Waves/Group Velocity

Waves : 1 Dimensional Waves
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Group Velocity

We now ask the following question: How fast do wave packets move? Surprisingly, we often find that wave packets move at a speed very different from the phase speed, $c=\omega /k$, of the wave composing the wave packet.

We shall find that the speed of motion of wave packets, referred to as the group velocity, is given by

$u=\ 0.1{\mbox{m}}^{-1}$. The phase speed of deep ocean waves is $c=(g/k)^{1/2}$. However, since $c\equiv \omega /k$, we find the frequency of deep ocean waves to be $\omega =(gk)^{1/2}$. The group velocity is therefore $u\equiv d\omega /dk=(g/k)^{1/2}/2=c/2$. For the specified central wavenumber, we find that $u\approx (9.8{\mbox{ m}}{\mbox{ s}}^{-2}/0.1{\mbox{ m}}^{-1})^{1/2}/2\approx 5{\mbox{ m}}{\mbox{ s}}^{-1}$. By contrast, the phase speed of deep ocean waves with this wavelength is $c\approx 10{\mbox{ m}}{\mbox{ s}}^{-1}$.

Dispersive waves are waves in which the phase speed varies with wavenumber. It is easy to show that dispersive waves have unequal phase and group velocities, while these velocities are equal for non-dispersive waves.

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

$A=2\sin(k_{0}x-\omega _{0}t)\cos(\Delta kx-\Delta \omega t),$ (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:

$\Delta kx-\Delta \omega t=const.$ (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

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

which is the desired result.

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Waves : 1 Dimensional Waves
1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13
Examples - Problems - Solutions - Terminology