Wave Phenomena Physics 15c. Lecture 11 Dispersion

Transcription

1 Wave Phenomena Physics 15c Lecture 11 Dispersion

2 What We Did Last Time Defined Fourier transform f (t) = F(ω)e iωt dω F(ω) = 1 2π f(t) and F(w) represent a function in time and frequency domains Analyzed pulses and wave packets Time resolution Δt and bandwidth Δω related by Can be proven for arbitrary waveform Rate of information transmission bandwidth Dirac s δ(t) a limiting case of infinitely fast pulse δ(t) = 1 2π e iωt dω f (t)e iωt dt ΔtΔω 1 2

3 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?

4 Mass-Spring Transmission Line In Lecture #4, we had ξ n-1 ξ n ξ n+1 m d 2 dt 2 ξ n = k s (ξ n ξ n 1 ) k s (ξ n ξ n+1 ) We ignored the gravity by making the strings very long mg L ξ n L 0 What if we didn t make this approximation?

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

6 Solution 2 ξ(x,t) t ξ(x,t) = ω 2 x 2 0 ξ(x,t) Assume Wave eqn. ξ(x,t) = a(x)e iωt ω 2 a(x)e iωt = 2 d 2 a(x) dx 2 e iωt ω 0 2 a(x)e iωt d 2 a(x) = ω 2 ω 0 a(x) dx 2 2 SHO-like if ω 2 ω 2 c 0 > 0 w As before, we can write the solution as but the wavenumber k is given by 2 k = ω 2 ω 0 2 ξ(x,t) = Ae i(kx±ωt ) No longer proportional to ω We have dispersive waves

7 Dispersion Relation Normal-mode solutions are still ξ(x,t) = e i(kx±ωt ) What changed is the relationship between k and ω a.k.a. dispersion relation k(ω) = ω Non-dispersive waves k(ω) = ω 2 2 ω 0 Dispersive waves 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

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

9 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

10 Dispersion Relation k k = ω c p c p = ω ω 2 ω 0 2 k = ω 2 ω 0 2 ω 0 ω ω 0 ω 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! What happened to Relativity?

11 Special Relativity Nothing travels faster than the speed of light, with the possible exception of bad news, which obeys its own set of laws. Douglas Adams Special Relativity does not allow anything to travel faster than the speed of light c What, exactly? Lorentz transformation would lead to problems with causality if signals can be transmitted faster than c Signals = Information Relativity forbids superluminal transfer of information

12 Finite-Length Signal Phase velocity c p is the speed of pure sine waves But pure sine waves don t carry information Let s think about a finite-length pulse from last lecture f (t) F(ω) t T Problem: this medium can t carry waves with 0 ω < ω 0 We need to make a pulse that does not contain frequencies below the cut-off Solution: wave packet

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

14 Making a Wave Packet g(t) = f (t)e iω c t Forward-going wave packet is generated at x = 0 as ξ(0,t) = g(t) = G(ω)e iωt dω We know how each normal mode travels The total waves should travel as e iωt at x = 0 e i(kx ωt ) ξ(x,t) = G(ω)e i(kx ωt ) dω G(ω ) 0 only near ω c

15 Traveling Wave Packet Recall that k is a function of ω ξ(x,t) = + ω c + Δω ω c Δω i( k(ω )x ωt) G(ω)e dω = F(ω ω c )e + Δω = F( ω )e i k c + Δω = e i ( k c x ω c t ) + Δω Δω i( k(ω )x ωt) dω dk dω ω x (ω c + ω )t F( ω )e i ω = e i ( k c x ω c t ) f (t dk x) dω dk dω x t d ω d ω ω = ω c + ω k c k(ω c ) Taylor expansion of k(ω ) around ω c Carrier waves Shape of the wave packet

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

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

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

19 Beats Consider beats between two cosine waves ξ(x,t) = cos(k 1 x ω 1 t) + cos(k 2 x ω 2 t) Frequencies ω 1 and ω 2 are close to each other So are the wavenumbers k 1 and k 2 ξ(x,t) = 2cos k + k x ω 1 + ω 2 2 = 2cos(kx ωt)cos(δkx Δωt) t cos k k x ω 1 ω 2 2 t Carrier waves with average frequency Beats move with Shape of the beats Δω Δk dω dk = c g

20 Quantum Mechanics Already mentioned that momentum p is related to k p = k Similarly, energy E is related to ω as E = ω Consider a moving object of mass m and velocity v p = k = mv E = ω = 1 2 mv 2 eliminate v ω = k 2 From this dispersion relation, 2m c g = dω dk = k m = v In QM, objects are wave packets Classical velocity is given by the group velocity of the waves

21 Cut-Off Frequency Waves can t exist below the cut-off frequency ω 0 Exactly what is happening there? Look at the wave equation Substitute We can t have solutions that goes to infinity at This leaves us with ξ(x,t) = a(x)e iω 0 t ω 0 2 a(x)e iω 0 t = 2 d 2 a(x) dx 2 e iω 0t 2 ξ(x,t) t 2 ω 0 2 a(x)e iω 0 t ξ(x,t) = Ae iω 0 t 2 2 ξ(x,t) = ω 2 x 2 0 ξ(x,t) solution x ± No x dependence a(x) = A + Bx

22 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(0,t) = re iωt ω < ω 0 What happens? As usual, we write the solution as ξ(x,t) = re i(kx ωt ) Wave equation We find imaginary k ω 2 re iωt = 2 k 2 re iωt ω 0 2 re iω 0 t k = ± ω 2 2 ω 0 = ±i ω 0 2 ω 2 = ±iγ Γ > 0 What is the physical meaning?

23 Below Cut-Off Frequency We have found k = ±i Γ where The solution becomes Since e +Γx diverges to, we can only take e Γx i.e., the solution shrinks exponentially with x Our waves never go much further than 1 Γ = We have covered all bases Γ = ω 2 ω 2 0 ξ(x,t) = re i(kx ωt ) = re Γx e iωt ω > ω 0 Traveling waves described by dispersion relation ω = ω 0 Uniform oscillation over entire space ω < ω 0 Exponentially attenuating with distance ω 0 2 ω 2

24 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 c g = dω dk Next: multi-dimensional waves