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
=