# Frequency dependent permittivity

1. Nov 16, 2008

### LocationX

I am given the permittivity to be $$\epsilon(\omega) = 1+ \frac{\omega_p^2}{\omega_0^2 -\omega^2}$$

I am asked to sketch $k$ vs $\omega$ using the dispersion relation $$k^2 =\frac{\omega^2 \epsilon(\omega)}{c^2}$$

here is what I have:

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

plotting this, i get something like: link to plot

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 $\omega_p$ 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 $\omega_p$.

The second part of the question asks for the plot of $\omega$ vs $k$

In the link above, I have rotated and inversed the graph to get $\omega$ vs $k$

I need to show that for k>0, there are two allowed values of $\omega$. 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. $\omega = v_p k$ where $v_p$ is weakly dependent on k and that the other mode has a frequency $\omega$ 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?

2. Nov 17, 2008

### LocationX

any ideas?

3. Nov 17, 2008

### gabbagabbahey

Why are you including negative values of $\omega$ in your plot? Are such values actually physical?

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

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

Again-- negative omega?

I'm not sure pointing at the graph will be enough for your prof....you might also want to solve the equation
$$k=\frac{\omega}{c} \sqrt{1+ \frac{\omega_p^2}{\omega_0^2 -\omega^2}}$$ for $\omega$....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?

Last edited: Nov 17, 2008
4. Nov 17, 2008

### LocationX

$$\omega^4 + \omega^2 \left( -\omega_0^2 - \omega_p^2 -c^2 k^2 \right) + c^2 k^2 \omega_0^2 =0$$

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

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

thus:

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

I dont see where to go from here? How can one determine the which is real?

5. Nov 17, 2008

### gabbagabbahey

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

$$\omega^2=\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}$$

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

$$\omega^2=\frac{\kappa \pm \sqrt{\kappa^2 -4( c^2 k^2 \omega_0^2)}}{2}$$

we see that $\sqrt{\kappa^2 -4( c^2 k^2 \omega_0^2)}}$ will always be less than or equal to $\kappa$ (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?

6. Nov 18, 2008

### LocationX

omega squared is real when $\kappa^2 >4( c^2 k^2 \omega_0^2)$?

7. Nov 18, 2008

### gabbagabbahey

Now that I think about it a little bit, I think it's safe to just assume $\omega \in \Re$ and $\omega \geq 0$ on physical grounds....that rules out your $\omega_2$ and $\omega_4$ solutions since they are always negative, and leaves you with two solutions at most....if $$\kappa^2-4c^2k^2\omega_0^2 = 0$$, 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.)

8. Nov 18, 2008

### LocationX

ok, I'm going to try and taylor expand this:

$$\omega^2=\frac{\kappa \pm \sqrt{\kappa^2 -4( c^2 k^2 \omega_0^2)}}{2}$$

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

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?

9. Nov 18, 2008

### gabbagabbahey

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

$$\sqrt{\omega_0^4+2\omega_0^2\omega_p^2+\omega_p^4}=\sqrt{(\omega_0^2+\omega_p^2)^2}=\omega_0^2+\omega_p^2$$

You should get:

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

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:

$$\frac{\omega_0 c k}{\sqrt{\omega_0^2+\omega_p^2}}+\ldots$$

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
10. Nov 18, 2008

### LocationX

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 $(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2$. 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:

$$\omega_1=\sqrt{\frac{c^2 k^2 + \sqrt{(c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}}$$

would this work?

11. Nov 18, 2008

### gabbagabbahey

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
12. Nov 19, 2008

### gabbagabbahey

For large k, start by factoring out a k:

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

where $$\delta \equiv \frac{1}{k}$$....for large k, delta is small....so taylor expand this in powers of delta and keep the smallest order non-zero term....

13. Nov 19, 2008

### LocationX

I tried using mathematica and I cant seem to get the result of $$\frac{\omega_0 c k}{\sqrt{\omega_0^2+\omega_p^2}}+\ldots$$

I get:

For large k, I get $$\omega_+ = kc$$ and I was going to use mathematica to get the $$\delta$$ dependence of $$\omega_-$$.

14. Nov 19, 2008

### gabbagabbahey

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,$\omega_p$>0,$\omega_0$>0},Series[...]]
That tells mathematica to evaluate the series while making those assumptions

15. Nov 19, 2008

### gabbagabbahey

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

$$f_{\pm}(k)\equiv \sqrt{\frac{\omega_0^2 + \omega_p^2 +c^2 k^2 \pm \sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}}{2}}$$

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

$$f(k)=f(0)+\frac{f'(0)}{1!}k+\frac{f''(0)}{2!}k^2 + \ldots$$

what are you getting for $f_{\pm}'(0)$ and$f_{\pm}''(0)$?

16. Nov 19, 2008

### LocationX

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

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

I get f'(0)=0
f''_-=0
$$f''_+=\frac{-c^2}{4(\omega_0^2+\omega_p^2)^{3/2}}$$

17. Nov 19, 2008

### gabbagabbahey

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

$$f_{-}'(0)=\infty$$.... ($0^{-1/2}$ is undefined)

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

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

....that should make thing easier for you!

Last edited: Nov 19, 2008
18. Nov 19, 2008

### LocationX

$$\sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)}$$
$$= (\omega_0^2 + \omega_p^2 +c^2 k^2) \sqrt{1 -\frac{4( c^2 k^2 \omega_0^2)}{(\omega_0^2 + \omega_p^2 +c^2 k^2)}}$$
$$\approx (\omega_0^2 + \omega_p^2 +c^2 k^2) \left( 1 -\frac{2( c^2 k^2 \omega_0^2)}{(\omega_0^2 + \omega_p^2 +c^2 k^2)} \right)$$

I dont see where $$\omega_0^2 + \omega_p^2$$ appears in the denominator

19. Nov 19, 2008

### gabbagabbahey

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

$$g(k)=\sqrt{(-\omega_0^2 - \omega_p^2 -c^2 k^2)^2 -4( c^2 k^2 \omega_0^2)} \approx g(0)+g'(0)k+\frac{g''(0)}{2!} k^2$$

20. Nov 19, 2008

### LocationX

$$g(0)=\omega_0^2 + \omega_p^2$$
$$g'(0)k=0$$
$$\frac{g''(0)}{2!} k^2 =c^2k^2 \frac{\omega_p^2-\omega_0^2}{\omega_0^2 + \omega_p^2}$$

Used mathematica for second derivative: