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 = ∫ 1 ∞ ∞ ∫ ω − ω ω ω = ω i t i t f t ( ) F ( ) e d F ( ) f t e ( ) dt π 2 −∞ −∞ � f ( t ) and F ( ω ) represent a function in time/frequency domains � Analyzed pulses and wave packets 1 � Time resolution ∆ t and bandwidth ∆ ω related by ∆ ∆ t ω > 2 � 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

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 ξ n − 1 ξ n ξ n+ 1 � In Lecture #5, we had 2 d dt ξ = − ξ − ξ − ξ − ξ m k ( ) k ( ) − + n s n n 1 s n n 1 2 � We ignored the gravity by making the strings very long → ∞ mg L What if we didn’t make − ξ  → 0 n this approximation? L

Wave Equation � Equations of motion is now 2 d mg ξ = − ξ − ξ − ξ − ξ − ξ m k ( ) k ( ) − + n s n n 1 s n n 1 n 2 dt L � Usual Taylor-expansion trick ∂ ξ ∂ ξ 2 2 ( , ) x t ( , x t ) ( m g = ∆ − ξ 2 m k x ) ( , ) x t s ∂ ∂ 2 2 t x L � Divide by ( ∆ x ) K ρ ∂ ξ ∂ ξ ρ 2 2 ( , ) x t ( , ) x t g = c = − ξ K l ( , ) x t w ρ l ∂ ∂ 2 2 t x L l ∂ ξ ∂ ξ 2 2 ( , ) x t ( , ) x t g � Wave equation: = − ω ξ 2 2 c ( , ) x t ω = w 0 ∂ ∂ 2 2 0 t x L Natural frequency of pendulum

∂ ξ ∂ ξ 2 2 ( , ) x t ( , ) x t Solution = − ω ξ 2 2 c ( , ) x t w 0 ∂ ∂ 2 2 t x ξ = a x e ω � Assume i t ( , ) x t ( ) 2 d a x ( ) − ω ω = ω − ω ω 2 i t 2 i t 2 i t a x e ( ) c e a x e ( ) Wave eqn. w 0 2 dx ω − ω 2 2 2 d a x ( ) = − 0 a x ( ) ω − ω > 2 2 0 SHO-like if 2 2 dx c 0 w � As before, we can write the solution as ω − ω 2 2 ξ = ± ω i kx ( t ) = 0 ( , ) x t Ae but with k c w This is the difference

Dispersion Relation � Normal-mode solutions are still ξ = ± ω i kx ( t ) ( , ) x t e � What changed is the relationship between k and ω � A.k.a. dispersion relation ω ω = k ( ) Non-dispersive waves c w ω − ω 2 2 ω = 0 k ( ) Dispersive waves c w � 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 We’ll study how…

Phase Velocity ξ = ξ ± ω � To calculate the propagation velocity of i kx ( t ) ( , ) x t e 0 � We follow the point where the phase kx ± ω t is constant ω ω C t dx m Phase velocity c p = = m ± ω = x kx t C k dt k � Phase velocity is the velocity of pure sine waves � Easily calculated from the dispersion relation ω Non- ω = = ω = c ( ) c const. k ( ) p w dispersive c No longer w constant! ω − ω 2 2 ω ω = ω = 0 k ( ) c ( ) c Dispersive p w c ω − ω 2 2 w 0

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 c p � On dispersive medium, the pulse shape must change � The pulse gets dispersed � Hence the name: dispersion � Dispersion makes poor media for communication

Dispersion Relation c k ω c p = c w p ω − ω 2 2 = ω c k 0 w = ω − ω 2 2 c k c w 0 w ω ω ω ω 0 0 � Dispersive waves have no solution for ω < ω 0 � It has a low frequency cut-off at ω 0 � Phase velocity goes to infinity at cut-off � Wait! Isn’t it unphysical? What happened to Relativity?

Finite-Length Signal � Phase velocity c p 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 F ω f t ( ) ( ) ω t 0 T ω < ω � Problem: this medium can’t carry waves with 0 � We need to make a pulse that does not contain frequencies below the cut-off Solution: wave packet

General Wave Packet f t ( ) g t ( ) � Consider a wave packet − ω = i t g t ( ) f t e ( ) t c − ω i t � Modulate carrier wave e c with a pulse f ( t ) � Fourier integral of such wave packet is 1 1 ∞ ∞ ∫ ∫ − ω ω = ω = ω − ω ω = ω i t i t i t G ( ) f t e ( ) e dt F ( ) F ( ) f t e ( ) dt c c π π 2 2 −∞ −∞ � G ( ω ) has the same shape as F ( ω ), G ω ( ) but centered around ω c ω � Now we examine how g ( t ) travels ω c in space

Making a Wave Packet − ω = i t g t ( ) f t e ( ) c � Forward-going wave packet is generated at x = 0 as = ∫ ∞ ξ = ω − ω ω i t (0, ) t g t ( ) G ( ) e d G ( ω ) ≠ 0 only near ω c −∞ � We know how each normal mode travels − ω − ω i t i kx ( t ) e e at x = 0 = ∫ ∞ ξ ω − ω ω i kx ( t ) � The total waves should travel as ( , ) x t G ( ) e d −∞ k = k ( ω )!

Traveling Wave Packet +∞ ( ) ∫ ω − ω ξ = ω i k ( ) x t ω ( , ) x t G ( ) e d −∞ ω +∆ ω ( ) ∫ ω − ω i k ( ) x t = c ω − ω ω F ( ) e d c ω −∆ ω c ≡ k ω k ( )   dk ′ ′ + ω − ω + ω i k x ( ) t +∆ ω c c   ∫ c c ′ ′ = ω ω ω  d  F ( ) e d −∆ ω Taylor expansion of k ( ω )  dk  ′ ω − i x t +∆ ω   ( ) ∫ − ω ′ i k x t = ω ω ω  d  e F ( ) e d c c −∆ ω ( ) − ω i k x t = − e f t ( dk x ) c c ω d Shape of the wave packet travels this way Carrier waves

Traveling Wave Packet − f t ( x ) dk ω d t x ( ) − ω i k x t e c c ξ ( x , t ) x x

Group Velocity     dk � Wave packet travels as −   f t x     ω  d    ω ω = � Velocity is given by c ω     dk dx d − = = t x C     ω  d  dt  dk  ω ω = ω ω = c c � We call it the group velocity c g � Now we have two definitions of propagation velocity ω ω � Phase velocity c p for sine waves d = = c c p g � Group velocity c g for wave packets k dk � How do they change with frequency?

Phase and Group Velocities ω = ω + 2 2 2 ( ) k c k ω ω 2 ω 0 w = = + 2 c c 0 p w 2 k k c w ω ω 2 d c 0 = = c w g dk + ω 2 2 2 c k w 0 k k Slope = c p Slope = c g � c g remains less than c w for the wave packet � Information never travels faster than light

Cut-Off Frequency � Waves can’t exist below the cut-off frequency ω 0 � Exactly what is happening there? ∂ ξ ∂ ξ 2 2 ( , ) x t ( , ) x t � Look at the wave equation = − ω ξ 2 2 c ( , ) x t w 0 ∂ ∂ 2 2 t x a x e ω ξ = i t ( , ) x t ( ) � Throw in 0 2 d a x ( ) − ω ω = ω − ω ω = + 2 i t 2 i t 2 i t a x e ( ) c e a x e ( ) a x ( ) A Bx solution 0 w 0 2 dx x → ±∞ � We can’t have solutions that goes to infinity at Ae ω ξ = i t ( , ) x t � This leaves us with No x dependence 0

Below Cut-Off Frequency � Waves can’t exist below the cut-off frequency ω 0 � But we can attach a motor and run it at any frequency ξ = − ω i t (0, ) t re What happens? ω < ω 0 ξ = − ω � As usual, we write the solution as i kx ( t ) ( , ) x t re � The wave equation gives us imaginary k ω − ω ω − ω 2 2 2 2 = ± = ± 0 0 k i c c w w Does this make physical sense?

Below Cut-Off Frequency � For an imaginary k , we define = ± Γ Γ > k i 0 ξ = − ω = Γ − ω i kx ( t ) x i t � The solution becomes ( , ) x t re re m e + Γ x goes infinity, so we pick −Γ x � I.e., the solution shrinks exponentially with x � Your “waves” never go much further than 1/ Γ � We have covered all bases ω > ω � Traveling waves described by dispersion relation � 0 ω = ω � Uniform oscillation over entire space � 0 ω < ω � Exponentially attenuating with distance � 0

LC Transmission Line � Consider a coaxial cable ω ∆ ∆ x x 1 = = = const. εµ k L C Non-dispersive � Life isn’t that easy ε > ε µ = µ � Insulating material in the cable has 0 0 � Where does the permittivity ε come from? � You did this in Physics 15b