# Group speed in a dielectric

tomwilliam2

## Homework Statement

I'm given the refractive index of a piece of glass:
$$n(\omega)=A+B\omega$$
And I have to find the speed at which a pulse of radiation will travel through the glass at an angular frequency $$\omega = 1.2 \times 10^{15} s^{-1}$$
I also have A = 1.4, B=3.00 x 10^-17.

## Homework Equations

I know that the pulse speed should be given by group speed, where

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

as opposed to

$$v_{phase}=\omega / k$$

## The Attempt at a Solution

$$k = \frac{n(\omega)\omega}{c}$$
$$v_{group}^{-1}=\frac{dk}{d\omega} = \frac{n(\omega)+n^{'}(\omega)\omega}{c}$$
$$v_{group}=\frac{c}{(A+B\omega)+B\omega}=\frac{c}{A+2B\omega}$$

Then I just put the figures in and get an answer. However, I have checked this answer by considering:

$$v_{phase} \times v_{group} = c^2$$

Using the formula for $$v_{phase}$$ above, and I don't seem to get the right answer. Is this identity above not always true?

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Staff Emeritus
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I've never seen that relationship between the group and phase velocities before. I don't see why it would be true in general.

Homework Helper
Gold Member
$$v_{phase} \times v_{group} = c^2$$ ... Is this identity above not always true?

No, it's not true in most situations. For example, suppose you had a nondispersive medium where the index of refraction, n, is a constant. What would $v_{phase} \times v_{group}$ equal?

You might be thinking of deBroglie waves.

The quantity $n(\omega) + \omega n'(\omega)$ is sometimes called the "group index" $n_g(\omega)$. See http://www.rp-photonics.com/group_index.html . In your post you showed that the group velocity is generally equal to $c/n_g(\omega)$. So, $v_{phase} \times v_{group} = c^2/(n(\omega)n_g(\omega))$

As an exercise, you could work out the specific dispersion relation $n(\omega)$ that would make $v_{phase} \times v_{group} = c^2$.

tomwilliam2
Rereading my textbook I think I picked up the identity
$$v_{phase} \times v_{group} = c^2$$
As being true for TE waves in a waveguide, which means that I was applying it wrongly.

Presumably the dispersion relation required for the identity to hold otherwise would be
$$n(\omega)n_g(\omega) = 1$$

Could someone just confirm that for the situation described in my question, I really should be using the group speed? The only really difficult bit to the question is knowing whether the physical situation requires group or phase speed, and that is of course the most important point for my forthcoming exam.
If there is a "pulse" of radiation, this suggests group speed to me, but a colleague has argued that because a specific frequency is given, it must be a single wave and therefore it is the phase speed that's asked for. It's very confusing to a poor student.

Thanks again

Homework Helper
Gold Member
Just my opinion: I think the group velocity is what you want since you are dealing with a pulse. The pulse can be thought of as formed from a superposition of sinusoidal waves of different frequencies. The frequency given in the problem would be the dominant frequency of the pulse and the group velocity would be evaluated at that frequency. Due to the dispersion, the pulse would change shape as it propagates and the time it takes the pulse to travel a specified distance would be somewhat ambiguous.

tomwilliam2
Thanks, much appreciated.