Wave Phenomena Physics 15c Lecture 12 Dispersion (H&L - - PowerPoint PPT Presentation

Wave Phenomena

Physics 15c

Lecture 12 Dispersion

(H&L Sections 2.6)

What We Did Last Time

Defined Fourier integral

f(t) and F(ω) represent a function in time/frequency domains

Analyzed pulses and wave packets

Time resolution ∆t and bandwidth ∆ω related by

Proved for arbitrary waveform

Rate of information transmission ∝ bandwidth Dirac’s δ(t) a limiting case of infinitely fast pulse Connection with Heisenberg’s Uncertainty Principle in QM

( ) ( )

i t

f t F e d

ω

ω ω

∞ − −∞

= ∫ 1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

1 2 t ω ∆ ∆ >

Goals For Today

Discuss dispersive waves

When velocity is not constant for different ω Waveform changes as it travels Dispersion relation: dependence of k on ω

Define group velocity

How fast can you send signals if the wave velocity is not

constant?

Mass-Spring Transmission Line

ξn−1 ξn ξn+1

In Lecture #5, we had

We ignored the gravity by making the strings very long

2 1 1 2

( ) ( )

n s n n s n n

d m k k dt ξ ξ ξ ξ ξ

− +

= − − − −

n

mg L L ξ → ∞ −  →

What if we didn’t make this approximation?

Wave Equation

Equations of motion is now

Usual Taylor-expansion trick Divide by (∆x)

Wave equation:

2 1 1 2

( ) ( )

n s n n s n n n

d m k k dt mg L ξ ξ ξ ξ ξ ξ

− +

− = − − − −

2 2 2 2 2

( , ) ( , ( , ) ) ( )

s

x t x t m k x t x m L x g t ξ ξ ξ ∂ ∂ = ∆ ∂ ∂ −

2 2 2 2

( , ) ( , ) ( , )

l l

x t x t K g t x t x L ρ ξ ξ ξ ρ ∂ ∂ = ∂ ∂ −

2 2 2 2 2 2

( , ) ( , ) ( , )

w

x t x t c x t t x ξ ξ ω ξ ∂ ∂ = − ∂ ∂

w l

K c ρ = g L ω =

Natural frequency of pendulum

2 2 2 2 2 2

( , ) ( , ) ( , )

w

x t x t c x t t x ξ ξ ω ξ ∂ ∂ = − ∂ ∂

Solution

Assume

As before, we can write the solution as

( , ) ( )

i t

x t a x e ω ξ =

2 2 2 2 2

( ) ( ) ( )

i t i t i t w

d a x a x e c e a x e dx

ω ω ω

ω ω − = −

Wave eqn.

2 2 2 2 2

( ) ( )

w

d a x a x dx c ω ω − = −

SHO-like if

2 2

ω ω − >

( )

( , )

i kx t

x t Ae

ω

ξ

±

=

2 2 w

k c ω ω − = but with

This is the difference

Dispersion Relation

Normal-mode solutions are still

What changed is the relationship between k and ω

A.k.a. dispersion relation NB: there are different types of dispersive waves

We are looking at just one example here

Dispersion relation determines how the waves

propagate in time and space

( )

( , )

i kx t

x t e

ω

ξ

±

= ( )

w

k c ω ω =

2 2

( )

w

k c ω ω ω − =

Non-dispersive waves Dispersive waves We’ll study how…

Phase Velocity

To calculate the propagation velocity of

We follow the point where the phase kx ± ωt is constant

Phase velocity is the velocity of pure sine waves

Easily calculated from the dispersion relation

( )

( , )

i kx t

x t e

ω

ξ ξ

±

= kx t C ω ± = C t x k ω = m dx dt k ω = m

Phase velocity cp

( ) const.

p w

c c ω = = ( )

w

k c ω ω =

2 2

( )

w

k c ω ω ω − =

Non- dispersive Dispersive

2 2

( )

p w

c c ω ω ω ω = −

No longer constant!

Dispersing Pulses

Imagine a pulse being sent over a distance

On non-dispersive medium, the pulse shape is unchanged

That was because all normal modes had the same cp

On dispersive medium, the pulse shape must change

The pulse gets dispersed Hence the name: dispersion

Dispersion makes poor media for communication

Dispersion Relation

Dispersive waves have no solution for ω < ω0

It has a low frequency cut-off at ω0

Phase velocity goes to infinity at cut-off

Wait! Isn’t it unphysical? What happened to Relativity?

k ω

w

c k ω =

2 2 w

c k ω ω = − ω

p

c ω ω

w

c

2 2 w p

c c ω ω ω = −

Finite-Length Signal

Phase velocity cp is the speed of pure sine waves

But pure sine waves don’t carry information Relativity forbids superluminal transfer of information

Let’s think about a finite-length pulse

Problem: this medium can’t carry waves with We need to make a pulse that does not contain frequencies

below the cut-off ( ) f t ( ) F ω t T ω ω ω <

Solution: wave packet

General Wave Packet

Consider a wave packet

Modulate carrier wave

with a pulse f(t)

Fourier integral of such wave packet is

G(ω) has the same shape as F(ω),

but centered around ωc

Now we examine how g(t) travels

in space ( ) f t t ( ) ( )

c

i t

g t f t e

ω −

= 1 ( ) ( ) ( ) 2

c

i t i t c

G f t e e dt F

ω ω

ω ω ω π

∞ − −∞

= = −

∫

( ) G ω ω

c

ω ( ) g t

c

i t

e

ω −

1 ( ) ( ) 2

i t

F f t e dt

ω

ω π

∞ −∞

=

∫

Making a Wave Packet

Forward-going wave packet is generated at x = 0 as

We know how each normal mode travels The total waves should travel as

(0, ) ( ) ( )

i t

t g t G e d

ω

ξ ω ω

∞ − −∞

= = ∫

i t

e

ω − ( ) i kx t

e

ω −

at x = 0

( )

( , ) ( )

i kx t

x t G e d

ω

ξ ω ω

∞ − −∞

= ∫ ( ) ( )

c

i t

g t f t e

ω −

= G(ω) ≠ 0 only near ωc

k = k(ω)!

Traveling Wave Packet

( ) ( ) ( ) ( )

( ) ( ) ( )

( , ) ( ) ( ) ( ) ( ) ( )

c c c c c c c c

i k x t i k x t c dk i k x t d dk i x t i k x t d i k x t dk d

f t x t G e d F e d F e d F e e x e d

ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω

ξ ω ω ω ω ω ω ω ω ω

+∞ − −∞ +∆ − −∆   ′ ′ + − + +∆     −∆   ′ − +∆   −   −∆ −

− = = − ′ ′ = ′ = =

∫ ∫ ∫ ∫

( )

c c

k k ω ≡

Taylor expansion of k(ω) Shape of the wave packet travels this way Carrier waves

Traveling Wave Packet

t ( )

dk d

f t x

ω

−

ξ(x, t)

( )

c c

i k x t

e

ω −

x x x

Group Velocity

Wave packet travels as

Velocity is given by We call it the group velocity cg

Now we have two definitions of propagation velocity

Phase velocity cp for sine waves Group velocity cg for wave packets

How do they change with frequency?

c

dk f t x d

ω ω

ω

=

    −          

c

dk t x C d

ω ω

ω

=

  − =    

c

d dx dt dk

ω ω

ω

=

      =

g

d c dk ω =

p

c k ω =

Phase and Group Velocities

cg remains less than cw for the wave packet

Information never travels faster than light

k ω ω k

w

c

2 2 2 p w

c c k k ω ω = = +

2 2 2

( )

w

k c k ω ω = +

2 2 2 2 w g w

c d c dk c k ω ω = = +

Slope = cp Slope = cg

Cut-Off Frequency

Waves can’t exist below the cut-off frequency ω0

Exactly what is happening there?

Look at the wave equation

Throw in We can’t have solutions that goes to infinity at This leaves us with

2 2 2 2 2 2

( , ) ( , ) ( , )

w

x t x t c x t t x ξ ξ ω ξ ∂ ∂ = − ∂ ∂ ( , ) ( )

i t

x t a x e ω ξ =

2 2 2 2 2

( ) ( ) ( )

i t i t i t w

d a x a x e c e a x e dx

ω ω ω

ω ω − = − ( ) a x A Bx = +

solution

x → ±∞ ( , )

i t

x t Ae ω ξ =

No x dependence

Below Cut-Off Frequency

Waves can’t exist below the cut-off frequency ω0

But we can attach a motor and run it at any frequency

As usual, we write the solution as

The wave equation gives us imaginary k

(0, )

i t

t re

ω

ξ

−

= ω ω <

What happens?

2 2 2 2 w w

k i c c ω ω ω ω − − = ± = ±

( )

( , )

i kx t

x t re

ω

ξ

−

=

Does this make physical sense?

Below Cut-Off Frequency

For an imaginary k, we define

The solution becomes I.e., the solution shrinks exponentially with x

Your “waves” never go much further than 1/Γ

We have covered all bases

• Traveling waves described by dispersion relation
• Uniform oscillation over entire space
• Exponentially attenuating with distance

( )

( , )

i kx t x i t

x t re re e

ω ω

ξ

− Γ −

= =

m

k i = ± Γ Γ >

+Γx goes infinity, so we pick −Γx

ω ω > ω ω = ω ω <

LC Transmission Line

Consider a coaxial cable

Life isn’t that easy

Insulating material in the cable has

Where does the permittivity ε come from?

You did this in Physics 15b

1 const. x x k L C ω εµ ∆ ∆ = = =

Non-dispersive

ε ε > µ µ =

Dielectric Material

Most insulator is made of molecules that can polarize

Imagine +q and –q are connected by a spring ks Equation of motion:

We are interested in changing E

This is a forced oscillator We know how to do this

Apply E field

E q + q − x E ∝ qE

s

k x 2

s

mx qE k x = − &&

Each half moves only x/2

m m ( )

i t

E t E e

ω −

= ( )

i t

x t x e

ω −

=

2 2 2

2 2 2 ( )

s

qE qE x k m m ω ω ω = = − − 2

s

k m ω =

Capacitor

Consider a parallel-plate capacitor with area S

Electric field polarizes the insulator Charge appears on top/bottom Induced charge partially cancels Q The field inside the capacitor is

Q + Q − E

polarize

Q qnSx ′ =

Density of molecules

2 2

2 ( ) q Q Q Q qnS E m ω ω ′ − = − − Q Q E S ε ′ − =

Solve for Q

2 2 2

2 ( ) q n Q SE m ε ω ω   = +   −  

This is ε

LC Transmission Line

For a coaxial cable, the dispersion relation is

Using Now we can calculate the velocities

1 k ω εµ =

2 2 2 2 2 2

2 1 ( ) q n m ω ε ε ε ρ ω ω ω ω   = + = +   − −  

2 2 2 2 2 2

( ) 1 1 k c ω ω ω ω ω ε ρ µ ρ ω ω ω ω   = + = +   − −  

2 2 2

1

p

c c k

ω ω ω

ω ρ

−

= = +

( )

2 2 2 4 2 2 2

1 1

g

c d c dk

ω ω ω ω ω ω

ρ ω ρ

− −

+ = = +

Dispersion Relation

ε ω k ω k c ω =

1

ω

imaginary

ω ω

1 0 1

ω ω ρ = +

1

ω ε

ε goes to infinity at the resonance frequency No wave solution between ω0 and ω1

Wave Velocities

ω

p

c ω c

1

ω ω

g

c ω c

1

ω

Phase velocity cp greater than c above the forbidden band Group velocity cg is always slower than c

Quantum Mechanics

Already mentioned that momentum p is related to k

Similarly, energy E is related to ω as

Consider a moving object of mass m and velocity v

From this dispersion relation,

p k = h E ω = h p k mv = = h

2

1 2 E mv ω = = h

eliminate v

2

2 k m ω = h

g

d k c v dk m ω = = = h In QM, objects are wave packets Classical velocity is given by the group velocity of the waves

Summary

Discussed dispersive waves

Dispersion relation = dependence between k and ω

Determines how the waves are transmitted

Normal modes propagate with different velocities

Waveforms are not conserved

Defined group velocity

Velocity of wave packets Represents how fast information can travel in space Never faster than light

Next: multi-dimensional waves

g

d c dk ω =