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Homework Help: Frequency dependent permittivity

  1. Nov 16, 2008 #1
    I am given the permittivity to be [tex] \epsilon(\omega) = 1+ \frac{\omega_p^2}{\omega_0^2 -\omega^2} [/tex]

    I am asked to sketch [itex]k[/itex] vs [itex]\omega[/itex] using the dispersion relation [tex]k^2 =\frac{\omega^2 \epsilon(\omega)}{c^2} [/tex]

    here is what I have:

    [tex]k=\frac{\omega}{c} \sqrt{\epsilon(\omega)}[/tex]
    [tex]k=\frac{\omega}{c} \sqrt{1+ \frac{\omega_p^2}{\omega_0^2 -\omega^2}}[/tex]
    [tex]\sqrt{1+x} = 1+\frac{x}{2} +...[/tex]
    [tex]k=\frac{\omega}{c} \left( 1+ \frac{1}{2} \left( \frac{\omega_p^2}{\omega_0^2 -\omega^2}}\right) \right)[/tex]

    plotting this, i get something like: http://sites.google.com/site/question1site/" [Broken]

    I'm not sure if I should expand this or not. Or what exactly accounts for the differences between the two (expanded vs not expanded). Also, I do not see how [itex]\omega_p[/itex] will play a role in the graphs. As in, I know that the asymtotes of the graph are the resonate frequency, but I do not know where to place [itex]\omega_p[/itex].

    The second part of the question asks for the plot of [itex]\omega[/itex] vs [itex]k[/itex]

    In the link above, I have rotated and inversed the graph to get [itex]\omega[/itex] vs [itex]k[/itex]

    I need to show that for k>0, there are two allowed values of [itex]\omega[/itex]. We see this clearly as there are two angular frequencies for when k>0 (since graph is not one to one).

    The question then asks to show that at small k and large k, one of the two modes will have a dispersion relation similar to an EM wave in the vacuum, i.e. [itex]\omega = v_p k[/itex] where [itex]v_p[/itex] is weakly dependent on k and that the other mode has a frequency [itex]\omega[/itex] that is (to lowest order) independent of k. I'm not sure what to do here. Am I supposed to see this from the graph? Or should I solve for ω in terms for k to answer the questions above?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 17, 2008 #2
    any ideas?
     
  4. Nov 17, 2008 #3

    gabbagabbahey

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    Why are you including negative values of [itex]\omega[/itex] in your plot? Are such values actually physical?:wink:

    I see no reason to Taylor expand this unless you are asked to examine the behavior of this function near a specific point.

    [itex]\omega_p[/itex] seems to just be a scaling factor that determines how quickly/slowly [itex]k[/itex] increases/decreases with changing omega.

    Again-- negative omega? :confused:

    I'm not sure pointing at the graph will be enough for your prof....you might also want to solve the equation
    [tex]k=\frac{\omega}{c} \sqrt{1+ \frac{\omega_p^2}{\omega_0^2 -\omega^2}}[/tex] for [itex]\omega[/itex]....you will of course have to solve a quadratic, which you can show gives two real values by looking at the discriminant of said quadratic.

    Once you solve for omega, then taylor expand about k=0 to look at the behavior for small k....what do you get for your two solutions in this case? :wink:
     
    Last edited by a moderator: May 3, 2017
  5. Nov 17, 2008 #4
    Instead of a quadratic equation, I get a quartic equation.

    [tex]\omega^4 + \omega^2 \left( -\omega_0^2 - \omega_p^2 -c^2 k^2 \right) + c^2 k^2 \omega_0^2 =0[/tex]

    If we let omega^2 = omega', we can solve it quadratically:

    [tex]\omega'=\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2} [/tex]

    thus:

    [tex]\omega_1=\sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 + \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}} [/tex]
    [tex]\omega_2=-\sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 + \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}} [/tex]
    [tex]\omega_3=\sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 - \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}} [/tex]
    [tex]\omega_4=-\sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 - \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}} [/tex]


    I dont see where to go from here? How can one determine the which is real?
     
  6. Nov 17, 2008 #5

    gabbagabbahey

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    Let's take a look at [itex]\omega^2[/itex] first....clearly as long as it is positive and real, omega will be real....so let's see when [itex]\omega^2[/itex] is positive and real:

    [tex]\omega^2=\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}
    [/tex]

    Let's call [itex]\kappa \equiv \omega_0^2 + \omega_p^2 +c^2 k^2[/itex] to make things easier...clearly as long as [itex]\omega_p[/itex], [itex]\omega_0[/itex] and [itex]k[/itex] are assumed to be real, [itex]\kappa>0[/itex]....and looking at omega square again:

    [tex]\omega^2=\frac{\kappa \pm \sqrt{\kappa^2 -4( c^2 k^2 \omega_0^2)}}{2}
    [/tex]

    we see that [itex]\sqrt{\kappa^2 -4( c^2 k^2 \omega_0^2)}}[/itex] will always be less than or equal to [itex]\kappa[/itex] (as long as this quantity is real), so omega squared will never be real and negative....that means we only need to find where omega squared is real.....When is that?
     
  7. Nov 18, 2008 #6
    omega squared is real when [itex]\kappa^2 >4( c^2 k^2 \omega_0^2)[/itex]?
     
  8. Nov 18, 2008 #7

    gabbagabbahey

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    Now that I think about it a little bit, I think it's safe to just assume [itex]\omega \in \Re[/itex] and [itex]\omega \geq 0[/itex] on physical grounds....that rules out your [itex]\omega_2[/itex] and [itex]\omega_4[/itex] solutions since they are always negative, and leaves you with two solutions at most....if [tex]\kappa^2-4c^2k^2\omega_0^2 = 0[/tex], then you will only have one solution, but you should be able to show that this never happens for k>0. (just solve the resulting quadratic for k^2 (its a quartic in k)...you'll find that there are no real solutions, so k>0 precludes the possibility of there being only one solution for omega.)
     
  9. Nov 18, 2008 #8
    ok, I'm going to try and taylor expand this:

    [tex]
    \omega^2=\frac{\kappa \pm \sqrt{\kappa^2 -4( c^2 k^2 \omega_0^2)}}{2}
    [/tex]


    I am getting that both frequencies do not depend on k (to lowest order):

    http://sites.google.com/site/question2site/

    I'm not sure where to get a frequency that is weakly dependent on k since both frequencies are independent of k for first order. And what exactly does it mean to have omenga weakly dependent on k?
     
  10. Nov 18, 2008 #9

    gabbagabbahey

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    Ugghh....you can cleanup your expressions Out[73] and Out[74] quite a bit by noting that

    [tex]\sqrt{\omega_0^4+2\omega_0^2\omega_p^2+\omega_p^4}=\sqrt{(\omega_0^2+\omega_p^2)^2}=\omega_0^2+\omega_p^2[/tex]

    You should get:

    [tex]\text{Out[73]}=\sqrt{\omega_0^2+\omega_p^2}+\frac{c^2 \omega_p^2}{2(\omega_0^2+\omega_p^2)^{3/2}}k^2 +\ldots \approx \sqrt{\omega_0^2+\omega_p^2} [/tex]

    Which gives you your constant term....And for Out[74], you need to explicitly assume that all your variables are >0 before calculating the series with mathematica (or you can do it by hand)....you should get:

    [tex]\frac{\omega_0 c k}{\sqrt{\omega_0^2+\omega_p^2}}+\ldots[/tex]

    which is dependent on k; and since omega_p is usually MUCH bigger than omega_0 for most mediums, your coefficient is typically pretty small, and so that frequency is 'weakly dependent' on k....try doing the series again while assuming the c>0,omega_0>0, omega_p>0 and k>0.
     
    Last edited: Nov 18, 2008
  11. Nov 18, 2008 #10
    Sorry, I'm doing the expansion by hand and see if I can get what you got.

    I'm also asked to show that the two modes exchange their characteristic behavior as one crosses from small k to large k. I think I was supposed to do the expansion for both small k and large k. For each case, what can be ignored?

    I was thinking for small k, the k^4 term can be ignored, but that means I would have to expand [itex](-\omega_0^2 - \omega_p^2 -c^2 k^2)^2[/itex]. Is there a better way of doing the approximation? If we ignored k^2 then the whole term would just be a constant.

    For large k, I was thinking about ignoring the omega_0 and omega_p terms:

    [tex]
    \omega_1=\sqrt{\frac{c^2 k^2 + \sqrt{(c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}} [/tex]

    would this work?
     
  12. Nov 18, 2008 #11

    gabbagabbahey

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    For large k, just Taylor expand about the point (1/k)=0

    Don't ignore any terms until after you have calculated the series....in each case, just keep the lowest order non-zero term in the series.
     
    Last edited: Nov 18, 2008
  13. Nov 19, 2008 #12

    gabbagabbahey

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    For large k, start by factoring out a k:

    [tex]\omega_{\pm}=k\sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2k^2}}=k\sqrt{\frac{\omega_0^2\delta^2 + \omega_p^2\delta^2 +c^2 \pm \sqrt{(-\omega_0^2\delta^2 - \omega_p^2\delta^2 -c^2)^2 -4( c^2 \delta^2 \omega_0^2)}}{2}}[/tex]

    where [tex]\delta \equiv \frac{1}{k}[/tex]....for large k, delta is small....so taylor expand this in powers of delta and keep the smallest order non-zero term....
     
  14. Nov 19, 2008 #13
    I tried using mathematica and I cant seem to get the result of [tex]
    \frac{\omega_0 c k}{\sqrt{\omega_0^2+\omega_p^2}}+\ldots[/tex]

    I get:

    http://sites.google.com/site/question2site/

    For large k, I get [tex]\omega_+ = kc[/tex] and I was going to use mathematica to get the [tex]\delta[/tex] dependence of [tex]\omega_-[/tex].
     
  15. Nov 19, 2008 #14

    gabbagabbahey

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    The way you've written your mathematica code, your telling it to make the assumptions after calculating the series....instead try doing
    Code (Text):
    Assuming[{k>0,c>0,[itex]\omega_p[/itex]>0,[itex]\omega_0[/itex]>0},Series[...]]
    That tells mathematica to evaluate the series while making those assumptions
     
  16. Nov 19, 2008 #15

    gabbagabbahey

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    To do it by hand, for small k, just let

    [tex]f_{\pm}(k)\equiv \sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}}[/tex]

    Then use the formula for a taylor expansion about k=0:

    [tex]f(k)=f(0)+\frac{f'(0)}{1!}k+\frac{f''(0)}{2!}k^2 + \ldots[/tex]

    what are you getting for [itex]f_{\pm}'(0)[/itex] and[itex]f_{\pm}''(0)[/itex]?
     
  17. Nov 19, 2008 #16
    [tex]
    f_{\pm}'=\frac{1}{2} \left( \frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2} \right)^{-\frac{1}{2}} \left( c^2 k \pm \frac{1}{2} \left( \frac{1}{2} \left( \omega_0^2 + \omega_p^2 +c^2 k^2 - 4 c^2 k^2 \omega_0^2 \right)^{-\frac{1}{2}} \left(2c^2k-8c^2k \omega_0^2\right) \right) [/tex]

    [tex]
    f_{\pm}''=-\frac{1}{4} \left( \frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2} \right)^{-\frac{3}{2}} \left( c^2 \pm \frac{1}{2} \left( \frac{-1}{4} \left( \omega_0^2 + \omega_p^2 +c^2 k^2 - 4 c^2 k^2 \omega_0^2 \right)^{-\frac{3}{2}} \left(2c^2k-8c^2k \omega_0^2\right)\right)\left(2c^2-8c^2 \omega_0^2\right) \right) [/tex]

    I get f'(0)=0
    f''_-=0
    [tex]f''_+=\frac{-c^2}{4(\omega_0^2+\omega_p^2)^{3/2}}[/tex]
     
  18. Nov 19, 2008 #17

    gabbagabbahey

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    Hmmm...now that I think about, I see why mathematica is having a problem...when I use this method by hand I get:

    [tex]f_{-}'(0)=\infty[/tex].... ([itex]0^{-1/2}[/itex] is undefined)

    Try expanding [tex]\sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}[/tex] first and keep terms up to order k^2....that should give you:

    [tex]\omega_{\pm}(k) \approx \sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \left( \omega_0^2 + \omega_p^2+c^2k^2 -\frac{2c^2k^2\omega_0^2}{\omega_0^2 + \omega_p^2}\right) }{2}}[/tex]

    ....that should make thing easier for you!
     
    Last edited: Nov 19, 2008
  19. Nov 19, 2008 #18
    [tex]
    \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}[/tex]
    [tex]
    = (\omega_0^2 + \omega_p^2 +c^2 k^2) \sqrt{1 -\frac{4( c^2 k^2 \omega_0^2)}{(\omega_0^2 + \omega_p^2 +c^2 k^2)}}[/tex]
    [tex]
    \approx (\omega_0^2 + \omega_p^2 +c^2 k^2) \left( 1 -\frac{2( c^2 k^2 \omega_0^2)}{(\omega_0^2 + \omega_p^2 +c^2 k^2)} \right)[/tex]

    I dont see where [tex]\omega_0^2 + \omega_p^2[/tex] appears in the denominator
     
  20. Nov 19, 2008 #19

    gabbagabbahey

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    What you've done is expand it to first order in powers of [tex]\frac{4( c^2 k^2 \omega_0^2)}{(\omega_0^2 + \omega_p^2 +c^2 k^2)}[/tex] not to second order in powers of k...instead use taylor expansion:

    [tex]g(k)=\sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)} \approx g(0)+g'(0)k+\frac{g''(0)}{2!} k^2[/tex]
     
  21. Nov 19, 2008 #20
    [tex]g(0)=\omega_0^2 + \omega_p^2[/tex]
    [tex]g'(0)k=0[/tex]
    [tex]\frac{g''(0)}{2!} k^2 =c^2k^2 \frac{\omega_p^2-\omega_0^2}{\omega_0^2 + \omega_p^2}[/tex]

    Used mathematica for second derivative:
    http://sites.google.com/site/question2site/

    not sure if this matches up
     
    Last edited: Nov 19, 2008
  22. Nov 19, 2008 #21

    gabbagabbahey

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    Your first two are correct, but I get:

    [tex]\frac{g''(0)}{2!} k^2 =c^2k^2 -\frac{2c^2k^2\omega_0^2}{\omega_0^2 + \omega_p^2}[/tex]

    Mathematica agrees with me, so I would check your math.

    Edit- our two expressions are equivalent :smile:
     
    Last edited: Nov 19, 2008
  23. Nov 19, 2008 #22
    [tex]
    \omega_{\pm}=k\sqrt{\frac{\omega_0^2\delta ^2 + \omega_p^2\delta^2 +c^2 \pm \sqrt{(-\omega_0^2\delta^2 - \omega_p^2\delta^2 -c^2)^2 -4( c^2 \delta^2 \omega_0^2)}}{2}}[/tex]

    Here's what I get:
    [tex]
    \omega_+ \approx kc[/tex]

    [tex]\sqrt{(-\omega_0^2\delta^2 - \omega_p^2\delta^2 -c^2)^2 -4( c^2 \delta^2 \omega_0^2)} \approx c^2+\delta^2(\omega_p^2 - \omega_0^2) [/tex]

    [tex]
    \omega_{-}\approx k\sqrt{\frac{\omega_0^2\delta ^2 + \omega_p^2\delta^2 +c^2 -(c^2+\delta^2(\omega_p^2 - \omega_0^2))}{2}}[/tex]

    [tex] \omega_{-} \approx k \delta \omega_0 = \omega_0 [/tex]

    In summary:

    small k:

    [tex]\omega_+ \approx \sqrt{\omega_0^2 + \omega_p^2} [/tex]

    [tex]\omega_- \approx \frac{\omega_0 c k}{\sqrt{\omega_0^2 + \omega_p^2}} [/tex]

    large k:

    [tex]\omega_+ \approx kc [/tex]

    [tex]\omega_- \approx \omega_0 [/tex]

    When the question asks to to show that the two modes exchange their characteristic behavior as one crosses from small k to large k, is it enough to say that:

    [tex]\omega_+[/tex] at small k is independent of k, and for large k, there is a k dependence, and vice versa for [tex]\omega_-[/tex] ?
     
  24. Nov 19, 2008 #23

    gabbagabbahey

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    Looks good to me, but I would maybe say "At small k, the [itex]\omega_{-}[/itex] mode is a constant and [itex]\omega_{+}[/itex] is linearly dependent on k; while for large k, these modes switch and the [itex]\omega_{+}[/itex] mode is a constant and [itex]\omega_{-}[/itex] is linearly dependent on k."
     
  25. Nov 20, 2008 #24
    Thanks for the check. The last part of the question asks to show that at a given value of k, the higher frequency mode has a phase velocity [itex]v_p > c[/itex] and the lower frequency mode [itex]vp < c[/itex]. Lastly, the question asks to show for the high and low frequency modes that the group velocity is always [itex]vg < c[/itex]. For this part, should this be done for frequencies of small k and large k as well? And what about the frequencies that do not depend on k? Or should the original (non-expanded) form of omega be used?
     
    Last edited: Nov 20, 2008
  26. Nov 20, 2008 #25
    I can answer the question above for:
    [tex]
    \omega_- \approx \frac{\omega_0 c k}{\sqrt{\omega_0^2 + \omega_p^2}} [/tex]

    and

    [tex]

    \omega_+ \approx kc[/tex]

    however, for the constant omegas, I am having trouble. Proving the above becomes hard specifically for small k high frequency, and large k low frequency, since the values do not depend on k.

    As an example, here is what I tried for [tex]\omega_- \approx \omega_0 [/tex]
    Since this is a constant, dividing that by k to get the phase velocity doesnt tell us if it's greater or less than c. What I tried was to get higher order terms:

    The expansion of [tex]\sqrt{(-\omega_0^2\delta^2 - \omega_p^2\delta^2 -c^2)^2 -4( c^2 \delta^2 \omega_0^2)}[/tex] gives:

    [tex]g(0)=c^2[/tex]
    [tex]g'(0)k=0[/tex]
    [tex]g''(0)k^2/2=\delta^2 (\omega_p^2-\omega_0^2) [/tex]
    [tex]g'''(0)=0[/tex]
    [tex]g^{(4)} k^4/4! = \frac{2 \omega_0^2 \omega_p^2 \delta^4}{c^2} [/tex]


    [tex]

    \omega_{-}\approx k\sqrt{\frac{\omega_0^2\delta ^2 + \omega_p^2\delta^2 +c^2 -\left(c^2+\delta^2(\omega_p^2 - \omega_0^2)+\frac{2 \omega_0^2 \omega_p^2 \delta^4}{c^2} \right)}{2}}[/tex]

    [tex]=\omega_0 \sqrt{1+\left( \frac{\omega_p}{ck} \right) ^2} [/tex]

    [tex]\approx \omega_0+\frac{\omega_0 \omega_p^2}{2k^2} [/tex]

    I did not see how the extra term gave any help to see whether vp was greater or less than c.

    My final idea was to just say this:

    for large k, [tex]\omega_-[/tex] is a constant. The phase velocity is [tex]\frac{\omega_0}{k} [/tex] so that if k increases large enough, it will reach a value when [tex]\frac{\omega_0}{k}<c [/tex] thus, proving [itex]vp < c[/itex] for loq frequencies. Similarly for [tex]\omega_+[/tex], it's a constant divided by k, so that if k is small enough, the phase velocity will be greater than c.

    I am not sure if there is any other way to prove this. Are there any other options?
     
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