Question

7. (i) The dispersion relation for surface water waves on deep water is given by w? = gk. Show that the group velocity is Vg = g/(2w). Derive an expression relating the phase velocity Vp and the group velocity vg and hence determine if the dispersion is normal or anomalous. What are the physical significances of Vg and vp? [7 marks] Now consider a propagating wavepacket with initial length Lo. (ii) Use the bandwidth theorem to show that the minimum range of angular frequen- cies present in the wavepacket is approximately Aw ~ [4 marks] (iii) Describe what is meant by group velocity dispersion and give an example of its consequences. [3 marks] (iv) The range of group velocities present in the wavepacket is Avg = |dvg/dwlAw. Use this to obtain an expression for the spreading AL of the wavepacket upon propagation and hence show that when AL » Lo the length of the resulting wavepacket is approximately La dvolx > dw to where x is the propagation distance. [5 marks] (v) A wavepacket of surface water waves has a centre wavelength 1 = 0.5 m and its initial length is Lo = 1 m. Calculate the time for the wavepacket to propagate 200 m and the length of the packet at this distance. [6 marks]

Answer 1

(i)The given dispersion relation is

$\omega^{2}=gk$

The group velocity is the velocity with which the envelope of a pulse propagates in the medium. This is the significance of group velocity and is given by,

$v_{g}=\frac{d\omega}{dk}$

So differentiation of the dispersion relation with respect to k, we get,

$2\omega\frac{d\omega}{dk}=g$

$2\omega v_{g}=g$

$v_{g}=\frac{g}{2\omega}$

Hence Proved.

The phase velocity which is the rate at which phase of a wave propagates in space. This is the significance of phase velocity and is given by,

$v_{p}=\frac{\omega}{k}$

Dividing our given dispersion relation by k,

$\frac{\omega^{2}}{k}=g$

using the value for phase velocity we rewrite it as,

$v_{p}=\frac{g}{\omega}$

and $v_{g}=\frac{g}{2\omega}$

$v_{g}=\frac{v_{p}}{2}$

This is the relation between phase velocity and group velocity.

To check for normal or anomalous dispersion we need to check the value of $\frac{\partial^2 k}{\partial \omega^2}$

If $\frac{\partial^2 k}{\partial \omega^2}>0$ it is normal dispersion

if $\frac{\partial^2 k}{\partial \omega^2}<0$ it is anomalous dispersion

Ꭷk _ Ꭴ ) aw

$\frac{\partial^2 k}{\partial \omega^2}=\frac{2}{g}$

This is greater than 0 so it is normal dispersion.

(ii) The bandwidth theorem is

        $\Delta \omega \Delta t> 2\pi$

This is similar to Heisenberg's uncertainty principle.

We know that phase velocity is given by,

$v_{g}=\frac{\Delta \omega }{\Delta k}$

But $Time=\frac{Distance}{Velocity}$ so,

$\Delta t=\frac{\Delta l }{\Delta v_{g}}$

So substituting in the bandwidth theorem,

$\Delta \omega>\frac{2\pi \Delta v_{g}}{\Delta l}$

We now make an approximation

$\Delta v_{g}\approx v_{g} and \Delta l=L_{0}$

So,

$\Delta \omega \approx \frac{v_{g}}{L_{0}}$

(iii) Group velocity dispersion is the phenomena in which the group velocity in a transparent medium is dependent on the optical frequency. As above the group velocity dispersion is given by $\frac{\partial^2 k}{\partial \omega^2}$ .

There are two types of dispersion, normal and anomalous. In normal dispersion, group velocity dispersion decreases for increasing optical frequency.

One example of it's consequence is that it causes temporal broadening which is a common dispersion effect.

(iv) As per the question,

$\Delta v_{g}\approx\left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right |\Delta \omega$

But since $Velocity=\frac{Distance}{Time}$ so,

$\Delta v_{g}=\frac{\Delta L}{\Delta t}$

$\Delta L\approx \left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right |\Delta \omega \Delta t$

But $\Delta \omega \Delta t> 2\pi$ from (ii) so,

$\Delta L\approx \left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right | 2\pi$

To prove the length of the resulting packet we use,

$\Delta L\approx \left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right |\Delta \omega \Delta t$ and $\Delta \omega \approx \frac{v_{g}}{L_{0}}$ to get

$\Delta L\approx \left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right |\frac{v_{g}\Delta t}{L_{0}}$

But $v_{g}=\frac{x}{t}$ and $\Delta t\approx t$ to get

$L-L_{0}\approx \left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right |\frac{x}{L_{0}}$

since change in L=Final L-Initial L

However from question we assume$L>>L_{0}$ so,

$L\approx \left | \frac{\mathrm{d} v_{g}}{\mathrm{d} \omega} \right |\frac{x}{L_{0}}$

Hence Proved.