Derivation of Group Velocity Formula

We now derive equation (1.36). It is easiest to do this for the simplest wave packets, namely those constructed out of the superposition of just two sine waves. We will proceed by adding two waves with full space and time dependence:

$$A = \sin ( k_1 x - \omega_1 t ) + \sin ( k_2 x - \omega_2 t )$$ (2.37)

After algebraic and trigonometric manipulations familiar from earlier sections, we find

$$A = 2 \sin ( k_0 x - \omega_0 t ) \cos ( \Delta k x - \Delta \omega t ) ,$$ (2.38)

where as before we have $k_0 = (k_1 + k_2 )/2$, $\omega_0 = ( \omega_1 + \omega_2 )/2$, $\Delta k = (k_2 - k_1 )/2$, and $\Delta \omega = ( \omega_2 - \omega_1 )/2$. Again think of this as a sine wave of frequency $\omega_0$ and wavenumber $k_0$ modulated by a cosine function. In this case the modulation pattern moves with a speed so as to keep the argument of the cosine function constant:

$$\Delta k x - \Delta \omega t = const.$$ (2.39)

Differentiating this with respect to $t$ while holding $\Delta k$ and $\Delta \omega$ constant yields

$$u \equiv \frac{dx}{dt} = \frac{\Delta \omega}{\Delta k} .$$ (2.40)

In the limit in which the deltas become very small, this reduces to the derivative

$$u = \frac{d \omega}{d k} ,$$ (2.41)

which is the desired result.