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Group Velocity of Waves in Gas Problem

  1. Sep 18, 2013 #1
    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 [itex]k = n^{2}[/itex].

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

    [itex]\frac{c}{\sqrt{k}}(1 - \frac{ω}{2k}\frac{dk}{dω}[/itex]

    where c is the speed of light in vacuum.

    2. Relevant equations

    [itex]v_{g} = v_{p}(1 + \frac{λ}{n}\frac{dn}{dλ})[/itex] (1)
    [itex]v_{p} = \frac{c}{n}[/itex] (2)

    3. The attempt at a solution

    Plugging (2) into one

    [itex]v_{g} = \frac{c}{n}(1 + \frac{λ}{n}\frac{dn}{dλ})[/itex] (3)

    Taking the given information and solving for n

    [itex]k = n^{2}, n = \sqrt{k}[/itex]

    Plugging this into (3)

    [itex]v_{g} = \frac{c}{sqrt(k)}(1 + \frac{λ}{sqrt(k)}\frac{dn}{dλ})[/itex]

    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. jcsd
  3. Sep 18, 2013 #2
    This is a chain rule exercise.
     
  4. Sep 19, 2013 #3
    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.
     
  5. Sep 19, 2013 #4
    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.
     
  6. Sep 19, 2013 #5
    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).
     
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