Waves, dispersion, group speed

In summary, the conversation discusses the difficulty in finding understandable information on the topic of wave packets, dispersion, phase and group velocity. The surface tension governs the motion of waves with small wavelength, and the phase velocity is given by v_p = ... The group velocity of a superposition of waves with comparable wavelength is v_g = 3v_p/2, which means that the modulation of amplitude moves faster than the waves themselves. However, this result only applies to two waves and it is unclear how it can be generalized. The group velocity is the speed at which the envelope of the wave propagates and it can be derived from the expression v_g = dω/dk. The group velocity is only meaningful when the wavenumber distribution has
  • #1

quasar987

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I'm having extreme difficulties finding understandable informations on the why and how on the topic of wave paquets, dispersion, phase and group velocity. My question is this.

The motion of waves of small wavelenght at the surface of water is governed by the surface tension. The phase velocity of such waves is givent by the expression v_p = ...

a) Show that the group velocity of a superposition of waves of comparable wavelenght is v_g = 3v_p/2. (I've done that, though I don't really understand why what I've done is true)

b) What is the implication of this result on the observation of the motion of such a superposition of waves at the surface of water?

My answer would be: we would observe that the modulation of amplitude of the waves move faster than the waves themselves.

Some clarification and any info whatesoever would be warmly welcomed. I can't believe how unlclearly the teacher has bombared us with this stuff!
 
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  • #2
No one interested in giving out a little lecture about dispersion and group velocity? o:)
 
  • #3
Let me ask this then...

Everywhere I look, book, website, whatever, they always introduce the group velocity by saying something like "the group velocity, defined by

[tex]v_g = \frac{d\omega}{dk}[/tex]

is the speed at which the envelope of the wave propagates."

Nobody actually show that this is true, or derive it directly from an equation (except in the very particular case of a superposition of just 2 progressive waves).

So, if anyone could provide a derivation (or at least an explication), I would be eternally grateful.
 
  • #4
well by definition phase velocity is the velocity at which surfaces of equal phase propagate in a normal form wave. In a modulated wave where two normal waves with nearly equal frequency interfere the energy carried by the wave is separated into small packages which propagate with the modulation wave's phase velocity. If we go to the limit where the the difference in frequency and wave vectors between the two interfering waves goes to zero we get the group velocity to be

[tex] v_g={d\omega \over dk}[/tex],
 
  • #5
The formula for vg comes from a Taylor expansion of n in omega that is cut off at the linear term. When this is put into the integral for f(x,t), the formula results.
There is a simple derivation in "Classical Electromagnetism" by Franklin.
 
  • #6
inha said:
well by definition phase velocity is the velocity at which surfaces of equal phase propagate in a normal form wave. In a modulated wave where two normal waves with nearly equal frequency interfere the energy carried by the wave is separated into small packages which propagate with the modulation wave's phase velocity. If we go to the limit where the the difference in frequency and wave vectors between the two interfering waves goes to zero we get the group velocity to be

[tex] v_g={d\omega \over dk}[/tex],

Again, this is good in the context of two waves only.
 
  • #7
ah sorry I didn't read your 3rd post well enough. I can't think of a reason why it couldn't be generalized for more waves under the same conditions of the top of my head but I'll think about it.
 
  • #8
What about the explanation : a wavepacket is defined by a superposition of partial waves of the type : A(kx-w(k)t)...

Consider a point which is fixed on all the partial waves...it's by defintion a point moving the velocity of the group...hence let see

A(kx-w(k)t)=cste forall k...derive towards k :

A'*(x-w'(k)t)=0 forall k and A' (this is valid for all shapes of partial waves)

hence x=w'(k)t hence the group velocity is v_g=w'(k)

Is this acceptable ?
 
  • #9
Hey kleinwolf,

This development is very seducing, but I don't understand the first step. :grumpy:

A wave packet is a superposition of progressive waves of the form

[tex]y(x,t) = A(k)cos(kx-\omega(k)t)[/tex]

I can't imagine a single point that is fixed on all the partial waves simultaneously. I can imagine a fixed point on each particular wave, but those would be moving with the particular phase speed of the wave they're attacjed to.
 
  • #10
Here's what it says in chapter 7 of Classical Electrodynamics by John "Action Jackson" Jackson. Suppose you have a wavepacket F(x,t) given by

[tex]F(x,t) = \int_{-\infty}^{\infty} f(k) e^{ikx - i \omega (k) t} dk[/tex]

where

[tex]f(k) = \frac{1}{2 \pi}\int_{-\infty}^{\infty} F(x, 0) e^{-ikx} dx[/tex]

is the wavenumber distribution of partial waves that make up F(x, 0). Suppose that this distribution is peaked about a dominant wavenumber k_0, with a fairly small width. Then as Meir Achuz was saying, we can write

[tex]\omega (k) \simeq \omega_{0} + \omega^{\prime}(k_0) (k - k_0)[/tex]

put this in for F(x,t) to get

[tex]F(x,t) \simeq e^{-i(\omega_0 - k_0\omega^\prime (k_0))t}\int_{-\infty}^{\infty} f(k) e^{i(x - \omega^\prime (k_0)t)k} dk[/tex]

If you squint your eyes and look real hard, you will notice that the last integral is in the form of F(x,0), except with x replaced by x - w'(k_0)t. So we have

[tex]F(x,t) \simeq e^{-i(\omega_0 - k_0\omega^\prime (k_0))t} F(x - \omega^\prime (k_0)t, 0)[/tex]

Therefore F(x-w'(k_0)t) approximately represents the envelope of the wavepacket. The factor out in front is just the rapid oscillation. This treatment does not take dispersion into account. You need to include second order terms of omega for that. Also note that since the energy density is proportional to the absolute value squared of F(x,t), energy travels at group velocity. So apparently, the group velocity is only meaningful when the wavenumber distribution is fairly sharply peaked.
 
  • #11
I think it's only sensible to talk about group velocity if the wavenumber distribution itself has a more or less definite peak (like PBRM. said).
So if we write the wave packet it the general form:

[tex]f(x,t)=\int_{-\infty}^{+\infty} g(k)e^{i(kx-\omega(k)t)}dk[/tex]

so at time t=0:

[tex]f(x,0)=\int_{-\infty}^{+\infty} g(k)e^{ikx}dk[/tex]

then |g(k)| should have a narrow peak, or else your wave packet will change form very rapidly and the notion of a group of waves with a well defined velocity will then have no meaning. Suppose |g(k)| has a maximum around [itex]k_0[/itex].
[itex]|f(x,0)|[/itex] is maximum when the different plane waves with big amplitudes interfere constructively. This happens when the k-dependent phases of these waves vary slightly around [itex]k_0[/itex]. So the idea is to set the derivative of the phase wrt. k at zero when [itex]k=k_0[/itex]. Writing [itex]g(k)=|g(k)|\exp(i\alpha(k))[/itex] the expression for the wave packet becomes:

[tex]f(x,0)=\int_{-\infty}^{+\infty} |g(k)|e^{i[kx+\alpha(k)]}dk[/tex]

So the phase of the wave corresponding to k is [itex]kx+\alpha(k)[/itex]. Its derivative wrt k is: [itex]x+\alpha'[/itex].
Let's call this point where they interferere constructively the maximum or 'center' of the wavepacket at t=0 [itex]x_M(0)[/itex].

[tex]x_M(0)=-\frac{d \alpha}{dk}|_{k=k_0}[/tex]

A little while later the wavepacket will be:

[tex]f(x,t)=\int_{-\infty}^{+\infty} |g(k)|e^{i(kx+\alpha(k)-\omega(k)t)}dk[/tex]

so now the maximum is at:

[tex]x_M(t)=\frac{d\omega}{dk}|_{k=k_0}t-\frac{d\alpha}{dk}|_{k=k_0}[/tex]

From which you can easily see the speed at which x_M travels is [itex]\frac{d\omega}{dk}[/itex] (evaluated at [itex]k=k_0[/itex]).
 
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  • #12
PBRMEASAP said:
Here's what it says in chapter 7 of Classical Electrodynamics by John "Action Jackson" Jackson. Suppose you have a wavepacket F(x,t) given by

[tex]F(x,t) = \int_{-\infty}^{\infty} f(k) e^{ikx - i \omega (k) t} dk[/tex]

where

[tex]f(k) = \frac{1}{2 \pi}\int_{-\infty}^{\infty} F(x, 0) e^{-ikx} dx[/tex]

is the wavenumber distribution of partial waves that make up F(x, 0). Suppose that this distribution is peaked about a dominant wavenumber k_0, with a fairly small width. Then as Meir Achuz was saying, we can write

[tex]\omega (k) \simeq \omega_{0} + \omega^{\prime}(k_0) (k - k_0)[/tex]

put this in for F(x,t) to get

[tex]F(x,t) \simeq e^{-i(\omega_0 - k_0\omega^\prime (k_0))t}\int_{-\infty}^{\infty} f(k) e^{i(x - \omega^\prime (k_0)t)k} dk[/tex]

If you squint your eyes and look real hard, you will notice that the last integral is in the form of F(x,0), except with x replaced by x - w'(k_0)t. So we have

[tex]F(x,t) \simeq e^{-i(\omega_0 - k_0\omega^\prime (k_0))t} F(x - \omega^\prime (k_0)t, 0)[/tex]

Therefore F(x-w'(k_0)t) approximately represents the envelope of the wavepacket. The factor out in front is just the rapid oscillation. This treatment does not take dispersion into account. You need to include second order terms of omega for that. Also note that since the energy density is proportional to the absolute value squared of F(x,t), energy travels at group velocity. So apparently, the group velocity is only meaningful when the wavenumber distribution is fairly sharply peaked.

Let's see if I read that last equation correctly: It says that F(x,t) is a progressive sine wave whose amplitude is modulated by the progressive wave [itex]F(x-v_gt,0)[/itex]. This is hard to visualize!



Galileo, I apreciate the effort, maybe I will be able to fully understand your perspective when I have a better understanding of this all.
 
  • #13
The factor out in front is just [itex]e^{-i \Omega t}[/itex], where [itex]\Omega = \omega_0 - k_0 \omega^\prime(k_0)[/itex] is a constant. It's not really the "progressive" part of the wave, since it doesn't depend on x or k. It modulates the progressive waveform [itex]F(x - v_g t, 0)[/itex]. Since [itex]\Omega[/itex] is often very large, the traveling wavepacket looks like a pulse of high frequency squiggles whose envelope is [itex]F(x - v_g t, 0)[/itex]. A common example for F(x, 0) is a gaussian (bell curve) pulse. According to the derivation above, this overall gaussian pulse F(x-v_g t, 0) moves along at group velocity with its shape intact. But in an actual dispersive medium, where the phase velocity of an individual partial wave depends on its frequency, the wavepacket spreads out as it moves along. The faster partial waves get farther and farther ahead, while the slower ones lag farther behind.

Actually Galileo's derivation is the one I wanted to post, but I couldn't remember how it went. So I had to look up the one I gave.
 
  • #14
What about the explanation : the group velocity is +/- the average velocity ?

hence : [tex] v_g\approx\frac{d<x>}{dt}=\frac{d}{dt}\int xf(x,t)dx [/tex]

Then you can use the Fourier transform of f and do like in the Jackson a series expansion keeping only the linear term :

[tex] f(x,t)=\int g(k)e^{i(kx-w(k)t)}dk [/tex]

hence

[tex]\frac{d}{dt}\int xf(x,t)dx=\int dk g(k)e^{-iw(k)t}(-i\frac{d}{dk}\delta(k))[/tex]

aso...

I got something proportional to w'(k0) but it actually depends on the shape around the k0 value...
 
  • #15
Galileo,

I already understand much better your idea and what (the heck) you're doing in your post and it makes a lot of sense. But I still don't understand the crucial step [itex]g(k)=|g(k)|\exp(i\alpha(k))[/itex]... How can there be such an alpha? In my sense [itex]g(k) = \pm |g(k)|[/itex], which would limit alpha to a value such that [itex]\exp(i\alpha(k)) = \pm 1[/itex]. So what's the deal with alpha?
 
  • #16
quasar987 said:
How can there be such an alpha? In my sense [itex]g(k) = \pm |g(k)|[/itex]

This is only true when g(k) is a real function.
 
  • #17
I though it were. Isn't g(k) the spectral distribution of amplitudes? I see no need for it to be complex.
 
  • #18
quasar987 said:
I though it were. Isn't g(k) the spectral distribution of amplitudes? I see no need for it to be complex.

Yes, but that doesn't make it real: g(k) is the *complex* amplitude of the partial wave with wavenumber k. That means that g(k) contains both magnitude AND phase information for each partial wave. That alpha in the "phase factor" exp(i alpha(k)) is the relative phase of the partial wave with wavenumber k.
 
  • #19
Often g(k) is real, so you can set [itex]\alpha=0[/itex] or [itex]\alpha=\pi[/itex] depending on whether g(k) is positive or negative. The result still holds as it is just a special case.

alpha carries the phase information of each wave at t=0 as you can see from f(x,0). If alpha=0, then all the waves are positioned 'on top of each other' at the origin at t=0. (consequently, x_M(0)=0).
 
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  • #20
In which cases is g(k) not real?
 
  • #21
Remember that g(k) is determined by f(x,0). Mathematically, g(k) is (maybe apart from a constant) just the Fourier-transform of f(x,0). If you've done Fourier transforms before, you know they don't have to be real, even if f(x,0) is.

Physically, you are giving the wave corresponding to k a phase difference wrt to other waves. Try experimenting with different g(k)'s (make it discrete for simplicity) to get a feel for it.
 
  • #22
What is the exact definition of k_0?

At first I thought it was the k of highest (absolute) amplitude, but suppose you have an amplitude spectrum with many extrema. This means you'd have many k with highest amplitudes. What would the k_0 be in that case?
 
  • #23
"In which cases is g(k) not real?"
g(k) is the FT of f(x,0).
g(k)=\int f(x,0)exp(-ikx)dx (omitting factors of 2pi).
Only if f(x,0)*=f(-x,0) is g(k) real.
The constant k_0 depends on how it appears in f(x,0).
In many cases there is no specific k_0.
 
  • #25
You guys,

with all your help and references, I have finally understood what wave packets, group and phase velocity are. (at only 4 AM, 3 months later) Hurray! I love you guys. :smile:
 
  • #27
[tex]f(x-\omega_0 't)=e^0 \int _{-\infty} ^{+\infty} e^{ik(x-\omega_0 't-0)}=\int _{-\infty} ^{+\infty} e^{ik(x-\omega_0 't)}[/tex]
the required term
 
  • #28
Huuuh? I know that. I'm asking about the term

[tex]e^{ik_0\omega_0't}[/tex]

Where did it disapear to in the step between equation 12 and the equation right before it?
 
  • #29
Seriously, is it a mistake in the pdf file?
 
  • #30
It could be incoorporated in the function g, I don't know if they introduced it earlier. But my guess is that g(x)=f(x,t=0) so that it is a typo

The second 'x' in the equation should be [itex] \omega _0 '[/itex]. Which is an understandable error as k and x often go side by side in such formula.
 
  • #31
I should have added the first part of the "tutorial" too so you see what is g(x) and can actually do the calculations involving equ.2 and 4 to get to equ.12. here it is:

https://www.physicsforums.com/attachment.php?attachmentid=3928

Personally, I see no mistake up to equ. 12 where the exponential term just dispears. It can't have been "incorporated" in g(x), that'd make no sense.
 
  • #32
quasar987 said:
I should have added the first part of the "tutorial" too so you see what is g(x) and can actually do the calculations involving equ.2 and 4 to get to equ.12. here it is:

https://www.physicsforums.com/attachment.php?attachmentid=3928

Aha...if that is the definition of g, it exactly accounts for eq 12.

Personally, I see no mistake up to equ. 12 where the exponential term just dispears. It can't have been "incorporated" in g(x), that'd make no sense.

It does, and it is...

[tex]g(x-\omega '_0 t) e^{i(k_0 x -\omega _0 t)} = f(x-\omega '_0 t,0) e^{ik_0 (\omega '_0 -x)} e^{i(k_0 x -\omega _0 t)}[/tex]

where you see two terms cancel and the result is the given equation.
 
  • #33
OOOOOOOOOOOOOOOOOOOOOOOH. My mistake was that I though

[tex]f(x, 0) = g(x)e^{ik_0x} \ \ \Rightarrow \ \ f(x-\omega_0't,0) = g(x-\omega_0't)e^{ik_0x}[/tex]

I forgot to substitute the x for [itex]x-\omega_0't[/itex] in the exponential! :yuck:

thx willem.
 
  • #34
I heard that wave packets are important in QM. 2 questions:

1) How important? Do they arise very often?

2) Are the notion of group velocity and dispersion important for quantum mechanical wave packets? I would guess that 'no', since the waves are probabilistic and not physical, so they don't propagate in a medium that we could call "dispersive".
 
  • #35
The group velocity is very important in QM (in my opinion at least) since the group velocity of a wave packet allows us to recover the classical expression of a free particle's kinetic energy.

From the Schrodinger equation of a free particle (V=0 everywhere):

[tex]i\hbar \frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2 \Psi[/tex]
You can see it admits a plane wave solution of the form:

[tex]\Psi(\vec r)=A\exp\left(\vec k \cdot \vec r - \omega t\right)[/tex]
if the dispersion relation is:
[tex]\omega = \frac{\hbar |\vec k|^2}{2m}[/tex]

I'll use the Planck-Einstein relations: [itex]E=\hbar \omega[/itex] and [itex]\vec p = \hbar \vec k[/itex].
Applying them above we immediately see that [itex]E=p^2/2m[/itex] as it should classically.

Now (let's take it in 1D), the phase velocity is:
[tex]\frac{\omega}{k}=\frac{\hbar k}{2m}=\frac{p}{2m}[/tex]
so it goes at half the speed the it is supposed to go classically. You need the GROUP velocity though:

[tex]\frac{d\omega}{dk}=\frac{\hbar k}{m}=\frac{p}{m}[/tex]
which corresponds to the classical speed of the particle.
 
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