1.1 The wave equation for a string

Phys 460 Fall 2006 Susan M. Lea

1 Wavesw basics

1.1 The wave equation for a string

Waves occur in many physical systems. A wave is a disturbance that propagates through the system at a well-determined speed that depends on the physical properties of the system. All wave disturbances may be decomposed into a sum of simpler waves- sinusoidal waves. In these waves, the shape of the disturbance, captured by taking a snapshot of the disturbance in the medium at a ¤xed time, has the shape of a sine wave. Every point of the system oscillates in time. Thus we might expect that the system has restoring forces that try to return the system to its initial equilibrium. Wezll begin by studying a simple mechanical systemw a string. The restoring force is the tension in the string.

In equilibrium the string lies along the x¡axis: We look at a di¤erential piece of the displaced string, as shown in the diagram.

The net force on the string has components:

dF x = T (x + dx) cos (µ + dµ) ¡ T (x) cos µ dF y = T (x + dx) sin µ ¡ T (x) sin µ

Now let the displacement of the string be small everywhere, fo that µ ¿ 1 everywhere. Then

cos µ = 1 ¡ µ ²

2 + ¢ ¢ ¢ ' 1 sin µ = µ ¡ µ ³

6 + ¢ ¢ ¢ ' µ and

dF x = T (x + dx) ¡ T (x)

dF y = T (x + dx) (µ + dµ) ¡ T (x) µ

Each piece of the string moves vertically but not horizontally, so dF x = 0 ) T (x + dx) = T (x)

The tension is constant along the string. Then the y¡component is dF y = T dµ = (dm) @ ² y

@t ²

The time derivatives are partial derivatives: we are looking at a ¤xed value of x: The mass of the string segment is

dm = ¹dx

The string has to stretch a bit in order to curve, but the mass per unit length

¹ is measured along the undisturbed string. Thus T dµ = ¹dx @ ² y

@t ²

or @ ² y

@t ² = T

¹

@µ

@x

The x¡derivatives are also partial because the picture shown is a snapshot taken at a ¤xed time. Now we work on the angle µ: The slope of the string is

tan µ = @y

@x ' µ for µ ¿ 1:

Thus @ ² y

@t ² = T

¹

@ ² y

@x ² (1)

This is the wave equation for the string. The wave speed is given by v ² = T

¹

and v is roughly equal to the square root of the restoring force/inertia. The general solution to this equation has the form

y = f (x § vt)

as you can easily check by di¤erentiating and stu¢ng in. With the minus sign, the solution represents a wave propagating in the direction of increasing x;

with the plus sign the pulse propagates in the direction of decreasing x: Since the wave equation is linear, we can have superpositions of such solutions, for example

y = f (x + vt) + g (x ¡ vt) Sinusoidal waves occur when f is a sine (or cosine) wave.

y = A cos [k (x ¡ vt) + Á] = A cos (kx ¡ !t + Á)

At ¤xed t; the cosine repeats after a distance ¢x where k¢x = 2¼

or

¢x = 2¼ k = ¸

where ¸ is the wavelength of the wave. Similarly, at ¤xed x; the cosine repeats after a time

¢t = 2¼

! = T where T is the wave period. Also

! k = v

A is the wave amplitude. It is the maximum value of the disturbance anywhere in the wave. Á is a phase constant. It tells us where the maximum displacement occurs at t = 0:

The speed v is the phase speed of the wave. The wave phase is the argument of the cosine. So a ¤xed phase Á ₀ corresponds to

kx ¡ !t + Á = Á 0

or

x = !

k t + Á ₀ ¡ Á

k = vt + Á ₀ ¡ Á k

so if k is positive, the value of x corresponding to phase Á ₀ increases at speed v:

Standing waves occur when equal amplitude waves travel in opposite direc- tions:

y stand = A cos (kx ¡ !t) + A cos (kx + !t + Á)

= 2A cos µ

kx + Á 2

¶ cos !t

Now we have a ¤xed spatial function that oscillates in time.

1.1.1 More math makes it easier

We can use complex numbers to make the math easier. Eulerzs formula tells us e ^iµ = cos µ + i sin µ

Thus

A cos (kx ¡ !t + Á) = Re A exp (i [kx ¡ !t + Á])

= Re n¡

Ae ^iÁ ¢ ³

e ^i(kx¡!t) ´o

= Re ³

Ae e ^ikx¡i!t ´

where

A = Ae e ^iÁ

is the complex amplitude of the wave. With this notation, the general solution of the wave equation may be written:

f (x; t) = Z ₊₁

¡1

A (k) e e ^ikx¡i!t dk where

! = kv

The "Re " is implied but is not always written. Real physical quantities are always the real part of complex numbers.

1.2 Boundary conditions: re¢ection and transmission

When a wave reaches a point where the properties of the system change, part of the wave energy will be re¢ected and part transmitted into the new region. In the case of 1-D waves on a string, the mass/unit length ¹ is the critical system property, As ¹ changes, so does v :

v 1 = s T

¹ ₁ ; v 2 = s T

¹ ₂

and thus ! 1

k 1 = v 1 ; ! 2

k 2 = v 2

The disturbance on the string has three components: the incident wave y i = A exp (ik i x ¡ i!t), the re¢ected wave travelling in the negative x¡direction, y r = A r exp (¡ik r x ¡ i! r t) and the transmitted wave y t = A t exp (ik t x ¡ i! t t).

The ¤rst two travel at speed v 1 while the transmitted wave travels at speed v 2 : The amplitudes of the re¢ected and transmitted waves are determined by the boundary conditions at the junction between the strings. To determine the two amplitudes, we need two boundary conditions. They are:

(1) The string displacement is continuous across the junction. If this were not true, the string would have an unphysical "break" at the junction.

For simplicity, we place the origin at the junction between the strings.

y i (0; t) + y r (0; t) = y t (0; t) A exp (¡i!t) + A r exp (¡i! r t) = A t exp (¡i! t t)

This boundary condition must hold for all times t: Thus it must be true that

! = ! r = ! t

The frequency of all the waves is the same. Then it follows that

A + A r = A t (2)

and also

k r = ! v 1

= k i ; k t = ! v 2

(2) The slope of the string is continuous across the junction. If this were not true, there would be unbalanced forces on the junction that would cause rapid acceleration, rapidly restoring the force balance and the continuity of the slope. (Be careful herex if the strings are tied together so that there is a massive knot at the junction, then there may be a discontinuity of slope. This is not the usual case.)

ik i Ae ^¡i!t ¡ ik ^r A t e ^¡i!t = ik t A t e ^¡i!t The exponentials cancel, leaving

A v 1 ¡ A r

v 1

= A t

v 2

(3a) From equations (2) and (3a) we can solve for the amplitudes:

A ¡ A r = v 1

v 2

(A + A r ) A

µ 1 ¡ v 1

v 2

¶

= A r

µ 1 + v 1

v 2

¶

A r = A v 2 ¡ v ¹ v 2 + v 1

(4) and

A t = A µ

1 + v 2 ¡ v 1

v 2 + v 1

¶

= A

µ 2v 2

v 2 + v 1

¶

(5) Note that if v 2 = v 1 ; there is no junction, and A r = 0; A t = A; as expected.

If the second string is heavier than the ¤rst, ¹ ₂ > ¹ ₁ ; then v 2 < v 1 and A r is negative. Recall that these amplitudes are complex numbers, so

A r = ¡A v 1 ¡ v 2

v 2 + v 1

= e ^i¼ A v 1 ¡ v 2

v 2 + v 1

The re¢ected amplitude has a phase di¤erence of ¼ compared with the incident amplitude. Taking the real part,

y r = jAj v 1 ¡ v ²

v 2 + v 1 cos (¡ik ^r x ¡ !t + ¼)

and the wave has a phase change of ¼: The re¢ected wave is ¢ipped "upside down".

Boundary conditions like this (continuity of function, continuity of deriva-

tive) apply in most physical systems.

1.3 Polarization

The waves on the string are transverse waves: the string itself moves perpen- dicular (in the y¡direction) to the direction that the wave itself moves (the x¡direction). The wave is said to be polarized in the vertical (y) direction.

We could also set the string vibrating so that it moves in the z¡direction and we would say the wave is polarized in the z¡direction. Or we could choose any direction in the y ¡ z plane. These are linearly polarized waves. Now suppose we set the string vibrating in the y¡direction, and a quarter period later we also set it going with equal amplitude in the z¡direction. The total displacement is

~s = A

½

y exp (ikx ¡ i!t) + ^zexp ^

·µ

ikx ¡ i!

µ t ¡ T

4

¶¶¸¾

= Ae ^ikx¡i!t ³

^

y + ^ ze ^i¼=2 ´

= Ae ^ikx¡i!t (^ y + i ^ z) Now we take the real part:

~

s = A [ ^ y cos (kx ¡ !t) ¡ ^z sin (kx ¡ !t)]

Fix attention at one point of the string w say x = 0; then we have

~

s (0; t) = A (^ y cos !t + ^ z sin !t)

Now the string moves around a circle of radius A in the y ¡ z¡plane. This is circular polarization.

Contrast the string with waves on a springw these are longitudinal wavesw the spring coils travel back and forth along the spring- the same direction that the wave moves. The concept of polarization does not arise for these waves.

1.4 Energy transmission

As the wave travels along the string, it transmits energy. We can see how the energy is related to the wave properties by looking at the rate at which the string to the left of a point x = x 0 does work on the string to its right. The string to the left exerts a force ~T on the string to the right, and at the point of contact the string has velocity

~

v = ^ y @y (x 0 ; t)

@t Thus

P = ~ T ¢ ~v = ¡T sin µ @y (x 0 ; t)

@t But since the string has small displacement, µ ¿ 1;

sin µ ' tan µ = @y

@x

So, with y = A cos (kx ¡ !t)

P = ¡T @y

@x

¯ ¯

¯ ¯ _x

0

@y (x 0 ; t)

@t

= T kA sin (kx 0 ¡ !t) !A sin (kx ⁰ ¡ !t)

= ¹v!A ² sin ² (kx 0 ¡ !t)

Since the sine function is squared, the power is always positive; energy is trans- mitted to the right continuouslyw this is the direction in which the wave is propagating. The power depends on the square of the wave ampliudew doubling the amplitude quadruples the energy transmittedwand also depends on the wave speed and the wave frequency.

2 Electromagnetic waves in vacuum

Maxwellzs equations in vacuum are (½ = 0; ~j = 0)

r ¢ ~ ~ E = 0 (6)

r ¢ ~ ~ B = 0 (7)

r £ ~ ~ E = ¡ @ ~ B

@t (8)

r £ ~ ~ B = " 0 ¹ ₀ @ ~ E

@t (9)

Even though ½ and ~j are zero, the ¤elds are not necessarily zero, because chang- ing ~ B acts as a source of ~ E; and changing ~ E acts as a source of ~ E: The resulting

¤elds are wave ¤elds that travel at the speed of light. To demonstrate this, letzs try to eliminate one of the ¤elds. Start with Amperezs law (9) and take the curl:

r £ ~ ³

r £ ~ ~ B ´

= 1

c ² r £ ~ @ ~ E

@t r ~ ³

r ¢ ~ ~ B ´

¡ r ² B ~ = 1 c ²

@

@t

³ r £ ~ ~ E ´

Now use equation (7) on the left and Faradayzs law (8) on the right:

¡r ² B = ¡ ~ 1 c ²

@ ² B ~

@ ² t so we obtain the wave equation with wave speed c :

r ² B = ~ 1 c ²

@ ² B ~

@ ² t (10)

The equation for waves on a string (1) was 1-dimensional in the space variables, but here we have the r ² operator, so we have derivatives in all 3 space variables.

µ @ ²

@x ² + @ ²

@y ² + @ ²

@z ²

¶ B = ~ 1

c ²

@ ² B ~

@ ² t

The waves are not constrained to move in one direction, but can move along any line in space, described by the wave vector ~k: The magnitude of this vector is related to the wavelength, as with 1¡D waves.

k =

¯ ¯

¯~k

¯ ¯

¯ = ! c = 2¼

¸ The speed of the waves in vacuum is

c = 1 p " 0 ¹ 0

= 3 £ 10 ⁸ m/s

EM waves occur at all frequencies. The names we give the waves are historical and re¢ect the methods we use to detect them. They range from gamma rays at very high frequencies to low frequency radio. See G page 377 and LB page 533.

Visible light occupies a small fraction of the spectrum around º =10 ¹⁴ ¡ 10 ¹⁵ Hz (or, ¸ = 400-700 nm)

The electric ¤eld satis¤es the same equation, as we can see by starting with Faradayzs law:

r £ ~ ³

r £ ~ ~ E ´

= ¡ ~r £ @ ~ B

@t r ~ ³

r ¢ ~ ~ E ´

¡ r ² E ~ = ¡ @

@t

³ r £ ~ ~ B ´

= ¡ @

@t Ã

" 0 ¹ 0

@ ~ E

@t

!

(11)

r ² E ~ = 1 c ²

@ ² E ~

@t ²

2.1 Monochromatic plane waves

The term "monochromatic" or "single-color" means that the wave has a single frequency !: This is an idealization, of course, but it greatly simpli¤es our math. The wave is plane if a surface of constant phase is a ¢at plane. Letzs choose to put the z¡axis along the vector ~k; so that the wave is travelling in the z¡direction. Then we can write the waves as:

B ~ = B ~ 0 exp (ikz ¡ i!t) E ~ = E ~ 0 exp (ikz ¡ i!t) Stu¢ng into Gaussz law, we have

r ¢ ~ ~ E = ikE 0;z exp (ikz ¡ i!t) = 0

This can be satis¤ed only if k = 0; which makes no sense, or E 0;z = 0: Thus E ~ 0 has no z¡component, and ~E is perpendicular to ~k: Since ~r ¢ ~B is also zero, we get the same result for ~ B 0 : Since both ¤eld vectors are perpendicular to the direction of wave propagation, EM waves in vacuum are transverse waves. Now letzs put the x¡axis along the directionof ~E 0 : From Faradayzs law, we get

r £ ~ ~ E = r £ [E ~ 0 x exp (ikz ¡ i!t)] ^

= y ^ @

@z [E 0 exp (ikz ¡ i!t)] = ikE ⁰ y exp (ikz ¡ i!t) ^

= ¡ @

@t B ~ 0 exp (ikx ¡ i!t) = i! ~ B 0 exp (ikx ¡ i!t) Thus we get

B ~ 0 = k

! E 0 y = ^ E 0

c ^ y (12)

Thus the magnetic ¤eld is perpendicular to the electric ¤eld, has magnitude E 0 =c; and the wave amplitude has the same phase as the ~ E¡wave amplitude.

Since ~ E is perpendicular to ~ B; by convention we choose the electric ¤eld vector to give the direction of polarization. This wave is polarized in the x¡direction:

We can write all of this in a coordinate-free form. If the wave propagates in direction ^k; the dot product ^k ¢ ~r gives the distance along the direction of ^k;

so we replace kz with k^k ¢ ~r = ~k ¢ ~r: Thus E ~ = E ~ 0 exp ³

i~k ¢ ~r ¡ i!t ´ B ~ = B ~ 0 exp ³

i~ k ¢ ~r ¡ i!t ´ where, from the divergence equations,

~k ¢ ~E ⁰ = ~k ¢ ~ B 0 = 0 and from the curl equation

r £ ~ ~ E 0 exp ³

i~k ¢ ~r ¡ i!t ´

= x ^ µ @

@y E z ¡ @

@z E y

¶ + ^ y

µ @

@z E x ¡ @

@x E z

¶ + ^ z

µ @

@x E y ¡ @

@y E x

¶

= [ ^ x (ik y E 0;z ¡ ¡k z E 0y ) + ^ y (ik z E 0;x ¡ ik x E 0;z ) +^ z (ik x E 0;y ¡ ik y E 0x )] exp ³

i~k ¢ ~r ¡ i!t ´

= i~k £ ~ E 0 exp ³

i~ k ¢ ~r ¡ i!t ´

= ¡ @

@t B ~ 0 exp ³

i~k ¢ ~r ¡ i!t ´

= i! ~ B 0 exp ³

i~k ¢ ~r ¡ i!t ´

Thus ~k £ ~E ⁰ = ! ~ B 0 (13)

2.2 Energy and momentum in EM waves

The energy density in the waves ¤elds is

u = 1

2 µ

" 0 E ² + B ²

¹ ₀

¶

= " 0

2

¡ E ² + c ² B ² ¢

But since we already showed that B = E=c; there is equal energy in the electric and magnetic components of the wave.

u = " 0 E ² = " 0 E ₀ ² cos ² ³

~k ¢ ~r ¡ !t + Á ´ The Poynting vector describes the ¢ux of energy:

S ~ = 1

¹ 0

E £ ~ ~ B = 1

¹ 0

E £ ~

à ~ k £ ~ E

!

!

= 1

!¹ ₀

h ~ kE ² ¡ ~ E ³

~k ¢ ~E ´i

But ~k ¢ ~E = 0; so

S = ~ E ²

c¹ ₀ k = ^ " 0 E ² c¹ ₀ " 0

k = cu^ ^ k

This shows explicitly that the energy density in the wave is carried along at the wave speed c:

If we average over several (or many) wave periods, we get the time averaged transmitted power per unit area of the wavefront:

< P > = 1 2c " 0 E ₀ ²

As with waves on a string, the power goes as the square of the wave amplitude, depends on the wave speed c; and also on the properties of the medium, here

" 0 : Physicists call this quantity the wave intensity. Astronomers beware w the term intensity in astronomy means something di¤erent.

The waves carry momentum as well as energy: From Notes 2 equation 6, the average momentum ¢ux density is

< ~ P E M > = < ~ S >

c

Now if the wave energy is absorbed at a ¢at screen, momentum is also aborbed, and thus there is a force exerted on the screen. The force, from Newtonzs second law, is

F = ~ d~ p

dt

and the pressure on the screen is the normal force per units area. If the wave strikes the screen at normal incidence, then

P = jF j

A =

¯ ¯

¯< ~S >

¯ ¯

¯

c = E ²

2c ² ¹ ₀ = 1 2 " 0 E ²

If the screen re¢ects all the energy and momentum, then the momentum im- parted to the screen is twice as big:

Before re¢ection after re¢ection Momentum in wave ~ p in p ~ ou t = ¡~p in

Momentum in screen 0 p ~ _screen

Total ~ p in p ~ _screen ¡ ~p in

Setting the total before equal to the total after, we get

~

p s creen = 2~ p in

Things get more interesting when the wave is incident at an angle µ: (See LB

"Digging Deeper" pg 1049.)

Momentum in direction ^k at angle µ to the surface normal ^n is carried by the wave. Only the normal component is absorbed or re¢ected:

p norm al = p cos µ

The momentum in area dA _? of the wave impacts area dA of the surface, where

dA _? = dA cos µ

Thus the momentum absorbed per unit area is

¢p norm

¢A = p cos µ

dA _? = cos µ = p

dA _? cos ² µ Thus the radiation pressure is

P = f

2 " 0 E ² cos ² µ

where the factor f = 1 for total absorption and = 2 for total re¢ection.

3 Electromagnetic waves in matter

When waves propagate in matter, there are non-zero sources ½ and ~j that must be included in Maxwellzs equations. These sources a¤ect the speed and other properties of the waves. Here we will discuss LIH materials that can be de- scribed by the ¤elds ~ D and ~ H; thus allowing us to "bury" the bound charges.

Then if the free charge density and free current density are zero, the equations are:

r ¢ ~ ~ D = 0 = " ~ r ¢ ~ E (14) r ¢ ~ ~ B = 0

r £ ~ ~ E = ¡ @ ~ B

@t r £ ~ B ~

¹ = r £ ~ ~ H = " @ ~ E

@t (15)

We ¤nd the wave equation using the same methods as before. Comparing these equations with equations (6) to (9), we see that the only di¤erence is that " 0

has been replaced by " and ¹ ₀ has been releced by ¹: Thus the wave equation for ~ E takes the form

r ² E = "¹ ~ @ ² E ~

@t ² = 1 v ²

@ ² E ~

@t ² (16)

where the wave phase speed is

v = 1 p "¹

In most ordinary materials like glass, ¹ ' ¹ 0 and " > " 0 ; so that v

c = r " 0 ¹ ₀

"¹ ' r " 0

" < 1

and light travels more slowly than in vacuum. The index of refraction of the material is

n = v c '

r "

" 0

> 1

The Poynting vector is now

S = ~ 1

¹ E £ ~ ~ B

The relation between the amplitudes of the ~ E and ~ B ¤elds may be found from Faradayzs law:

~ k £ ~ E 0 = ! ~ B 0 = k ~ B 0

v so that

B 0 = E 0

v (17)

The ¤elds are still perpendicular to each other and to ~k; and vary in phase.

4 Re¢ection and transmission of waves at a bound- ary

4.1 Normal incidence

When a wave reaches a boundary between two di¤erent, LIH media, the wave is partially transmitted and partially re¢ected. As with waves on a string, each wave has the same frequency. This ensures that the boundary conditions are satis¤ed at all times, not just at one time. The boundary conditions on the

¤elds are the same ones that we found in Physics 360:

normal ~ D is continuous (18)

tangential ~ E is continuous (19)

normal ~ B is continuous (20)

tangential ~ H is continuous (21) We begin by writing expressions for ~ E and ~ B in all three waves. For simplicity, let the boundary be the x; y¡plane, with medium 1 occupying the region z < 0: The wave travels in the +^z direction (normal incidence). First E : We put the x¡axis parallel to ~ ~ E ; so

E ~ i = E ~ i0 exp (ik 1 z ¡ i!t) = E i0 ^ x exp (ik 1 z ¡ i!t) E ~ t = E ~ i0 exp (ik 2 z ¡ i!t) = E t0 x exp (ik ^ 2 z ¡ i!t) E ~ r = E ~ r0 exp (¡ik ¹ z ¡ i!t) = E r0 x exp (¡ik ^ ¹ z ¡ i!t)

The minus sign in front of the k 1 z term in the re¢ected wave signi¤es that this wave is going in the ¡z direction The wave numbers are

k 1 = ! v 1

= ! p " 1 ¹ ₁ and

k 2 = ! v 2

= ! p " 2 ¹ ₂

Now in writing the magnetic ¤eld vectors, we must remember that

~k £ ~E ⁰ = ! ~ B 0

So

B ~ i = B ~ i0 exp (ik 1 z ¡ i!t) = B i0 ^ z £ ^x exp (ik 1 z ¡ i!t) = E i0

v 1

^

y exp (ik 1 z ¡ i!t) B ~ t = B ~ t0 exp (ik 2 z ¡ i!t) = E _t0

v 2 y exp (ik ^ 2 z ¡ i!t)

B ~ r = B ~ r0 exp (¡ik ¹ z ¡ i!t) = B r0 (¡^z) £ ^x exp (ik ¹ z ¡ i!t) = ¡ E r0

v 1 y exp (¡ik ^ ¹ z ¡ i!t) Now we apply the boundary conditions (18) to (21).

Continuity of normal ~ D :

0 = 0 Continuity of tangential ~ E :

E i0 + E r0 = E t0 (22)

Continuity of normal ~ B :

0 = 0 Continuity of tangential ~ H :

B _i0 + B _r0

¹ 1

= B _to

¹ 2

E i0 ¡ E r0

v 1 ¹ ₁ = E t0

v 2 ¹ ₂ (23)

Now we solve. First multiply eqn (23) by v 1 ¹ ₁ : E i0 ¡ E r0 = v 1 ¹ 1

E r0

v 2 ¹ ₂ = E t0 n 2

n 1

¹ 1

¹ ₂ and now add to equation (22):

2E i0 = E t 0

µ 1 + n 2

n 1

¹ 1

¹ ₂

¶

Thus

E t0 = 2E i0 n 1 ¹ ₂

n 1 ¹ ₂ + n 2 ¹ ₁ (24)

and then from (22),

E r0 = E i0

µ

2 n 1 ¹ ₂

n 1 ¹ 2 + n 2 ¹ 1 ¡ 1

¶

= E i0

µ n 1 ¹ ₂ ¡ n 2 ¹ ₁ n 1 ¹ ₂ + n 2 ¹ ₁

¶

(25)

This result is interesting, because it can be negative if n 2 ¹ ₁ > n 1 ¹ ₂ : As in the case of the waves on a string, it means that the ~ E vector has a phase change of

¼; or, equivalently, it changes direction. We may simplify these results if, as is usual, ¹ ₁ ' ¹ 2 ' ¹ 0 : Then

E t0 = E i0 2n 1

n 1 + n 2

E r0 = E i0

µ n 1 ¡ n 2

n 1 + n 2

¶

From these results we may calculate the re¢ected and transmitted intensities:

< S inc >= 1

2¹ ₁ E i0 B i0 = E _i0 ² 2¹ ₁ v 1

= 1 2

r " 1

¹ ₁ E _i0 ²

< S tr >= 1 2

r " 2

¹ 2

E _t0 ² = 1 2

r " 2

¹ 2

4E _i0 ²

µ n 1 ¹ ₂ n 1 ¹ 2 + n 2 ¹ 1

¶ 2

< S ref >= 1 2

r " 1

¹ 1

E _r0 ² = 1 2

r " 1

¹ 1

E _i0 ²

µ n 1 ¹ 2 ¡ n ² ¹ 1

n 1 ¹ 2 + n 2 ¹ 1

¶ 2

and

< S t r > + < S ref >= 1 2

r " 1

¹ 1

E _i0 ²

"r " 2 ¹ ₁

¹ 2 " 1 4

µ n 1 ¹ ₂ n 1 ¹ 2 + n 2 ¹ 1

¶ 2

+

µ n 1 ¹ ₂ ¡ n 2 ¹ ₁ n 1 ¹ 2 + n 2 ¹ 1

¶ 2 #

= 1

2 r " 1

¹ ₁ E _i0 ²

"

¹ ₁

¹ ₂ r " 2 ¹ ₂

" 1 ¹ ₁ 4

µ n 1 ¹ ₂ n 1 ¹ ₂ + n 2 ¹ ₁

¶ 2

+

µ n 1 ¹ ₂ ¡ n 2 ¹ ₁ n 1 ¹ ₂ + n 2 ¹ ₁

¶ 2 #

= 1

2 r " 1

¹ ₁ E _i0 ²

"

n 2 ¹ 1

n 1 ¹ ₂

4 (n 1 ¹ 2 ) ² (n 1 ¹ ₂ + n 2 ¹ ₁ ) ² +

µ n 1 ¹ 2 ¡ n ² ¹ 1

n 1 ¹ ₂ + n 2 ¹ ₁

¶ 2 #

= 1

2 r " 1

¹ ₁ E _i0 ²

"

4n 1 ¹ ₂ n 2 ¹ ₁

(n 1 ¹ ₂ + n 2 ¹ ₁ ) ² + (n 1 ¹ ₂ ) ² ¡ 2n 1 ¹ ₂ n 2 ¹ ₁ + (n 2 ¹ ₁ ) ² (n 1 ¹ ₂ + n 2 ¹ ₁ ) ²

#

= 1

2 r " 1

¹ ₁ E _i0 ²

"

(n 1 ¹ ₂ ) ² + 2n 1 ¹ ₂ n 2 ¹ ₁ + (n 2 ¹ ₁ ) ² (n 1 ¹ 2 + n 2 ¹ 1 ) ²

#

= 1 2

r " 1

¹ ₁ E _i0 ² = < S inc >

This result is required by energy conservation.

4.2 Non-normal incidence

When a wave is incident at an angle µ 6= 0; we have to tbe aware that the polarization of the wave enters into the boundary conditions. The plane that contains the normal to the surface and the incident ray is called the plane of incidence. (The ray is normal to the wavefront- see LB p547) If the electric

¤eld vector is perpendicular to the plane of incidence, as in the diagram below,

we say that the wave is polarized perpendicular to the plane of incidence. If

the electric ¤eld vector lies in the plane of incidence, then the wave is polarized

in the plane of incidence.

4.2.1 Polarization perpendicular to the plane of incidence

In this case the electric ¤eld vector is entirely tangential to the boundary. We choose coordinates so that the boundary is the x ¡ y plane. Then

E ~ inc = ¡E ⁰ ^ y exp ³

i~k ¢ ~r ¡ i!t ´ E ~ _ref = ¡E r0 y exp ^ ³

i~k _r ¢ ~r ¡ i!t ´

and E ~ tra ns = ¡E t0 y exp ^ ³

i~k t ¢ ~r ¡ i!t ´ As usual each wave has the same frequency, and

¯ ¯

¯~k ^r

¯ ¯

¯ =

¯ ¯

¯~k

¯ ¯

¯ = ! v 1

; ¯ ¯ ¯~k ^t

¯ ¯

¯ = ! v 2

Further, for points on the boundary,

~ k ¢ ~r = kr sin µ

Thus the boundary condition (19wcontinuity of tangential ~ E ) becomes E ~ inc + ~ E ref = E ~ tra ns

¡E ⁰ exp ³

i~k ¢ ~r ¡ i!t ´

¡ E r0 exp ³

i~ k r ¢ ~r ¡ i!t ´

= ¡E t 0 exp ³

i~k t ¢ ~r ¡ i!t ´ The factors exp (¡i!t) cancel, leaving

¡E 0 exp (ikr sin µ) ¡ E r0 exp (ik r r sin µ r ) = ¡E t0 exp (ik t r sin µ t ) (26) This relation must hold at all points on the boundary. Thus we must have

k sin µ = k r sin µ r = k t sin µ t

Since k = k r ; we must have sin µ = sin µ r ; and further, since the angles of incidence and re¢ection always lie in the range · µ · ¼=2; we must have

µ = µ r (27)

This is the law of re¢ection. Similarly, since k = !=v 1 and k _t = !=v 2 ; sin µ

v 1

= sin µ t

v 2

or, equivalently,

n 1 sin µ = n 2 sin µ _t (28)

This is Snellzs Law.

It is important to note that these relations arise from the fact that we have to satisfy a boundary condition at every point of the boundary. They do not depend on the details of the boundary condition at all. Thus they hold for plane waves of any kind, not just EM waves.

With equations (27) and (28) satis¤ed, boundary condition (26) becomes:

E 0 + E r0 = E t0 (29)

Now we look at the conditions on ~ B and ~ H: These cannot give 2 independent relations, because we only need two equations to ¤nd the two unknowns E r0 and E t0 : Letzs look at them:

B ~ inc = ~ k £ ~ E inc = ¡~k £ ~yE ⁰ exp ³

i~k ¢ ~r ¡ i!t ´

= ¡k (cos µ ^z + sin µ ^x) £ ~yE 0 exp ³

i~k ¢ ~r ¡ i!t ´

= k (cos µ ^ x ¡ sin µ ^z) E ⁰ exp ³

i~k ¢ ~r ¡ i!t ´

B ~ ref = ~ k r £ ~ E ref = ¡~k r £ ^yE r0 exp ³

i~k r ¢ ~r ¡ i!t ´

= ¡k (¡ cos µ ^z + sin µ ^x) £ ~yE r0 exp ³

i~k ¢ ~r ¡ i!t ´

= k (¡ cos µ ^x ¡ sin µ ^z) E r0 exp ³

i~ k ¢ ~r ¡ i!t ´ and

B ~ trans = ~k t £ ~ E trans = ¡E t0 ~k t £ ^yE t0 exp ³

i~k t ¢ ~r ¡ i!t ´

= k t (cos µ t x ¡ sin µ ^ t z) E ^ 0 exp ³

i~ k ¢ ~r ¡ i!t ´ Normal ~ B :

¡k sin µE 0 ¡ k sin µE r0 = ¡k t sin µ _t E _t0

Using Snellzs law, we get back equation (29). We do get an independent equa- tion from the boundary condition for tangential ~ H :

k

¹ ₁ cos µ (E 0 ¡ E r0 ) = k t

¹ ₂ cos µ t E t0

(E 0 ¡ E r0 ) = v 1 ¹ 1

v 2 ¹ ₂ E t 0

p 1 ¡ sin ² µ t

cos µ

(E 0 ¡ E r0 ) = n 2 ¹ ₁ n 1 ¹ ₂ E t0

q

1 ¡ n ² 1 sin ² µ=n ² ₂

cos µ (30)

We will simplify by taking ¹ ₁ = ¹ ₂ = ¹ ₀ : Then combining equations (29) and (30), we have

2E 0 = 0

@1 + q

n ² ₂ ¡ n ² 1 sin ² µ n 1 cos µ

1 A E _t0

E t0 = 2E 0 n 1 cos µ n 1 cos µ +

q

n ² ₂ ¡ n ² 1 sin ² µ

(31)

and then

E r0 = E 0

n 1 cos µ ¡ q

n ² ₂ ¡ n ² 1 sin ² µ n 1 cos µ +

q

n ² ₂ ¡ n ² 1 sin ² µ

(32)

4.2.2 Polarization parallel to the plane of incidence

Now the boundary condition on tangential ~ H (again with all ¹s equal) is B 0 + B r0 = B t0

n 1 (E ₀ + E _r0 ) = n 2 E _t0 (33) while for tangential ~ E we get

E 0 cos µ ¡ E r0 cos µ = E t0 cos µ t

E 0 ¡ E r0 = E t0

p 1 ¡ sin ² µ t

cos µ (34)

Again the third non-trivial condition (for normal D) does not give an indepen- dent relation. Combining these, we get

2E 0 = E t0

0

@ q

1 ¡ n ² 1 sin ² µ=n ² ₂ cos µ + n 2

n 1

1 A

E t0 = 2E 0 n 1 cos µ n 2 cos µ + n 1

q

1 ¡ n ² 1 sin ² µ=n ² ₂

= 2E 0 n 1 n 2 cos µ n ² ₂ cos µ + n 1

q

n ² ₂ ¡ n ² 1 sin ² µ

(35)

and then

E r0 = E 0

n ² ₂ cos µ ¡ n 1

q

n ² ₂ ¡ n ² 1 sin ² µ n ² ₂ cos µ + n 1

q

n ² ₂ ¡ n ² 1 sin ² µ

(36)

In both polarizations we ¤nd that the re¢ected and transmitted amplitudes depend on the angle of incidence, as well as the properties of the two materials.

In fact, the refelected amplitude (36) may actually be zero at an angle given by n ² ₂ cos µ = n 1

q

n ² ₂ ¡ n ² 1 sin ² µ square both sides:

n ⁴ ₂ cos ² µ = n ² ₁ ¡

n ² ₂ ¡ n ² 1 sin ² µ ¢ Now use a trick: write 1 = cos ² µ + sin ² µ

n ⁴ 2

n ² ₁ cos ² µ = n ² 2

¡ cos ² µ + sin ² µ ¢

¡ n ² 1 sin ² µ n ² ₂

n ² ₁

¡ n ² ₂ ¡ n ² 1

¢ cos ² µ = ¡ n ² ₂ ¡ n ² 1

¢ sin ² µ

Then, provided n 2 6= n ¹ ;

tan µ B = n 2

n 1

(37)

This angle is called Brewsterzs angle. At this angle of incidence, in this po- larization, all of the wave energy is transmitted and none is re¢ected. (Wezll discuss why this happens later- see also LB §33.3.2). Note, however, that nothing special happens at this angle for waves polarized perpendicular to the plane of incidence. In fact, E r0 in equation (32) is never zero for any µ: Check this for yourself. What this means is that if we have unpolarized incident light, equal amplitudes in the two polarizations, at Brewsterzs angle only one of the two polarizations is re¢ected. The re¢ected light is completely polarized per- pendicular to the plane of incidence. This phenomenon is called "polarization by re¢ection".

Under what conditions do we get a phase change of the re¢ected wave? That is, when is the sign of E r0 opposite that of E 0 ? Starting from equation (32), we want to ¤nd when

n 1 cos µ <

q

n ² ₂ ¡ n ² 1 sin ² µ n ² ₁ cos ² µ < n ² ₂ ¡ n ² 1 sin ² µ This is true for

n 1 < n 2

independent of µ; as we also found for normal incidence.

For the other polarization, Start with equation (36) n ² ₂ cos µ ¡ n 1

q

n ² ₂ ¡ n ² 1 sin ² µ < 0 or, equivalently,

n ⁴ ₂ cos ² µ < n ² ₁ ¡

n ² ₂ ¡ n ² 1 sin ² µ ¢

Using the same trick as in ¤nding Brewsterzs angle, we ¤nd this reduces to n ² 2

n ² ₁

¡ n ² 2 ¡ n ² 1

¢ cos ² µ < ¡ n ² 2 ¡ n ² 1

¢ sin ² µ

or

µ < µ B if n 2 > n 1

and

µ > µ B if n 1 < n 2

4.2.3 Re¢ection and transmission coe¢cients The incident power striking unit area of the interface is

S ~ in ¢ ^z = ³

E ~ in £ ~ H in

´

¢ ^z

Now we time average to get I in = 1

2 E 0 ²

"

^ e £ Ã ~k

! £ ^e

!#

¢ ^z

= 1

2v 1 ¹ ₁ E ² ₀ ^ k ¢ ^z

= 1

2 " 1 v 1 E ² ₀ cos µ where

E ~ 0 = E 0 e ^

and ^e is the polarization vector. We can compute the re¢ected and transmitted intensities similarly. Then the re¢ection coe¢cient is de¤ned as

R = I R

I in = E _{r 0} ² E ₀ ² and the transmission coe¢cient is

T = I _T I in = v 1

v 2

E _{t 0} ² E ₀ ²

cos µ _t cos µ = n 2

n 1

E _t0 ² E ² ₀

cos µ _t cos µ For polarization perpendicular to the plane of incidence, we get

R = 0

@ n 1 cos µ ¡ q

n ² ₂ ¡ n ² 1 sin ² µ n 1 cos µ +

q

n ² ₂ ¡ n ² 1 sin ² µ 1 A

2

and

T =

0

@ 2n 1 cos µ n 1 cos µ +

q

n ² ₂ ¡ n ² 1 sin ² µ 1 A

2

n 2 cos µ t

n 1 cos µ

= 4n 1 n 2 cos µ cos µ t

µ

n 1 cos µ + q

n ² 2 ¡ n ² 1 sin ² µ

¶ 2

and

R + T = µ

n 1 cos µ ¡ q

n ² ₂ ¡ n ² 1 sin ² µ

¶ 2

+ 4n 1 n 2 cos µ cos µ t

³ n 1 cos µ + p

n ² ₂ ¡ n ² 1 sin µ ´ ²

= n ² ₁ cos ² µ + n ² ₂ ¡ n ² 1 sin ² µ ¡ 2n 1 cos µ q

n ² ₂ ¡ n ² 1 sin ² µ + 4n 1 cos µ q

n ² ₂ ¡ n ² 1 sin ² µ µ

n 1 cos µ + q

n ² 2 ¡ n ² 1 sin ² µ

¶ 2

= 1

Verify that this result also holds, as it must, for the other polarization.

5 Waves in conductors

An electric ¤eld in a conductor drives a current.

~j = ¾ ~ E = E ~

½

and thus some of the ¤eld energy is converted to kinetic energy of electrons in the conductor. This leads to absorption of the wave energy. In order to investigate this, we must include the current in Maxwellzs equations. The only equation to change is the Ampere-Maxwell law.

r £ ~ ~ B = ¹~j + ¹" @ ~ E

@t = ¹¾ ~ E + ¹" @ ~ E

@t

We may still take the free charge density ½ _f to be zero. For suppose ½ _f is not zero. Then from charge conservation,

@½ _f

@t = ¡ ~ r ¢ ~j = ¡~r ¢ ¾ ~ E

= ¡¾ ³½ _f

"

´

This di¤erential equation has the solution

½ _f = ½ ₀ exp ³

¡ ¾

" t ´ so the charge density dies away with a time scale

¿ = "

¾

which is very small for good conductors. As we discussed before, even if ¾ ! 1;

the timescale does not go to zero, as the dominant factor is then the timescale for the ¤elds to change, of order (scale of system)/(speed of light).

Now we may rederive the wave equation with the extra term in Amperezs law. Starting from equation (11), we have

r ~ ³ r ¢ ~ ~ E ´

¡ r ² E = ¡ ~ @

@t

³ r £ ~ ~ B ´

= ¡ @

@t Ã

¹¾ ~ E + ¹" @ ~ E

@t

!

which becomes

r ² E = ~ @

@t ¹¾ ~ E + ¹" @ ² E ~

@t ²

The wave equation now has an extra term. We may still ¤nd a plane wave soution of the form:

E ~ = E ~ 0 exp ³

i~k ¢ ~r ¡ i!t ´ B ~ = B ~ 0 exp ³

i~ k ¢ ~r ¡ i!t ´

Stu¢ng these expressions into the wave equation and Maxwellzs equations, we have

¡k ² = ¡i!¹¾ ¡ ¹"! ² (38)

~k ¢ ~E 0 = 0

~ k ¢ ~ B 0 = 0

~k £ ~E 0 = ! ~ B 0 (39)

Thus the wave has the same transverse nature as before. The only thing that has changed is the relation of k to !: Equation (38) requires that k be complex if ! is real. So let k = · + i°: Then

k ² = · ² ¡ ° ² + 2i·° = i!¹¾ + ¹"! ²

So we have two equations obtained by setting the real part on the left equal to the real part on the right, and also by setting the two imaginary parts equal.

· ² ¡ ° ² = ¹"! ² 2·° = !¹¾ From the second, we get

° = !¹¾

2· (40)

and then from the ¤rst

· ² ¡ ³!¹¾ 2·

´ 2

¡ ¹"! ² = 0 This is a quadratic equation for · ² with solution

· ² = ¹"! ² § q

(¹"! ² ) ² + (!¹¾) ² 2

= ¹"! ² 1 § q

1 + (¾=!") ² 2

We must get back our previous result k = !p¹" as ¾ ! 0; so we need the plus sign:

· = ! p ¹"

v u u t 1 +

q

1 + (¾=!") ² 2

and then from (40), we have

° = ¾ 2

r ¹

"

v u

u t 2

1 + q

1 + (¾=!") ²

The imaginary part of k shows that the wave is damped:

exp i~k ¢ ~r = exp h

i^ k ¢ ~r (· + i°) i

= exp ³ i·^ k ¢ ~r ´

exp ³

¡°^k ¢ ~r ´ and taking the real part, we have

cos ³

·^ k ¢ ~r ´ exp ³

¡°^k ¢ ~r ´

The wave attenuates exponentially as it propagates. The distance travelled before the wave amplitude drops to 1=e of its original value is called the skin depth

± = 1

°

Faradayzs law (equation (39)) shows that ~ E and ~ B no longer oscillate in phase.

We may write ~k in exponential form as

~k = ^k (· + i°) = ^k p

· ² + ° ² exp iÁ; tan Á = °

· where

k = p

· ² + ° ² = v u u u

t¹"! ² 1 + q

1 + (¾=!") ²

2 + ¾ ²

2

¹

"

1 1 +

q

1 + (¾=!") ²

= ! p ¹"

v u u u u t

2 + (¾=!") ² + 2 q

1 + (¾=!") ² + ¾ ² =! ² " ² 2

µ 1 +

q

1 + (¾=!") ²

¶

= ! p ¹" ³

1 + (¾=!") ² ´ 1=4

and

tan Á = ¾

"!

µ 1 +

q

1 + (¾=!") ²

¶

Then equation (39) becomes

^ k £ ^eE ⁰ ¡ ke ^iÁ ¢

= B ~ 0

B 0 = ! p ¹" ³

1 + (¾=!") ² ´ 1=4

E 0 e ^iÁ

A material is a good conductor as far as wave propagation is concerned if

¾

!" À 1 in which case we ¤nd

B 0 ' p ¾!¹E 0 e ^iÁ

so that

B 0 À E 0

v

and the phase shift ± is almost ¼=4: (tan ± ! 1): The ¤elds in the conductor are primarily magnetic in character. In this limit we also have

· ' ! p ¹"

r 1 + ¾=!"

2 '

r !¾¹ 2 ' °

and the wave damps within one wavelength. On the other hand, if ¾=!" ¿ 1;

we have a poor conductor, the phase shift is small, and the amplitudes of the

¤elds are not much di¤erent from those in a non-conducting medium.

5.1 Re¢ection at a conducting surface

There may be a free surface charge density on the surface of a conductor. In fact, there usually is. But there cannot be a free surface current density as this would require an in¤nite electric ¤eld (~j = ¾ ~ E ; and if ~j = ~ K± (z ¡ z surfa ce ) then E = ~ ^{K ~} _¾ ± (z ¡ z surf ace ) : Such an electric ¤eld cannot exist. Thus we choose to use the boundary conditions on tangential ~ E; normal ~ B and tangential ~ H: Letzs consider normal incidence for simplicity. Then we have

E 0 + E r = E t

and

H 0 ¡ H r = H t

! v 1

(E 0 ¡ E r ) = r ¾!

¹ 2

E _t e ^iÁ (E 0 ¡ E r ) = v 1

r ¾

!¹ ₂ E t e ^iÁ Thus we have

2E 0 = E t

µ 1 + v 1

r ¾

!¹ ₂ e ^iÁ

¶

E t = 2E 0

µ 1 + v 1

q _¾

!¹

₂

e ^iÁ

¶

and then

E r = E 0

0

@ 1 ¡ v 1

q _¾

!¹

₂

e ^iÁ 1 + v 1

q _¾

!¹

₂

e ^iÁ 1

A (41)

As ¾ ! 1; E t ! 0 and E r ! ¡1:

5.2 Note on computing power using complex numbers

When our amplitudes are complex, as in (41), we must be careful when evalu- ating physical quantities that involve the square of these numbers. Letzs look at the power.

S = ~ 1

¹ ₀ E £ ~ ~ B Since

Re (z 1 z 2 ) 6= Re (z ¹ ) Re (z 2 ) we must take the real part before multiplying. Thus

S ~ phys = 1

¹ ₀ E ~ 0 cos ³

~k ¢ ~r ¡ !t + Á E

´

£ ~ B 0 cos ³

~k ¢ ~r ¡ !t + Á B

´

= 1

2¹ ₀ E ~ 0 £ ~ B 0

h cos ³

2~k ¢ ~r ¡ 2!t + Á E + Á _B ´

+ cos (Á _E ¡ Á B ) i Now we time average, to get

< ~ S phys > = 1 2¹ 0

E ~ 0 £ ~ B 0 cos (Á E ¡ Á B ) We can get the same result more easily as follows:

1

2¹ ₀ E £ ~ ~ B ^¤ = 1

2¹ ₀ E ~ 0 £ ~ B 0 exp (iÁ _E ¡ iÁ B ) and so

< ~ S phys > = Re µ 1

2¹ ₀ E £ ~ ~ B ^¤

¶

(42)

6 Dispersive media

In general the properties of a medium, such as " and ¹; are frequency dependent.

This means that waves of di¤erent frequencies travel at di¤erent speeds. A plane wave carries no information. To transmit a signal we have to vary either the amplitude of the frequency of the wave- the resulting signal is called AM or FM ( as on your radio dial). The wave crests travel at the phase speed v Á = !=k but the envelope that carries the information travels at the group speed v g = d!=dk: It is v g; not v Á; that must be less than c: In order to get the signal, many di¤erent frequencies make up the wave packet, and because they do not all travel at the same speed, over time the signal becomes less well de¤nedw the pulse is dispersed. Those of you who plan to take Phys 704 in the spring will learn more about these e¤ects then. For now, letzs see if we can understand how the frequency dependence arises.

We begin with a simple classical model of an electron in an atom. The electron behaves as an oscillator with a restoring force

F ~ res = ¡m! ² 0 ~s

where ! 0 is the natural oscillation frequency and ~s is the electron displacement.

There is also damping due to radiation reaction, which we may write as F ~ _{dam p} = ¡m° d~s

dt

Now when an incident EM wave reaches the electron, it will experience a force F ~ EM = ¡e ³

E + ~ ~ v £ ~ B ´

We have already seen that, for waves in a medium with phase speed v Á , B 0 = E 0

v Á

so the magnetic force is smaller than the electric force by a factor v=v Á which is typically ¿ 1 since v Á » c: So we will ignore the magnetic force. Now we may write the equation of motion for the electron:

F ~ to t = m~a

¡e ~ E 0 cos !t ¡ m! ² 0 ~ s ¡ m° d~s

dt = m d ² ~ s dt ²

The electron oscillates with ~s parallel to ~ E 0 : We may put the x¡axis along this direction, to get

x + ° _ ¢¢ x + ! ² ₀ x = ¡eE 0 cos !t = Re ³

¡ e

m E 0 exp (¡i!t) ´ The solution will be of the form

x = Re ¡

x 0 e ^¡i!t ¢ Stu¤ this in and check:

¡ ¡! ² ¡ i°! + ! ² 0

¢ x 0 = ¡ e m E 0

so

x 0 = eE 0 =m

! ² + i°! ¡ ! ² 0

(43) and the electron contributes a dipole moment about its original equilibrium position of

~ p = ¡e~s = ¡ e ² m

E ~ 0

! ² + i°! ¡ ! ² 0

Summing up over all the particles in the material, where there are n molecules per unit volume, each of which has f j electrons with frequency ! j and damping

° j ; we get a polarization

P = n~ ~ p = ¡n e ² m

X

j

f j E ~ 0

! ² + i° j ! ¡ ! ² j

So we now have a model for the susceptibilty of the material P = " ~ 0 Â _e E ~

Thus

 _e = ¡n e ² m

X

j

f j

! ² + i° _j ! ¡ ! ² j

Because  is a complex number, the polarization is out of phase with the driving

¤eld ~ E : The dielectric constant is than

" = " 0

0

@1 + ne ²

" 0 m X

j

f j

! ² _j ¡ i° j ! ¡ ! ² 1

A (44)

and consequently the wave equation (16) has plane wave solutions of the form E = ~ ~ E 0 exp ³

i~ k ¢ ~r ¡ i!t ´ where

¡k ² E ~ 0 = "¹! ² E ~ 0 = " 0 ¹! ² E ~ 0

0

@1 + ne ²

" 0 m X

j

f j ¡

! ² _j + i° _j ! ¡ ! ² ¢

¡ ! ² j ¡ ! ² ¢ 2

+ ¡

° j ! ¢ 2

1 A

and because " is complex, k must be complex too.

k = k _r + ik _i Putting the z¡axis along ^k; we have

E = ~ ~ E 0 exp (¡k i z) exp (ik r z ¡ i!t)

The wave is damped. The wave intensity is proportional to E ² ; and so it de- creases as exp (¡2k i z) = exp (¡®z) where ® is the absorption coe¢cient.

Letzs look at the special case where the second term in equation (44) is small, for example, in gases. Then we can approximate the square root using the binomal series: p

1 + x ' 1 + x

2 if x ¿ 1 So

k = p "¹! = ! c

0

@1 + ne ² 2" 0 m

X

j

f j

¡ ! ² _j + i° _j ! ¡ ! ² ¢

¡ ! ² _j ¡ ! ² ¢ 2

+ ¡

° _j ! ¢ 2

1 A so

® = ! ² c

ne ²

" 0 m X

j

f j ° _j

¡ ! ² _j ¡ ! ² ¢ ² + ¡

° _j ! ¢ ² (45)

and the refective index is

n = ck r

! = 1 + ne ² 2" 0 m

X

j

f j

¡ ! ² _j ¡ ! ² ¢

¡ ! ² _j ¡ ! ² ¢ 2

+ ¡

° _j ! ¢ 2 (46) The denominator of the fraction in these equations becomes very small when

! ' ! j because ° _j ¿ ! j ; and thus the fraction becomes large. There are resonances at the fundamental frequencies of the system. These correspond to atomic and molecular line transition frequencies. To display the behavior more clearly, look at a system with only one resonance, and let

ne ²

" 0 m = ! ² _p ; !=! j = x; ° j =! j = ´ Then

® = ! ²

! ² _j c

! ² _p

! ² _j

f j ° _j

¡ 1 ¡ ! ² =! ² _j ¢ 2

+ ¡

° _j ! ¢ 2

=! ⁴ _j = ! ² _p c! j

´ f j

(1 ¡ x ² ) ² + (´x) ² and

n = 1 + ! ² _p 2! ² _j

f j

¡ 1 ¡ x ² ¢ (1 ¡ x ² ) ² + (´x) ²

The plot shows c®=! j (solid line) and n ¡ 1 (dashed line) with ´ = 0:1 (much larger than actual values, for clarity in the plot) and f j ! ² p =! ² j = 1

0.5 1 1.5 x 2 2.5 3

The absorption coe¢cient peaks strongly around x = 1; and is very small away

from that region. The width of the peak is determined by ´; which is usually

much less than the value of 1/10 used in these plots. The index of refraction n is greater than 1 for x < 1 but becomes <1 for x > 1: Thus the phase speed of the waves is actually greater than c for frequencies slightly greater than resonance.

The group speed, however, remains less than c:The refractive index n drops steeply at the resonancew this e¤ect is called anomalous dispersion. For real materials, we have to sum the contributions from all the resonances, but graphs like this describe the behavior near each one.

For frequencies well away from resonance, the e¤ect of damping is negligible and we have the simpler relation

n = 1 + ! ² p

2! ² _j f j

(1 ¡ x ² )

Further, if we are well below resonance, x = !=! j ¿ 1 (in transparent materials like glass ! j is usually in the UV), we have

n = 1 + ! ² p

2! ² _j f j

¡ 1 + x ² ¢

and we see that n increases with frequency. (See LB §16.5.5).

7 Wave guides

Wezll close this discussion of waves by looking at happens when a wave is con-

¤ned within a set of boundaries. The wave ¤elds have to satisfy boundary conditions on the walls as well as the wave equation. Consider a wave guide made of of a long, straight, perfectly conducting tube with a constant cross- section, whose shape, for the moment, is arbitrary. The electric ¤eld has to be zero inside the perfectly conducting walls, and since wave ¤elds are time- varying, that means that ~ B must also be zero. Otherwise, the non-zero @ ~ B=@t would create an ~ E through Faradayzs law, and we know there can be no ~ E : Then making use of the boundary conditions, we must have

B ¢ ^n = 0 ~ on the guide walls, and

n £ ~ ^ E = 0

as well. We do not use the boundary condition on ~ D ¢ ^n because there could be a non-zero surface charge density on the wall.

We are interested in waves propagating along the guide, in the z¡direction.

The waves can start o¤ propagating at an angle to the z¡axis, in which case

they will re¢ect at the wall, and propagate down the guide by bouncing back

and forth. The superposition of these waves forms the total disturbance in

the guide. As the waves superpose, they will interfere, and only the waves

that interfere constructively contribute. It is easier to solve a boundary-value

problem than to try to sum the appropriate set of re¢ecting waves. (But see LB Digging Deeper, page 1060 for a simple example of how it goes.) We should note that each of the re¢ecting waves has the form we have come to expect ( ~ E perpendicular to ~ B perpendicular to ~ k), but when we superpose them to get a disturbance travelling parallel to the guide (z) axis, the resulting ¤elds can have a z¡component. (See LB Figure 33.28, for example.) So we have to allow for this possibility. We start o¤ with the following assumed form for ~ E and ~ B (as usual, real part is implied).

E ~ = E ~ 0 exp (ikz ¡ i!t) = h

E ~ _? (x; y) + E z (x; y) ^ z i

exp (ikz ¡ i!t) B ~ = B ~ 0 exp (ikz ¡ i!t) = h

B ~ _? (x; y) + B z (x; y) ^ z i

exp (ikz ¡ i!t) The amplitudes must depend on x; y so as to satisfy the boundary conditions.

Now we stu¤ these into Maxwellzs equations. For simplicity let the guide interior be vacuum.

r ¢ ~ ~ E = ³

r ~ ? ¢ ~ E _? + ikE z

´

exp (ikz ¡ i!t) = 0 r ¢ ~ ~ B = ³

r ~ ? ¢ ~ B _? + ikB z

´ exp (ikz ¡ i!t) = 0

From these ¤rst two equations we can already see that the z¡components of the ¤elds act as sources for the perpendicular components:

r ~ ? ¢ ~ E _? = @E x

@x + @E y

@y = ¡ikE z (47)

r ~ ? ¢ ~ B _? = @B x

@x + @B y

@y = ¡ikB ^z (48)

From the last two equations, we have

³ r £ ~ ~ E ´

x = @E z

@y ¡ @E y

@z = i!B x = @E z

@y ¡ ikE ^y (49)

³ r £ ~ ~ E ´

y = ikE x ¡ @E z

@x = i!B y (50)

³ r £ ~ ~ E ´

z = @E y

@x ¡ @E x

@y = i!B z (51)

and similarly

³ r £ ~ ~ B ´

x = @B z

@y ¡ ikB y = ¡i !

c ² E x (52)

³ r £ ~ ~ B ´

y = ikB x ¡ @B z

@x = ¡i !

c ² E y (53)

³ r £ ~ ~ B ´

z = @B y

@x ¡ @B x

@y = ¡i !

c ² E z (54)

We can solve these equations to get the x¡ and y¡components in terms of the z¡components. From (49)

E y = 1 ik

µ @E z

@y ¡ i!B ^x

¶

and then from (53)

B x = 1 ik

µ @B z

@x ¡ i!

c ² E y

¶

Combining these, we have E y = 1

ik

· @E z

@y ¡ i! 1 ik

µ @B z

@x ¡ i!

c ² E y

¶¸

= 1

ik

@E z

@y + i! 1 k ²

@B z

@x + ! ² k ² c ² E y

So

E y

µ 1 ¡ ! ²

c ² k ²

¶

= i µ !

k ²

@B z

@x ¡ 1 k

@E z

@y

¶

E y = i

! ² =c ² ¡ k ² µ

k @E z

@y ¡ ! @B z

@x

¶

(55) Similarly, we get for the other components:

E x = i

! ² =c ² ¡ k ² µ

k @E z

@x + ! @B z

@y

¶

(56)

B x = i

! ² =c ² ¡ k ² µ

k @B z

@x ¡ ! c ²

@E z

@y

¶

(57)

B y = i

! ² =c ² ¡ k ² µ

k @B z

@y + ! c ²

@E z

@x

¶

(58) Now if we can just ¤nd the z¡components, we are done! So go back to equations (47) and (48) and insert the results for the x¡and y¡components.

We simplify the notation by writing

° ² ´ ! ² c ² ¡ k ²

@

@x

· i

° ² µ

k @E z

@x + ! @B z

@y

¶¸

+ @

@y

· i

° ² µ

k @E z

@y ¡ ! @B z

@x

¶¸

= ¡ikE z

@ ² E z

@x ² + @ ² E z

@y ² = ¡° ² E z (59) and we get the same equation for B z :

@ ² B z

@x ² + @ ² B z

@y ² = ¡° ² B z (60)

These are decoupled equations: the solution for E z is independent of B z and vice versa. That means we can solve for E z assuming B z = 0 w these are the TM or transverse magnetic modes, and then solve for B z assuming E z = 0 w these are the transverse electric modes, and then get a general solution as a linear combination

E = ® ~ ~ E T E + ¯ ~ E TM

If both E z and B z are zero we have TEM, or transverse electro-magnetic modes. For these modes, equation (51) becomes

µ @E y

@x ¡ @E x

@y

¶

^

z = 0 = ~ r £ ~ E ) ~ E = ¡~rª and equation (47) becomes

@E x

@x + @E y

@y = ~ r ¢ ~ E = 0 = ¡r ² ª

We know that the maxima and minima of the function ª, a solution of Laplacezs equation, must occur on the boundary of the region. But ~ E £ ^n is zero on the boundary, which means that ª is constant on the boundary, and thus it is constant throughout the guide. But if ª is constant then ~ E = 0: The only exception to this is if the guide has two separate pieces to its boundaryw as in a coaxial cable.

Now it is time to tackle the boundary conditions. Tangential ~ E must be zero, and one of the tangential components is E z ; so for the TM mode we must have

E z = 0 on the walls. (61)

The boundary condition for B z is a bit more complicated, so wezll wait for a speci¤c example.

7.1 Waves in a rectangular guide

Let the guide have a rectangular cross-section measuring a £ b:

7.1.1 TM modes

The E z component satis¤es the di¤erential equation (59) together with the boundary condition (61). We may solve using separation of variables:

E z = X (x) Y (y) Then the di¤erential equation is

X ⁰⁰ Y + XY ⁰⁰ = ¡° ² XY

or X ⁰⁰

X + Y ⁰⁰

Y = ¡° ²