Frequency dependent permittivity

[LocationX]

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/"

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?

 

[LocationX]

any ideas?

 

[gabbagabbahey]

[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/"

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

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].

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.

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]

Again-- negative omega?

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).

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.

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?

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?

 

[LocationX]

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.

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 don't see where to go from here? How can one determine the which is real?

 

[gabbagabbahey]

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?

 

[LocationX]

omega squared is real when [itex]\kappa^2 >4( c^2 k^2 \omega_0^2)[/itex]?

 

[gabbagabbahey]

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.)

 

[LocationX]

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?

 

[gabbagabbahey]

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.

 

[LocationX]

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]

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?

 

[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.

 

[gabbagabbahey]

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...

 

[LocationX]

I tried using mathematica and I can't 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].

 

[gabbagabbahey]

The way you've written your mathematica code, your telling it to make the assumptions after calculating the series...instead try doing

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

 

[gabbagabbahey]

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]?

 

[LocationX]

[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]

 

[gabbagabbahey]

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!

 

[LocationX]

[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 don't see where [tex]\omega_0^2 + \omega_p^2[/tex] appears in the denominator

 

[gabbagabbahey]

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]

 

[LocationX]

[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

 

[gabbagabbahey]

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

 

[LocationX]

[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] ?

 

[gabbagabbahey]

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."

 

[LocationX]

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?

 

[LocationX]

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 doesn't 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?

 

[gabbagabbahey]

I think the higher frequency mode only has a phase velocity greater than c for small k; for that mode [tex]v_p \approx \frac{\sqrt{\omega_0^2 + \omega_p^2}}{k}[/tex] which can be greater than c for small enough k...For large k, I think it's safe to just assume [tex]\frac{\omega_0}{k}<c[/tex]

For the group velocity, I would take the derivative of the general formula for [itex]\omega_{\pm}[/itex] and just make a hand waving type argument that it is always less than c.