Wave PhenomenaPhysics 15c

Lecture 12Dispersion

(H&L Sections 2.6)

What We Did Last Time

Defined Fourier integral

f(t) and F(ω) represent a function in time/frequency domainsAnalyzed pulses and wave packets

Time resolution ∆t and bandwidth ∆ω related byProved for arbitrary waveform

Rate of information transmission ∝ bandwidthDirac’s δ(t) a limiting case of infinitely fast pulseConnection with Heisenberg’s Uncertainty Principle in QM

( ) ( ) i tf t F e dωω ω∞ −

−∞= ∫

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

12

t ω∆ ∆ >

Goals For Today

Discuss dispersive wavesWhen velocity is not constant for different ωWaveform changes as it travelsDispersion relation: dependence of k on ω

Define group velocityHow 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 12 ( ) ( )n s n n s n ndm k kdt

ξ ξ ξ ξ ξ− += − − − −

0nmgL

Lξ →∞− → 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 12 ( ) ( )n s n n s n n ndm k kdt

mgL

ξ ξ ξξ ξ ξ− + −= − − − −

2 22

2 2

( , ) ( , ( , )) ( )sx t x tm k x

t xmL

x g tξ ξξ∂ ∂= ∆

∂ ∂−

2 2

2 2

( , ) ( , ) ( , )llx t x tK g

tx t

x Lρ ξξ ξρ ∂ ∂

=∂ ∂

−

2 22 2

02 2

( , ) ( , ) ( , )wx t x tc x t

t xξ ξ ω ξ∂ ∂

= −∂ ∂

wl

Kcρ

=

0gL

ω =

Natural frequency of pendulum

2 22 2

02 2

( , ) ( , ) ( , )wx t x tc x t

t xξ ξ ω ξ∂ ∂

= −∂ ∂Solution

Assume

As before, we can write the solution as

( , ) ( ) i tx t a x e ωξ =2

2 2 202

( )( ) ( )i t i t i tw

d a xa x e c e a x edx

ω ω ωω ω− = −Wave eqn.

2 220

2 2

( ) ( )w

d a x a xdx c

ω ω−= − SHO-like if 2 2

0 0ω ω− >

( )( , ) i kx tx t Ae ωξ ±=2 2

0

w

kc

ω ω−=but with

This is the difference

Dispersion Relation

Normal-mode solutions are stillWhat changed is the relationship between k and ω

A.k.a. dispersion relation

NB: there are different types of dispersive wavesWe are looking at just one example here

Dispersion relation determines how the waves propagate in time and space

( )( , ) i kx tx t e ωξ ±=

( )w

kcωω =

2 20( )

w

kc

ω ωω

−=

Non-dispersive waves

Dispersive waves

We’ll study how…

Phase Velocity

To calculate the propagation velocity ofWe follow the point where the phase kx ± ωt is constant

Phase velocity is the velocity of pure sine wavesEasily calculated from the dispersion relation

( )0( , ) i kx tx t e ωξ ξ ±=

kx t Cω± =C tx

kω

=m dx

dt kω

= m Phase velocity cp

( ) const.p wc cω = =( )w

kcωω =

2 20( )

w

kc

ω ωω

−=

Non-dispersive

Dispersive2 2

0

( )p wc c ωωω ω

=−

No longerconstant!

Dispersing Pulses

Imagine a pulse being sent over a distanceOn 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 dispersedHence the name: dispersion

Dispersion makes poor media for communication

Dispersion Relation

Dispersive waves have no solution for ω < ω0It has a low frequency cut-off at ω0

Phase velocity goes to infinity at cut-offWait! Isn’t it unphysical? What happened to Relativity?

k

0ω

wc k ω=

2 20wc k ω ω= −

ω

pc

0ωω

wc

2 20

wp

cc ω

ω ω=

−

Finite-Length Signal

Phase velocity cp is the speed of pure sine wavesBut pure sine waves don’t carry informationRelativity 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

ω0

0ω ω<

Solution: wave packet

General Wave Packet

Consider a wave packet

Modulate carrier wavewith 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) travelsin space

( )f t

t( ) ( ) ci tg t f t e ω−=

1( ) ( ) ( )2

ci t i tcG f t e e dt Fω ωω ω ω

π∞ −

−∞= = −∫

( )G ω

ωcω

( )g t

ci te ω−

1( ) ( )2

i tF f t e dtωωπ

∞

−∞= ∫

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 tt g t G e dωξ ω ω∞ −

−∞= = ∫

i te ω− ( )i kx te ω−at x = 0( )( , ) ( ) i kx tx t G e dωξ ω ω

∞ −

−∞= ∫

( ) ( ) ci tg 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 tc

dki k x td

dki x ti k x t d

i k x t dkdf 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 ωω ω

ω

ωωω

ω

ω

ω

ω

ξ ω ω

ω ω ω

ω ω

ω ω

+∞ −

−∞

+∆ −

−∆

′ ′+ − ++∆

−∆

′ −+∆ −

−∆

− −

=

= −

′ ′=

′=

=

∫∫

∫

∫

( )c ck k ω≡

Taylor expansion of k(ω)

Shape of the wave packet travels this wayCarrier waves

Traveling Wave Packet

t( )dk

df t xω−

ξ(x, t)

( )c ci k x te ω−

x

x

x

Group Velocity

Wave packet travels asVelocity is given by

We call it the group velocity cg

Now we have two definitions of propagation velocityPhase velocity cp for sine wavesGroup velocity cg for wave packets

How do they change with frequency?

c

dkf t xd ω ωω =

−

c

dkt x Cd ω ωω =

− = c

ddxdt dk ω ω

ω

=

=

gdcdkω

=pckω

=

Phase and Group Velocities

cg remains less than cw for the wave packetInformation never travels faster than light

k

0ω

ω

k

wc

22 0

2p wc ck k

ωω= = +

2 2 20( ) wk c kω ω= +

2

2 2 20

wg

w

cdcdk c kω

ω= =

+

Slope = cp Slope = cg

Cut-Off Frequency

Waves can’t exist below the cut-off frequency ω0Exactly what is happening there?

Look at the wave equationThrow in

We can’t have solutions that goes to infinity atThis leaves us with

2 22 2

02 2

( , ) ( , ) ( , )wx t x tc x t

t xξ ξ ω ξ∂ ∂

= −∂ ∂

0( , ) ( ) i tx t a x e ωξ =2

2 2 20 02

( )( ) ( )i t i t i tw

d a xa x e c e a x edx

ω ω ωω ω− = − ( )a x A Bx= +solution

x →±∞0( , ) i tx t Ae ωξ = No x dependence

Below Cut-Off Frequency

Waves can’t exist below the cut-off frequency ω0But we can attach a motor and run it at any frequency

As usual, we write the solution asThe wave equation gives us imaginary k

(0, ) i tt re ωξ −=

0ω ω<What

happens?

2 2 2 20 0

w w

k ic c

ω ω ω ω− −= ± = ±

( )( , ) i kx tx t re ωξ −=

Does this make physical sense?

Below Cut-Off Frequency

For an imaginary k, we defineThe solution becomes

I.e., the solution shrinks exponentially with xYour “waves” never go much further than 1/Γ

We have covered all basesTraveling waves described by dispersion relationUniform oscillation over entire space

Exponentially attenuating with distance

( )( , ) i kx t x i tx t re re eω ωξ − Γ −= = m

k i= ± Γ 0Γ >

+Γx goes infinity, so we pick −Γx

0ω ω>

0ω ω=

0ω ω<

LC Transmission Line

Consider a coaxial cable

Life isn’t that easy

Insulating material in the cable hasWhere does the permittivity ε come from?

You did this in Physics 15b

1 const.x xk L Cω

εµ∆ ∆

= = =

Non-dispersive

0ε ε> 0µ 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 EThis is a forced oscillator We know how to do this

Apply E field Eq+q− x E∝

qEsk x

2 smx qE k x= −&&

Each half moves only x/2

m m

0( ) i tE t E e ω−= 0( ) i tx t x e ω−=

0 00 2 2 2

0

2 22 ( )s

qE qExk m mω ω ω

= =− − 0

2 skm

ω =

Capacitor

Consider a parallel-plate capacitor with area SElectric field polarizes the insulatorCharge appears on top/bottom

Induced charge partially cancels Q

The field inside the capacitor is

Q+

Q−E

polarizeQ qnSx′ =

Density of molecules

2 20

2( )

qQ Q Q qnS Em ω ω

′− = −−

0

Q QESε′−

= Solvefor Q

2

0 2 20

2( )

q nQ SEm

εω ω

= + −

This is ε

LC Transmission Line

For a coaxial cable, the dispersion relation is

Using

Now we can calculate the velocities

1kω

εµ=

220

0 02 2 2 20 0

2 1( )

q nm

ωε ε ε ρω ω ω ω

= + = + − −

2 20 0

0 02 2 2 20 0

( ) 1 1kc

ω ωωω ω ε ρ µ ρω ω ω ω

= + = + − −

20

2 20

1p

cck ω

ω ω

ω

ρ−

= =+ ( )

20

2 2040

22 20

1

1g

cdcdk

ωω ω

ω

ω ω

ρωρ

−

−

+= =

+

Dispersion Relationε

ω

k

0ω

kcω

=

1ω

imaginary

ω0ω

1 0 1ω ω ρ= +1ω

0ε

ε goes to infinity at the resonance frequencyNo wave solution between ω0 and ω1

Wave Velocities

ω

pc

0ω

c

1ωω

gc

0ω

c

1ωPhase velocity cp greater than c above the forbidden bandGroup 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 asConsider a moving object of mass m and velocity v

From this dispersion relation,

p k= h

E ω= h

p k mv= =h 212

E mvω= =h eliminate v2

2km

ω =h

gd kc vdk mω

= = =h In QM, objects are wave packets

Classical velocity is given by the group velocity of the waves

Summary

Discussed dispersive wavesDispersion relation = dependence between k and ω

Determines how the waves are transmittedNormal modes propagate with different velocities

Waveforms are not conservedDefined group velocity

Velocity of wave packetsRepresents how fast information can travel in spaceNever faster than light

Next: multi-dimensional waves

gdcdkω

=