waveguide5497

E1

⌘ E, B1

⌘ B, D1

⌘ D, H1

⌘ H

and5498

E2

= 0, B2

= 0, D2

= 0, H2

= 0.

The field intensities and inductions are related by constitutive relations D = "E and B = µH .

Figure 17.11: A waveguide cross section.

(a) (b)

Figure 17.12: Waveguides (a) and coaxial cable (b)

410

17.4 Waveguides

5499

Let n be a normal vector to the surface of a waveguide pointing outside. According to Maxwell’s5500

equations the fields satisfy the boundary conditions at the waveguide surface S, namely5501

D · n = �4⇡�, B · n = 0, n⇥E = 0, n⇥H = �4⇡

cK. (17.4.238)

Maxwell’s equations imply that electric and magnetic field obey wave equation ⇤E = 0 and5502

⇤B = 0 where ⇤ :=

"µc2@2t � r2. We choose the axis x3 as being parallel to the longitudinal5503

axis of the waveguide. The electric and magnetic field can be considered in the following form5504

E(t, r) = E(x1, x2)ei(±kx3�!t) (17.4.239)

B(t, r) = B(x1, x2)ei(±kx3�!t). (17.4.240)

After plugging this expression to wave equation we get equations for E(x1, x2) and B(x1, x2).5505

We get5506

�⇤E(t, r) =

hr2

? + @33

� "µ

c2@2t

i ⇣E(x1, x2)ei(±kx3�!t)

⌘=

⇥r2

? + �2⇤E(x1, x2)ei(±kx3�!t) (17.4.241)

where5507

�2 := "µ!2

c2� k2. (17.4.242)

It gives5508 ⇥r2

? + �2⇤E(x1, x2) = 0,

⇥r2

? + �2⇤B(x1, x2) = 0. (17.4.243)

17.4.1 Parallel and transversal components5509

Electric and magnetic field amplitudes E(x1, x2), B(x1, x2) can be decomposed on a perpen-5510

dicular and parallel part to the axis x35511

E = E1e1

+ E2e2| {z }

E?

+E3e3| {z }

Ek

B = B1e1

+B2e2| {z }

B?

+B3e3| {z }

Bk

(17.4.244)

This decomposition is very useful. In some cases (transverse electric and transverse magnetic5512

waves) the perpendicular E?, B? and parallel components satisfy some algebraic relations.It5513

follows that the perpendicular components can be directly determined from the parallel ones.5514

The parallel components E3 and B3 are determined by equations (17.4.243). We shall rewrite5515

411


Maxwell’s equations in terms of perpendicular and parallel components splitting the grad oper-5516

ator as r = r? + e3

@3

.5517

Faraday’s law takes the form5518

r? ⇥E(t, r) + e3

⇥ @3

E(t, r)| {z }±ike3⇥E(t,r)

= �1

c@tB(t, r)| {z }�i!B(t,r)

(17.4.245)

It follows that expression ei(±kx3�!t) became just an multiplicative factor, then the amplitude5519

vectors E(x1, x2) and B(x1, x2) satisfy equation5520

r? ⇥E ± ik e3

⇥E = i!

cB

then plugging (17.4.244) one gets5521

r? ⇥Ek +r? ⇥E? ± ik e3

⇥E? = i!

c(Bk +B?). (17.4.246)


and considering that e3

· (r? ⇥ Ek) = 0 and5522

e3

· (e3

⇥E?) = 0 we obtin5523

e3

· (r? ⇥E?) = i!

cB3. (17.4.247)

Next, taking a cross product of (17.4.246) with e3

and taking into account relations e3

⇥5524

(r? ⇥E?) = 0 and e3

⇥ (e3

⇥E?) = �E? and5525

e3

⇥ (r? ⇥Ek) = e3

⇥ (e1

⇥ e3

@1

E3

+ e2

⇥ e3

@2

E3

)

= e3

⇥ (�e2

@1

E3

+ e1

@2

E3

) = e1

@1

E3

+ e2

@2

E3

= r?E3

we get equation5526

r?E3 ⌥ ikE? = i

!

ce3

⇥B?. (17.4.248)

For Ampere-Maxwell’s law we obtain5527

r? ⇥B(t, r) + e3

⇥ @3

B(t, r)| {z }±ike3⇥B(t,r)

=

"µ

c@tE(t, r)| {z }�i!E(t,r)

(17.4.249)

then5528

r? ⇥Bk +r? ⇥B? ± ik e3

⇥B? = �i"µ!

c(Ek +E?). (17.4.250)

412

17.4 Waveguides


one gets equation5529

e3

· (r? ⇥B?) = �i"µ!

cE3. (17.4.251)

whereas for cross vector product one obtain5530

r?B3 ⌥ ikB? = �i"µ

!

ce3

⇥E?. (17.4.252)

Decomposition of Gauss laws for electric and magnetic fields gives5531

r? ·E? = ⌥ik E3, r? ·B? = ⌥ik B3. (17.4.253)

17.4.1.1 Relations between amplitudes for �2 6= 05532

In order to calculate E? we multiply (17.4.252) by e3

from left what gives5533

e3

⇥ (r?B3

)⌥ ik e3

⇥B? = �i"µ!

ce3

⇥ (e3

⇥E?)| {z }�E?

. (17.4.254)

Substituting e3

⇥B? =

ci! (r?E3⌥ikE?) from (17.4.248) and multiplying by �i!c we obtain5534

E? =

i

�2

h±kr?E

3 � !

ce3

⇥r?B3

i(17.4.255)

where in the last step divide by �2. This relation is true unless �2 = 0 i.e. when "µ!2

c2= k2.5535

Similar result can be obtained for B?. This time we multiply (17.4.248) by e3

from left5536

e3

⇥ (r?E3

)⌥ ik e3

⇥E? = i!

ce3

⇥ (e3

⇥B?)| {z }�B?

. (17.4.256)

and substitute e3

⇥E? = � ci!"µ(r?B3 ⌥ ikB?) from (17.4.252). Multiplying the result by5537

i"µ!c and finally dividing by �2 we obtain5538

B? =

i

�2

h±kr?B

3

+ "µ!

ce3

⇥r?E3

i. (17.4.257)

Let us observe that in this case the amplitudes E? and B? are uniquely determined from E3

5539

and B3.5540

413


17.4.2 TE and TM waves5541

The electromagnetic field takes non-zero values only inside the waveguide. Electric field E5542

satisfy equation (r2

? + �2)E = 0 and similarly for B. Moreover, since the waveguide surface5543

S is a conductor then tangential component of the electric field must vanish at this surface5544

E3|S = 0. Taking a scalar product of the equation (17.4.252) with the vector n normal to the5545

waveguide surface we obtain5546

n · (r?B3

)| {z }@B

3

@n

⌥ik n ·B?| {z }n·B

= �i"µ!

cn · (e

3

⇥E?)| {z }�e3·(n⇥E)

. (17.4.258)

At the surface S where n ·B|S = 0 and n⇥E|S = 0 the equations imply that5547

@B3

@n

��S

= 0.

Any electromagnetic wave solution that propagates in waveguide must satisfy the boundary5548

conditions5549

n ·B|S = 0, n⇥E|S = 0, E3|S = 0,@B3

@n

��S

= 0. (17.4.259)

Transverse electric waves TE are electromagnetic waves such that E3

= 0 everywhere. Rela-5550

tion between E? and B? for TE waves follows directly from (17.4.248) and it reads5551

E? = ⌥ !

kce3

⇥B? , B? = ±kc

!e3

⇥E?. (17.4.260)

From equation (17.4.257) we obtain5552

B? = ± ik

�2r?B

3 (17.4.261)

where B3 is a solution of equation (r2

? + �2)B3

= 0. This solution must to satisfy the Neu-5553

mann boundary conditions @B3

@n |S = 0. The details of the solution depends on geometry of the5554

waveguide.5555

Transverse magnetic waves TM are solutions with B3

= 0 everywhere. From equation5556

(17.4.252)5557

B? = ±"µ !kc

e3

⇥E? , E? = ⌥ kc

"µ!e3

⇥B? (17.4.262)

where E? follows from (17.4.255) and it reads5558

E? = ± ik

�2r?E

3. (17.4.263)

The transverse component of the electric field is a solution of the equation (r2

? + �2)E3

= 05559

with the Dirichlet boundary condition E3|S = 0.5560

414

17.4 Waveguides

17.4.2.1 TEM waves5561

Transverse electromagnetic waves TEM are solutions such that E3

= 0 and B3

= 0 every-5562

where. Equations (17.4.247) and (17.4.251) became5563

e3

· (r? ⇥E?) = 0, e3

· (r? ⇥B?) = 0. (17.4.264)

From (17.4.253) we have5564

r? ·E? = 0 r? ·B? = 0. (17.4.265)

The equations for electric field in Cartesian coordinates read5565

@1

E2 � @2

E1

= 0, @1

E1

+ @2

E2

= 0 (17.4.266)

then differentiating a first equation with respect to x1 and using the second equation we obtain5566

(@21

+ @22

)E2

= 0. If one takes derivative with respect to x2 instead of x1 then the resulting5567

equations became (@21

+ @22

)E1

= 0. The same steps can be repeated for a magnetic field. We5568

conclude that5569

r2

?E? = 0, r2

?B? = 0. (17.4.267)

If one compare these equation with equations (17.4.243) then it follows that5570

�2 = "µ!2

c2� k2 = 0

We conclude that for TEM waves the dispersion relation is exactly the same as for waves in5571

infinite space5572

k = n!

c. (17.4.268)

The algebraic relations between electric and magnetic fields follow directly from (17.4.248) aor5573

(17.4.252) and they read5574

B = ±n e3

⇥E? E = ⌥ 1

ne3

⇥B?. (17.4.269)

Another important property of TEM waves is fact that such waves cannot exist in hollow5575

waveguides. The equation e3

· (r? ⇥ E?) = 0 is equivalent to r? ⇥ E? = 0 because5576

for E3

= 0 the expression r? ⇥E? has only component parallel to x3. However, since E? is5577

a function of two spatial variables then5578

r? ⇥E? = 0 ) E? = �r'(x1, x2). (17.4.270)

415


Plugging this expression to r? ·E? = 0 we obtain Laplace equation for '5579

r2

?' = 0. (17.4.271)

Since the electric field vanishes on S then the potential ' = const on S. The Laplace equation5580

has only a trivial constant solution in such a case. Then E? = 0 inside the waveguide. For this5581

reason there are no electromagnetic TEM waves in hollow waveguides.5582

416

17 electromagnetic waves - paweł klimasklimas.paginas.ufsc.br/files/2017/05/chapter17-1.pdf ·...

Documents