# Group Velocity of Waves in Gas Problem

1. Sep 18, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

The dielectric constant k of a gas is related to its index of refraction by the relation $k = n^{2}$.

a. Show that the group velocity for waves traveling in the gas may be expressed in terms of the dielectric constant by

$\frac{c}{\sqrt{k}}(1 - \frac{ω}{2k}\frac{dk}{dω}$

where c is the speed of light in vacuum.

2. Relevant equations

$v_{g} = v_{p}(1 + \frac{λ}{n}\frac{dn}{dλ})$ (1)
$v_{p} = \frac{c}{n}$ (2)

3. The attempt at a solution

Plugging (2) into one

$v_{g} = \frac{c}{n}(1 + \frac{λ}{n}\frac{dn}{dλ})$ (3)

Taking the given information and solving for n

$k = n^{2}, n = \sqrt{k}$

Plugging this into (3)

$v_{g} = \frac{c}{sqrt(k)}(1 + \frac{λ}{sqrt(k)}\frac{dn}{dλ})$

I'm not really sure where to go from here. I would imagine I need to some how find λ as a function of n and take the derivative of this function. I would imagine that this function is also a function of ω and k in some way. I'm not sure of what this equation is though. I have looked through my book in the chapter in which this problem was given and can find no equation.

Thanks for any help.

2. Sep 18, 2013

### voko

This is a chain rule exercise.

3. Sep 19, 2013

### GreenPrint

I'm not really sure how because I don't have an equation for ω and am not really sure how it related to the problem. Is it rotational speed in this case or another variable? I don't see how rotational speed relates to k in this case and I believe is where I'm getting stuck.

4. Sep 19, 2013

### nasu

ω is the angular frequency.
ω=2∏f where f is the frequency in Hz.
Now you should know a relationship between λ and ω.

5. Sep 19, 2013

### voko

Even though nasu explained how you should proceed, I would like to point out that you should have said that you do not understand the meaning of $omega$ in the original post. Obviously, if one does not even understand the specification of the problem, trying anything could solve it only by chance (unlikely).