# Homework Help: Group velocity for a discrete superposition of waves

1. May 31, 2005

### quasar987

For a superposition of two since waves of equal amplitude in a dispersive media, we find that the group velocity is given exactly by

$$v_g = \frac{\omega_2-\omega_1}{k_2-k_1}$$

and approximately by $d\omega / dk|_{k=k_0}$.

How do we show that this approximation holds for any type of waves (not just sine), and for a superposition of any number of them?

Here's the developement that allowed me to conclude to the statement above. You may not read this.

In the 2 sine wave superposition problem (beats), we have two progressive waves of equal amplitude, $y_1(x,t) = Asin(k_1x - \omega_1t)$ and $y_2(x,t) = Asin(k_2x - \omega_2t)$ and their superposition is

$$y(x,t)=2Acos\left(\frac{k_2-k_1}{2}x-\frac{\omega_2-\omega_1}{2}t\right) sin\left(\frac{k_2+k_1}{2}x-\frac{\omega_2+\omega_1}{2}t\right)$$

The speed $v_g$ of the modulation envelope is found by setting $x=x(t)=v_g t$ in $cos(\frac{k_2-k_1}{2}x-\frac{\omega_2-\omega_1}{2}t)$ and arguing that since x(t) moves at the same speed as the cos wave, the phase is constant. I.e.,

$$\frac{k_2-k_1}{2}v_g t -\frac{\omega_2-\omega_1}{2}t = \mbox{cst}$$

and taking the derivative wrt t gives

$$v_g = \frac{\omega_2-\omega_1}{k_2-k_1}$$

This gives $v_g$ exactly. Here, the distribution of amplitude wrt k is discrete rather than continuous as in the case of the wave packet, but extending the definition of $k_0$ to the discrete case we have,

$$k_0 = \frac{|k_2+k_1|}{2}$$

So evidently, as $k_1$ approaches $k_2$, by definition of derivative, $v_g$ approaches $d\omega / dk|_{k=k_0}$, but the equality $v_g = d\omega / dk|_{k=k_0}$ is not exact! For nonlinear dispersion relations, this is only an approximation.

Last edited by a moderator: Jan 14, 2014
2. Jun 14, 2005

### Claude Bile

First and foremost, group velocity is dw/dk by definition.

There is confusion here as to why two discrete frequencies, when analysed, do not give the correct group velocity. The reason why, is because it is not intended to. Group velocity is a term we apply to continuous spectra in recognition of the fact that no light source has a single frequency, but is comprised of a frequency bandwidth. Thus in the case where k1 approaches k2 (i.e. where the seperate frequencies merge into a continuous spectra), the solution is equal to dw/dk by definition of a derivative. Now that the expression for the group velocity is a derivative, it applies to any continuous spectrum.

For nonlinear dispersion relations, the definition of group velocity is retained, however a distortion parameter is introduced to provide the required corrections.

Claude.

3. Jun 17, 2005

### Meir Achuz

I am attaching the simplest derivation I know of.
As you can see, the validity of dw/dk as a velocity of the envelope depends on how good an approximation it is to truncate the Taylor expansion.
The next term in the expansion leads to a spread of the wave packet.
If higher terms are needed, distortion of the packet is rapid.

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4. Jun 17, 2005

### Meir Achuz

"But what if the waves were not sine-like? And what if the envelope was not Gaussian? Isn't there a totally general way to approach the notion of group velocity? Or do we have to treat case by case?
Does v_g = dw/dk hold for more than 2 superposed sin waves? Does it hold for non-sinusoidal sine waves? If so, how can this be shown?"

The Gaussian envelope was just one simple example. The derivation for v_g is the same (totally general way) for any shape of envelope. The expansion in sine waves (really exponential waves) is also completely general. The simple case of two sine waves in elementary books is a special case where the Fourier integral becomes a Fourier sum. You should study Fourier sums and integrals in a book like Boas, Butkov, or Arfken.

5. Jun 17, 2005

### quasar987

Thx for the recomendations.