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Homework Help: Group velocity of EMW

  1. Dec 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Electromagnetic wave, with the frequency [itex]\omega[/itex], travels through the medium for which:
    [itex]\epsilon=\epsilon_0\left(1+\frac{\omega_p^2}{\omega_0^2-\omega^2}\right)[/itex]
    where [itex]\omega_p[/itex] and [itex]\omega_0[/itex] are constants. How does the expression for the group velocity [itex]v_g(\omega)[/itex] looks like? What condition do the constants need to satisfy so that the relativity principle [itex]v_g\leq c[/itex] should hold?


    2. Relevant equations
    The expression for group velocity of the travelling wave is [itex]v_g=\frac{d\omega}{dk}[/itex]
    where [itex]\omega[/itex] is angular frequency and k is wave number (vector in 3D).


    3. The attempt at a solution

    The problem is I don't really know where to start. I have the expression for electric permittivity, and I don't know how to connect that with group velocity of the emw. Does any one know how to start, and what to look? Thnx!
     
  2. jcsd
  3. Dec 16, 2009 #2
    You (should) know that the phase velocity is also found from

    [tex]
    v_p=c\cdot n(\omega)
    [/tex]

    Do you know of any relations between the dielectric permittivity you are given and the real refractive index [itex]n(\omega)[/itex]?
     
  4. Dec 16, 2009 #3
    We have learned that [itex]v_p=\frac{c}{n(k)}[/itex], and our substitute assistant said sth about:

    [itex]k^2=\mu\epsilon\frac{\omega^2}{c^2}[/itex]

    And he said that that was form our general physics class when we learned about electricity and magnetism, but I really can't recall when and where did we learn that :\

    EDIT: No, that first term I found in Jacson: Classical electrodynamics, and this problem is from general physics class (sth like 8.03 in MIT)
     
  5. Dec 16, 2009 #4

    gabbagabbahey

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    This is all you need to solve this problem...For most media, [itex]\mu\approx\mu_0[/itex], so I think that is a fair assumption to make here (Or at least assume that [itex]\mu[/itex] is independent of [itex]\omega[/itex]). After that, just substitute in your expression for [itex]\epsilon[/itex] and find [itex]\frac{d\omega}{d k}[/itex] through implicit differentiation of the above expression.

    Side note: The above expression can be easily derived by looking at the fields of a plane wave (with wavenumber [itex]k[/itex] and angular frequency [itex]\omega[/itex]) and seeing what conditions are necessary for such fields to satisfy Maxwell's equations (or, equivalently, the wave equation) in a medium with permittivity [itex]\epsilon[/itex].
     
  6. Dec 16, 2009 #5
    Whoops, I was thinking phase velocity and you wanted group velocity....
     
  7. Dec 16, 2009 #6
    Thank you very much! I'll do that! :)
     
  8. Dec 16, 2009 #7

    gabbagabbahey

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    Sorry, I just noticed your expression is actually incorrect....it should really be:

    [tex]k^2=\mu\epsilon \omega^2=\mu_r\epsilon_r \frac{\omega^2}{c^2}[/tex]

    (And of course, this is only true for a non-conducting, source-free medium)
     
  9. Dec 17, 2009 #8
    Yeah, because [itex]\mu =\mu_0\mu_r[/itex]? Right? And [itex]\mu_0\epsilon_0=\frac{1}{c^2}[/itex]...

    EDIT: After I put my original eq. into the [tex]k^2=\mu\epsilon \omega^2=\mu_r\epsilon_r \frac{\omega^2}{c^2}[/tex] I get quadratic equation for [tex]\omega_{1,2}^2[/tex]. So I get my solutions and derive those with the respect to k?

    [tex]v_g=\frac{d\omega}{dk}=\frac{1}{2\omega}\frac{d\omega^2}{dk}[/tex]?
     
    Last edited: Dec 17, 2009
  10. Dec 17, 2009 #9

    gabbagabbahey

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    Yup.

    That's one way to do it (although [itex]\omega[/itex] is positive by definition, so I suspect you can throw away one of your solutions), but I suggest an easier method is as follows:

    [tex]k^2=f(\omega)\implies 2k=f'(\omega)\frac{d\omega}{d k}\implies \frac{d\omega}{d k}=2\frac{\sqrt{f(\omega)}}{f'(\omega)} [/tex]
     
  11. Dec 17, 2009 #10
    Oh, ok! :D So I put my expression for [tex]\epsilon[/tex] in the [tex]k^2[/tex] expression and derive it, and I get my [tex]v_g=\frac{d\omega}{dk}[/tex] immediately? That's great!

    Thank you very much!!

    P.S. I thought about asking you about that last expression you got, but then I wrote it down and I figured how you got it :) Silly me :D

    EDIT: I have solved it and it came out as:

    [tex]v_g=\frac{d\omega}{dk}=\frac{(\omega_0^2-\omega^2)\sqrt{\varepsilon_0\mu\omega^2\left(1+\frac{\omega_p^2}{\omega_0^2-\omega^2}\right)}}{\varepsilon_0\mu\omega(\omega_0^4+\omega^4+\omega_0^2(\omega_p^2-2\omega^2))}[/tex]

    What conditions should [tex]\omega_p[/tex] and [tex]\omega_0[/tex] satisfy so that my group velocity is smaller or equal than c? Do I solve the inequality?
     
    Last edited: Dec 17, 2009
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